How many different unit fractions are lessorequal than all unit
fractions? The correct answer is: one unit fraction. If you claim more
than one (two or three or infintely many), then these more must be
equal. But different unit fractions are different and not equal to each other.
Another answer is that no unit fraction is lessorequal than all unit fractions. That means the function NUF(x)
Number of UnitFractions between 0 and x > 0
with NUF(0) = 0 will never increase but stay at 0. There are no unit fractions existing at all.
Therefore there is only the one correct answer given above.
Regards, WM
On 9/2/24 1:07 PM, WM wrote:
How many different unit fractions are lessorequal than all unitNope, because there does not exist AHY unit fraction that is less than
fractions? The correct answer is: one unit fraction. If you claim more
than one (two or three or infintely many), then these more must be
equal. But different unit fractions are different and not equal to
each other.
Another answer is that no unit fraction is lessorequal than all unit
fractions. That means the function NUF(x)
Number of UnitFractions between 0 and x > 0
with NUF(0) = 0 will never increase but stay at 0. There are no unit
fractions existing at all.
Therefore there is only the one correct answer given above.
or equal to ALL Unit fractions,
as any unit fraction you might claim to
be that one has a unit fraction smaller than itself, so it wasn't the smallest.
The problem with your NUF, is that it is trying to count something from
and uncountable end, one that doesn't actually have an end.
On 02.09.2024 19:19, Richard Damon wrote:
On 9/2/24 1:07 PM, WM wrote:
How many different unit fractions are lessorequal than all unitNope, because there does not exist AHY unit fraction that is less than
fractions? The correct answer is: one unit fraction. If you claim
more than one (two or three or infintely many), then these more must
be equal. But different unit fractions are different and not equal to
each other.
Another answer is that no unit fraction is lessorequal than all unit
fractions. That means the function NUF(x)
Number of UnitFractions between 0 and x > 0
with NUF(0) = 0 will never increase but stay at 0. There are no unit
fractions existing at all.
Therefore there is only the one correct answer given above.
or equal to ALL Unit fractions,
Impossible because then NUF will never increase. Then there are no unit fractions.
as any unit fraction you might claim to be that one has a unit
fraction smaller than itself, so it wasn't the smallest.
Your argument stems from visible unit fractions but becomes invalid in
the dark domain.
The problem with your NUF, is that it is trying to count something
from and uncountable end, one that doesn't actually have an end.
The unit fractions end before zero.
Regards, WM
On 02.09.2024 19:19, Richard Damon wrote:
as any unit fraction you might claim to be
that one has a unit fraction smaller than itself,
so it wasn't the smallest.
Your argument stems from visible unit fractions
but becomes invalid in the dark domain.
The problem with your NUF, is that
it is trying to count something
from and uncountable end,
one that doesn't actually have an end.
The unit fractions end before zero.
On 9/2/24 4:37 PM, WM wrote:
Which just shows the error in the "definition" of NUF.
as any unit fraction you might claim to be that one has a unit
fraction smaller than itself, so it wasn't the smallest.
Your argument stems from visible unit fractions but becomes invalid in
the dark domain.
But all the unit fractions are visible.
Thus, there is no smallest visible unit fraction as there can't be a
last one.
On 9/2/2024 4:37 PM, WM wrote:
On 02.09.2024 19:19, Richard Damon wrote:
as any unit fraction you might claim to be
that one has a unit fraction smaller than itself,
so it wasn't the smallest.
Your argument stems from visible unit fractions
but becomes invalid in the dark domain.
The darkᵂᴹ domain
between 0 and visibleᵂᴹ unit.fractions
is empty.
Each positive point is undercut by
visibleᵂᴹ unit.fractions,
⎛ Assume otherwise.
The unit fractions end before zero.
The lower.end of unit fractions
is not.
On 9/3/24 6:17 AM, WM wrote:
On 02.09.2024 23:43, Richard Damon wrote:
On 9/2/24 4:37 PM, WM wrote:
Which just shows the error in the "definition" of NUF.
There is no error.
Sure there is,
Unit fractions are fixed points on the real line which differ from
each other. Therefore there is a first one. It cannot be seen.
Therefore there are dark unit fractions.
That is not correct logic.
each other
On 02.09.2024 23:43, Richard Damon wrote:
On 9/2/24 4:37 PM, WM wrote:
Which just shows the error in the "definition" of NUF.
There is no error.
as any unit fraction you might claim to be that one has a unit
fraction smaller than itself, so it wasn't the smallest.
Your argument stems from visible unit fractions but becomes invalid
in the dark domain.
But all the unit fractions are visible.
All which you can see, yes. But there are many which you cannot see.
Thus, there is no smallest visible unit fraction as there can't be a
last one.
Unit fractions are fixed points on the real line which differ from each other. Therefore there is a first one. It cannot be seen. Therefore
there are dark unit fractions.
Regards, WM
On 03.09.2024 06:25, Jim Burns wrote:
On 9/2/2024 4:37 PM, WM wrote:
On 02.09.2024 19:19, Richard Damon wrote:
as any unit fraction you might claim to be
that one has a unit fraction smaller than itself,
so it wasn't the smallest.
Your argument stems from visible unit fractions
but becomes invalid in the dark domain.
The darkᵂᴹ domain
between 0 and visibleᵂᴹ unit.fractions
is empty.
Then you could see the smallest unit fraction.
Remember that they are fixed points with
non-empty gaps on the real line.
Hence there is a first one.
Each positive point is undercut by
visibleᵂᴹ unit.fractions,
No.
Only each visible positive point is undercut by
visible unit.fractions.
On 03.09.2024 13:29, Richard Damon wrote:
On 9/3/24 6:17 AM, WM wrote:
On 02.09.2024 23:43, Richard Damon wrote:
On 9/2/24 4:37 PM, WM wrote:
Which just shows the error in the "definition" of NUF.
There is no error.
Sure there is,
What is it? Note that better mathematicians than you have accepted the definition of NUF(x).
Unit fractions are fixed points on the real line which differ from
each other. Therefore there is a first one. It cannot be seen.
Therefore there are dark unit fractions.
That is not correct logic.
What?
Unit fractions are fixed points on the real line?
They differ from eaxh other?
Regards, WM
each other
How many different unit fractions are
lessorequal than all unit fractions?
The correct answer is: one unit fraction.
Another answer is that
no unit fraction is
lessorequal than all unit fractions.
Another answer is that
no unit fraction is
lessorequal than all unit fractions.
That means the function NUF(x)
Number of UnitFractions between 0 and x > 0
with NUF(0) = 0 will never increase but stay at 0.
That means the function NUF(x)
Number of UnitFractions between 0 and x > 0
with NUF(0) = 0 will never increase but stay at 0.
There are no unit fractions existing at all.
Therefore
there is only the one correct answer given above.
On 9/3/2024 6:22 AM, WM wrote:No, that is impossible by your argument:
If the smallest unit.fraction existed,
you could see it
positive and undercut byEither ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is wrong or Peano is wrong.
a visibleᵂᴹ smaller.than.smallest unit.fraction.
But you can't see that.
The smallest unit fraction doesn't exist.
On 9/2/2024 1:07 PM, WM wrote:
How many different unit fractions are
lessorequal than all unit fractions?
The correct answer is: one unit fraction.
Another answer is that
no unit fraction is
lessorequal than all unit fractions.
Not all sets are finite.
On 03.09.2024 19:50, Jim Burns wrote:
If the smallest unit.fraction existed,
you could see it
No, that is impossible by your argument:
positive and undercut by
a visibleᵂᴹ smaller.than.smallest unit.fraction.
But you can't see that.
The smallest unit fraction doesn't exist.
Either
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or Peano is wrong.
Peano has been generalized from
the small natural numbers.
"All different unit fractions are different"
however is a basic truth.
Therefore I accept the latter.
On 04.09.2024 20:08, Jim Burns wrote:
On 9/2/2024 1:07 PM, WM wrote:
How many different unit fractions are
lessorequal than all unit fractions?
The correct answer is: one unit fraction.
Another answer is that
no unit fraction is
lessorequal than all unit fractions.
Then NUF(x) will never leave the value 0.
Not all sets are finite.
But all different unit fractions are different,
i.e., they sit at different positive x.
But all different unit fractions are different,
i.e., they sit at different positive x.
Therefore only one can sit closest to zero.
On 04.09.2024 20:08, Jim Burns wrote:
On 9/2/2024 1:07 PM, WM wrote:Then NUF(x) will never leave the value 0.
How many different unit fractions are
lessorequal than all unit fractions?
The correct answer is: one unit fraction.
Another answer is that
no unit fraction is
lessorequal than all unit fractions.
Not all sets are finite.
But all different unit fractions are different, i.e., they sit at
different positive x. Therefore only one can sit closest to zero.
Regards, WM
On 03.09.2024 19:50, Jim Burns wrote:
On 9/3/2024 6:22 AM, WM wrote:
If the smallest unit.fraction existed,No, that is impossible by your argument:
you could see it
positive and undercut by
a visibleᵂᴹ smaller.than.smallest unit.fraction.
But you can't see that.Either ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is wrong or Peano is wrong.
The smallest unit fraction doesn't exist.
Peano has been generalized from the small natural numbers.
"All different unit fractions are different" however is a basic truth. Therefore I accept the latter.
Regards, WM
On 9/4/2024 3:10 PM, WM wrote:
Either
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or Peano is wrong.
...or natural numbers aren't what you think they are.
Peano has been generalized from
the small natural numbers.
Peano describes the finite natural numbers.
'Finite' doesn't need to be 'small'.
"All different unit fractions are different"
however is a basic truth.
Therefore I accept the latter.
You also accept quantifier shifts,
which breaks your logicᵂᴹ.
Quantifier shifts are unreliable.
On 9/4/2024 4:23 PM, WM wrote:
But all different unit fractions are different,
i.e., they sit at different positive x.
Yes,
each two sit at two points,
because ∀n ∈ ℕ: ⅟n-⅟(n+1) > 0
Therefore only one can sit closest to zero.
No,
each sits not.closest to zero,
because ∀n ∈ ℕ: ⅟n-⅟(n+1) > 0
On 9/4/24 3:10 PM, WM wrote:
Either ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is wrong or Peano is wrong.
Why?
Peano has been generalized from the small natural numbers.
"All different unit fractions are different" however is a basic truth.
Therefore I accept the latter.
And all ARE different as there is always a space between them, but that
space gets arbitrary small (but still finite.)
On 9/4/24 4:23 PM, WM wrote:
But all different unit fractions are different, i.e., they sit atnope, NONE sit closest to zero, as there are always more that are closer.
different positive x. Therefore only one can sit closest to zero.
That is what UNBOUNDED means.
Am 05.09.2024 um 09:30 schrieb Chris M. Thomasson:
On 9/4/2024 5:38 PM, Richard Damon wrote:
one [unit fraction] sit[s] closest to zero. [WM]
nope, NONE [no unit fraction] sit[s] closest to zero, as there are
always more that are closer.
Right.
On 9/4/2024 5:38 PM, Richard Damon wrote:
one [unit fraction] sit[s] closest to zero. [WM]
nope, NONE [no unit fraction] sit[s] closest to zero, as there are always more that are closer.
That is what UNBOUNDED means.
[...] WM must be trolling all night long
Le 04/09/2024 à 22:52, Jim Burns a écrit :
On 9/4/2024 3:10 PM, WM wrote:
Either
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or Peano is wrong.
...or
natural numbers aren't what you think they are.
That is possible.
My arguments hold only under
the premise of actual infinity
My arguments hold only under
the premise of actual infinity
showing that Hilbert's hotel is nonsense
because the set of natural numbers cannot be extended.
If all rooms are occupied
than no guest can leave his room for a not occupied room.
(When I was in USA or the first time,
I asked in a Hilton whether they had free rooms.
They laughed.)
Peano has been generalized from
the small natural numbers.
Peano describes the finite natural numbers.
'Finite' doesn't need to be 'small'.
Finite is much larger than
Peano or you could/can imagine.
Finite is much larger than
Peano or you could/can imagine.
"All different unit fractions are different"
however is a basic truth.
Therefore I accept the latter.
You also accept quantifier shifts,
which breaks your logicᵂᴹ.
Quantifier shifts are unreliable.
Do you believe that it needs a shift to state:
All different unit fractions are different.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
I can see no shift.
Le 04/09/2024 à 23:26, Jim Burns a écrit :
On 9/4/2024 4:23 PM, WM wrote:
But all different unit fractions are different,
i.e., they sit at different positive x.
Yes,
each two sit at two points,
because ∀n ∈ ℕ: ⅟n-⅟(n+1) > 0
Therefore only one can sit closest to zero.
No,
each sits not.closest to zero,
because ∀n ∈ ℕ: ⅟n-⅟(n+1) > 0
NUF(x) increases from 0 to more.
It cannot increase to 2 or more
before having accepted 1.
It cannot increase to ℵo without
having accepted 1, 2, 3, ...
On 9/5/2024 9:53 AM, WM wrote:
Insisting that ω-1 exists and that,
for b ≠ 0 and β < ω, β-1 exists
is
insisting that ω is finite.
The most frugal explanation of your claim is that
you simply do not know what 'finite' means.
Do you believe that it needs a shift to state:
All different unit fractions are different.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
I can see no shift.
It needs a shift to conclude from
( for each ⅟j: there is ⅟k≠⅟j: ⅟k < ⅟j
that
( there is ⅟k: for each ⅟j≠⅟k: ⅟k < ⅟j
Have you evolved on that topic?
On 05.09.2024 20:56, Jim Burns wrote:
On 9/5/2024 9:53 AM, WM wrote:
Insisting that ω-1 exists and that,
for b ≠ 0 and β < ω, β-1 exists
is
insisting that ω is finite.
No.
The most frugal explanation of your claim is that
you simply do not know what 'finite' means.
Finite means that you can count from one end to the other. Infinite
means that it is impossible to count from one end to the other.
Do you believe that it needs a shift to state:
All different unit fractions are different.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
I can see no shift.
It needs a shift to conclude from
( for each ⅟j: there is ⅟k≠⅟j: ⅟k < ⅟j
that
( there is ⅟k: for each ⅟j≠⅟k: ⅟k < ⅟j
Have you evolved on that topic?
You are mistaken. I do not conclude the latter from the former. I
conclude the latter from the fact that NUF(0) = 0 and NUF(x>0) > 0 and
never, at no x, NUF can increase by more than 1.
Le 05/09/2024 à 22:44, crank Wolfgang Mückenheim, aka WM a écrit :
[...] and never, at no x, NUF can increase by more than 1.
What the Hell could mean "to increase at an x" ?
Le 05/09/2024 à 22:44, crank Wolfgang Mückenheim, aka WM a écrit :
[...] and never, at no x, NUF can increase by more than 1.
What the Hell could mean "to increase at an x" ?
Am 06.09.2024 um 00:36 schrieb Python:
Le 05/09/2024 à 22:44, crank Wolfgang Mückenheim, aka WM a écrit :
[...] and never, at no x, NUF can increase by more than 1.
What the Hell could mean "to increase at an x" ?
No one knows. :-P
(We had the same "discussion" in dsm. To no end.)
Am 06.09.2024 um 00:36 schrieb Python:
Le 05/09/2024 à 22:44, crank Wolfgang Mückenheim, aka WM a écrit :
[...] and never, at no x [e IR, x > 0], NUF can increase by more than 1.
What the Hell could mean "to increase at an x" ?
No one knows. :-P
(We had the same "discussion" in dsm. To no end.)
On 05.09.2024 20:56, Jim Burns wrote:
On 9/5/2024 9:53 AM, WM wrote:
Le 04/09/2024 à 22:52, Jim Burns a écrit :
On 9/4/2024 3:10 PM, WM wrote:
Either
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or Peano is wrong.
...or
natural numbers aren't what you think they are.
That is possible.
Insisting that ω-1 exists and that,
for b ≠ 0 and β < ω, β-1 exists
is
insisting that ω is finite.
No.
Either
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or Peano is wrong.
The most frugal explanation of your claim is that
you simply do not know what 'finite' means.
Finite means that
you can count from one end to the other.
Infinite means that
it is impossible to count from one end to the other.
Do you believe that it needs a shift to state:
All different unit fractions are different.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
I can see no shift.
It needs a shift to conclude from
( for each ⅟j: there is ⅟k≠⅟j: ⅟k < ⅟j
that
( there is ⅟k: for each ⅟j≠⅟k: ⅟k < ⅟j
Have you evolved on that topic?
You are mistaken.
I do not conclude the latter from the former.
I conclude the latter from the fact that
NUF(0) = 0 and NUF(x>0) > 0
and never, at no x, NUF can increase by more than 1.
Try to understand that.
On 05.09.2024 16:22, Moebius wrote:
Am 05.09.2024 um 09:30 schrieb Chris M. Thomasson:
On 9/4/2024 5:38 PM, Richard Damon wrote:
one [unit fraction] sit[s] closest to zero. [WM]
nope, NONE [no unit fraction] sit[s] closest to zero, as there are
always more that are closer.
Right.
Wrong.
NUF(x) increases from 0 to ℵo. But it cannot increase to ℵo at one point x > 0 because at every point there is at most one unit fraction. Before ℵo there come 1, 2, 3, ...
Regards, WM
Le 05/09/2024 à 02:38, Richard Damon a écrit :
On 9/4/24 4:23 PM, WM wrote:
NUF(x) must grow. It cannot grow by more than 1 at any x.But all different unit fractions are different, i.e., they sit atnope, NONE sit closest to zero, as there are always more that are closer.
different positive x. Therefore only one can sit closest to zero.
That is what UNBOUNDED means.
You are in error. That was the ancient idea of infinity. In fact
unbounded means that you cannot see the end. But you can see a point
behind the end, namly zero.
Regards, WM
Le 05/09/2024 à 02:35, Richard Damon a écrit :
On 9/4/24 3:10 PM, WM wrote:
Either ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is wrong or Peano is wrong.
Why?
Peano has been generalized from the small natural numbers.
"All different unit fractions are different" however is a basic
truth. Therefore I accept the latter.
And all ARE different as there is always a space between them, but
that space gets arbitrary small (but still finite.)
NUF(x) increases from 0 to ℵo. But it cannot increase to ℵo at one point x > 0 because at every point there is at most one unit fraction. .Before
ℵo there come 1, 2, 3, ...
Regards, WM
On 9/5/24 10:08 AM, WM wrote:
NUF(x) must grow. It cannot grow by more than 1 at any x.
This is NOT the "ancient" idea of infinity,
On 9/5/2024 9:59 AM, WM wrote:
NUF(x) increases from 0 to more.
...at 0.
It cannot increase to 2 or more
before having accepted 1.
NUFᵈᵉᶠ(x) cannot increase to 2
without having already been ≥ 2
It cannot increase to ℵo without
having accepted 1, 2, 3, ...
x > 0 ⇒
0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x
Le 05/09/2024 à 22:44, crank Wolfgang Mückenheim, aka WM a écrit :
You are mistaken. I do not conclude the latter from the former. I
conclude the latter from the fact that NUF(0) = 0 and NUF(x>0) > 0 and
never, at no x, NUF can increase by more than 1.
What the Hell could mean "to increase at an x" ?
On 06.09.2024 00:36, Python wrote:What exactly happens at those points?
Le 05/09/2024 à 22:44, crank Wolfgang Mückenheim, aka WM a écrit :
Example: The function f(x) = [x] increases at every x ∈ ℕ by 1.You are mistaken. I do not conclude the latter from the former. IWhat the Hell could mean "to increase at an x" ?
conclude the latter from the fact that NUF(0) = 0 and NUF(x>0) > 0 and
never, at no x, NUF can increase by more than 1.
The function NUF(x) increases at every x = unit fraction 1/n by 1. It
does not increase at 0 because 0 is not a unit fraction.
Am Fri, 06 Sep 2024 14:22:00 +0200 schrieb WM:
Example: The function f(x) = [x] increases at every x ∈ ℕ by 1.What exactly happens at those points?
The function NUF(x) increases at every x = unit fraction 1/n by 1. It
does not increase at 0 because 0 is not a unit fraction.
On 05.09.2024 21:57, Jim Burns wrote:Therefore no x can be the least unit fraction.
On 9/5/2024 9:59 AM, WM wrote:
No. NUF(x) counts the unit fractions between 0 and x. Between 0 and 0NUF(x) increases from 0 to more....at 0.
there is no unit fraction. NUF(0) = 0.
Impossible. NUF(0) = 0. There is a first increase in linear order.It cannot increase to 2 or more before having accepted 1.NUFᵈᵉᶠ(x) cannot increase to 2 without having already been ≥ 2
0 is smaller than all that. Therefore there is no increase at 0.It cannot increase to ℵo without having accepted 1, 2, 3, ...x > 0 ⇒
0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x
x is larger than all that. Therefore your x is not the least one
posiible.
On the other hand, since NUF is constant on (0, oo)
On 06.09.2024 14:26, joes wrote:
Am Fri, 06 Sep 2024 14:22:00 +0200 schrieb WM:
Example: The function f(x) = [x] increases at every x ∈ ℕ by 1.What exactly happens at those points?
The function NUF(x) increases at every x = unit fraction 1/n by 1. It
does not increase at 0 because 0 is not a unit fraction.
The simplest action possible in mathematics: f --> f + 1
On 05.09.2024 21:57, Jim Burns wrote:
On 9/5/2024 9:59 AM, WM wrote:
NUF(x) increases from 0 to more.
...at 0.
No.
NUF(x) counts the unit fractions between 0 and x.
Between 0 and 0
there is no unit fraction.
NUF(0) = 0.
It cannot increase to 2 or more
before having accepted 1.
NUFᵈᵉᶠ(x) cannot increase to 2
without having already been ≥ 2
Impossible.
NUF(0) = 0.
There is a first increase in linear order.
It cannot increase to ℵo without
having accepted 1, 2, 3, ...
x > 0 ⇒
0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x
0 is smaller than all that.
Therefore there is no increase at 0.
x is larger than all that.
Therefore your x is not the least one posiible.
Le 06/09/2024 à 14:29, Crank Wolfgang Mückenheim, aka WM a écrit :
On 06.09.2024 14:26, joes wrote:
Am Fri, 06 Sep 2024 14:22:00 +0200 schrieb WM:
Example: The function f(x) = [x] increases at every x ∈ ℕ by 1.What exactly happens at those points?
The function NUF(x) increases at every x = unit fraction 1/n by 1. It
does not increase at 0 because 0 is not a unit fraction.
The simplest action possible in mathematics: f --> f + 1
What you wrote above is a function associating a function to a function.
Am Fri, 06 Sep 2024 14:12:29 +0200 schrieb WM:
Therefore no x can be the least unit fraction.0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x0 is smaller than all that. Therefore there is no increase at 0.
x is larger than all that. Therefore your x is not the least one
posiible.
WM submitted this idea :
What the Hell could mean "to increase at an x" ?
Example: The function f(x) = [x] increases at every x ∈ ℕ by 1.
Make up your mind, is x real or natural.
On 9/6/2024 8:12 AM, WM wrote:
On 05.09.2024 21:57, Jim Burns wrote:
On 9/5/2024 9:59 AM, WM wrote:
NUF(x) increases from 0 to more.
...at 0.
No.
NUF(x) counts the unit fractions between 0 and x.
Between 0 and 0 there is no unit fraction.
NUF(0) = 0.
Between 0 and x
there are more than 0 unit fractions
Between 0 and x
there are more.than.any.k<ℵ₀ unit fractions.
ε = 1. But that is not related to the topic.NUF(0) = 0.
There is a first increase in linear order.
∀ᴿx ∈ (-1,0]: ⌈x⌉ = 0
∀ᴿx ∈ (0,1]: ⌈x⌉ = 1
Is there a first ε at which 5⋅⌈ε⌉ = 5 ?
0 is smaller than all that.
Therefore there is no increase at 0.
Increases and decreases involve nearby points
from which increases and decreases increase and decrease.
Can you answer
whether f(x) increases or decreases at 0?
No, you can't answer without information about
nearby points.
x is larger than all that.
"All that" are more.than.any.k<ℵ₀ unit.fractions.
NUF(x) = ℵ₀
Therefore your x is not the least one posiible.
Yes.
x is NOT the first point > 0 after
more.than.any.k<ℵ₀ unit.fractions.
Generalize.
What do we know about x ?
x > 0
What else?
On 9/5/2024 7:22 AM, Moebius wrote:
Am 05.09.2024 um 09:30 schrieb Chris M. Thomasson:
On 9/4/2024 5:38 PM, Richard Damon wrote:
one [unit fraction] sit[s] closest to zero. [WM]
nope, NONE [no unit fraction] sit[s] closest to zero, as there are
always more that are closer.
Right.
If u is a unit fraction, 1/(1/u + 1) is a unit fraction which is
closer to 0 than s.
That is what UNBOUNDED means.
Nope. The set of unit fractions is BOUNDED. It's lower bound is 0 and
its upper bound (which also is its maximum) is 1.
But it's "unbounded" wrt the infinity of unit fractions that get closer
and closer to zero? Fair enough?
[...] WM must be trolling all night long
Well, he's just a mathematical crank.
On 06.09.2024 18:41, Jim Burns wrote:
On 9/6/2024 8:12 AM, WM wrote:
On 05.09.2024 21:57, Jim Burns wrote:
On 9/5/2024 9:59 AM, WM wrote:
NUF(x) increases from 0 to more.
...at 0.
No.
NUF(x) counts the unit fractions between 0 and x.
Between 0 and 0 there is no unit fraction.
NUF(0) = 0.
Between 0 and x
there are more than 0 unit fractions
Between 0 and your defined x or epsilon,
not between 0 and every possible x.
Between 0 and x
there are more.than.any.k<ℵ₀ unit fractions.
Betwee 0 and every epsilon you can define.
NUF(0) = 0.
There is a first increase in linear order.
∀ᴿx ∈ (-1,0]: ⌈x⌉ = 0
∀ᴿx ∈ (0,1]: ⌈x⌉ = 1
Is there a first ε at which 5⋅⌈ε⌉ = 5 ?
ε = 1.
But that is not related to the topic.
0 is smaller than all that.
Therefore there is no increase at 0.
Increases and decreases involve nearby points
from which increases and decreases increase and decrease.
Can you answer
whether f(x) increases or decreases at 0?
It is constant 0.
No, you can't answer without information about
nearby points.
I have information about
nearby points on the negative axis.
Only nearby points less than x are relevant
for judging about an increase in x.
Nearby points larger than x are irrelevant.
x is larger than all that.
"All that" are more.than.any.k<ℵ₀ unit.fractions.
NUF(x) = ℵ₀
Therefore your x is not the least one posiible.
Yes.
x is NOT the first point > 0 after
more.than.any.k<ℵ₀ unit.fractions.
But there is an x immediately after 0.
Generalize.
What do we know about x ?
x > 0
What else?
We do not know anything.
But we know that
all unit fractions differ from each other.
That is sufficient to know that
NUF(x) increases by 1 only at any unit fraction.
On 06.09.2024 05:08, Richard Damon wrote:
On 9/5/24 10:08 AM, WM wrote:
NUF(x) must grow. It cannot grow by more than 1 at any x.
This is NOT the "ancient" idea of infinity,
NUF(x) must grow. It cannot grow by more than 1 at any x.
Right or wrong in your opinion?
Regards, WM
On 06.09.2024 15:42, Python wrote:
Le 06/09/2024 à 14:29, Crank Wolfgang Mückenheim, aka WM a écrit :
On 06.09.2024 14:26, joes wrote:
Am Fri, 06 Sep 2024 14:22:00 +0200 schrieb WM:
Example: The function f(x) = [x] increases at every x ∈ ℕ by 1.What exactly happens at those points?
The function NUF(x) increases at every x = unit fraction 1/n by 1. It >>>>> does not increase at 0 because 0 is not a unit fraction.
The simplest action possible in mathematics: f --> f + 1
What you wrote above is a function associating a function to a function.
No it is a value changing to this value + 1.
Le 06/09/2024 à 22:42, Crank Wolfgang Mückenheim, aka WM a écrit :
On 06.09.2024 15:42, Python wrote:
Le 06/09/2024 à 14:29, Crank Wolfgang Mückenheim, aka WM a écrit :
On 06.09.2024 14:26, joes wrote:
Am Fri, 06 Sep 2024 14:22:00 +0200 schrieb WM:
Example: The function f(x) = [x] increases at every x ∈ ℕ by 1. >>>>>> The function NUF(x) increases at every x = unit fraction 1/n by 1. It >>>>>> does not increase at 0 because 0 is not a unit fraction.What exactly happens at those points?
The simplest action possible in mathematics: f --> f + 1
What you wrote above is a function associating a function to a function.
No it is a value changing to this value + 1.
So it is another value. A value does not change.
On 9/6/2024 4:52 PM, WM wrote:
Between 0 and your defined x or epsilon,
not between 0 and every possible x.
If x > 0 then there is ⅟⌊1+⅟x⌋ between 0 and x
A quantifier shift is not reliable.
Between 0 and x
there are more.than.any.k<ℵ₀ unit fractions.
Between 0 and every epsilon you can define.
Between 0 and any epsilon satisfying my description.
I haven't made a claim for other epsilons.
Le 06/09/2024 à 22:42, Crank Wolfgang Mückenheim, aka WM a écrit :
[...] it is a value changing to this value + 1.
[...] A value does not change.
Meanwhile you are still unable to provide a sound algebraic definition
of "increasing at a point" [...].
On 07.09.2024 11:19, Python wrote:We are getting closer to a definition, but the difference is still
Le 06/09/2024 à 22:42, Crank Wolfgang Mückenheim, aka WM a écrit :The value is a function of x. It can change when x changes.
On 06.09.2024 15:42, Python wrote:So it is another value. A value does not change.
Le 06/09/2024 à 14:29, Crank Wolfgang Mückenheim, aka WM a écrit : >>>>> On 06.09.2024 14:26, joes wrote:No it is a value changing to this value + 1.
What you wrote above is a function associating a function to aAm Fri, 06 Sep 2024 14:22:00 +0200 schrieb WM:The simplest action possible in mathematics: f --> f + 1
Example: The function f(x) = [x] increases at every x ∈ ℕ by 1. >>>>>>> The function NUF(x) increases at every x = unit fraction 1/n by 1. >>>>>>> It does not increase at 0 because 0 is not a unit fraction.What exactly happens at those points?
function.
Am 07.09.2024 um 11:19 schrieb Python:
Le 06/09/2024 à 22:42, Crank Wolfgang Mückenheim, aka WM a écrit :
[...] it is a value changing to this value + 1.
Where exactly does it "chance"?
Am 07.09.2024 um 11:19 schrieb Python:
Le 06/09/2024 à 22:42, Crank Wolfgang Mückenheim, aka WM a écrit :
[...] it is a value changing to this value + 1.
Where exactly does it "chance"?
[...] A value does not change.
Meanwhile you are still unable to provide a sound algebraic definition
of "increasing at a point" [...].
He rejects the notion of a "jump" (at some point). :-)
On 07.09.2024 01:51, Jim Burns wrote:
On 9/6/2024 4:52 PM, WM wrote:
Between 0 and your defined x or epsilon,
not between 0 and every possible x.
If x > 0 then there is ⅟⌊1+⅟x⌋ between 0 and x
Then identical unit fractions differ.
They are identical because NUF(x) counts them at the same x, but they
differ because NUF(x) counts more than 1 at this x.
A quantifier shift is not reliable.
That is no quantifier shift but simplest logic.
Between 0 and x
there are more.than.any.k<ℵ₀ unit fractions.
Between 0 and every epsilon you can define.
Between 0 and any epsilon satisfying my description.
I haven't made a claim for other epsilons.
Fine. There I agree. Every of your epsilons has ℵo smaller unit
fractions in (0, eps). This proves dark numbers.
Regards, WM
On 07.09.2024 13:29, Moebius wrote:And thus is always has a value of aleph_0 for all x > 0, since there is
Am 07.09.2024 um 11:19 schrieb Python:
Le 06/09/2024 à 22:42, Crank Wolfgang Mückenheim, aka WM a écrit :
[...] it is a value changing to this value + 1.
Where exactly does it "chance"?
It, NUF(x), changes at every x = 1/n for n ∈ ℕ.
NUF(x) changes by 1 because a change by more at any x would count more different unit fractions 1/n, 1/m, 1/k, ... which are identical because they are the same x = 1/n = 1/m = 1/k = ... .
Regards, WM
On 9/6/24 8:12 AM, WM wrote:
On 05.09.2024 21:57, Jim Burns wrote:
On 9/5/2024 9:59 AM, WM wrote:
NUF(x) increases from 0 to more.
...at 0.
No. NUF(x) counts the unit fractions between 0 and x. Between 0 and 0
there is no unit fraction. NUF(0) = 0.
But it CAN'T for the values of x that are finite, as for ANY finite
number, there is an infinite number of unit factions below it, so it
can't ever have a finite value for any finite x.
On 09/06/2024 07:01 PM, Richard Damon wrote:
On 9/6/24 8:07 AM, WM wrote:
On 06.09.2024 05:08, Richard Damon wrote:
On 9/5/24 10:08 AM, WM wrote:
NUF(x) must grow. It cannot grow by more than 1 at any x.
This is NOT the "ancient" idea of infinity,
NUF(x) must grow. It cannot grow by more than 1 at any x.
Right or wrong in your opinion?
Regards, WM
Only if it exists.
If it does, it must be counting some sub-finite values as "unit
fractions" that are not the reciprocal of the Natural Numbers (since
there is no smallest of those unit fractions to count from).
So, either it is counting some sub-finite values (actually a lot of them
a countable infinity of them) or it just doesn't exist.
Maybe that is your dark numbers, these sub-finite numbers that are
reciprocals of some post-finite values above the infinite set of Natural
Numbers (which have no upper bound) and are below Omega.
That's a remarkable supposition, I wonder how you'd imagine
both to satisfy to yourself and others that thusly is a
"consistent" form, of course which only requires "internal
consistency" for its own sake, then besides, to suffer the
running of the gauntlet, of those who'd insist it contradicted
theirs. For, their are simple inductive arguments that nothing
ever happens or is, at all.
I sort of appreciate the sentiment, though, that "infinite"
is big enough to have quite a range.
For all x > 0: NUF(x) = aleph_0 (no "change" at any x)
On 9/7/24 8:04 AM, WM wrote:
NUF(x) changes by 1 because a change by more at any x would count moreAnd thus is always has a value of aleph_0 for all x > 0, since there is always aleph_0 unit fractions below and finite positive x value.
different unit fractions 1/n, 1/m, 1/k, ... which are identical
because they are the same x = 1/n = 1/m = 1/k = ... .
On 07.09.2024 04:01, Richard Damon wrote:
On 9/6/24 8:12 AM, WM wrote:
On 05.09.2024 21:57, Jim Burns wrote:
On 9/5/2024 9:59 AM, WM wrote:
NUF(x) increases from 0 to more.
...at 0.
No. NUF(x) counts the unit fractions between 0 and x. Between 0 and 0
there is no unit fraction. NUF(0) = 0.
But it CAN'T for the values of x that are finite, as for ANY finite
number, there is an infinite number of unit factions below it, so it
can't ever have a finite value for any finite x.
This idea must be dropped.
Regards, WM
On 9/6/24 8:07 AM, WM wrote:
NUF(x) must grow. It cannot grow by more than 1 at any x.
Right or wrong in your opinion?
Only if it exists.
If it does, it must be counting some sub-finite values as "unitfractions" that are not the reciprocal of the Natural Numbers
(since there is no smallest of those unit fractions to count from).
Maybe that is your dark numbers, these sub-finite numbers that arereciprocals of some post-finite values above the infinite set of Natural Numbers (which have no upper bound) and are below Omega.
On 9/6/24 4:40 PM, WM wrote:
On 06.09.2024 14:38, joes wrote:
Am Fri, 06 Sep 2024 14:12:29 +0200 schrieb WM:
Therefore no x can be the least unit fraction.0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x0 is smaller than all that. Therefore there is no increase at 0.
x is larger than all that. Therefore your x is not the least one
posiible.
No definable x. No epsilon.
Which means no "Unit Fraction" as the reciprical of a Natural Number,
since they are all definable.
Thus, your NUF must also be counting some sub-finite values, or it just doesn't exist.
On 07.09.2024 14:37, Richard Damon wrote:
On 9/7/24 8:04 AM, WM wrote:
NUF(x) changes by 1 because a change by more at any x would countAnd thus is always has a value of aleph_0 for all x > 0, since there
more different unit fractions 1/n, 1/m, 1/k, ... which are identical
because they are the same x = 1/n = 1/m = 1/k = ... .
is always aleph_0 unit fractions below and finite positive x value.
Stop that nonsense. ℵo unit fractions cannot fit into every interval (0, x). The distance between two unit fractions already is larger. And if
you claim that ℵo unit fractions exist, then you must accept that their distances do exist too.
By the way this is independent of the existence of NUF.
Regards, WM
On 9/7/24 8:55 AM, WM wrote:
On 07.09.2024 04:01, Richard Damon wrote:
But it CAN'T for the values of x that are finite, as for ANY finite
number, there is an infinite number of unit factions below it, so it
can't ever have a finite value for any finite x.
This idea must be dropped.
Then the idea that NUF(x) "increases by one" at the values of "unit fractions" must be dropped.
Sorry, you can't have one without the other.
On 07.09.2024 13:36, Moebius wrote:
For all x > 0: NUF(x) = aleph_0 (no "change" at any x)
Stop that nonsense. ℵo unit fractions and their distances do not fit
into every interval (0, x). The distance between any pair of unit
fractions is larger.
Regards, WM
On 07.09.2024 04:01, Richard Damon wrote:
On 9/6/24 4:40 PM, WM wrote:
On 06.09.2024 14:38, joes wrote:
Am Fri, 06 Sep 2024 14:12:29 +0200 schrieb WM:
Therefore no x can be the least unit fraction.0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x0 is smaller than all that. Therefore there is no increase at 0.
x is larger than all that. Therefore your x is not the least one
posiible.
No definable x. No epsilon.
Which means no "Unit Fraction" as the reciprical of a Natural Number,
since they are all definable.
All natural numbers which you recognize are definable.
Thus, your NUF must also be counting some sub-finite values, or it
just doesn't exist.
It exists. It counts definable numbers as well as sub-finitely definale numbers.
Regards, WM
On 9/7/24 9:07 AM, WM wrote:
On 07.09.2024 13:36, Moebius wrote:
For all x > 0: NUF(x) = aleph_0 (no "change" at any x)
Stop that nonsense. ℵo unit fractions and their distances do not fit
into every interval (0, x). The distance between any pair of unit
fractions is larger.
Of course they do.
On 07.09.2024 01:51, Jim Burns wrote:What?
On 9/6/2024 4:52 PM, WM wrote:Then identical unit fractions differ.
If x > 0 then there is ⅟⌊1+⅟x⌋ between 0 and x
Between 0 and your defined x or epsilon, not between 0 and every
possible x.
They are identical because NUF(x) counts them at the same x, but they
differ because NUF(x) counts more than 1 at this x.
And all mathematics is well. Only WM doesn’t understand infinity.A quantifier shift is not reliable.That is no quantifier shift but simplest logic.
Fine. There I agree. Every of your epsilons has ℵo smaller unitBetween 0 and any epsilon satisfying my description.Between 0 and x there are more.than.any.k<ℵ₀ unit fractions.Between 0 and every epsilon you can define.
I haven't made a claim for other epsilons.
fractions in (0, eps). This proves dark numbers.
On 06.09.2024 18:41, Jim Burns wrote:Which are yet to be defined.
On 9/6/2024 8:12 AM, WM wrote:Between 0 and your defined x or epsilon, not between 0 and every
On 05.09.2024 21:57, Jim Burns wrote:Between 0 and x there are more than 0 unit fractions
On 9/5/2024 9:59 AM, WM wrote:
No.NUF(x) increases from 0 to more....at 0.
NUF(x) counts the unit fractions between 0 and x.
Between 0 and 0 there is no unit fraction. NUF(0) = 0.
possible x.
Enough for me.Between 0 and x there are more.than.any.k<ℵ₀ unit fractions.Betwee 0 and every epsilon you can define.
First, not last.ε = 1. But that is not related to the topic.NUF(0) = 0.∀ᴿx ∈ (-1,0]: ⌈x⌉ = 0 ∀ᴿx ∈ (0,1]: ⌈x⌉ = 1
There is a first increase in linear order.
Is there a first ε at which 5⋅⌈ε⌉ = 5 ?
So what?But here we have always 2^ℵo points where the function is constant0 is smaller than all that.Increases and decreases involve nearby points from which increases and
Therefore there is no increase at 0.
decreases increase and decrease.
before it increases by 1.
What about the negative of Dirac’s delta?Can you answer whether f(x) increases or decreases at 0?It is constant 0.
No, you can't answer without information about nearby points.I have information about nearby points on the negative axis. Only nearby points less than x are relevant for judging about an increase in x.
Nearby points larger than x are irrelevant.
You in particular should know the reals are dense. There is noBut there is an x immediately after 0. What else should be there?Therefore your x is not the least one posiible.Yes.
x is NOT the first point > 0 after more.than.any.k<ℵ₀ unit.fractions.
Generalize.We do not know anything. But we know that all unit fractions differ from
What do we know about x ?
x > 0 What else?
each other. That is sufficient to know that NUF(x) increases by 1 only
at any unit fraction.
On 07.09.2024 04:01, Richard Damon wrote:
On 9/6/24 8:07 AM, WM wrote:
NUF(x) must grow. It cannot grow by more than 1 at any x.
Right or wrong in your opinion?
Only if it exists.
There is no reason to deny its existence. If there is zero and all real numbers x > 0 and if there are unit fractions, then we can ask how many
unit fractions are between 0 and any x.
If it does, it must be counting some sub-finite values as "unitfractions" that are not the reciprocal of the Natural Numbers
Only reciprocals of natural numbers are counted.
(since there is no smallest of those unit fractions to count from).
Perhaps you consider only definable natural numbers. They have no
largest element.
Maybe that is your dark numbers, these sub-finite numbers that arereciprocals of some post-finite values above the infinite set of Natural Numbers (which have no upper bound) and are below Omega.
Dark numbers are post finitely definable.
Regards, WM
On 07.09.2024 14:37, Richard Damon wrote:Why can they not „fit”?
On 9/7/24 8:04 AM, WM wrote:
Stop that nonsense. ℵo unit fractions cannot fit into every interval (0, x). The distance between two unit fractions already is larger. And ifNUF(x) changes by 1 because a change by more at any x would count moreAnd thus is always has a value of aleph_0 for all x > 0, since there is
different unit fractions 1/n, 1/m, 1/k, ... which are identical
because they are the same x = 1/n = 1/m = 1/k = ... .
always aleph_0 unit fractions below and finite positive x value.
you claim that ℵo unit fractions exist, then you must accept that their distances do exist too.
By the way this is independent of the existence of NUF.
Am Fri, 06 Sep 2024 22:52:37 +0200 schrieb WM:
But there is an x immediately after 0. What else should be there?
You in particular should know the reals are dense. There is no
„immediately after”.
Am Sat, 07 Sep 2024 15:03:35 +0200 schrieb WM:
By the way this is independent of the existence of NUF.Why can they not „fit”?
On 07.09.2024 04:01, Richard Damon wrote:You seem to be denying the existence of an infinite step function
On 9/6/24 8:07 AM, WM wrote:
There is no reason to deny its existence. If there is zero and all real numbers x > 0 and if there are unit fractions, then we can ask how manyNUF(x) must grow. It cannot grow by more than 1 at any x.Only if it exists.
unit fractions are between 0 and any x.
Which are all visible/definable/light/finite.If it does, it must be counting some sub-finite values as "unitOnly reciprocals of natural numbers are counted.
fractions" that are not the reciprocal of the Natural Numbers
Thank you.(since there is no smallest of those unit fractions to count from).Perhaps you consider only definable natural numbers. They have no
largest element.
Uh what now?Maybe that is your dark numbers, these sub-finite numbers that areDark numbers are post finitely definable.
reciprocals of some post-finite values above the infinite set of
Natural Numbers (which have no upper bound) and are below Omega.
What exactly is the proof?
On 07.09.2024 04:01, Richard Damon wrote:
On 9/6/24 8:12 AM, WM wrote:
On 05.09.2024 21:57, Jim Burns wrote:
On 9/5/2024 9:59 AM, WM wrote:
NUF(x) increases from 0 to more.
...at 0.
No. NUF(x) counts the unit fractions between 0 and x. Between 0 and 0
there is no unit fraction. NUF(0) = 0.
But it CAN'T for the values of x that are finite, as for ANY finite
number, there is an infinite number of unit factions below it, so it
can't ever have a finite value for any finite x.
This idea must be dropped.
On 07.09.2024 14:37, Richard Damon wrote:
On 9/7/24 8:04 AM, WM wrote:
NUF(x) changes by 1 because a change by more at any x would countAnd thus is always has a value of aleph_0 for all x > 0, since there
more different unit fractions 1/n, 1/m, 1/k, ... which are identical
because they are the same x = 1/n = 1/m = 1/k = ... .
is always aleph_0 unit fractions below and finite positive x value.
Stop that nonsense. ℵo unit fractions cannot fit into every interval (0, x).
The distance between two unit fractions already is larger.
On 07.09.2024 15:21, Richard Damon wrote:
On 9/7/24 9:07 AM, WM wrote:
On 07.09.2024 13:36, Moebius wrote:
For all x > 0: NUF(x) = aleph_0 (no "change" at any x)
Stop that nonsense. ℵo unit fractions and their distances do not fit
into every interval (0, x). The distance between any pair of unit
fractions is larger.
Of course they do.
The distance is made of 2^ℵo points. Every x > 0 can mean 3 points.
Regards, WM
On 07.09.2024 15:36, joes wrote:
What exactly is the proof?In your opinion ℵo unit fractions are identical because they sit at the same x, but they differ because they are ℵo different unit fractions.
Your opinion is the death of logic.
Regards, WM
On 07.09.2024 15:37, joes wrote:
Am Sat, 07 Sep 2024 15:03:35 +0200 schrieb WM:
By the way this is independent of the existence of NUF.Why can they not „fit”?
Because ℵo unit fractions have ℵo gaps. We can use one of the gaps as x
0.
Regards, WM
On 07.09.2024 15:33, joes wrote:
Am Fri, 06 Sep 2024 22:52:37 +0200 schrieb WM:
But there is an x immediately after 0. What else should be there?
You in particular should know the reals are dense. There is no
„immediately after”.
In actual infinity all points are there. If you believe that between
every pair of points points can be inserted - and that infinitely often
- then you apply potential infinity. In actual infinity all points are
there from the beginning without infinite processing.
Regards, WM
On 9/7/24 9:45 AM, WM wrote:
On 07.09.2024 15:37, joes wrote:
Am Sat, 07 Sep 2024 15:03:35 +0200 schrieb WM:
By the way this is independent of the existence of NUF.Why can they not „fit”?
Because ℵo unit fractions have ℵo gaps. We can use one of the gaps as
x > 0.
Regards, WM
But you started with that x, so it was fixed to begin with.
From that x, we can build aleph_0 unit fractions below it.
You are just showing you don't understand what you are talking about.
On 07.09.2024 14:58, Richard Damon wrote:
On 9/7/24 8:55 AM, WM wrote:
On 07.09.2024 04:01, Richard Damon wrote:
But it CAN'T for the values of x that are finite, as for ANY finite
number, there is an infinite number of unit factions below it, so it
can't ever have a finite value for any finite x.
This idea must be dropped.
Then the idea that NUF(x) "increases by one" at the values of "unit
fractions" must be dropped.
Sorry, you can't have one without the other.
Then logic breaks down because identical elements differ.
The unit fractions are identical because they sit at the same x, but
they differ because they are ℵo different unit fractions.
Logic or the successor axiom? Without the former the latter would be
useless.
Regards, WM
On 09/07/2024 08:34 AM, Ross Finlayson wrote:
On 09/07/2024 05:57 AM, Richard Damon wrote:
On 9/6/24 11:00 PM, Ross Finlayson wrote:
On 09/06/2024 07:01 PM, Richard Damon wrote:
On 9/6/24 8:07 AM, WM wrote:
On 06.09.2024 05:08, Richard Damon wrote:
On 9/5/24 10:08 AM, WM wrote:
NUF(x) must grow. It cannot grow by more than 1 at any x.
This is NOT the "ancient" idea of infinity,
NUF(x) must grow. It cannot grow by more than 1 at any x.
Right or wrong in your opinion?
Regards, WM
Only if it exists.
If it does, it must be counting some sub-finite values as "unit
fractions" that are not the reciprocal of the Natural Numbers (since >>>>> there is no smallest of those unit fractions to count from).
So, either it is counting some sub-finite values (actually a lot of
them
a countable infinity of them) or it just doesn't exist.
Maybe that is your dark numbers, these sub-finite numbers that are
reciprocals of some post-finite values above the infinite set of
Natural
Numbers (which have no upper bound) and are below Omega.
That's a remarkable supposition, I wonder how you'd imagine
both to satisfy to yourself and others that thusly is a
"consistent" form, of course which only requires "internal
consistency" for its own sake, then besides, to suffer the
running of the gauntlet, of those who'd insist it contradicted
theirs. For, their are simple inductive arguments that nothing
ever happens or is, at all.
I sort of appreciate the sentiment, though, that "infinite"
is big enough to have quite a range.
The problem with "consistancy" is that WM's mathematicss isn't
consistent with the full Natural Numbers, and unlikely to be helped with >>> the addition of something even more esoteric.
His work doesn't define the set well enough to actually define how it
must work, and the best answer is likly to just adopt one of the
existing set of sub-finite number, it just needs to have a countably
infinite subset of values that can be reasonable defined as "unit
fractions".
Maybe it'd do better with less.
How about this, imagine it was your duty to convince a panel of
mathematicians that something that made for the most properties
possible of the notion of an iota-value or least-positive-rational,
had a way to define this thing. Is it any different than f(1) for
n/d, d-> oo, n -> d, modeling a not-a-real-function as a limit of
real functions?
Or does that just mean crazy-town to you? The crazy-town here
is actually sort of crazy-town, like, you walk out on the streets
and at various intervals encounter derelict indigents who are
entirely insane, on most given Tuesdays.
How do you keep your sanity in crazy-town, or help rehabilitate
crazy-town? Among the ideas that nothing can be crazy if it's
all consistent, in the infinite, get into things like Hausdorff's
constructible universe, and Skolem's countable universe, or,
"model of ZF", if "universe" makes no sound to you.
You might aver "nobody does that" and that
would be wrong - every day there's Dirichlet
and Poincare and Dirac and there's time-ordering
and making the derivation of Fourier series and
most usually an equi-partitioning of a unit interval
as according to a large number N in a sort of
reverse delta-epsilonics, in fact it suffices to
say that without such a notion of equi-partitioning
infinitely a unit interval there'd be no modularity
of distance at all.
In fact it was in the physics courses where it was
introduced "not-a-real-function yet with real
analytical character as modeled as a (continuum)
limit of real functions", Dirac delta, which is only
so _in the limit_ and just like calculus only perfect
and so _in the limit_, that it's first little bit is just
like its last little bit.
And non-zero, ....
And if you've never heard of that then congratulations,
here's a new word for your vocabulary.
On 07.09.2024 01:51, Jim Burns wrote:
On 9/6/2024 4:52 PM, WM wrote:
Between 0 and your defined x or epsilon,
not between 0 and every possible x.
If x > 0 then there is ⅟⌊1+⅟x⌋ between 0 and x
Then identical unit fractions differ.
They are identical because
NUF(x) counts them at the same x,
but they differ because
NUF(x) counts more than 1
at this x.
A quantifier shift is not reliable.
That is no quantifier shift but simplest logic.
Between 0 and x
there are more.than.any.k<ℵ₀ unit fractions.
Between 0 and every epsilon you can define.
Between 0 and any epsilon satisfying my description.
I haven't made a claim for other epsilons.
Fine.
There I agree.
Every of your epsilons has
ℵo smaller unit fractions in (0, eps).
This proves dark numbers.
The unit fractions are identical because they sit at the same x, but
they differ because they are ℵo different unit fractions.
On 09/07/2024 12:01 PM, Jim Burns wrote:
[...]
Aristotle has both _prior_ and _posterior_ analytics.
So, when you give him
a perfectly good syllogism with which he disagrees,
he has either of prior or posterior to deconstruct
either posterior or prior,
thusly not allowing himself to be fooled
by otherwise perfectly and as-far-as-the-eye-can-see
linear induction,
because that would leave a fool of him.
The unit fractions are identical because they sit at the same x, but
they differ because they are ℵo different unit fractions.
On 07.09.2024 04:01, Richard Damon wrote:Great, it all works out in the end.
On 9/6/24 4:40 PM, WM wrote:All natural numbers which you recognize are definable.
On 06.09.2024 14:38, joes wrote:Which means no "Unit Fraction" as the reciprical of a Natural Number,
Am Fri, 06 Sep 2024 14:12:29 +0200 schrieb WM:No definable x. No epsilon.
Therefore no x can be the least unit fraction.0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x0 is smaller than all that. Therefore there is no increase at 0.
x is larger than all that. Therefore your x is not the least one
posiible.
since they are all definable.
As we knew, you’re not working in the reals but with some extension.Thus, your NUF must also be counting some sub-finite values, or it justIt exists. It counts definable numbers as well as sub-finitely definale numbers.
doesn't exist.
On 09/07/2024 07:29 PM, Ross Finlayson wrote:
On 09/07/2024 05:29 PM, Jim Burns wrote:
On 9/7/2024 3:13 PM, Ross Finlayson wrote:
Aristotle has both _prior_ and _posterior_ analytics.
So, when you give him
a perfectly good syllogism with which he disagrees,
he has either of prior or posterior to deconstruct
either posterior or prior,
Wikipedia seems to say that
syllogisms are prior, and
use of syllogisms is posterior.
They don't seem to be 'either.or', but 'both.and'.
So, being constructive, constructive criticism,
when I look at the outcome of
otherwise a proof by contradiction to be rejected,
that a "strongly constructivist" view requires that
it's immaterial the order of the introduction of
any stipulations,
where
in the usual syllogism's proof by contradiction,
whatever non-logical term is introduced last
sort of wins,
when
if the terms are discovered and evaluated
in an arbitrary order,
it's arbitrary which decides and which is decided.
Then,
that Russell's retro-thesis is
simply not a fact, logically, [...]
On 9/7/2024 7:02 AM, WM wrote:
Different unit fractions are different.
They are identical because
NUF(x) counts them at the same x,
Counting.at.x a unit.fraction.in.⅟ℕᵈᵉᶠ∩(0,x)
does not mean the unit fraction is at x
Le 07/09/2024 à 15:03, WM a écrit :
Stop that nonsense. ℵo unit fractions cannot fit into every interval
(0, x).
Of course they can.
All those add up to less than x, so they fit.neighbour. Call its size x. Then ℵo unit fractions cannot fit into the interval (0, x), independent of the actual size.
Select any gap between one of the first ℵo unit fractions and its
On 9/7/24 9:45 AM, WM wrote:
Because ℵo unit fractions have ℵo gaps. We can use one of the gaps asBut you started with that x, so it was fixed to begin with.
x > 0.
On 07.09.2024 16:05, Python wrote:What’s your point? Of course the distances decrease.
Le 07/09/2024 à 15:03, WM a écrit :
Select any gap between one of the first ℵo unit fractions and its neighbour. Call its size x. Then ℵo unit fractions cannot fit into the interval (0, x), independent of the actual size.Stop that nonsense. ℵo unit fractions cannot fit into every interval
(0, x).
On 07.09.2024 16:49, Richard Damon wrote:
All those add up to less than x, so they fit.neighbour. Call its size x. Then ℵo unit fractions cannot fit into the interval (0, x), independent of the actual size.
Select any gap between one of the first ℵo unit fractions and its
Regards, WM
On 07.09.2024 15:37, joes wrote:What does this mean?
Am Sat, 07 Sep 2024 15:03:35 +0200 schrieb WM:
Because ℵo unit fractions have ℵo gaps. We can use one of the gaps asBy the way this is independent of the existence of NUF.Why can they not „fit”?
x > 0.
On 07.09.2024 21:01, Jim Burns wrote:
Different unit fractions are different.
Therefore there is only one the smallest one.
They are identical because
NUF(x) counts them at the same x,
Counting.at.x a unit.fraction.in.⅟ℕᵈᵉᶠ∩(0,x)
does not mean the unit fraction is at x
NUF counts only unit fractions at their positions.
On 07.09.2024 16:49, Richard Damon wrote:Why?
All those add up to less than x, so they fit.neighbour. Call its size x. Then ℵo unit fractions cannot fit into the interval (0, x), independent of the actual size.
Select any gap between one of the first ℵo unit fractions and its
Am Sun, 08 Sep 2024 21:48:11 +0200 schrieb WM:
On 07.09.2024 16:49, Richard Damon wrote:
All those add up to less than x, so they fit.
Select any gap between one of the first ℵo unit fractions and itsneighbour. Call its size x. Then ℵo unit fractions cannot fit into the
interval (0, x), independent of the actual size.
Why?
On 07.09.2024 16:05, Python wrote:
Le 07/09/2024 à 15:03, WM a écrit :
Stop that nonsense. ℵo unit fractions cannot fit into every interval
(0, x).
Of course they can.
Select any gap between one of the first ℵo unit fractions and its neighbour. Call its size x.
Then ℵo unit fractions cannot fit into the
interval (0, x), independent of the actual size.
On 09/08/2024 12:13 PM, Jim Burns wrote:
[...]
It's already been thoroughly elaborated and
as attached to formalistic symbolry,
that "Russell's thesis, of an antinomy" is that
the set of
the finite sets that don't contain themselves,
exactly like the ordinals are mostly simply modeled to be,
does and doesn't contain itself,
blowing wide open that
there is a class of propositions
external [to] "tertium non datur".
Then,
"Russell's retro-thesis", is
"forget I, Russell I, said that,
and now that Frege's out of the way,
think me and Whitehead made 1 + 1 = 2,
after this brief 0 = 1 as it were",
asking you simply ignore his stated antinomy,
and furthermore your own conscience
as with regards to these matters.
Am 08.09.2024 um 22:25 schrieb joes:
Am Sun, 08 Sep 2024 21:48:11 +0200 schrieb WM:
On 07.09.2024 16:49, Richard Damon wrote:
All those add up to less than x, so they fit.
Select any gap between one of the first ℵo unit fractions and itsneighbour. Call its size x. Then ℵo unit fractions cannot fit into the >>> interval (0, x), independent of the actual size.
Why?
Weiß niemand.
Am Sun, 08 Sep 2024 21:28:44 +0200 schrieb WM:
On 07.09.2024 16:05, Python wrote:What’s your point? Of course the distances decrease.
Le 07/09/2024 à 15:03, WM a écrit :Select any gap between one of the first ℵo unit fractions and its
Stop that nonsense. ℵo unit fractions cannot fit into every interval >>>> (0, x).
neighbour. Call its size x. Then ℵo unit fractions cannot fit into the
interval (0, x), independent of the actual size.
Le 08/09/2024 à 21:28, Crank Mückenheim, aka WM a écrit :
On 07.09.2024 16:05, Python wrote:
Le 07/09/2024 à 15:03, WM a écrit :
Stop that nonsense. ℵo unit fractions cannot fit into every interval >>>> (0, x).
Of course they can.
Select any gap between one of the first ℵo unit fractions and its
neighbour. Call its size x.
x = 1/k - 1/(k+1) = 1/[k*(k+1)] > 0
Then ℵo unit fractions cannot fit into the interval (0, x),
independent of the actual size.
It can and it does
Am Sat, 07 Sep 2024 15:45:08 +0200 schrieb WM:
On 07.09.2024 15:37, joes wrote:What does this mean?
Am Sat, 07 Sep 2024 15:03:35 +0200 schrieb WM:Because ℵo unit fractions have ℵo gaps. We can use one of the gaps as
By the way this is independent of the existence of NUF.Why can they not „fit”?
x > 0.
On 9/8/24 3:48 PM, WM wrote:
Select any gap between one of the first ℵo unit fractions and itsBut that is changing the value of x in the middle of the problem which
neighbour. Call its size x. Then ℵo unit fractions cannot fit into the
interval (0, x), independent of the actual size.
isn't allowed.
Given that new x, we can choose a new set of Aleph_0 unit fractions
below that x.
On 9/8/2024 3:39 PM, WM wrote:
On 07.09.2024 21:01, Jim Burns wrote:
Different unit fractions are different.
Therefore there is only one the smallest one.
There aren't two smallest unit fractions,
and no one has said otherwise.
There isn't one smallest unit fraction because,
for each ⅟k, ⅟(k+1) disproves ⅟k being smallest.
There aren't two.
There isn't one.
On 9/8/24 3:52 PM, WM wrote:
On 07.09.2024 16:51, Richard Damon wrote:
On 9/7/24 9:45 AM, WM wrote:
Because ℵo unit fractions have ℵo gaps. We can use one of the gaps >>>> as x > 0.But you started with that x, so it was fixed to begin with.
No, it is the size of one of the gaps between two adjacent unit
fractions, ℵo of which are claimed to fit into (0, x).
So it is the size of *ONE* of the gaps, you need to choose which one, as
all the gaps are different sizes.
Remember, just as there is no smallest unit fraction, there is no
smallest gap, so you can't specify choosing the smallest.
The fact that the choice is for ANY, it requires you to actually CHOOSE
one, and then we can find the Aleph_0 smaller.
On 08.09.2024 22:29, Moebius wrote:
Am 08.09.2024 um 22:25 schrieb joes:
Am Sun, 08 Sep 2024 21:48:11 +0200 schrieb WM:
On 07.09.2024 16:49, Richard Damon wrote:
All those add up to less than x, so they fit.
Select any gap between one of the first ℵo unit fractions and itsneighbour. Call its size x. Then ℵo unit fractions cannot fit into the >>>> interval (0, x), independent of the actual size.
Why?
Weiß niemand.
ℵo unit fractions cannot fit into one of the ℵo intervals between two of them, because ℵo unit fractions occupy ℵo intervals.
Regards, WM
On 08.09.2024 22:11, Richard Damon wrote:
On 9/8/24 3:48 PM, WM wrote:
Select any gap between one of the first ℵo unit fractions and itsBut that is changing the value of x in the middle of the problem which
neighbour. Call its size x. Then ℵo unit fractions cannot fit into
the interval (0, x), independent of the actual size.
isn't allowed.
No.
Given that new x, we can choose a new set of Aleph_0 unit fractions
below that x.
ℵo unit fractions are claimed to be smaller than every x > 0. If that is true then I can choose as the x one of the ℵo intervals between two of them.
Regards, WM
On 08.09.2024 22:11, Richard Damon wrote:
On 9/8/24 3:52 PM, WM wrote:
On 07.09.2024 16:51, Richard Damon wrote:
On 9/7/24 9:45 AM, WM wrote:
Because ℵo unit fractions have ℵo gaps. We can use one of the gaps >>>>> as x > 0.But you started with that x, so it was fixed to begin with.
No, it is the size of one of the gaps between two adjacent unit
fractions, ℵo of which are claimed to fit into (0, x).
So it is the size of *ONE* of the gaps, you need to choose which one,
as all the gaps are different sizes.
And all gaps are occupied by the unit fractions. Hence every gap is too small.
Remember, just as there is no smallest unit fraction, there is no
smallest gap, so you can't specify choosing the smallest.
The fact that the choice is for ANY, it requires you to actually
CHOOSE one, and then we can find the Aleph_0 smaller.
No. First you claim the unit fractions with their gaps. Then I choose
one of them, irrelevant which one. Each one is smaller than all.
Regards, WM
On 08.09.2024 22:29, Moebius wrote:What? It’s about the size, not the number of intervals.
Am 08.09.2024 um 22:25 schrieb joes:ℵo unit fractions cannot fit into one of the ℵo intervals between two of them, because ℵo unit fractions occupy ℵo intervals.
Am Sun, 08 Sep 2024 21:48:11 +0200 schrieb WM:
On 07.09.2024 16:49, Richard Damon wrote:Why?
Select any gap between one of the first ℵo unit fractions and itsneighbour. Call its size x. Then ℵo unit fractions cannot fit into
the interval (0, x), independent of the actual size.
On 08.09.2024 22:11, Richard Damon wrote:More precisely: every positive x has infinitely many smaller unit
On 9/8/24 3:48 PM, WM wrote:
ℵo unit fractions are claimed to be smaller than every x > 0. If that is true then I can choose as the x one of the ℵo intervals between two of them.Select any gap between one of the first ℵo unit fractions and itsBut that is changing the value of x in the middle of the problem which
neighbour. Call its size x. Then ℵo unit fractions cannot fit into the >>> interval (0, x), independent of the actual size.
isn't allowed.
Given that new x, we can choose a new set of Aleph_0 unit fractions
below that x.
On 08.09.2024 22:11, Richard Damon wrote:In which sense are the gaps „occupied”?
On 9/8/24 3:52 PM, WM wrote:And all gaps are occupied by the unit fractions. Hence every gap is too small.
On 07.09.2024 16:51, Richard Damon wrote:So it is the size of *ONE* of the gaps, you need to choose which one,
On 9/7/24 9:45 AM, WM wrote:No, it is the size of one of the gaps between two adjacent unit
Because ℵo unit fractions have ℵo gaps. We can use one of the gaps >>>>> as x > 0.But you started with that x, so it was fixed to begin with.
fractions, ℵo of which are claimed to fit into (0, x).
as all the gaps are different sizes.
Each gap is smaller than all gaps?Remember, just as there is no smallest unit fraction, there is noNo. First you claim the unit fractions with their gaps. Then I choose
smallest gap, so you can't specify choosing the smallest.
The fact that the choice is for ANY, it requires you to actually CHOOSE
one, and then we can find the Aleph_0 smaller.
one of them, irrelevant which one. Each one is smaller than all.
On 08.09.2024 22:16, joes wrote:
Am Sat, 07 Sep 2024 15:45:08 +0200 schrieb WM:
On 07.09.2024 15:37, joes wrote:What does this mean?
Am Sat, 07 Sep 2024 15:03:35 +0200 schrieb WM:Because ℵo unit fractions have ℵo gaps. We can use one of the gaps as >>> x > 0.
By the way this is independent of the existence of NUF.Why can they not „fit”?
ℵo unit fractions are claimed to be smaller than every x > 0. If they
are existing then I can choose as the x one of the ℵo intervals between
two of them. Irrelevant which one, ℵo unit fractions do not fit into it.
Regards, WM
Am Mon, 09 Sep 2024 12:22:21 +0200 schrieb WM:
ℵo unit fractions cannot fit into one of the ℵo intervals between two of >> them, because ℵo unit fractions occupy ℵo intervals.What? It’s about the size, not the number of intervals.
On 9/9/24 6:31 AM, WM wrote:
And all gaps are occupied by the unit fractions. Hence every gap is
too small.
But there is always room at the bottom, where the gaps keep getting smaller
How can a claim a set of values less then a SPECIFIC NUMBER if I don't
have the number?
On 08.09.2024 22:34, Python wrote:
Le 08/09/2024 à 21:28, Crank Mückenheim, aka WM a écrit :
On 07.09.2024 16:05, Python wrote:
Le 07/09/2024 à 15:03, WM a écrit :
Stop that nonsense. ℵo unit fractions cannot fit into every
interval (0, x).
Of course they can.
Select any gap between one of the first ℵo unit fractions and its
neighbour. Call its size x.
x = 1/k - 1/(k+1) = 1/[k*(k+1)] > 0
Then ℵo unit fractions cannot fit into the interval (0, x),
independent of the actual size.
It can and it does
Nonsense. ℵo unit fractions cannot fit into one of the ℵo intervals between two of them.
No, because the claim is GIVEN an x, we can do "Y", that is make the
Aleph_0 unit fractions below it.
Until you have chosen your x, we don't
need to provide those unit fractions, so, you can't use them to create
your x.
To follow YOUR idea, then *YOU* get stuck in the infinite loop of every
time you change your x, the unit fractions change so you need to change
to another x, and the unit fractions change again.
Am Mon, 09 Sep 2024 12:27:47 +0200 schrieb WM:
On 08.09.2024 22:11, Richard Damon wrote:More precisely: every positive x has infinitely many smaller unit
On 9/8/24 3:48 PM, WM wrote:ℵo unit fractions are claimed to be smaller than every x > 0. If that is >> true then I can choose as the x one of the ℵo intervals between two of
Select any gap between one of the first ℵo unit fractions and itsBut that is changing the value of x in the middle of the problem which
neighbour. Call its size x. Then ℵo unit fractions cannot fit into the >>>> interval (0, x), independent of the actual size.
isn't allowed.
Given that new x, we can choose a new set of Aleph_0 unit fractions
below that x.
them.
fractions (mind the quantifier order).
A number is not an interval.
On 09.09.2024 13:46, joes wrote:What does this mean? This is way too terse.
Am Mon, 09 Sep 2024 12:22:21 +0200 schrieb WM:
One of ℵo intervals is smaller than ℵo intervals.ℵo unit fractions cannot fit into one of the ℵo intervals between two >>> of them, because ℵo unit fractions occupy ℵo intervals.What? It’s about the size, not the number of intervals.
On 09.09.2024 13:47, Richard Damon wrote:How do you do that?
On 9/9/24 6:31 AM, WM wrote:
They all are present from the start. I simply choose a gap that is tooAnd all gaps are occupied by the unit fractions. Hence every gap isBut there is always room at the bottom, where the gaps keep getting
too small.
smaller
small to contain ℵo unit fractions.
Those first points are on the right, obviously.How can a claim a set of values less then a SPECIFIC NUMBER if I don'tI claim that all unit fractions are existing as real points of the real
have the number?
line. Therefore there is a first one. NUF(x) cannot increase without
passing 1 when real points are counted. Your ℵo points cannot exist
without including 1, 2, 3 first points.
Le 09/09/2024 à 12:19, Prof. Dr. Mückenheim, aka WM a écrit :
ℵo unit fractions cannot fit into one of the ℵo intervals
between two of them.
(O, x) is NOT an interval between two unit fractions.
Am Mon, 09 Sep 2024 17:01:35 +0200 schrieb WM:
They all are present from the start. I simply choose a gap that is tooHow do you do that?
small to contain ℵo unit fractions.
Those first points are on the right, obviously.How can a claim a set of values less then a SPECIFIC NUMBER if I don'tI claim that all unit fractions are existing as real points of the real
have the number?
line. Therefore there is a first one. NUF(x) cannot increase without
passing 1 when real points are counted. Your ℵo points cannot exist
without including 1, 2, 3 first points.
Am Mon, 09 Sep 2024 12:31:55 +0200 schrieb WM:
Al gaps are occupied by the unit fractions. Hence every gap is tooIn which sense are the gaps „occupied”?
small.
First you claim the unit fractions with their gaps. Then I chooseEach gap is smaller than all gaps?
one of them, irrelevant which one. Each one is smaller than all.
Am Mon, 09 Sep 2024 16:53:32 +0200 schrieb WM:
You claim that ℵo unit fractions are smaller than ANY x > 0.Yes. Not all the same ones of course.
On 09.09.2024 16:32, Python wrote:
Le 09/09/2024 à 12:19, Crank Mückenheim, aka WM a écrit :
ℵo unit fractions cannot fit into one of the ℵo intervals between two >>> of them.
(O, x) is NOT an interval between two unit fractions.
1/n - 1/(n+1) = x is an interval between two unit fraction.
This
interval is shifted to the origin, yielding the interval (0, x). It does
not contain ℵo unit fractions. It does not contain 1/n.
[snip more nonsense]
On 09.09.2024 17:15, joes wrote:That is a different question (quantifier order). Of course no unit
Am Mon, 09 Sep 2024 16:53:32 +0200 schrieb WM:
My question concerns same unit fractions only. Do ℵo unit fractionsYou claim that ℵo unit fractions are smaller than ANY x > 0.Yes. Not all the same ones of course.
exist smaller than any x > 0? If not, how many same unit fractions exist smaller than any x > 0? How many are smalleror equal than all unit
fractions?
On 09.09.2024 17:14, joes wrote:All unit fractions? How do you choose that interval?
Am Mon, 09 Sep 2024 17:01:35 +0200 schrieb WM:
I know that all exist.They all are present from the start. I simply choose a gap that is tooHow do you do that?
small to contain ℵo unit fractions.
Why should that be so? It is contradicted by the infinity of unitIf ℵo points are there, then one is on the left-hand side.Those first points are on the right, obviously.How can a claim a set of values less then a SPECIFIC NUMBER if II claim that all unit fractions are existing as real points of the
don't have the number?
real line. Therefore there is a first one. NUF(x) cannot increase
without passing 1 when real points are counted. Your ℵo points cannot
exist without including 1, 2, 3 first points.
On 09.09.2024 16:32, Python wrote:I don’t follow this step.
Le 09/09/2024 à 12:19, Prof. Dr. Mückenheim, aka WM a écrit :
1/n - 1/(n+1) = x is an interval between two unit fraction. Thisℵo unit fractions cannot fit into one of the ℵo intervals between two >>> of them.(O, x) is NOT an interval between two unit fractions.
interval is shifted to the origin, yielding the interval (0, x).
It does not contain ℵo unit fractions. It does not contain 1/n.Nor does it contain (finitely) many others. So what?
Note that unit fractions are points on the real line. Therefore there isHow does that follow?
a beginning.
On 09.09.2024 13:49, joes wrote:The wrong quantifier order is: „There is a fixed infinite set of unit fractions, which are all less than any positive x.” The right one is:
Am Mon, 09 Sep 2024 12:27:47 +0200 schrieb WM:The quantifier order related to the problem is this: NUF(x) = ℵo means: There exist ℵo unit fractions smaller than any x > 0. If this is not
On 08.09.2024 22:11, Richard Damon wrote:More precisely: every positive x has infinitely many smaller unit
On 9/8/24 3:48 PM, WM wrote:ℵo unit fractions are claimed to be smaller than every x > 0. If that
Select any gap between one of the first ℵo unit fractions and itsBut that is changing the value of x in the middle of the problem
neighbour. Call its size x. Then ℵo unit fractions cannot fit into >>>>> the interval (0, x), independent of the actual size.
which isn't allowed.
Given that new x, we can choose a new set of Aleph_0 unit fractions
below that x.
is true then I can choose as the x one of the ℵo intervals between two >>> of them.
fractions (mind the quantifier order).
true, then there are fewer. How many unit fractions are smaller than any
x > 0. THAT is the question. None. But all are differente. Hence there
must be a first one smaller than all other unit fractions.
But still infinitely many, since only finitely many are missing.A number is not an interval.An interval has a length that can be expressed by a real number:
1/n - 1/(n+1) = x .
Then the interval (0, x) contains not all unit fractions, for instance
not 1/n.
Le 09/09/2024 à 17:15, Crank Mückenheim, aka WM a écrit :
On 09.09.2024 16:32, Python wrote:
Le 09/09/2024 à 12:19, Crank Mückenheim, aka WM a écrit :
ℵo unit fractions cannot fit into one of the ℵo intervals between
two of them.
(O, x) is NOT an interval between two unit fractions.
1/n - 1/(n+1) = x is an interval between two unit fraction. [WM}
No. It is a number.
This interval is shifted to the origin, yielding the interval (0, x).
It does not contain ℵo unit fractions. It does not contain 1/n.
1/n is not in (0, x). Sure. So what? Nevertheless there are Aleph_0 unit fractions in (0, x). No need for 1/n to be there, there far enough other fractions.
Am Mon, 09 Sep 2024 17:08:49 +0200 schrieb WM:
The quantifier order related to the problem is this: NUF(x) = ℵo means:
There exist ℵo unit fractions smaller than any x > 0.
I still don’t understand. You can choose any size of interval and slide
it around to include an arbitrary number of unit fractions.
Le 09/09/2024 à 17:15, Prof. Dr. Mückenheim, aka WM a écrit :
On 09.09.2024 16:32, Python wrote:
Le 09/09/2024 à 12:19, Prof. Dr. Mückenheim, aka WM a écrit :
1/n is not in (0, x). Sure. So what? Nevertheless there are Aleph_0 unit fractions in (0, x). No need for 1/n to be there, there far enough other fractions.
On 09/08/2024 09:59 PM, Jim Burns wrote:
2.
It is an essential aspect of these discussions
that
whatever is not being discussed
is not being discussed.
That essential aspect seems to be
what you (RF) consider
hypocritically ignoring
whatever is not being discussed.
Consider that it is possible for there to be
many discussions, with many different topics.
It is true at the same time
that
nonstandard analysis can be pursued
and
the complete ordered field doesn't hold
anything other than elements of
the complete ordered field.
As context expands there's nothing left out.
The objects of mathematics all live in
one mathematical universe,
1.
I wonder what your own conscientious response
will be to the infiniteness of the set of
all finite non.self.membered sets.
Am Mon, 09 Sep 2024 17:32:11 +0200 schrieb WM:
On 09.09.2024 17:15, joes wrote:That is a different question
Am Mon, 09 Sep 2024 16:53:32 +0200 schrieb WM:My question concerns same unit fractions only. Do ℵo unit fractions
You claim that ℵo unit fractions are smaller than ANY x > 0.Yes. Not all the same ones of course.
exist smaller than any x > 0? If not, how many same unit fractions exist
smaller than any x > 0? How many are smalleror equal than all unit
fractions?
Of course no unit
fraction is smaller than every other unit fraction.
On 08.09.2024 22:21, Jim Burns wrote:
There aren't two.
There isn't one.
There is no smallest unit fraction.
Then there is no unit fraction.
ℵo unit fractions cannot exist without
1, 2, 3, ... unit fractions before.
Am Mon, 09 Sep 2024 17:26:12 +0200 schrieb WM:
On 09.09.2024 17:14, joes wrote:
If ℵo points are there, then one is on the left-hand side.Why should that be so?
On 9/9/2024 6:39 AM, WM wrote:
ℵo unit fractions cannot exist without
1, 2, 3, ... unit fractions before.
Each unit.fraction has 1,2,3,... unit.fraction before.
It is incorrect (it is a quantifier shift)
to conclude from that correct claim
that
🛇 there are 1,2,3,... unit.fractions before each
Am Mon, 09 Sep 2024 17:15:23 +0200 schrieb WM:
Note that unit fractions are points on the real line. Therefore there isHow does that follow?
a beginning.
On 09.09.2024 17:15, joes wrote:
I still don’t understand.
You can choose any size of interval and
slide it around to include
an arbitrary number of unit fractions.
ℵo unit fractions exist invariably
and require a minimum length d.
On 09.09.2024 17:55, joes wrote:An infinite chain beginning at 1.
Am Mon, 09 Sep 2024 17:15:23 +0200 schrieb WM:
What configuration could be the alternative?Note that unit fractions are points on the real line. Therefore thereHow does that follow?
is a beginning.
Am Mon, 09 Sep 2024 17:08:49 +0200 schrieb WM:
On 09.09.2024 13:49, joes wrote:
Am Mon, 09 Sep 2024 12:27:47 +0200 schrieb WM:
[Such] an interval has a length that can be expressed by a real number:
1/n - 1/(n+1) = x .
Then the interval (0, x) contains not all unit fractions, for instance
not 1/n.
But still infinitely many, since only finitely many are missing.
On 09.09.2024 17:15, joes wrote:SO WHAT
I still don’t understand. You can choose any size of interval and slideℵo unit fractions exist invariably and require a minimum length d. Take
it around to include an arbitrary number of unit fractions.
one of the ℵo gaps. It is smaller than d.
Simplest argument: If a chain of real points exists on the real axis,Proof?
then it has a beginning.
On 09.09.2024 21:05, Jim Burns wrote:
On 9/9/2024 6:39 AM, WM wrote:
ℵo unit fractions cannot exist without
1, 2, 3, ... unit fractions before.
Each unit.fraction has 1,2,3,... unit.fraction before.
It is incorrect (it is a quantifier shift)
to conclude from that correct claim
that
🛇 there are 1,2,3,... unit.fractions before each
It is a property of
real points on the positive real line,
that every sequence has a beginning.
What alternative configuration can the points have
in your opinion.
This is the first question:
Do ℵo unit fractions exist smaller than any x > 0,
as Fritsche claims?
The second question concerns a weaker claim:
Do ℵo unit fractions exist smaller than any 1/n > 0?
Am Mon, 09 Sep 2024 20:44:35 +0200 schrieb WM:
Simplest argument: If a chain of real points exists on the real axis,
then it has a beginning.
Am 10.09.2024 um 01:08 schrieb Chris M. Thomasson:
Take the gap between:
1/1 and 1/2. There are infinitely many unit fractions that are small
enough to fit within that gap.
Oh, really?! Could you name (just) o n e? :-)
I mean ... a unit fraction u such that 1/2 < u < 1/1. :-o
Take the gap between:
1/1 and 1/2. There are infinitely many unit fractions that are small
enough to fit within that gap.
Am Mon, 09 Sep 2024 20:44:35 +0200 schrieb WM:
ℵo unit fractions [...] require a minimum length d.
On 09.09.2024 17:51, joes wrote:They do in fact.
Am Mon, 09 Sep 2024 17:26:12 +0200 schrieb WM:
On 09.09.2024 17:14, joes wrote:
This is so because every sequence of points on the positive real axisIf ℵo points are there, then one is on the left-hand side.Why should that be so?
has either one smallest point or more than one. In any case it has a beginning because real points of a sequence do not appear after an
opaque cloud.
On 09/09/2024 11:23 AM, Jim Burns wrote:
On 9/9/2024 12:59 PM, Ross Finlayson wrote:
On 09/08/2024 09:59 PM, Jim Burns wrote:
On 9/8/2024 6:34 PM, Ross Finlayson wrote:
It's already been thoroughly elaborated and
as attached to formalistic symbolry,
that "Russell's thesis, of an antinomy" is that
the set of
the finite sets that don't contain themselves,
exactly like the ordinals are mostly simply modeled to be,
does and doesn't contain itself,
I wonder what your own conscientious response
will be to the infiniteness of the set of
all finite non.self.membered sets.
To your question, "are the finites infinite",
well yeah.
On 09.09.2024 17:15, joes wrote:
Am Mon, 09 Sep 2024 16:53:32 +0200 schrieb WM:
You claim that ℵo unit fractions are smaller than ANY x > 0.Yes. Not all the same ones of course.
My question concerns same unit fractions only. Do ℵo unit fractions
exist smaller than any x > 0? If not, how many same unit fractions exist smaller than any x > 0? How many are smalleror equal than all unit
fractions?
Regards, WM
On 09.09.2024 13:32, Richard Damon wrote:
No, because the claim is GIVEN an x, we can do "Y", that is make the
Aleph_0 unit fractions below it.
You are in error. MY CLAIM is: Given ℵo unit fractions smaller than
every x > 0, then I choose any of their gaps.
Until you have chosen your x, we don't need to provide those unit
fractions, so, you can't use them to create your x.
You claim that ℵo unit fractions are smaller than ANY x > 0. That is
simply fool's crap.
To follow YOUR idea, then *YOU* get stuck in the infinite loop of
every time you change your x, the unit fractions change so you need to
change to another x, and the unit fractions change again.
I do not change anything. There are unit fractions as real points on the
real line. These real points are really real and therefore must start somewhere. Hence there is a beginning, one ore more first points.
Regards, WM
On 09.09.2024 13:46, joes wrote:
Am Mon, 09 Sep 2024 12:22:21 +0200 schrieb WM:
ℵo unit fractions cannot fit into one of the ℵo intervals between two ofWhat? It’s about the size, not the number of intervals.
them, because ℵo unit fractions occupy ℵo intervals.
One of ℵo intervals is smaller than ℵo intervals.
Unit fractions are real points on the real line. Therefore there is a beginning, one or more smallest unit fractions.
Regards, WM
On 09.09.2024 13:47, Richard Damon wrote:
On 9/9/24 6:31 AM, WM wrote:
And all gaps are occupied by the unit fractions. Hence every gap is
too small.
But there is always room at the bottom, where the gaps keep getting
smaller
They all are present from the start. I simply choose a gap that is too
small to contain ℵo unit fractions.
How can a claim a set of values less then a SPECIFIC NUMBER if I don't
have the number?
I claim that all unit fractions are existing as real points of the real
line. Therefore there is a first one. NUF(x) cannot increase without
passing 1 when real points are counted. Your ℵo points cannot exist
without including 1, 2, 3 first points.
Regards, WM
On 09.09.2024 17:14, joes wrote:
Am Mon, 09 Sep 2024 17:01:35 +0200 schrieb WM:
They all are present from the start. I simply choose a gap that is tooHow do you do that?
small to contain ℵo unit fractions.
I know that all exist.
Those first points are on the right, obviously.How can a claim a set of values less then a SPECIFIC NUMBER if I don't >>>> have the number?I claim that all unit fractions are existing as real points of the real
line. Therefore there is a first one. NUF(x) cannot increase without
passing 1 when real points are counted. Your ℵo points cannot exist
without including 1, 2, 3 first points.
If ℵo points are there, then one is on the left-hand side. Note that
they are real points on the real line. There cannot be more points
unless they start with 1. Everything else is belief in ghosts of matheology
Regards, WM
On 09.09.2024 17:51, joes wrote:
Am Mon, 09 Sep 2024 17:26:12 +0200 schrieb WM:
On 09.09.2024 17:14, joes wrote:
If ℵo points are there, then one is on the left-hand side.Why should that be so?
This is so because every sequence of points on the positive real axis
has either one smallest point or more than one. In any case it has a beginning because real points of a sequence do not appear after an
opaque cloud.
Regards, WM
On 09.09.2024 13:51, joes wrote:
Am Mon, 09 Sep 2024 12:31:55 +0200 schrieb WM:
Al gaps are occupied by the unit fractions. Hence every gap is tooIn which sense are the gaps „occupied”?
small.
ℵo unit fractions cover a distance d which is the sum of the gaps
between them.
First you claim the unit fractions with their gaps. Then I chooseEach gap is smaller than all gaps?
one of them, irrelevant which one. Each one is smaller than all.
The sum of all gaps is larger than one of them.
Regards, WM
On 09.09.2024 17:15, joes wrote:
I still don’t understand. You can choose any size of interval and slideℵo unit fractions exist invariably and require a minimum length d. Take
it around to include an arbitrary number of unit fractions.
one of the ℵo gaps. It is smaller than d.
Simplest argument: If a chain of real points exists on the real axis,
then it has a beginning.
Regards, WM
On 9/9/2024 2:44 PM, WM wrote:
The unit fractions
before x > 0 and before x′ > 0
vary.
For both x and x′, they are ℵ₀.many,
but not the same ℵ₀.many.
and require a minimum length d.
No.
There is a greatest.lower.bound 0
but there isn't a _minimum_
so
each d > 0 is not.minimum.
Am Mon, 09 Sep 2024 21:08:42 +0200 schrieb WM:
On 09.09.2024 17:55, joes wrote:An infinite chain beginning at 1.
Am Mon, 09 Sep 2024 17:15:23 +0200 schrieb WM:What configuration could be the alternative?
Note that unit fractions are points on the real line. Therefore thereHow does that follow?
is a beginning.
Am Mon, 09 Sep 2024 20:44:35 +0200 schrieb WM:
Simplest argument: If a chain of real points exists on the real axis,Proof?
then it has a beginning.
Why do they need a beginning (at that end)?
There is a boundry, and that is 0, but that isn't the "first" of them,
as it is outside the set.
On 9/9/24 10:57 AM, WM wrote:
Unit fractions are real points on the real line. Therefore there is aWhere do you get that from?
beginning, one or more smallest unit fractions.
A GIVEN infinte set of unit fractions will have a total. length of d,
but by just removing a finite number of the largest elements, you still
have an infinite set, and the size of that new set can be made as small
as you want (as long as that value is actually > 0)
The size of that new set can be made as small as you want
It seems that any non-zero gap can have unit fractions
small enough to fit in it...
On 9/9/2024 5:32 PM, Moebius wrote:
Am 10.09.2024 um 01:08 schrieb Chris M. Thomasson:
Take the gap between:
1/1 and 1/2. There are infinitely many unit fractions that are small
enough to fit within that gap.
Oh, really?! Could you name (just) o n e? :-)
1/4?
I mean ... a unit fraction u such that 1/2 < u < 1/1. :-o
On 9/10/2024 11:42 AM, Chris M. Thomasson wrote:
On 9/10/2024 4:09 AM, FromTheRafters wrote:
Chris M. Thomasson brought next idea :
On 9/9/2024 11:49 AM, WM wrote:
On 09.09.2024 17:27, Python wrote:
Le 09/09/2024 à 17:15, Prof. Dr. Mückenheim, aka WM a écrit : >>>>>>> On 09.09.2024 16:32, Python wrote:If you cannot understand mathematics consider the simplest logic:
1/n is not in (0, x). Sure. So what? Nevertheless there areLe 09/09/2024 à 12:19, Prof. Dr. Mückenheim, aka WM a écrit : >>>>>
Aleph_0 unit
fractions in (0, x). No need for 1/n to be there, there far enough >>>>>> other
fractions.
If a sequence of different real points exists on the positive real
axis, then it has a beginning. Otherwise it could be a cloud but
not a sequence.
Oh wow. You need help! Or we do for even giving you the time of day.
YIKES!!!!!!!!!!
Actually, I think he is close to correct. The thing is, he wants the
first term of the sequence to be last. The reals are not a sequence,
but the unit fractions are and they start with 1/1 not at some
imagined other end. You need a first and likely a next to start.
So, I am thinking there is a unit fraction that fits in the simple gap
of 1/1 and 1/2:
1/1----->1/4------>1/2
?
Fair enough? Is this getting on track or going off the rails?
Actually, there
unit fractions in that gap[.]
On 9/10/2024 11:35 AM, Chris M. Thomasson wrote:
On 9/9/2024 5:32 PM, Moebius wrote:
Am 10.09.2024 um 01:08 schrieb Chris M. Thomasson:
Take the gap between:
1/1 and 1/2. There are infinitely many unit fractions that are small
enough to fit within that gap.
Oh, really?! Could you name (just) o n e? :-)
1/4?
1/4 is between 1/1 and 1/2:
1/1---->1/4---->1/2
So, it's not _strict_
Actually, I think he is close to correct. The thing is, he wants the
first term of the sequence to be last.
It isn't an "opaque cloud" that stops them from having a smallest, but a infinite accumulation of values at that bound (that is outside the set).
On 9/10/2024 12:31 PM, Moebius wrote:
Am 10.09.2024 um 21:05 schrieb Chris M. Thomasson:
On 9/10/2024 11:42 AM, Chris M. Thomasson wrote:
On 9/10/2024 4:09 AM, FromTheRafters wrote:
Chris M. Thomasson brought next idea :So, I am thinking there is a unit fraction that fits in the simple
On 9/9/2024 11:49 AM, WM wrote:
On 09.09.2024 17:27, Python wrote:
Le 09/09/2024 à 17:15, Prof. Dr. Mückenheim, aka WM a écrit : >>>>>>>>> On 09.09.2024 16:32, Python wrote:If you cannot understand mathematics consider the simplest logic: >>>>>>> If a sequence of different real points exists on the positive
1/n is not in (0, x). Sure. So what? Nevertheless there areLe 09/09/2024 à 12:19, Prof. Dr. Mückenheim, aka WM a écrit : >>>>>>>
Aleph_0 unit
fractions in (0, x). No need for 1/n to be there, there far
enough other
fractions.
real axis, then it has a beginning. Otherwise it could be a cloud >>>>>>> but not a sequence.
Oh wow. You need help! Or we do for even giving you the time of
day. YIKES!!!!!!!!!!
Actually, I think he is close to correct. The thing is, he wants
the first term of the sequence to be last. The reals are not a
sequence, but the unit fractions are and they start with 1/1 not at
some imagined other end. You need a first and likely a next to start. >>>>
gap of 1/1 and 1/2:
1/1----->1/4------>1/2
?
Fair enough? Is this getting on track or going off the rails?
Actually, there
is NO
unit fractions in that gap[.]
:-)
How about:
1/1----->(1/4 + 1/4 + 1/4)------>1/2
1----->.75----->1/2
?
There are infinite[ly many] unit fractions in the gap?
Am 10.09.2024 um 22:25 schrieb Chris M. Thomasson:
On 9/10/2024 12:31 PM, Moebius wrote:
Am 10.09.2024 um 21:05 schrieb Chris M. Thomasson:
On 9/10/2024 11:42 AM, Chris M. Thomasson wrote:
On 9/10/2024 4:09 AM, FromTheRafters wrote:
Chris M. Thomasson brought next idea :
On 9/9/2024 11:49 AM, WM wrote:
On 09.09.2024 17:27, Python wrote:
Le 09/09/2024 à 17:15, Prof. Dr. Mückenheim, aka WM a écrit : >>>>>>>>>> On 09.09.2024 16:32, Python wrote:If you cannot understand mathematics consider the simplest logic: >>>>>>>> If a sequence of different real points exists on the positive
1/n is not in (0, x). Sure. So what? Nevertheless there areLe 09/09/2024 à 12:19, Prof. Dr. Mückenheim, aka WM a écrit : >>>>>>>>
Aleph_0 unit
fractions in (0, x). No need for 1/n to be there, there far
enough other
fractions.
real axis, then it has a beginning. Otherwise it could be a
cloud but not a sequence.
Oh wow. You need help! Or we do for even giving you the time of
day. YIKES!!!!!!!!!!
Actually, I think he is close to correct. The thing is, he wants
the first term of the sequence to be last. The reals are not a
sequence, but the unit fractions are and they start with 1/1 not
at some imagined other end. You need a first and likely a next to
start.
So, I am thinking there is a unit fraction that fits in the simple
gap of 1/1 and 1/2:
1/1----->1/4------>1/2
?
Fair enough? Is this getting on track or going off the rails?
Actually, there
is NO
unit fractions in that gap[.]
:-)
How about:
1/1----->(1/4 + 1/4 + 1/4)------>1/2
1----->.75----->1/2
?
There are infinite[ly many] unit fractions in the gap?
Nope. There is NO unit fraction in te gap (between 1/2 and 1/2).
Hint: 1/4 + 1/4 + 1/4 = 3/4 is not a unit fraction. :-)
How many rounds do you want to... :-)
On 9/10/2024 12:30 PM, Moebius wrote:
Am 10.09.2024 um 20:38 schrieb Chris M. Thomasson:
On 9/10/2024 11:35 AM, Chris M. Thomasson wrote:
On 9/9/2024 5:32 PM, Moebius wrote:
Am 10.09.2024 um 01:08 schrieb Chris M. Thomasson:
Take the gap between:
1/1 and 1/2. There are infinitely many unit fractions that are
small enough to fit within that gap.
Oh, really?! Could you name (just) o n e? :-)
1/4?
1/4 is between 1/1 and 1/2:
1/1---->1/4---->1/2
lol. If you say so. :-)
So, it's not _strict_
Actually, it's not between 1/2 and 1/1 at all (if we presuppose the
usual order < on the rational numbers).
The LENGTH OF THE gap from 1/1 to 1/2 is 1/2 so there are infinity many unit fractions
that are smaller than 1/2.
On 09.09.2024 21:43, Jim Burns wrote:
The unit fractions
before x > 0 and before x′ > 0
vary.
For both x and x′, they are ℵ₀.many,
but not the same ℵ₀.many.
The same ℵ₀ many are existing in both cases,
in addition there are
finitely many further unit fractions.
and require a minimum length d.
No.
There is a greatest.lower.bound 0
but there isn't a _minimum_
There is a minimum larger than
a distance of countably many points.
so
each d > 0 is not.minimum.
A d of of countably many points is
less than the minimum.
On 10.09.2024 13:49, Richard Damon wrote:
On 9/9/24 10:57 AM, WM wrote:
Unit fractions are real points on the real line. Therefore there is aWhere do you get that from?
beginning, one or more smallest unit fractions.
What is the alternative?
Regards, WM
On 10.09.2024 13:41, Richard Damon wrote:
Why do they need a beginning (at that end)?
Because there is none in the negative and some in the prositive. At that
end.
There is a boundry, and that is 0, but that isn't the "first" of them,
as it is outside the set.
Either the set starts with one unit fraction or with a cloud of unit fractions.
Regards, WM
On 10.09.2024 13:59, Richard Damon wrote:
It isn't an "opaque cloud" that stops them from having a smallest, but
a infinite accumulation of values at that bound (that is outside the
set).
One or finitely many or a cloud. More is not available.
Regards, WM
On 09.09.2024 21:57, joes wrote:
Am Mon, 09 Sep 2024 20:44:35 +0200 schrieb WM:
Simplest argument: If a chain of real points exists on the real axis,Proof?
then it has a beginning.
What else could there be? Any idea?
Regards, WM
On 09.09.2024 16:32, Python wrote:
Le 09/09/2024 à 12:19, Prof. Dr. Mückenheim, aka WM a écrit :
ℵo unit fractions cannot fit into one of the ℵo intervals between two >>> of them.
(O, x) is NOT an interval between two unit fractions.
1/n - 1/(n+1) = x is an interval between two unit fraction. This
interval is shifted to the origin, yielding the interval (0, x). It does
not contain ℵo unit fractions. It does not contain 1/n.
Note that unit fractions are points on the real line. Therefore there is
a beginning. How many unit fractions can be smallerorequal than all unit fractions. This question proves the existence of a smallest unit fraction.
Regards, WM
On 10.09.2024 14:07, Richard Damon wrote:
A GIVEN infinte set of unit fractions will have a total. length of d,
but by just removing a finite number of the largest elements, you
still have an infinite set, and the size of that new set can be made
as small as you want (as long as that value is actually > 0)
Do countably many points exist as a subdistance of every distance of uncountably many points, like the gaps between two unit fractions?
The size of that new set can be made as small as you want
Even when the size is only countably many points?
Regards, WM
On 10.09.2024 13:09, FromTheRafters wrote:
Actually, I think he is close to correct. The thing is, he wants the
first term of the sequence to be last.
Real points can be addressed from every side.
Regards, WM
On 09.09.2024 17:55, joes wrote:
Am Mon, 09 Sep 2024 17:15:23 +0200 schrieb WM:
Note that unit fractions are points on the real line. Therefore there is >>> a beginning.How does that follow?
What configuration could be the alternative?
Regards, WM
On 09.09.2024 21:51, joes wrote:
Am Mon, 09 Sep 2024 21:08:42 +0200 schrieb WM:
On 09.09.2024 17:55, joes wrote:An infinite chain beginning at 1.
Am Mon, 09 Sep 2024 17:15:23 +0200 schrieb WM:What configuration could be the alternative?
Note that unit fractions are points on the real line. Therefore there >>>>> is a beginning.How does that follow?
What begins after zero? Does there something related to unit fractions
exist?
Regards, WM
On 9/10/24 2:59 PM, WM wrote:
Real points can be addressed from every side.
But only indexed from an actual end point.
On 9/10/24 12:18 PM, WM wrote:
What begins after zero? Does there something related to unit fractions
exist?
And infinite set that approaches it arbitrarily close. It is an
unbounded end.
On 9/10/2024 12:16 PM, WM wrote:
On 09.09.2024 21:43, Jim Burns wrote:
The unit fractions
before x > 0 and before x′ > 0
vary.
For both x and x′, they are ℵ₀.many,
but not the same ℵ₀.many.
The same ℵ₀ many are existing in both cases,
No.
They are not the same for ⌊⅟x⌋ < ⌊⅟x′⌋
_Minimum distances of countably.many.points does not exist_
On 11.09.2024 04:09, Richard Damon wrote:What about open intervals?
On 9/10/24 2:59 PM, WM wrote:
Every real point is an end point (of its subinterval).Real points can be addressed from every side.But only indexed from an actual end point.
On 10.09.2024 23:12, Jim Burns wrote:
On 9/10/2024 12:16 PM, WM wrote:
On 09.09.2024 21:43, Jim Burns wrote:
The unit fractions
before x > 0 and before x′ > 0
vary.
For both x and x′, they are ℵ₀.many,
but not the same ℵ₀.many.
The same ℵ₀ many are existing in both cases,
No.
They are not the same for ⌊⅟x⌋ < ⌊⅟x′⌋
The unit fractions are fixed at the real line.
If you cut some more or some less by x than by x'
is irrelevant for the fact that
the same ℵo unit fractions remain in both cases.
0, ..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.
_Minimum distances of countably.many.points does not exist_
Every uncountable set has a countable subset,
in fact uncoubtably many.
Therefore (0, d) c (0, x).
(0, d) is coubtable and contains at most one
unit fraction.
On 11.09.2024 04:11, Richard Damon wrote:
On 9/10/24 12:18 PM, WM wrote:
What begins after zero?
Does there
something related to unit fractions
exist?
And infinite set that approaches it
arbitrarily close.
It is an unbounded end.
Real points on the real line
either are there or are not there.
In particular
they do not "approach" any other point.
Am Wed, 11 Sep 2024 16:37:30 +0200 schrieb WM:
On 11.09.2024 04:09, Richard Damon wrote:What about open intervals?
On 9/10/24 2:59 PM, WM wrote:Every real point is an end point (of its subinterval).
Real points can be addressed from every side.But only indexed from an actual end point.
On 9/11/2024 10:48 AM, WM wrote:
The unit fractions are fixed at the real line.
If you cut some more or some less by x than by x'
is irrelevant for the fact that
the same ℵo unit fractions remain in both cases.
⅟⌊k+⅟x⌋ ⟷ ⅟⌊k+⅟x′⌋
is a bijection between different visibleᵂᴹ unit.fractions
⅟⌊k+⅟x⌋ ≠ ⅟⌊k+⅟x′⌋
0, ..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.
0 is not a unit.fraction.
..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.
holds only visibleᵂᴹ unit fractions
On 9/11/2024 10:35 AM, WM wrote:
Real points on the real line
either are there or are not there.
Sets of real.line.points and
real.line.points
are not the same things.
In particular
they do not "approach" any other point.
Define β as
the greatest.lower.bound of visiblesᵂᴹ ⅟ℕᵈᵉᶠ
β = glb.⅟ℕᵈᵉᶠ
Positive β requires impossibilities.
It requires that ½.β
both IS and IS NOT
undercut by a visibleᵂᴹ unit.fraction.
⅟ℕᵈᵉᶠ does not change. (Sets do not change.)
We use "approach" to say "no gap".
On 11.09.2024 19:29, joes wrote:
Am Wed, 11 Sep 2024 16:37:30 +0200 schrieb WM:
On 11.09.2024 04:09, Richard Damon wrote:What about open intervals?
On 9/10/24 2:59 PM, WM wrote:Every real point is an end point (of its subinterval).
Real points can be addressed from every side.But only indexed from an actual end point.
Open intervals are intervals which have dark endpoints. Note that
between any pair of visible real numbers there are many dark real numbers.
Regards, WM
On 11.09.2024 04:09, Richard Damon wrote:
On 9/10/24 2:59 PM, WM wrote:
Real points can be addressed from every side.
But only indexed from an actual end point.Every real point is an end point (of its subinterval).
Regards, WM
On 11.09.2024 20:52, Jim Burns wrote:
⅟ℕᵈᵉᶠ does not change. (Sets do not change.)
Because their elements,
here: the points
do not change.
β = glb.⅟ℕᵈᵉᶠ
Positive β requires impossibilities.
It requires that ½.β
both IS and IS NOT
undercut by a visibleᵂᴹ unit.fraction.
In some dark cases ½.β does not exist.
But you claim that if β exists, then ½.β exists.
And if ½.β exists, then β/4 exists, and so on.
That is potential infinity.
If all exists, then a smallest exists.
We use "approach" to say "no gap".
And infinite set that approaches it
arbitrarily close.
It is an unbounded end.
Real points on the real line
either are there or are not there.
Sets of real.line.points and
real.line.points
are not the same things.
But we need not use sets at all
if we use the points.
On 9/11/24 3:40 PM, WM wrote:
Open intervals are intervals which have dark endpoints. Note that
between any pair of visible real numbers there are many dark real
numbers.
No, Open intervals are intervals that exclude the endpoint from the
interval, and thus don't HAVE an "endpoint".
Chris M. Thomasson laid this down on his screen :Grey points are dark points which can become visible. But the endpoints of
On 9/11/2024 12:40 PM, WM wrote:
Open intervals are intervals which have dark endpoints.
What about the gray ones?
On 9/11/2024 3:53 PM, WM wrote:
On 11.09.2024 20:52, Jim Burns wrote:
⅟ℕᵈᵉᶠ does not change. (Sets do not change.)
Because their elements,
here: the points
do not change.
Mathematical sets do not change
because
mathematical sets are immaterial;
thus, a different set merely to hold
slightly different elements
weighs nothing and costs nothing.
Mathematical sets do not change
because,
in an all.not.first.false finite claim.sequence,
claims about
mathematical sets both with and without some element
are NOT reliably.true.
Mathematical sets do not change
because,
in an all.not.first.false finite claim.sequence,
claims about
mathematical sets NOT both with and without some element
ARE reliably.true,
even if they're about infinitely.many never.seen
-- and, in that way,
unchanging sets can serve as a finite tool with which
to explore infinity.
β = glb.⅟ℕᵈᵉᶠ
Positive β requires impossibilities.
It requires that ½.β
both IS and IS NOT
undercut by a visibleᵂᴹ unit.fraction.
In some dark cases ½.β does not exist.
For each β, ½⋅β exists
among points (in ℝ) situating
splits (of ℚ) of differences of ratios of
of points (in ℕ) countable.to from 1,
which is to say well.ordered,
with each point successored and predecessored
except the first point, 1, only successored.
If darkᵂᴹ β allows ½⋅β to not.exist,
then darkᵂᴹ β not.exists among the points of ℝ
as described here.
For each of those β, ½⋅β exists.
However,
no ½⋅β exists for positive β = glb.⅟ℕᵈᵉᶠ
Therefore,
none of those β are positive β = glb.⅟ℕᵈᵉᶠ
0 = glb.⅟ℕᵈᵉᶠ
No gap exists between 0 and ⅟ℕᵈᵉᶠ
No points not undercut by points of ⅟ℕᵈᵉᶠ
But you claim that if β exists, then ½.β exists.
And if ½.β exists, then β/4 exists, and so on.
There's a proof.
That is potential infinity.
You (WM) apparently only say "potential infinity"
when you've recognized that you've lost an argument.
I accept your concession.
If all exists, then a smallest exists.
In ℝ.points situating ℚ.splits of
differences.of.ratios of countable.to ℕ.points,
no smallest exists.
We use "approach" to say "no gap".
For example,
⎛ ⅟ℕᵈᵉᶠ approaches 0
⎜ 0 = glb.⅟ℕᵈᵉᶠ
⎜ 0 ≤ᵉᵃᶜʰ ⅟ℕᵈᵉᶠ
⎝ ¬∃ᴿx: 0 < x ≤ᵉᵃᶜʰ ⅟ℕᵈᵉᶠ
And infinite set that approaches it
arbitrarily close.
It is an unbounded end.
Real points on the real line
either are there or are not there.
Sets of real.line.points and
real.line.points
are not the same things.
But we need not use sets at all
if we use the points.
If you use points as sets, expect gibberish.
For each point x in (0,1]: 0 < x/2 < x
No point exists between 0 and (0,1]
On 9/11/2024 3:53 PM, WM wrote:
But you claim that if β exists, then ½.β exists.
And if ½.β exists, then β/4 exists, and so on.
There's a proof.
But we need not use sets at all
if we use the points.
If you use points as sets, expect gibberish.
For each point x in (0,1]: 0 < x/2 < x
No point exists between 0 and (0,1]
So, you can't "index" an unbounded set of unit fractions from 0, as
there isn't a "first" unit fraction from that end.
We can "address" those unit fractions with the value, but we can not
"index" them from 0, only from 1/1.
On 12.09.2024 03:04, Richard Damon wrote:
On 9/11/24 3:40 PM, WM wrote:
Open intervals are intervals which have dark endpoints. Note that
between any pair of visible real numbers there are many dark real
numbers.
No, Open intervals are intervals that exclude the endpoint from the
interval, and thus don't HAVE an "endpoint".
Intervals consist of points. Where no point is, there is no interval.
Where an interval is, there is a point. The belif in an interval-end
without points is pure matheology.
Regards, WM
Le 12/09/2024 à 03:00, Richard Damon a écrit :
So, you can't "index" an unbounded set of unit fractions from 0, as
there isn't a "first" unit fraction from that end.
We can "address" those unit fractions with the value, but we can not
"index" them from 0, only from 1/1.
If you can index all unit fractions, then you can index them from every
side.
Fact is that NUF(x) increases from 0, but at no point it can inc4rease
by more than 1 because of
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Regards, WM
On 12.09.2024 07:03, Jim Burns wrote:
On 9/11/2024 3:53 PM, WM wrote:
It is based on an axiom which does not reproduce real mathematics.But you claim that if β exists, then ½.β exists.
And if ½.β exists, then β/4 exists, and so on.
There's a proof.
But we need not use sets at all
if we use the points.
If you use points as sets, expect gibberish.
Use points *as points* and investigate their properties.
For each point x in (0,1]: 0 < x/2 < x
No point exists between 0 and (0,1]
Nevertheless (0,1] has a smallest point. It is dark. Where no point is,
there is no interval. Interval-ends without points are matheology.
Regards, WM
After serious thinking WM wrote :
Le 11/09/2024 à 23:13, FromTheRafters a écrit :
Chris M. Thomasson laid this down on his screen :Grey points are dark points which can become visible.
On 9/11/2024 12:40 PM, WM wrote:
Open intervals are intervals which have dark endpoints.
What about the gray ones?
Points which change? What function causes a point to change?
But the endpoints of open intervals will remain dark forever.
Open intervals simply don't contain endpoints, dark or otherwise.
On 9/12/24 7:24 AM, WM wrote:
On 12.09.2024 03:04, Richard Damon wrote:
On 9/11/24 3:40 PM, WM wrote:
Open intervals are intervals which have dark endpoints. Note that
between any pair of visible real numbers there are many dark real
numbers.
No, Open intervals are intervals that exclude the endpoint from the
interval, and thus don't HAVE an "endpoint".
Intervals consist of points. Where no point is, there is no interval.
Where an interval is, there is a point. The belief in an interval-end
without points is pure matheology.
But there isn't a spot on the line without a "point", so there is no
spot with "no point"
On 9/12/24 7:18 AM, WM wrote:
Le 12/09/2024 à 03:00, Richard Damon a écrit :
So, you can't "index" an unbounded set of unit fractions from 0, as
there isn't a "first" unit fraction from that end.
We can "address" those unit fractions with the value, but we can not
"index" them from 0, only from 1/1.
If you can index all unit fractions, then you can index them from
every side.
Fact is that NUF(x) increases from 0, but at no point it can increase
by more than 1 because of
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Nope, that ASSUMPTION just means you can't actually have an infinite
set, as you can't get to the upper end to let you count down.
1/n - 1/(n+1) > 0 thus says that no 1/n is the smallest, as there will
exist a 1/(n+1) that is smaller that that.
On 09/10/2024 02:21 AM, Jim Burns wrote:
On 9/9/2024 11:11 PM, Ross Finlayson wrote:
On 09/09/2024 11:23 AM, Jim Burns wrote:
On 9/9/2024 12:59 PM, Ross Finlayson wrote:
On 09/08/2024 09:59 PM, Jim Burns wrote:
On 9/8/2024 6:34 PM, Ross Finlayson wrote:
It's already been thoroughly elaborated and
as attached to formalistic symbolry,
that "Russell's thesis, of an antinomy" is that
the set of
the finite sets that don't contain themselves,
exactly like the ordinals are mostly simply modeled to be,
does and doesn't contain itself,
I wonder what your own conscientious response
will be to the infiniteness of the set of
all finite non.self.membered sets.
To your question, "are the finites infinite",
well yeah.
It seems to me (JB) that
⎛
⎜ Russell's thesis is that
⎜ the set of
⎜ the finite sets that don't contain themselves
⎜ does and doesn't contain itself
⎝
is and is not what you (RF) are claiming.
Oh, perhaps, maybe, in a sense, weighing alternatives,
it's what I'm claiming is that it isn't and not is
what BR Bertrand Russell is claiming.
Of course, overgeneralization is generally unsound,
while there are absolutes in logic,
and for example
the unbounded and completions and closures in logic,
which you make BR a dupe and dupe you to unfasten.
Now, maybe this message will reach you or
maybe it won't, ...,
maybe it won't.
On 12.09.2024 14:29, FromTheRafters wrote:
After serious thinking WM wrote :
Le 11/09/2024 à 23:13, FromTheRafters a écrit :
Chris M. Thomasson laid this down on his screen :Grey points are dark points which can become visible.
On 9/11/2024 12:40 PM, WM wrote:
Open intervals are intervals which have dark endpoints.
What about the gray ones?
Points which change? What function causes a point to change?
For an eartworm all numbers are dark, for a dove numbers 1 to 7 are
visible, for your pocket calculator numbers 1 to 10^99 are visible. If
you couple some calculators, you get farther.
But the endpoints of open intervals will remain dark forever.
Open intervals simply don't contain endpoints, dark or otherwise.
You simply don't know about them. No spot of an interval is free of
points. No point of an interval is free of points.
Regards, WM
On 12.09.2024 14:29, Richard Damon wrote:
On 9/12/24 7:24 AM, WM wrote:
On 12.09.2024 03:04, Richard Damon wrote:
On 9/11/24 3:40 PM, WM wrote:
Open intervals are intervals which have dark endpoints. Note that
between any pair of visible real numbers there are many dark real
numbers.
No, Open intervals are intervals that exclude the endpoint from the
interval, and thus don't HAVE an "endpoint".
Intervals consist of points. Where no point is, there is no interval.
Where an interval is, there is a point. The belief in an interval-end
without points is pure matheology.
But there isn't a spot on the line without a "point", so there is no
spot with "no point"
Therefore the end of an open interval consist of points.
Regards, WM
On 12.09.2024 14:35, Richard Damon wrote:
On 9/12/24 7:18 AM, WM wrote:
Le 12/09/2024 à 03:00, Richard Damon a écrit :
So, you can't "index" an unbounded set of unit fractions from 0, as
there isn't a "first" unit fraction from that end.
We can "address" those unit fractions with the value, but we can not
"index" them from 0, only from 1/1.
If you can index all unit fractions, then you can index them from
every side.
Fact is that NUF(x) increases from 0, but at no point it can increase
by more than 1 because of
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Nope, that ASSUMPTION just means you can't actually have an infinite
set, as you can't get to the upper end to let you count down.
Wrong. Dark numbers prevent counting to the end. Dark numbers establish
the existence of a set where no end can be seen. That is the only way to
make infinity and completeness compatible.
1/n - 1/(n+1) > 0 thus says that no 1/n is the smallest, as there will
exist a 1/(n+1) that is smaller that that.
That is not a proof of existence. The formula says: If n and n+1 exist,
then they differ.
Regards, WM
On 12.09.2024 07:03, Jim Burns wrote:
On 9/11/2024 3:53 PM, WM wrote:
But you claim that if β exists, then ½.β exists.
And if ½.β exists, then β/4 exists, and so on.
There's a proof.
It is based on an axiom which
does not reproduce real mathematics.
But we need not use sets at all
if we use the points.
If you use points as sets, expect gibberish.
Use points *as points* and
investigate their properties.
For each point x in (0,1]: 0 < x/2 < x
No point exists between 0 and (0,1]
Nevertheless (0,1] has a smallest point.
It is dark.
Where no point is, there is no interval.
Interval-ends without points are matheology.
On 12.09.2024 14:29, FromTheRafters wrote:
After serious thinking WM wrote :
But the endpoints of open intervals
will remain dark forever.
Open intervals simply don't contain endpoints,
dark or otherwise.
You simply don't know about them.
On 9/2/2024 10:07 AM, WM wrote:
How many different unit fractions are lessorequal than all unit
fractions? [A wrong/nonsensical] answer is: one [or more] unit fraction[s]. >>
[The correct] answer is that no unit fraction is lessorequal than all unit >> fractions. [...]
[In math] there is only the one correct answer given above.
Is this your Dog?
https://youtu.be/ADRGgyhX4YE?t=43
On 09/10/2024 02:21 AM, Jim Burns wrote:
(re-sent)
Oh, perhaps, maybe, in a sense, weighing alternatives,
it's what I'm claiming is that it isn't and not is
what BR Bertrand Russell is claiming.
On 12.09.2024 14:29, Richard Damon wrote:The "end" of an interval is a number such that no element is larger/
On 9/12/24 7:24 AM, WM wrote:Therefore the end of an open interval consist of points.
On 12.09.2024 03:04, Richard Damon wrote:But there isn't a spot on the line without a "point", so there is no
On 9/11/24 3:40 PM, WM wrote:Intervals consist of points. Where no point is, there is no interval.
Open intervals are intervals which have dark endpoints. Note thatNo, Open intervals are intervals that exclude the endpoint from the
between any pair of visible real numbers there are many dark real
numbers.
interval, and thus don't HAVE an "endpoint".
Where an interval is, there is a point. The belief in an interval-end
without points is pure matheology.
spot with "no point"
On 9/12/24 1:44 PM, WM wrote:
Therefore the end of an open interval consist of points.
The "end" of the open interval isn't in the interval, that is why it is called "open".
There is no "first" point of that interval, just an infinite set of
points
approaching that point.
On 9/12/24 1:48 PM, WM wrote:
No spot of an interval is free of
points. No point of an interval is free of points.
Right, and that includes the space between the point you think is the
first point of the open interval and the end point of that interval
On 12.09.2024 20:15, Richard Damon wrote:
On 9/12/24 1:48 PM, WM wrote:
No spot of an interval is free of points. No point of an interval is
free of points.
Right, and that includes the space between the point you think is the
first point of the open interval and the end point of that interval
Between [0, 1] and (0, 1] there is nothing, there is not a spot or point
of the interval.
Regards, WM
If you think there is some n that exist that doesn't have an n+1 that
exist, then your idea of the Natural Number system is
On 12.09.2024 20:16, Richard Damon wrote:
On 9/12/24 1:44 PM, WM wrote:
Therefore the end of an open interval consist of points.
The "end" of the open interval isn't in the interval, that is why it
is called "open".
The end of the interval is a point of the interval. It is called open
because next to a definable point there are always dark points.
There is no "first" point of that interval, just an infinite set of
points
Yes, dark points.
approaching that point.
Points do not approach. They are fixed.
Regards, WM
On 12.09.2024 20:20, Richard Damon wrote:
If you think there is some n that exist that doesn't have an n+1 that
exist, then your idea of the Natural Number system is
much more advanced than you can understand.
Regards, WM
The "end" of an interval is a number such that no element is larger/
smaller, defined as the limit, which may or may not be included.
On 9/12/2024 7:39 AM, WM wrote:
⎛ Expecting infinity to be unsurprising
⎜ is like
⎜ traveling to Outer Mongolia and
⎝ expecting everyone there to speak perfect German.
Infinity is foreign.
Foreign isn't uninvestigatable,
Nevertheless (0,1] has a smallest point.
It is dark.
Where no point is, there is no interval.
Interval-ends without points are matheology.
"Obvious" has devolved into gibberish.
On 13.09.2024 08:40, joes wrote:
The "end" of an interval is a number such that no element is larger/
smaller, defined as the limit, which may or may not be included.
No, the end is a point, in case of open intervals a dark point. There is
no gap on the real axis. All is full of points.
Regards, WM
On 9/12/2024 1:48 PM, WM wrote:
On 12.09.2024 14:29, FromTheRafters wrote:
After serious thinking WM wrote :
But the endpoints of open intervals
will remain dark forever.
Open intervals simply don't contain endpoints,
dark or otherwise.
You simply don't know about them.
The boundary ∂A of set A is
Am 12.09.2024 um 22:36 schrieb Chris M. Thomasson:
On 9/2/2024 10:07 AM, WM wrote:
How many different unit fractions are lessorequal than all unit
fractions?
[The correct] answer is that one unit fraction is lessorequal than all
unit fractions. [...]
[In math] there is only the one correct answer given above.
Indeed,
On 12.09.2024 23:40, Moebius wrote:
Am 12.09.2024 um 22:36 schrieb Chris M. Thomasson:
On 9/2/2024 10:07 AM, WM wrote:
How many different unit fractions are lessorequal than all unit
fractions?
[The correct] answer is that one unit fraction is lessorequal than
all unit fractions. [...]
[In math] there is only the one correct answer given above.
Indeed,
Mathematical proof: NUF grows from 0 to more. At no point it can grow by
more than 1.
Regards, WM
On 12.09.2024 21:12, Jim Burns wrote:
On 9/12/2024 7:39 AM, WM wrote:
⎛ Expecting infinity to be unsurprising
⎜ is like
⎜ traveling to Outer Mongolia and
⎝ expecting everyone there to speak perfect German.
Expecting infinity to obey logic is justified.
Infinity is foreign.
Foreign isn't uninvestigatable,
I investigate it. I am sure that at every finite step Bob remains. And
there are no other steps, even if there are infinitely many.
Nevertheless (0,1] has a smallest point.
It is dark.
Where no point is, there is no interval.
Interval-ends without points are matheology.
"Obvious" has devolved into gibberish.
The real axis has no gaps, not even a single one at the end of (0, 1]
after zero.
Regards, WM
On 12.09.2024 22:54, Jim Burns wrote:
On 9/12/2024 1:48 PM, WM wrote:
On 12.09.2024 14:29, FromTheRafters wrote:
After serious thinking WM wrote :
But the endpoints of open intervals
will remain dark forever.
Open intervals simply don't contain endpoints,
dark or otherwise.
You simply don't know about them.
The boundary ∂A of set A is
not a result of understanding points.
Fact is that the real axis is nothing but its points. There is no gap.
Every point has a next point but next to defined points are dark points.
Next to every defined points there are ℵo dark points. This
configuration cannot be changed.
Regards, WM
On 9/13/24 11:41 AM, WM wrote:
Between [0, 1] and (0, 1] there is nothing, there is not a spot orBut that doesn't mean there is a lowest most point in (0, 1] as any
point of the interval.
point you might want to call it will have another point between it and 0.
Note, I said between the point your THINK is the first, there is no such point, and thus you are agreeing to that fact.
You can only have a first point in the open interval if the interval has
only a finite number of points,
there is no such thing as two consecutive points in a dense system
Thus, we have no points are adjacent, as between them is always more
points.
On 9/13/24 11:43 AM, WM wrote:
Points do not approach. They are fixed.
Right, the points don't move, they just all exist
between any to points (which have no width) is an
infinite number of other points to fill that gap.
On 9/13/24 1:16 PM, WM wrote:
On 12.09.2024 23:40, Moebius wrote:
Am 12.09.2024 um 22:36 schrieb Chris M. Thomasson:
On 9/2/2024 10:07 AM, WM wrote:
How many different unit fractions are lessorequal than all unit
fractions?
[The correct] answer is that one unit fraction is lessorequal than
all unit fractions. [...]
[In math] there is only the one correct answer given above.
Indeed,
Mathematical proof: NUF grows from 0 to more. At no point it can grow
by more than 1.
And at no point can it grow by ONE
so it doesn't actually exist.
On 12.09.2024 20:16, Richard Damon wrote:
On 9/12/24 1:44 PM, WM wrote:
Therefore the end of an open interval consist of points.
The "end" of the open interval isn't in the interval,
that is why it is called "open".
The end of the interval is a point of the interval.
It is called open because next to a definable point
there are always dark points.
There is no "first" point of that interval,
just an infinite set of points
Yes, dark points.
just an infinite set
of points approaching that point.
Points do not approach. They are fixed.
On 9/13/2024 11:43 AM, WM wrote:
An open set holds none of its boundary.
A closed set holds all of its boundary.
If intervals holding endpoints are open,
then single.point interval.intersections are open
and arbitrary unions of single.point.sets are open
and all point.sets are open.
"All point.sets open" is the devolution
of
"Obviously, Bob cannot disappear".
There is no "first" point of that interval,
just an infinite set of points
Yes, dark points.
Points in ℝ[ℚ[ℤ[ℕ[⟨0,…,n⟩]]]] have no
🛇 first in (0,1]
On 12.09.2024 14:35, Richard Damon wrote:When do they exist?
On 9/12/24 7:18 AM, WM wrote:Wrong. Dark numbers prevent counting to the end. Dark numbers establish
Le 12/09/2024 à 03:00, Richard Damon a écrit :Nope, that ASSUMPTION just means you can't actually have an infinite
So, you can't "index" an unbounded set of unit fractions from 0, asIf you can index all unit fractions, then you can index them from
there isn't a "first" unit fraction from that end.
We can "address" those unit fractions with the value, but we can not
"index" them from 0, only from 1/1.
every side.
set, as you can't get to the upper end to let you count down.
the existence of a set where no end can be seen. That is the only way to
make infinity and completeness compatible.
1/n - 1/(n+1) > 0 thus says that no 1/n is the smallest, as there willThat is not a proof of existence. The formula says: If n and n+1 exist,
exist a 1/(n+1) that is smaller that that.
then they differ.
On 13.09.2024 17:52, Richard Damon wrote:
On 9/13/24 11:41 AM, WM wrote:
Between [0, 1] and (0, 1] there is nothing, there is not a spot orBut that doesn't mean there is a lowest most point in (0, 1] as any
point of the interval.
point you might want to call it will have another point between it
and 0.
I will not call any point but consider all points. There is no point
smaller than all points in the open interval but a smallest one. Only 0
is smaller than all.
Note, I said between the point your THINK is the first, there is no
such point, and thus you are agreeing to that fact.
You can only have a first point in the open interval if the interval
has only a finite number of points,
No, that is your big mistake. In the interval [0, 1] there is a point
next to 0 and a point next to 1, and infinitely many are beteen them.
Regards, WM
On 13.09.2024 17:54, Richard Damon wrote:
On 9/13/24 11:43 AM, WM wrote:
Points do not approach. They are fixed.
Right, the points don't move, they just all exist
including one which is the smallest.
Regards, WM
On 13.09.2024 19:27, Richard Damon wrote:
between any to points (which have no width) is an infinite number of
other points to fill that gap.
Yes, they are dark until you define them.
Regards, WM
On 13.09.2024 19:12, Richard Damon wrote:
there is no such thing as two consecutive points in a dense system
There are consecutive points constituting the real line. But most are dark.
Thus, we have no points are adjacent, as between them is always more
points.
Yes, dark points some of which can become visible. How else could you
find more than you see initially?
Regards, WM
On 13.09.2024 19:32, Richard Damon wrote:
On 9/13/24 1:16 PM, WM wrote:
On 12.09.2024 23:40, Moebius wrote:
Am 12.09.2024 um 22:36 schrieb Chris M. Thomasson:
On 9/2/2024 10:07 AM, WM wrote:
How many different unit fractions are lessorequal than all unit
fractions?
[The correct] answer is that one unit fraction is lessorequal than >>>>>> all unit fractions. [...]
[In math] there is only the one correct answer given above.
Indeed,
Mathematical proof: NUF grows from 0 to more. At no point it can grow
by more than 1.
And at no point can it grow by ONE
It must.
so it doesn't actually exist.
There is no question for mathematicians that NUF does exist.
Regards, WM
On 12.09.2024 22:54, Jim Burns wrote:
On 9/12/2024 1:48 PM, WM wrote:
On 12.09.2024 14:29, FromTheRafters wrote:
After serious thinking WM wrote :
But the endpoints of open intervals
will remain dark forever.
Open intervals simply don't contain endpoints,
dark or otherwise.
You simply don't know about them.
The boundary ∂A of set A is
not a result of understanding points.
Fact is that the real axis is nothing but its points.
There is no gap.
Every point has a next point but
next to defined points are dark points.
Next to every defined points there are ℵo dark points.
This configuration cannot be changed.
WM explained :
No, that is your big mistake. In the interval [0, 1] there is a point
next to 0 and a point next to 1, and infinitely many are beteen them.
Define 'next' in this context.
On 9/13/2024 1:20 PM, WM wrote:
Fact is that the real axis is nothing but its points.
Yes.
ℝ[ℚ[ℤ[ℕ[⟨0,…,n⟩]]]] holds
points.between ratios of countable.to numbers.
There is no gap.
Yes.
Every point has a next point but
next to defined points are dark points.
No.
Next.to.defined points are not.in ℝ[ℚ[ℤ[ℕ[⟨0,…,n⟩]]]]
Next to every defined points there are ℵo dark points.
This configuration cannot be changed.
Whatever it is you are talking about,
we are talking about ℝ[ℚ[ℤ[ℕ[⟨0,…,n⟩]]]]
On 14.09.2024 01:05, FromTheRafters wrote:
WM explained :
No, that is your big mistake. In the interval [0, 1] there is a point
next to 0 and a point next to 1, and infinitely many are beteen them.
Define 'next' in this context.
Two points are next to each other means that no point is between them.
Regards, WM
On 13.09.2024 23:13, Jim Burns wrote:
On 9/13/2024 1:20 PM, WM wrote:
Fact is that
the real axis is nothing but its points.
Yes.
ℝ[ℚ[ℤ[ℕ[⟨0,…,n⟩]]]] holds
points.between ratios of countable.to numbers.
There is no gap.
Yes.
Every point has a next point but
next to defined points are dark points.
No.
Next.to.defined points are not.in
ℝ[ℚ[ℤ[ℕ[⟨0,…,n⟩]]]]
Not in what you know about these sets,
but existing,
because:
⎛ WM: There is no gap.
⎝ JB: Yes
Next to every defined points
there are ℵo dark points.
This configuration cannot be changed.
Whatever it is you are talking about,
we are talking about ℝ[ℚ[ℤ[ℕ[⟨0,…,n⟩]]]]
That's not enough.
On 14.09.2024 01:05, FromTheRafters wrote:Which is the case for no two (different) reals.
WM explained :
Two points are next to each other means that no point is between them.No, that is your big mistake. In the interval [0, 1] there is a pointDefine 'next' in this context.
next to 0 and a point next to 1, and infinitely many are beteen them.
On 9/14/2024 10:30 AM, FromTheRafters wrote:
WM presented the following explanation :
On 14.09.2024 01:05, FromTheRafters wrote:
WM explained :
No, that is your big mistake. In the interval [0, 1] there is a
point next to 0 and a point next to 1, and infinitely many are
beteen them.
Define 'next [to]' in this context.
Two points are next to each other means that no point is between them.
So which point is next to zero. Is it rational, or is it irrational?
In either case it is greater than zero and there is again more room
for an even closer number.
WM says .00001 is next to zero... Ahhh shit, what about .000001?
On 9/14/2024 11:35 AM, joes wrote:
Am Sat, 14 Sep 2024 16:01:02 +0200 schrieb WM:Two points on the real line that are different from one another have infinite***ly manY*** points between them.
On 14.09.2024 01:05, FromTheRafters wrote:Which is the case for no two (different) reals.
WM explained :Two points are next to each other means that no point is between them.
No, that is your big mistake. In the interval [0, 1] there is a point >>>>> next to 0 and a point next to 1, and infinitely many are beteen them. >>>> Define 'next' in this context.
Two points that are not different are the same point... ;^D
On 9/14/2024 5:07 PM, Moebius wrote:
Am 15.09.2024 um 01:54 schrieb Chris M. Thomasson:
On 9/14/2024 10:30 AM, FromTheRafters wrote:
WM presented the following explanation :
On 14.09.2024 01:05, FromTheRafters wrote:So which point is next to zero. Is it rational, or is it irrational?
WM explained :
No, that is your big mistake. In the interval [0, 1] there is a
point next to 0 and a point next to 1, and infinitely many are
beteen them.
Define 'next [to]' in this context.
Two points are next to each other means that no point is between them. >>>>
In either case it is greater than zero and there is again more room
for an even closer number.
If r e IR, r > 0, then 0 < r/2 < r.
Hence there is no real number > 0 "next to 0".
WM says .00001 is next to zero... Ahhh shit, what about .000001?
.(0)1 is close to zero? ;^D
On 9/14/2024 11:35 AM, joes wrote:
Am Sat, 14 Sep 2024 16:01:02 +0200 schrieb WM:
On 14.09.2024 01:05, FromTheRafters wrote:Which is the case for no two (different) reals.
WM explained :Two points are next to each other means that no point is between them.
No, that is your big mistake. In the interval [0, 1] there is a point >>>>> next to 0 and a point next to 1, and infinitely many are beteen them. >>>> Define 'next' in this context.
Two points on the real line that are different from one another have
infinite points between them, and so on and so forth. :^)
It might be worth pointing out that any non-trivial interval [a, b] on
the real line (i.e. with b > a) contains an uncountable number of
points.
Am Sat, 14 Sep 2024 16:01:02 +0200 schrieb WM:
Two points are next to each other means that no point is between them.Which is the case for no two (different) reals.
WM presented the following explanation :
Two points are next to each other means that no point is between them.
So which point is next to zero. Is it rational, or is it irrational?
In
either case it is greater than zero and there is again more room for an
even closer number.
On 9/14/24 10:01 AM, WM wrote:
Two points are next to each other means that no point is between them.
And, since for ANY two points on the real number line, x, and y, the
point (x+y)/2 is another real point between them,
there are absolutely
NO points on the real line next to each other.
On 14.09.2024 18:33, Richard Damon wrote:Sure, inbetween (x+y)/2 and x is (3x+y)/4. There is no end.
On 9/14/24 10:01 AM, WM wrote:
Why don't you start with (x+y)/2? Or better with the end of theTwo points are next to each other means that no point is between them.And, since for ANY two points on the real number line, x, and y, the
point (x+y)/2 is another real point between them,
sequence? In actual infinity you could.
I.e. „neighbouring” points can’t be defined. There are no gaps.there are absolutely NO points on the real line next to each other.No definable points, to be precise. If there is no point next to 0 then
there is a gap. I do not accept gaps on the real line.
On 15.09.2024 18:38, Ben Bacarisse wrote:Why do you believe there must be a total order?
It might be worth pointing out that any non-trivial interval [a, b] onThat proves that small intervals cannot be defined (they are dark). An uncountable number cannot be completed without a finite entry.
the real line (i.e. with b > a) contains an uncountable number of
points.
On 14.09.2024 18:33, Richard Damon wrote:
On 9/14/24 10:01 AM, WM wrote:
Two points are next to each other means that no point is between them.
And, since for ANY two points on the real number line, x, and y, the
point (x+y)/2 is another real point between them,
Why don't you start with (x+y)/2? Or better with the end of the
sequence? In actual infinity you could.
there are absolutely NO points on the real line next to each other.No definable points, to be precise. If there is no point next to 0 then
there is a gap. I do not accept gaps on the real line.
Regards, WM
On 15.09.2024 18:38, Ben Bacarisse wrote:
It might be worth pointing out that any non-trivial interval [a, b] on
the real line (i.e. with b > a) contains an uncountable number of
points.
That proves that small intervals cannot be defined (they are dark). An uncountable number cannot be completed without a finite entry.
Regards, WM
On 14.09.2024 20:35, joes wrote:
Am Sat, 14 Sep 2024 16:01:02 +0200 schrieb WM:
Two points are next to each other
means that
no point is between them.
Which is the case for no two (different) reals.
For no two different visible numbers,
to be precise.
Next to a visible number
there is a dark number or a gap,
i.e., nothing.
I don't believe in gaps on the real line.
On 09/15/2024 03:07 PM, FromTheRafters wrote:
on 9/15/2024, Ross Finlayson supposed :
On 09/15/2024 11:03 AM, FromTheRafters wrote:
After serious thinking Ross Finlayson wrote :
"What, no witty rejoinder?"
What you said has no relation to
the 'nextness' of elements in discrete sets.
What is 'next' to Pi+2 in the reals?
In the, "hyper-reals", it's its neighbors,
in the line-reals, put's previous and next,
in the field-reals, there's none,
and in the signal-reals, there's nothing.
What is the successor function on the reals?
Give me that, and maybe we can find the
'next' number greater than Pi.
Ah, good sir, then I'd like you to consider
a representation of real numbers as
with an integer part and a non-integer part,
the integer part of the integers, and
the non-integer part a value in [0,1],
where the values in [0,1], are as of
this model of (a finite segment of a) continuous domain,
these iota-values, line-reals,
as so established as according to the properties of
extent, density, completeness, and measure,
fulfilling implementing the Intermediate Value Theorem,
thus for
if not being the complete-ordered-field the field-reals,
yet being these iota-values a continuous domain [0,1]
these line-reals.
I wonder what you think of something like Hilbert's
"postulate of continuity" for geometry, as with
regards to that in the course-of-passage of
the growth of a continuous quantity, it encounters,
in order, each of the points in the line.
On 9/15/2024 9:31 PM, Ross Finlayson wrote:
On 09/15/2024 03:07 PM, FromTheRafters wrote:
on 9/15/2024, Ross Finlayson supposed :
On 09/15/2024 11:03 AM, FromTheRafters wrote:
After serious thinking Ross Finlayson wrote :
"What, no witty rejoinder?"
What you said has no relation to
the 'nextness' of elements in discrete sets.
What is 'next' to Pi+2 in the reals?
In the, "hyper-reals", it's its neighbors,
in the line-reals, put's previous and next,
in the field-reals, there's none,
and in the signal-reals, there's nothing.
What is the successor function on the reals?
Give me that, and maybe we can find the
'next' number greater than Pi.
Ah, good sir, then I'd like you to consider
a representation of real numbers as
with an integer part and a non-integer part,
the integer part of the integers, and
the non-integer part a value in [0,1],
where the values in [0,1], are as of
this model of (a finite segment of a) continuous domain,
these iota-values, line-reals,
as so established as according to the properties of
extent, density, completeness, and measure,
fulfilling implementing the Intermediate Value Theorem,
thus for
if not being the complete-ordered-field the field-reals,
yet being these iota-values a continuous domain [0,1]
these line-reals.
As n → ∞, (ι=⅟n), ⟨0,ι,2⋅ι,...,n⋅ι⟩ → ℚ∩[0,1]
ℚ∩[0,1] is not complete.
ℚ∩[0,1] has one connected component,
being what you (RF) call "continuous".
ℚ∩[0,1] has no points next to each other.
I wonder what you think of something like Hilbert's
"postulate of continuity" for geometry, as with
regards to that in the course-of-passage of
the growth of a continuous quantity, it encounters,
in order, each of the points in the line.
That sounds like the Intermediate Value Theorem,
where "encounters each" == 'no skips".
The Intermediate Value Theorem
is equivalent to
Dedekind completeness.
On 9/15/2024 9:38 AM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 9/14/2024 11:35 AM, joes wrote:It might be worth pointing out that any non-trivial interval [a, b] on
Am Sat, 14 Sep 2024 16:01:02 +0200 schrieb WM:
On 14.09.2024 01:05, FromTheRafters wrote:
WM explained :Two points are next to each other means that no point is between them. >>>> Which is the case for no two (different) reals.
No, that is your big mistake. In the interval [0, 1] there is a point >>>>>>> next to 0 and a point next to 1, and infinitely many are beteen them. >>>>>> Define 'next' in this context.
Two points on the real line that are different from one another have
infinite points between them, and so on and so forth. :^)
the real line (i.e. with b > a) contains an uncountable number of
points. Constructing the mid point and the quarter points and so on
only shows a countably infinite number of internal points, but giving a
bijection between [a, b] and [0, 1] shows that they have the same
cardinality.
Agreed. Afaict, one way to cover all the points is to draw a solid line between two different points p0 and p1 where p0 does not equal p1. We can
say the line covers them all? Fair enough? Put two different points on a piece of paper and draw a line from p0 to p1. That line contains infinitely dense points.
On 9/16/2024 10:30 AM, Jim Burns wrote:
On 9/15/2024 3:47 PM, WM wrote:
On 14.09.2024 20:35, joes wrote:
Am Sat, 14 Sep 2024 16:01:02 +0200 schrieb WM:
Two points are next to each other
means that
no point is between them.
Which is the case for no two (different) reals.
For no two different visible numbers,
to be precise.
⎛ WM: Two points are next to each other
⎜ means that
⎝ no point is between them.
WM is a strange one.
If no point is between them,
then they are the same.
On 09/15/2024 09:38 AM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 9/14/2024 11:35 AM, joes wrote:
Am Sat, 14 Sep 2024 16:01:02 +0200 schrieb WM:
On 14.09.2024 01:05, FromTheRafters wrote:
WM explained :Two points are next to each other means that no point is between them. >>>> Which is the case for no two (different) reals.
No, that is your big mistake. In the interval [0, 1] there is a point >>>>>>> next to 0 and a point next to 1, and infinitely many are beteen them. >>>>>> Define 'next' in this context.
Two points on the real line that are different from one another have
infinite points between them, and so on and so forth. :^)
It might be worth pointing out that any non-trivial interval [a, b] on
the real line (i.e. with b > a) contains an uncountable number of
points. Constructing the mid point and the quarter points and so on
only shows a countably infinite number of internal points, but giving a
bijection between [a, b] and [0, 1] shows that they have the same
cardinality.
Perhaps you might care to define "function" and "topology".
On 09/16/2024 11:24 AM, Jim Burns wrote:
On 9/16/2024 2:13 PM, Jim Burns wrote:
On 9/15/2024 9:31 PM, Ross Finlayson wrote:
On 09/15/2024 03:07 PM, FromTheRafters wrote:
What is the successor function on the reals?
Give me that, and maybe we can find the
'next' number greater than Pi.
Ah, good sir, then I'd like you to consider
a representation of real numbers as
with an integer part and a non-integer part,
the integer part of the integers, and
the non-integer part a value in [0,1],
where the values in [0,1], are as of
this model of (a finite segment of a) continuous domain,
these iota-values, line-reals,
as so established as according to the properties of
extent, density, completeness, and measure,
fulfilling implementing the Intermediate Value Theorem,
thus for
if not being the complete-ordered-field the field-reals,
yet being these iota-values a continuous domain [0,1]
these line-reals.
As n → ∞, (ι=⅟n), ⟨0,ι,2⋅ι,...,n⋅ι⟩ → ℚ∩[0,1]
It's shewn that [0,1] has no points not in ran(f).
About 2014, ....
On 09/12/2024 11:05 AM, Jim Burns wrote:
[...]
Excuse this delay,
where as
with regards to
why Russell's paradox applies to
just a plain old inductive set
when the merely-finite sets are
all the sets in the theory
after reverse mathematics
before infinity's axiomatized,
that the same telling blow
that Russell used to submarine Frege
is just declared gone away,
is what it is.
I.e., in ZF minus Infinity,
comprehending the usual set of v.N. ordinals,
results this.
On 09/17/2024 10:59 AM, Jim Burns wrote:
[...]
I do seem to recall that
your account was around when it was set out,
for example,
that least-upper-bound is nice neat trivial next,
On 09/17/2024 01:11 PM, Jim Burns wrote:
On 9/17/2024 2:57 PM, Ross Finlayson wrote:
Unlike ℕ and ℤ, ℚ and ℝ do not 'next'.
Then, for initial segments or n-sets of naturals,
the LUB of {f{n < m)} is "next": f(m+1).
Am Sun, 15 Sep 2024 21:52:30 +0200 schrieb WM:
No definable points, to be precise. If there is no point next to 0 thenI.e. „neighbouring” points can’t be defined. There are no gaps.
there is a gap. I do not accept gaps on the real line.
On 9/15/24 3:39 PM, WM wrote:
On 15.09.2024 18:38, Ben Bacarisse wrote:
It might be worth pointing out that any non-trivial interval [a, b] on
the real line (i.e. with b > a) contains an uncountable number of
points.
That proves that small intervals cannot be defined (they are dark). An
uncountable number cannot be completed without a finite entry.
But arbirary small intevals CAN be defined.
Try to name one that can't.
There aren't gaps and there aren't next.numbers
Am Sun, 15 Sep 2024 21:39:46 +0200 schrieb WM:
That proves that small intervals cannot be defined (they are dark). AnWhy do you believe there must be a total order?
uncountable number cannot be completed without a finite entry.
On 09/17/2024 03:20 PM, Jim Burns wrote:
On 9/17/2024 4:16 PM, Ross Finlayson wrote:
On 09/17/2024 01:11 PM, Jim Burns wrote:
On 9/17/2024 2:57 PM, Ross Finlayson wrote:
Unlike ℕ and ℤ, ℚ and ℝ do not 'next'.
Then, for initial segments or n-sets of naturals,
the LUB of {f{n < m)} is "next": f(m+1).
Yes.
(Presumably, you mean lub.{n:n≤m} = m. f()=? )
You might enjoy this:
⎛ Define ℕ as well.ordered and nexted.
⎜ well.ordered (A ⊆ ℕ holds min.A or is empty)
⎝ nexted (m ∈ ℕ has m+1 m-1 next, except 0=min.ℕ)
⎛ In a finite order,
⎝ each nonempty subset is 2.ended.
Consider upper.bounded nonempty A ⊆ ℕ and
its set UB[A] ⊆ ℕ of upper.bounds
A ᵉᵃᶜʰ≤ᵉᵃᶜʰ UB[A]
A is upper.bounded.
UB[A] ⊆ ℕ is nonempty.
UB[A] holds min.UB[A]
A holds min.UB[A]
Otherwise,
(min.UB[A])-1 is a less.than.least upper.bound
(that is, what.it.is is gibberish)
A holds min.UB[A] which upper.bounds A
min.UB[A] = max.A
Upper.bounded nonempty A holds max.A
Upper.bounded nonempty A ⊆ ℕ holds min.A
(well.order)
Upper.bounded nonempty A is 2.ended.
And, similarly,
each (also.bounded) nonempty S ⊆ A is 2.ended.
Upper.bounded nonempty A ⊆ ℕ is finite,
because
ℕ is well.ordered and nexted.
Yet,
didn't you just reject, "infinite middle"?
On 16.09.2024 19:30, Jim Burns wrote:
On 9/15/2024 3:47 PM, WM wrote:
I don't believe in gaps on the real line.
There aren't gaps and there aren't next.numbers
in numbers.situating.splits of rationals with
countable.to.numerators.and.denominators
So what is next instead?
On 9/18/2024 5:44 AM, WM wrote:
On 16.09.2024 03:16, Richard Damon wrote:
On 9/15/24 3:39 PM, WM wrote:
On 15.09.2024 18:38, Ben Bacarisse wrote:
It might be worth pointing out that any non-trivial interval [a, b] on >>>>> the real line (i.e. with b > a) contains an uncountable number of
points.
That proves that small intervals cannot be defined (they are dark).
But arbirary small intervals CAN be defined.
Try to name one that can't.
Define an interval comprising 9182024 points, starting at zero.
On 09/18/2024 07:02 AM, Jim Burns wrote:
On 9/17/2024 10:58 PM, Ross Finlayson wrote:
On 09/17/2024 03:20 PM, Jim Burns wrote:
On 9/17/2024 4:16 PM, Ross Finlayson wrote:
On 09/17/2024 01:11 PM, Jim Burns wrote:
On 9/17/2024 2:57 PM, Ross Finlayson wrote:
Unlike ℕ and ℤ, ℚ and ℝ do not 'next'.
Then, for initial segments or n-sets of naturals,
the LUB of {f{n < m)} is "next": f(m+1).
Yes.
(Presumably, you mean lub.{n:n≤m} = m. f()=? )
You might enjoy this:
⎛ Define ℕ as well.ordered and nexted.
⎜ well.ordered (A ⊆ ℕ holds min.A or is empty)
⎝ nexted (m ∈ ℕ has m+1 m-1 next, except 0=min.ℕ)
⎛ In a finite order,
⎝ each nonempty subset is 2.ended.
Consider upper.bounded nonempty A ⊆ ℕ and
its set UB[A] ⊆ ℕ of upper.bounds
A ᵉᵃᶜʰ≤ᵉᵃᶜʰ UB[A]
A is upper.bounded.
UB[A] ⊆ ℕ is nonempty.
UB[A] holds min.UB[A]
A holds min.UB[A]
Otherwise,
(min.UB[A])-1 is a less.than.least upper.bound
(that is, what.it.is is gibberish)
A holds min.UB[A] which upper.bounds A
min.UB[A] = max.A
Upper.bounded nonempty A holds max.A
Upper.bounded nonempty A ⊆ ℕ holds min.A
(well.order)
Upper.bounded nonempty A is 2.ended.
And, similarly,
each (also.bounded) nonempty S ⊆ A is 2.ended.
Upper.bounded nonempty A ⊆ ℕ is finite,
because
ℕ is well.ordered and nexted.
Yet,
didn't you just reject, "infinite middle"?
Please include in your posts
more of what you must consider
goes without saying;
you must, as you post without saying it.
I can't track you without footprints
or candy wrappers or something
to hint at where you're going (metaphorically).
----
Yes,
I am still recognizing the falsehood of
"infinite middle" (0:ω symmetry).
The post to which you respond is pretty much
a Festshrift to rejecting infinite middle.
I cannot guess why you need to hear that.
And I have tried.
ℕ is uniformly well.ordered and nexted
apart from 0=min.ℕ
That is what ℕ is.
From the uniform well.ordering and nexting
it is derivable (see prev.) that,
uniformly, finitely.many are before and,
uniformly, infinitely many are after.
ℕ is uniformly on the near side of
infinite not.the.middle.
ℕ is lacking an infinite middle.
Yet,
didn't you just reject, "infinite middle"?
Yet,
in the infinite limit, d -> infinity.
Please consider this as a tendered reserve and
not so much a purposeful omission of any sort,
where as with regards to
my tens of thousands of posts and
brief essays to usenet
there are more than enough crumbs to go around.
On 15.09.2024 23:07, joes wrote:
Am Sun, 15 Sep 2024 21:52:30 +0200 schrieb WM:
No definable points, to be precise. If there is no point next to 0 thenI.e. „neighbouring” points can’t be defined. There are no gaps.
there is a gap. I do not accept gaps on the real line.
So it is.These points are dark.
Regards, WM
Am 18.09.2024 um 21:35 schrieb Chris M. Thomasson:
On 9/18/2024 5:44 AM, WM wrote:
On 16.09.2024 03:16, Richard Damon wrote:
On 9/15/24 3:39 PM, WM wrote:
On 15.09.2024 18:38, Ben Bacarisse wrote:
It might be worth pointing out that any non-trivial interval [a,
b] on
the real line (i.e. with b > a) contains an uncountable number of
points.
That proves that small intervals cannot be defined (they are dark).
But arbirary small intervals CAN be defined.
Try to name one that can't.
Define an interval comprising 9182024 points, starting at zero.
Mückenheim, Du bist wirklich der größte Spinner unter der Sonne!
Hint: The IS NO "interval comprising 9182024 points", hence NOTHING TO
DEFINE HERE, you silly crank.
Am 18.09.2024 um 21:35 schrieb Chris M. Thomasson:
On 9/18/2024 5:44 AM, WM wrote:
On 16.09.2024 03:16, Richard Damon wrote:
On 9/15/24 3:39 PM, WM wrote:
On 15.09.2024 18:38, Ben Bacarisse wrote:
It might be worth pointing out that any non-trivial interval [a,
b] on
the real line (i.e. with b > a) contains an uncountable number of
points.
That proves that small intervals cannot be defined (they are dark).
But arbirary small intervals CAN be defined.
Try to name one that can't.
Define an interval comprising 9182024 points, starting at zero.
Mückenheim, Du bist wirklich der größte Spinner unter der Sonne!
Hint: The IS NO "interval comprising 9182024 points", hence NOTHING TO
DEFINE HERE, you silly crank.
On 9/18/2024 5:44 AM, WM wrote:
Define an interval comprising [exactly] 9182024 points, starting at zero.
p0 = (0, 0)
p1 = (1, 0)
pdif = p1 - p0;
pnormal_base = 1.f / 9182024;
// add one more point to make it all the way to p1...
for (i = 0; i < 9182024 + 1; ++i)
{
pnormal = pnormal_base * i;
p2 = p0 + pdif * pnormal;
plot(p2);
}
?
On 9/18/2024 8:39 AM, WM wrote:
On 16.09.2024 19:30, Jim Burns wrote:
On 9/15/2024 3:47 PM, WM wrote:
I don't believe in gaps on the real line.
There aren't gaps and there aren't next.numbers
in numbers.situating.splits of rationals with
countable.to.numerators.and.denominators
So what is next instead?
What is between one and the next?
A gap.
There is no gap in the real line.
There is no next in the real line.
If there were, there'd be a gap.
On 19.09.2024 00:50, Richard Damon wrote:
On 9/18/24 8:33 AM, WM wrote:
On 15.09.2024 23:07, joes wrote:
Am Sun, 15 Sep 2024 21:52:30 +0200 schrieb WM:
No definable points, to be precise. If there is no point next to 0I.e. „neighbouring” points can’t be defined. There are no gaps.
then
there is a gap. I do not accept gaps on the real line.
So it is.These points are dark.
Nope, they just aren't.
To be distinct points, there must be a gap, so they can not be
neighboring,
There are infinitely many points with no extension filling the space.
the space is filled by more points, creating smaller and smaller gaps
That is potential infinity. In actual infinity all points are there at
once.
Regards, WM
, that only disappear in the infinite.
On 9/18/24 8:33 AM, WM wrote:
On 15.09.2024 23:07, joes wrote:
Am Sun, 15 Sep 2024 21:52:30 +0200 schrieb WM:
No definable points, to be precise. If there is no point next to 0 then >>>> there is a gap. I do not accept gaps on the real line.I.e. „neighbouring” points can’t be defined. There are no gaps.
So it is.These points are dark.
Nope, they just aren't.
To be distinct points, there must be a gap, so they can not be
neighboring,
the space is filled by more points, creating smaller and
smaller gaps
On 19.09.2024 00:50, Richard Damon wrote:Yes, and they are not neighbours.
On 9/18/24 8:33 AM, WM wrote:There are infinitely many points with no extension filling the space.
On 15.09.2024 23:07, joes wrote:Nope, they just aren't.
Am Sun, 15 Sep 2024 21:52:30 +0200 schrieb WM:So it is.These points are dark.
No definable points, to be precise. If there is no point next to 0I.e. „neighbouring” points can’t be defined. There are no gaps.
then there is a gap. I do not accept gaps on the real line.
To be distinct points, there must be a gap, so they can not be
neighboring,
They are there.the space is filled by more points, creating smaller and smaller gapsThat is potential infinity. In actual infinity all points are there at
once.
, that only disappear in the infinite.--
On 18.09.2024 16:31, Jim Burns wrote:If there were a point next to zero, there would be a gap inbetween.
On 9/18/2024 8:39 AM, WM wrote:Try to use logic. Either there is a point next to zero or there is no
On 16.09.2024 19:30, Jim Burns wrote:
On 9/15/2024 3:47 PM, WM wrote:
I don't believe in gaps on the real line.
There aren't gaps and there aren't next.numbers in
numbers.situating.splits of rationals with
countable.to.numerators.and.denominators
So what is next instead?
What is between one and the next? A gap.
There is no gap in the real line.
There is no next in the real line.
If there were, there'd be a gap.
point next to zero, that means there is nothing, i.e., a gap.
Am 18.09.2024 um 21:35 schrieb Chris M. Thomasson:
On 9/18/2024 5:44 AM, WM wrote:
On 16.09.2024 03:16, Richard Damon wrote:
On 9/15/24 3:39 PM, WM wrote:
On 15.09.2024 18:38, Ben Bacarisse wrote:
It might be worth pointing out that any non-trivial interval [a,
b] on
the real line (i.e. with b > a) contains an uncountable number of
points.
That proves that small intervals cannot be defined (they are dark).
But arbitrary small intervals CAN be defined.
Try to name one that can't.
Define an interval comprising 9182024 points, starting at zero.
There IS NO "interval comprising 9182024 points", hence NOTHING TO
DEFINE HERE,
On 9/19/24 7:22 AM, WM wrote:
On 19.09.2024 00:50, Richard Damon wrote:
On 9/18/24 8:33 AM, WM wrote:
On 15.09.2024 23:07, joes wrote:
Am Sun, 15 Sep 2024 21:52:30 +0200 schrieb WM:So it is.These points are dark.
No definable points, to be precise. If there is no point next to 0 >>>>>> thenI.e. „neighbouring” points can’t be defined. There are no gaps. >>>>
there is a gap. I do not accept gaps on the real line.
Nope, they just aren't.
To be distinct points, there must be a gap, so they can not be
neighboring,
There are infinitely many points with no extension filling the space.
Right, and none "next" to another, because there is always another in
between them.
the space is filled by more points, creating smaller and smaller gaps
That is potential infinity. In actual infinity all points are there at
once.
No, that is what infinity actually is. There are not two different
"kinds" of it. All the point do exist, and we can express any of them.
The Infinite Set of these values actually exists, and all the values are
in it,
Am Thu, 19 Sep 2024 12:38:43 +0200 schrieb WM:
If there were a point next to zero, there would be a gap inbetween.Try to use logic. Either there is a point next to zero or there is no
point next to zero, that means there is nothing, i.e., a gap.
Am Thu, 19 Sep 2024 13:22:45 +0200 schrieb WM:
There are infinitely many points with no extension filling the space.Yes, and they are not neighbours.
They are there.the space is filled by more points, creating smaller and smaller gapsThat is potential infinity. In actual infinity all points are there at
once.
On 9/19/2024 12:29 AM, Ross Finlayson wrote:
[...]
ℝ is what.we.mean.by the continuum because
ℚ is gapless (ℚ does not 'next')
⎛ ¬∃q,q″ ∈ ℚ: q < q″ ∧
⎝ ¬∃q′ ∈ ℚ: q < q′ < q″
and ℝ completes ℚ
⎛ ¬∃S ⊆ ℚ: {} ≠ S ᵉᵃᶜʰ<ᵉᵃᶜʰ ℚ\S ≠ {} ∧ ⎝ ¬∃r ∈ ℝ\ℚ: S ᵉᵃᶜʰ< r <ᵉᵃᶜʰ ℚ\S
such that
crossing ℝ×ℝ.curves intersect,
either in ℚ or in ℝ\ℚ
On 09/18/2024 01:51 PM, Ross Finlayson wrote:
On 09/18/2024 12:37 PM, Chris M. Thomasson wrote:
On 9/17/2024 7:58 PM, Ross Finlayson wrote:
On 09/17/2024 01:11 PM, Jim Burns wrote:
[...]Unlike ℕ and ℤ, ℚ and ℝ do not 'next'.
Put pencil to paper and draw a straight line,
each of the points were encountered in order.
No matter how fine it's sliced, ....
Sometimes called "Hilbert's Postulate of Continuity".
Which he says is required, ....
On 18.09.2024 16:31, Jim Burns wrote:
On 9/18/2024 8:39 AM, WM wrote:
On 16.09.2024 19:30, Jim Burns wrote:
On 9/15/2024 3:47 PM, WM wrote:
I don't believe in gaps on the real line.
There aren't gaps and there aren't next.numbers
in numbers.situating.splits of rationals with
countable.to.numerators.and.denominators
So what is next instead?
What is between one and the next?
A gap.
There is no gap in the real line.
There is no next in the real line.
If there were, there'd be a gap.
Try to use logic.
Either there is a point next to zero
or there is no point next to zero,
or there is no point next to zero,
that means there is nothing, i.e., a gap.
On 19.09.2024 13:44, Richard Damon wrote:
On 9/19/24 7:22 AM, WM wrote:
There are infinitely many points with no extension
filling the space.
Right,
and none "next" to another, because
there is always another in between them.
"always another" is potential infinity.
I am discussing actual infinity
where all are there at once and no "always" is used.
On 9/19/2024 5:55 AM, WM wrote:
On 18.09.2024 22:49, Moebius wrote:
Am 18.09.2024 um 21:35 schrieb Chris M. Thomasson:
On 9/18/2024 5:44 AM, WM wrote:
On 16.09.2024 03:16, Richard Damon wrote:
On 9/15/24 3:39 PM, WM wrote:
On 15.09.2024 18:38, Ben Bacarisse wrote:But arbitrary small intervals CAN be defined.
It might be worth pointing out that any non-trivial interval [a, >>>>>>>> b] on
the real line (i.e. with b > a) contains an uncountable number of >>>>>>>> points.
That proves that small intervals cannot be defined (they are dark). >>>>>>
Try to name one that can't.
Define an interval comprising 9182024 points, starting at zero.
There IS NO "interval comprising 9182024 points", hence NOTHING TO
DEFINE HERE, you fucking asshole full of shit.
How can infinitely many points be accumulated without a first one?
The real line is infinitely long
and infinitely dense, or granular if you will...
Then the reals that are infinitely dense. They are uncountable because
they are so dense.
On 19.09.2024 14:38, joes wrote:
Am Thu, 19 Sep 2024 13:22:45 +0200 schrieb WM:
There are infinitely many points with no extension filling the space.Yes, and they are not neighbours.
Either they are next to each other ot somethig else is in between.
They are there.the space is filled by more points, creating smaller and smaller gapsThat is potential infinity. In actual infinity all points are there at
once.
But cannot be found. For instance the smallest positive point.
Regards, WM
On 19.09.2024 13:44, Richard Damon wrote:
On 9/19/24 7:22 AM, WM wrote:
On 19.09.2024 00:50, Richard Damon wrote:
On 9/18/24 8:33 AM, WM wrote:
On 15.09.2024 23:07, joes wrote:
Am Sun, 15 Sep 2024 21:52:30 +0200 schrieb WM:So it is.These points are dark.
No definable points, to be precise. If there is no point next to >>>>>>> 0 thenI.e. „neighbouring” points can’t be defined. There are no gaps. >>>>>
there is a gap. I do not accept gaps on the real line.
Nope, they just aren't.
To be distinct points, there must be a gap, so they can not be
neighboring,
There are infinitely many points with no extension filling the space.
Right, and none "next" to another, because there is always another in
between them.
"always another" is potential infinity. I am discussing actual infinity
where all are there at once and no "always" is used.
the space is filled by more points, creating smaller and smaller gaps
That is potential infinity. In actual infinity all points are there
at once.
No, that is what infinity actually is. There are not two different
"kinds" of it. All the point do exist, and we can express any of them.
Then you need no "always" but can point directly to the first point next
to another.
The Infinite Set of these values actually exists, and all the values
are in it,
that includes the first point of the interval and the second and so on.
Regards, WM
On 18.09.2024 22:49, Moebius wrote:
Am 18.09.2024 um 21:35 schrieb Chris M. Thomasson:
On 9/18/2024 5:44 AM, WM wrote:
On 16.09.2024 03:16, Richard Damon wrote:
On 9/15/24 3:39 PM, WM wrote:
On 15.09.2024 18:38, Ben Bacarisse wrote:But arbitrary small intervals CAN be defined.
It might be worth pointing out that any non-trivial interval [a, >>>>>>> b] on
the real line (i.e. with b > a) contains an uncountable number of >>>>>>> points.
That proves that small intervals cannot be defined (they are dark). >>>>>
Try to name one that can't.
Define an interval comprising 9182024 points, starting at zero.
There IS NO "interval comprising 9182024 points", hence NOTHING TO
DEFINE HERE,
How can infinitely many points be accumulated without a first one? This belief proves that matheology is mistaken.
Regards, WM
On 09/18/2024 01:51 PM, Ross Finlayson wrote:
On 09/18/2024 12:37 PM, Chris M. Thomasson wrote:
[...]
Put pencil to paper and draw a straight line,
each of the points were encountered in order.
No matter how fine it's sliced, ....
Sometimes called "Hilbert's Postulate of Continuity".
Which he says is required, ....
On 9/19/24 9:02 AM, WM wrote:
"always another" is potential infinity. I am discussing actual
infinity where all are there at once and no "always" is used.
So, then how do you describe that fact that there IS always another.
On 9/19/24 8:55 AM, WM wrote:
How can infinitely many points be accumulated without a first one?
This belief proves that matheology is mistaken.
How can you have infinity many unit fractions if there IS a smallest
one.
From that smallest one, take its reciprical, and that tells you how
many unit fractions you have.
Put pencil to paper and draw two curves which cross.
There is a point at which the curves intersect.
On 9/19/2024 9:02 AM, WM wrote:
On 19.09.2024 13:44, Richard Damon wrote:
On 9/19/24 7:22 AM, WM wrote:
There are infinitely many points with no extension
filling the space.
Right,
and none "next" to another, because
there is always another in between them.
"always another" is potential infinity.
For each finite ordinal ξ,
always another finite ordinal ξ∪{ξ} > ξ exists.
The set ω of all and only finite ordinals
holds all and only finite ordinals.
Am 19.09.2024 um 21:13 schrieb Chris M. Thomasson:
On 9/19/2024 5:55 AM, WM wrote:
How can infinitely many points be accumulated without a first one?
@WM: There's no need for them "to be accumulated"
On 19.09.2024 21:15, Jim Burns wrote:The other ordinal exists regardless.
On 9/19/2024 9:02 AM, WM wrote:Potential infinity.
On 19.09.2024 13:44, Richard Damon wrote:For each finite ordinal ξ,
On 9/19/24 7:22 AM, WM wrote:
"always another" is potential infinity.There are infinitely many points with no extension filling theRight,
space.
and none "next" to another, because there is always another in
between them.
always another finite ordinal ξ∪{ξ} > ξ exists.
There is an infinite gap, if you will.The set ω of all and only finite ordinals holds all and only finiteAnd there is no gap before ω.
ordinals.
On 9/19/2024 6:38 AM, WM wrote:
Either there is a point next to zero
or there is no point next to zero,
Consider ABCDEFGHIJKLMNOPQRSTUVWXYZ
Q is between A and Z
A and Z are not next to each other.
Nothing is between P and Q
P and Q are next to each other.
⎛ There is a gap between P and Q
⎜ A gap isn't some _presence_
⎜ A gap is an _absence_ of whatever.
Consider the real line.
For any two points x and y,
(x+y)/2 is between x and y.
x and y are not next to each other.
On 20.09.2024 19:51, Jim Burns wrote:
Put pencil to paper and draw two curves which cross.
There is a point at which the curves intersect.
This proves that no line has gaps.
On 20.09.2024 05:39, Richard Damon wrote:
From that smallest one,
take its reciprical, and that tells you
how many unit fractions you have.
Dark numbers don't tell anything.
On 19.09.2024 20:31, Jim Burns wrote:
On 9/19/2024 6:38 AM, WM wrote:
Either there is a point next to zero
or there is no point next to zero,
Consider ABCDEFGHIJKLMNOPQRSTUVWXYZ
Q is between A and Z
A and Z are not next to each other.
Nothing is between P and Q
P and Q are next to each other.
⎛ There is a gap between P and Q
No.
⎜ A gap isn't some _presence_
⎜ A gap is an _absence_ of whatever.
A gap in the real line is the absence of points.
Consider the real line.
For any two points x and y,
(x+y)/2 is between x and y.
x and y are not next to each other.
That is true for definable points only.
On 19.09.2024 21:15, Jim Burns wrote:
On 9/19/2024 9:02 AM, WM wrote:
On 19.09.2024 13:44, Richard Damon wrote:
On 9/19/24 7:22 AM, WM wrote:
There are infinitely many points with no extension
filling the space.
Right,
and none "next" to another, because
there is always another in between them.
"always another" is potential infinity.
For each finite ordinal ξ,
always another finite ordinal ξ∪{ξ} > ξ exists.
Potential infinity.
The set ω of all and only finite ordinals
holds all and only finite ordinals.
And there is no gap before ω.
On 20.09.2024 05:35, Richard Damon wrote:
On 9/19/24 9:02 AM, WM wrote:
"always another" is potential infinity.
I am discussing actual infinity
where all are there at once
and no "always" is used.
So, then how do you describe that fact that
there IS always another.
The reason is that only
a potentially infinite collection of elements
can be utilzed.
On 19.09.2024 23:13, Moebius wrote:
Am 19.09.2024 um 21:13 schrieb Chris M. Thomasson:
On 9/19/2024 5:55 AM, WM wrote:
How can infinitely many points be accumulated without a first one?
@WM: There's no need for them "to be accumulated"
How can 10 points exist in linear order without a first one?
Regards, WM
On 20.09.2024 05:35, Richard Damon wrote:
On 9/19/24 9:02 AM, WM wrote:
"always another" is potential infinity. I am discussing actual
infinity where all are there at once and no "always" is used.
So, then how do you describe that fact that there IS always another.
The reason is that only a potentially infinite collection of elements
can be utilzed.
Regards, WM
On 20.09.2024 05:39, Richard Damon wrote:
On 9/19/24 8:55 AM, WM wrote:
How can infinitely many points be accumulated without a first one?
This belief proves that matheology is mistaken.
How can you have infinity many unit fractions if there IS a smallest one.
That is accomplished by dark numbers. Look, the interval [0, 1] has a smallest point and nevetheless infinitely many points.
From that smallest one, take its reciprical, and that tells you how
many unit fractions you have.
Dark numbers don't tell anything.
Regards, WM
On 9/19/2024 2:13 PM, Moebius wrote:
Am 19.09.2024 um 21:13 schrieb Chris M. Thomasson:
On 9/19/2024 5:55 AM, WM wrote:
On 18.09.2024 22:49, Moebius wrote:
Am 18.09.2024 um 21:35 schrieb Chris M. Thomasson:
On 9/18/2024 5:44 AM, WM wrote:
On 16.09.2024 03:16, Richard Damon wrote:
On 9/15/24 3:39 PM, WM wrote:
On 15.09.2024 18:38, Ben Bacarisse wrote:
It might be worth pointing out that any non-trivial interval >>>>>>>>>> [a, b] on
the real line (i.e. with b > a) contains an uncountable number of >>>>>>>>>> points.
That proves that small intervals cannot be defined (they are >>>>>>>>> dark).
But arbitrary small intervals CAN be defined.
For any eps e IR, eps > 0: [0, eps] is an interval (@WM: you see, I
just defined it) of length eps and it countains an uncountable number
of points. Hint: eps may be arbitrarily small, as long as it is > 0.
Try to name one that can't.
Define an interval comprising 9182024 points, starting at zero.
There IS NO "interval comprising 9182024 points", hence NOTHING TO
DEFINE HERE, you fucking asshole full of shit.
How can infinitely many points be accumulated without a first one?
@WM: There's no need for them "to be accumulated" (whatever this may
mean), you fucking asshole full of shit.
The real line is infinitely long
WM is talking about some interval of finite length here, it seems.
and infinitely dense, or granular if you will...
Sorta.
Well, its infinitely long...
...(-1)------(0)--------(+1)...
It has no end just like there is no end to the signed integers. Also,
its infinitely dense due to the nature of the reals.
?
On 9/20/2024 2:10 PM, WM wrote:
On 20.09.2024 19:51, Jim Burns wrote:
Put pencil to paper and draw two curves which cross.
There is a point at which the curves intersect.
This proves that no line has gaps.
A point (hypothetically) next to 0
has an absence of points between it and 0
No such absence of points exists.
On 9/20/2024 2:27 PM, WM wrote:
And there is no gap before ω.
ω-1 requires impossibilities:
a gap between ω and (hypothetical) ω-1
Am Fri, 20 Sep 2024 20:27:32 +0200 schrieb WM:
There is an infinite gap, if you will.The set ω of all and only finite ordinals holds all and only finiteAnd there is no gap before ω.
ordinals.
On 9/20/24 2:33 PM, WM wrote:
How can 10 points exist in linear order without a first one?
Who says they can't.
Its just when they become infinite that there might be ends that don't
exist.
On 20.09.2024 22:02, Jim Burns wrote:
On 9/20/2024 2:27 PM, WM wrote:
And there is no gap before ω.
ω-1 requires impossibilities:
a gap between ω and (hypothetical) ω-1
No.
A gap is where something could be
but is not.
And there is no gap before ω.
A gap is where something could be
but is not.
On 20.09.2024 21:26, Jim Burns wrote:
On 9/20/2024 2:10 PM, WM wrote:
On 20.09.2024 19:51, Jim Burns wrote:
Put pencil to paper and draw two curves which cross.
There is a point at which the curves intersect.
This proves that no line has gaps.
A point (hypothetically) next to 0
has an absence of points between it and 0
No such absence of points exists.
You cannot see the point because it is dark.
On 9/20/24 2:13 PM, WM wrote:infinity where all are there at once and no "always" is used.
On 20.09.2024 05:35, Richard Damon wrote:
On 9/19/24 9:02 AM, WM wrote:
"always another" is potential infinity. I am discussing actual
elements can be utilzed.
So, then how do you describe that fact that there IS always another.
The reason is that only a potentially infinite collection of
But what keeps you from actually utilizing any of them?
On 9/20/2024 2:24 PM, WM wrote:
On 19.09.2024 20:31, Jim Burns wrote:
On 9/19/2024 6:38 AM, WM wrote:
Either there is a point next to zero
or there is no point next to zero,
Consider ABCDEFGHIJKLMNOPQRSTUVWXYZ
Q is between A and Z
A and Z are not next to each other.
Nothing is between P and Q
P and Q are next to each other.
⎛ There is a gap between P and Q
No.
There is no letter between P and Q
in ABCDEFGHIJKLMNOPQRSTUVWXYZ
There is an absence of letters between P and Q
in ABCDEFGHIJKLMNOPQRSTUVWXYZ
On 9/21/2024 9:47 AM, WM wrote:
On 20.09.2024 22:02, Jim Burns wrote:
On 9/20/2024 2:27 PM, WM wrote:
And there is no gap before ω.
ω-1 requires impossibilities:
a gap between ω and (hypothetical) ω-1
No.
A gap is where something could be
but is not.
Gaps are where you locate your darkᵂᴹ numbers.
However,
'things always well.ordered' is
what we mean by 'ordinals'.
There are no non.well.ordered WWMB ordinals.
There is no finite WWMB ordinal α not.before
the first transfinite ordinal, WWMB ω
There is no finite WWMB ordinal α without
its WWMB successor α+1 before ω
There is no transfinite WWMB ordinal ξ before
the first transfinite ordinal, WWMB ω
On 21.09.2024 19:26, Jim Burns wrote:
On 9/21/2024 9:47 AM, WM wrote:
On 20.09.2024 22:02, Jim Burns wrote:
On 9/20/2024 2:27 PM, WM wrote:
And there is no gap before ω.
ω-1 requires impossibilities:
a gap between ω and (hypothetical) ω-1
No.
A gap is where something could be
but is not.
Gaps are where you locate your darkᵂᴹ numbers.
Yes.
However,
'things always well.ordered' is
what we mean by 'ordinals'.
That is a potentially infinite collection.
There are no non.well.ordered WWMB ordinals.
That may be.
There is no finite WWMB ordinal α not.before
the first transfinite ordinal, WWMB ω
There is no finite WWMB ordinal α without
its WWMB successor α+1 before ω
There is no transfinite WWMB ordinal ξ before
the first transfinite ordinal, WWMB ω
That may be. But then
there is no complete set of natural numbers
and no ω.
there is no complete set of natural numbers
On 21.09.2024 01:06, Richard Damon wrote:
On 9/20/24 2:33 PM, WM wrote:
How can 10 points exist in linear order without a first one?
Who says they can't.
In order to count to 10, you have to start at 1.
Its just when they become infinite that there might be ends that don't
exist.
In order to count a countable set, you have to start at 1.
Regards, WM
On 21.09.2024 01:04, Richard Damon wrote:
On 9/20/24 2:13 PM, WM wrote:infinity where all are there at once and no "always" is used.
On 20.09.2024 05:35, Richard Damon wrote:
On 9/19/24 9:02 AM, WM wrote:
"always another" is potential infinity. I am discussing actual
elements can be utilzed.
So, then how do you describe that fact that there IS always another.
The reason is that only a potentially infinite collection of
But what keeps you from actually utilizing any of them?
The fact that every used number belongs to a finite initial segment upon which almost all numbers are following.
Regards, WM
[...] so the countable numbers have ONE end that can be used to count
from. [RD]
What about the signed integers?
On 9/21/2024 4:11 PM, WM wrote:
There is no transfinite WWMB ordinal ξ before
the first transfinite ordinal, WWMB ω
That may be. But then
there is no complete set of natural numbers
and no ω.
WWMB ω is the first transfinite ordinal.
Existing or not.existing, WWMB ω isn't
anything other than the first transfinite ordinal.
As long as it's WWMB ω which you discuss,
WWMB ω-1 doesn't exist.
On 9/21/24 4:02 PM, WM wrote:
On 21.09.2024 01:04, Richard Damon wrote:
On 9/20/24 2:13 PM, WM wrote:infinity where all are there at once and no "always" is used.
On 20.09.2024 05:35, Richard Damon wrote:
On 9/19/24 9:02 AM, WM wrote:;
"always another" is potential infinity. I am discussing actual
elements can be utilzed.;The reason is that only a potentially infinite collection of
So, then how do you describe that fact that there IS always another. >> >>
;
But what keeps you from actually utilizing any of them?
The fact that every used number belongs to a finite initial segment
upon which almost all numbers are following.
So, we can just choose the finite initial segment that includes that
number, and thus can use it,
On 9/21/24 9:57 AM, WM wrote:
On 21.09.2024 01:06, Richard Damon wrote:
On 9/20/24 2:33 PM, WM wrote:
How can 10 points exist in linear order without a first one?
Who says they can't.
In order to count to 10, you have to start at 1.
Its just when they become infinite that there might be ends that
don't exist.
In order to count a countable set, you have to start at 1.
Right, so the countable numbers have ONE end that can be used to count
from.
You can count UP to infinity by counting up.
You can't count DOWN from infinity, as there is no place to start.
On 20.09.2024 22:02, Jim Burns wrote:This is too easy. ω is defined such that there is no gap.
On 9/20/2024 2:27 PM, WM wrote:
No. A gap is where something could be but is not.And there is no gap before ω.ω-1 requires impossibilities:
a gap between ω and (hypothetical) ω-1
On 22.09.2024 01:24, Richard Damon wrote:So you can’t make general statements?
On 9/21/24 4:02 PM, WM wrote:
On 21.09.2024 01:04, Richard Damon wrote:
On 9/20/24 2:13 PM, WM wrote:infinity where all are there at once and no "always" is used.
On 20.09.2024 05:35, Richard Damon wrote:
On 9/19/24 9:02 AM, WM wrote:;
"always another" is potential infinity. I am discussing actual
I don’t care, infinite is infinite.elements can be utilzed.So, then how do you describe that fact that there IS alwaysThe reason is that only a potentially infinite collection of
another.
Why are you talking about „potential infinity”?But what keeps you from actually utilizing any of them?The fact that every used number belongs to a finite initial segment
upon which almost all numbers are following.
Of course you can.So, we can just choose the finite initial segment that includes thatBut you cannot choose the infinite rest.
number, and thus can use it,
Am Sun, 22 Sep 2024 15:27:01 +0200 schrieb WM:
Of course you can.So, we can just choose the finite initial segment that includes thatBut you cannot choose the infinite rest.
number, and thus can use it,
On 22.09.2024 00:32, Jim Burns wrote:
On 9/21/2024 4:11 PM, WM wrote:
On 21.09.2024 19:26, Jim Burns wrote:
There is no transfinite WWMB ordinal ξ before
the first transfinite ordinal, WWMB ω
That may be. But then
there is no complete set of natural numbers
and no ω.
WWMB ω is the first transfinite ordinal.
Existing or not.existing, WWMB ω isn't
anything other than the first transfinite ordinal.
As long as it's WWMB ω which you discuss,
WWMB ω-1 doesn't exist.
Then ω does not exist.
On 20.09.2024 21:51, Jim Burns wrote:
On 9/20/2024 2:24 PM, WM wrote:
On 19.09.2024 20:31, Jim Burns wrote:
On 9/19/2024 6:38 AM, WM wrote:
Either there is a point next to zero
or there is no point next to zero,
Consider ABCDEFGHIJKLMNOPQRSTUVWXYZ
Q is between A and Z
A and Z are not next to each other.
Nothing is between P and Q
P and Q are next to each other.
⎛ There is a gap between P and Q
No.
There is no letter between P and Q
in ABCDEFGHIJKLMNOPQRSTUVWXYZ
There is an absence of
letters between P and Q
in ABCDEFGHIJKLMNOPQRSTUVWXYZ
No, there is nothing.
An absence requires a space which
could be occupied by the absent.
Either there is a point next to zero
or there is no point next to zero,
Either there is a point next to zero
or there is no point next to zero,
There is no point next to 0.
On 9/22/2024 11:37 AM, WM wrote:
On 22.09.2024 19:44, Jim Burns wrote:
There is no point next to 0.This is
There is a smallest unit fraction
because there are
no unit fractions without a first one when counting from zero.
Huh? Wow... Hummm... You suffer from some sort of learning disorder? Or,
pure troll? Humm...
There is no smallest unit fraction.
On 9/22/2024 6:27 AM, WM wrote:
On 22.09.2024 01:24, Richard Damon wrote:
On 9/21/24 4:02 PM, WM wrote:
On 21.09.2024 01:04, Richard Damon wrote:
On 9/20/24 2:13 PM, WM wrote:infinity where all are there at once and no "always" is used.
On 20.09.2024 05:35, Richard Damon wrote:
On 9/19/24 9:02 AM, WM wrote:;
"always another" is potential infinity. I am discussing actual
another.;
So, then how do you describe that fact that there IS always
elements can be utilzed.;
The reason is that only a potentially infinite collection of
;
But what keeps you from actually utilizing any of them?
The fact that every used number belongs to a finite initial segment
upon which almost all numbers are following.
So, we can just choose the finite initial segment that includes that
number, and thus can use it,
But you cannot choose the infinite rest.
If a tree falls in a forest... ;^D
On 22.09.2024 00:32, Jim Burns wrote:
On 9/21/2024 4:11 PM, WM wrote:
There is no transfinite WWMB ordinal ξ before
the first transfinite ordinal, WWMB ω
That may be. But then
there is no complete set of natural numbers
and no ω.
WWMB ω is the first transfinite ordinal.
Existing or not.existing, WWMB ω isn't
anything other than the first transfinite ordinal.
As long as it's WWMB ω which you discuss,
WWMB ω-1 doesn't exist.
Then ω does not exist.
Regards, WM
On 22.09.2024 19:44, Jim Burns wrote:
There is no point next to 0.
There is a smallest unit fraction
because
there are no unit fractions without a first one
when counting from zero.
On 22.09.2024 19:44, Jim Burns wrote:
There is no point next to 0.This is definite: There is a smallest unit fraction because there are no
unit fractions without a first one when counting from zero.
Regards, WM
On 22.09.2024 16:18, joes wrote:
Am Sun, 22 Sep 2024 15:27:01 +0200 schrieb WM:
No, I cannot. But if you can, please show it. Choose a number havingOf course you can.So, we can just choose the finite initial segment that includes thatBut you cannot choose the infinite rest.
number, and thus can use it,
more predecessors than successors.
Regards, WM
On 22.09.2024 01:24, Richard Damon wrote:
On 9/21/24 4:02 PM, WM wrote:
On 21.09.2024 01:04, Richard Damon wrote:
On 9/20/24 2:13 PM, WM wrote:infinity where all are there at once and no "always" is used.
On 20.09.2024 05:35, Richard Damon wrote:
On 9/19/24 9:02 AM, WM wrote:;
"always another" is potential infinity. I am discussing actual
another.;
So, then how do you describe that fact that there IS always
elements can be utilzed.;
The reason is that only a potentially infinite collection of
;
But what keeps you from actually utilizing any of them?
The fact that every used number belongs to a finite initial segment
upon which almost all numbers are following.
So, we can just choose the finite initial segment that includes that
number, and thus can use it,
But you cannot choose the infinite rest.
Regards, WM
[...} each number has infiniteLY MANY predecessors and infiniteLY MANY successors.
Am 23.09.2024 um 01:28 schrieb Chris M. Thomasson:
Concerning the integers:
[...} each number has infiniteLY MANY predecessors and infiniteLY MANY
successors.
...
On 22.09.2024 00:32, Jim Burns wrote:
On 9/21/2024 4:11 PM, WM wrote:
There is no transfinite WWMB ordinal ξ before
the first transfinite ordinal, WWMB ω
That may be. But then
there is no complete set of natural numbers
and no ω.
WWMB ω is the first transfinite ordinal.
Existing or not.existing, WWMB ω isn't
anything other than the first transfinite ordinal.
As long as it's WWMB ω which you discuss,
WWMB ω-1 doesn't exist.
Then ω does not exist.
Regards, WM
Count down from infinity please, what is the actual first number you
use, (not w-1, that isn't a number).
The problem is it doesn't exist, because the numbers don't have a bound
on that end.
On 9/22/2024 2:37 PM, WM wrote:
On 22.09.2024 19:44, Jim Burns wrote:
There is no point next to 0.
There is a smallest unit fraction
because
there are no unit fractions without a first one
when counting from zero.
Each visibleᵂᴹ unit.fraction ⅟k has
a counter.example ¼⋅⅟k to its being smallest.
No positive point, unit.fraction or otherwise,
is NOT undercut by some visible unit fraction.
On 23.09.2024 00:53, Richard Damon wrote:
Count down from infinity please, what is the actual first number you
use, (not w-1, that isn't a number).
The problem is it doesn't exist, because the numbers don't have a bound
on that end.
Chosse the greates existing number m. Then double it to get 2m.
Why didn't you choose 2m originally? Obviously it exists.
Am 23.09.2024 14:56:03 WM drivels:
Chosse the greates [...] number m. Then [...]
On 22.09.2024 22:22, Jim Burns wrote:
On 9/22/2024 2:37 PM, WM wrote:
On 22.09.2024 19:44, Jim Burns wrote:
There is no point next to 0.
There is a smallest unit fraction
because
there are no
unit fractions without a first one
when counting from zero.
Each visibleᵂᴹ unit.fraction ⅟k has
a counter.example ¼⋅⅟k to its being smallest.
Yes.
Therefore
the smallest unit fraction must be dark.
No positive point, unit.fraction or otherwise,
is NOT undercut by some visible unit fraction.
Wrong.
NUF(x) grows from 0 to more.
This increase cannot avoid 1
because
of ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
On 23.09.2024 00:53, Richard Damon wrote:
Count down from infinity please,
what is the actual first number you use,
(not w-1, that isn't a number).
The problem is it doesn't exist,
because the numbers don't have a bound
on that end.
The problem is not non-existence.
Proof:
Chosse the greates existing number m.
Then double it to get 2m.
Why didn't you choose 2m originally?
Obviously it exists.
On 23.09.2024 00:53, Richard Damon wrote:
Count down from infinity please, what is the actual first number you
use, (not w-1, that isn't a number).
The problem is it doesn't exist, because the numbers don't have a
bound on that end.
The problem is not non-existence. Proof: Chosse the greates existing
number m. Then double it to get 2m. Why didn't you choose 2m originally? Obviously it exists.
Regards, WM
On 22.09.2024 22:22, Jim Burns wrote:
On 9/22/2024 2:37 PM, WM wrote:
On 22.09.2024 19:44, Jim Burns wrote:
There is no point next to 0.
There is a smallest unit fraction
because
there are no unit fractions without a first one
when counting from zero.
Each visibleᵂᴹ unit.fraction ⅟k has
a counter.example ¼⋅⅟k to its being smallest.
Yes. Therefore the smallest unit fraction must be dark.
No positive point, unit.fraction or otherwise,
is NOT undercut by some visible unit fraction.
Wrong. NUF(x) grows from 0 to more. This increase cannot avoid 1 because
of ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Regards, WM
On 22.09.2024 01:25, Richard Damon wrote:What is „an end”?
On 9/21/24 9:57 AM, WM wrote:
On 21.09.2024 01:06, Richard Damon wrote:
On 9/20/24 2:33 PM, WM wrote:
Really existing sets of real unit fractions have two ends.Right, so the countable numbers have ONE end that can be used to countIts just when they become infinite that there might be ends thatIn order to count a countable set, you have to start at 1.
don't exist.
from.
On 23.09.2024 00:53, Richard Damon wrote:What screwy reasoning. If m were the greatest natural, you couldn’t
Count down from infinity please, what is the actual first number youThe problem is not non-existence. Proof: Chosse the greates existing
use, (not w-1, that isn't a number).
The problem is it doesn't exist, because the numbers don't have a bound
on that end.
number m. Then double it to get 2m. Why didn't you choose 2m originally? Obviously it exists.
Am Mon, 23 Sep 2024 14:56:03 +0200 schrieb WM:
The problem is not non-existence. Proof: Chosse the greates existing
number m. Then double it to get 2m. Why didn't you choose 2m originally?
What screwy reasoning. If m were the greatest natural, you couldn’t
choose 2m.
Obviously it exists.
Am Sun, 22 Sep 2024 15:28:38 +0200 schrieb WM:
On 22.09.2024 01:25, Richard Damon wrote:
On 9/21/24 9:57 AM, WM wrote:
On 21.09.2024 01:06, Richard Damon wrote:
Its just when they become infinite that
there might be ends that don't exist.
In order to count a countable set,
you have to start at 1.
Right, so
the countable numbers have ONE end that
can be used to count from.
Really existing sets of real unit fractions
have two ends.
What is „an end”?
On 09/20/2024 12:26 PM, Jim Burns wrote:
On 9/20/2024 2:10 PM, WM wrote:
On 20.09.2024 19:51, Jim Burns wrote:
Put pencil to paper and
draw two curves which cross.
There is a point at which
the curves intersect.
Theorems or axioms?
"Drawing" a line, "tire en regle", or curve,
has that when you put pencil to paper,
and draw a line, or curve if you will,
and life the pencil and put it back down,
and draw another one, intersecting the first:
the _curves_ cross.
... At a point, of for example where
they're incident, they coincide.
Then these lines-reals these iota-values
are about the only "standard infinitesimals"
there are: with extent you observe, density
you observe, least-upper-bound as trivial,
and measure as assigned, length assignment.
On 9/23/2024 8:57 AM, WM wrote:
a smallest unit.fraction, visibleᵂᴹ or darkᵂᴹ,
is gibberish.
On 9/23/24 8:57 AM, WM wrote:
Yes. Therefore the smallest unit fraction must be dark.
Which is something that doesn't exist.
Am Sun, 22 Sep 2024 15:28:38 +0200 schrieb WM:
Really existing sets of real unit fractions have two ends.What is „an end”?
WM's argument goes something like this:
🛇⎛ The set of unit.fractions exists.
🛇⎜ No _identifiable_ unit.fraction is its second end.
🛇⎜ (axiom) All sets have two ends.
🛇⎜ The second end of the unit.fractions exists
🛇⎝ but it is _not identifiable_
There is plenty to correct in that,
but I think WM's cornerstone.error is
how he thinks axioms work.
Because largest.number m is gibberish.
On 23.09.2024 19:58, Jim Burns wrote:
On 9/23/2024 8:57 AM, WM wrote:
a smallest unit.fraction, visibleᵂᴹ or darkᵂᴹ,
is gibberish.
But
the increase of NUF(x) from 0 to infinity
without intermediate steps
is not gibberish?
On 9/24/2024 12:30 PM, WM wrote:
On 23.09.2024 20:53, Jim Burns wrote:
Because largest.number m is gibberish.
The largest number
**** that you can choose **** depends on your facilities.
On 24.09.2024 10:00, joes wrote:Does it need to be a member of the set?
Am Sun, 22 Sep 2024 15:28:38 +0200 schrieb WM:
An end is where nothing follows.Really existing sets of real unit fractions have two ends.What is „an end”?
On 9/24/2024 3:28 PM, WM wrote:
Thus
increasing NUF(x) from 0 to infinity
WITH intermediate steps
is gibberish,
Of many suitable definitions of natural numbers,
one is:
they are well.ordered (subsets minimummed or empty)
they continue (have successors)
they are reached by a step (≠0 have predecessors)
The natural numbers are our Paradigm of Finite.
There is no first unreachable natural number.
By that and by its well.order,
there is no unreachable natural number.
ω is not a natural number.
⎛ Each before ω can be reached.
⎝ Each which can be reached is before ω.
If ω-1 existed such that (ω-1)+1 = ω
then ω could be reached
On 24.09.2024 22:19, Jim Burns wrote:
On 9/24/2024 3:28 PM, WM wrote:
On 23.09.2024 19:58, Jim Burns wrote:
On 9/23/2024 8:57 AM, WM wrote:
a smallest unit.fraction, visibleᵂᴹ or darkᵂᴹ,
is gibberish.
But
the increase of NUF(x) from 0 to infinity
without intermediate steps is not gibberish?
Anything which
can be reached by intermediate steps
is not infinite.
(You (WM) apparently mean something different.)
Thus
increasing NUF(x) from 0 to infinity
WITH intermediate steps
is gibberish,
The only alternative would by
infinitely many unit fractions at one point.
The only alternative would by
infinitely many unit fractions at one point.
That is not gibberish but wrong.
The only alternative would by
infinitely many unit fractions at one point.
That is not gibberish but wrong.
Of many suitable definitions of natural numbers,
one is:
they are well.ordered (subsets minimummed or empty)
they continue (have successors)
they are reached by a step (≠0 have predecessors)
The natural numbers are our Paradigm of Finite.
There is no first unreachable natural number.
By that and by its well.order,
there is no unreachable natural number.
The natural numbers n belonging to
the first infinitely many unit fractions 1/n, i.e.
there where NUF(x) increases at one point
from 0 to infinity,
cannot be distinguished in your opinion.
Thus they are unreachable.
By that and by its well.order,
there is no unreachable natural number.
ω is not a natural number.
⎛ Each before ω can be reached.
⎝ Each which can be reached is before ω.
If ω-1 existed such that (ω-1)+1 = ω
then ω could be reached
If ω-1 could be seen. But it cannot.
On 24.09.2024 18:49, Jim Burns wrote:
WM's argument goes something like this:
🛇⎛ The set of unit.fractions exists.
🛇⎜ No _identifiable_ unit.fraction is its second end.
🛇⎜ (axiom) All sets have two ends.
Not an axiom but the fact that below zero there is no unit fraction
proves the lower end.
🛇⎜ The second end of the unit.fractions exists
🛇⎝ but it is _not identifiable_
There is plenty to correct in that,
but I think WM's cornerstone.error is
how he thinks axioms work.
I think that mathematics of fractions is correct.
It contains ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 from which follows that never two different5 fractions sit upon each other. From NUF(0) = 0 the smallest
unit fraction follows immeditely, From it the largest natural number.
Regards WM
On 24.09.2024 10:00, joes wrote:
Am Sun, 22 Sep 2024 15:28:38 +0200 schrieb WM:
Really existing sets of real unit fractions have two ends.What is „an end”?
An end is where nothing follows.
Regards, WM
On 23.09.2024 19:58, Jim Burns wrote:
On 9/23/2024 8:57 AM, WM wrote:
a smallest unit.fraction, visibleᵂᴹ or darkᵂᴹ,
is gibberish.
But the increase of NUF(x) from 0 to infinity without intermediate steps
is not gibberish?
Regards, WM
On 24.09.2024 00:53, Richard Damon wrote:
On 9/23/24 8:57 AM, WM wrote:
Yes. Therefore the smallest unit fraction must be dark.
Which is something that doesn't exist.
The function NUF(0) increases. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 proves that it can gain only one unit fraction at any point x ∈ ℝ, hence NUF has to
pass the value 1.
Regards, WM
On 24.09.2024 22:19, Jim Burns wrote:
On 9/24/2024 3:28 PM, WM wrote:
Thus
increasing NUF(x) from 0 to infinity
WITH intermediate steps
is gibberish,
The only alternative would by infinitely many unit fractions at one
point. That is not gibberish but wrong.
Of many suitable definitions of natural numbers,
one is:
they are well.ordered (subsets minimummed or empty)
they continue (have successors)
they are reached by a step (≠0 have predecessors)
The natural numbers are our Paradigm of Finite.
There is no first unreachable natural number.
The natural numbers n belonging to the first infinitely many unit
fractions 1/n, i.e. there where NUF(x) increases at one point from 0 to infinity, cannot be distinguished in your opinion. Thus they are
unreachable.
By that and by its well.order,
there is no unreachable natural number.
ω is not a natural number.
⎛ Each before ω can be reached.
⎝ Each which can be reached is before ω.
If ω-1 existed such that (ω-1)+1 = ω
then ω could be reached
If ω-1 could be seen. But it cannot.
Regards, WM
Am Tue, 24 Sep 2024 21:38:17 +0200 schrieb WM:
On 24.09.2024 10:00, joes wrote:Does it need to be a member of the set?
Am Sun, 22 Sep 2024 15:28:38 +0200 schrieb WM:An end is where nothing follows.
Really existing sets of real unit fractions have two ends.What is „an end”?
on 9/24/2024, WM supposed :
The function NUF(0) increases. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 proves that >> it can gain only one unit fraction at any point x ∈ ℝ, hence NUF has
to pass the value 1.
It means "For all"
and says nothing about starting somewhere, moving, and ending somewhere
else.
On 9/24/24 3:43 PM, WM wrote:
I think that mathematics of fractions is correct.
It contains ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 from which follows that never
two different5 fractions sit upon each other. From NUF(0) = 0 the
smallest unit fraction follows immediately, From it the largest natural
number.
Except that there doesn't exist a largest natural number, BY DEFINITION,
On 9/24/24 3:38 PM, WM wrote:
On 24.09.2024 10:00, joes wrote:
Am Sun, 22 Sep 2024 15:28:38 +0200 schrieb WM:
Really existing sets of real unit fractions have two ends.What is „an end”?
An end is where nothing follows.
And thus the lower end of the numbers x > 0 would be 0, a value not in
the set, as the set doesn't have a lower end in it.
That is why there is no smallest positive real or unit fraction.
It shows that NUF(x) itself is the source of the giberish, as it
presumes things that don't actually exist,
On 9/24/24 4:37 PM, WM wrote:
The alternative is that the unit fractions keep getting closer and
closer without limit.
On 9/24/24 3:35 PM, WM wrote:
Yes, it increases, but there is no point that in can increase to 1, so
it just jumps to infinity.
On 9/24/2024 4:37 PM, WM wrote:
Thus
increasing NUF(x) from 0 to infinity
WITH intermediate steps
is gibberish,
The only alternative would by
infinitely many unit fractions at one point.
The correct (different) alternative is:
infinitely.many unit.fractions
at one point per unit.fraction,
with infinitely.many points in all.
Although
no more than finitely.many points
can be stepped.through end.to.end
we don't require these points to do more than _exist_
More.than.finitely.many can _exist_
On 25.09.2024 13:55, Richard Damon wrote:How could it see them?
It shows that NUF(x) itself is the source of the giberish, as it
presumes things that don't actually exist,
It is a mathematical function. It is assumed to see dark numbers. You
believe that infinitely many of the smallest fractions cannot be distinguished. So you believe in dark numbers too.
On 25.09.2024 06:59, Jim Burns wrote:Thus it increases to infinity at every positive point.
On 9/24/2024 4:37 PM, WM wrote:
That means NUF increases by 1 at every point occupied by a unitThe correct (different) alternative is: infinitely.manyThus increasing NUF(x) from 0 to infinity WITH intermediate steps
is gibberish,
unit.fractions at one point per unit.fraction,
fraction.
How does it distinguish dark points?with infinitely.many points in all.NUF(x) distinguishes all points.
Right, and there is no point "next to" 0, so it doesn't actuallyAlthough no more than finitely.many points can be stepped.through end.to.end we don't require these points to do more than _exist_ More.than.finitely.many can _exist_Yes, but they are dark.
NUF increases. At no point it can increase by more than 1.
Even if most mathematicians are far too stupid to understand this, IYou should try explaining it a different way.
will repeat it on and on, maybe that sometime some will get it.
On 25.09.2024 13:53, Richard Damon wrote:"Actual" doesn't mean "finite".
On 9/24/24 3:43 PM, WM wrote:
That may be. Then there are no complete sets, no actual infinity. GoodI think that mathematics of fractions is correct.Except that there doesn't exist a largest natural number, BY
It contains ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 from which follows that never >>> two different5 fractions sit upon each other. From NUF(0) = 0 the
smallest unit fraction follows immediately, From it the largest
natural number.
DEFINITION,
for maths, bad for matheology.
On 9/24/24 3:43 PM, WM wrote:
I think that [the] mathematics of fractions is correct.
It contains ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 from which follows that never >>>> two different5 fractions sit upon each other. From NUF(0) = 0 the
[existence of the] smallest unit fraction follows immediately
On 25.09.2024 13:53, Richard Damon wrote:
On 9/24/24 3:43 PM, WM wrote:
I think that mathematics of fractions is correct.
It contains ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 from which follows that never >>> two different5 fractions sit upon each other. From NUF(0) = 0 the
smallest unit fraction follows immediately, From it the largest
natural number.
Except that there doesn't exist a largest natural number, BY DEFINITION,
That may be. Then there are no complete sets, no actual infinity. Good
for maths, bad for matheology.
Regards, WM
On 24.09.2024 22:51, joes wrote:
Am Tue, 24 Sep 2024 21:38:17 +0200 schrieb WM:The end is a member of the set. But in many cases it is invisible and
On 24.09.2024 10:00, joes wrote:Does it need to be a member of the set?
Am Sun, 22 Sep 2024 15:28:38 +0200 schrieb WM:An end is where nothing follows.
Really existing sets of real unit fractions have two ends.What is „an end”?
only proved by the fact that there is a smaller infimum or a larger
supremum.
Regards, WM
On 25.09.2024 13:54, Richard Damon wrote:
On 9/24/24 3:38 PM, WM wrote:
On 24.09.2024 10:00, joes wrote:
Am Sun, 22 Sep 2024 15:28:38 +0200 schrieb WM:
Really existing sets of real unit fractions have two ends.What is „an end”?
An end is where nothing follows.
And thus the lower end of the numbers x > 0 would be 0, a value not in
the set, as the set doesn't have a lower end in it.
If it is a set (actual infinity), then the set has a smallest member.
But that is hard to understand.
That is why there is no smallest positive real or unit fraction.
It is easier to understand with unit fractions. But most mathematicians
are too stupid even for this obvious fact.
Regards, WM
On 25.09.2024 13:58, Richard Damon wrote:
On 9/24/24 4:37 PM, WM wrote:
The alternative is that the unit fractions keep getting closer and
closer without limit.
Either they occupy one point or NUF will distinguish them.
Regards, WM
On 09/24/2024 02:47 PM, Ross Finlayson wrote:
On 09/24/2024 01:35 PM, Ross Finlayson wrote:
On 09/24/2024 10:22 AM, Jim Burns wrote:
On 9/20/2024 5:15 PM, Ross Finlayson wrote:
Then these lines-reals these iota-values
are about the only "standard infinitesimals"
there are: with extent you observe, density
you observe, least-upper-bound as trivial,
and measure as assigned, length assignment.
Lines with the least.upper.bound property
(equivalent to "crossing must intersect")
do not have infinitesimals.
For example,
there are no infinitesimals
between 0 and all the _finite_ unit.fractions.
⎛ Each positive point has
⎜ a finite unit.fraction between it and 0
⎜
⎜⎛ Otherwise,
⎜⎜ greatest.lower.bound β of finite unit.fractions
⎜⎜ is positive, and
⎜⎜ not.bounding 2⋅β > finite ⅟k
⎜⎜ ½⋅β > ¼⋅⅟k
⎜⎜ β > ½⋅β > ¼⋅⅟k
⎜⎜ greatest.lower.bound β is not.bounding,
⎝⎝ which is gibberish.
Well now, there are as many kinds infinitesimals
as there are infinities,
and all in a general sense differing in
differences quite clustered about zero,
make for that Peano, Dodgson, Veronese,
Stolz, Leibniz, MacLaurin, Price,
the entire field of infinitesimal analysis as
what real analysis was named for hundreds of years,
make for that even Robinson's
rather modest and of no analytical character
the hyper-reals, or
as among Conway's surreal numbers,
has that most people's ideas of infinitesimals
are exactly as an infinite of them in [0,1],
constant monotone strictly increasing,
as with regards to "asymptotic equipartitioning"
and other aspects of higher, and lower, mathematics.
Newton's "fluxions", Aristotle's contemplations and
deliberations about atoms, Zeno's classical expositions,
quite a few of these have infinitesimals all quite
throughout every region of the linear continuum.
Maybe Hardy's pure mathematics makes for conflating
the objects of geometry, points and lines, with
a descriptive set theory's, a theory with only
one relation and only one-way, point-sets, yet
for making a theory with them all together,
makes for that since antiquity and through
today, notions like Bell's smooth analysis,
and Nelson's Internal Set Theory, if you
didn't know, each have that along the linear
continuum: are not "not infinitesimals".
Here these "iota-values" are considered
"standard infinitesimals".
Then, in the complete ordered field,
there's nothing to say about them
except nothing, well, some have that
its properties of least-upper-bound
and measure are actually courtesy already
a more fundamental continuum, in the theory,
as a constant, and not just stipulated
to match expectations.
The MacLaurin's infinitesimals and then for
Price's textbook "Infinitesimal Analsysis",
from the mid 1700's through the late 1800's
and fin-de-siecle, probably most closely match
the fluxion and Leibniz's notions, our notions,
while, "iota-values" are after the particular
special character of the special function,
the natural/unit equivalency function, in
as with regards to plural: laws of large numbers,
models of real numbers, definitions of continuity,
models of Cantor space, and this as with being
sets in a set theory, obviously extra-ordinary.
Or, iota-values are consistent, and constructive,
and their (relevant) properties decide-able,
in descriptive set theories about a linear continuum,
like today's most well-known, ZFC, and its models
of a continuous domain: extent density completeness measure.
There's also Cavalieri to consider,
and Bradwardine from the Mertonian school
about De Continuo, where sometimes it's
said that Cavalieri in the time of Galileo
formalized infinitesimals.
https://www.youtube.com/watch?v=EyWpZQny5cY&t=1590
"Moment and Motion: meters, seconds, orders, inverses"
Of course most people's usual ideas about
infinitesimals are what's called "atomism".
This is Democritus vis-a-vis Eudoxus.
Wow, it's like I just mentioned the conversation
here where was defined "continuous topology"
as "own initial and final topology".
Cantor of course had an oft-repeated opinion
on infinitesimals: "bacteria". This was after the
current theory of the day of bacteria vis-a-vis miasma
as the scientific source of disease, while these days
it's known that there are symbiotic bacteria,
while miasmas are still usually considered bad.
He though was happy to ride Russell's retro-thesis,
after borrowing Heine's result in trigonometric series,
which though has some reasonings where it's not so,
and collecting the anti-diagonal from duBois-Reymond,
nested intervals from Pythagoras, and this kind of thing.
Poincare didn't much concur.
Euler'd been kind of like "notice I move things around
in my infinite series", with regards to it's sort of like
he took the maxim of the lever as that he was the origin.
Yet, the resulting Euler's identity and Eulerian/Gaussian
analysis has its own sort of crazing as with regards
to the veneer of the analysis. It's considered
"standard", though.
Leibniz treated the differential as an algebraic
quantity, MacLaurin sort of righted that with
regards to fluent and fluxion, yet, sometimes
it's so, varies how and why it's so.
Then, for something like George Berkley's
infinitesimals as "ghosts of a departed quantity",
you can read about as much into that as something
like George Carlin and "infinities are so profound I'm profane".
Then after Leibniz there's nil-potent and particularly
nil-square, as what descends these days to that "the
only standard infinitesimal is zero", or as with regards
to that otherwise what results are tiny, yet standard
quantities, what in an infinite series may appreciate.
Of course for things like non-linear analysis and where
what otherwise the nil-square washes out as obviated
by triangle, Cauchy-Schwartz, Holder inequality and so on,
it's part of the field about where it's resulted non-negligeable,
as with regards to usually nil-potent and nil-square.
Anyways you can often find that if something like
Hilbert said that infinitary reasoning is the finest
creation of the human mind, it's in a wide variety.
So, yeah, pretty much any matters of "non-" standard
analysis, of course in no way contradict standard real
analysis at all, instead their being "super-" standard.
... Because Eudoxus/Cauchy/Dedekind is insufficient, or,
at least it's known "at best: incomplete".
Anyways: Democritus and atomism, and Eudoxus and the
Pythagorean and the Archimedean and the field, have
at least that Aristotle describes both as theoretically
so, like line-reals and field-reals, then though he
picks field-reals as he simply wasn't an atomist, for
Aristotle's substances and forms and Platonism.
Which here is a retro-Heraclitan dual-monism so that
increment and equi-partition build arithmetic together.
On 25.09.2024 13:55, Richard Damon wrote:
It shows that NUF(x) itself is the source of the giberish, as it
presumes things that don't actually exist,
It is a mathematical function. It is assumed to see dark numbers. You
believe that infinitely many of the smallest fractions cannot be distinguished. So you believe in dark numbers too.
Regards, WM
On 24.09.2024 23:47, FromTheRafters wrote:
on 9/24/2024, WM supposed :
The function NUF(0) increases. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 proves that >>> it can gain only one unit fraction at any point x ∈ ℝ, hence NUF has >>> to pass the value 1.
It means "For all" and says nothing about starting somewhere, moving,
and ending somewhere else.
Whatever it means. I see that there are no unit fractions below 0 and
many above 0. Therefore NUF increases. At no point it can increase by
more than 1.
Even if most mathematicians are far too stupid to understand this, I
will repeat it on and on, maybe that sometime some will get it.
Regards, WM
On 25.09.2024 13:59, Richard Damon wrote:
On 9/24/24 3:35 PM, WM wrote:
Yes, it increases, but there is no point that in can increase to 1, so
it just jumps to infinity.
That means infinitely many cannot be distinguished. That is what I call
dark numbers.
Regards, WM
On 23.09.2024 20:53, Jim Burns wrote:
Because largest.number m is gibberish.
The largest number that you can choose depends on your facilities.
Consider the largest number available on your pocket calculator.
Regards, WM
On 25.09.2024 06:59, Jim Burns wrote:
On 9/24/2024 4:37 PM, WM wrote:
On 24.09.2024 22:19, Jim Burns wrote:
Thus
increasing NUF(x) from 0 to infinity
WITH intermediate steps
is gibberish,
The only alternative would by
infinitely many unit fractions at one point.
The correct (different) alternative is:
infinitely.many unit.fractions
at one point per unit.fraction,
That means NUF increases by 1
at every point occupied by a unit fraction.
That means NUF increases by 1
at every point occupied by a unit fraction.
with infinitely.many points in all.
NUF(x) distinguishes all points.
Although
no more than finitely.many points
can be stepped.through end.to.end
we don't require these points to do more than _exist_
More.than.finitely.many can _exist_
Yes, but they are dark.
[...] I will repeat it on and on, [...]
On 9/25/2024 8:26 AM, WM wrote:
If it is a set (actual infinity), then the set has a smallest member.
But that is hard to understand.
It is easier to understand with unit fractions. But most
mathematicians are too stupid even for this obvious fact.
On 09/25/2024 10:11 AM, Jim Burns wrote:
[...]
How would you define "atom"
the otherwise "infinitely-divisible"?
It's seems quite Aristotlean to be against atomism,
yet, at the same time
it's a very useful theory,
for example, with Democritan chemistry, atomic chemistry,
and stoichiometry.
This is foundations under consideration here,
not merely "pre-calc".
On 09/25/2024 02:00 PM, Jim Burns wrote:
On 9/25/2024 2:44 PM, Ross Finlayson wrote:
How would you define "atom"
the otherwise "infinitely-divisible"?
It's seems quite Aristotlean to be against atomism,
yet, at the same time
it's a very useful theory,
for example, with Democritan chemistry, atomic chemistry,
and stoichiometry.
This is foundations under consideration here,
not merely "pre-calc".
I think we don't choose foundations which
choose for us what is to be built on them.
(I think they shouldn't, so, Yay!)
ℝ is anti-atomic, that is, without infinitesimals.
And yet, ℝ is very useful for describing solutions to
the hydrogen.atom Hamiltonian ̂H = ̂p²/2m - e²/̂r
The periodic table and the complete ordered field
seem to connect differently from
the way i which you (RF) think they connect.
"you know I wrote a paper about the
infinitesimals with measurable character
in Nelson's Internal Set Theory, IST,
that IST, was co-consistent, with ZFC,
Zermelo-Fraenkel set theory".
About foundations, then it seems
that reflects on what Leibniz and others call,
"the fundamental question of meta-physics",
then why there's any one theory at all.
(Truth, ....)
On 25.09.2024 19:12, Richard Damon wrote:
The problem is that it turns out the NUF(x) NEVER actually
"increments" by 0ne at any finite point, it jumps from 0 to infinity
(Aleph_0) in the unboundedly small gap between 0 and all x > 0,
How do you distinguish them?
Regards, WM
On 25.09.2024 20:40, Jim Burns wrote:
On 9/25/2024 11:51 AM, WM wrote:
That means NUF increases by 1
at every point occupied by a unit fraction.
There are numbers (cardinalities) which increase by 1
and other numbers (cardinalities), which
don't increase by 1.
No. Every countable set is countable, i.e., it increases one by one.
For each positive point x
for each number (cardinality) k which can increase by 1
there are more.than.k unit.fractions between 0 and x
That is a misinterpretation of the law valid for small numbers.
For each positive point x
the number (cardinality) of unit.fractions between 0 and x
is not
any number (cardinality) which increases by 1
Instead, it is
a number (cardinality) which doesn't increase by 1.
For every x NUF increases by not more than 1.
Each positive point is undercut by
some finite.unit.fraction.
Repetition is apparently what you (JB) think mathematics is!
Prove that ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is wrong or agree.
Regards, WM
On 25.09.2024 19:19, Richard Damon wrote:
On 9/25/24 11:32 AM, WM wrote:
I do NOT believe that there ANY "smallest" factions that cannot be
distingusished, I know such a thing does not exist, and that ALL unit
fractions, like all rationals and reals can be distinguished,
The claim of all requires a last one.
Regards, WM
On 25.09.2024 19:12, Richard Damon wrote:
The problem is that it turns out the NUF(x) NEVER actually
"increments" by 0ne at any finite point, it jumps from 0 to infinity
(Aleph_0) in the unboundedly small gap between 0 and all x > 0,
How do you distinguish them?
Regards, WM
On 25.09.2024 19:22, Richard Damon wrote:
On 9/25/24 11:14 AM, WM wrote:
NUF increases. At no point it can increase by more than 1.
Why not?
Because there is a finite gap between two unit fractions.
Regards, WM
On 25.09.2024 19:25, Richard Damon wrote:
What values, that EXIST, can't be distinguished.
You cannot distinguish the ℵo smallest unit fractions. You cannot select
a unit fractions with less smaller unit fractions.
Regards, WM
On 25.09.2024 19:28, Richard Damon wrote:
On 9/24/24 3:30 PM, WM wrote:
The largest number that you can choose depends on your facilities.
Consider the largest number available on your pocket calculator.
Mathematics isn't based on what WE can do, but on what the numbers
themselves can do,
They cannot do anything. They simply are created in potential infinity
or are there in actual infinity.
Regards, WM
On 9/27/2024 11:57 AM, WM wrote:
On 25.09.2024 19:28, Richard Damon wrote:
On 9/24/24 3:30 PM, WM wrote:
The largest number that you can choose depends on your facilities.
Consider the largest number available on your pocket calculator.
Mathematics isn't based on what ___we can do___, but on
On 9/27/2024 11:42 AM, WM wrote:
On 25.09.2024 18:39, joes wrote:
Am Wed, 25 Sep 2024 17:51:56 +0200 schrieb WM:
NUF(x) distinguishes all points.How does it distinguish dark points?
By our mathematical knowledge about ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
NUF increases. At no point it can increase by more than 1.Right, and there is no point "next to" 0
There is a unit fraction next to 0. [WM]
Next as in closer and closer forevermore? ;^)
On 9/27/2024 3:13 PM, FromTheRafters wrote:
WM expressed precisely :
On 25.09.2024 19:22, Richard Damon wrote:
On 9/25/24 11:14 AM, WM wrote:
NUF increases
At no point it can increase by more than 1.Why not?
Because there is a finite gap between two [adjacent] unit fractions.
On 9/27/2024 3:13 PM, FromTheRafters wrote:
WM expressed precisely :
On 25.09.2024 19:22, Richard Damon wrote:
On 9/25/24 11:14 AM, WM wrote:
NUF increases
At no point it can increase by more than 1.Why not?
Because there is a finite gap between two [adjacent] unit fractions.
Here, I'm thinking of gaps as meaning places where a cauchy sequence
does not converge to a member of the set but somehow would fit between
them in an extended domain which included that element.
[In this sense] there are [...] gaps [in Q]?
It's infinitely dense.
Here, I'm thinking of gaps as meaning places where a cauchy sequence
does not converge to a member of the set but somehow would fit between
them in an extended domain which included that element.
Wrt to R, there are [...] gaps [in Q]?
It's infinitely dense.
Am 28.09.2024 um 06:26 schrieb Chris M. Thomasson:
On 9/27/2024 3:13 PM, FromTheRafters wrote:
WM expressed precisely :
On 25.09.2024 19:22, Richard Damon wrote:
On 9/25/24 11:14 AM, WM wrote:
NUF increases
No, it "jumps at 0".
MEANING: NUF(0) = 0 and Ax > 0: NUF(x) = aleph_0.
At no point it can increase by more than 1.Why not?
Because there is a finite gap between two [adjacent] unit fractions.
Using WM's mantra: An e IN: 1/n - 1/(n + 1) > 0.
@Mückenheim: Ax > 0: NUF(x) = aleph_0 besagt GENAU, dass für jedes x e
IR mit x > 0 _abzählbar unendlich_ viele Stammbrüche in (0, x) enthalten sind. Und das ist auch richtig (denn es ist so).
Der Umstand, dass trivialerweise An e IN: 1/n - 1/(n + 1) > 0 gilt, tut
dem keinen Abbruch; auch wenn Du zu doof und zu blöde bist, um das zu verstehen.
Actual completed infinity, to him, means both a first element and a last element must be present in the set.
Am Fri, 27 Sep 2024 21:08:46 +0200 schrieb WM:
On 25.09.2024 18:53, Moebius wrote:Duh, that is your premise.
Between two unit fractions there is always a finite gap.Proof?On 9/24/24 3:43 PM, WM wrote:
It contains ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 from which follows that never
two different5 fractions sit upon each other. From NUF(0) = 0 the >>>>>>> [existence of the] smallest unit fraction follows immediately
Am Fri, 27 Sep 2024 20:42:06 +0200 schrieb WM:
On 25.09.2024 18:39, joes wrote:How can you talk about a dark n and it’s (maybe) n+1?
Am Wed, 25 Sep 2024 17:51:56 +0200 schrieb WM:By our mathematical knowledge about ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
NUF(x) distinguishes all points.How does it distinguish dark points?
Is it closer to zero or to the next UF?There is a unit fraction next to 0.NUF increases. At no point it can increase by more than 1.Right, and there is no point "next to" 0
On 09/26/2024 06:48 AM, Jim Burns wrote:
On 9/25/2024 9:50 PM, Ross Finlayson wrote:
On 09/25/2024 02:00 PM, Jim Burns wrote:
On 9/25/2024 2:44 PM, Ross Finlayson wrote:
How would you define "atom"
the otherwise "infinitely-divisible"?
It's seems quite Aristotlean to be against atomism,
I am anti.atomized.ℝ (complete ordered field)
You seem to read that as anti.atomized.anything.
How do I make claims to you (RF) which
are only about the things I intend?
For example, how do I claim to you (RF),
for a right triangles but not for any triangle, that
⎛ the square of its longest side equals
⎝ the sum of the squares of its two other sides
?
I enjoy reading that, Jim, if I may be so familiar,
I enjoy reading that because it sounds _sincere_,
and, it reflects a "conscientiousness", given what
there is, given the milieu, then, for given the surrounds.
And for a countably INFINITE set, like the Natural Numbers, there is
only one end you can count from.
They all have different values, so why can't you distinguish them.
THey are all based on different Natural Numbers, so why can't you
distinguish them?
On 9/27/24 3:06 PM, WM wrote:
On 25.09.2024 19:12, Richard Damon wrote:
The problem is that it turns out the NUF(x) NEVER actually
"increments" by 0ne at any finite point, it jumps from 0 to infinity
(Aleph_0) in the unboundedly small gap between 0 and all x > 0,
How do you distinguish them?
They have different values, so why can't you?
After serious thinking WM wrote :
On 28.09.2024 12:10, FromTheRafters wrote:
Actual completed infinity, to him, means both a first element and a
last element must be present in the set.
There must be all elements. In a linear order this implies a first and
a last one. Without knowing that there is a last element, you cannot
reasonable claim that all are there.
Yes we can.
WM presented the following explanation :
Between two unit fractions there is always a finite gap.
How big?
On 25.09.2024 20:40, Jim Burns wrote:
On 9/25/2024 11:51 AM, WM wrote:
That means NUF increases by 1
at every point occupied by a unit fraction.
There are numbers (cardinalities) which increase by 1
and other numbers (cardinalities), which
don't increase by 1.
No.
Every countable set is countable,
i.e., it increases one by one.
For each positive point x
for each number (cardinality) k which can increase by 1
there are more.than.k unit.fractions between 0 and x
That is a misinterpretation of
the law valid for small numbers.
For each positive point x
the number (cardinality) of
unit.fractions between 0 and x
is not
any number (cardinality) which increases by 1
Instead, it is
a number (cardinality) which doesn't increase by 1.
For every x NUF increases by not more than 1.
Each positive point is undercut by
some finite.unit.fraction.
Repetition
is apparently what you (JB) think mathematics is!
Each positive point is
undercut by some finite.unit.fraction.
0 is not.undercut bysome finite.unit.fraction.
Prove that
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or agree.
@Mückenheim: Ax > 0: NUF(x) = aleph_0 besagt GENAU, dass für jedes x e
IR mit x > 0 _abzählbar unendlich_ viele Stammbrüche in (0, x) enthalten sind. Und das ist auch richtig (denn es ist so).
Der Umstand, dass trivialerweise An e IN: 1/n - 1/(n + 1) > 0 gilt, tut
dem keinen Abbruch; auch wenn Du zu doof und zu blöde bist, um das zu verstehen.
@Mückenheim: Ax > 0: NUF(x) = aleph_0 besagt GENAU, dass für jedes x e
IR mit x > 0 _abzählbar unendlich_ viele Stammbrüche in (0, x) enthalten sind. Und das ist auch richtig (denn es ist so).
Der Umstand, dass trivialerweise An e IN: 1/n - 1/(n + 1) > 0 gilt, tut
dem keinen Abbruch; auch wenn Du zu doof und zu blöde bist, um das zu verstehen.
On 9/29/2024 12:30 PM, WM wrote:
On 29.09.2024 21:21, FromTheRafters wrote:
After serious thinking WM wrote :
On 28.09.2024 12:10, FromTheRafters wrote:
Actual completed infinity, to him, means both a first element and a
last element must be present in the set.
There must be all elements. In a linear order this implies a first
and a last one. [WM]
Without knowing that there is a last element, you cannot reasonable claim that all are there.
Yes we can.
Like Obama, the war president, making a laughing stock of the Nobel
prize?
Huh? What does one of our former presidents have to do with any of this
shit?
On 28.09.2024 00:08, FromTheRafters wrote:
WM presented the following explanation :
Between two unit fractions there is always a finite gap.
How big?
In terms of set theory: uncountably many points.
Regards, WM
On 27.09.2024 22:12, joes wrote:
Am Fri, 27 Sep 2024 20:42:06 +0200 schrieb WM:
On 25.09.2024 18:39, joes wrote:How can you talk about a dark n and it’s (maybe) n+1?
Am Wed, 25 Sep 2024 17:51:56 +0200 schrieb WM:By our mathematical knowledge about ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
NUF(x) distinguishes all points.How does it distinguish dark points?
I assume that dark n exist and follow the rules of visible n, with one exception.
Is it closer to zero or to the next UF?There is a unit fraction next to 0.NUF increases. At no point it can increase by more than 1.Right, and there is no point "next to" 0
That is unknown.
Regards, WM
On 28.09.2024 13:24, Richard Damon wrote:
And for a countably INFINITE set, like the Natural Numbers, there is
only one end you can count from.
There is no reason to believe so if all elements are there.
Regards, WM
On 28.09.2024 13:30, Richard Damon wrote:
They all have different values, so why can't you distinguish them.
THey are all based on different Natural Numbers, so why can't you
distinguish them?
Because always infinitely many remain undistinguished. They are dark.
Regards, WM
On 28.09.2024 14:58, Richard Damon wrote:
On 9/27/24 3:06 PM, WM wrote:
On 25.09.2024 19:12, Richard Damon wrote:
The problem is that it turns out the NUF(x) NEVER actually
"increments" by 0ne at any finite point, it jumps from 0 to infinity
(Aleph_0) in the unboundedly small gap between 0 and all x > 0,
How do you distinguish them?
They have different values, so why can't you?
Then distinguish the first one.
Regards, WM
On 28.09.2024 12:10, FromTheRafters wrote:
Actual completed infinity, to him, means both a first element and a
last element must be present in the set.
There must be all elements. In a linear order this implies a first and a
last one. Without knowing that there is a last element, you cannot
reasonable claim that all are there.
Regards, WM
On 9/27/2024 2:54 PM, WM wrote:
What you want is
to tell us we've been wrong about sets.
For each positive point x
for each number (cardinality) k which can increase by 1
there are more.than.k unit.fractions between 0 and x
That is a misinterpretation of
the law valid for small numbers.
| "If the law supposes that," said Mr. Bumble,
| squeezing his hat emphatically in both hands,
| "the law is a ass — a idiot."
|
-- Charles Dickens, "Oliver Twist"
For every x NUF increases by not more than 1.
For every x>0 and x′>0
NUF increases by not more and not less than 0.
🎜 Aleph.naught bottles of beer on the wall,
WM explained on 9/29/2024 :
On 28.09.2024 00:08, FromTheRafters wrote:
WM presented the following explanation :
Between two unit fractions there is always a finite gap.;
How big?
In terms of set theory: uncountably many points.
How wide are these points?
D. h. (abzählbar) unendlich viele Stammbrüche haben in [0, 1] Platz.
On 9/29/24 3:16 PM, WM wrote:
On 28.09.2024 14:58, Richard Damon wrote:There isn't a first one.
On 9/27/24 3:06 PM, WM wrote:
On 25.09.2024 19:12, Richard Damon wrote:
The problem is that it turns out the NUF(x) NEVER actually
"increments" by 0ne at any finite point, it jumps from 0 to
infinity (Aleph_0) in the unboundedly small gap between 0 and all x
0,
How do you distinguish them?
They have different values, so why can't you?
Then distinguish the first one.
On 29.09.2024 21:56, Jim Burns wrote:
On 9/27/2024 2:54 PM, WM wrote:
On 25.09.2024 20:40, Jim Burns wrote:
There are numbers (cardinalities) which increase by 1
and other numbers (cardinalities), which
don't increase by 1.
No.
You invoke _axiom.1_
Every countable set is countable,
i.e., it increases one by one.
Axiom.1 _describes_
what you are currently discussing.
Axiom.1 means
⎛ If
⎜ the set of unit fraction can't increase by 1
⎜ then
⎝ we aren't discussing that set.
Axiom.1 does not mean
⎛ If
⎜ we are discussing the set of unit fractions
⎜ then
⎝ that set can increase by 1
What you want is
to tell us we've been wrong about sets.
You have been wrong about sets.
You assume that sets are invariable
but you don't assume that all elements,
here unit fractions, can be detected.
For every x NUF increases by not more than 1.
For every x>0 and x′>0
NUF increases by not more and not less than 0.
⅟⌈1+⅟x⌉ → ⅟⌈1+⅟x′⌉
⅟⌈2+⅟x⌉ → ⅟⌈2+⅟x′⌉
⅟⌈3+⅟x⌉ → ⅟⌈3+⅟x′⌉
⅟⌈4+⅟x⌉ → ⅟⌈4+⅟x′⌉
...
Wrong.
🎜 Aleph.naught bottles of beer on the wall,
ℵo unit fractions cannot come into being
without a first one.
On 9/30/2024 11:12 AM, WM wrote:
What you are talking about aren't _our_ sets.
We have no more reason to care about _your_ "sets".
Unit fractions do not come into being.
On 30.09.2024 20:33, Jim Burns wrote:
On 9/30/2024 11:12 AM, WM wrote:
On 29.09.2024 21:56, Jim Burns wrote:
On 9/27/2024 2:54 PM, WM wrote:
On 25.09.2024 20:40, Jim Burns wrote:
There are
numbers (cardinalities) which increase by 1
and other numbers (cardinalities),
which don't increase by 1.
No.
You invoke _axiom.1_
Every countable set is countable,
i.e., it increases one by one.
What you are talking about aren't _our_ sets.
NUF(0) = 0 and NUF(1) = ℵo.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
shows that at
no point x NUF can increase by
more than one step 1.
It is fact with your set too.
I am not responsible.
I only made the discovery.
We have no more reason to care about _your_ "sets".
No reason even to care about
mathematical basic truths like
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
Unit fractions do not come into being.
But they come into sight.
On 30.09.2024 13:13, Richard Damon wrote:
On 9/29/24 3:16 PM, WM wrote:
On 28.09.2024 14:58, Richard Damon wrote:There isn't a first one.
On 9/27/24 3:06 PM, WM wrote:
On 25.09.2024 19:12, Richard Damon wrote:
The problem is that it turns out the NUF(x) NEVER actually
"increments" by 0ne at any finite point, it jumps from 0 to
infinity (Aleph_0) in the unboundedly small gap between 0 and all
x > 0,
How do you distinguish them?
They have different values, so why can't you?
Then distinguish the first one.
NUF(0) = 0 and NUF(1) = ℵo. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that at no
point x NUF can increase by more than one step 1. It is fact. I am not responsible. I only made the discovery.
Regards, WM
On 30.09.2024 20:33, Jim Burns wrote:
On 9/30/2024 11:12 AM, WM wrote:
What you are talking about aren't _our_ sets.
NUF(0) = 0 and NUF(1) = ℵo. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that at no
point x NUF can increase by more than one step 1. It is fact with your
set too. I am not responsible. I only made the discovery.
We have no more reason to care about _your_ "sets".
No reason even to care about mathematical basic truths like
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
Unit fractions do not come into being.
But they come into sight.
Regards, WM
On 30.09.2024 00:29, FromTheRafters wrote:
WM explained on 9/29/2024 :
On 28.09.2024 00:08, FromTheRafters wrote:
WM presented the following explanation :
Between two unit fractions there is always a finite gap.;
How big?
In terms of set theory: uncountably many points.
How wide are these points?
More than nothing.
Regards, WM
On 9/30/2024 4:13 AM, Richard Damon wrote:
On 9/29/24 3:16 PM, WM wrote:
On 28.09.2024 14:58, Richard Damon wrote:
On 9/27/24 3:06 PM, WM wrote:
On 25.09.2024 19:12, Richard Damon wrote:
The problem is that it turns out the NUF(x) NEVER actually
"increments" by 0ne at any finite point, it jumps from 0 to
infinity (Aleph_0) in the unboundedly small gap between 0 and all
x > 0,
How do you distinguish them?
They have different values, so why can't you?
Then distinguish the first one.
Regards, WM
There isn't a first one.
Show me a circle with 4 sides.
;^) Humm, an n-gon where n is taken to infinity is a circle?
You are just proving your logic is non-sense.
On 09/29/2024 12:02 PM, Jim Burns wrote:
On 9/26/2024 4:50 PM, Ross Finlayson wrote:
On 09/26/2024 06:48 AM, Jim Burns wrote:
On 9/25/2024 9:50 PM, Ross Finlayson wrote:
On 09/25/2024 02:00 PM, Jim Burns wrote:
On 9/25/2024 2:44 PM, Ross Finlayson wrote:
How would you define "atom"
the otherwise "infinitely-divisible"?
It's seems quite Aristotlean to be against atomism,
I am anti.atomized.ℝ (complete ordered field)
You seem to read that as anti.atomized.anything.
How do I make claims to you (RF) which
are only about the things I intend?
For example, how do I claim to you (RF),
for a right triangles but not for any triangle, that
⎛ the square of its longest side equals
⎝ the sum of the squares of its two other sides
?
I enjoy reading that, Jim, if I may be so familiar,
I enjoy reading that because it sounds _sincere_,
and, it reflects a "conscientiousness", given what
there is, given the milieu, then, for given the surrounds.
Thank you.
A right triangle: either has one right angle,
or two angles that sum to a right angle.
The sum of the angles is the angle of the sum.
Then, as with regards to atoms,
On 9/30/24 2:54 PM, WM wrote:
NUF(0) = 0 and NUF(1) = ℵo. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that at no
point x NUF can increase by more than one step 1. It is fact with your
set too. I am not responsible. I only made the discovery.
Actually, it shows that at no point CAN it increase by 1.
For any finite number x, NUF(x) will be Aleph_0, and Aleph_0 when you
attempt to "increment" it, doesn't change.
Since there is no finite value of x where NUF(x) can be 1
For ANY finite x, NUF(x) will be Aleph_0, since there is an infinite
number of unit fractions smaller than it.
On 9/30/24 11:15 AM, WM wrote:
On 30.09.2024 00:29, FromTheRafters wrote:
WM explained on 9/29/2024 :
On 28.09.2024 00:08, FromTheRafters wrote:
WM presented the following explanation :
Between two unit fractions there is always a finite gap.;
How big?
In terms of set theory: uncountably many points.
How wide are these points?
More than nothing.
But it must be next to nothing,
which becomes nothing when we get to the
actual infinite set.
On 9/30/2024 2:54 PM, WM wrote:
NUF(0) = 0 and NUF(1) = ℵo.
ℵ₀ is the cardinality of
cardinalities which increase by 1
If ℵ₀ is any
cardinality which increases by 1
Therefore,
ℵ₀ does not increase by 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
shows that at
...no unit.fraction...
no point x NUF can increase by
more than one step 1.
0 is not a unit.fraction.
It is fact with your set too.
I am not responsible.
Also,
you are not correct.
No reason even to care about
mathematical basic truths like
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
∀n ∈ ℕ: 1/n > 1/(n+1) > 0
∀n ∈ ℕ: 1/n isn't first unit.fraction.
On 10/01/2024 10:00 AM, Jim Burns wrote:
I hope this will help me understand you better.
Please accept or reject each claim and
-- this is important --
replace rejected claims with
what you _would_ accept.
⎛ ℝ, the complete ordered field, is
⎝ the consensus theory in 2024 of the continuum.
⎛ ℝ contains ℚ the rationals and
⎜ the least upper bound of
⎝ each bounded nonempty subset of ℚ and of ℝ
( The greatest lower bound of ⅟ℕ unit fractions is 0
⎛ A unit fraction is reciprocal to a natural>0
⎜
⎜ A set≠{} ⊆ ℕ naturals holds a minimum
⎜ A natural≠0 has a predecessor.natural.
⎜ A natural has a successor.natural.
⎜
⎜ The sum of two naturals is a natural
⎝ the product of two naturals is a natural.
⎛ There are no points in ℝ
⎜ between 0 and all the unit fractions
⎝ (which is what I mean here by 'infinitesimal').
Thank you in advance.
Here it's that "Eudoxus/Dedekind/Cauchy is
_insufficient_ to represent the character
of the real numbers".
Then, that there are line-reals and signal-reals
besides field-reals, has that of course there are
also models of line-reals and signal-reals in the
mathematics today, like Jordan measure and the ultrafilter,
and many extant examples where a simple deliberation
of continuity according to the definitions of
line-reals or signal-reals, results any contradictions
you might otherwise see as arriving their existence.
Then, besides noting how it's broken, then also
there's given a reasoning how it's repaired,
resulting "less insufficient", or at least making
it so that often found approaches in the applied,
and their success, make the standard linear curriculum,
unsuited.
Then, I think it's quite standard how I put it,
really very quite standard.
On 30.09.2024 21:51, Jim Burns wrote:
On 9/30/2024 2:54 PM, WM wrote:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
shows that at
...no unit.fraction...
no point x NUF can increase by
more than one step 1.
0 is not a unit.fraction.
That proves NUF(0) = 0.
It is fact with your set too.
I am not responsible.
Also,
you are not correct.
What is incorrect?
No reason even to care about
mathematical basic truths like
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
∀n ∈ ℕ: 1/n > 1/(n+1) > 0
∀n ∈ ℕ: 1/n isn't first unit.fraction.
NUF cannot increase by more than 1 and
cannot start with more than 0 at 0.
You should understand that
Do you understand that?
an n-gon as n goes to infinity approaches a circle? Fair enough?
For every nonzero distance d from 0
there are ℵ₀.many unit.fractions closer than d
On 10/1/2024 1:29 PM, WM wrote:
What is incorrect?
This is incorrect:
🛇⎛ ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that
🛇⎜ at no point x
🛇⎝ NUF can increase by more than one step 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 doesn't show that.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows
∀n ∈ ℕ: 1/n > 1/(n+1) > 0
which shows
each unit fraction 1/n is not first.
On 01.10.2024 01:07, Richard Damon wrote:
On 9/30/24 2:54 PM, WM wrote:
NUF(0) = 0 and NUF(1) = ℵo. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that at >>> no point x NUF can increase by more than one step 1. It is fact with
your set too. I am not responsible. I only made the discovery.
Actually, it shows that at no point CAN it increase by 1.
Wrong because it cannot start with ℵo.
For any finite number x, NUF(x) will be Aleph_0, and Aleph_0 when you
attempt to "increment" it, doesn't change.
If NUF is ℵo at all x > 0 then it must count ℵo unit fractions at 0. Wrong.
Since there is no finite value of x where NUF(x) can be 1
Wrong presupposition.
Regards, WM
On 01.10.2024 01:14, Richard Damon wrote:
On 9/30/24 11:15 AM, WM wrote:
On 30.09.2024 00:29, FromTheRafters wrote:
WM explained on 9/29/2024 :
On 28.09.2024 00:08, FromTheRafters wrote:
WM presented the following explanation :
Between two unit fractions there is always a finite gap.;
How big?
In terms of set theory: uncountably many points.
How wide are these points?
More than nothing.
But it must be next to nothing,
No, between them and nothing there are infinitely many countable sets
and then finite sets of points.
which becomes nothing when we get to the actual infinite set.
Not in mathematics.
Regards, WM
On 01.10.2024 01:11, Richard Damon wrote:
For ANY finite x, NUF(x) will be Aleph_0, since there is an infinite
number of unit fractions smaller than it.
You claim that ℵo unit fractions are already there, whatever x > 0 you choose. That means they cannot be chosen. They are dark.
Regards, WM
WM pretended :
No. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 does not prove that n+1 is a natural
number. Note the infinite sequence
1, 2, 3, ..., ω-2, ω-1, ω.
Omega minus one or two is undefined
and n plus one closure is axiomatic.
On 10/1/24 1:39 PM, WM wrote:
On 01.10.2024 01:14, Richard Damon wrote:
On 9/30/24 11:15 AM, WM wrote:
On 30.09.2024 00:29, FromTheRafters wrote:
WM explained on 9/29/2024 :
On 28.09.2024 00:08, FromTheRafters wrote:
WM presented the following explanation :
Between two unit fractions there is always a finite gap.;
How big?
In terms of set theory: uncountably many points.
How wide are these points?
More than nothing.
But it must be next to nothing,
No, between them and nothing there are infinitely many countable sets
and then finite sets of points.
So your "smallest" wasn't the smallest,
which becomes nothing when we get to the actual infinite set.
Not in mathematics.
Sure it does.
On 10/1/24 1:37 PM, WM wrote:
On 01.10.2024 01:11, Richard Damon wrote:
For ANY finite x, NUF(x) will be Aleph_0, since there is an infinite
number of unit fractions smaller than it.
You claim that ℵo unit fractions are already there, whatever x > 0 you
choose. That means they cannot be chosen. They are dark.
Nope.
On 01.10.2024 22:05, Jim Burns wrote:Nobody believes that. It just doesn’t follow.
On 10/1/2024 1:29 PM, WM wrote:
This is incorrect:You believe that more than one unit fractions can occupy one and the
🛇⎛ ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that 🛇⎜ at no point x 🛇⎝ NUF can
increase by more than one step 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 doesn't show that.
same point nevertheless? That would make the distance 0, but it is > 0. Therefore you are wrong.
Those last numbers are not finite, and this is not a sequence (it is∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows ∀n ∈ ℕ: 1/n > 1/(n+1) > 0 whichNo. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 does not prove that n+1 is a natural number. Note the infinite sequence 1, 2, 3, ..., ω-2, ω-1, ω.
shows each unit fraction 1/n is not first.
It consists of infinitely many finite numbers.
Am Wed, 02 Oct 2024 13:10:09 +0200 schrieb WM:
On 01.10.2024 22:05, Jim Burns wrote:Nobody believes that. It just doesn’t follow.
On 10/1/2024 1:29 PM, WM wrote:You believe that more than one unit fractions can occupy one and the
This is incorrect:
🛇⎛ ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that 🛇⎜ at no point x 🛇⎝ NUF can
increase by more than one step 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 doesn't show that.
same point nevertheless? That would make the distance 0, but it is > 0.
Therefore you are wrong.
Am Wed, 02 Oct 2024 13:10:09 +0200 schrieb WM:
On 01.10.2024 22:05, Jim Burns wrote:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows ∀n ∈ ℕ: 1/n > 1/(n+1) [...] >>> which shows each unit fraction 1/n is not first.
No.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 does not prove that n+1 is a natural
number.
Note the infinite sequence 1, 2, 3, ..., ω-2, ω-1, ω. It consists of infinitely many finite numbers.
Am 02.10.2024 um 14:56 schrieb joes:
Am Wed, 02 Oct 2024 13:10:09 +0200 schrieb WM:
On 01.10.2024 22:05, Jim Burns wrote:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows ∀n ∈ ℕ: 1/n > 1/(n+1) [...] >>>> which shows each unit fraction 1/n is not first.
Right.
No.
Yes.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 does not prove that n+1 is a natural
number.
@Mückenheim:
Right. But the following PEANO-AXIOM
An e IN: S(n) e IN
zusammen mit der Definition
n+1 := s(n)
beweist es, Du Depp.
Since from them we get the theorem:
An e IN: n+1 e IN (*).
Hint: Jim DIDN'T claim that your trivial statement proves that n+1 is a natural number, but that it proves that for each and every unit fraction there's a SMALLER one [which it does together with (*)].
[Well, actually, for showing this we would also have to refer to a
definition of "unit fraction": u is a /unit fraction/ iff there's an n e
IN such that u = 1/n.]
Note the infinite sequence 1, 2, 3, ..., ω-2, ω-1, ω. It consists of
infinitely many finite numbers.
@Mückenheim:
Red' doch keine solche Scheiße daher, Mückenheim!
1. Ist ω GANZ GEWISS keine "finite number", sondern /the smallest
infinite ordinal number/.
2. Sind die Ausdrücke "ω-2", "ω-1" nicht definiert. Du redest also
wieder mal saudummen Scheißdreck daher.
3. Ist 1, 2, 3, ... ω "technisch gesehen" keine "unendliche
Folge" (jedenfalls keine mit Indexmenge IN).
Kurz und gut: Nichts als purer Sachwachsinn, der auch nicht das
Geringste mit dem in Rede stehenden Sachverhalt zu tun hat.
Am Wed, 02 Oct 2024 13:10:09 +0200 schrieb WM:
Note the infinite sequence 1, 2, 3, ..., ω-2, ω-1, ω.
It consists of infinitely many finite numbers.
Those last numbers are not finite [...]
On 10/01/2024 01:13 PM, Jim Burns wrote:
On 10/1/2024 2:02 PM, Ross Finlayson wrote:
Here it's that "Eudoxus/Dedekind/Cauchy is
_insufficient_ to represent the character
of the real numbers".
Then, that there are line-reals and signal-reals
besides field-reals, has that of course there are
also models of line-reals and signal-reals in the
mathematics today, like Jordan measure and the ultrafilter,
and many extant examples where a simple deliberation
of continuity according to the definitions of
line-reals or signal-reals, results any contradictions
you might otherwise see as arriving their existence.
Then, besides noting how it's broken, then also
there's given a reasoning how it's repaired,
resulting "less insufficient", or at least making
it so that often found approaches in the applied,
and their success, make the standard linear curriculum,
unsuited.
Then, I think it's quite standard how I put it,
really very quite standard.
I hope this will help me understand you better.
Please accept or reject each claim and
-- this is important --
replace rejected claims with
what you _would_ accept.
⎛ ℝ, the complete ordered field, is
⎝ the consensus theory in 2024 of the continuum.
⎛ ℝ contains ℚ the rationals and
⎜ the least upper bound of
⎝ each bounded nonempty subset of ℚ and of ℝ
( The greatest lower bound of ⅟ℕ unit fractions is 0
⎛ A unit fraction is reciprocal to a natural>0
⎜
⎜ A set≠{} ⊆ ℕ naturals holds a minimum
⎜ A natural≠0 has a predecessor.natural.
⎜ A natural has a successor.natural.
⎜
⎜ The sum of two naturals is a natural
⎝ the product of two naturals is a natural.
⎛ There are no points in ℝ
⎜ between 0 and all the unit fractions
⎝ (which is what I mean here by 'infinitesimal').
Thank you in advance.
Well, first of all there's a quibble that
R is not usually said to contain Q as much as that
there's that in real-values that
there's a copy of Q embedded in R.
The, "1/N unit fractions", what is that,
that does not have a definition.
Is that some WM-speak?
I suppose that
if it means the set 1/n for n in N
then the g-l-b is zero.
Then otherwise what you have there appear facts
about N and R.
Then,
where there exists a well-ordering of R,
then to take the well-ordering it results that
first there's a well-ordering of [0,1]
for both simplicity and necessity,
and it's as the range of the function n/d
with 0 <= n < d and as d -> oo,
i.e., only in the infinite limit,
that the properties of the range of naturals,
apply to the properties of the range of [0,1].
On 01.10.2024 22:05, Jim Burns wrote:
For every nonzero distance d from 0
there are ℵ₀.many unit.fractions closer than d
That means
there is no d by what
you can distinguish ℵ₀ unit fractions?
On 01.10.2024 22:05, Jim Burns wrote:
On 10/1/2024 1:29 PM, WM wrote:
What is incorrect?
This is incorrect:
🛇⎛ ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that
🛇⎜ at no point x
🛇⎝ NUF can increase by more than one step 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 doesn't show that.
You [JB] believe that
You believe that
more than one unit fractions
can occupy one and the same point
nevertheless?
That would make the distance 0,
but it is > 0.
Therefore you are wrong.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
shows
∀n ∈ ℕ: 1/n > 1/(n+1) > 0
which shows
each unit fraction 1/n is not first.
No.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
does not prove that n+1 is a natural number.
Note the infinite sequence
1, 2, 3, ..., ω-2, ω-1, ω.
It consists of infinitely many finite numbers.
Note the infinite sequence
1, 2, 3, ..., ω-2, ω-1, ω
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
does not prove that n+1 is a natural number.
On 02.10.2024 13:17, Richard Damon wrote:
On 10/1/24 1:39 PM, WM wrote:
On 01.10.2024 01:14, Richard Damon wrote:
On 9/30/24 11:15 AM, WM wrote:
On 30.09.2024 00:29, FromTheRafters wrote:
WM explained on 9/29/2024 :
On 28.09.2024 00:08, FromTheRafters wrote:
WM presented the following explanation :
Between two unit fractions there is always a finite gap.;
How big?
In terms of set theory: uncountably many points.
How wide are these points?
More than nothing.
But it must be next to nothing,
No, between them and nothing there are infinitely many countable sets
and then finite sets of points.
So your "smallest" wasn't the smallest,
Of course all gaps between unit fractions are made of more than finitely
many points. I never denied that.
which becomes nothing when we get to the actual infinite set.
Not in mathematics.
Sure it does.
Try to learn the basics and to understand ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Regards, WMK
On 02.10.2024 13:30, FromTheRafters wrote:
WM pretended :
No. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 does not prove that n+1 is a natural >>> number. Note the infinite sequence
1, 2, 3, ..., ω-2, ω-1, ω.
Omega minus one or two is undefined
Yes in so far as these natural numbers are dark and cannot be reached by
a FISON.
and n plus one closure is axiomatic.
But it is in contradiction with NUF(x) passing 1. Do you understand that NUF(x) can nowhere increase by more than 1?
Regards, WM
On 02.10.2024 14:56, joes wrote:
Am Wed, 02 Oct 2024 13:10:09 +0200 schrieb WM:
On 01.10.2024 22:05, Jim Burns wrote:Nobody believes that. It just doesn’t follow.
On 10/1/2024 1:29 PM, WM wrote:You believe that more than one unit fractions can occupy one and the
This is incorrect:
🛇⎛ ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows that 🛇⎜ at no point x 🛇⎝ NUF can
increase by more than one step 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 doesn't show that.
same point nevertheless? That would make the distance 0, but it is > 0.
Therefore you are wrong.
It follows that either all unit fractions are at different places (then
NUF grows one by one) or not. Then at least two are at the same place.
There are no further alternatives.
Regards, WM
On 02.10.2024 13:21, Richard Damon wrote:
On 10/1/24 1:37 PM, WM wrote:
On 01.10.2024 01:11, Richard Damon wrote:
For ANY finite x, NUF(x) will be Aleph_0, since there is an infinite
number of unit fractions smaller than it.
You claim that ℵo unit fractions are already there, whatever x > 0
you choose. That means they cannot be chosen. They are dark.
Nope.
Show a counter example leaving less than ℵo smaller unit fractions undefined. Fail.
Regards, WM
On 10/02/2024 11:57 AM, Jim Burns wrote:
[...]
Such casual extensionality as
"these Q's are those Q's"
is certainly usual,
it's considered a lax sort of accommodation
that thereafter
that "all proofs reflecting on a model of Q
will be as entirely agnostic to
the actual set modeling Q
as any other may do,
only fulfilling the role of the model of Q
of representing structurally all the relations
of all the elements of Q
with all the elements
of all the elements
of other structures so related,
model theory".
it's considered a lax sort of accommodationwho is it who is doing this considering?
Then about how you find
properties of the function _in_ the limit
_as_ a limit,
it's that:
those properties don't exist at all,
that function doesn't exist at all
except _in the limit_.
Now, you're talking about
a family of functions that model
this not-a-real-function, in the limit,
yet, they are not it, in the limit.
This way the extent, density, completeness,
and measure, aren't from being the union
of ranges of functions that model it,
they're the infinite integers in it.
On 10/03/2024 09:14 AM, Jim Burns wrote:
On 10/2/2024 10:01 PM, Ross Finlayson wrote:
Such casual extensionality as
"these Q's are those Q's"
is certainly usual,
Since we aren't being lax here,
you should say
⎛ Such casual isomorphistry as
⎜ "these Q's are those Q's"
⎝ is certainly usual,
The "up to isomorphism" like "the reals are
the complete ordered field 'up to isomorphism'",
does _not_ have that R = C, and indeed it is
so that R =/= C, so,
no, we do not say "isomorphistry".
Having the properties of a complete ordered field,
and, being analytic under continuous functions,
then, maintaining analyticity under
transforms of continuous functions,
is _not_ something that R and C both have,
so, the usual idea that
"the reals R are unique up to isomorphism
the complete ordered field",
gets broken
and is not considered thorough.
The extensionality is
the term of model theory with regards to that
a) it's assumed that models are faithful, and
b) it's assumed that models are equivalent
in all interpretations, thusly
c) model Q_1 and Q_2 each a, b are
extensionally equivalent and considered equals,
while of course
no properties of their structure that isn't
equi-interpretable and bi-relatable
is considered true for both, and
such aspects of their consideration are
with regards to a different theoretical object.
About "isomorphistry", getting into stuff
like "equals, 'almost' everywhere", has
that: "equals, a.e., is _not_, 'equals'".
And R, or R^2, and C, "up to isomorphism",
are _not_ equals.
WM formulated the question :
But it is in contradiction with NUF(x) passing 1. Do you understand
that NUF(x) can nowhere increase by more than 1?
Nowhere but in your mind is that 'passing' a necessity.
On 10/2/2024 7:12 AM, WM wrote:
On 01.10.2024 22:05, Jim Burns wrote:
For every nonzero distance d from 0
there are ℵ₀.many unit.fractions closer than d
That means
there is no d by what
you can distinguish ℵ₀ unit fractions?
You (WM) require us to guess
what you mean by 'distinguish'.
There is no d between
ℵ₀.many.smaller.unit.fractions and slightly.fewer.than.ℵ₀.many.smaller.unit.fractions.
On 10/2/24 7:57 AM, WM wrote:
Of course all gaps between unit fractions are made of more than
finitely many points. I never denied that.
But that also means that no point are "next to" each other.
Try to learn the basics and to understand ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Which means that for all n that are in the Natural numbers, for the unit fraction 1/n, theree DOES EXIST another unit fraction 1/(n+1) that is smaller than it.
NUF only grows by one if it is at a finite value. Since it doesn't have
a finite value at ANY value of x > 0,
On 10/2/24 8:00 AM, WM wrote:
Show a counter example leaving less than ℵo smaller unit fractions
undefined. Fail.
That isn't a valid request.
On 10/2/2024 7:10 AM, WM wrote:
You (WM) believe in quantifier shifts,
the opposite of that.
You believe that
more than one unit fractions
can occupy one and the same point
nevertheless?
No.
1, 2, 3, ..., ω-2, ω-1, ω
...is a _finite_ sequence.
As you have written it,
ω and each non.0 α < ω has a predecessor,
which makes ω finite.
WM used his keyboard to write :
On 02.10.2024 21:44, FromTheRafters wrote:
WM formulated the question :
But it is in contradiction with NUF(x) passing 1. Do you understand
that NUF(x) can nowhere increase by more than 1?
Nowhere but in your mind is that 'passing' a necessity.
If all fractions are separated real points on the real axis, then
there must be a first one.
Still wrong. Because you say so is not enough to convince anyone.
Am Wed, 25 Sep 2024 17:51:56 +0200 schrieb WM:
NUF(x) distinguishes all points.How does it distinguish dark points?
NUF increases. At no point it can increase by more than 1.Right, and there is no point "next to" 0
Even if most mathematicians are far too stupid to understand this, IYou should try explaining it a different way.
will repeat it on and on, maybe that sometime some will get it.
On 9/24/24 3:30 PM, WM wrote:
The largest number that you can choose depends on your facilities.
Consider the largest number available on your pocket calculator.
Mathematics isn't based on what WE can do, but on what the numbers
themselves can do,
Am Wed, 25 Sep 2024 17:32:15 +0200 schrieb WM:
It is a mathematical function. It is assumed to see dark numbers. You
believe that infinitely many of the smallest fractions cannot be
distinguished. So you believe in dark numbers too.
We can "distinguish" them.
On 9/25/2024 11:51 AM, WM wrote:
That means NUF increases by 1
at every point occupied by a unit fraction.
There are numbers (cardinalities) which increase by 1
and other numbers (cardinalities), which
don't increase by 1.
For each positive point x
for each number (cardinality) k which can increase by 1
there are more.than.k unit.fractions between 0 and x
For each positive point x
the number (cardinality) of unit.fractions between 0 and x
is not
any number (cardinality) which increases by 1
Instead, it is
a number (cardinality) which doesn't increase by 1.
Each positive point is undercut by
some finite.unit.fraction.
What values, that EXIST, can't be distinguished.
On 9/25/24 11:14 AM, WM wrote:
NUF increases. At no point it can increase by
more than 1.
Why not?
The problem is that it turns out the NUF(x) NEVER actually "increments"
by 0ne at any finite point, it jumps from 0 to infinity (Aleph_0) in the unboundedly small gap between 0 and all x > 0,
On 9/25/24 11:32 AM, WM wrote:
I do NOT believe that there ANY "smallest" factions that cannot be distingusished, I know such a thing does not exist, and that ALL unit fractions, like all rationals and reals can be distinguished,
On 9/24/24 3:43 PM, WM wrote:
I think that [the] mathematics of fractions is correct.
It contains ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 from which follows that never >>>>> two different5 fractions sit upon each other. From NUF(0) = 0 the
[existence of the] smallest unit fraction follows immediately
Proof?
On 25.09.2024 19:19, Richard Damon wrote:No. Can you not conceptualise in infinite whole?
On 9/25/24 11:32 AM, WM wrote:
I do NOT believe that there ANY "smallest" factions that cannot beThe claim of all requires a last one.
distingusished, I know such a thing does not exist, and that ALL unit
fractions, like all rationals and reals can be distinguished,
On 25.09.2024 19:25, Richard Damon wrote:Which ones are those? Clearly you can „distinguish” (please define)
What values, that EXIST, can't be distinguished.You cannot distinguish the ℵo smallest unit fractions. You cannot select
a unit fractions with less smaller unit fractions.
On 25.09.2024 18:39, joes wrote:How can you talk about a dark n and it’s (maybe) n+1?
Am Wed, 25 Sep 2024 17:51:56 +0200 schrieb WM:
By our mathematical knowledge about ∀n ∈ ℕ: 1/n - 1/(n+1) > 0NUF(x) distinguishes all points.How does it distinguish dark points?
Is it closer to zero or to the next UF?There is a unit fraction next to 0.NUF increases. At no point it can increase by more than 1.Right, and there is no point "next to" 0
--∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is invincible.Even if most mathematicians are far too stupid to understand this, IYou should try explaining it a different way.
will repeat it on and on, maybe that sometime some will get it.
Am 27.09.2024 um 22:12 schrieb joes:
Am Fri, 27 Sep 2024 20:42:06 +0200 schrieb WM:
On 25.09.2024 18:39, joes wrote:
You should try explaining it a different way.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is invincible. [W. Mückenheim]
Well put!
Am Fri, 27 Sep 2024 20:42:06 +0200 schrieb WM:
On 25.09.2024 18:39, joes wrote:
You should try explaining it a different way.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is invincible. [W. Mückenheim]
On 25.09.2024 18:53, Moebius wrote:
On 9/24/24 3:43 PM, WM wrote:
I think that [the] mathematics of fractions is correct.
It contains ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 from which follows that never
two different5 fractions sit upon each other. From NUF(0) = 0 the
[existence of the] smallest unit fraction follows immediately
Proof?
Between two unit fractions there is always a finite gap.
Regards, WM
On 25.09.2024 18:39, joes wrote:
Am Wed, 25 Sep 2024 17:51:56 +0200 schrieb WM:
NUF(x) distinguishes all points.How does it distinguish dark points?
By our mathematical knowledge about ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
NUF increases. At no point it can increase by more than 1.Right, and there is no point "next to" 0
There is a unit fraction next to 0.
Even if most mathematicians are far too stupid to understand this, IYou should try explaining it a different way.
will repeat it on and on, maybe that sometime some will get it.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is invincible.
Regards, WM
On 03.10.2024 01:03, Richard Damon wrote:
On 10/2/24 7:57 AM, WM wrote:
Of course all gaps between unit fractions are made of more than
finitely many points. I never denied that.
But that also means that no point are "next to" each other.
Therefore NUF can increase at most by 1 at every real point.
Which means that for all n that are in the Natural numbers, for the
Try to learn the basics and to understand ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. >>
unit fraction 1/n, theree DOES EXIST another unit fraction 1/(n+1)
that is smaller than it.
That is true for all visible and most dark numbers bu not for all.
Regards, WM
On 03.10.2024 20:40, FromTheRafters wrote:
WM used his keyboard to write :
On 02.10.2024 21:44, FromTheRafters wrote:
WM formulated the question :
But it is in contradiction with NUF(x) passing 1. Do you understand
that NUF(x) can nowhere increase by more than 1?
Nowhere but in your mind is that 'passing' a necessity.
If all fractions are separated real points on the real axis, then
there must be a first one.
Still wrong. Because you say so is not enough to convince anyone.
Everyone I consider worthy to talk to knows it.
Regards, WM
On 03.10.2024 01:06, Richard Damon wrote:
NUF only grows by one if it is at a finite value. Since it doesn't
have a finite value at ANY value of x > 0,
It has 0 at 0. And it cannot increase at any x > 0 by more than 1.
Regards, WM
On 03.10.2024 01:27, Richard Damon wrote:
On 10/2/24 8:00 AM, WM wrote:
Show a counter example leaving less than ℵo smaller unit fractions
undefined. Fail.
That isn't a valid request.
Why? Because you cannot? That shows the dark numbers.
Regards, WM
On 02.10.2024 22:07, Jim Burns wrote:
On 10/2/2024 7:12 AM, WM wrote:
On 01.10.2024 22:05, Jim Burns wrote:
For every nonzero distance d from 0
there are ℵ₀.many unit.fractions closer than d
That means
there is no d by what
you can distinguish ℵ₀ unit fractions?
You (WM) require us to guess
what you mean by 'distinguish'.
You distinguish two unit fractions 1/n and 1/m if you place a point in distance d from 0 between them: 1/n < d < 1/m.
There is no d between
ℵ₀.many.smaller.unit.fractions and
slightly.fewer.than.ℵ₀.many.smaller.unit.fractions.
That means you cannot distinguish ℵ₀ smaller unit fractions.
Regards, WM
On 02.10.2024 23:06, Jim Burns wrote:That is not the same thing, but an invalid deduction.
On 10/2/2024 7:10 AM, WM wrote:
You (WM) believe in quantifier shifts,I believe that if every n that can be chosen has ℵ₀ successors, then all n that can be chosen have ℵ₀ successors. Proof by your inability to contradict this.
the opposite of that.
No, the infinity is at the end, not in the middle.Then NUF can increase at every point by at most 1.You believe that more than one unit fractions can occupy one and theNo.
same point nevertheless?
The predecessors in the dark realm cannot be known. You can subtract1, 2, 3, ..., ω-2, ω-1, ω...is a _finite_ sequence.
As you have written it,
ω and each non.0 α < ω has a predecessor,
which makes ω finite.
every natural number from ω without traversing the dark realm. You can
add every natural number to every natural number without traversing the
dark realm. That is the essence of infinity!
Am Thu, 03 Oct 2024 20:31:41 +0200 schrieb WM:You'd rather read Peter Suber's "Infinite Reflections", Mückenheim.
[Bla bla bla] That is the essence of infinity!
Am Thu, 03 Oct 2024 20:20:52 +0200 schrieb WM:
If all fractions are separated real points on the real axis, then there
must be a first one.
No, they get denser toward zero, because you can always fit points
inbetween.
On 02.10.2024 21:44, FromTheRafters wrote:No, they get denser toward zero, because you can always fit points
WM formulated the question :
If all fractions are separated real points on the real axis, then thereBut it is in contradiction with NUF(x) passing 1. Do you understandNowhere but in your mind is that 'passing' a necessity.
that NUF(x) can nowhere increase by more than 1?
must be a first one.
On 02.10.2024 22:07, Jim Burns wrote:Which is always possible.
On 10/2/2024 7:12 AM, WM wrote:You distinguish two unit fractions 1/n and 1/m if you place a point in distance d from 0 between them: 1/n < d < 1/m.
On 01.10.2024 22:05, Jim Burns wrote:You (WM) require us to guess what you mean by 'distinguish'.
For every nonzero distance d from 0 there are ℵ₀.many unit.fractions >>>> closer than dThat means there is no d by what you can distinguish ℵ₀ unit
fractions?
Misconception: there is no „slightly fewer than ℵ₀”. An finiteThere is no d between ℵ₀.many.smaller.unit.fractions andThat means you cannot distinguish ℵ₀ smaller unit fractions.
slightly.fewer.than.ℵ₀.many.smaller.unit.fractions.
On 03.10.2024 01:27, Richard Damon wrote:Or because it is impossible.
On 10/2/24 8:00 AM, WM wrote:
Why? Because you cannot? That shows the dark numbers.Show a counter example leaving less than ℵo smaller unit fractionsThat isn't a valid request.
undefined. Fail.
On 03.10.2024 01:03, Richard Damon wrote:What is the distance between the last natural and its successor?
On 10/2/24 7:57 AM, WM wrote:
Therefore NUF can increase at most by 1 at every real point.Of course all gaps between unit fractions are made of more thanBut that also means that no point are "next to" each other.
finitely many points. I never denied that.
That is true for all visible and most dark numbers bu not for all.Try to learn the basics and to understand ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. >> Which means that for all n that are in the Natural numbers, for theunit fraction 1/n, theree DOES EXIST another unit fraction 1/(n+1)
that is smaller than it.
On 10/3/24 2:25 PM, WM wrote:
You distinguish two unit fractions 1/n and 1/m if you place a point in
distance d from 0 between them: 1/n < d < 1/m.
That means you cannot distinguish ℵ₀ smaller unit fractions.
Sure we can, we just can't find any that have less than Aleph_0 numbers
below them since those numbers don't exist.
Am Thu, 03 Oct 2024 20:20:52 +0200 schrieb WM:
On 02.10.2024 21:44, FromTheRafters wrote:No, they get denser toward zero, because you can always fit points
WM formulated the question :If all fractions are separated real points on the real axis, then there
But it is in contradiction with NUF(x) passing 1. Do you understandNowhere but in your mind is that 'passing' a necessity.
that NUF(x) can nowhere increase by more than 1?
must be a first one.
inbetween.
On 10/3/24 2:34 PM, WM wrote:
On 03.10.2024 01:03, Richard Damon wrote:
But that also means that no point are "next to" each other.
Therefore NUF can increase at most by 1 at every real point.
Why?
Am Thu, 03 Oct 2024 20:31:41 +0200 schrieb WM:
On 02.10.2024 23:06, Jim Burns wrote:That is not the same thing, but an invalid deduction.
On 10/2/2024 7:10 AM, WM wrote:I believe that if every n that can be chosen has ℵ₀ successors, then all >> n that can be chosen have ℵ₀ successors. Proof by your inability to
You (WM) believe in quantifier shifts,
the opposite of that.
contradict this.
No, the infinity is at the end, not in the middle.The predecessors in the dark realm cannot be known. You can subtract1, 2, 3, ..., ω-2, ω-1, ω...is a _finite_ sequence.
As you have written it,
ω and each non.0 α < ω has a predecessor,
which makes ω finite.
every natural number from ω without traversing the dark realm. You can
add every natural number to every natural number without traversing the
dark realm. That is the essence of infinity!
Am 04.10.2024 um 10:11 schrieb joes:
Am Thu, 03 Oct 2024 20:31:41 +0200 schrieb WM:
You can subtract every natural number from ω without traversing the dark realm. You can add every natural number to every natural number without traversing the dark realm. That is the essence of infinity!
You'd rather read Peter Suber's "Infinite Reflections".
Am 04.10.2024 um 10:17 schrieb joes:
Am Thu, 03 Oct 2024 20:20:52 +0200 schrieb WM:
If all fractions are separated real points on the real axis, then there
must be a first one.
Nope.
Hint: The harmonic series is divergent [at least in mathematics].
Am Thu, 03 Oct 2024 20:37:47 +0200 schrieb WM:
On 03.10.2024 01:27, Richard Damon wrote:Or because it is impossible.
On 10/2/24 8:00 AM, WM wrote:Why? Because you cannot? That shows the dark numbers.
Show a counter example leaving less than ℵo smaller unit fractionsThat isn't a valid request.
undefined. Fail.
Am Thu, 03 Oct 2024 20:34:20 +0200 schrieb WM:
On 03.10.2024 01:03, Richard Damon wrote:What is the distance between the last natural and its successor?
On 10/2/24 7:57 AM, WM wrote:Therefore NUF can increase at most by 1 at every real point.
Of course all gaps between unit fractions are made of more thanBut that also means that no point are "next to" each other.
finitely many points. I never denied that.
That is true for all visible and most dark numbers but not for all.Try to learn the basics and to understand ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. >>> Which means that for all n that are in the Natural numbers, for theunit fraction 1/n, theree DOES EXIST another unit fraction 1/(n+1)
that is smaller than it.
Am Thu, 03 Oct 2024 20:25:07 +0200 schrieb WM:
On 02.10.2024 22:07, Jim Burns wrote:Which is always possible.
On 10/2/2024 7:12 AM, WM wrote:You distinguish two unit fractions 1/n and 1/m if you place a point in
On 01.10.2024 22:05, Jim Burns wrote:You (WM) require us to guess what you mean by 'distinguish'.
For every nonzero distance d from 0 there are ℵ₀.many unit.fractions >>>>> closer than dThat means there is no d by what you can distinguish ℵ₀ unit
fractions?
distance d from 0 between them: 1/n < d < 1/m.
Misconception: there is no „slightly fewer than ℵ₀”.There is no d between ℵ₀.many.smaller.unit.fractions andThat means you cannot distinguish ℵ₀ smaller unit fractions.
slightly.fewer.than.ℵ₀.many.smaller.unit.fractions.
on 10/4/2024, WM supposed :
The sum of all natural numbers is larger than ω.
Wrong, it doesn't sum in the normal sense because it is not convergent.
You could use the Zeta function for complex numbers and achieve -1/12 as
a 'sum' in that sense.
Nevertheless ω-1 is the last natural number.
Because you say so? Explain what minus one means in reference to omega.
On 04.10.2024 00:55, Richard Damon wrote:
On 10/3/24 2:34 PM, WM wrote:
On 03.10.2024 01:03, Richard Damon wrote:
But that also means that no point are "next to" each other.
Therefore NUF can increase at most by 1 at every real point.
Why?
This is the precondition: The unit fractions are all existing! If so,
then they are different fixed points on the positive real line above
zero. Hence one is the first after zero.
Regards, WM
On 04.10.2024 11:00, joes wrote:Trick question, it has none (or it is infinite).
Am Thu, 03 Oct 2024 20:34:20 +0200 schrieb WM:Unknown.
On 03.10.2024 01:03, Richard Damon wrote:What is the distance between the last natural and its successor?
On 10/2/24 7:57 AM, WM wrote:Therefore NUF can increase at most by 1 at every real point.
Of course all gaps between unit fractions are made of more thanBut that also means that no point are "next to" each other.
finitely many points. I never denied that.
That is true for all visible and most dark numbers but not for all.Try to learn the basics and to understand ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.Which means that for all n that are in the Natural numbers, for the
unit fraction 1/n, theree DOES EXIST another unit fraction 1/(n+1)
that is smaller than it.
On 04.10.2024 10:11, joes wrote:I don’t believe anything. The rules of logic which you are so fond of
Am Thu, 03 Oct 2024 20:31:41 +0200 schrieb WM:It is fact. What you believe valid or invalid is not of interest to me.
On 02.10.2024 23:06, Jim Burns wrote:That is not the same thing, but an invalid deduction.
On 10/2/2024 7:10 AM, WM wrote:I believe that if every n that can be chosen has ℵ₀ successors, then >>> all n that can be chosen have ℵ₀ successors. Proof by your inability >>> to contradict this.
You (WM) believe in quantifier shifts,
the opposite of that.
I know that all numbers which you can choose have ℵ₀ successors. Therefore the collection containing all numbers which you can choose has ℵ₀ successors.
--A common misbelief about actual infinity based on potential infinity.No, the infinity is at the end, not in the middle.The predecessors in the dark realm cannot be known. You can subtract1, 2, 3, ..., ω-2, ω-1, ω...is a _finite_ sequence.
As you have written it,
ω and each non.0 α < ω has a predecessor,
which makes ω finite.
every natural number from ω without traversing the dark realm. You can
add every natural number to every natural number without traversing
the dark realm. That is the essence of infinity!
On 04.10.2024 03:23, Richard Damon wrote:
On 10/3/24 2:25 PM, WM wrote:
You distinguish two unit fractions 1/n and 1/m if you place a point
in distance d from 0 between them: 1/n < d < 1/m.
That means you cannot distinguish ℵ₀ smaller unit fractions.
Sure we can, we just can't find any that have less than Aleph_0
numbers below them since those numbers don't exist.
They do. 0 has no unit fraction below it. If the next unit fraction has
unit fractions below it, then it is not the next. It has ℵ₀ smaller unit fractions below it which cannot be distinguished.
Regards, WM
On 04.10.2024 10:52, joes wrote:
Am Thu, 03 Oct 2024 20:37:47 +0200 schrieb WM:
On 03.10.2024 01:27, Richard Damon wrote:Or because it is impossible.
On 10/2/24 8:00 AM, WM wrote:Why? Because you cannot? That shows the dark numbers.
Show a counter example leaving less than ℵo smaller unit fractions >>>>> undefined. Fail.That isn't a valid request.
Of course it is impossible. The reason is this: dark numbers cannot be
chosen individually.
Regards, WM
On 04.10.2024 10:29, joes wrote:
Am Thu, 03 Oct 2024 20:25:07 +0200 schrieb WM:
On 02.10.2024 22:07, Jim Burns wrote:Which is always possible.
On 10/2/2024 7:12 AM, WM wrote:You distinguish two unit fractions 1/n and 1/m if you place a point in
On 01.10.2024 22:05, Jim Burns wrote:You (WM) require us to guess what you mean by 'distinguish'.
For every nonzero distance d from 0 there are ℵ₀.many unit.fractions >>>>>> closer than dThat means there is no d by what you can distinguish ℵ₀ unit
fractions?
distance d from 0 between them: 1/n < d < 1/m.
For a finite collection of unit fractions. For almost all unit fractions
it is impossible.
Misconception: there is no „slightly fewer than ℵ₀”.There is no d between ℵ₀.many.smaller.unit.fractions andThat means you cannot distinguish ℵ₀ smaller unit fractions.
slightly.fewer.than.ℵ₀.many.smaller.unit.fractions.
Of course there is fewer, namely zero. If slightly fewer is not
possible, then ℵ₀ unit fractions sit at one point.
Regards, WM
On 04.10.2024 00:55, Richard Damon wrote:What kind of fucked up condition is that? It doesn’t even make sense
On 10/3/24 2:34 PM, WM wrote:
On 03.10.2024 01:03, Richard Damon wrote:
This is the precondition: The unit fractions are all existing! If so,Why?But that also means that no point are "next to" each other.Therefore NUF can increase at most by 1 at every real point.
then they are different fixed points on the positive real line above
zero. Hence one is the first after zero.
Am Fri, 04 Oct 2024 11:23:13 +0200 schrieb WM:
This is the precondition: The unit fractions are all existing! If so,
then they are different fixed points on the positive real line above
zero. Hence one is the first after zero.
What kind of fucked up condition is that? It doesn’t even make sense
for a UF to not exist.
on 10/4/2024, WM supposed :
On 04.10.2024 12:02, FromTheRafters wrote:
on 10/4/2024, WM supposed :
The sum of all natural numbers is larger than ω.
Wrong, it doesn't sum in the normal sense because it is not convergent.
Yes, I cannot calculate the sum, but I know that already ω-1 + 1 = ω.
Omega minus one is not defined.
natural number one, but only signifies the "next" ordinal.
You could use the Zeta function for complex numbers and achieve -1/12
as a 'sum' in that sense.
Nonsense.
Your use of the word nonsense simply means that you don't understand something.
On 10/4/24 5:27 AM, WM wrote:
Sure we can, we just can't find any that have less than Aleph_0
numbers below them since those numbers don't exist.
They do. 0 has no unit fraction below it. If the next unit fraction
has unit fractions below it, then it is not the next. It has ℵ₀
smaller unit fractions below it which cannot be distinguished.
But there is not "next" unit fraction above zero
Am 04.10.2024 um 17:10 schrieb joes:
Am Fri, 04 Oct 2024 11:23:13 +0200 schrieb WM:
This is the precondition: The unit fractions are all existing! If so,
then they are different fixed points on the positive real line above
zero. Hence one is the first after zero.
But the MAIN PROBLEM is his usage of "hence", which clearly is not
correct. Why on earth should there be a "first after zero".
Am Fri, 04 Oct 2024 11:23:13 +0200 schrieb WM:
On 04.10.2024 00:55, Richard Damon wrote:What kind of fucked up condition is that? It doesn’t even make sense
On 10/3/24 2:34 PM, WM wrote:This is the precondition: The unit fractions are all existing! If so,
On 03.10.2024 01:03, Richard Damon wrote:
Why?But that also means that no point are "next to" each other.Therefore NUF can increase at most by 1 at every real point.
then they are different fixed points on the positive real line above
zero. Hence one is the first after zero.
for a UF to not exist.
On 10/4/2024 5:53 AM, WM wrote:
If slightly fewer is not possible,
then ℵ₀ unit fractions sit at one point.
No.
Slightly fewer is not possible.
You (WM) think that that's wrong because
you (WM) think that a quantifier shift is reliable.
However, a quantifier shift is unreliable.
Am Fri, 04 Oct 2024 11:33:02 +0200 schrieb WM:
I know that all numbers which you can choose have ℵ₀ successors.I don’t believe anything. The rules of logic which you are so fond of simply do not admit that deduction in general (i.e. it is wrong).
Therefore the collection containing all numbers which you can choose has
ℵ₀ successors.
What do you mean with the successor of a collection anyway?
On 04.10.2024 10:29, joes wrote:
Am Thu, 03 Oct 2024 20:25:07 +0200 schrieb WM:
On 02.10.2024 22:07, Jim Burns wrote:
On 10/2/2024 7:12 AM, WM wrote:
On 01.10.2024 22:05, Jim Burns wrote:
For every nonzero distance d from 0
there are ℵ₀.many unit.fractions
closer than d
That means there is no d by what
you can distinguish ℵ₀ unit fractions?
You (WM) require us to guess
what you mean by 'distinguish'.
You distinguish two unit fractions 1/n and 1/m
if you place a point in distance d from 0
between them: 1/n < d < 1/m.
Which is always possible.
For a finite collection of unit fractions.
For almost all unit fractions it is impossible.
There is no d between
ℵ₀.many.smaller.unit.fractions and
slightly.fewer.than.ℵ₀.many.smaller.unit.fractions.
That means you cannot distinguish
ℵ₀ smaller unit fractions.
Misconception:
there is no „slightly fewer than ℵ₀”.
If slightly fewer is not possible,
then ℵ₀ unit fractions sit at one point.
If slightly fewer is not possible,
then ℵ₀ unit fractions sit at one point.
On 04.10.2024 17:36, Moebius wrote:
Am 04.10.2024 um 17:10 schrieb joes:
Am Fri, 04 Oct 2024 11:23:13 +0200 schrieb WM:
This is the precondition: The unit fractions are all existing! If so,
then they are different fixed points on the positive real line above
zero. Hence one is the first after zero.
But the MAIN PROBLEM is his usage of "hence", which clearly is not
correct. Why on earth should there be a "first after zero".
Points on the real line are real points. One is the first.
Regards, WM
On 04.10.2024 17:10, joes wrote:
Am Fri, 04 Oct 2024 11:23:13 +0200 schrieb WM:
On 04.10.2024 00:55, Richard Damon wrote:What kind of fucked up condition is that? It doesn’t even make sense
On 10/3/24 2:34 PM, WM wrote:This is the precondition: The unit fractions are all existing! If so,
On 03.10.2024 01:03, Richard Damon wrote:
Why?But that also means that no point are "next to" each other.Therefore NUF can increase at most by 1 at every real point.
then they are different fixed points on the positive real line above
zero. Hence one is the first after zero.
for a UF to not exist.
Learn what potential infinity is.
Regards, WM
On 04.10.2024 13:00, joes wrote:
Am Fri, 04 Oct 2024 11:33:02 +0200 schrieb WM:
hasI know that all numbers which you can choose have ℵ₀ successors.
Therefore the collection containing all numbers which you can choose
ℵ₀ successors.I don’t believe anything. The rules of logic which you are so fond of simply do not admit that deduction in general (i.e. it is wrong).
It is prove correct by the impossibility to circumvent it: all numbers
which you can choose have ℵ₀ successors. That is essential, not your "rules".
What do you mean with the successor of a collection anyway?
Between every defined element of the collection and ω, there are ℵ₀ undefinable successors.
Regards, WM
On 04.10.2024 13:03, Richard Damon wrote:
On 10/4/24 5:27 AM, WM wrote:
Sure we can, we just can't find any that have less than Aleph_0
numbers below them since those numbers don't exist.
They do. 0 has no unit fraction below it. If the next unit fraction
has unit fractions below it, then it is not the next. It has ℵ₀
smaller unit fractions below it which cannot be distinguished.
But there is not "next" unit fraction above zero
Of course there is.
a concept that you don't seem to be able to understaend.
Nevertheless it is unavoidable.
Regards, WM
On 04.10.2024 21:37, Jim Burns wrote:
On 10/4/2024 5:53 AM, WM wrote:
If slightly fewer is not possible,
then ℵ₀ unit fractions sit at one point.
No.
Otherwise
we could enter the gaps between them.
Slightly fewer is not possible.
That proves one point.
You (WM) think that that's wrong because
you (WM) think that a quantifier shift is reliable.
However, a quantifier shift is unreliable.
It is more reliable than your unfounded twaddle.
But it is no quantifier shift but simplest logic:
If ℵ₀ unit fractions do not sit at one point
(and are not dark) then
they can be subdivided into smaller parts.
On 04.10.2024 12:38, FromTheRafters wrote:That is clearly an infinite number.
on 10/4/2024, WM supposed :It is defined by ω-1 + 1 = ω.
On 04.10.2024 12:02, FromTheRafters wrote:Omega minus one is not defined.
on 10/4/2024, WM supposed :Yes, I cannot calculate the sum, but I know that already ω-1 + 1 = ω.
The sum of all natural numbers is larger than ω.Wrong, it doesn't sum in the normal sense because it is not
convergent.
--The 'plus one' here is not plus theI understand that the sum 1+2+3+... > 1. More is not required.
natural number one, but only signifies the "next" ordinal.
Your use of the word nonsense simply means that you don't understandYou could use the Zeta function for complex numbers and achieve -1/12Nonsense.
as a 'sum' in that sense.
something.
On 04.10.2024 21:37, Jim Burns wrote:We can.
On 10/4/2024 5:53 AM, WM wrote:
Otherwise we could enter the gaps between them.If slightly fewer is not possible,No.
then ℵ₀ unit fractions sit at one point.
Apparently you have no idea what a quantifier shift even is.Slightly fewer is not possible.That proves one point.
You (WM) think that that's wrong because you (WM) think that aIt is more reliable than your unfounded twaddle.
quantifier shift is reliable.
However, a quantifier shift is unreliable.
But it is no quantifier shift but simplest logic:Which they can.
If ℵ₀ unit fractions do not sit at one point (and are not dark) then
they can be subdivided into smaller parts.
Am Fri, 04 Oct 2024 22:09:41 +0200 schrieb WM:
On 04.10.2024 21:37, Jim Burns wrote:
You (WM) think that that's wrong because you (WM) think that a
quantifier shift is reliable. However, a quantifier shift is unreliable.
It is more reliable than your unfounded twaddle.
Apparently you have no idea what a quantifier shift even is.
Have you looked at the Wikipedia page?
But it is no quantifier shift but simplest logic:
If ℵ₀ unit fractions [bla]
Am Fri, 04 Oct 2024 20:25:35 +0200 schrieb WM:
On 04.10.2024 12:38, FromTheRafters wrote:
on 10/4/2024, WM supposed :
On 04.10.2024 12:02, FromTheRafters wrote:
on 10/4/2024, WM supposed :
The sum of all natural numbers is larger than ω.
That is clearly an infinite number.It is defined by ω-1 + 1 = ω.Wrong, it doesn't sum in the normal sense because it is notYes, I cannot calculate the sum, but I know that already ω-1 + 1 = ω. >>> Omega minus one is not defined.
convergent.
I understand that the sum 1+2+3+... > 1.
After serious thinking Chris M. Thomasson wrote :
Now WM is trying to say there is a first real number?
...and it is dark.
On 10/4/24 2:36 PM, WM wrote:
Points on the real line are real points. One is the first.
Nope, and that is your problem, that property only hold at both ends for finite sets.
WM formulated the question :
Points on the real line are real points. One is the first.
Sounds intuitive
, but for each first there is a firster.
On 10/4/24 2:34 PM, WM wrote:
Learn what potential infinity is.
WHich is just terms that it seems YOU invented
On 10/4/24 4:10 PM, WM wrote:
Between every defined element of the collection and ω, there are ℵ₀
undefinable successors.
No, every term is defined,
On 10/4/2024 4:09 PM, WM wrote:
Otherwise
we could enter the gaps between them.
You (WM) think that we can't enter gaps because
you (WM) think that a quantifier shift is reliable.
However, a quantifier shift is unreliable.
If ℵ₀ unit fractions do not sit at one point
(and are not dark) then they can be subdivided into smaller parts.
∀ᴿx>0:
⎛ uₓ(k) = ⅟⌈k+⅟x⌉
⎜ uₓ: ℕ → (0,x]: one.to.one and not.at.one.point
⎝ NUF(x) = |uₓ(ℕ)| = |ℕ| = ℵ₀
a quantifier shift is NOT reliable und wird daher in
der Mathematik tunlichst vermieden (und nicht nur dort).
Am Fri, 04 Oct 2024 22:09:41 +0200 schrieb WM:
But it is no quantifier shift but simplest logic:Which they can.
If ℵ₀ unit fractions do not sit at one point (and are not dark) then
they can be subdivided into smaller parts.
Am 05.10.2024 um 10:10 schrieb joes:
Am Fri, 04 Oct 2024 20:25:35 +0200 schrieb WM:
Hint: There is no ordinal number o such that o + 1 = ω. Hence there is especially no ordinal number denoted by "ω-1" such that ω-1 + 1 = ω.
I understand that the sum 1+2+3+... > 1.
Die gewöhnliche Summe von "1 + 2 + 3 + ... " IM
KONTEXT DER REELLEN ZAHLEN ist NICHT DEFINIERT,
Allerdings kann man IR auch etwas erweitern, nämlich um die beiden
Elemente {-oo, oo} um auf diese Weise die "extended reals" IR* zu erhalten.
IN DIESEM Kontext kann man nun tatsächlich schreiben:
"1 + 2 + 3 + ... = oo"
mit oo > 1 .
IN DIESEM KONTEXT kann man also
"1 + 2 + 3 + ... > 1"
tatsächlich hinschreiben und es ist dann sinnvoll und korrekt.
On 05.10.2024 00:46, Richard Damon wrote:
On 10/4/24 2:34 PM, WM wrote:
Learn what potential infinity is.
WHich is just terms that it seems YOU invented
Hahaha. Read Aristotle!
Regards, WM
On 05.10.2024 00:45, Richard Damon wrote:
On 10/4/24 2:36 PM, WM wrote:
Points on the real line are real points. One is the first.
Nope, and that is your problem, that property only hold at both ends
for finite sets.
Every point is a finite set.
Regards, WM
On 05.10.2024 10:08, joes wrote:
Am Fri, 04 Oct 2024 22:09:41 +0200 schrieb WM:
But it is no quantifier shift but simplest logic:Which they can.
If ℵ₀ unit fractions do not sit at one point (and are not dark) then >>> they can be subdivided into smaller parts.
Do it. Fail because slightly fewer is not possible.
Regards, WM
On 05.10.2024 00:47, Richard Damon wrote:
On 10/4/24 4:10 PM, WM wrote:
Between every defined element of the collection and ω, there are ℵ₀ >>> undefinable successors.
No, every term is defined,
Find the largest natural number. It is a finite set, a so-called so singleton.
Regards, WM
On 05.10.2024 10:38, Moebius wrote:
Am 05.10.2024 um 10:10 schrieb joes:
Am Fri, 04 Oct 2024 20:25:35 +0200 schrieb WM:
Hint: There is no ordinal number o such that o + 1 = ω. Hence there is
especially no ordinal number denoted by "ω-1" such that ω-1 + 1 = ω.
May be. But in actual infinity there is no gap before ω.
I understand that the sum 1+2+3+... > 1.
Die gewöhnliche Summe von "1 + 2 + 3 + ... " IM KONTEXT DER REELLEN
ZAHLEN ist NICHT DEFINIERT,
Jedenfalls ist sie nicht negativ.
Allerdings kann man IR auch etwas erweitern, nämlich um die beiden
Elemente {-oo, oo} um auf diese Weise die "extended reals" IR* zu
erhalten.
IN DIESEM Kontext kann man nun tatsächlich schreiben:
"1 + 2 + 3 + ... = oo"
mit oo > 1 .
In actual infinity ω-2 is a natural number and 3 + ω-2 = ω+1.
IN DIESEM KONTEXT kann man also
"1 + 2 + 3 + ... > 1"
tatsächlich hinschreiben und es ist dann sinnvoll und korrekt.
Na also.
Gruß, WM
On 05.10.2024 10:46, Moebius wrote:
a quantifier shift is NOT reliable und wird daher in der Mathematik
tunlichst vermieden (und nicht nur dort).
I many cases it is correct. For instance if every definable natural
number has ℵo natural successors, then there are ℵo natural numbers larger than all definable natural numbers. They are dark however and
cannot be specified.
Regards, WM
But actual infinity doesn't exist.
On 10/5/24 5:18 AM, WM wrote:
On 05.10.2024 00:45, Richard Damon wrote:
And the set of *A* point has no "next" point, as it is just one.Nope, and that is your problem, that property only hold at both ends
for finite sets.
Every point is a finite set.
But actual infinity doesn't exist.
On 10/5/24 5:20 AM, WM wrote:
On 05.10.2024 00:46, Richard Damon wrote:
On 10/4/24 2:34 PM, WM wrote:
Learn what potential infinity is.
WHich is just terms that it seems YOU invented
Hahaha. Read Aristotle!
Regards, WM
Maybe you should, since you don't understand his concepts, and made up
your own to match the terms.
Remember, Aristotle said Actual Infinities cannot exist.
So, looking at there properties is futile.
That is your ignorance.
On 05.10.2024 01:41, Jim Burns wrote:You constantly make that incorrect deduction. At one point you seemed
On 10/4/2024 4:09 PM, WM wrote:
they are dark.Otherwise we could enter the gaps between them.You (WM) think that we can't enter gaps because
Don't parrot that nonsense. There is no quantifier shift. Between manyyou (WM) think that a quantifier shift is reliable.
However, a quantifier shift is unreliable.
unit fractions there are many gaps. But we cannot find them.
What about the gap between the last definable and the first dark UF?What shall that prove? Between n unit fractions there are n-1 gaps. EachIf ℵ₀ unit fractions do not sit at one point (and are not dark) then >>> they can be subdivided into smaller parts.∀ᴿx>0:
⎛ uₓ(k) = ⅟⌈k+⅟x⌉
⎜ uₓ: ℕ → (0,x]: one.to.one and not.at.one.point ⎝ NUF(x) = |uₓ(ℕ)| =
|ℕ| = ℵ₀
gap can be found for definable unit fractions.
Note also that every pair of unit fractions is a finite set.
On 05.10.2024 10:08, joes wrote:What do you mean? All the unit fractions have, as you say,
Am Fri, 04 Oct 2024 22:09:41 +0200 schrieb WM:Do it. Fail because slightly fewer is not possible.
But it is no quantifier shift but simplest logic:Which they can.
If ℵ₀ unit fractions do not sit at one point (and are not dark) then >>> they can be subdivided into smaller parts.
On 05.10.2024 13:01, Richard Damon wrote:
On 10/5/24 5:18 AM, WM wrote:
On 05.10.2024 00:45, Richard Damon wrote:
And the set of *A* point has no "next" point, as it is just one.Nope, and that is your problem, that property only hold at both ends
for finite sets.
Every point is a finite set.
But it has a position, like the point 1/3 - if it is not dark.
By the way: Also a pair of points is a finite set. And it has a gap and
every member has a next point.
Regards, WM
On 05.10.2024 13:12, Richard Damon wrote:
How can bijections between infinite sets exist then.
But actual infinity doesn't exist.
Regards, WM
Richard Damon <richard@damon-family.org> wrote:
[ .... ]
But actual infinity doesn't exist.
What does it mean for a mathematical concept not to exist?
You seem to be falling into the same trap as John Gabriel used to when he
was still posting here. He asserted things like "irrational numbers
don't exist" and "negative numbers don't exist". He was never able to
state exactly what he meant by this non-existence.
[ .... ]
On 10/5/24 8:58 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
[ .... ]
But actual infinity doesn't exist.
What does it mean for a mathematical concept not to exist?
That it doesn't create a usable (non-contradictory) logical system.
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago. I'm not convinced there is any mathematically valid distinction between them.
If there were, I would have heard of it back then.
Does "actual infinity" create a logical system?
classical mathematics = ZFC (or something like that) + classical logic.
Am Sat, 05 Oct 2024 11:32:44 +0200 schrieb WM: [nosnense]
What do you mean?
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago. I'm not convinced there is any mathematically valid distinction between them.
If there were, I would have heard of it back then.
Does "actual infinity" create a logical system?
classical mathematics = ZFC (or something like that) + classical logic.
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago. I'm not convinced there is any mathematically valid distinction between them.
If there were, I would have heard of it back then.
Does "actual infinity" create a logical system?
classical mathematics = ZFC (or something like that) + classical logic.
Am 05.10.2024 um 15:02 schrieb joes:
Am Sat, 05 Oct 2024 11:32:44 +0200 schrieb WM: [nosnense]
What do you mean?
He does not /mean/ anything.
Falls Du es noch nicht bemerkt haben solltest: WM ist nicht ganz dicht
in der Birne (=mad).
He's just talking NONSENSE.
Am 05.10.2024 um 15:57 schrieb Alan Mackenzie:
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago. I'm not
convinced there is any mathematically valid distinction between them.
Actually, there is.
But in classical mathematics "infinity" means "actual infinity", and "potential infinity" is of no significance here.
"Cantor's work was well received by some of the prominent
mathematicians of his day, such as Richard Dedekind. But his
willingness to regard infinite sets as objects to be treated in
much the same way as finite sets was bitterly attacked by others, particularly Kronecker. There was no objection to a 'potential
infinity' in the form of an unending process, but an 'actual
infinity' in the form of a completed infinite set was harder to
accept."
(Herb Enderton, Elements of Set Theory)
There was no objection to a 'potential infinity' in the form of an
unending process, but an 'actual infinity' in the form of a completed infinite set was harder to accept." [H.B. Enderton: "Elements of set
theory", Academic Press, New York (1977) p. 14f]
If there were, I would have heard of it back then.
Right.
Does "actual infinity" create a logical system?
classical mathematics = ZFC (or something like that) + classical logic.
So why is Richard writing that "actual infinity" doesn't exist?
Am 05.10.2024 um 18:05 schrieb Alan Mackenzie:
So why is Richard writing that "actual infinity" doesn't exist?
Don't ask. Maybe it helps to note that he isn't a mathematician. :-P
Moebius <invalid@example.invalid> wrote:
Am 05.10.2024 um 18:05 schrieb Alan Mackenzie:
So why is Richard writing that "actual infinity" doesn't exist?
Don't ask. Maybe it helps to note that he isn't a mathematician. :-P
But certainly he's more of a mathematician than Wolfgang Mückenheim.
So why is Richard writing that "actual infinity" doesn't exist?
Am 05.10.2024 um 15:57 schrieb Alan Mackenzie:
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago. I'm not
convinced there is any mathematically valid distinction between them.
Actually, there is.
But in classical mathematics "infinity" means "actual infinity", and "potential infinity" is of no significance her.
"Cantor's work was well received by some of the prominent
mathematicians of his day, such as Richard Dedekind. But his
willingness to regard infinite sets as objects to be treated in
much the same way as finite sets was bitterly attacked by others, particularly Kronecker. There was no objection to a 'potential
infinity' in the form of an unending process, but an 'actual
infinity' in the form of a completed infinite set was harder to
accept."
(Herb Enderton, Elements of Set Theory)
If there were, I would have heard of it back then.
Right.
Am 05.10.2024 um 15:57 schrieb Alan Mackenzie:
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago. I'm not
convinced there is any mathematically valid distinction between them.
Actually, there is.
On 05.10.2024 10:46, Moebius wrote:
a quantifier shift is NOT reliable
und wird daher in der Mathematik
tunlichst vermieden (und nicht nur dort).
[In] many cases it is correct.
For instance
if every definable natural number has
ℵo natural successors,
then there are ℵo natural numbers larger than
all definable natural numbers.
if every definable natural number has
ℵo natural successors,
then there are ℵo natural numbers larger than
all definable natural numbers.
They are dark however and cannot be specified.
What about the gap between the last definable and the first dark UF?
Am 05.10.2024 um 16:58 schrieb Moebius:
Am 05.10.2024 um 15:57 schrieb Alan Mackenzie:
I first came across the terms "potential infinity" and "actual infinity" >>> on this newsgroup, not in my degree course a few decades ago. I'm not
convinced there is any mathematically valid distinction between them.
Actually, there is.
"Nearly all research-level mathematicians today (I would guess 99.99% of them) take for granted both "potential" and "completed" [aka "actual"] infinity, and most probably do not even know the distinction indicated
by those two terms."
Source: https://math.vanderbilt.edu/schectex/courses/thereals/
potential.html
Am Sat, 05 Oct 2024 11:32:44 +0200 schrieb WM:But these numbers and these unit fractions cannot be found. It is
On 05.10.2024 10:08, joes wrote:What do you mean? All the unit fractions have, as you say,
Am Fri, 04 Oct 2024 22:09:41 +0200 schrieb WM:Do it. Fail because slightly fewer is not possible.
But it is no quantifier shift but simplest logic:Which they can.
If ℵ₀ unit fractions do not sit at one point (and are not dark) then >>>> they can be subdivided into smaller parts.
finite distances, which means there are numbers inbetween.
On 10/5/24 8:15 AM, WM wrote:
On 05.10.2024 13:01, Richard Damon wrote:
On 10/5/24 5:18 AM, WM wrote:
On 05.10.2024 00:45, Richard Damon wrote:
And the set of *A* point has no "next" point, as it is just one.Nope, and that is your problem, that property only hold at both
ends for finite sets.
Every point is a finite set.
But it has a position, like the point 1/3 - if it is not dark.
By the way: Also a pair of points is a finite set. And it has a gap
and every member has a next point.
So why does that say it has a point next to it?
On 10/5/24 8:18 AM, WM wrote:
On 05.10.2024 13:12, Richard Damon wrote:
How can bijections between infinite sets exist then.
But actual infinity doesn't exist.
Because the sets exist as what you think of as "potential" infinity,
we just don't need to generate the
full set before we use them (since that is the impossible step).
Am 05.10.2024 um 16:58 schrieb Moebius:
Am 05.10.2024 um 15:57 schrieb Alan Mackenzie:
I first came across the terms "potential infinity" and "actual infinity" >>> on this newsgroup, not in my degree course a few decades ago. I'm not
convinced there is any mathematically valid distinction between them.
Actually, there is.
"Nearly all research-level mathematicians today (I would guess 99.99% of them) take for granted both "potential" and "completed" [aka "actual"] infinity, and most probably do not even know the distinction indicated
by those two terms."
Source: https://math.vanderbilt.edu/schectex/courses/thereals/potential.html
On 10/5/2024 5:43 AM, WM wrote:
[In] many cases it is correct.
"Many cases" is insufficient when
the argument requires "all cases".
Yes! At least, sort of. My understanding of "doesn't exist" is either
the concept is not (yet?) developed mathematically, or it leads to contradictions. WM's "dark numbers" certainly fall into the first
category, and possibly the second, too.
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago.
WM thinks the natural numbers are not infinite?
On 05.10.2024 15:57, Alan Mackenzie wrote:
Yes! At least, sort of. My understanding of "doesn't exist" is either
the concept is not (yet?) developed mathematically, or it leads to
contradictions. WM's "dark numbers" certainly fall into the first
category, and possibly the second, too.
Definition: A natural number is "named" or "addressed" or "identified"
or "(individually) defined" or "instantiated" ....
.... if it can be communicated, necessarily by a finite amount of information, in the sense of Poincaré, such that sender and receiver understand the same and can link it by a finite initial segment (1, 2,
3, ..., n) of natural numbers to the origin 0.
.... All other natural numbers are called dark natural numbers. Dark
numbers are numbers that cannot be chosen as individuals.
Communication can occur
- by direct description in the unary system like ||||||| or as many
beeps, raps, or flashes,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7),
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
Only when a number n is identified we can use it in mathematical
discourse ....
.... and can determine the trichotomy properties of n and of every
multiple k*n or power n^k or power tower k_^n with respect to every identified number k. ℕdef contains all defined natural numbers as
elements – and nothing else. ℕdef is a potentially infinite set; therefore henceforth it will be called a collection.
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago.
It is carefully avoided because closer inspection shows contradictions.
Therefore set theorists use just what they can defend. If actual
infinity is shown self contradictory (without dark numbers), then they
evade to potential infinity temporarily which has no completed sets and cannot complete bijections.
"You use terms like completed versus potential infinity, which are not
part of the modern vernacular." [P.L. Clark in "Physicists can be
wrong", tea.MathOverflow (2 Jul 2010)] This is the typical reproach to
be expected when the different kinds of infinity are analyzed and taught.
Here the difference is clearly stated:
"Should we briefly characterize the new view of the infinite introduced
by Cantor, we could certainly say: In analysis we have to deal only with
the infinitely small and the infinitely large as a limit-notion, as something becoming, emerging, produced, i.e., as we put it, with the potential infinite. But this is not the proper infinite. That we have
for instance when we consider the entirety of the numbers 1, 2, 3, 4,
... itself as a completed unit, or the points of a line as an entirety
of things which is completely available. That sort of infinity is named actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische
Annalen 95 (1925) p. 167]
Regards, WM
On 05.10.2024 15:32, Richard Damon wrote:
On 10/5/24 8:18 AM, WM wrote:
On 05.10.2024 13:12, Richard Damon wrote:
How can bijections between infinite sets exist then.
But actual infinity doesn't exist.
Because the sets exist as what you think of as "potential" infinity,
"the integers separately as well as in their actually infinite totality
exist as eternal ideas in intellectu Divino in the highest degree of reality." [G. Cantor, letter to C. Hermite (30 Nov 1895)]
we just don't need to generate the full set before we use them (since
that is the impossible step).
Is it impossible for the intellectu Divino too? Then Cantor has erred
and you have discovered his error.
Regards, WM
On 05.10.2024 15:28, Richard Damon wrote:
On 10/5/24 8:15 AM, WM wrote:
On 05.10.2024 13:01, Richard Damon wrote:
On 10/5/24 5:18 AM, WM wrote:
On 05.10.2024 00:45, Richard Damon wrote:
And the set of *A* point has no "next" point, as it is just one.Nope, and that is your problem, that property only hold at both
ends for finite sets.
Every point is a finite set.
But it has a position, like the point 1/3 - if it is not dark.
By the way: Also a pair of points is a finite set. And it has a gap
and every member has a next point.
So why does that say it has a point next to it?
A point between both could be chosen unless it was dark.
Regards, WM
Richard Damon <richard@damon-family.org> wrote:
On 10/5/24 8:58 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
[ .... ]
But actual infinity doesn't exist.
What does it mean for a mathematical concept not to exist?
That it doesn't create a usable (non-contradictory) logical system.
Yes! At least, sort of. My understanding of "doesn't exist" is either
the concept is not (yet?) developed mathematically, or it leads to contradictions. WM's "dark numbers" certainly fall into the first
category, and possibly the second, too.
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago. I'm not convinced there is any mathematically valid distinction between them. If there were, I would have heard of it back then.
Does "actual infinity" create a logical system? If so, what is unusable
or contradictory about that system?
[ .... ]
On 05.10.2024 15:02, joes wrote:
Am Sat, 05 Oct 2024 11:32:44 +0200 schrieb WM:But these numbers and these unit fractions cannot be found. It is
On 05.10.2024 10:08, joes wrote:What do you mean? All the unit fractions have, as you say,
Am Fri, 04 Oct 2024 22:09:41 +0200 schrieb WM:Do it. Fail because slightly fewer is not possible.
But it is no quantifier shift but simplest logic:Which they can.
If ℵ₀ unit fractions do not sit at one point (and are not dark) then >>>>> they can be subdivided into smaller parts.
finite distances, which means there are numbers inbetween.
impossible to define a unit fraction having less than ℵo smaller unit fractions. It is impossible to define a unit fraction being closer to
zero although it is obvious that there are points closer to zero than ℵo*2^ℵo points.
Regards, WM
On 10/05/2024 02:57 PM, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.10.2024 15:57, Alan Mackenzie wrote:
I first came across the terms "potential infinity" and "actual infinity" >>>> on this newsgroup, not in my degree course a few decades ago.
If these terms had any significance, they would still be taught in
mathematics degree courses. Otherwise, bright students would become
aware of them and catch out their teachers in inconsistencies. Some
such students are almost incredibly bright, and catching out teachers
is something in the nature of a sport. It happens rarely, but is
satisfying for all concerned when it does happen.
The "actual" and "potential" is rather common usage
and is usually used to explain "bounded" and "unbounded".
You never heard of "actual infinity and potential infinity"?
I must most surely assume you rather have, ....
.... then thinking that your students ....
.... don't need it to understand the surrounds, have that it very much
is what it is, the distinction between the possible and the realized.
I mean, you do know a difference between potential and actual as it
relates to infinity? You agree that there is such a thing.
On 10/5/24 9:57 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/5/24 8:58 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
[ .... ]
But actual infinity doesn't exist.
What does it mean for a mathematical concept not to exist?
That it doesn't create a usable (non-contradictory) logical system.
Yes! At least, sort of. My understanding of "doesn't exist" is either
the concept is not (yet?) developed mathematically, or it leads to
contradictions. WM's "dark numbers" certainly fall into the first
category, and possibly the second, too.
I first came across the terms "potential infinity" and "actual infinity"
on this newsgroup, not in my degree course a few decades ago. I'm not
convinced there is any mathematically valid distinction between them. If
there were, I would have heard of it back then.
Does "actual infinity" create a logical system? If so, what is unusable
or contradictory about that system?
[ .... ]
After a bit of reseach, there does seem to be indications that Aristotle
did do some reasoning with the terms. I am not sure on the exact
definitions, but the indications are that "potential" infinity was generative, where the numbers are realized as they are needed, and you
can keep creating more and more of them as you go.
Actual infinity presumed that somehow all the values were created up
front and none could be added, and he found that logic done on this definition was too full of contradictions to be usable, so he concluded
that "actual infinity" did not really exist.
My guess is that WM doesn't understand this conclusion, or thinks that
he is somehow smarter than Aristotle and can make it work (when he can't)
or just thinks that since the name given was "actual infinity" that fact
that it doesn't work just means that infinity can't actually exist.
My guess, from what I have seen from WM, one of the problems with
"actual infinity" is that it makes it at least seem possible to apply
the rules of "finite" logic to an infinite logic, and that just breaks
it. To our finite minds, the rules of infinite logic just don't make intuitive sense.
On 05.10.2024 15:28, Richard Damon wrote:Many points could even be chosen!
On 10/5/24 8:15 AM, WM wrote:A point between both could be chosen unless it was dark.
On 05.10.2024 13:01, Richard Damon wrote:So why does that say it has a point next to it?
On 10/5/24 5:18 AM, WM wrote:But it has a position, like the point 1/3 - if it is not dark.
On 05.10.2024 00:45, Richard Damon wrote:
And the set of *A* point has no "next" point, as it is just one.Nope, and that is your problem, that property only hold at bothEvery point is a finite set.
ends for finite sets.
By the way: Also a pair of points is a finite set. And it has a gap
and every member has a next point.
On 05.10.2024 15:57, Alan Mackenzie wrote:That is possible for all natural numbers.
Yes! At least, sort of. My understanding of "doesn't exist" is eitherDefinition: A natural number is "named" or "addressed" or "identified"
the concept is not (yet?) developed mathematically, or it leads to
contradictions. WM's "dark numbers" certainly fall into the first
category, and possibly the second, too.
or "(individually) defined" or "instantiated" if it can be communicated, necessarily by a finite amount of information, in the sense of Poincaré, such that sender and receiver understand the same and can link it by a
finite initial segment (1, 2, 3, ..., n) of natural numbers to the
origin 0. All other natural numbers are called dark natural numbers.
Dark numbers are numbers that cannot be chosen as individuals.
Communication can occur - by direct description in the unary system like ||||||| or as many beeps, raps, or flashes,Where did you get this idea from?
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7),
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
Only when a number n is identified we can use it in mathematicalSeems sensible not to use the contradictory distinction between
discourse and can determine the trichotomy properties of n and of every multiple k*n or power n^k or power tower k_^n with respect to every identified number k. ℕdef contains all defined natural numbers as
elements – and nothing else. ℕdef is a potentially infinite set; therefore henceforth it will be called a collection.
I first came across the terms "potential infinity" and "actualIt is carefully avoided because closer inspection shows contradictions. Therefore set theorists use just what they can defend. If actual
infinity"
on this newsgroup, not in my degree course a few decades ago.
infinity is shown self contradictory (without dark numbers), then they
evade to potential infinity temporarily which has no completed sets and cannot complete bijections.
"You use terms like completed versus potential infinity, which are notThey are not taught anymore.
part of the modern vernacular." [P.L. Clark in "Physicists can be
wrong", tea.MathOverflow (2 Jul 2010)] This is the typical reproach to
be expected when the different kinds of infinity are analyzed and
taught.
Here the difference is clearly stated:--
"Should we briefly characterize the new view of the infinite introduced
by Cantor, we could certainly say: In analysis we have to deal only with
the infinitely small and the infinitely large as a limit-notion, as
something becoming, emerging, produced, i.e., as we put it, with the potential infinite. But this is not the proper infinite. That we have
for instance when we consider the entirety of the numbers 1, 2, 3, 4,
... itself as a completed unit, or the points of a line as an entirety
of things which is completely available. That sort of infinity is named actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische
Annalen 95 (1925) p. 167]
On 05.10.2024 15:02, joes wrote:Yes, the can be found, for example by the arithmetic mean. There simply
Am Sat, 05 Oct 2024 11:32:44 +0200 schrieb WM:But these numbers and these unit fractions cannot be found. It is
On 05.10.2024 10:08, joes wrote:What do you mean? All the unit fractions have, as you say,
Am Fri, 04 Oct 2024 22:09:41 +0200 schrieb WM:Do it. Fail because slightly fewer is not possible.
But it is no quantifier shift but simplest logic:Which they can.
If ℵ₀ unit fractions do not sit at one point (and are not dark) then >>>>> they can be subdivided into smaller parts.
finite distances, which means there are numbers inbetween.
impossible to define a unit fraction having less than ℵo smaller unit fractions. It is impossible to define a unit fraction being closer to
zero although it is obvious that there are points closer to zero than ℵo*2^ℵo points.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.10.2024 15:57, Alan Mackenzie wrote:
Yes! At least, sort of. My understanding of "doesn't exist" is either
the concept is not (yet?) developed mathematically, or it leads to
contradictions. WM's "dark numbers" certainly fall into the first
category, and possibly the second, too.
Definition: A natural number is "named" or "addressed" or "identified"
or "(individually) defined" or "instantiated" ....
That's five terms for the same thing. Four of them (at least) are thus redundant. It is unmathematical to have such redundancy.
.... if it can be communicated, necessarily by a finite amount of
information, in the sense of Poincaré, such that sender and receiver
understand the same and can link it by a finite initial segment (1, 2,
3, ..., n) of natural numbers to the origin 0.
This is ridiculous! It is so far removed from the austere simplicity of, for example, Peano's axioms as to be thoroughly unmathematical.
.... All other natural numbers are called dark natural numbers. Dark
numbers are numbers that cannot be chosen as individuals.
Is "chosen" a sixth redundant word for "named", "addressed", ....?
"Chosen as individuals" isn't a mathemtical concept. This phrase, as it
is written, makes it sound like the choice is being made by a conscious individual person, according to something unspecified. That doesn't
belong in mathematics.
Communication can occur
- by direct description in the unary system like ||||||| or as many
beeps, raps, or flashes,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7),
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
The existence of natural numbers is independent of their communication by people.
Only when a number n is identified we can use it in mathematical
discourse ....
This is something you haven't proved.
Besides, mathematicians routinely use "unidentified" numbers in
discourse. For example "If p is a prime number of the form 4m + 1, it is the sum of two squares.". That is a statement about an infinite number
of numbers, none of which are "identified".
.... and can determine the trichotomy properties of n and of every
multiple k*n or power n^k or power tower k_^n with respect to every
identified number k. ℕdef contains all defined natural numbers as
elements – and nothing else. ℕdef is a potentially infinite set;
therefore henceforth it will be called a collection.
All natural numbers are "defined" in your sense of that word.
proof, we only need note that every non-empty subset of N has a least member. Suppose there is a non-empty set of "undefined" natural numbers. Then there is a least such number.
infinity"I first came across the terms "potential infinity" and "actual
on this newsgroup, not in my degree course a few decades ago.
It is carefully avoided because closer inspection shows contradictions.
There are no such contradictions.
Therefore set theorists use just what they can defend. If actual
infinity is shown self contradictory (without dark numbers), then they
evade to potential infinity temporarily which has no completed sets and
cannot complete bijections.
Not really. There is simply no need for "actual" and "potential"
infinity.
They are relics from the past, from before the time when
mathematicians understood infinity as they do today.
taught."You use terms like completed versus potential infinity, which are not
part of the modern vernacular." [P.L. Clark in "Physicists can be
wrong", tea.MathOverflow (2 Jul 2010)] This is the typical reproach to
be expected when the different kinds of infinity are analyzed and
Here the difference is clearly stated:
"Should we briefly characterize the new view of the infinite introduced
by Cantor, we could certainly say: In analysis we have to deal only with
the infinitely small and the infinitely large as a limit-notion, as
something becoming, emerging, produced, i.e., as we put it, with the
potential infinite. But this is not the proper infinite. That we have
for instance when we consider the entirety of the numbers 1, 2, 3, 4,
... itself as a completed unit, or the points of a line as an entirety
of things which is completely available. That sort of infinity is named
actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische
Annalen 95 (1925) p. 167]
That's from 1925. It is not a modern understanding of the infinite.
If these terms had any significance, they would still be taught in mathematics degree courses.No, the teachers of such courses are too stupid or too dishonest.
Otherwise, bright students would become
aware of them and catch out their teachers in inconsistencies.
On 10/5/24 2:58 PM, WM wrote:
And that point CAN be found, Given two point x and y,
So why does that say it has a point next to it?
A point between both could be chosen unless it was dark.
On 04.10.2024 13:00, joes wrote:Those are the rules of logic, you have to follow them.
Am Fri, 04 Oct 2024 11:33:02 +0200 schrieb WM:
It is prove correct by the impossibility to circumvent it: all numbersI know that all numbers which you can choose have ℵ₀ successors.I don’t believe anything. The rules of logic which you are so fond of simply do not admit that deduction in general (i.e. it is wrong).
Therefore the collection containing all numbers which you can choose
has ℵ₀ successors.
which you can choose have ℵ₀ successors. That is essential, not your "rules".
--What do you mean with the successor of a collection anyway?
On 10/5/24 3:04 PM, WM wrote:
On 05.10.2024 15:32, Richard Damon wrote:
On 10/5/24 8:18 AM, WM wrote:
On 05.10.2024 13:12, Richard Damon wrote:
How can bijections between infinite sets exist then.
But actual infinity doesn't exist.
Because the sets exist as what you think of as "potential" infinity,
"the integers separately as well as in their actually infinite
totality exist as eternal ideas in intellectu Divino in the highest
degree of reality." [G. Cantor, letter to C. Hermite (30 Nov 1895)]
but that isn't what your "actual infinity" is, that is just protential infinity.
Am Fri, 04 Oct 2024 22:10:35 +0200 schrieb WM:
On 04.10.2024 13:00, joes wrote:Those are the rules of logic, you have to follow them.
> Am Fri, 04 Oct 2024 11:33:02 +0200 schrieb WM:
>> I know that all numbers which you can choose have ℵ₀ successors.
>> Therefore the collection containing all numbers which you can choose
>> has ℵ₀ successors.
> I don’t believe anything. The rules of logic which you are so fond of >> > simply do not admit that deduction in general (i.e. it is wrong).
It is prove correct by the impossibility to circumvent it: all numbers
which you can choose have ℵ₀ successors. That is essential, not your
"rules".
ANY finite value above 0 will have Aleph_0 unit fractions below it.
Your math and logic just can't handle that fact, because
On 10/5/24 2:58 PM, WM wrote:
Every point is a finite set.
The fact that we can keep doing that indefinitely, and never reach a
point we can't continue, is proof that the concept of "next point"
doesn't exist
You mean, that there is a difference? I remain unconvinced.
Am Sat, 05 Oct 2024 21:15:43 +0200 schrieb WM:
Dark numbers are numbers that cannot be chosen as individuals.That is possible for all natural numbers.
Communication can occur - by direct description in the unary system likeWhere did you get this idea from?
||||||| or as many beeps, raps, or flashes,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7),
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
"You use terms like completed versus potential infinity, which are notThey are not taught anymore.
part of the modern vernacular." [P.L. Clark in "Physicists can be
wrong", tea.MathOverflow (2 Jul 2010)] This is the typical reproach to
be expected when the different kinds of infinity are analyzed and
taught.
A point between both could be chosen unless it was dark.Many points could even be chosen!
On 05.10.2024 23:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.10.2024 15:57, Alan Mackenzie wrote:
Yes! At least, sort of. My understanding of "doesn't exist" is either >>>> the concept is not (yet?) developed mathematically, or it leads to
contradictions. WM's "dark numbers" certainly fall into the first
category, and possibly the second, too.
Definition: A natural number is "named" or "addressed" or "identified"
or "(individually) defined" or "instantiated" ....
That's five terms for the same thing. Four of them (at least) are thus
redundant. It is unmathematical to have such redundancy.
The vocabulary of my readers is very different. So the chance of understanding is increased.
.... if it can be communicated, necessarily by a finite amount of
information, in the sense of Poincaré, such that sender and receiver
understand the same and can link it by a finite initial segment (1, 2,
3, ..., n) of natural numbers to the origin 0.
This is ridiculous! It is so far removed from the austere simplicity of,
for example, Peano's axioms as to be thoroughly unmathematical.
Peano's axioms are invalid for large numbers.
.... All other natural numbers are called dark natural numbers. Dark
numbers are numbers that cannot be chosen as individuals.
Is "chosen" a sixth redundant word for "named", "addressed", ....?
Yes, all these words have the same meaning, but not all readers know all words.
"Chosen as individuals" isn't a mathemtical concept. This phrase, as it
is written, makes it sound like the choice is being made by a conscious
individual person, according to something unspecified. That doesn't
belong in mathematics.
It does not belong to the simple but inconsistent present mathematics.
Communication can occur
- by direct description in the unary system like ||||||| or as many
beeps, raps, or flashes,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7),
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
The existence of natural numbers is independent of their communication by
people.
You are a believer in God having created them?
Only when a number n is identified we can use it in mathematical
discourse ....
This is something you haven't proved.
Try to use another number.
Besides, mathematicians routinely use "unidentified" numbers in
discourse. For example "If p is a prime number of the form 4m + 1, it is
the sum of two squares.". That is a statement about an infinite number
of numbers, none of which are "identified".
It is not a statement about an individual number but about a set, some
of which can be defined.
.... and can determine the trichotomy properties of n and of every
multiple k*n or power n^k or power tower k_^n with respect to every
identified number k. ℕdef contains all defined natural numbers as
elements – and nothing else. ℕdef is a potentially infinite set;
therefore henceforth it will be called a collection.
All natural numbers are "defined" in your sense of that word.
No, after every identified number almost all are following, almost all
of them are dark.
As a proof, we only need note that every non-empty subset of N has a
least member. Suppose there is a non-empty set of "undefined" natural
numbers. Then there is a least such number.
No. The identified numbers are a potentially infinite collection.
Therefore there is no least dark number.
I first came across the terms "potential infinity" and "actual
infinity" on this newsgroup, not in my degree course a few decades
ago.
It is carefully avoided because closer inspection shows contradictions.
There are no such contradictions.
There are many. For instance: All unit fractions are separate points on
the positive real axis, but there are infinitely many for every x > 0.
That can only hold for definable x, not for all.
Therefore set theorists use just what they can defend. If actual
infinity is shown self contradictory (without dark numbers), then they
evade to potential infinity temporarily which has no completed sets and
cannot complete bijections.
Not really. There is simply no need for "actual" and "potential"
infinity.
You have not learnt about it during study and obviously not afterwards.
They are relics from the past, from before the time when
mathematicians understood infinity as they do today.
You cannot judge because you don't know that topic ....
.... and as fellow traveler can only parrot the words of matheologians
who are either too stupid to recognize or too dishonest to confess the
truth.
"You use terms like completed versus potential infinity, which are
not part of the modern vernacular." [P.L. Clark in "Physicists can be
wrong", tea.MathOverflow (2 Jul 2010)] This is the typical reproach
to be expected when the different kinds of infinity are analyzed and
taught.
Here the difference is clearly stated:
"Should we briefly characterize the new view of the infinite introduced
by Cantor, we could certainly say: In analysis we have to deal only with >>> the infinitely small and the infinitely large as a limit-notion, as
something becoming, emerging, produced, i.e., as we put it, with the
potential infinite. But this is not the proper infinite. That we have
for instance when we consider the entirety of the numbers 1, 2, 3, 4,
... itself as a completed unit, or the points of a line as an entirety
of things which is completely available. That sort of infinity is named
actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische
Annalen 95 (1925) p. 167]
That's from 1925. It is not a modern understanding of the infinite.
But a correct one. The modern understanding is pure deceit.
If these terms had any significance, they would still be taught in
mathematics degree courses.
No, the teachers of such courses are too stupid or too dishonest.
Otherwise, bright students would become aware of them and catch out
their teachers in inconsistencies.
They do. But every publishing is intercepted by the leading liars.
Regards, WM
There simply
are no unit fractions with a finite number of lesser UFs,
You
cannot start from zero, count a finite number of UFs, and reach an
infinity in ever greater steps.
Am Fri, 04 Oct 2024 22:10:35 +0200 schrieb WM:
all numbersThose are the rules of logic, you have to follow them
which you can choose have ℵ₀ successors. That is essential, not your
"rules".
> What do you mean with the successor of a collection anyway?
Who do you think you are to accuse others of being stupid or dishonest?
On 06.10.2024 12:16, Alan Mackenzie wrote:
You mean, that there is a difference? I remain unconvinced.
"A potential infinity is a quantity which is finite but indefinitely
large. For instance, when we enumerate the natural numbers as 0, 1, 2,
..., n, n+1, ..., the enumeration is finite at any point in time, ....
.... but it grows indefinitely and without bound. [...] An actual
infinity is a completed infinite totality. Examples: , , C[0, 1],
L2[0, 1], etc. Other examples: gods, devils, etc." [S.G. Simpson:
"Potential versus actual infinity: Insights from reverse mathematics"
(2015)]
"Potential infinity refers to a procedure that gets closer and closer
to, but never quite reaches, an infinite end. For instance, the sequence
of numbers 1, 2, 3, 4, ... gets higher and higher, but it has no end; it never gets to infinity. Infinity is just an indication of a direction – it's 'somewhere off in the distance'. Chasing this kind of infinity is
like chasing a rainbow or trying to sail to the edge of the world – you may think you see it in the distance, but when you get to where you
thought it was, you see it is still further away. Geometrically, imagine
an infinitely long straight line; then 'infinity' is off at the 'end' of
the line. Analogous procedures are given by limits in calculus, whether
they use infinity or not. For example, limx0(sinx)/x = 1. This means
that when we choose values of x that are closer and closer to zero, but never quite equal to zero, then (sinx)/x gets closer and closer to one.
Completed infinity, or actual infinity, is an infinity that one actually reaches; the process is already done. For instance, let's put braces around that sequence mentioned earlier: {1, 2, 3, 4, ...}. With
this notation, we are indicating the set of all positive integers. This
is just one object, a set. But that set has infinitely many members. By
that I don't mean that it has a large finite number of members and it
keeps getting more members. Rather, I mean that it already has
infinitely many members.
We can also indicate the completed infinity geometrically. For instance, the diagram at right shows a one-to-one correspondence between points on an infinitely long line and points on a semicircle. There are
no points for plus or minus infinity on the line, but it is natural to attach those 'numbers' to the endpoints of the semicircle.
Isn't that 'cheating', to simply add numbers in this fashion? Not really; it just depends on what we want to use those numbers for. For instance, f(x) = 1/(1 + x2) is a continuous function defined for all
real numbers x, and it also tends to a limit of 0 when x 'goes to' plus
or minus infinity (in the sense of potential infinity, described
earlier). Consequently, if we add those two 'numbers' to the real line,
to get the so-called 'extended real line', and we equip that set with
the same topology as that of the closed semicircle (i.e., the semicircle including the endpoints), then the function f is continuous everywhere
on the extended real line." [E. Schechter: "Potential versus completed infinity: Its history and controversy" (5 Dec 2009)]
Regards, WM
Am Sun, 06 Oct 2024 12:32:16 +0200 schrieb WM:
On 06.10.2024 11:49, Moebius wrote:Komisch auch, dass ω/2*2 (ist die Reihenfolge wichtig?)
Am 06.10.2024 um 10:40 schrieb WM:
On 06.10.2024 05:35, Moebius wrote:die Menge IN (und deren Elemente), wie schon gesagt.
Am 05.10.2024 um 22:38 schrieb WM:Das gilt für
On 05.10.2024 22:13, Moebius wrote:Doch, weil IN gegenüber der Multiplikation ABGESCHLOSSEN ist. (Das
Am 05.10.2024 um 22:01 schrieb WM:Nein.
ω/2 * 2 e IN,
kannst Du sogar in deinem Bestseller nachlesen, falls Du es
inzwischen vergessen haben solltest.) Mit ω/2 e IN wäre auch ω/2 * 2 >>>>> e IN (weil 2 e IN ist).
In Zeichen: An,m e IN: n * m e IN.
nicht in N ist,
obwohl ω/2 es ja sein soll.
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} mit ω oder ω+1 mittendrin.
Korrektheit hat nichts mit Ästhetik zu tun.Wenn Du es sagstAber man kann die offensichtlich falschen zurückweisen. Dazu gehört,
Leider kann man sich in der Mathematik /die Ergebnisse/ nicht
aussuchen.
dass bei Multiplikation mit 2 die Realität der Menge halbiert wird.
Was ist diese „Realität”?
Dass G zu N gleichmächtig ist, mit der
Bijektion n->2n?
Und doch hast du keine Alternative gebotenDas wiederum gilt nur für das potentiell Unendliche. Im aktuales bleiben nach wie vor |ℕ| Zahlen,DAS ist wiederum richtig! :-)
Hint: card(IN) = card(G).
Unendlichen ist Card Unsinn.
(außer, dass jede Menge
ihre eigene Kardinalität hätte).
WM <wolfgang.mueckenheim@tha.de> wrote:
"A potential infinity is a quantity which is finite but indefinitely
large. For instance, when we enumerate the natural numbers as 0, 1, 2,
..., n, n+1, ..., the enumeration is finite at any point in time, ....
That is a mistake. The ellipses indicate the enumeration. There is no
time. If one must consider time, then the enumeration happens instantaneously.
This idea of time may be what misleads the mathematically less adept
into believing that 0.999... < 1.
9*0.999...999 < 9 as it should be.
.... but it grows indefinitely and without bound. [...] An actual
infinity is a completed infinite totality. Examples: , , C[0, 1],
L2[0, 1], etc. Other examples: gods, devils, etc." [S.G. Simpson:
"Potential versus actual infinity: Insights from reverse mathematics"
(2015)]
"Potential infinity refers to a procedure that gets closer and closer
to, but never quite reaches, an infinite end. For instance, the sequence
of numbers 1, 2, 3, 4, ... gets higher and higher, but it has no end; it
never gets to infinity. Infinity is just an indication of a direction –
it's 'somewhere off in the distance'. Chasing this kind of infinity is
like chasing a rainbow or trying to sail to the edge of the world – you
may think you see it in the distance, but when you get to where you
thought it was, you see it is still further away. Geometrically, imagine
an infinitely long straight line; then 'infinity' is off at the 'end' of
the line. Analogous procedures are given by limits in calculus, whether
they use infinity or not. For example, limx0(sinx)/x = 1. This means
that when we choose values of x that are closer and closer to zero, but
never quite equal to zero, then (sinx)/x gets closer and closer to one.
Completed infinity, or actual infinity, is an infinity that one
actually reaches; the process is already done. For instance, let's put
braces around that sequence mentioned earlier: {1, 2, 3, 4, ...}. With
this notation, we are indicating the set of all positive integers. This
is just one object, a set. But that set has infinitely many members. By
that I don't mean that it has a large finite number of members and it
keeps getting more members. Rather, I mean that it already has
infinitely many members.
We can also indicate the completed infinity geometrically. For
instance, the diagram at right shows a one-to-one correspondence between
points on an infinitely long line and points on a semicircle. There are
no points for plus or minus infinity on the line, but it is natural to
attach those 'numbers' to the endpoints of the semicircle.
Isn't that 'cheating', to simply add numbers in this fashion? Not
really; it just depends on what we want to use those numbers for. For
instance, f(x) = 1/(1 + x2) is a continuous function defined for all
real numbers x, and it also tends to a limit of 0 when x 'goes to' plus
or minus infinity (in the sense of potential infinity, described
earlier). Consequently, if we add those two 'numbers' to the real line,
to get the so-called 'extended real line', and we equip that set with
the same topology as that of the closed semicircle (i.e., the semicircle
including the endpoints), then the function f is continuous everywhere
on the extended real line." [E. Schechter: "Potential versus completed
infinity: Its history and controversy" (5 Dec 2009)]
The above is all very poetic, this supposed difference between "actual"
and "potential" infinite, but it is not mathematical. There are no mathematical theorems which depend for their theoremhood on the supposed distinction between "actual" and "potential" infinite.
WM <wolfgang.mueckenheim@tha.de> wrote:
All unit fractions are separate points on
the positive real axis, but there are infinitely many for every x > 0.
That can only hold for definable x, not for all.
Poppycock! You'll have to do better than that to provide such a contradiction.
Hint: Skilled mathematicians have worked on trying to
prove the inconsistency of maths, without success.
You cannot judge because you don't know that topic ....
I am a graduate in maths ....
.... and as fellow traveler can only parrot the words of matheologians
who are either too stupid to recognize or too dishonest to confess the
truth.
.... and able to understand and follow mathematical argument,
If these terms had any significance, they would still be taught in
mathematics degree courses.
No, the teachers of such courses are too stupid or too dishonest.
Who do you think you are to accuse others of being stupid or dishonest?
Otherwise, bright students would become aware of them and catch out
their teachers in inconsistencies.
They do. But every publishing is intercepted by the leading liars.
<Sigh> When I was an undergraduate, students published lots of
magazines, some of them about maths. I'm sure they still do, though they are likely to be online these days. The "deceit" you think happens would
be exposed in these magazines, and thus become known,
It's generated by an infinite process
It just is.Right.
On 06.10.2024 16:52, Alan Mackenzie wrote:In actual infinity, there is no last 9 (that would not be infinite).
WM <wolfgang.mueckenheim@tha.de> wrote:
In potential infinity there is time or at least a sequence of steps."A potential infinity is a quantity which is finite but indefinitelyThat is a mistake. The ellipses indicate the enumeration. There is no
large. For instance, when we enumerate the natural numbers as 0, 1, 2,
..., n, n+1, ..., the enumeration is finite at any point in time, ....
time. If one must consider time, then the enumeration happens
instantaneously.
This idea of time may be what misleads the mathematically less adeptThat is true even in actual infinity.
into believing that 0.999... < 1.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying 0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==> 9*0.999...999 < 9 as it should be.
Your argument is not particular to Canter, you merely picked him toThe above is all very poetic, this supposed difference between "actual"Set theory depends on actual infinity. Bijections must be complete. But Cantor's bijections never are complete. Cantor's list must be completely enumerated by natural numbers. The diagonal number must be complete such
and "potential" infinite, but it is not mathematical. There are no
mathematical theorems which depend for their theoremhood on the
supposed distinction between "actual" and "potential" infinite.
that no digit is missing in order to be distinct from every listed real number. Impossible. All that is nonsense.
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
"A potential infinity is a quantity which is finite but indefinitely
large. For instance, when we enumerate the natural numbers as 0, 1, 2,
..., n, n+1, ..., the enumeration is finite at any point in time, ....
That is a mistake. The ellipses indicate the enumeration. There is no
time. If one must consider time, then the enumeration happens
instantaneously.
In potential infinity there is time or at least a sequence of steps.
This idea of time may be what misleads the mathematically less adept
into believing that 0.999... < 1.
That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying 0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999
9*0.999...999 < 9 as it should be.
.... but it grows indefinitely and without bound. [...] An actual
infinity is a completed infinite totality. Examples: , , C[0, 1],
L2[0, 1], etc. Other examples: gods, devils, etc." [S.G. Simpson:
"Potential versus actual infinity: Insights from reverse mathematics"
(2015)]
"Potential infinity refers to a procedure that gets closer and closer
to, but never quite reaches, an infinite end. For instance, the sequence >>> of numbers 1, 2, 3, 4, ... gets higher and higher, but it has no end; it >>> never gets to infinity. Infinity is just an indication of a direction – >>> it's 'somewhere off in the distance'. Chasing this kind of infinity is
like chasing a rainbow or trying to sail to the edge of the world – you >>> may think you see it in the distance, but when you get to where you
thought it was, you see it is still further away. Geometrically, imagine >>> an infinitely long straight line; then 'infinity' is off at the 'end' of >>> the line. Analogous procedures are given by limits in calculus, whether
they use infinity or not. For example, limx0(sinx)/x = 1. This means
that when we choose values of x that are closer and closer to zero, but
never quite equal to zero, then (sinx)/x gets closer and closer to one.
Completed infinity, or actual infinity, is an infinity that one
actually reaches; the process is already done. For instance, let's put
braces around that sequence mentioned earlier: {1, 2, 3, 4, ...}. With
this notation, we are indicating the set of all positive integers. This
is just one object, a set. But that set has infinitely many members. By
that I don't mean that it has a large finite number of members and it
keeps getting more members. Rather, I mean that it already has
infinitely many members.
We can also indicate the completed infinity geometrically. For
instance, the diagram at right shows a one-to-one correspondence between >>> points on an infinitely long line and points on a semicircle. There are
no points for plus or minus infinity on the line, but it is natural to
attach those 'numbers' to the endpoints of the semicircle.
Isn't that 'cheating', to simply add numbers in this fashion? Not >>> really; it just depends on what we want to use those numbers for. For
instance, f(x) = 1/(1 + x2) is a continuous function defined for all
real numbers x, and it also tends to a limit of 0 when x 'goes to' plus
or minus infinity (in the sense of potential infinity, described
earlier). Consequently, if we add those two 'numbers' to the real line,
to get the so-called 'extended real line', and we equip that set with
the same topology as that of the closed semicircle (i.e., the semicircle >>> including the endpoints), then the function f is continuous everywhere
on the extended real line." [E. Schechter: "Potential versus completed
infinity: Its history and controversy" (5 Dec 2009)]
The above is all very poetic, this supposed difference between "actual"
and "potential" infinite, but it is not mathematical. There are no
mathematical theorems which depend for their theoremhood on the supposed
distinction between "actual" and "potential" infinite.
Set theory depends on actual infinity.
Bijections must be complete.
But Cantor's bijections never are complete. Cantor's list must be
completely enumerated by natural numbers. The diagonal number must be complete such that no digit is missing in order to be distinct from
every listed real number. Impossible. All that is nonsense.
Regards, WM
On 06.10.2024 14:01, joes wrote:Interessant. ω*2 /2 ist ja eindeutig unendlich, aber ω/2 *2 nicht?
Am Sun, 06 Oct 2024 12:32:16 +0200 schrieb WM:Nein.
On 06.10.2024 11:49, Moebius wrote:Komisch auch, dass ω/2*2 (ist die Reihenfolge wichtig?)
Am 06.10.2024 um 10:40 schrieb WM:
On 06.10.2024 05:35, Moebius wrote:die Menge IN (und deren Elemente), wie schon gesagt.
Am 05.10.2024 um 22:38 schrieb WM:Das gilt für
On 05.10.2024 22:13, Moebius wrote:Doch, weil IN gegenüber der Multiplikation ABGESCHLOSSEN ist. (Das >>>>>> kannst Du sogar in deinem Bestseller nachlesen, falls Du es
Am 05.10.2024 um 22:01 schrieb WM:Nein.
ω/2 * 2 e IN,
inzwischen vergessen haben solltest.) Mit ω/2 e IN wäre auch ω/2 * >>>>>> 2 e IN (weil 2 e IN ist).
In Zeichen: An,m e IN: n * m e IN.
Das nennt man Teilmenge.nicht in N ist, obwohl ω/2 es ja sein soll.
Ist auch so. Siehe hier:
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} mit ω oder ω+1
mittendrin.
Realität ist, dass es mehr ganze als gerade Zahlen gibt, weil jedeKorrektheit hat nichts mit Ästhetik zu tun.Wenn Du es sagst Leider kann man sich in der Mathematik /dieAber man kann die offensichtlich falschen zurückweisen. Dazu gehört,
Ergebnisse/ nicht aussuchen.
dass bei Multiplikation mit 2 die Realität der Menge halbiert wird.
Was ist diese „Realität”?
gerade ganz, aber nicht jede ganze gerade ist.
Schwach.Dass G zu N gleichmächtig ist, mit der Bijektion n->2n?Ist Unsinn.
Doch, s. unten.
Und doch hast du keine Alternative gebotenDas wiederum gilt nur für das potentiell Unendliche. Im aktuales bleiben nach wie vor |ℕ| Zahlen,DAS ist wiederum richtig! :-)
Hint: card(IN) = card(G).
Unendlichen ist Card Unsinn.
(außer, dass jede Menge ihre eigene Kardinalität hätte).Das ist doch auch richtig (wobei man besser von Anzahl sprechen sollte).
Not all infinite sets can be compared by size, but we can establish some useful rules.
The rule of subset proves that every proper subset has fewer elements than its superset. So there are more natural numbers than prime numbers,1
|N| > |P|, and more complex numbers than real numbers, |C| > |R|. Even finitely many exceptions from the subset-relation are admitted for
infinite subsets. Therefore there are more odd numbers than prime
numbers |O| > |P|.
The rule of construction yields the numbers of integers |Z| = 2|N| +
and the number of fractions |Q| = 2|N|^2 + 1 (there are fewer rational numbers Q# ). Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers thanHow small an error? Surely there is one more even number.
rational numbers |X| > |Q#|.
The rule of symmetry yields precisely the same number of real
geometric points in every interval (n, n+1] and with at most a small
error same number of odd numbers and of even numbers in every finite
interval and in the whole real line.
This theory makes the number of natural numbers (and of course of otherI don’t understand this argument. What is the cardinality of the set
sets too) depending on the numerical representation. The set {1, 11,
111, ...} of natural numbers has only comparatively few elements.
Therefore the set of natural numbers in unary or binary notation has
fewer, in hexadecimal notation more than |N| elements. The set {10, 20,
30, ...} has |N|/10 elements, but if the zeros are only applied as decoration, this set, like the set {1', 2', 3', ...}, has |N| elements.
It will be a matter of future research to investigate the effect of
different numerical systems in detail.
On 06.10.2024 15:59, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
All unit fractions are separate points on
the positive real axis, but there are infinitely many for every x > 0.
That can only hold for definable x, not for all.
Poppycock! You'll have to do better than that to provide such a
contradiction.
It is good enough, but you can't understand.
Hint: Skilled mathematicians have worked on trying to
prove the inconsistency of maths, without success.
What shall that prove? Try to understand.
You cannot judge because you don't know that topic ....
I am a graduate in maths ....
Here is not discussed what you have studied. Remember, not even infinity
has been taught. Therefore you cannot judge.
.... and as fellow traveler can only parrot the words of matheologians
who are either too stupid to recognize or too dishonest to confess the
truth.
.... and able to understand and follow mathematical argument, and to
distinguish mathematical notions from pure hogwash. [ Citation
restored after snipping. ]
Try only to understand my argument. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. How can infinitely many unit fractions appear before every x > 0?
If these terms had any significance, they would still be taught in
mathematics degree courses.
No, the teachers of such courses are too stupid or too dishonest.
Who do you think you are to accuse others of being stupid or dishonest?
I know that I have understood that topic better than the stupids.
Otherwise, bright students would become aware of them and catch out
their teachers in inconsistencies.
They do. But every publishing is intercepted by the leading liars.
<Sigh> When I was an undergraduate, students published lots of
magazines, some of them about maths. I'm sure they still do, though they
are likely to be online these days. The "deceit" you think happens would
be exposed in these magazines, and thus become known,
You cannot believe that I am right, therefore you don't wish that I am right, and you try to dismiss my argument.
Regards, WM
On 06.10.2024 04:51, Richard Damon wrote:
On 10/5/24 2:58 PM, WM wrote:
And that point CAN be found, Given two point x and y,
So why does that say it has a point next to it?
A point between both could be chosen unless it was dark.
They are not given, but dark. Discrete points on the positive axis have
a minimum.
Regards, WM
On 06.10.2024 13:35, joes wrote:
But ℵo unit fractions and the points between them cannot.A point between both could be chosen unless it was dark.Many points could even be chosen!
Regards, WM
On 06.10.2024 04:51, Richard Damon wrote:
On 10/5/24 2:58 PM, WM wrote:
Every point is a finite set.
The fact that we can keep doing that indefinitely, and never reach a
point we can't continue, is proof that the concept of "next point"
doesn't exist
The concept of point however does exist. Every single point can be found unless it is dark.
Regards, WM
On 05.10.2024 23:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.10.2024 15:57, Alan Mackenzie wrote:
Yes! At least, sort of. My understanding of "doesn't exist" is either
the concept is not (yet?) developed mathematically, or it leads to
contradictions. WM's "dark numbers" certainly fall into the first
category, and possibly the second, too.
Definition: A natural number is "named" or "addressed" or "identified"
or "(individually) defined" or "instantiated" ....
That's five terms for the same thing. Four of them (at least) are thus redundant. It is unmathematical to have such redundancy.
The vocabulary of my readers is very different. So the c hance of understanding is increased.
.... if it can be communicated, necessarily by a finite amount of
information, in the sense of Poincaré, such that sender and receiver
understand the same and can link it by a finite initial segment (1, 2,
3, ..., n) of natural numbers to the origin 0.
This is ridiculous! It is so far removed from the austere simplicityof,
for example, Peano's axioms as to be thoroughly unmathematical.
Peano's axioms are invalid fpr large numbers.
.... All other natural numbers are called dark natural numbers. Dark
numbers are numbers that cannot be chosen as individuals.
Is "chosen" a sixth redundant word for "named", "addressed", ....?
Yes, all these words have the same meaning, but not all readers know all words.
"Chosen as individuals" isn't a mathemtical concept. This phrase, as it is written, makes it sound like the choice is being made by a conscious individual person, according to something unspecified. That doesn't belong in mathematics.
It does not belong to the simple but inconsistent present mathematics.
Communication can occur
- by direct description in the unary system like ||||||| or as many
beeps, raps, or flashes,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7),
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
The existence of natural numbers is independent of theircommunication by
people.
You are a believer in God having created them?
Only when a number n is identified we can use it in mathematical
discourse ....
This is something you haven't proved.
Try to use another number.
Besides, mathematicians routinely use "unidentified" numbers in discourse. For example "If p is a prime number of the form 4m + 1,it is
the sum of two squares.". That is a statement about an infinite number of numbers, none of which are "identified".
It is not a statement about an individual number but about a set, some
of which can be defined.
.... and can determine the trichotomy properties of n and of every
multiple k*n or power n^k or power tower k_^n with respect to every
identified number k. ℕdef contains all defined natural numbers as
elements – and nothing else. ℕdef is a potentially infinite set;
therefore henceforth it will be called a collection.
All natural numbers are "defined" in your sense of that word.
No, after every identified number almost all are following, almost all
of them are dark.
As a
proof, we only need note that every non-empty subset of N has a least member. Suppose there is a non-empty set of "undefined" naturalnumbers.
Then there is a least such number.
No. The identified numbers are a potentially infinite collection.
Therefore there is no least dark number.
infinity"I first came across the terms "potential infinity" and "actual
on this newsgroup, not in my degree course a few decades ago.
It is carefully avoided because closer inspection shows contradictions.
There are no such contradictions.
There are many. For instance: All unit fractions are separate points on
the positive real axis, but there are infinitely many for every x > 0.
That can only hold for definable x, not for all.
Therefore set theorists use just what they can defend. If actual
infinity is shown self contradictory (without dark numbers), then they
evade to potential infinity temporarily which has no completed sets and
cannot complete bijections.
Not really. There is simply no need for "actual" and "potential" infinity.
You have not learnt about it during study and obviously not afterwards.
They are relics from the past, from before the time when
mathematicians understood infinity as they do today.
You cannot judge because you don't know that topic and as fellow
traveler can only parrot the words of matheologians who are either too
stupid to recognize or too dishonest to confess the truth.
taught."You use terms like completed versus potential infinity, which are not
part of the modern vernacular." [P.L. Clark in "Physicists can be
wrong", tea.MathOverflow (2 Jul 2010)] This is the typical reproach to
be expected when the different kinds of infinity are analyzed and
withHere the difference is clearly stated:
"Should we briefly characterize the new view of the infinite introduced
by Cantor, we could certainly say: In analysis we have to deal only
the infinitely small and the infinitely large as a limit-notion, as
something becoming, emerging, produced, i.e., as we put it, with the
potential infinite. But this is not the proper infinite. That we have
for instance when we consider the entirety of the numbers 1, 2, 3, 4,
... itself as a completed unit, or the points of a line as an entirety
of things which is completely available. That sort of infinity is named
actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische
Annalen 95 (1925) p. 167]
That's from 1925. It is not a modern understanding of the infinite.
But a correct one. The modern understanding is pure deceit.
If these terms had any significance, they would still be taught in mathematics degree courses.No, the teachers of such courses are too stupid or too dishonest.
Otherwise, bright students would become
aware of them and catch out their teachers in inconsistencies.
They do. But every publishing is intercepted by the leading liars.
Regards, WM
On 06.10.2024 13:45, joes wrote:
Am Fri, 04 Oct 2024 22:10:35 +0200 schrieb WM:
all numbersThose are the rules of logic, you have to follow them
which you can choose have ℵ₀ successors. That is essential, not your >>> "rules".
I follow the reality of mathematics.
> What do you mean with the successor of a collection anyway?
Beyond the collection there are ℵo elements that cannot be reached individually.
Regards, WM
On 06.10.2024 12:16, Alan Mackenzie wrote:
You mean, that there is a difference? I remain unconvinced.
"Numerals constitute a potential infinity. Given any numeral, we can construct a new numeral by prefixing it with S. Now imagine this
potential infinity to be completed. Imagine the inexhaustible process of constructing numerals somehow to have been finished, and call the result
the set of all numbers, denoted by . Thus is thought to be an actual infinity or a completed infinity. This is curious terminology, since the etymology of 'infinite' is 'not finished'." [E. Nelson: "Hilbert's
mistake" (2007) p. 3]
"A potential infinity is a quantity which is finite but indefinitely
large. For instance, when we enumerate the natural numbers as 0, 1,
2, ..., n, n+1, ..., the enumeration is finite at any point in time, but
it grows indefinitely and without bound. [...] An actual infinity is a completed infinite totality. Examples: , , C[0, 1], L2[0, 1], etc. Other examples: gods, devils, etc." [S.G. Simpson: "Potential versus
actual infinity: Insights from reverse mathematics" (2015)]
"Potential infinity refers to a procedure that gets closer and closer
to, but never quite reaches, an infinite end. For instance, the sequence
of numbers 1, 2, 3, 4, ... gets higher and higher, but it has no end; it never gets to infinity. Infinity is just an indication of a direction – it's 'somewhere off in the distance'. Chasing this kind of infinity is
like chasing a rainbow or trying to sail to the edge of the world – you
may think you see it in the distance, but when you get to where you
thought it was, you see it is still further away. Geometrically, imagine
an infinitely long straight line; then 'infinity' is off at the 'end' of
the line. Analogous procedures are given by limits in calculus, whether
they use infinity or not. For example, limx0(sinx)/x = 1. This means
that when we choose values of x that are closer and closer to zero, but
never quite equal to zero, then (sinx)/x gets closer and closer to one.
Completed infinity, or actual infinity, is an infinity that one actually reaches; the process is already done. For instance, let's put
braces around that sequence mentioned earlier: {1, 2, 3, 4, ...}. With
this notation, we are indicating the set of all positive integers. This
is just one object, a set. But that set has infinitely many members. By
that I don't mean that it has a large finite number of members and it
keeps getting more members. Rather, I mean that it already has
infinitely many members.
We can also indicate the completed infinity geometrically. For instance, the diagram at right shows a one-to-one correspondence between points on an infinitely long line and points on a semicircle. There are
no points for plus or minus infinity on the line, but it is natural to
attach those 'numbers' to the endpoints of the semicircle.
Isn't that 'cheating', to simply add numbers in this fashion? Not really; it just depends on what we want to use those numbers for. For instance, f(x) = 1/(1 + x2) is a continuous function defined for all
real numbers x, and it also tends to a limit of 0 when x 'goes to' plus
or minus infinity (in the sense of potential infinity, described
earlier). Consequently, if we add those two 'numbers' to the real line,
to get the so-called 'extended real line', and we equip that set with
the same topology as that of the closed semicircle (i.e., the semicircle including the endpoints), then the function f is continuous everywhere
on the extended real line." [E. Schechter: "Potential versus completed infinity: Its history and controversy" (5 Dec 2009)]
Regards, WM
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
"A potential infinity is a quantity which is finite but indefinitely
large. For instance, when we enumerate the natural numbers as 0, 1, 2,
..., n, n+1, ..., the enumeration is finite at any point in time, ....
That is a mistake. The ellipses indicate the enumeration. There is no
time. If one must consider time, then the enumeration happens
instantaneously.
In potential infinity there is time or at least a sequence of steps.
This idea of time may be what misleads the mathematically less adept
into believing that 0.999... < 1.
That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying 0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999
9*0.999...999 < 9 as it should be.
.... but it grows indefinitely and without bound. [...] An actual
infinity is a completed infinite totality. Examples: , , C[0, 1],
L2[0, 1], etc. Other examples: gods, devils, etc." [S.G. Simpson:
"Potential versus actual infinity: Insights from reverse mathematics"
(2015)]
"Potential infinity refers to a procedure that gets closer and closer
to, but never quite reaches, an infinite end. For instance, the sequence >>> of numbers 1, 2, 3, 4, ... gets higher and higher, but it has no end; it >>> never gets to infinity. Infinity is just an indication of a direction – >>> it's 'somewhere off in the distance'. Chasing this kind of infinity is
like chasing a rainbow or trying to sail to the edge of the world – you >>> may think you see it in the distance, but when you get to where you
thought it was, you see it is still further away. Geometrically, imagine >>> an infinitely long straight line; then 'infinity' is off at the 'end' of >>> the line. Analogous procedures are given by limits in calculus, whether
they use infinity or not. For example, limx0(sinx)/x = 1. This means
that when we choose values of x that are closer and closer to zero, but
never quite equal to zero, then (sinx)/x gets closer and closer to one.
Completed infinity, or actual infinity, is an infinity that one
actually reaches; the process is already done. For instance, let's put
braces around that sequence mentioned earlier: {1, 2, 3, 4, ...}. With
this notation, we are indicating the set of all positive integers. This
is just one object, a set. But that set has infinitely many members. By
that I don't mean that it has a large finite number of members and it
keeps getting more members. Rather, I mean that it already has
infinitely many members.
We can also indicate the completed infinity geometrically. For
instance, the diagram at right shows a one-to-one correspondence between >>> points on an infinitely long line and points on a semicircle. There are
no points for plus or minus infinity on the line, but it is natural to
attach those 'numbers' to the endpoints of the semicircle.
Isn't that 'cheating', to simply add numbers in this fashion?
Not
really; it just depends on what we want to use those numbers for. For
instance, f(x) = 1/(1 + x2) is a continuous function defined for all
real numbers x, and it also tends to a limit of 0 when x 'goes to' plus
or minus infinity (in the sense of potential infinity, described
earlier). Consequently, if we add those two 'numbers' to the real line,
to get the so-called 'extended real line', and we equip that set with
the same topology as that of the closed semicircle (i.e., the semicircle >>> including the endpoints), then the function f is continuous everywhere
on the extended real line." [E. Schechter: "Potential versus completed
infinity: Its history and controversy" (5 Dec 2009)]
The above is all very poetic, this supposed difference between "actual"
and "potential" infinite, but it is not mathematical. There are no
mathematical theorems which depend for their theoremhood on the supposed
distinction between "actual" and "potential" infinite.
Set theory depends on actual infinity. Bijections must be complete. But Cantor's bijections never are complete. Cantor's list must be completely enumerated by natural numbers. The diagonal number must be complete such
that no digit is missing in order to be distinct from every listed real number. Impossible. All that is nonsense.
Regards, WM
On 06.10.2024 05:35, Richard Damon wrote:
ANY finite value above 0 will have Aleph_0 unit fractions below it.
Impossible. Among them there must be a first. But they cannot be seen.
They are dark.
Your math and logic just can't handle that fact, because
it is nonsense. Every point is a singleton, a finite set. Every point
that is not dark can be seen.
Regards, WM
On 06.10.2024 04:57, Richard Damon wrote:
On 10/5/24 3:04 PM, WM wrote:
On 05.10.2024 15:32, Richard Damon wrote:
On 10/5/24 8:18 AM, WM wrote:
On 05.10.2024 13:12, Richard Damon wrote:
How can bijections between infinite sets exist then.
But actual infinity doesn't exist.
Because the sets exist as what you think of as "potential" infinity,
"the integers separately as well as in their actually infinite
totality exist as eternal ideas in intellectu Divino in the highest
degree of reality." [G. Cantor, letter to C. Hermite (30 Nov 1895)]
but that isn't what your "actual infinity" is, that is just protential
infinity.
No.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.10.2024 15:59, Alan Mackenzie wrote:
Who do you think you are to accuse others of being stupid or dishonest?
I know that I have understood that topic better than the stupids. [WM]
Everybody else understands that the "stupids" are very bright indeed, and collectively understand the topic far better than any non-specialist.
On 06.10.2024 17:48, Alan Mackenzie wrote:Nor does it get shorter, it stays infinite.
WM <wolfgang.mueckenheim@tha.de> wrote:
No, even an unbounded sequence does not get longer when shifted by oneTotally irrelevant to my point. I was talking about the unboundedThis idea of time may be what misleads the mathematically less adeptWe can add 9 to 0.999...999 to obtain 9.999...999. But multiplying
into believing that 0.999... < 1.
0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves
it constant: 9.99...9990.
sequence 0.999.... You have replied about a bounded finite sequence of
9's.
step.
Bijection is not about completeness, countability is. Of course stoppingBijection means completeness. Potential infinity is never complete. But potential infinity is used in fact, best seen with Hilbert's hotel orSet theory depends on actual infinity.How would it go wrong if there were merely potential infinity?
mapping of natural numbers on even natural numbers. It is the reason why
all countable sets are countable. I actual infinity there are more
natural numbers than even natural numbers.
Then you are talking about actual infinity.No. Completeness is required. Cantor claims it: "The infinite sequenceBijections must be complete.The word "complete" is misleading when talking about infinite things.
thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place."
[G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
*between infinite sets, for youThen bijections are impossible.But Cantor's bijections never are complete. Cantor's list must beYes I agree with that last sentiment. Talking about "completely" with
completely enumerated by natural numbers. The diagonal number must be
complete such that no digit is missing in order to be distinct from
every listed real number. Impossible. All that is nonsense.
regard to infinite sets is nonsense.
It is completely infinite.It isn't even clear what you mean by saying the diagonal number must beIf it is never complete, then always more is following and the diagonal number is never excluded from the list.
"complete". It's generated by an infinite process, but remember
there's no time involved. It just is.
WM <wolfgang.mueckenheim@tha.de> wrote:
This idea of time may be what misleads the mathematically less adept
into believing that 0.999... < 1.
That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying
0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves it
constant: 9.99...9990.
Totally irrelevant to my point. I was talking about the unbounded
sequence 0.999.... You have replied about a bounded finite sequence of
9's.
Again, totally missing the point.
Set theory depends on actual infinity.
How would it go wrong if there were merely potential infinity?
Bijections must be complete.
The word "complete" is misleading when talking about infinite things.
Bijections are just as complete with "potential infinity" as with "actual infinity".
But Cantor's bijections never are complete. Cantor's list must be
completely enumerated by natural numbers. The diagonal number must be
complete such that no digit is missing in order to be distinct from
every listed real number. Impossible. All that is nonsense.
Yes I agree with that last sentiment. Talking about "completely" with
regard to infinite sets is nonsense.
It isn't even clear what you mean
by saying the diagonal number must be "complete". It's generated by an infinite process, but remember there's no time involved. It just is.
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:In potential infinity there is time or at least a sequence of steps.
"A potential infinity is a quantity which is finite but indefinitelyThat is a mistake. The ellipses indicate the enumeration. There is no
large. For instance, when we enumerate the natural numbers as 0, 1, 2, >>>> ..., n, n+1, ..., the enumeration is finite at any point in time, ....
time. If one must consider time, then the enumeration happens
instantaneously.
In actual infinity, there is no last 9 (that would not be infinite).This idea of time may be what misleads the mathematically less adeptThat is true even in actual infinity.
into believing that 0.999... < 1.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying
0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves it
constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==>
9*0.999...999 < 9 as it should be.
Am 06.10.2024 um 17:48 schrieb Alan Mackenzie:
It's generated by an infinite process
No, it isn't.
It just is.Right.
A set is a collection of well-defined objects,
meaning we must be able to determine if an element belongs to a
particulr set.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.10.2024 15:59, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
All unit fractions are separate points on
the positive real axis, but there are infinitely many for every x > 0. >>>> That can only hold for definable x, not for all.
Poppycock! You'll have to do better than that to provide such a
contradiction.
It is good enough, but you can't understand.
I do understand. I understand that what you are writing is not maths.
I'm trying to explain to you why. I've already proved that there are no "undefinable" natural numbers. So assertions about them can not make any sense.
Hint: Skilled mathematicians have worked on trying to
prove the inconsistency of maths, without success.
What shall that prove? Try to understand.
It shows that any such results are vanishingly unlikely to be found by non-specialists such as you and I.
Try only to understand my argument. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. How can >> infinitely many unit fractions appear before every x > 0?
You are getting confused with quantifiers, here. For each such x, there
is an infinite set of fractions less than x. For different x's that set varies. There is no such infinite set which appears before every x > 0.
Am Sun, 06 Oct 2024 16:15:27 +0200 schrieb WM:
On 06.10.2024 14:01, joes wrote:Interessant. ω*2 /2 ist ja eindeutig unendlich, aber ω/2 *2 nicht?
Am Sun, 06 Oct 2024 12:32:16 +0200 schrieb WM:Nein.
On 06.10.2024 11:49, Moebius wrote:Komisch auch, dass ω/2*2 (ist die Reihenfolge wichtig?)
Am 06.10.2024 um 10:40 schrieb WM:
On 06.10.2024 05:35, Moebius wrote:die Menge IN (und deren Elemente), wie schon gesagt.
Am 05.10.2024 um 22:38 schrieb WM:Das gilt für
On 05.10.2024 22:13, Moebius wrote:Doch, weil IN gegenüber der Multiplikation ABGESCHLOSSEN ist. (Das >>>>>>> kannst Du sogar in deinem Bestseller nachlesen, falls Du es
Am 05.10.2024 um 22:01 schrieb WM:Nein.
ω/2 * 2 e IN,
inzwischen vergessen haben solltest.) Mit ω/2 e IN wäre auch ω/2 * >>>>>>> 2 e IN (weil 2 e IN ist).
In Zeichen: An,m e IN: n * m e IN.
Not all infinite sets can be compared by size, but we can establish someSchwach.
useful rules.
The rule of subset proves that every proper subset has fewer elements1
than its superset. So there are more natural numbers than prime numbers,
|N| > |P|, and more complex numbers than real numbers, |C| > |R|. Even
finitely many exceptions from the subset-relation are admitted for
infinite subsets. Therefore there are more odd numbers than prime
numbers |O| > |P|.
The rule of construction yields the numbers of integers |Z| = 2|N| +
and the number of fractions |Q| = 2|N|^2 + 1 (there are fewer rationalHow small an error? Surely there is one more even number.
numbers Q# ). Since all products of rational numbers with an irrational
number are irrational, there are many more irrational numbers than
rational numbers |X| > |Q#|.
The rule of symmetry yields precisely the same number of real
geometric points in every interval (n, n+1] and with at most a small
error same number of odd numbers and of even numbers in every finite
interval and in the whole real line.
This theory makes the number of natural numbers (and of course of otherI don’t understand this argument. What is the cardinality of the set
sets too) depending on the numerical representation. The set {1, 11,
111, ...} of natural numbers has only comparatively few elements.
Therefore the set of natural numbers in unary or binary notation has
fewer, in hexadecimal notation more than |N| elements. The set {10, 20,
30, ...} has |N|/10 elements, but if the zeros are only applied as
decoration, this set, like the set {1', 2', 3', ...}, has |N| elements.
It will be a matter of future research to investigate the effect of
different numerical systems in detail.
{1, 11, 111, …} ? What of {0, 1, 10, 11, 100, …} in binary and decimal? What is a decoration different from?
On 10/6/24 7:45 AM, WM wrote:
On 06.10.2024 04:51, Richard Damon wrote:Then you are just proving yourself to be a liar, as you said you could
On 10/5/24 2:58 PM, WM wrote:
And that point CAN be found, Given two point x and y,
So why does that say it has a point next to it?
A point between both could be chosen unless it was dark.
They are not given, but dark. Discrete points on the positive axis
have a minimum.
choose them.
The actual points are not "dark" but defined.
On 10/6/24 9:42 AM, WM wrote:
On 06.10.2024 04:51, Richard Damon wrote:And every single point that exists can be found, and that includes every point on the "Number Line"
On 10/5/24 2:58 PM, WM wrote:
Every point is a finite set.
The fact that we can keep doing that indefinitely, and never reach a
point we can't continue, is proof that the concept of "next point"
doesn't exist
The concept of point however does exist. Every single point can be
found unless it is dark.
Except the "Unit Fractions" or the "Natural Numbers" those have gaps
between the elements, with an "accumulation point" at 0 where the Unit Fractions become dense there.
Nope, Set Theory can work on potential infinity.
Am Mon, 07 Oct 2024 09:41:23 +0200 schrieb WM:
On 06.10.2024 17:48, Alan Mackenzie wrote:Nor does it get shorter, it stays infinite.
even an unbounded sequence does not get longer when shifted by one
step.
Bijection is not about completeness, countability is.
Of course stopping
after a finite number, which potential infinity seems to mean, is not „complete” in that sense. Hilbert’s Hotel is actually infinite, it already holds infinite guests.
All of them can at once move to the next
room,
WM formulated the question :
On 06.10.2024 19:03, FromTheRafters wrote:
A set is a collection of well-defined objects, meaning we must be
able to determine if an element belongs to a particulr set.
But you can't determine the smallest unit fraction although it is a
singleton set, a point on the real axis.
There is no smallest unit fraction.
WM wrote :
On 06.10.2024 18:48, Alan Mackenzie wrote:
You are getting confused with quantifiers, here. For each such x, there >>> is an infinite set of fractions less than x. For different x's that set >>> varies. There is no such infinite set which appears before every x > 0. >>The set varies but infinitely many elements remain the same. A
shrinking infinite set which remains infinite has an infinite core.
Wow, your shrinking sets again? Sets don't change.
On 06.10.2024 19:11, Richard Damon wrote:„that exists”
On 10/6/24 9:42 AM, WM wrote:Find the smallest unit fraction.
On 06.10.2024 04:51, Richard Damon wrote:And every single point that exists can be found, and that includes
On 10/5/24 2:58 PM, WM wrote:The concept of point however does exist. Every single point can be
The fact that we can keep doing that indefinitely, and never reach aEvery point is a finite set.
point we can't continue, is proof that the concept of "next point"
doesn't exist
found unless it is dark.
every point on the "Number Line"
That is not the meaning of „dense”.Except the "Unit Fractions" or the "Natural Numbers" those have gapsNo. They nowhere become dense: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
between the elements, with an "accumulation point" at 0 where the Unit
Fractions become dense there.
On 06.10.2024 17:55, joes wrote:Actually infinite means infinite, which doesn’t change when you add or subtract a finite number.
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actual means all, but not more. This implies a last before ω. TheIn actual infinity, there is no last 9 (that would not be infinite).This idea of time may be what misleads the mathematically less adeptThat is true even in actual infinity.
into believing that 0.999... < 1.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying
0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves
it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==>
9*0.999...999 < 9 as it should be.
infinity means an end cannot be determined. It is produced by the dark numbers.
On 06.10.2024 19:11, Richard Damon wrote:
On 10/6/24 9:42 AM, WM wrote:
On 06.10.2024 04:51, Richard Damon wrote:And every single point that exists can be found, and that includes
On 10/5/24 2:58 PM, WM wrote:
Every point is a finite set.
The fact that we can keep doing that indefinitely, and never reach a
point we can't continue, is proof that the concept of "next point"
doesn't exist
The concept of point however does exist. Every single point can be
found unless it is dark.
every point on the "Number Line"
Find the smallest unit fraction.
Except the "Unit Fractions" or the "Natural Numbers" those have gaps
between the elements, with an "accumulation point" at 0 where the Unit
Fractions become dense there.
No. They nowhere become dense: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
Regards, WM
On 06.10.2024 19:05, Richard Damon wrote:
On 10/6/24 7:45 AM, WM wrote:
On 06.10.2024 04:51, Richard Damon wrote:Then you are just proving yourself to be a liar, as you said you could
On 10/5/24 2:58 PM, WM wrote:
And that point CAN be found, Given two point x and y,
So why does that say it has a point next to it?
A point between both could be chosen unless it was dark.
They are not given, but dark. Discrete points on the positive axis
have a minimum.
choose them.
I said: a point between both could be chosen unless it was dark!
The actual points are not "dark" but defined.
All unit fractios are such points. Each one is a singleton adhering to
the rules of geometry. Therefore each one including the smallest oe
could be found unless it was dark.
Regards, WM
On 06.10.2024 18:14, Moebius wrote:
Am 06.10.2024 um 17:48 schrieb Alan Mackenzie:
It's generated by an infinite process
No, it isn't.
That is potential infiity.
It just is.Right.
And therefore the number of elements or digits does not increase by
changing the positions.
Regards, WM
On 06.10.2024 18:48, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.10.2024 15:59, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
All unit fractions are separate points on
the positive real axis, but there are infinitely many for every x > 0. >>>>> That can only hold for definable x, not for all.
Poppycock! You'll have to do better than that to provide such a
contradiction.
It is good enough, but you can't understand.
I do understand. I understand that what you are writing is not maths.
I'm trying to explain to you why. I've already proved that there are no
"undefinable" natural numbers. So assertions about them can not make any
sense.
You have not understood that all unit fractions are separate points on
the positive axis.
Every point is a singleton set and could be seen as such, but it
cannot. Hence it is dark.
Hint: Skilled mathematicians have worked on trying to
prove the inconsistency of maths, without success.
What shall that prove? Try to understand.
It shows that any such results are vanishingly unlikely to be found by
non-specialists such as you and I.
Unlikely is not impossible.
Try only to understand my argument. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. How can
infinitely many unit fractions appear before every x > 0?
You are getting confused with quantifiers, here. For each such x, there
is an infinite set of fractions less than x. For different x's that set
varies. There is no such infinite set which appears before every x > 0.
The set varies but infinitely many elements remain the same.
A shrinking infinite set which remains infinite has an infinite core.
Regards, WM
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you add orActual means all, but not more. This implies a last before ω. TheIn actual infinity, there is no last 9 (that would not be infinite).This idea of time may be what misleads the mathematically less adept >>>>>> into believing that 0.999... < 1.That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying
0.999...999 by 10 or, what is the same, shifting the digits 9 by one >>>>> step to the left-hand side, does not increase their number but leaves >>>>> it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==> >>>>> 9*0.999...999 < 9 as it should be.
infinity means an end cannot be determined. It is produced by the dark
numbers.
subtract a finite number.
Actual infinity doesn't exist for us finite beings.
On 07.10.2024 11:36, FromTheRafters wrote:
WM formulated the question :
On 06.10.2024 19:03, FromTheRafters wrote:
A set is a collection of well-defined objects, meaning we must be
able to determine if an element belongs to a particulr set.
But you can't determine the smallest unit fraction although it is a
singleton set, a point on the real axis.
There is no smallest unit fraction.
If there are only fixed points, then there is a point such that between
it and zero there is no further point.
Regards, WM
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you add or subtract a finite number.Actual means all, but not more. This implies a last before ω. TheIn actual infinity, there is no last 9 (that would not be infinite).This idea of time may be what misleads the mathematically less adept >>>>> into believing that 0.999... < 1.That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying
0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves
it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==>
9*0.999...999 < 9 as it should be.
infinity means an end cannot be determined. It is produced by the dark
numbers.
On 06.10.2024 17:48, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
This idea of time may be what misleads the mathematically less adept
into believing that 0.999... < 1.
That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying
0.999...999 by 10 or, what is the same, shifting the digits 9 by one
step to the left-hand side, does not increase their number but leaves it >>> constant: 9.99...9990.
Totally irrelevant to my point. I was talking about the unbounded
sequence 0.999.... You have replied about a bounded finite sequence of
9's.
No, even an unbounded sequence does not get longer when shifted by one step.
Again, totally missing the point.
You don't understand that actual infinity is a fixed quantity.
Set theory depends on actual infinity.
How would it go wrong if there were merely potential infinity?
Bijection means completeness.
Potential infinity is never complete. But potential infinity is used in
fact, best seen with Hilbert's hotel or mapping of natural numbers on
even natural numbers. It is the reason why all countable sets are
countable. In actual infinity there are more natural numbers than even natural numbers.
Bijections are just as complete with "potential infinity" as with "actual
infinity".
No, that is wrong.
But Cantor's bijections never are complete. Cantor's list must be
completely enumerated by natural numbers. The diagonal number must be
complete such that no digit is missing in order to be distinct from
every listed real number.
Impossible. All that is nonsense.
Yes I agree with that last sentiment. Talking about "completely" with
regard to infinite sets is nonsense.
Then bijections are impossible.
It isn't even clear what you mean
by saying the diagonal number must be "complete". It's generated by an
infinite process, but remember there's no time involved. It just is.
If it is never complete, then always more is following and the diagonal number is never excluded from the list.
Regards, WM
On 06.10.2024 18:48, Alan Mackenzie wrote:Why can some points not be „seen” as a singleton set?
WM <wolfgang.mueckenheim@tha.de> wrote:You have not understood that all unit fractions are separate points on
On 06.10.2024 15:59, Alan Mackenzie wrote:I do understand. I understand that what you are writing is not maths.
WM <wolfgang.mueckenheim@tha.de> wrote:
It is good enough, but you can't understand.All unit fractions are separate points on the positive real axis,Poppycock! You'll have to do better than that to provide such a
but there are infinitely many for every x > 0.
That can only hold for definable x, not for all.
contradiction.
I'm trying to explain to you why. I've already proved that there are
no "undefinable" natural numbers. So assertions about them can not
make any sense.
the positive axis. Every point is a singleton set and could be seen as
such, but it cannot. Hence it is dark.
Nothing is impossible…Unlikely is not impossible.It shows that any such results are vanishingly unlikely to be found byHint: Skilled mathematicians have worked on trying toWhat shall that prove? Try to understand.
prove the inconsistency of maths, without success.
non-specialists such as you and I.
Here is your essential misunderstanding: there is no mysterious SomethingThe set varies but infinitely many elements remain the same. A shrinking infinite set which remains infinite has an infinite core.Try only to understand my argument. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. How canYou are getting confused with quantifiers, here. For each such x,
infinitely many unit fractions appear before every x > 0?
there is an infinite set of fractions less than x. For different x's
that set varies. There is no such infinite set which appears before
every x > 0.
On 06.10.2024 19:03, FromTheRafters wrote:The proof of its existence you leave us wanting.
A set is a collection of well-defined objects,But you can't determine the smallest unit fraction although it is a
meaning we must be able to determine if an element belongs to a
particulr set.
singleton set, a point on the real axis.
On 07.10.2024 11:36, FromTheRafters wrote:How do you imagine that? It has a finite distance from 0.
WM formulated the question :If there are only fixed points, then there is a point such that between
On 06.10.2024 19:03, FromTheRafters wrote:There is no smallest unit fraction.
A set is a collection of well-defined objects, meaning we must be
able to determine if an element belongs to a particulr set.
But you can't determine the smallest unit fraction although it is a
singleton set, a point on the real axis.
it and zero there is no further point.
Alan Mackenzie wrote :
The idea of one countable set being "bigger" than another countable set is >> simply nonsense.
Oops. Finite sets are countable too. :)
On 07.10.2024 10:05, joes wrote:„More” being a different kind of infinity, namely at least uncountable.
Am Mon, 07 Oct 2024 09:41:23 +0200 schrieb WM:It keeps all its elements but not more.
On 06.10.2024 17:48, Alan Mackenzie wrote:Nor does it get shorter, it stays infinite.
even an unbounded sequence does not get longer when shifted by one
step.
Exactly, which is the case for every finite subset.Bijection is not about completeness, countability is.Bijective means injective and surjective.
It can, if I begin numbering with 2. The cardinality of N\{1} can’t be finite.Of course stopping after a finite number, which potential infinityName them by all the natural numbers. Then no further guest can appear.
seems to mean, is not „complete” in that sense. Hilbert’s Hotel is
actually infinite, it already holds infinite guests.
Huh? They are all fixed, we can move them „rigidly”.All of them can at once move to the next room,All rooms are enumerated by all the natural numbers. Hence there is no
chance to move. Only in potential infinity there is.
On 05.10.2024 20:12, Jim Burns wrote:
On 10/5/2024 5:43 AM, WM wrote:
[In] many cases it is correct.
"Many cases" is insufficient when
the argument requires "all cases".
My argument requires only one case,
My argument requires only one case,
My argument requires only one case,
best demonstrated with endsegments E(n).
The intersection of all *infinite* endsegments
is infinite,
because
they all contain the same natural numbers which
have not yet been eliminated by
the process E(n+1) = E(n) \ {n}.
On 10/7/2024 4:17 AM, Richard Damon wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:Actually infinite means infinite, which doesn’t change when you add or >>> subtract a finite number.
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actual means all, but not more. This implies a last before ω. TheIn actual infinity, there is no last 9 (that would not be infinite).This idea of time may be what misleads the mathematically less adept >>>>>>> into believing that 0.999... < 1.That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying >>>>>> 0.999...999 by 10 or, what is the same, shifting the digits 9 by one >>>>>> step to the left-hand side, does not increase their number but leaves >>>>>> it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==> >>>>>> 9*0.999...999 < 9 as it should be.
infinity means an end cannot be determined. It is produced by the dark >>>> numbers.
Actual infinity doesn't exist for us finite beings.
What do you mean? Are you trying to go the WM route where there is a
largest natural number... We just can't see it yet because it's dark? Oh shit.
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you add or >>> subtract a finite number.Actual means all, but not more. This implies a last before ω. TheIn actual infinity, there is no last 9 (that would not be infinite).This idea of time may be what misleads the mathematically less adept >>>>>>> into believing that 0.999... < 1.That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying >>>>>> 0.999...999 by 10 or, what is the same, shifting the digits 9 by one >>>>>> step to the left-hand side, does not increase their number but leaves >>>>>> it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==> >>>>>> 9*0.999...999 < 9 as it should be.
infinity means an end cannot be determined. It is produced by the dark >>>> numbers.
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not attached to
the word "infinite", it meant something like "This is really true:".
Anyhow, what do you mean when you say that "actual infinity doesn't
exist"? I think we established over the weekend that for a mathematical entity not to exist, it must cause a contradiction. Or something like
that.
So what contradiction would the existence of actual infinity cause?
On 10/7/2024 4:20 AM, Richard Damon wrote:
On 10/7/24 5:51 AM, WM wrote:
On 07.10.2024 11:36, FromTheRafters wrote:
WM formulated the question :
On 06.10.2024 19:03, FromTheRafters wrote:
A set is a collection of well-defined objects, meaning we must be
able to determine if an element belongs to a particulr set.
But you can't determine the smallest unit fraction although it is a
singleton set, a point on the real axis.
There is no smallest unit fraction.
If there are only fixed points, then there is a point such that
between it and zero there is no further point.
Regards, WM
Nope, not if you have an INFINITE set of fixed points.
The problem is we can't have an infinite set of fixed points, as we
are finite.
If each one of these is "fixed" in your line of thinking:
1/1, 1/2, 1/3, 1/4, 1/5, ...
Well, there are infinitely many of them... ;^)
So, your "actual infinity" is something beyond what we can have, so it
doesn't exist for us, and logic that assumes it is just breaks.
On 10/7/2024 4:01 PM, Richard Damon wrote:
On 10/7/24 4:27 PM, Chris M. Thomasson wrote:
On 10/7/2024 4:17 AM, Richard Damon wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:Actually infinite means infinite, which doesn’t change when you add or >>>>> subtract a finite number.
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
darkIn actual infinity, there is no last 9 (that would not be infinite). >>>>>> Actual means all, but not more. This implies a last before ω. The >>>>>> infinity means an end cannot be determined. It is produced by theThis idea of time may be what misleads the mathematically less >>>>>>>>> adeptThat is true even in actual infinity.
into believing that 0.999... < 1.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying >>>>>>>> 0.999...999 by 10 or, what is the same, shifting the digits 9 by >>>>>>>> one
step to the left-hand side, does not increase their number but >>>>>>>> leaves
it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 >>>>>>>> ==>
9*0.999...999 < 9 as it should be.
numbers.
Actual infinity doesn't exist for us finite beings.
What do you mean? Are you trying to go the WM route where there is a
largest natural number... We just can't see it yet because it's dark?
Oh shit.
I am allowing that an INFINITE being MIGHT be able to comprehend
something like an actual infinity. But this can not possibly be done
by a finite being.
Well, us as finite beings know that there is not a largest natural
number... That right there is a basic understanding of the infinite:
Fair enough?
On 10/7/2024 3:58 PM, Richard Damon wrote:
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you add or >>>>> subtract a finite number.darkIn actual infinity, there is no last 9 (that would not be infinite). >>>>>> Actual means all, but not more. This implies a last before ω. The >>>>>> infinity means an end cannot be determined. It is produced by theThis idea of time may be what misleads the mathematically less >>>>>>>>> adeptThat is true even in actual infinity.
into believing that 0.999... < 1.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying >>>>>>>> 0.999...999 by 10 or, what is the same, shifting the digits 9 by >>>>>>>> one
step to the left-hand side, does not increase their number but >>>>>>>> leaves
it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 >>>>>>>> ==>
9*0.999...999 < 9 as it should be.
numbers.
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not
attached to
the word "infinite", it meant something like "This is really true:".
But all his reference to the word "Actually" are part of his trying to
define the term "Actual Infinity".
Anyhow, what do you mean when you say that "actual infinity doesn't
exist"? I think we established over the weekend that for a mathematical >>> entity not to exist, it must cause a contradiction. Or something like
that.
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number or
unit fraction for one (at least the way WM uses it).
A first unit fraction would be 1/1? :^) However this is in the very
_strict_ realm of unit fractions. WM seems to like to mix and match
realms. Not sure why he thinks there is a smallest unit fraction...
Anyway, shit happens.
On 10/7/2024 4:08 PM, Richard Damon wrote:
On 10/7/24 4:05 PM, Chris M. Thomasson wrote:
On 10/7/2024 4:20 AM, Richard Damon wrote:
On 10/7/24 5:51 AM, WM wrote:
On 07.10.2024 11:36, FromTheRafters wrote:
WM formulated the question :
On 06.10.2024 19:03, FromTheRafters wrote:
A set is a collection of well-defined objects, meaning we must >>>>>>>> be able to determine if an element belongs to a particulr set.
But you can't determine the smallest unit fraction although it is >>>>>>> a singleton set, a point on the real axis.
There is no smallest unit fraction.
If there are only fixed points, then there is a point such that
between it and zero there is no further point.
Regards, WM
Nope, not if you have an INFINITE set of fixed points.
The problem is we can't have an infinite set of fixed points, as we
are finite.
If each one of these is "fixed" in your line of thinking:
1/1, 1/2, 1/3, 1/4, 1/5, ...
Well, there are infinitely many of them... ;^)
The problem with his "Actual infinity" is you need to get rid of
the ... (as that is generative) and put in the full list of the number.
Because you have a full list of the numbers, he think that means there
is a last one to that list.
Odd to me! He must think 1/6 is dark wrt the list, right? Am I getting
closer to WM's strange way of thinking? Well, 1/6 is not so dark anymore because I wrote it here... ;^) Wow.
The problem is that the full list can not be presented to a finite
being, and thus, that "Actual infinity" can't be comprehended by it.
A fill list:
all of the natural numbers.
That is a finite term for all of them. WM, well, he might explode?
You need to b infinite, to actually have a fully generated infinite.
We can handle infinities by generative processes (his potential
infinity) because we don't need to actually get to the end, we just
know the process can continue forever and there will be no end, except
after infinite work is done.
So, your "actual infinity" is something beyond what we can have, so
it doesn't exist for us, and logic that assumes it is just breaks.
On 10/7/2024 4:17 AM, Richard Damon wrote:
Actual infinity doesn't exist for us finite beings.
It exists in our math...
On 10/7/2024 3:58 PM, Richard Damon wrote:at.
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number or
unit fraction
On 10/7/2024 4:01 PM, Richard Damon wrote:
I am allowing that an INFINITE being MIGHT be able to comprehend
something like an actual infinity. But this can not possibly be done
by a finite being.
Well, us as finite beings know that there is not a largest natural
number... That right there is a basic understanding of the infinite:
Fair enough?
Am 08.10.2024 um 02:38 schrieb Chris M. Thomasson:
On 10/7/2024 4:01 PM, Richard Damon wrote:
I am allowing that an INFINITE being MIGHT be able to comprehend
something like an actual infinity. But this can not possibly be done
by a finite being.
It can.
This idiot should read Peter Suber's Infinite Reflections:
"Conclusion
Properly understood, the idea of a completed infinity is no longer a
problem in mathematics or philosophy. It is perfectly intelligible and coherent. Perhaps it cannot be imagined but it can be conceived; it is
not reserved for infinite omniscience, but knowable by finite humanity;
it may contradict intuition, but it does not contradict itself. To
conceive it adequately we need not enumerate or visualize infinitely
many objects, but merely understand self-nesting. We have an actual,
positive idea of it, or at least with training we can have one; we are
not limited to the idea of finitude and its negation. In fact, it is at
least as plausible to think that we understand finitude as the negation
of infinitude as the other way around. The world of the infinite is not barred to exploration by the equivalent of sea monsters and tempests; it
is barred by the equivalent of motion sickness. The world of the
infinite is already open for exploration, but to embark we must unlearn
our finitistic intuitions which instill fear and confusion by making
some consistent and demonstrable results about the infinite literally counter-intuitive. Exploration itself will create an alternative set of intuitions which make us more susceptible to the feeling which Kant
called the sublime. Longer acquaintance will confirm Spinoza's
conclusion that the secret of joy is to love something infinite."
Source: http://legacy.earlham.edu/~peters/writing/infinity.htm
Well, us as finite beings know that there is not a largest natural number... That right there is a basic understanding of the infinite:
Fair enough?
Right.
Am 08.10.2024 um 09:29 schrieb Moebius:
Am 08.10.2024 um 02:38 schrieb Chris M. Thomasson:
On 10/7/2024 4:01 PM, Richard Damon wrote:
I am allowing that an INFINITE being MIGHT be able to comprehend
something like an actual infinity. But this can not possibly be done
by a finite being.
It can.
This idiot should read Peter Suber's Infinite Reflections:
"Perhaps it cannot be imagined but it can be conceived; it is not
reserved for infinite omniscience, but knowable by finite humanity..."
"Conclusion
Properly understood, the idea of a completed infinity is no longer a
problem in mathematics or philosophy. It is perfectly intelligible and
coherent. Perhaps it cannot be imagined but it can be conceived; it is
not reserved for infinite omniscience, but knowable by finite
humanity; it may contradict intuition, but it does not contradict
itself. To conceive it adequately we need not enumerate or visualize
infinitely many objects, but merely understand self-nesting. We have
an actual, positive idea of it, or at least with training we can have
one; we are not limited to the idea of finitude and its negation. In
fact, it is at least as plausible to think that we understand finitude
as the negation of infinitude as the other way around. The world of
the infinite is not barred to exploration by the equivalent of sea
monsters and tempests; it is barred by the equivalent of motion
sickness. The world of the infinite is already open for exploration,
but to embark we must unlearn our finitistic intuitions which instill
fear and confusion by making some consistent and demonstrable results
about the infinite literally counter-intuitive. Exploration itself
will create an alternative set of intuitions which make us more
susceptible to the feeling which Kant called the sublime. Longer
acquaintance will confirm Spinoza's conclusion that the secret of joy
is to love something infinite."
Source: http://legacy.earlham.edu/~peters/writing/infinity.htm
Well, us as finite beings know that there is not a largest natural
number... That right there is a basic understanding of the infinite:
Fair enough?
On 10/7/24 4:53 AM, WM wrote:
Nope, the problem is you "rule of geometry" that there must be aThe actual points are not "dark" but defined.
All unit fractios are such points. Each one is a singleton adhering to
the rules of geometry. Therefore each one including the smallest oe
could be found unless it was dark.
smallest (that isn't 0) doesn't actually exist.
Note, "Geometry" says nothing about "adjacent points",
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
Actual means all, but not more. This implies a last before ω. TheActually infinite means infinite, which doesn’t change when you add or subtract a finite number.
infinity means an end cannot be determined. It is produced by the dark
numbers.
On 10/7/24 4:56 AM, WM wrote:
On 06.10.2024 19:11, Richard Damon wrote:
On 10/6/24 9:42 AM, WM wrote:
On 06.10.2024 04:51, Richard Damon wrote:And every single point that exists can be found, and that includes
On 10/5/24 2:58 PM, WM wrote:
Every point is a finite set.
The fact that we can keep doing that indefinitely, and never reach
a point we can't continue, is proof that the concept of "next
point" doesn't exist
The concept of point however does exist. Every single point can be
found unless it is dark.
every point on the "Number Line"
Find the smallest unit fraction.
A concept that doesn't exist.
Except the "Unit Fractions" or the "Natural Numbers" those have gaps
between the elements, with an "accumulation point" at 0 where the
Unit Fractions become dense there.
No. They nowhere become dense: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
So, what is the maximum density of unit fractions?
The fact that 1/n - 1/(n+1) > 0 just means that 1/n > 1/(n+1) and thus
there is always a unit fraction smaller than the one you have.
The is no lower limit to the value of 1/n - 1/(n+1) except 0
On 10/7/24 5:51 AM, WM wrote:
On 07.10.2024 11:36, FromTheRafters wrote:
WM formulated the question :
On 06.10.2024 19:03, FromTheRafters wrote:
A set is a collection of well-defined objects, meaning we must be
able to determine if an element belongs to a particulr set.
But you can't determine the smallest unit fraction although it is a
singleton set, a point on the real axis.
There is no smallest unit fraction.
If there are only fixed points, then there is a point such that
between it and zero there is no further point.
Nope, not if you have an INFINITE set of fixed points.
The problem is we can't have an infinite set of fixed points, as we are finite.
So, your "actual infinity" is something beyond what we can have, so it doesn't exist for us, and logic that assumes it is just breaks.
WM <wolfgang.mueckenheim@tha.de> wrote:
You have not understood that all unit fractions are separate points on
the positive axis.
I understand that full well.
A shrinking infinite set which remains infinite has an infinite core.
Again, no. There is no such thing as a "core", here. Each of these sets
has an infinitude of elements. No element is in all of these sets.
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you add or >>>> subtract a finite number.infinity means an end cannot be determined. It is produced by the dark >>>>> numbers.In actual infinity, there is no last 9 (that would not be infinite). >>>>> Actual means all, but not more. This implies a last before ω. TheThis idea of time may be what misleads the mathematically less adept >>>>>>>> into believing that 0.999... < 1.That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying >>>>>>> 0.999...999 by 10 or, what is the same, shifting the digits 9 by one >>>>>>> step to the left-hand side, does not increase their number but leaves >>>>>>> it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==> >>>>>>> 9*0.999...999 < 9 as it should be.
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not attached to
the word "infinite", it meant something like "This is really true:".
But all his reference to the word "Actually" are part of his trying to define the term "Actual Infinity".
Anyhow, what do you mean when you say that "actual infinity doesn't
exist"? I think we established over the weekend that for a mathematical
entity not to exist, it must cause a contradiction. Or something like
that.
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number or
unit fraction for one (at least the way WM uses it).
WM <wolfgang.mueckenheim@tha.de> wrote:
No, even an unbounded sequence does not get longer when shifted by one step.
The concept of "length" appropriate for finite sets doesn't apply to
infinite sets.
infinite means "without end" - unendlich.
You don't understand that actual infinity is a fixed quantity.
It may be "fixed" whatever that might mean, but to regard it as a
"quantity" is more than questionable.
Bijection means completeness.
No. Bijection just means a 1-1 correspondence between the elements of
two sets. > Nowhere in that definition is any mention of completeness.
The diagonal number _is_ complete. Remember, we are not constructing it tiringly one digit per second, or anything like that. The entire number
is defined and simply exists.
Am 08.10.2024 um 09:29 schrieb Moebius:
Properly understood, the idea of a completed infinity is no longer a
problem in mathematics or philosophy. It is perfectly intelligible and
coherent.
It is infinite if it is not finite.
On 07.10.2024 14:11, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
You have not understood that all unit fractions are separate points on
the positive axis.
I understand that full well.
Then you understand that every point, if existing, is independent of the others.
All unit fractions are points with uncounably many points between each
pair.
Hence all must be visible including the point next to zero, but they
are not.
A shrinking infinite set which remains infinite has an infinite core.
Again, no. There is no such thing as a "core", here. Each of these sets
has an infinitude of elements. No element is in all of these sets.
Try to think better. A function of sets which are losing some elements
but remain infinite, have the same infinite core.
That argument is absolutely definite, a logical necessity. If you
cannot understand it, ....
.... then it is useless to continue this discussion.
Regards, WM
What I should have
written (WM please take note) is:
The idea of one countably infinite set being "bigger" than another
countably infinite set is simply nonsense.
On 10/5/2024 3:33 PM, WM wrote:
On 05.10.2024 20:12, Jim Burns wrote:
On 10/5/2024 5:43 AM, WM wrote:
[In] many cases it is correct.
"Many cases" is insufficient when
the argument requires "all cases".
My argument requires only one case,
Your argument,
in order to _be an argument_
needs to _show_ its result is true.
My argument requires only one case,
best demonstrated with endsegments E(n).
The intersection of all *infinite* endsegments
is infinite, because
they all contain the same natural numbers which
have not yet been eliminated by
the process E(n+1) = E(n) \ {n}.
The intersection of all end.segments is
the set of natural numbers in each end.segment.
Each natural number is not.in one or more end.segment.
That creates a problem (but for only you).
Your attempt to fix the problem involves
a change of the definition of intersection, or
of natural number, or of something else.
Changing what you think "infinite" means
from what we mean by "very large finite"
to what we mean by "infinite"
Am Mon, 07 Oct 2024 10:47:25 +0200 schrieb WM:
The set varies but infinitely many elements remain the same. A shrinkingHere is your essential misunderstanding: there is no mysterious Something that makes a set infinite. It is infinite because it is not finite, has
infinite set which remains infinite has an infinite core.
no natural number as its size.
WM <wolfgang.mueckenheim@tha.de> wrote:
All unit fractions are points with uncounably many points between each
pair.
Yes, OK.
Hence all must be visible including the point next to zero, but they
are not.
There is no point next to zero.
A shrinking infinite set which remains infinite has an infinite core.
Again, no. There is no such thing as a "core", here. Each of these sets >>> has an infinitude of elements. No element is in all of these sets.
Try to think better. A function of sets which are losing some elements
but remain infinite, have the same infinite core.
That is untrue. For any element which you assert is in the "core", I
can give one of these sets which does not contain that element.
The
"core" is thus empty.
That argument is absolutely definite, a logical necessity. If you
cannot understand it, ....
It wasn't an argument, it was a bare statement, devoid of any supporting argument.
I understand it full well, and I understand that it's
mistaken.
Am Mon, 07 Oct 2024 11:08:33 +0200 schrieb WM:
On 07.10.2024 10:05, joes wrote:„More” being a different kind of infinity, namely at least uncountable.
Am Mon, 07 Oct 2024 09:41:23 +0200 schrieb WM:It keeps all its elements but not more.
On 06.10.2024 17:48, Alan Mackenzie wrote:Nor does it get shorter, it stays infinite.
even an unbounded sequence does not get longer when shifted by one
step.
All ω+k are equally infinite.
Exactly, which is the case for every finite subset.Bijection is not about completeness, countability is.Bijective means injective and surjective.
It can, if I begin numbering with 2. The cardinality of N\{1} can’t be finite.Of course stopping after a finite number, which potential infinityName them by all the natural numbers. Then no further guest can appear.
seems to mean, is not „complete” in that sense. Hilbert’s Hotel is >>> actually infinite, it already holds infinite guests.
Huh? They are all fixed, we can move them „rigidly”.All of them can at once move to the next room,All rooms are enumerated by all the natural numbers. Hence there is no
chance to move. Only in potential infinity there is.
Am Mon, 07 Oct 2024 11:51:43 +0200 schrieb WM:
How do you imagine that?There is no smallest unit fraction.If there are only fixed points, then there is a point such that between
it and zero there is no further point.
It has a finite distance from 0.
His view is that if the values of the existing unit fractions exist, and
we travel up the line from the negative side in increasing value, it
only make sense that there should be a "first" point we reach.
The concept, on the face of it, seems logical,
it just is a fact that it
doesn't work, and that is in part because of some of the strange
properties that trying to imagine a realized infinity creates.
On 07.10.2024 13:08, Richard Damon wrote:
On 10/7/24 4:53 AM, WM wrote:
Nope, the problem is you "rule of geometry" that there must be aThe actual points are not "dark" but defined.
All unit fractios are such points. Each one is a singleton adhering
to the rules of geometry. Therefore each one including the smallest
oe could be found unless it was dark.
smallest (that isn't 0) doesn't actually exist.
The rule of geometry is that every point exists.
Note, "Geometry" says nothing about "adjacent points",
So it is. It simply says that a single point exists or does not exist, independent of any other point. Hence the first existing point can be
seen unless it is dark.
Regards, WM
On 08.10.2024 01:26, FromTheRafters wrote:
It is infinite if it is not finite.
But does it go on and on or is it fixed and completed?
Regards, WM
On 07.10.2024 18:11, Alan Mackenzie wrote:
What I should have
written (WM please take note) is:
The idea of one countably infinite set being "bigger" than another
countably infinite set is simply nonsense.
The idea is supported by the fact that set A as a superset of set B is
bigger than B. Simply nonsense is the claim that there are as many
algebraic numbers as prime numbers. For Cantor's enumeration of all
fractions I have given a simple disproof.
Regards, WM
On 07.10.2024 15:19, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
No, even an unbounded sequence does not get longer when shifted by
one step.
The concept of "length" appropriate for finite sets doesn't apply to
infinite sets.
It is the concept of number of elements. It is appropriate in actual infinity.
infinite means "without end" - unendlich.
Actual infinity means complete. That implies a fixed number.
You don't understand that actual infinity is a fixed quantity.
It may be "fixed" whatever that might mean, but to regard it as a
"quantity" is more than questionable.
Fixed means that no element can be added and no element can be lost. The number of nines is fixed. That is an assumption only, but necessary for bijections.
Bijection means completeness.
No. Bijection just means a 1-1 correspondence between the elements of
two sets. > Nowhere in that definition is any mention of completeness.
It also means surjectivity. The preimage is complete by definition.
The diagonal number _is_ complete. Remember, we are not constructing it
tiringly one digit per second, or anything like that. The entire number
is defined and simply exists.
Only when the set of indexes is complete.
Regards, WM
On 07.10.2024 13:12, Richard Damon wrote:
On 10/7/24 4:56 AM, WM wrote:
On 06.10.2024 19:11, Richard Damon wrote:
On 10/6/24 9:42 AM, WM wrote:
On 06.10.2024 04:51, Richard Damon wrote:And every single point that exists can be found, and that includes
On 10/5/24 2:58 PM, WM wrote:
Every point is a finite set.
The fact that we can keep doing that indefinitely, and never reach >>>>>> a point we can't continue, is proof that the concept of "next
point" doesn't exist
The concept of point however does exist. Every single point can be
found unless it is dark.
every point on the "Number Line"
Find the smallest unit fraction.
A concept that doesn't exist.
Except the "Unit Fractions" or the "Natural Numbers" those have gaps
between the elements, with an "accumulation point" at 0 where the
Unit Fractions become dense there.
No. They nowhere become dense: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
So, what is the maximum density of unit fractions?
It is zero because every unit fraction is a point, every gap has
uncountable may points.
The fact that 1/n - 1/(n+1) > 0 just means that 1/n > 1/(n+1) and thus
there is always a unit fraction smaller than the one you have.
The is no lower limit to the value of 1/n - 1/(n+1) except 0
This limit is what you can see because you cannot see the reality:
Between all points there are giant gaps of uncountable many points.
Regards, WM
On 10/7/2024 5:53 PM, Richard Damon wrote:
On 10/7/24 8:35 PM, Chris M. Thomasson wrote:
On 10/7/2024 3:58 PM, Richard Damon wrote:
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you >>>>>>> add orActual means all, but not more. This implies a last before ω. The >>>>>>>> infinity means an end cannot be determined. It is produced byIn actual infinity, there is no last 9 (that would not beThis idea of time may be what misleads the mathematically >>>>>>>>>>> less adeptThat is true even in actual infinity.
into believing that 0.999... < 1.
We can add 9 to 0.999...999 to obtain 9.999...999. But
multiplying
0.999...999 by 10 or, what is the same, shifting the digits 9 >>>>>>>>>> by one
step to the left-hand side, does not increase their number but >>>>>>>>>> leaves
it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 +
0.999...999 ==>
9*0.999...999 < 9 as it should be.
infinite).
the dark
numbers.
subtract a finite number.
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not
attached to
the word "infinite", it meant something like "This is really true:".
But all his reference to the word "Actually" are part of his trying
to define the term "Actual Infinity".
Anyhow, what do you mean when you say that "actual infinity doesn't
exist"? I think we established over the weekend that for a
mathematical
entity not to exist, it must cause a contradiction. Or something like >>>>> that.
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number
or unit fraction for one (at least the way WM uses it).
A first unit fraction would be 1/1? :^) However this is in the very
_strict_ realm of unit fractions. WM seems to like to mix and match
realms. Not sure why he thinks there is a smallest unit fraction...
Anyway, shit happens.
No, because he IS specifying the direction of counting, and is trying
to count from the "end" that doesn't have an end.
Very strange. Since there is no smallest unit fraction, trying to count
from the "smallest" up is moronic to me. Humm...
His view is that if the values of the existing unit fractions exist,
and we travel up the line from the negative side in increasing value,
it on.y make sense that there should be a "first" point we reach.
The concept, on the face of it, seems logical, it just is a fact that
it doesn't work, and that is in part because of some of the strange
properties that trying to imagine a realized infinity creates.
On 08.10.2024 02:53, Richard Damon wrote:
His view is that if the values of the existing unit fractions exist,
and we travel up the line from the negative side in increasing value,
it only make sense that there should be a "first" point we reach.
The concept, on the face of it, seems logical,
of course, it is logical.
it just is a fact that it doesn't work, and that is in part because of
some of the strange properties that trying to imagine a realized
infinity creates.
This property is darkness. Without dark elements actual or realized
infinity cannot exist.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.10.2024 14:11, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
A shrinking infinite set which remains infinite has an infinite core.
Again, no. There is no such thing as a "core", here. Each of these sets >>> has an infinitude of elements. No element is in all of these sets.
Try to think better. A function of sets which are losing some elements
but remain infinite, have the same infinite core.
That is untrue. For any element which you assert is in the "core", I
can give one of these sets which does not contain that element. The
"core" is thus empty.
That argument is absolutely definite, a logical necessity.
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you add or >>>>> subtract a finite number.In actual infinity, there is no last 9 (that would not be infinite). >>>>>> Actual means all, but not more. This implies a last before ω. The >>>>>> infinity means an end cannot be determined. It is produced by the dark >>>>>> numbers.This idea of time may be what misleads the mathematically less adept >>>>>>>>> into believing that 0.999... < 1.That is true even in actual infinity.
We can add 9 to 0.999...999 to obtain 9.999...999. But multiplying >>>>>>>> 0.999...999 by 10 or, what is the same, shifting the digits 9 by one >>>>>>>> step to the left-hand side, does not increase their number but leaves >>>>>>>> it constant: 9.99...9990.
10*0.999...999 = 9.99...9990 = 9 + 0.99...9990 < 9 + 0.999...999 ==> >>>>>>>> 9*0.999...999 < 9 as it should be.
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not attached to >>> the word "infinite", it meant something like "This is really true:".
But all his reference to the word "Actually" are part of his trying to
define the term "Actual Infinity".
OK, maybe you're right, there. The semantics are a bit ambiguous. Joes
is not a native English speaker. Apologies to Joes.
Anyhow, what do you mean when you say that "actual infinity doesn't
exist"? I think we established over the weekend that for a mathematical >>> entity not to exist, it must cause a contradiction. Or something like
that.
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number or
unit fraction for one (at least the way WM uses it).
Whoa! There're rather a lot of argument steps missing there. Just
because WM asserts the existence of both actual infinity and a first
strictly positive unit fraction doesn't mean the one implies the other.
Am 08.10.2024 um 02:38 schrieb Chris M. Thomasson:
On 10/7/2024 4:01 PM, Richard Damon wrote:
I am allowing that an INFINITE being MIGHT be able to comprehend
something like an actual infinity. But this can not possibly be done
by a finite being.
It can.
This idiot should read Peter Suber's Infinite Reflections:
"Conclusion
Properly understood, the idea of a completed infinity is no longer a
problem in mathematics or philosophy. It is perfectly intelligible and coherent. Perhaps it cannot be imagined but it can be conceived; it is
not reserved for infinite omniscience, but knowable by finite humanity;
it may contradict intuition, but it does not contradict itself. To
conceive it adequately we need not enumerate or visualize infinitely
many objects, but merely understand self-nesting. We have an actual,
positive idea of it, or at least with training we can have one; we are
not limited to the idea of finitude and its negation. In fact, it is at
least as plausible to think that we understand finitude as the negation
of infinitude as the other way around. The world of the infinite is not barred to exploration by the equivalent of sea monsters and tempests; it
is barred by the equivalent of motion sickness. The world of the
infinite is already open for exploration, but to embark we must unlearn
our finitistic intuitions which instill fear and confusion by making
some consistent and demonstrable results about the infinite literally counter-intuitive. Exploration itself will create an alternative set of intuitions which make us more susceptible to the feeling which Kant
called the sublime. Longer acquaintance will confirm Spinoza's
conclusion that the secret of joy is to love something infinite."
Source: http://legacy.earlham.edu/~peters/writing/infinity.htm
Well, us as finite beings know that there is not a largest natural number... That right there is a basic understanding of the infinite:
Fair enough?
Right.
On 10/8/24 5:42 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when youActual means all, but not more. This implies a last before ω.In actual infinity, there is no last 9 (that would not beThis idea of time may be what misleads the mathematicallyThat is true even in actual infinity. We can add 9 to
less adept into believing that 0.999... < 1.
0.999...999 to obtain 9.999...999. But multiplying 0.999...999 >>>>>>>>> by 10 or, what is the same, shifting the digits 9 by one step >>>>>>>>> to the left-hand side, does not increase their number but
leaves it constant: 9.99...9990. 10*0.999...999 = 9.99...9990 >>>>>>>>> = 9 + 0.99...9990 < 9 + 0.999...999 ==> 9*0.999...999 < 9 as >>>>>>>>> it should be.
infinite).
The infinity means an end cannot be determined. It is produced
by the dark numbers.
add or subtract a finite number.
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not attached to >>>> the word "infinite", it meant something like "This is really true:".
But all his reference to the word "Actually" are part of his trying to
define the term "Actual Infinity".
OK, maybe you're right, there. The semantics are a bit ambiguous. Joes
is not a native English speaker. Apologies to Joes.
Anyhow, what do you mean when you say that "actual infinity doesn't
exist"? I think we established over the weekend that for a mathematical >>>> entity not to exist, it must cause a contradiction. Or something like >>>> that.
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number or
unit fraction for one (at least the way WM uses it).
Whoa! There're rather a lot of argument steps missing there. Just
because WM asserts the existence of both actual infinity and a first
strictly positive unit fraction doesn't mean the one implies the other.
It does in his logic, which is all that matters to him.
Yes, it is a wrong conclusion, but that error is based on his initial assumption that something could be used that isn't available as a understandable entity to us finite beings.
On 07.10.2024 17:18, joes wrote:Only according to your broken concept of cardinality, by which N u {a}
Am Mon, 07 Oct 2024 11:08:33 +0200 schrieb WM:Nonsense.
On 07.10.2024 10:05, joes wrote:„More” being a different kind of infinity, namely at least uncountable.
Am Mon, 07 Oct 2024 09:41:23 +0200 schrieb WM:It keeps all its elements but not more.
On 06.10.2024 17:48, Alan Mackenzie wrote:Nor does it get shorter, it stays infinite.
even an unbounded sequence does not get longer when shifted by one
step.
I believe it is called the order type.All ω+k are equally infinite.Nonsense.
Isn’t N\{1} finite? It has ω-1 elements.Bijection is not about completeness, countability is.
Cardinality is nonsense.It can, if I begin numbering with 2. The cardinality of N\{1} can’t beOf course stopping after a finite number, which potential infinityName them by all the natural numbers. Then no further guest can
seems to mean, is not „complete” in that sense. Hilbert’s Hotel is >>>> actually infinite, it already holds infinite guests.
appear.
finite.
You have not indicated what this notation means. Where does the zero comeThat shows my point. Infinite sets can be moved. 0.999...999 moved gives 9.99...9990.Huh? They are all fixed, we can move them „rigidly”.All of them can at once move to the next room,All rooms are enumerated by all the natural numbers. Hence there is no
chance to move. Only in potential infinity there is.
Another point is this: [0, 1) moved gives (0, 1].Can you generalise this?
On 07.10.2024 17:19, joes wrote:Why is there no closer point?
Am Mon, 07 Oct 2024 11:51:43 +0200 schrieb WM:
One of discrete points is always next to zero.How do you imagine that?There is no smallest unit fraction.If there are only fixed points, then there is a point such that
between it and zero there is no further point.
What does this mean, it cannot be known?It has a finite distance from 0.Of course, but this point cannot be found.
On 07.10.2024 17:10, joes wrote:Because naturals are not infinite.
Am Mon, 07 Oct 2024 10:47:25 +0200 schrieb WM:
Why has it no such number?The set varies but infinitely many elements remain the same. AHere is your essential misunderstanding: there is no mysterious
shrinking infinite set which remains infinite has an infinite core.
Something that makes a set infinite. It is infinite because it is not
finite, has no natural number as its size.
Because infinitely many natural numbers areWhat does „they” refer to in the last sentence? There is no natural number which has no successors.
contained. This is true for all infinite sets of the function. Therefore
they cannot have lost all numbers.
Richard Damon <richard@damon-family.org> wrote:
On 10/8/24 5:42 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you >>>>>>> add or subtract a finite number.Actual means all, but not more. This implies a last before ω. >>>>>>>> The infinity means an end cannot be determined. It is produced >>>>>>>> by the dark numbers.In actual infinity, there is no last 9 (that would not beThis idea of time may be what misleads the mathematically >>>>>>>>>>> less adept into believing that 0.999... < 1.That is true even in actual infinity. We can add 9 to
0.999...999 to obtain 9.999...999. But multiplying 0.999...999 >>>>>>>>>> by 10 or, what is the same, shifting the digits 9 by one step >>>>>>>>>> to the left-hand side, does not increase their number but
leaves it constant: 9.99...9990. 10*0.999...999 = 9.99...9990 >>>>>>>>>> = 9 + 0.99...9990 < 9 + 0.999...999 ==> 9*0.999...999 < 9 as >>>>>>>>>> it should be.
infinite).
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not attached to >>>>> the word "infinite", it meant something like "This is really true:".
But all his reference to the word "Actually" are part of his trying to >>>> define the term "Actual Infinity".
OK, maybe you're right, there. The semantics are a bit ambiguous. Joes >>> is not a native English speaker. Apologies to Joes.
Anyhow, what do you mean when you say that "actual infinity doesn't
exist"? I think we established over the weekend that for a mathematical >>>>> entity not to exist, it must cause a contradiction. Or something like >>>>> that.
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number or >>>> unit fraction for one (at least the way WM uses it).
Whoa! There're rather a lot of argument steps missing there. Just
because WM asserts the existence of both actual infinity and a first
strictly positive unit fraction doesn't mean the one implies the other.
It does in his logic, which is all that matters to him.
Were we talking about WM's "logic", just there? I don't think I was.
Yes, it is a wrong conclusion, but that error is based on his initial
assumption that something could be used that isn't available as a
understandable entity to us finite beings.
I think infinity is understandable. I think I understand it. My
position is that the distinction between "potential infinity" and "actual infinity" is bogus. It makes no difference in mathematics, which is
probably why the terms have vanished from mathematical discourse.
On 08.10.2024 09:30, Moebius wrote:
Am 08.10.2024 um 09:29 schrieb Moebius:
Properly understood, the idea of a completed infinity is no longer a
problem in mathematics or philosophy. It is perfectly intelligible
and coherent.
Yes, but it is completed and therefore fixed. The number of nines in
0.999... does not change when shifted by one step.
Regards, WM
On 08.10.2024 12:04, Alan Mackenzie wrote:But not the point inbetween?
WM <wolfgang.mueckenheim@tha.de> wrote:
Points either are or are not. The points that are include one point nextHence all must be visible including the point next to zero, but theyThere is no point next to zero.
are not.
to zero.
In the limit, it is empty.A shrinking infinite set which remains infinite has an infinite
core.
These are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}.Of course, the core is dark.That is untrue. For any element which you assert is in the "core", IAgain, no. There is no such thing as a "core", here. Each of theseTry to think better. A function of sets which are losing some elements
sets has an infinitude of elements. No element is in all of these
sets.
but remain infinite, have the same infinite core.
can give one of these sets which does not contain that element.
The "core" is thus empty.The infinite sets contain what? No natural numbers? Natural numbers
dancing around, sometimes being in a set, sometimes not? An empty intersection requires that the infinite sets have different elements.
This goes for every single of these sets, but not for their infinite(!) intersection. If you imagine this as potential infinity, you areShrinking sets which remain infinite have not lost all elements.That argument is absolutely definite, a logical necessity. If youIt wasn't an argument, it was a bare statement, devoid of any
cannot understand it, ....
supporting argument.
That is not very difficult to understand. We are not talking about anyI understand it full well, and I understand that it's mistaken.Impossible. You don't understand that all sets are infinite and cannot
have lost all elements.
On 07.10.2024 13:13, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
Actual means all, but not more. This implies a last before ω. TheActually infinite means infinite, which doesn’t change when you add or
infinity means an end cannot be determined. It is produced by the dark
numbers.
subtract a finite number.
It doesn’t change when you add or subtract a definable number.
Regards, WM
On 07.10.2024 17:10, joes wrote:
Am Mon, 07 Oct 2024 10:47:25 +0200 schrieb WM:
The set varies but infinitely many elements remain the same. A shrinking >>> infinite set which remains infinite has an infinite core.Here is your essential misunderstanding: there is no mysterious Something
that makes a set infinite. It is infinite because it is not finite, has
no natural number as its size.
Why has it no such number? Because infinitely many natural numbers are contained. This is true for all infinite sets of the function. Therefore
they cannot have lost all numbers.
Regards, WM
On 07.10.2024 14:11, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
You have not understood that all unit fractions are separate points on
the positive axis.
I understand that full well.
Then you understand that every point, if existing, is independent of the others. All unit fractions are points with uncounably many points
between each pair. Hence all must be visible including the point next to zero, but they are not.
A shrinking infinite set which remains infinite has an infinite core.
Again, no. There is no such thing as a "core", here. Each of these sets >> has an infinitude of elements. No element is in all of these sets.
Try to think better. A function of sets which are losing some elements
but remain infinite, have the same infinite core. That argument is
absolutely definite, a logical necessity. If you cannot understand it,
then it is useless to continue this discussion.
Regards, WM
On 07.10.2024 13:20, Richard Damon wrote:
On 10/7/24 5:51 AM, WM wrote:
On 07.10.2024 11:36, FromTheRafters wrote:
WM formulated the question :
On 06.10.2024 19:03, FromTheRafters wrote:
A set is a collection of well-defined objects, meaning we must be
able to determine if an element belongs to a particulr set.
But you can't determine the smallest unit fraction although it is a
singleton set, a point on the real axis.
There is no smallest unit fraction.
If there are only fixed points, then there is a point such that
between it and zero there is no further point.
Nope, not if you have an INFINITE set of fixed points.
The individual point is independent of how many others are existing.
The problem is we can't have an infinite set of fixed points, as we
are finite.
That is just under investigation.
So, your "actual infinity" is something beyond what we can have, so it
doesn't exist for us, and logic that assumes it is just breaks.
Then set theory is outdated.
Regards, WM
On 07.10.2024 20:05, Jim Burns wrote:
On 10/5/2024 3:33 PM, WM wrote:
On 05.10.2024 20:12, Jim Burns wrote:
On 10/5/2024 5:43 AM, WM wrote:
[In] many cases it is correct.
"Many cases" is insufficient when
the argument requires "all cases".
My argument requires only one case,
Your argument,
in order to _be an argument_
needs to _show_ its result is true.
It is easy: Consider a function of shrinking sets which remain infinite.
Then there is an infinite subset or core common to all of them.
My argument requires only one case,
best demonstrated with endsegments E(n).
The intersection of all *infinite* endsegments
is infinite, because
they all contain the same natural numbers which
have not yet been eliminated by
the process E(n+1) = E(n) \ {n}.
The intersection of all end.segments is
the set of natural numbers in each end.segment.
Each natural number is not.in one or more end.segment.
That is true but wrong in case of infinite endsegments which are
infinite because they have not lost all natural numbers. Why else should
they be infinite?
That creates a problem (but for only you).
Your attempt to fix the problem involves
a change of the definition of intersection, or
of natural number, or of something else.
Not at all.
Changing what you think "infinite" means
from what we mean by "very large finite"
to what we mean by "infinite"
No. Infinite means infinite. All infinite endsegments cotain more than
any finite set of numbers.
Regards, WM
On 10/8/24 9:11 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/8/24 5:42 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you >>>>>>>> add or subtract a finite number.Actual means all, but not more. This implies a last before ω. >>>>>>>>> The infinity means an end cannot be determined. It is produced >>>>>>>>> by the dark numbers.In actual infinity, there is no last 9 (that would not beThis idea of time may be what misleads the mathematically >>>>>>>>>>>> less adept into believing that 0.999... < 1.That is true even in actual infinity. We can add 9 to
0.999...999 to obtain 9.999...999. But multiplying 0.999...999 >>>>>>>>>>> by 10 or, what is the same, shifting the digits 9 by one step >>>>>>>>>>> to the left-hand side, does not increase their number but >>>>>>>>>>> leaves it constant: 9.99...9990. 10*0.999...999 = 9.99...9990 >>>>>>>>>>> = 9 + 0.99...9990 < 9 + 0.999...999 ==> 9*0.999...999 < 9 as >>>>>>>>>>> it should be.
infinite).
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not
attached to the word "infinite", it meant something like "This is
really true:".
But all his reference to the word "Actually" are part of his trying to >>>>> define the term "Actual Infinity".
OK, maybe you're right, there. The semantics are a bit ambiguous.
Joes is not a native English speaker. Apologies to Joes.
Anyhow, what do you mean when you say that "actual infinity
doesn't exist"? I think we established over the weekend that for
a mathematical entity not to exist, it must cause a contradiction. >>>>>> Or something like that.
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number or >>>>> unit fraction for one (at least the way WM uses it).
Whoa! There're rather a lot of argument steps missing there. Just
because WM asserts the existence of both actual infinity and a first
strictly positive unit fraction doesn't mean the one implies the other.
It does in his logic, which is all that matters to him.
Were we talking about WM's "logic", just there? I don't think I was.
But my quote came from his logic.
Yes, it is a wrong conclusion, but that error is based on his initial
assumption that something could be used that isn't available as a
understandable entity to us finite beings.
I think infinity is understandable. I think I understand it. My
position is that the distinction between "potential infinity" and "actual
infinity" is bogus. It makes no difference in mathematics, which is
probably why the terms have vanished from mathematical discourse.
The difference is that in his "actual infinity" the generation process
is complete and nothing can change.
The problem is then that the objects are "fixed", The problem here is
that our concepts of such don't really handle "infinite" objects, we can think about them "going on to infinity" and it sort of fades out of view
in the distance, but since we have never really sensed "infinity" we
have no way to actually fully understand it.
When we understand the "potential infinity" we see that it goes on
forever, and don't feel a need to get there to see it, since we know we can't.
In presuming it is fixed, he presumes he can get to that infinite point, because everything he knows is finite (just like all we have actually
seen is finite) and his logic is based on that intuition.
WM's logic seems to be based on the presumption that:
If actual infinity exists ... (with the assumption that if it exists, it follows the logic we know)
Since such a thing does NOT exist, at least as far as our logic can
handle, NONE of his conclusions apply, since the logic was based on a
untrue premise.
In WM's mind, anything that is "fixed" has ends, but an infinite thing doesn't have all those ends.
Richard Damon <richard@damon-family.org> wrote:
On 10/8/24 9:11 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/8/24 5:42 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when you >>>>>>>>> add or subtract a finite number.Actual means all, but not more. This implies a last before ω. >>>>>>>>>> The infinity means an end cannot be determined. It is produced >>>>>>>>>> by the dark numbers.In actual infinity, there is no last 9 (that would not be >>>>>>>>>>> infinite).This idea of time may be what misleads the mathematically >>>>>>>>>>>>> less adept into believing that 0.999... < 1.That is true even in actual infinity. We can add 9 to >>>>>>>>>>>> 0.999...999 to obtain 9.999...999. But multiplying 0.999...999 >>>>>>>>>>>> by 10 or, what is the same, shifting the digits 9 by one step >>>>>>>>>>>> to the left-hand side, does not increase their number but >>>>>>>>>>>> leaves it constant: 9.99...9990. 10*0.999...999 = 9.99...9990 >>>>>>>>>>>> = 9 + 0.99...9990 < 9 + 0.999...999 ==> 9*0.999...999 < 9 as >>>>>>>>>>>> it should be.
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not
attached to the word "infinite", it meant something like "This is >>>>>>> really true:".
But all his reference to the word "Actually" are part of his trying to >>>>>> define the term "Actual Infinity".
OK, maybe you're right, there. The semantics are a bit ambiguous.
Joes is not a native English speaker. Apologies to Joes.
Anyhow, what do you mean when you say that "actual infinity
doesn't exist"? I think we established over the weekend that for >>>>>>> a mathematical entity not to exist, it must cause a contradiction. >>>>>>> Or something like that.
So what contradiction would the existence of actual infinity cause?
It implies that there exists a first positive real, rational number or >>>>>> unit fraction for one (at least the way WM uses it).
Whoa! There're rather a lot of argument steps missing there. Just
because WM asserts the existence of both actual infinity and a first >>>>> strictly positive unit fraction doesn't mean the one implies the other.
It does in his logic, which is all that matters to him.
Were we talking about WM's "logic", just there? I don't think I was.
But my quote came from his logic.
Ah, OK.
Yes, it is a wrong conclusion, but that error is based on his initial
assumption that something could be used that isn't available as a
understandable entity to us finite beings.
I think infinity is understandable. I think I understand it. My
position is that the distinction between "potential infinity" and "actual >>> infinity" is bogus. It makes no difference in mathematics, which is
probably why the terms have vanished from mathematical discourse.
The difference is that in his "actual infinity" the generation process
is complete and nothing can change.
In a sense, the "actual infinity" _is_ the generation process, which is
fixed and complete.
The problem is then that the objects are "fixed", The problem here is
that our concepts of such don't really handle "infinite" objects, we can
think about them "going on to infinity" and it sort of fades out of view
in the distance, but since we have never really sensed "infinity" we
have no way to actually fully understand it.
There are lots of things we haven't really sensed that we can understand. Electromagnetic waves outside the visible range, speeds where special relativity is significant, black holes, .... I don't see that infinity
is all that different.
When we understand the "potential infinity" we see that it goes on
forever, and don't feel a need to get there to see it, since we know we
can't.
In presuming it is fixed, he presumes he can get to that infinite point,
because everything he knows is finite (just like all we have actually
seen is finite) and his logic is based on that intuition.
Yes.
WM's logic seems to be based on the presumption that:
If actual infinity exists ... (with the assumption that if it exists, it
follows the logic we know)
Since such a thing does NOT exist, at least as far as our logic can
handle, NONE of his conclusions apply, since the logic was based on a
untrue premise.
His conclusions largely don't apply whether or not "actual infinity"
exists. It is his deductive steps which are at fault rather than his
axioms.
In WM's mind, anything that is "fixed" has ends, but an infinite thing
doesn't have all those ends.
Indeed not.
Note, "Geometry" doesn't have the concept of the "open interval", that
is a concept created in mathematics. In Geometry, all lines have
endpoints,
So, the "first existing point" can't be seen, not because it is dark,
but because it doesn't exist
I think infinity is understandable. I think I understand it.
My position is that the distinction between "potential infinity" and "actual infinity" is bogus. It makes no difference in mathematics, which is
probably why the terms have vanished from mathematical discourse.
WM <wolfgang.mueckenheim@tha.de> wrote:
That argument is absolutely definite, a logical necessity.
Which argument?
On 10/8/24 7:33 AM, WM wrote:
On 08.10.2024 02:53, Richard Damon wrote:
His view is that if the values of the existing unit fractions exist,
and we travel up the line from the negative side in increasing value,
it only make sense that there should be a "first" point we reach.
The concept, on the face of it, seems logical,
of course, it is logical.
No, it is erroneous, but based on something that might SEEM logical, but
has a flaw in it due to an incorrect basic assumption
No, darkness doesn't exist, just as actual infinity is unreachable by
finite beings.
Your neglect of this fact
I think infinity is understandable. I think I understand it. My
position is that the distinction between "potential infinity" and "actual infinity" is bogus.
It makes no difference in mathematics,
Am Tue, 08 Oct 2024 13:29:46 +0200 schrieb WM:
On 07.10.2024 17:19, joes wrote:Why is there no closer point?
Am Mon, 07 Oct 2024 11:51:43 +0200 schrieb WM:One of discrete points is always next to zero.
How do you imagine that?There is no smallest unit fraction.If there are only fixed points, then there is a point such that
between it and zero there is no further point.
> It has a finite distance from 0.What does this mean, it cannot be known?
Of course, but this point cannot be found.
Am Tue, 08 Oct 2024 12:51:03 +0200 schrieb WM:
On 07.10.2024 17:18, joes wrote:Only according to your broken concept of cardinality, by which N u {a}
Am Mon, 07 Oct 2024 11:08:33 +0200 schrieb WM:
On 07.10.2024 10:05, joes wrote:„More” being a different kind of infinity, namely at least uncountable. >> Nonsense.
Am Mon, 07 Oct 2024 09:41:23 +0200 schrieb WM:It keeps all its elements but not more.
On 06.10.2024 17:48, Alan Mackenzie wrote:Nor does it get shorter, it stays infinite.
even an unbounded sequence does not get longer when shifted by one >>>>>> step.
is „bigger” than N, and N\{1} u {a} can’t even be compared.
I believe it is called the order type.All ω+k are equally infinite.Nonsense.
Bijection is not about completeness, countability is.
Isn’t N\{1} finite? It has ω-1 elements.Cardinality is nonsense.It can, if I begin numbering with 2. The cardinality of N\{1} can’t be >>> finite.Of course stopping after a finite number, which potential infinityName them by all the natural numbers. Then no further guest can
seems to mean, is not „complete” in that sense. Hilbert’s Hotel is >>>>> actually infinite, it already holds infinite guests.
appear.
That shows my point. Infinite sets can be moved. 0.999...999 moved givesYou have not indicated what this notation means. Where does the zero come from?
9.99...9990.
Another point is this: [0, 1) moved gives (0, 1].Can you generalise this?
Am Tue, 08 Oct 2024 12:46:01 +0200 schrieb WM:
Because infinitely many natural numbers areWhat does „they” refer to in the last sentence?
contained. This is true for all infinite sets of the function. Therefore
they cannot have lost all numbers.
There is no natural number
which has no successors.
On 10/8/24 6:03 AM, WM wrote:
On 08.10.2024 09:30, Moebius wrote:But it must be infinite, and thus not have an "end"
Am 08.10.2024 um 09:29 schrieb Moebius:
Properly understood, the idea of a completed infinity is no longer a
problem in mathematics or philosophy. It is perfectly intelligible
and coherent.
Yes, but it is completed and therefore fixed. The number of nines in
0.999... does not change when shifted by one step.
Am Tue, 08 Oct 2024 12:40:26 +0200 schrieb WM:
On 08.10.2024 12:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
But not the point inbetween?Points either are or are not. The points that are include one point nextHence all must be visible including the point next to zero, but theyThere is no point next to zero.
are not.
to zero.
The infinite sets contain what? No natural numbers? Natural numbersThese are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}.
dancing around, sometimes being in a set, sometimes not? An empty
intersection requires that the infinite sets have different elements.
They contain all naturals larger than a given one, and nothing else.
Every natural is part of a finite number of these sets (namely, its
own value is that number). The set {n+1, n+2, …} does not contain n
and is still infinite; there are (trivially) infinitely many further
such sets. All of them differ.
Shrinking sets which remain infinite have not lost all elements.This goes for every single of these sets, but not for their infinite(!) intersection.
If you imagine this as potential infinity,
Real finite things have ends. Real
infinite things are missing some of the ends, so assuming they are there causes the problems.
In WM's mind, anything that is "fixed" has ends, but an infinite thing
doesn't have all those ends.
On 07.10.2024 20:05, Jim Burns wrote:
That creates a problem (but for only you).
Your attempt to fix the problem involves
a change of the definition of intersection, or
of natural number, or of something else.
Not at all.
Changing what you think "infinite" means
from what we mean by "very large finite"
to what we mean by "infinite"
No.
Infinite means infinite.
All infinite endsegments cotain
more than any finite set of numbers.
My argument requires only one case,
best demonstrated with endsegments E(n).
The intersection of all *infinite* endsegments
is infinite, because
they all contain the same natural numbers which
have not yet been eliminated by
the process E(n+1) = E(n) \ {n}.
The intersection of all end.segments is
the set of natural numbers in each end.segment.
Each natural number is not.in one or more end.segment.
That is true but wrong
in case of infinite endsegments which are infinite
because they have not lost all natural numbers.
Why else should they be infinite?
On 08.10.2024 15:24, joes wrote:
Am Tue, 08 Oct 2024 12:51:03 +0200 schrieb WM:
On 07.10.2024 17:18, joes wrote:Only according to your broken concept of cardinality, by which N u {a}
Am Mon, 07 Oct 2024 11:08:33 +0200 schrieb WM:Nonsense.
On 07.10.2024 10:05, joes wrote:„More” being a different kind of infinity, namely at least uncountable.
Am Mon, 07 Oct 2024 09:41:23 +0200 schrieb WM:It keeps all its elements but not more.
On 06.10.2024 17:48, Alan Mackenzie wrote:Nor does it get shorter, it stays infinite.
even an unbounded sequence does not get longer when shifted by one >>>>>>> step.
is „bigger” than N, and N\{1} u {a} can’t even be compared.
I believe it is called the order type.All ω+k are equally infinite.Nonsense.
Isn’t N\{1} finite? It has ω-1 elements.Bijection is not about completeness, countability is.
Cardinality is nonsense.It can, if I begin numbering with 2. The cardinality of N\{1} can’t be >>>> finite.Of course stopping after a finite number, which potential infinity >>>>>> seems to mean, is not „complete” in that sense. Hilbert’s Hotel is >>>>>> actually infinite, it already holds infinite guests.Name them by all the natural numbers. Then no further guest can
appear.
But after the visble natural numbers the dark domain comes, and that is
what prevents to see the end (which is dark too).
That shows my point. Infinite sets can be moved. 0.999...999 moved gives >>> 9.99...9990.You have not indicated what this notation means. Where does the zero come
from?
The last natural index has lost its 9 by shifting to the left-hand side. Hence there is nothing remaining.
Another point is this: [0, 1) moved gives (0, 1].Can you generalise this?
What example do you have in mind?
Regards, WM
On 08.10.2024 15:36, joes wrote:Ah, then the former point wasn’t the one next to zero. Same goes for this one. There are always infinitely many points between any two reals.
Am Tue, 08 Oct 2024 12:40:26 +0200 schrieb WM:If it exists then this point is next to zero.
On 08.10.2024 12:04, Alan Mackenzie wrote:But not the point inbetween?
WM <wolfgang.mueckenheim@tha.de> wrote:
Points either are or are not. The points that are include one pointHence all must be visible including the point next to zero, but they >>>>> are not.There is no point next to zero.
next to zero.
Every *finite* intersection.All of them differ by a finite set of numbers (whoich is irrelevant) but contain an infinite set of numbers in common.The infinite sets contain what? No natural numbers? Natural numbersThese are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}.
dancing around, sometimes being in a set, sometimes not? An empty
intersection requires that the infinite sets have different elements.
They contain all naturals larger than a given one, and nothing else.
Every natural is part of a finite number of these sets (namely, its own
value is that number). The set {n+1, n+2, …} does not contain n and is
still infinite; there are (trivially) infinitely many further such
sets. All of them differ.
Uh. So the naturals don’t have successors?If every single set is infinite, then the intersection is infinite too.Shrinking sets which remain infinite have not lost all elements.This goes for every single of these sets, but not for their infinite(!)
intersection.
These sets have lost some natural numbers but have kept infinitely many.
If you imagine this as potential infinity,No, in potential infinity there are no endsegments.
On 08.10.2024 15:26, joes wrote:We are, again, not talking about an element of the sequence, which has a natural index, contains infinitely many successors and is missing a
Am Tue, 08 Oct 2024 12:46:01 +0200 schrieb WM:
All endsegments which have infinitely many natural numbers.Because infinitely many natural numbers are contained. This is trueWhat does „they” refer to in the last sentence?
for all infinite sets of the function. Therefore they cannot have lost
all numbers.
Can you explain to me what an infinite intersection is?There is no natural number which has no successors.Infinite endsegments contain infinitely many numbers and therefore an infinite intersection.
On 08.10.2024 15:36, Richard Damon wrote:In what sense are infinite sets incomplete?
On 10/8/24 6:03 AM, WM wrote:
On 08.10.2024 09:30, Moebius wrote:
Am 08.10.2024 um 09:29 schrieb Moebius:
But it must be complete and therefore in linear order must have an end.Yes, but it is completed and therefore fixed. The number of nines inBut it must be infinite, and thus not have an "end"
0.999... does not change when shifted by one step.
On 08.10.2024 15:24, joes wrote:Clearly they are the same „size”.
Am Tue, 08 Oct 2024 12:51:03 +0200 schrieb WM:
On 07.10.2024 17:18, joes wrote:Only according to your broken concept of cardinality, by which N u {a}
Am Mon, 07 Oct 2024 11:08:33 +0200 schrieb WM:Nonsense.
On 07.10.2024 10:05, joes wrote:„More” being a different kind of infinity, namely at least
Am Mon, 07 Oct 2024 09:41:23 +0200 schrieb WM:It keeps all its elements but not more.
On 06.10.2024 17:48, Alan Mackenzie wrote:Nor does it get shorter, it stays infinite.
even an unbounded sequence does not get longer when shifted by one >>>>>>> step.
uncountable.
is „bigger” than N, and N\{1} u {a} can’t even be compared.
This is your useless replacement of cardinality.All ω+k are equally infinite.Nonsense.
But ω-1 is a finite natural.But after the visble natural numbers the dark domain comes, and that isIsn’t N\{1} finite? It has ω-1 elements.Cardinality is nonsense.It can, if I begin numbering with 2. The cardinality of N\{1} can’tOf course stopping after a finite number, which potential infinity >>>>>> seems to mean, is not „complete” in that sense. Hilbert’s Hotel is >>>>>> actually infinite, it already holds infinite guests.Name them by all the natural numbers. Then no further guest can
appear.
be finite.
what prevents to see the end (which is dark too).
Moving a set does not remove elements. You cannot shift through theThe last natural index has lost its 9 by shifting to the left-hand side. Hence there is nothing remaining.That shows my point. Infinite sets can be moved. 0.999...999 movedYou have not indicated what this notation means. Where does the zero
gives 9.99...9990.
come from?
Any interval with both ends open/closed.What example do you have in mind?Another point is this: [0, 1) moved gives (0, 1].Can you generalise this?
When does an infinite process finally complete? If it does, well, it
was NOT infinite in any way shape, or form... Right?
On 07.10.2024 18:11, Alan Mackenzie wrote:
What I should have
written (WM please take note) is:
The idea of one countably infinite set being "bigger" than another
countably infinite set is simply nonsense.
The idea is supported by the fact that set A as a superset of set B is
bigger than B.
Simply nonsense is the claim that there are as many algebraic numbers
as prime numbers.
For Cantor's enumeration of all fractions I have given a simple
disproof.
Regards, WM
On 10/8/2024 5:35 AM, Richard Damon wrote:
On 10/8/24 5:54 AM, WM wrote:
On 07.10.2024 15:19, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
No, even an unbounded sequence does not get longer when shifted by
one step.
The concept of "length" appropriate for finite sets doesn't apply to
infinite sets.
It is the concept of number of elements. It is appropriate in actual
infinity.
Which has been shown to not exist for us finite beings, as it is too
big for us to see.
infinite means "without end" - unendlich.
Actual infinity means complete. That implies a fixed number.
Which has been shown to not exist for us finite beings, as it is too
big for us to see.
You don't understand that actual infinity is a fixed quantity.
It may be "fixed" whatever that might mean, but to regard it as a
"quantity" is more than questionable.
Fixed means that no element can be added and no element can be lost.
The number of nines is fixed. That is an assumption only, but
necessary for bijections.
Yes, it is fixed, at INFINITY, which means there is no end to it, and
thus we can't add a zero "at the end" which doesn't exist.
This is why finite beings can't use "actual infinity" because it is
too big for us to handle.
Ummm... Well, not sure what to think about that. Hummm... Any time you
use a number it is in actual infinity. Think of the number four. It is
in a pool of the infinitely many natural numbers, and we just used it...
Fair enough, or weasel words?
On 10/8/2024 2:21 PM, Richard Damon wrote:
On 10/8/24 4:17 PM, Chris M. Thomasson wrote:
On 10/8/2024 7:10 AM, Richard Damon wrote:
On 10/8/24 10:00 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/8/24 9:11 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/8/24 5:42 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:44 AM, Alan Mackenzie wrote:
Richard Damon <richard@damon-family.org> wrote:
On 10/7/24 7:13 AM, joes wrote:
Am Mon, 07 Oct 2024 10:13:21 +0200 schrieb WM:
On 06.10.2024 17:55, joes wrote:
Am Sun, 06 Oct 2024 17:26:07 +0200 schrieb WM:
On 06.10.2024 16:52, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Actually infinite means infinite, which doesn’t change when >>>>>>>>>>>>> youActual means all, but not more. This implies a last before ω. >>>>>>>>>>>>>> The infinity means an end cannot be determined. It is >>>>>>>>>>>>>> producedIn actual infinity, there is no last 9 (that would not be >>>>>>>>>>>>>>> infinite).This idea of time may be what misleads the mathematically >>>>>>>>>>>>>>>>> less adept into believing that 0.999... < 1.That is true even in actual infinity. We can add 9 to >>>>>>>>>>>>>>>> 0.999...999 to obtain 9.999...999. But multiplying >>>>>>>>>>>>>>>> 0.999...999
by 10 or, what is the same, shifting the digits 9 by one >>>>>>>>>>>>>>>> step
to the left-hand side, does not increase their number but >>>>>>>>>>>>>>>> leaves it constant: 9.99...9990. 10*0.999...999 = >>>>>>>>>>>>>>>> 9.99...9990
= 9 + 0.99...9990 < 9 + 0.999...999 ==> 9*0.999...999 < >>>>>>>>>>>>>>>> 9 as
it should be.
by the dark numbers.
add or subtract a finite number.
Actual infinity doesn't exist for us finite beings.
English language tip: The "Actually" in that sentence was not >>>>>>>>>>> attached to the word "infinite", it meant something like >>>>>>>>>>> "This is
really true:".
But all his reference to the word "Actually" are part of his >>>>>>>>>> trying to
define the term "Actual Infinity".
OK, maybe you're right, there. The semantics are a bit ambiguous. >>>>>>>>> Joes is not a native English speaker. Apologies to Joes.
Anyhow, what do you mean when you say that "actual infinity >>>>>>>>>>> doesn't exist"? I think we established over the weekend that >>>>>>>>>>> for
a mathematical entity not to exist, it must cause a
contradiction.
Or something like that.
So what contradiction would the existence of actual infinity >>>>>>>>>>> cause?
It implies that there exists a first positive real, rational >>>>>>>>>> number or
unit fraction for one (at least the way WM uses it).
Whoa! There're rather a lot of argument steps missing there. >>>>>>>>> Just
because WM asserts the existence of both actual infinity and a >>>>>>>>> first
strictly positive unit fraction doesn't mean the one implies >>>>>>>>> the other.
It does in his logic, which is all that matters to him.
Were we talking about WM's "logic", just there? I don't think I >>>>>>> was.
But my quote came from his logic.
Ah, OK.
Yes, it is a wrong conclusion, but that error is based on his
initial
assumption that something could be used that isn't available as a >>>>>>>> understandable entity to us finite beings.
I think infinity is understandable. I think I understand it. My >>>>>>> position is that the distinction between "potential infinity" and >>>>>>> "actual
infinity" is bogus. It makes no difference in mathematics, which is >>>>>>> probably why the terms have vanished from mathematical discourse.
The difference is that in his "actual infinity" the generation
process
is complete and nothing can change.
In a sense, the "actual infinity" _is_ the generation process,
which is
fixed and complete.
But only after that infinite process has completed. Finite beings
can't actually SEE that result.[...]
When does an infinite process finally complete? If it does, well, it
was NOT infinite in any way shape, or form... Right?
The infinite process completes in the infinite, thus beyond what we
can perceive, but only dimly imagine.
That is why we can't have actual infinity, we can't get to the
infinite to see it done.
since a step-by-step process for the naturals aka:
1, 1+1, 1+1+1, ...
Will never end, well... That does not mean there is a largest natural.
WM is strange on this aspect. See, this right here is pondering on the infinite from a finite being... ;^)
On 10/8/2024 6:36 AM, Richard Damon wrote:
This is why it is said that completed/actual infinity doesn't exist,
because [bla bla bla]
On 10/8/2024 6:18 AM, WM wrote:
All infinite endsegments contain
more than any finite set of numbers.
...still true if 'infinite' means "very large".
An end segment E is infinite because
each natural number has a successor,
so no element of E is max.E
so not all nonempty subsets are two.ended
so E is not finite.
Each natural number k
is followed by successor k+1
k is not.in infinite E(k+1)
There are no other end segments, none are finite.
The intersection of all end.segments is
the set of natural numbers in each end.segment.
Each natural number is not.in one or more end.segment.
That is true but wrong
in case of infinite endsegments which are infinite
because they have not lost all natural numbers.
The intersection of all end.segments is
the set of natural numbers in each end.segment.
Consider end segment E(k+1)
k+1 is in E(k+1)
Is k+1 in each end segment? Is k+1 in E(k+2)?
Is E(k+1)
the set of natural numbers in each end.segment?
Why else should they be infinite?
Because each natural number is followed by
a natural number,
because 'infinite' DOES NOT mean 'very large'.
On 10/8/24 11:16 AM, WM wrote:
No, darkness doesn't exist, just as actual infinity is unreachable by
finite beings.
Your neglect of this fact
is based on Cantor's work. But you are right in that actual infinity
and dark elements are indissolubly connected.
No, you are abusing Cantor's work because you don't understand what
infinity is
it just blows up.
On 10/8/24 11:49 AM, WM wrote:
But there is no "end" to shift.In WM's mind, anything that is "fixed" has ends, but an infinite thing >>>>> doesn't have all those ends.
Unless the infinity is shifted from the end to the centre.
On 10/8/24 11:26 AM, WM wrote:
But after the visble natural numbers the dark domain comes, and that
is what prevents to see the end (which is dark too).
Where?
The "visible" numbers, per you definition are ALL the numbers, as all of
them can be used individually and are selectable.
Thus, there aren't any left to be dark, except the ones that don't
actually exist.
That shows my point. Infinite sets can be moved. 0.999...999 movedYou have not indicated what this notation means. Where does the zero
gives
9.99...9990.
come
from?
The last natural index has lost its 9 by shifting to the left-hand
side. Hence there is nothing remaining.
How did it "lose" it,
On 10/8/24 11:34 AM, WM wrote:
On 08.10.2024 15:36, Richard Damon wrote:
On 10/8/24 6:03 AM, WM wrote:
On 08.10.2024 09:30, Moebius wrote:But it must be infinite, and thus not have an "end"
Am 08.10.2024 um 09:29 schrieb Moebius:
Properly understood, the idea of a completed infinity is no longer >>>>>> a problem in mathematics or philosophy. It is perfectly
intelligible and coherent.
Yes, but it is completed and therefore fixed. The number of nines in
0.999... does not change when shifted by one step.
But it must be complete and therefore in linear order must have an end.
And then it isn't infinte.
Seems you don't understand the problem with "completed infinity" that
being infinite, it can't be actually fully understood by the finite.
Having linear order does not imply having an end, unless you can assume
a finiteness.
On 10/8/2024 1:28 AM, WM wrote:
The rule of geometry is that every point exists.
Except your dark ones, right?
Asking you to define a dark point would
mean that its not dark anymore, wrt your line of thinking, right?
Am Tue, 08 Oct 2024 17:40:50 +0200 schrieb WM:
On 08.10.2024 15:36, joes wrote:Ah, then the former point wasn’t the one next to zero. Same goes for this one. There are always infinitely many points between any two reals.
Am Tue, 08 Oct 2024 12:40:26 +0200 schrieb WM:If it exists then this point is next to zero.
On 08.10.2024 12:04, Alan Mackenzie wrote:But not the point inbetween?
WM <wolfgang.mueckenheim@tha.de> wrote:
Points either are or are not. The points that are include one pointHence all must be visible including the point next to zero, but they >>>>>> are not.There is no point next to zero.
next to zero.
Every *finite* intersection.All of them differ by a finite set of numbers (which is irrelevant) butThe infinite sets contain what? No natural numbers? Natural numbersThese are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}. >>> They contain all naturals larger than a given one, and nothing else.
dancing around, sometimes being in a set, sometimes not? An empty
intersection requires that the infinite sets have different elements.
Every natural is part of a finite number of these sets (namely, its own
value is that number). The set {n+1, n+2, …} does not contain n and is >>> still infinite; there are (trivially) infinitely many further such
sets. All of them differ.
contain an infinite set of numbers in common.
Think about it this way: we are taking the limit of N\{0, 1, 2, …}.
If every single set is infinite, then the intersection is infinite too.Shrinking sets which remain infinite have not lost all elements.This goes for every single of these sets, but not for their infinite(!)
intersection.
These sets have lost some natural numbers but have kept infinitely many.
Uh. So the naturals don’t have successors?If you imagine this as potential infinity,No, in potential infinity there are no endsegments.
Am Tue, 08 Oct 2024 17:30:19 +0200 schrieb WM:
On 08.10.2024 15:26, joes wrote:We are, again, not talking about an element of the sequence, which has a natural index, contains infinitely many successors and is missing a
Am Tue, 08 Oct 2024 12:46:01 +0200 schrieb WM:All endsegments which have infinitely many natural numbers.
Because infinitely many natural numbers are contained. This is trueWhat does „they” refer to in the last sentence?
for all infinite sets of the function. Therefore they cannot have lost >>>> all numbers.
finite number of predecessors.
What we are talking about is the, pardon,
limit of whatever function.
Shouldn’t the limit of 1/n be 1/ω != 0 ?
Can you explain to me what an infinite intersection is?There is no natural number which has no successors.Infinite endsegments contain infinitely many numbers and therefore an
infinite intersection.
On 08.10.2024 20:01, Richard Damon wrote:
On 10/8/24 11:16 AM, WM wrote:
No, darkness doesn't exist, just as actual infinity is unreachable
by finite beings.
Your neglect of this fact
is based on Cantor's work. But you are right in that actual infinity
and dark elements are indissolubly connected.
No, you are abusing Cantor's work because you don't understand what
infinity is
I understand what Cantor has said.
, and are trying to apply FINITE logic to the infinite, and
it just blows up.
All logic is finite.
Regards, WM
On 08.10.2024 20:01, Richard Damon wrote:
On 10/8/24 11:49 AM, WM wrote:
But there is no "end" to shift.In WM's mind, anything that is "fixed" has ends, but an infinite
thing
doesn't have all those ends.
Unless the infinity is shifted from the end to the centre.
If there is all, then there is an end.
Regards, WM
On 08.10.2024 21:17, joes wrote:
Am Tue, 08 Oct 2024 17:40:50 +0200 schrieb WM:
On 08.10.2024 15:36, joes wrote:Ah, then the former point wasn’t the one next to zero. Same goes for
Am Tue, 08 Oct 2024 12:40:26 +0200 schrieb WM:If it exists then this point is next to zero.
On 08.10.2024 12:04, Alan Mackenzie wrote:But not the point inbetween?
WM <wolfgang.mueckenheim@tha.de> wrote:
Points either are or are not. The points that are include one pointHence all must be visible including the point next to zero, butThere is no point next to zero.
they are not.
next to zero.
this one. There are always infinitely many points between any two
reals.
WDYM, all numbers in the segments are indices.As long as infinitely many numbers are captivated in endsegments, only finitely many indices are available, and the intersection is betweenEvery *finite* intersection.All of them differ by a finite set of numbers (which is irrelevant)The infinite sets contain what? No natural numbers? Natural numbersThese are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}. >>>> They contain all naturals larger than a given one, and nothing else.
dancing around, sometimes being in a set, sometimes not? An empty
intersection requires that the infinite sets have different
elements.
Every natural is part of a finite number of these sets (namely, its
own value is that number). The set {n+1, n+2, …} does not contain n
and is still infinite; there are (trivially) infinitely many further
such sets. All of them differ.
but contain an infinite set of numbers in common.
finitely many infinite endsegments.
What does this mean for the infinite intersection?Think about it this way: we are taking the limit of N\{0, 1, 2, …}.In the limit not a single natural number remains, let alone infinitely
many.
What is the difference between the sef of successors and an endsegment?They have successors but endsegments are sets and must be complete.Uh. So the naturals don’t have successors?If every single set is infinite, then the intersection is infiniteShrinking sets which remain infinite have not lost all elements.This goes for every single of these sets, but not for their
infinite(!) intersection.
too. These sets have lost some natural numbers but have kept
infinitely many.
If you imagine this as potential infinity,No, in potential infinity there are no endsegments.
On 08.10.2024 21:23, joes wrote:Such an intersection is itself part of the sequence.
Am Tue, 08 Oct 2024 17:30:19 +0200 schrieb WM:I am talking about such endsegments. Their intersection is infinite.
On 08.10.2024 15:26, joes wrote:We are, again, not talking about an element of the sequence, which has
Am Tue, 08 Oct 2024 12:46:01 +0200 schrieb WM:All endsegments which have infinitely many natural numbers.
Because infinitely many natural numbers are contained. This is trueWhat does „they” refer to in the last sentence?
for all infinite sets of the function. Therefore they cannot have
lost all numbers.
a natural index, contains infinitely many successors and is missing a
finite number of predecessors.
Why?What we are talking about is the, pardon,The limit-endsegment is empty.
limit of whatever function.
And how many segments have been intersected?Shouldn’t the limit of 1/n be 1/ω != 0 ?
The intersection is infinite because all infinite endsegments containCan you explain to me what an infinite intersection is?There is no natural number which has no successors.Infinite endsegments contain infinitely many numbers and therefore an
infinite intersection.
the same infinite set. Some have lost more or less numbers but the core remains infinite in all infinite endsegments.
On 08.10.2024 19:28, Jim Burns wrote:With each, but not with all at once (cf. quantifier shift).
On 10/8/2024 6:18 AM, WM wrote:
No. Very large is not more than any finite set.All infinite endsegments contain more than any finite set of numbers....still true if 'infinite' means "very large".
An end segment E is infinite because each natural number has aTherefore every infinite endsegment has infinitely many elements with
successor,
so no element of E is max.E so not all nonempty subsets are two.ended
so E is not finite.
each predecessor in common. This is valid for all infinite endsegments.
Intersection with what?There are no other end segments, none are finite.They all have an infinite intersection.
Invalid quantifier shift.Consider end segment E(k+1)No, but by definition there are infinitely many numbers. They are dark.
k+1 is in E(k+1)
Is k+1 in each end segment? Is k+1 in E(k+2)?
Is E(k+1) the set of natural numbers in each end.segment?
not only one but infinitely manyWhy else should they be infinite?Because each natural number is followed by a natural number,
because 'infinite' DOES NOT mean 'very large'.Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with lessNo. Why do you think that?
numbers.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.10.2024 18:11, Alan Mackenzie wrote:
What I should have
written (WM please take note) is:
The idea of one countably infinite set being "bigger" than another
countably infinite set is simply nonsense.
The idea is supported by the fact that set A as a superset of set B is
bigger than B.
What do you mean by "bigger" as applied to two infinite sets when one of
them is not a subset of the other?
The standard definition for infinite (or finite) sets being the same
size is the existence of a 1-1 correspondence between them.
You seem to be rejecting that definition. What would you replace it by?
You have specified "bigger" for a special case. What is your definition
for the general case?
Simply nonsense is the claim that there are as many algebraic numbers
as prime numbers.
It is not nonsense. The prime numbers can be put into 1-1
correspondence with the algebraic numbers, therefore there are exactly
as many of each.
For Cantor's enumeration of all fractions I have given a simple
disproof.
Your "proofs" tend to be nonsense.
Regards, WM
Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:
With each, but not with all at once (cf. quantifier shift).An end segment E is infinite because each natural number has aTherefore every infinite endsegment has infinitely many elements with
successor,
so no element of E is max.E so not all nonempty subsets are two.ended
so E is not finite.
each predecessor in common. This is valid for all infinite endsegments.
Intersection with what?There are no other end segments, none are finite.They all have an infinite intersection.
Consider end segment E(k+1)No, but by definition there are infinitely many numbers. They are dark.
k+1 is in E(k+1)
Is k+1 in each end segment? Is k+1 in E(k+2)?
Is E(k+1) the set of natural numbers in each end.segment?
not only one but infinitely manyWhy else should they be infinite?Because each natural number is followed by a natural number,
Invalid quantifier shift.because 'infinite' DOES NOT mean 'very large'.Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with lessNo. Why do you think that?
numbers.
WM presented the following explanation :
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.
A conjecture is not a proof. This one is simply another non sequitur.
Am Wed, 09 Oct 2024 11:47:57 +0200 schrieb WM:
On 08.10.2024 21:23, joes wrote:Such an intersection is itself part of the sequence.
Am Tue, 08 Oct 2024 17:30:19 +0200 schrieb WM:I am talking about such endsegments. Their intersection is infinite.
On 08.10.2024 15:26, joes wrote:We are, again, not talking about an element of the sequence, which has
Am Tue, 08 Oct 2024 12:46:01 +0200 schrieb WM:All endsegments which have infinitely many natural numbers.
Because infinitely many natural numbers are contained. This is true >>>>>> for all infinite sets of the function. Therefore they cannot haveWhat does „they” refer to in the last sentence?
lost all numbers.
a natural index, contains infinitely many successors and is missing a
finite number of predecessors.
Why?What we are talking about is the, pardon,The limit-endsegment is empty.
limit of whatever function.
The intersection is infinite because all infinite endsegments containAnd how many segments have been intersected?
the same infinite set. Some have lost more or less numbers but the core
remains infinite in all infinite endsegments.
Am Wed, 09 Oct 2024 11:41:31 +0200 schrieb WM:
On 08.10.2024 21:17, joes wrote:
Am Tue, 08 Oct 2024 17:40:50 +0200 schrieb WM:
On 08.10.2024 15:36, joes wrote:Ah, then the former point wasn’t the one next to zero. Same goes for
Am Tue, 08 Oct 2024 12:40:26 +0200 schrieb WM:If it exists then this point is next to zero.
On 08.10.2024 12:04, Alan Mackenzie wrote:But not the point inbetween?
WM <wolfgang.mueckenheim@tha.de> wrote:
Points either are or are not. The points that are include one point >>>>>> next to zero.Hence all must be visible including the point next to zero, but >>>>>>>> they are not.There is no point next to zero.
this one. There are always infinitely many points between any two
reals.
WDYM, all numbers in the segments are indices.As long as infinitely many numbers are captivated in endsegments, onlyEvery *finite* intersection.All of them differ by a finite set of numbers (which is irrelevant)The infinite sets contain what? No natural numbers? Natural numbers >>>>>> dancing around, sometimes being in a set, sometimes not? An emptyThese are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}. >>>>> They contain all naturals larger than a given one, and nothing else. >>>>> Every natural is part of a finite number of these sets (namely, its
intersection requires that the infinite sets have different
elements.
own value is that number). The set {n+1, n+2, …} does not contain n >>>>> and is still infinite; there are (trivially) infinitely many further >>>>> such sets. All of them differ.
but contain an infinite set of numbers in common.
finitely many indices are available, and the intersection is between
finitely many infinite endsegments.
But what about the intersection between all infinitely many segments?
What does this mean for the infinite intersection?Think about it this way: we are taking the limit of N\{0, 1, 2, …}.In the limit not a single natural number remains, let alone infinitely
many.
What is the difference between the sef of successors and an endsegment?They have successors but endsegments are sets and must be complete.Uh. So the naturals don’t have successors?If you imagine this as potential infinity,No, in potential infinity there are no endsegments.
Why can’t the segments be potentially infinite, or the successors
actually inf.?
On 08.10.2024 23:08, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.10.2024 18:11, Alan Mackenzie wrote:
What I should have
written (WM please take note) is:
The idea of one countably infinite set being "bigger" than another
countably infinite set is simply nonsense.
The idea is supported by the fact that set A as a superset of set B is
bigger than B.
What do you mean by "bigger" as applied to two infinite sets when one of
them is not a subset of the other?
That is not in every case defined. But here are some rules:
Not all infinite sets can be compared by size, but we can establish some useful rules.
The rule of subset proves that every proper subset has fewer elements than its superset. So there are more natural numbers than prime numbers,
|N| > |P|, and more complex numbers than real numbers, |C| > |R|. Even finitely many exceptions from the subset-relation are admitted for
infinite subsets. Therefore there are more odd numbers than prime
numbers |O| > |P|.
The rule of construction yields the numbers of integers |Z| = 2|N| + 1 and the number of fractions |Q| = 2|N|^2 + 1 (there are fewer rational numbers Q# ). Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers than
rational numbers |X| > |Q#|.
The rule of symmetry yields precisely the same number of real
geometric points in every interval (n, n+1] and with at most a small
error same number of odd numbers and of even numbers in every finite interval and in the whole real line.
The standard definition for infinite (or finite) sets being the same
size is the existence of a 1-1 correspondence between them.
You seem to be rejecting that definition. What would you replace it by?
You have specified "bigger" for a special case. What is your definition
for the general case?
Simply nonsense is the claim that there are as many algebraic numbers
as prime numbers.
It is not nonsense. The prime numbers can be put into 1-1
correspondence with the algebraic numbers, therefore there are exactly
as many of each.
Nonsense. Only potential infinity is used. Never the main body is applied.
For Cantor's enumeration of all fractions I have given a simple
disproof.
Your "proofs" tend to be nonsense.
It appears to you because you are unable to understand. Here is the simplest:
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less numbers.
Note: The shrinking endsegments cannot acquire new numbers.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 08.10.2024 23:08, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.10.2024 18:11, Alan Mackenzie wrote:
What I should have
written (WM please take note) is:
The idea of one countably infinite set being "bigger" than another
countably infinite set is simply nonsense.
The idea is supported by the fact that set A as a superset of set B is >>>> bigger than B.
What do you mean by "bigger" as applied to two infinite sets when one of >>> them is not a subset of the other?
That is not in every case defined. But here are some rules:
Not all infinite sets can be compared by size, but we can establish some
useful rules.
Possibly. But these rules would require proof, which you haven't
supplied.
The rule of subset proves that every proper subset has fewer elements
than its superset. So there are more natural numbers than prime numbers,
|N| > |P|, and more complex numbers than real numbers, |C| > |R|. Even
finitely many exceptions from the subset-relation are admitted for
infinite subsets. Therefore there are more odd numbers than prime
numbers |O| > |P|.
This breaks down in a contradiction, as shown by Richard D in another
post: To repeat his idea:
The set {0, 2, 4, 6, ...} is a subset of the natural numbers N,
{0, 1, 2, 3, 4, 5, 6, ...}, thus is smaller than it.
We can replace the second set by one of the same "size" by multiplying
each of its members by 4. We then get the set
{0, 4, 8, ...}.
Now this third set is a subset of the first hence is smaller than it.
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
That is simply false. You cannot specify a single number which is in
all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.
... which is gobbledegook, not maths.
Note: The shrinking endsegments cannot acquire new numbers.
An end segment is what it is. It doesn't change.
WM pretended :
On 09.10.2024 12:12, FromTheRafters wrote:
WM presented the following explanation :
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.
A conjecture is not a proof. This one is simply another non sequitur.
Inclusion-monotony proves that all infinite endsegments have a common
infinite subset because only a loss of elements is possible. As long
as all endsegments are infinite, the loss has spared an infinite set
common to all.
If you can't understand try to find a counterexample.
Or use a finite example.
Diminish the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} to get
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
...
Finite sets again, I don't care how many.
As long as five numbers remain in all sets, they are a common subset
of all sets.
When you add all numbers following 10 to all sets, the situation
remains the same.
Try using infinite sets.
If there were a last as well as a first you
would be right, but there is no last to be in all endsegments,
so it is empty.
On 09.10.2024 16:13, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 08.10.2024 23:08, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.10.2024 18:11, Alan Mackenzie wrote:
What I should have
written (WM please take note) is:
The idea of one countably infinite set being "bigger" than another >>>>>> countably infinite set is simply nonsense.
The idea is supported by the fact that set A as a superset of set B is >>>>> bigger than B.
What do you mean by "bigger" as applied to two infinite sets when one of >>>> them is not a subset of the other?
That is not in every case defined. But here are some rules:
Not all infinite sets can be compared by size, but we can establish some >>> useful rules.
Possibly. But these rules would require proof, which you haven't
supplied.
These rules are self-evident.
The rule of subset proves that every proper subset has fewer elements >>> than its superset. So there are more natural numbers than prime numbers, >>> |N| > |P|, and more complex numbers than real numbers, |C| > |R|. Even
finitely many exceptions from the subset-relation are admitted for
infinite subsets. Therefore there are more odd numbers than prime
numbers |O| > |P|.
This breaks down in a contradiction, as shown by Richard D in another
post: To repeat his idea:
The set {0, 2, 4, 6, ...} is a subset of the natural numbers N,
{0, 1, 2, 3, 4, 5, 6, ...}, thus is smaller than it.
We can replace the second set by one of the same "size" by multiplying
each of its members by 4. We then get the set
{0, 4, 8, ...}.
Now this third set is a subset of the first hence is smaller than it.
No. When we *in actual infinity* multiply all |ℕ|natural numbers by 2, then we keep |ℕ| numbers but only half of them are smaller than ω, i.e., are natural numbers. The other half is larger than ω.
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
That is simply false. You cannot specify a single number which is in
all endsegments.
True. This proves dark numbers.
Note: The shrinking endsegments cannot acquire new numbers.
An end segment is what it is. It doesn't change.
But the terms of the sequence do. Here is a simple finite example:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets.
Quantifier shift: There is a subset of three elements common to all
TN-sets. Understood?
Now complete all sets by the natural numbers > 10 and complete the
sequence.
Then we get: All sets which have lost at most n elements have the
remainder in common. Note: All sets which are infinite have lost at most
a finite number of elements.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
No. When we *in actual infinity* multiply all |ℕ|natural numbers by 2,
then we keep |ℕ| numbers but only half of them are smaller than ω, i.e., >> are natural numbers. The other half is larger than ω.
Ha ha ha ha! This is garbage. If you think doubling some numbers gives results which are "larger than ω" you'd better be prepared to give an example of such a number. But you're surely going to tell me that these
are "dark numbers
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
That is simply false. You cannot specify a single number which is in
all endsegments.
True. This proves dark numbers.
Dark numbers don't exist, or at least they're not natural numbers. There
is no number in each and every end segment of N.
[ .... ]
Note: The shrinking endsegments cannot acquire new numbers.
An end segment is what it is. It doesn't change.
But the terms of the sequence do. Here is a simple finite example:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
Example of what? The reasoning you might do on finite sets mostly isn't applicable to infinite sets.
Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets.
Not even an ignorant schoolboy would maintain this. The two TN-sets {0,
1, 2} and {3, 4, 5} have no numbers in common.
Quantifier shift: There is a subset of three elements common to all
TN-sets. Understood?
Yes, I understand completely.
Now complete all sets by the natural numbers > 10 and complete the
sequence.
You can't "complete" a set.
On 09.10.2024 17:11, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
No. When we *in actual infinity* multiply all |ℕ|natural numbers by 2, >>> then we keep |ℕ| numbers but only half of them are smaller than ω, i.e., >>> are natural numbers. The other half is larger than ω.
Ha ha ha ha! This is garbage. If you think doubling some numbers gives
results which are "larger than ω" you'd better be prepared to give an
example of such a number. But you're surely going to tell me that these
are "dark numbers
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Should all places ω+2, ω+4, ω+6, ... remain empty?
Should the even numbers in spite of doubling remain below ω?
Then they must occupy places not existing before.
That means the original set had not contained all natural numbers. That
mans no actual or complete infinity.
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
That is simply false. You cannot specify a single number which is in
all endsegments.
True. This proves dark numbers.
Dark numbers don't exist, or at least they're not natural numbers. There
is no number in each and every end segment of N.
True. But those endsegments which have lost only finitely many numbers
and yet contain infinitely many, have an infinite intersection.
[ .... ]
Note: The shrinking endsegments cannot acquire new numbers.
An end segment is what it is. It doesn't change.
But the terms of the sequence do. Here is a simple finite example:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
Example of what? The reasoning you might do on finite sets mostly isn't
applicable to infinite sets.
Why not? The essence is that only finitely many numbers have been lost
and the rest is remaining.
Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets.
Not even an ignorant schoolboy would maintain this. The two TN-sets {0,
1, 2} and {3, 4, 5} have no numbers in common.
These sets do not belong to the above example. They are not TN-sets.
Quantifier shift: There is a subset of three elements common to all
TN-sets. Understood?
Yes, I understand completely.
No.
Now complete all sets by the natural numbers > 10 and complete the
sequence.
You can't "complete" a set.
You can add elements.
Regards, WM
It happens that WM formulated :
The essence is that only finitely many numbers have been lost
and the rest is remaining.
Now complete all sets by the natural numbers > 10 and complete the
sequence.
You can't "complete" a set.
You can add elements.
No, you cannot!
WM <wolfgang.mueckenheim@tha.de> wrote:
You've misunderstood the nature of N. The set is not
{1, 2, 3, ..., ω}, it is {1, 2, 3, ...}.
Should all places ω+2, ω+4, ω+6, ... remain empty?
It's not clear what you mean by this. There are no such "places".
Should the even numbers in spite of doubling remain below ω?
Yes, of course.
Then they must occupy places not existing before.
No. Remember the set is infinite, so you cannot use finite intuition to reason about it.
That means the original set had not contained all natural numbers. That
mans no actual or complete infinity.
Nonsense.
WM <wolfgang.mueckenheim@tha.de> wrote:
Dark numbers don't exist, or at least they're not natural numbers. There >>> is no number in each and every end segment of N.
True. But those endsegments which have lost only finitely many numbers
and yet contain infinitely many, have an infinite intersection.
End segments don't "lose" anything. They are what they are, namely well defined sets. Note that your "True" in your last paragraph, agrees that
the intersection of all end segments is empty, which you immediately contradict by asserting it is not empty.
[ .... ]
Note: The shrinking endsegments cannot acquire new numbers.
An end segment is what it is. It doesn't change.
But the terms of the sequence do. Here is a simple finite example:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets.
Now complete all sets by the natural numbers > 10 and complete the
sequence.
You can't "complete" a set.
You can add elements.
Then you get different sets, which weren't the ones you were trying to
reason about.
On 08.10.2024 23:08, Alan Mackenzie wrote:[copypasta]
WM <wolfgang.mueckenheim@tha.de> wrote:That is not in every case defined. But here are some rules:
On 07.10.2024 18:11, Alan Mackenzie wrote:What do you mean by "bigger" as applied to two infinite sets when one
What I should have written (WM please take note) is:
The idea of one countably infinite set being "bigger" than anotherThe idea is supported by the fact that set A as a superset of set B is
countably infinite set is simply nonsense.
bigger than B.
of them is not a subset of the other?
Not all infinite sets can be compared by size, but we can establish some useful rules.
What "main body"?Nonsense. Only potential infinity is used. Never the main body isSimply nonsense is the claim that there are as many algebraic numbersIt is not nonsense. The prime numbers can be put into 1-1
as prime numbers.
correspondence with the algebraic numbers, therefore there are exactly
as many of each.
applied.
I struggle to follow this illogic. Why should one segment have lessTheorem: If every endsegment has infinitely many numbers, thenFor Cantor's enumeration of all fractions I have given a simpleYour "proofs" tend to be nonsense.
disproof.
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.
Note: The shrinking endsegments cannot acquire new numbers.Not necessary, they already contain as many as needed.
On 09.10.2024 16:40, FromTheRafters wrote:Only as long as. Then we just get some E(n) for some finite n.
WM pretended :
On 09.10.2024 12:12, FromTheRafters wrote:
WM presented the following explanation :Inclusion-monotony proves that all infinite endsegments have a common
Theorem: If every endsegment has infinitely many numbers, thenA conjecture is not a proof. This one is simply another non sequitur.
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less >>>>> numbers.
infinite subset because only a loss of elements is possible. As long
as all endsegments are infinite, the loss has spared an infinite set
common to all.
Can you write that out for me?The completed sets are infinite.Or use a finite example.Finite sets again, I don't care how many.
Diminish the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} to get {2, 3, 4, 5,
6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
...
As long as five numbers remain in all sets, they are a common subset
of all sets.
Try using infinite sets.
Exactly. There is no last n.If there were a last as well as a first you would be right, but thereThe same numbers beyond any n are in all infinite endsegments. There are infinitely many because there is no last number.
is no last to be in all endsegments,
NB there are infinitely many endsegments (for each element...).so it is empty.Infinitely many numbers in all infinite endsegments do not make an empty intersection.
An infinite set can follow only on a finite number n.Never contradicted.
On 09.10.2024 16:13, Alan Mackenzie wrote:I.e. can be dismissed without comment.
WM <wolfgang.mueckenheim@tha.de> wrote:These rules are self-evident.
On 08.10.2024 23:08, Alan Mackenzie wrote:Possibly. But these rules would require proof, which you haven't
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.10.2024 18:11, Alan Mackenzie wrote:
What I should have written (WM please take note) is:
That is not in every case defined. But here are some rules:What do you mean by "bigger" as applied to two infinite sets when oneThe idea of one countably infinite set being "bigger" than another >>>>>> countably infinite set is simply nonsense.The idea is supported by the fact that set A as a superset of set B
is bigger than B.
of them is not a subset of the other?
Not all infinite sets can be compared by size, but we can establish
some useful rules.
supplied.
So 2N = G u {w, w+2, w+4, ..., w+w-2}?This breaks down in a contradiction, as shown by Richard D in another
post: To repeat his idea:
The set {0, 2, 4, 6, ...} is a subset of the natural numbers N,
{0, 1, 2, 3, 4, 5, 6, ...}, thus is smaller than it.
We can replace the second set by one of the same "size" by multiplying
each of its members by 4. We then get the set
{0, 4, 8, ...}.
Now this third set is a subset of the first hence is smaller than it.
No. When we *in actual infinity* multiply all |ℕ|natural numbers by 2,
then we keep |ℕ| numbers but only half of them are smaller than ω, i.e., are natural numbers. The other half is larger than ω.
It is astounding to see you jump from "there is no such number" toTrue. This proves dark numbers.Theorem: If every endsegment has infinitely many numbers, thenThat is simply false. You cannot specify a single number which is in
infinitely many numbers are in all endsegments.
all endsegments.
In the infinite case: All predecessors are supersets. But we are notBut the terms of the sequence do. Here is a simple finite example:Proof: If not, then there would be at least one endsegment with less... which is gobbledegook, not maths.
numbers.
Note: The shrinking endsegments cannot acquire new numbers.An end segment is what it is. It doesn't change.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets. Quantifier shift: There
is a subset of three elements common to all TN-sets.
Then we get: All sets which have lost at most n elements have theBut what about the limit case, the intersection of all endsegments,
remainder in common. Note: All sets which are infinite have lost at most
a finite number of elements.
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:What does this --------------^ specify exactly that distinguishes it from
WM <wolfgang.mueckenheim@tha.de> wrote:
You do not understand the least! The intersection of all endsegments is empty. The intersection of infinite endsegments is infinite.End segments don't "lose" anything. They are what they are, namelyDark numbers don't exist, or at least they're not natural numbers.True. But those endsegments which have lost only finitely many numbers
There is no number in each and every end segment of N.
and yet contain infinitely many, have an infinite intersection.
well defined sets. Note that your "True" in your last paragraph,
agrees that the intersection of all end segments is empty, which you
immediately contradict by asserting it is not empty.
You should really be more careful with your phrasing. Intersection withThe completion of the above sets does not change the principle:Then you get different sets, which weren't the ones you were trying toBut the terms of the sequence do. Here is a simple finite example:Note: The shrinking endsegments cannot acquire new numbers.An end segment is what it is. It doesn't change.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
Theorem: Every set that contains at least 3 numbers (call it TN-set) >>>>> holds these numbers in common with all TN-sets.
Now complete all sets by the natural numbers > 10 and complete the
sequence.
reason about.
Non-empty inclusion-monotonic sets like infinite endsegments have a
non-empty intersection. All endsegments have an empty intersection.
On 09.10.2024 17:11, Alan Mackenzie wrote:You haven't even written them down on the right. Are they included?
WM <wolfgang.mueckenheim@tha.de> wrote:
No. When we *in actual infinity* multiply all |ℕ|natural numbers by 2, >>> then we keep |ℕ| numbers but only half of them are smaller than ω,
i.e., are natural numbers. The other half is larger than ω.
Ha ha ha ha! This is garbage. If you think doubling some numbers
gives results which are "larger than ω" you'd better be prepared to
give an example of such a number. But you're surely going to tell me
that these are "dark numbers
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Should all places ω+2, ω+4, ω+6, ... remain empty?
Should the even numbers in spite of doubling remain below ω?What else would you expect?
Then they must occupy places not existing before. That means theOf course not. For every n, also n+1, n*2 and n^n are already included.
original set had not contained all natural numbers. That mans no actual
or complete infinity.
Not what we are discussing.True. But those endsegments which have lost only finitely many numbersDark numbers don't exist, or at least they're not natural numbers.True. This proves dark numbers.Theorem: If every endsegment has infinitely many numbers, thenThat is simply false. You cannot specify a single number which is in
infinitely many numbers are in all endsegments.
all endsegments.
There is no number in each and every end segment of N.
and yet contain infinitely many, have an infinite intersection.
Why should it.Note: The shrinking endsegments cannot acquire new numbers.
So, for an infinite number of infinite sets infinitely many numbersWhy not? The essence is that only finitely many numbers have been lostExample of what? The reasoning you might do on finite sets mostlyAn end segment is what it is. It doesn't change.But the terms of the sequence do. Here is a simple finite example:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
isn't applicable to infinite sets.
and the rest is remaining.
Which "these"?Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets.
LOLNo.Quantifier shift: There is a subset of three elements common to allYes, I understand completely.
TN-sets. Understood?
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
You've misunderstood the nature of N. The set is not
{1, 2, 3, ..., ω}, it is {1, 2, 3, ...}.
I use ℕ U {ω} for clarity.
Should all places ω+2, ω+4, ω+6, ... remain empty?
It's not clear what you mean by this. There are no such "places".
According to Cantor they are there in actual infinity.
Should the even numbers in spite of doubling remain below ω?
Yes, of course.
Then they must occupy places not existing before.
No. Remember the set is infinite, so you cannot use finite intuition to
reason about it.
Numbers multiplied by 2 do not remain unchanged. That is not intuition
but mathematics.
That means the original set had not contained all natural numbers. That
mans no actual or complete infinity.
Nonsense.
Sorry, you are a real believer but cannot discuss rational arguments.
Regards, WM
On 08.10.2024 19:28, Jim Burns wrote:
On 10/8/2024 6:18 AM, WM wrote:
All infinite endsegments contain
more than any finite set of numbers.
...still true if 'infinite' means "very large".
No.
Very large
is not more than any
finite set.
There are no other end segments,
none are finite.
They all have an infinite intersection.
The intersection of all end.segments is
the set of natural numbers in each end.segment.
Each natural number is not.in one or more end.segment.
That is true but wrong
in case of infinite endsegments which are infinite
because they have not lost all natural numbers.
The intersection of all end.segments is
the set of natural numbers in each end.segment.
Infinitely many are in every infinite endsegment.
Consider end segment E(k+1)
k+1 is in E(k+1)
Is k+1 in each end segment? Is k+1 in E(k+2)?
Is E(k+1)
the set of natural numbers in each end.segment?
No, but by definition
there are infinitely many numbers.
They are dark.
Why else should they be infinite?
Because each natural number is followed by
a natural number,
not only one but infinitely many
because 'infinite' DOES NOT mean 'very large'.
Fine.
All that blather does not contradict the fact that
infinite endsegments are infinite.
Theorem:
If every endsegment has infinitely many numbers,
then infinitely many numbers are in all endsegments.
Proof:
If not, then there would be
at least one endsegment with less numbers.
On 10/4/2024 3:37 PM, FromTheRafters wrote:
After serious thinking Chris M. Thomasson wrote :
On 10/4/2024 1:47 PM, FromTheRafters wrote:
WM formulated the question :
On 04.10.2024 17:36, Moebius wrote:
Am 04.10.2024 um 17:10 schrieb joes:
Am Fri, 04 Oct 2024 11:23:13 +0200 schrieb WM:
This is the precondition: The unit fractions are all existing! >>>>>>>> If so,
then they are different fixed points on the positive real line >>>>>>>> above
zero. Hence one is the first after zero.
But the MAIN PROBLEM is his usage of "hence", which clearly is not >>>>>> correct. Why on earth should there be a "first after zero".
Points on the real line are real points. One is the first. [WM]
Sounds intuitive, but for each first there is a firster.
Now WM is trying to say there is a first real number?
...and it is dark. :D
ROFL!!! ;^D
On 09.10.2024 14:29, joes wrote:Never can all numbers be used.
Am Wed, 09 Oct 2024 11:41:31 +0200 schrieb WM:
On 08.10.2024 21:17, joes wrote:
Am Tue, 08 Oct 2024 17:40:50 +0200 schrieb WM:
On 08.10.2024 15:36, joes wrote:
Am Tue, 08 Oct 2024 12:40:26 +0200 schrieb WM:
On 08.10.2024 12:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
All numbers n get indices of endsegments E(n).WDYM, all numbers in the segments are indices.As long as infinitely many numbers are captivated in endsegments, onlyEvery *finite* intersection.All of them differ by a finite set of numbers (which is irrelevant)The infinite sets contain what? No natural numbers? NaturalThese are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}. >>>>>> They contain all naturals larger than a given one, and nothing
numbers dancing around, sometimes being in a set, sometimes not? >>>>>>> An empty intersection requires that the infinite sets have
different elements.
else. Every natural is part of a finite number of these sets
(namely, its own value is that number). The set {n+1, n+2, …} does >>>>>> not contain n and is still infinite; there are (trivially)
infinitely many further such sets. All of them differ.
but contain an infinite set of numbers in common.
finitely many indices are available, and the intersection is between
finitely many infinite endsegments.
What about the core?But what about the intersection between all infinitely many segments?It is empty.
It cannot be empty because there are always numbers remaining.It is empty because all numbers are becoming indices and then get lost,What does this mean for the infinite intersection?Think about it this way: we are taking the limit of N\{0, 1, 2, …}.In the limit not a single natural number remains, let alone infinitely
many.
one by one.
Do you mean there are no potentially infinite sets?An endsegment is a set. All elements must exist. That requires actual infinity. In potential infinity numbers come into being - and never all.What is the difference between the sef of successors and an endsegment?They have successors but endsegments are sets and must be complete.Uh. So the naturals don’t have successors?If you imagine this as potential infinity,No, in potential infinity there are no endsegments.
Why can’t the segments be potentially infinite, or the successors
actually inf.?
On 09.10.2024 14:26, joes wrote:All endsegments are infinite.
Am Wed, 09 Oct 2024 11:47:57 +0200 schrieb WM:
On 08.10.2024 21:23, joes wrote:
Am Tue, 08 Oct 2024 17:30:19 +0200 schrieb WM:
On 08.10.2024 15:26, joes wrote:
Am Tue, 08 Oct 2024 12:46:01 +0200 schrieb WM:
Because infinitely many natural numbers are contained. This is
true for all infinite sets of the function. Therefore they cannot >>>>>>> have lost all numbers.
There are no others.What does „they” refer to in the last sentence?All endsegments which have infinitely many natural numbers.
Which intersection?We are, again, not talking about an element of the sequence, whichI am talking about such endsegments. Their intersection is infinite.
has a natural index, contains infinitely many successors and is
missing a finite number of predecessors.
What about the intersection of all infinitely many segments?Such an intersection is itself part of the sequence.Of course.
It is impossible to use up an infinity.Because every n has become an index and then is lost.Why?What we are talking about is the, pardon, limit of whatever function.The limit-endsegment is empty.
No. You seem to imagine them as finite but sharing a mysterious omega.The intersection is infinite because all infinite endsegments contain
the same infinite set. Some have lost more or less numbers but the
core remains infinite in all infinite endsegments.
And what if we intersect infinitely many?And how many segments have been intersected?(Potentially in-)finitely many because the collection of indices is
finite as long as an infinite set of numbers remains within the
endsegments.
Am Wed, 09 Oct 2024 16:40:21 +0200 schrieb WM:
When we *in actual infinity* multiply all |ℕ|natural numbers by 2,So 2N = G u {w, w+2, w+4, ..., w+w-2}?
then we keep |ℕ| numbers but only half of them are smaller than ω, i.e., >> are natural numbers. The other half is larger than ω.
But what about the limit case, the intersection of all endsegments,
or the set which has lost an infinite number of elements?
Am Wed, 09 Oct 2024 17:39:36 +0200 schrieb WM:
But those endsegments which have lost only finitely many numbersNot what we are discussing.
and yet contain infinitely many, have an infinite intersection.
So, for an infinite number of infinite sets infinitely many numbers
have been "lost", leaving nothing.
Am Wed, 09 Oct 2024 16:50:55 +0200 schrieb WM:
Only as long as. Then we just get some E(n) for some finite n.Inclusion-monotony proves that all infinite endsegments have a common
infinite subset because only a loss of elements is possible. As long
as all endsegments are infinite, the loss has spared an infinite set
common to all.
Can you write that out for me?The completed sets are infinite.Or use a finite example.Finite sets again, I don't care how many.
Diminish the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} to get {2, 3, 4, 5,
6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
...
As long as five numbers remain in all sets, they are a common subset
of all sets.
Try using infinite sets.
Exactly. There is no last n.If there were a last as well as a first you would be right, but thereThe same numbers beyond any n are in all infinite endsegments. There are
is no last to be in all endsegments,
infinitely many because there is no last number.
NB there are infinitely many endsegments (for each element...).so it is empty.Infinitely many numbers in all infinite endsegments do not make an empty
intersection.
Am Wed, 09 Oct 2024 14:48:17 +0200 schrieb WM:
Not all infinite sets can be compared by size, but we can establish some
useful rules.
That is a weakness of your notion of cardinality.
How do you compare finite sets?
Nonsense. Only potential infinity is used. Never the main body isWhat "main body"?
applied.
I struggle to follow this illogic. Why should one segment have lessTheorem: If every endsegment has infinitely many numbers, thenFor Cantor's enumeration of all fractions I have given a simpleYour "proofs" tend to be nonsense.
disproof.
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.
numbers?
Note: The shrinking endsegments cannot acquire new numbers.Not necessary, they already contain as many as needed.
SETS DON'T CHANGE.
WM <invalid@no.org> wrote:
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
You've misunderstood the nature of N. The set is not
{1, 2, 3, ..., ω}, it is {1, 2, 3, ...}.
I use ℕ U {ω} for clarity.
You would do better not to do so. It gives wrong results.
Numbers multiplied by 2 do not remain unchanged. That is not intuition
but mathematics.
True,
Am Wed, 09 Oct 2024 18:56:29 +0200 schrieb WM:
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:Especially since both all segments are infinite,
WM <wolfgang.mueckenheim@tha.de> wrote:
and there are infinitely many of them.
You should really be more careful with your phrasing. Intersection withThe completion of the above sets does not change the principle:But the terms of the sequence do. Here is a simple finite example: >>>>>> {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}Note: The shrinking endsegments cannot acquire new numbers.An end segment is what it is. It doesn't change.
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
Non-empty inclusion-monotonic sets like infinite endsegments have a
non-empty intersection. All endsegments have an empty intersection.
what?
Am Wed, 09 Oct 2024 15:35:10 +0200 schrieb WM:Then their intersection is infinite too.
All endsegments are infinite.
Which intersection?We are, again, not talking about an element of the sequence, whichI am talking about such endsegments. Their intersection is infinite.
has a natural index, contains infinitely many successors and is
missing a finite number of predecessors.
What about the intersection of all infinitely many segments?Such an intersection is itself part of the sequence.Of course.
It is impossible to use up an infinity.Because every n has become an index and then is lost.Why?What we are talking about is the, pardon, limit of whatever function. >>>> The limit-endsegment is empty.
And what if we intersect infinitely many?
WM <wolfgang.mueckenheim@tha.de> wrote:
If all natnumbers are there and if 2n is greater than n, then the
doubled numbers do not fit into ℕ.
For any finite n greater than zero, 2n is greater than n. The same
does not hold for infinite n.
Numbers multiplied by 2 do not remain unchanged. That is not intuition >>>> but mathematics.
True,
Fine, then you can follow the above discussion. Either doubling creates
new natural numbers. Then not all have been doubled. Or all have been
doubled, then some products fall outside of ℕ.
No. Not even close.
On 09.10.2024 21:13, Alan Mackenzie wrote:
WM <invalid@no.org> wrote:
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
You've misunderstood the nature of N. The set is not
{1, 2, 3, ..., ω}, it is {1, 2, 3, ...}.
I use ℕ U {ω} for clarity.
You would do better not to do so. It gives wrong results.
It gives unfamiliar results because you have no clear picture of actual infinity.
If all natnumbers are there and if 2n is greater than n, then the
doubled numbers do not fit into ℕ.
But note, that is only true if all natnumbers do exist.
If not all do exist, the doubling yields larger natnumbers, some of which have not existed before. But that means potential infinity.
Numbers multiplied by 2 do not remain unchanged. That is not intuition
but mathematics.
True,
Fine, then you can follow the above discussion. Either doubling creates
new natural numbers. Then not all have been doubled. Or all have been doubled, then some products fall outside of ℕ.
Regards, WM
Am Wed, 09 Oct 2024 15:29:17 +0200 schrieb WM:
On 09.10.2024 14:29, joes wrote:
Am Wed, 09 Oct 2024 11:41:31 +0200 schrieb WM:
On 08.10.2024 21:17, joes wrote:
Am Tue, 08 Oct 2024 17:40:50 +0200 schrieb WM:
On 08.10.2024 15:36, joes wrote:
Am Tue, 08 Oct 2024 12:40:26 +0200 schrieb WM:
On 08.10.2024 12:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Never can all numbers be used.All numbers n get indices of endsegments E(n).WDYM, all numbers in the segments are indices.As long as infinitely many numbers are captivated in endsegments, only >>>> finitely many indices are available, and the intersection is betweenEvery *finite* intersection.All of them differ by a finite set of numbers (which is irrelevant) >>>>>> but contain an infinite set of numbers in common.The infinite sets contain what? No natural numbers? NaturalThese are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}.
numbers dancing around, sometimes being in a set, sometimes not? >>>>>>>> An empty intersection requires that the infinite sets have
different elements.
They contain all naturals larger than a given one, and nothing
else. Every natural is part of a finite number of these sets
(namely, its own value is that number). The set {n+1, n+2, …} does >>>>>>> not contain n and is still infinite; there are (trivially)
infinitely many further such sets. All of them differ.
finitely many infinite endsegments.
What about the core?But what about the intersection between all infinitely many segments?It is empty.
It cannot be empty because there are always numbers remaining.It is empty because all numbers are becoming indices and then get lost,What does this mean for the infinite intersection?Think about it this way: we are taking the limit of N\{0, 1, 2, …}. >>>> In the limit not a single natural number remains, let alone infinitely >>>> many.
one by one.
An endsegment is a set. All elements must exist. That requires actualDo you mean there are no potentially infinite sets?
infinity. In potential infinity numbers come into being - and never all.
Do you actually understand what it means for a mathematical entity to
exist?
No, you don't understand what existence means.
WM <wolfgang.mueckenheim@tha.de> wrote:
It gives unfamiliar results because you have no clear picture of actual
infinity.
It is you who lacks a clear picture of infinity.
WM <wolfgang.mueckenheim@tha.de> wrote:
It gives unfamiliar results because you have no clear picture of actual
infinity.
It is you who lacks a clear picture of infinity.
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:This makes it clear that w-1, w-2 and so on are not included.
WM <wolfgang.mueckenheim@tha.de> wrote:
You've misunderstood the nature of N. The set is not {1, 2, 3, ...,I use ℕ U {ω} for clarity.
ω}, it is {1, 2, 3, ...}.
What are these "places"?According to Cantor they are there in actual infinity.Should all places ω+2, ω+4, ω+6, ... remain empty?It's not clear what you mean by this. There are no such "places".
They do, however, remain natural.Numbers multiplied by 2 do not remain unchanged. That is not intuitionShould the even numbers in spite of doubling remain below ω?Yes, of course.
Then they must occupy places not existing before.No. Remember the set is infinite, so you cannot use finite intuition
to reason about it.
but mathematics.
On 09.10.2024 12:12, FromTheRafters wrote:You really mean as long as only finitely many numbers are missing.
WM presented the following explanation :
Inclusion-monotony proves that all infinite endsegments have a common infinite subset because only a loss of elements is possible. As long asTheorem: If every endsegment has infinitely many numbers, thenA conjecture is not a proof. This one is simply another non sequitur.
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.
all endsegments are infinite, the loss has spared an infinite set common
to all.
If you can't understand try to find a counterexample.Notice anything? You should write it like this:
Or use a finite example.
On 09.10.2024 14:38, joes wrote:
Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:
With all infinite endsegments at once! Inclusion monotony. If you can't understand try to find a counterexample.With each, but not with all at once (cf. quantifier shift).An end segment E is infinite because each natural number has aTherefore every infinite endsegment has infinitely many elements with
successor,
so no element of E is max.E so not all nonempty subsets are two.ended
so E is not finite.
each predecessor in common. This is valid for all infinite
endsegments.
Only the intersection of finitely many segments is infinite.Intersection with what?There are no other end segments, none are finite.They all have an infinite intersection.
You haven't proved your logical rule in general. In particular, you can'tValid quantifier shift.Invalid quantifier shift.because 'infinite' DOES NOT mean 'very large'.Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
--The shrinking endsegments have all their elements in common with allProof: If not, then there would be at least one endsegment with lessNo. Why do you think that?
numbers.
their predecessors. As long as all are infinite, then all have an
infinite set in common.
On 10.10.2024 20:45, Alan Mackenzie wrote:Exactly! There are furthermore no infinite doubles of naturals (2n).
WM <wolfgang.mueckenheim@tha.de> wrote:
There are no infinite n = natural numbers.If all natnumbers are there and if 2n is greater than n, then theFor any finite n greater than zero, 2n is greater than n. The same
doubled numbers do not fit into ℕ.
does not hold for infinite n.
Additionally: if n is finite, so is 2n. It cannot go beyond w.Deplorable. But note that all natural numbers are finite and follow thisNo. Not even close.Either doublingNumbers multiplied by 2 do not remain unchanged.
creates new natural numbers. Then not all have been doubled. Or all
have been doubled, then some products fall outside of ℕ.
law: When doubled then 2n > n. If a set of natural numbers is doubled,
then the results cover a larger set than before.
On 09.10.2024 17:11, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
No.
When we *in actual infinity*
multiply all |ℕ|natural numbers by 2,
then we keep |ℕ| numbers
but only half of them are smaller than ω,
i.e., are natural numbers.
The other half is larger than ω.
Ha ha ha ha! This is garbage.
If you think doubling some numbers gives results
which are "larger than ω"
you'd better be prepared to give
an example of such a number.
But you're surely going to tell me that
these are "dark numbers
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Should all places ω+2, ω+4, ω+6, ... remain empty?
Should the even numbers in spite of doubling
remain below ω?
Then they must occupy places not existing before.
That means
the original set had not contained all natural numbers.
On 10/10/2024 12:24 PM, FromTheRafters wrote:
After serious thinking WM wrote :
On 09.10.2024 20:07, FromTheRafters wrote:
SETS DON'T CHANGE.
The set of French people has changed and will change.
Idiot!
On 10/10/2024 12:24 PM, FromTheRafters wrote:
After serious thinking WM wrote :
On 09.10.2024 20:07, FromTheRafters wrote:
SETS DON'T CHANGE.
The set of French people has changed and will change.
Idiot!
On 09.10.2024 19:08, joes wrote:How is that defined?
Am Wed, 09 Oct 2024 14:48:17 +0200 schrieb WM:
How do you compare finite sets?By their numbers of elements.
Those are of course also bijected, even though you cannot see them.The actually infinite numbers of dark elements.Nonsense. Only potential infinity is used. Never the main body isWhat "main body"?
applied.
Yes, all of them have |N|=Aleph_0 numbers, which is also the amountAll have the same numbers, namely ℕ. Some of the first numbers are transformed from contents to indices and than lost. But almost allI struggle to follow this illogic. Why should one segment have lessTheorem: If every endsegment has infinitely many numbers, thenFor Cantor's enumeration of all fractions I have given a simpleYour "proofs" tend to be nonsense.
disproof.
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.
numbers?
numbers, namely ℵo, remain (because after every definable natnumber n ℵo numbers follow). If the intersection is less than ℵo, at least one endsegment must have fewer than ℵo numbers.
Right. Even though all sets are infinite, no two are the same.It would be nessessary if all are infinite but their intersection isNote: The shrinking endsegments cannot acquire new numbers.Not necessary, they already contain as many as needed.
empty. Then the infinitely many numbers cannot be the same in all endsegments.
Satz: Jede Menge, die mindestens drei Elemente enthält (nennen wir diese Mengen DE), enthält drei Elemente gemeinsam mit allen DE-Mengen. Es gibt also drei Elemente, die in allen DE-Mengen enthalten sind. Das ist ein erlaubter Quantorentausch.A quantifier shift is never valid.
On 09.10.2024 19:30, joes wrote:You say w/2 were natural and comes after the darkness. What is the
Am Wed, 09 Oct 2024 16:40:21 +0200 schrieb WM:If all numbers are there initially and multiplied by 2. And if every
When we *in actual infinity* multiply all |ℕ|natural numbers by 2,So 2N = G u {w, w+2, w+4, ..., w+w-2}?
then we keep |ℕ| numbers but only half of them are smaller than ω,
i.e., are natural numbers. The other half is larger than ω.
number 2n is greater than n, then this is unavoidable.
Note the premise: If all are there. Actual infinity!
WDYM "causes"? There is no such segment.But what about the limit case, the intersection of all endsegments,The endsegment which has lost an infinite number of elements is empty
or the set which has lost an infinite number of elements?
and causes an empty intersection. But infinite endsegments have not lost
an infinite number of numbers.
On 10/10/2024 12:24 PM, FromTheRafters wrote:
After serious thinking WM wrote :
On 09.10.2024 20:07, FromTheRafters wrote:
SETS DON'T CHANGE.
The set of French people has changed and will change.
Idiot!
On 10.10.2024 20:45, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
If all natnumbers are there and if 2n is greater than n, then the
doubled numbers do not fit into ℕ.
For any finite n greater than zero, 2n is greater than n. The same
does not hold for infinite n.
There are no infinite n = natural numbers.
Numbers multiplied by 2 do not remain unchanged. That is not intuition >>>>> but mathematics.
True,
Fine, then you can follow the above discussion. Either doubling creates
new natural numbers. Then not all have been doubled. Or all have been
doubled, then some products fall outside of ℕ.
No. Not even close.
Deplorable. But note that all natural numbers are finite and follow this
law: When doubled then 2n > n. If a set of natural numbers is doubled,
then the results cover a larger set than before.
Regards, WM
On 10/10/2024 01:47 PM, Jim Burns wrote:
ω is the first (our) transfinite ordinal.
∀γ: γ ∈ ⟦0,ω⦆ ⇔
∀β ∈ ⦅0,γ⟧: ∃α ∈ ⟦0,β⦆: α+1=β
Halmos has for "infinite-dimensional vector spaces",
so not only is the Archimedean contrived
either "potential" or "un-bounded",
so is the matter of the count of dimensions
and the schema or quantification or
comprehension of the dimensions,
where there's a space like R^N in effect, or R^w,
then for a usual sort of idea that
"the first transfinite ordinal"
is only kind of after all those, ...,
like a "spiral space-filling curve".
On 10/10/2024 6:38 PM, Richard Damon wrote:
On 10/10/24 2:32 PM, WM wrote:
On 09.10.2024 21:13, Alan Mackenzie wrote:
WM <invalid@no.org> wrote:
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
You've misunderstood the nature of N. The set is not
{1, 2, 3, ..., ω}, it is {1, 2, 3, ...}.
I use ℕ U {ω} for clarity.
You would do better not to do so. It gives wrong results.
It gives unfamiliar results because you have no clear picture of
actual infinity. If all natnumbers are there and if 2n is greater
than n, then the doubled numbers do not fit into ℕ. But note, that is
only true if all natnumbers do exist.
No, *YOU* don't understand what actual infinity is like as you picture
it just like the finite, but bigger.
If not all do exist, te doubling yields larger natnumbers, some of
which have not existed before. But that means potential infinity.
But there aren't any natural numbers that have not existed before.
That is just showing that your "actual infinity" is a finite set that
you grabbed on the way to infinity, but didn't get there yet.
Actual infinity means you need to wait until you get there. The[...]
problem is finite creatures can't do that, so can't handle actual
infinity.
What do you mean? I can say all the natural numbers. That was pretty
fast for all of them. ;^)
On 10/10/24 11:19 PM, Chris M. Thomasson wrote:
On 10/10/2024 6:38 PM, Richard Damon wrote:
On 10/10/24 2:32 PM, WM wrote:
If not all do exist,
te doubling yields larger natnumbers,
some of which have not existed before.
But that means potential infinity.
But there aren't any natural numbers that
have not existed before.
That is just showing that
your "actual infinity" is a finite set that
you grabbed on the way to infinity,
but didn't get there yet.
Actual infinity means
you need to wait until you get there.
The problem is
finite creatures can't do that,
so can't handle actual infinity.
What do you mean?
I can say all the natural numbers.
That was pretty fast for all of them. ;^)
Which names the set,
but not actually list the contents of that set.
On 10/10/2024 09:22 PM, Jim Burns wrote:
What I mean by 'first transfinite ordinal' ω is that,
of all ordinals which are not.finite,
ω is the first such ordinal.
What I mean by 'ordinal' is that
each set of ordinals holds a minimum or is empty.
What I mean by 'finite ordinal' γ is that
it is first (ie, γ=0) or
its predecessor.ordinal γ-1 exists and,
for each non.0 prior ordinal β<γ
its predecessor.ordinal β-1 exists.
The mathematical objects of
a strong mathematical platonist's mathematical universe
are _discovered_, not _invented_.
and for strong mathematical platonists
these are the words we use for
"infinite-dimensional vector spaces":
infinite-dimensional vector spaces.
and for strong mathematical platonists
these are the words we use for
"infinite-dimensional vector spaces":
infinite-dimensional vector spaces.
On 10/11/2024 7:06 AM, Richard Damon wrote:
On 10/10/24 11:19 PM, Chris M. Thomasson wrote:
On 10/10/2024 6:38 PM, Richard Damon wrote:
On 10/10/24 2:32 PM, WM wrote:
If not all do exist,
te doubling yields larger natnumbers,
some of which have not existed before.
But that means potential infinity.
But there aren't any natural numbers that
have not existed before. That is just showing that
your "actual infinity" is a finite set that you grabbed on the way
to infinity,
but didn't get there yet.
Actual infinity means
you need to wait until you get there.
The problem is
finite creatures can't do that,
so can't handle actual infinity.
What do you mean?
I can say all the natural numbers.
That was pretty fast for all of them. ;^)
Which names the set,
but not actually list the contents of that set.
Is a set
without a finite list but
with a finite description
potential or actual?
It looks to me as though the consensus is:
we don't care what the answer is.
We have a description of ℕ and its elements,
free of mentions of 'potential' and 'actual'.
From its description,
we can reason about it and its elements.
"Potential ℕ" vs. "actual ℕ" leaves unchanged
which claims we reason to.
I see them as in the same vein as
weekday.mathematics and Sunday.mathematics.[1]
A distinction without a difference.
And I see that indifference as there _by design_
It is why unwelcome mathematical results are accepted
not because mathematicians are paragons of rectitude
(no offense)
but because there is no weasel.ability in mathematics.
⎛ Not everyone appreciates that lack of weasel.ability.
⎝ Those who don't self.select to be non.mathematicians.
[1]
⎛
⎜ Most writers on the subject seem to agree that
⎜ the typical “working mathematician” is
⎜ a Platonist on weekdays and a formalist on Sundays.
⎜ That is, when he is doing mathematics,
⎜ he is convinced that
⎜ he is dealing with an objective reality
⎜ whose properties he is attempting to determine.
⎜ But then, when challenged to give
⎜ a philosophical account of this reality,
⎜ he finds it easiest to pretend that
⎜ he does not believe in it after all.
⎝
https://core.ac.uk/download/pdf/82047627.pdf
Some Proposals for Reviving the Philosophy of Mathematics
REUBEN HERSH
Department of Mathematics, University of New Mexico,
Albuquerque, New Mexico 87131
On 10/11/2024 4:06 AM, Richard Damon wrote:
On 10/10/24 11:19 PM, Chris M. Thomasson wrote:
On 10/10/2024 6:38 PM, Richard Damon wrote:
On 10/10/24 2:32 PM, WM wrote:
On 09.10.2024 21:13, Alan Mackenzie wrote:
WM <invalid@no.org> wrote:
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
You've misunderstood the nature of N. The set is not
{1, 2, 3, ..., ω}, it is {1, 2, 3, ...}.
I use ℕ U {ω} for clarity.
You would do better not to do so. It gives wrong results.
It gives unfamiliar results because you have no clear picture of
actual infinity. If all natnumbers are there and if 2n is greater
than n, then the doubled numbers do not fit into ℕ. But note, that >>>>> is only true if all natnumbers do exist.
No, *YOU* don't understand what actual infinity is like as you
picture it just like the finite, but bigger.
If not all do exist, te doubling yields larger natnumbers, some of
which have not existed before. But that means potential infinity.
But there aren't any natural numbers that have not existed before.
That is just showing that your "actual infinity" is a finite set
that you grabbed on the way to infinity, but didn't get there yet.
Actual infinity means you need to wait until you get there. The[...]
problem is finite creatures can't do that, so can't handle actual
infinity.
What do you mean? I can say all the natural numbers. That was pretty
fast for all of them. ;^)
Which names the set, but not actually list the contents of that set.
Who's counting anyway? lol. Infinite is infinite and finite is, well, finite... ;^) Just because we are finite beings does not negate the
infinite, right?
Am Wed, 09 Oct 2024 15:49:27 +0200 schrieb WM:
As long asYou really mean as long as only finitely many numbers are missing.
all endsegments are infinite, the loss has spared an infinite set common
to all.
If you can't understand try to find a counterexample.Notice anything? You should write it like this:
Or use a finite example.
{1, 2, 3, ...}
{2, 3, 4, ...}
{3, 4, 5, ...}
...
Am Wed, 09 Oct 2024 18:47:39 +0200 schrieb WM:
Numbers multiplied by 2 do not remain unchanged. That is not intuitionThey do, however, remain natural.
but mathematics.
On 10.10.2024 21:48, joes wrote:
Am Wed, 09 Oct 2024 18:47:39 +0200 schrieb WM:
Numbers multiplied by 2 do not remain unchanged. That is not intuitionThey do, however, remain natural.
but mathematics.
They do not remain the same set as before. They cover more of the real
line. If they all are natnumbers, then there are more than at the
outset. That means potential infinity. If there are not more natumbers
than at the outset, then infinite numbers have been created. There is no
way to avoid one of these results.
Regards, WM
On 10.10.2024 21:48, joes wrote:Of course multiplying a (finite) number by 2 changes the value.
Am Wed, 09 Oct 2024 18:47:39 +0200 schrieb WM:
They do not remain the same set as before. They cover more of the realNumbers multiplied by 2 do not remain unchanged. That is not intuitionThey do, however, remain natural.
but mathematics.
line. If they all are natnumbers, then there are more than at the
outset. That means potential infinity. If there are not more natumbers
than at the outset, then infinite numbers have been created. There is no
way to avoid one of these results.
Am Wed, 09 Oct 2024 15:24:21 +0200 schrieb WM:
On 09.10.2024 14:38, joes wrote:
Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:With all infinite endsegments at once! Inclusion monotony. If you can't
With each, but not with all at once (cf. quantifier shift).An end segment E is infinite because each natural number has aTherefore every infinite endsegment has infinitely many elements with
successor,
so no element of E is max.E so not all nonempty subsets are two.ended >>>>> so E is not finite.
each predecessor in common. This is valid for all infinite
endsegments.
understand try to find a counterexample.
Only the intersection of finitely many segments is infinite.Intersection with what?There are no other end segments, none are finite.They all have an infinite intersection.
You haven't proved your logical rule in general. In particular, you can'tValid quantifier shift.Invalid quantifier shift.because 'infinite' DOES NOT mean 'very large'.Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
use it here to argue it is valid, that would be circular.
Can you explain what a quantifier shift is?
On 10/9/2024 11:39 AM, WM wrote:
On 09.10.2024 17:11, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
No.
When we *in actual infinity*
multiply all |ℕ|natural numbers by 2,
then we keep |ℕ| numbers
but only half of them are smaller than ω,
i.e., are natural numbers.
The other half is larger than ω.
You (WM) are treating ω as though it is (our) finite.
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Each γ≠0 preceding ω is (our) finite.
β < ω ∧ γ < ω ⇒ β+γ < ω
⎛ Assume a counterexample.
⎜ Assume
⎜ β < ω ∧ γ < ω ∧ β+γ ≥ ω
β < ω ∧ γ < ω ⇒ β×γ < ω
⎛ Assume a counterexample.
⎜ Assume
⎜ β < ω ∧ γ < ω ∧ β×γ ≥ ω
Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:
On 10.10.2024 20:45, Alan Mackenzie wrote:Exactly! There are furthermore no infinite doubles of naturals (2n).
WM <wolfgang.mueckenheim@tha.de> wrote:There are no infinite n = natural numbers.
If all natnumbers are there and if 2n is greater than n, then theFor any finite n greater than zero, 2n is greater than n. The same
doubled numbers do not fit into ℕ.
does not hold for infinite n.
Additionally: if n is finite, so is 2n. It cannot go beyond w.Deplorable. But note that all natural numbers are finite and follow thisNo. Not even close.Either doublingNumbers multiplied by 2 do not remain unchanged.
creates new natural numbers. Then not all have been doubled. Or all
have been doubled, then some products fall outside of ℕ.
law: When doubled then 2n > n. If a set of natural numbers is doubled,
then the results cover a larger set than before..
Am Thu, 10 Oct 2024 19:56:26 +0200 schrieb WM:
On 09.10.2024 19:08, joes wrote:
Am Wed, 09 Oct 2024 14:48:17 +0200 schrieb WM:
How is that defined?How do you compare finite sets?By their numbers of elements.
Satz: Jede Menge, die mindestens drei Elemente enthält (nennen wir dieseA quantifier shift is never valid.
Mengen DE), enthält drei Elemente gemeinsam mit allen DE-Mengen. Es gibt
also drei Elemente, die in allen DE-Mengen enthalten sind. Das ist ein
erlaubter Quantorentausch.
On 10/10/24 2:32 PM, WM wrote:
If not all do exist, the doubling yields larger natnumbers, some of
which have not existed before. But that means potential infinity.
But there aren't any natural numbers that have not existed before.
The SIZE of the set of natural numbers is infinite, and thus obeys the
laws of infinite numbers. An infinite number, which has a finite number, added to, multiplied by, or used as a power, is still that same infinite number. It may seem impossible, but that is the nature of infinite numbers.
On 10.10.2024 21:54, joes wrote:
Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:
On 10.10.2024 20:45, Alan Mackenzie wrote:Exactly! There are furthermore no infinite doubles of naturals (2n).
WM <wolfgang.mueckenheim@tha.de> wrote:There are no infinite n = natural numbers.
If all natnumbers are there and if 2n is greater than n, then theFor any finite n greater than zero, 2n is greater than n. The same
doubled numbers do not fit into ℕ.
does not hold for infinite n.
But the doubles are larger. Hence after doubling the set has a smaller density
"Potential ℕ" vs. "actual ℕ" leaves unchanged
which claims we reason to.
On 10/12/24 9:32 AM, WM wrote:
On 10.10.2024 21:48, joes wrote:
Am Wed, 09 Oct 2024 18:47:39 +0200 schrieb WM:
Numbers multiplied by 2 do not remain unchanged. That is not intuition >>>> but mathematics.They do, however, remain natural.
They do not remain the same set as before. They cover more of the real
line. If they all are natnumbers, then there are more than at the
outset. That means potential infinity. If there are not more natumbers
than at the outset, then infinite numbers have been created. There is
no way to avoid one of these results.
No, they do not cover "more" of the line, as they still cover exactly
Aleph_0 point within the range of the finite numbers below omega.
The problem is you are applying properties of FINITE sets
Am Sat, 12 Oct 2024 15:32:25 +0200 schrieb WM:
On 10.10.2024 21:48, joes wrote:Of course multiplying a (finite) number by 2 changes the value.
Am Wed, 09 Oct 2024 18:47:39 +0200 schrieb WM:They do not remain the same set as before. They cover more of the real
Numbers multiplied by 2 do not remain unchanged. That is not intuition >>>> but mathematics.They do, however, remain natural.
line. If they all are natnumbers, then there are more than at the
outset. That means potential infinity. If there are not more natumbers
than at the outset, then infinite numbers have been created. There is no
way to avoid one of these results.
If anything, the even numbers cover less, being a subset of the naturals.
If doubling a natural yielded omega, there would need to be an n = w/2.
That is not defined, nor finite.
But the doubles are larger. Hence after doubling the set has a smaller
density
Not with the (always) Dedekind-infinite sets
On 12.10.2024 20:26, Tom Bola wrote:How do you define that?
2n > n. That holds for all finite numbers. It halves the density.But the doubles are larger. Hence after doubling the set has a smallerNot with the (always) Dedekind-infinite sets
density
On 11.10.2024 01:01, joes wrote:What is counting? Seriously.
Am Thu, 10 Oct 2024 19:56:26 +0200 schrieb WM:How far can you count? Already until one hundred?
On 09.10.2024 19:08, joes wrote:How is that defined?
Am Wed, 09 Oct 2024 14:48:17 +0200 schrieb WM:
How do you compare finite sets?By their numbers of elements.
But that's too easy. I have been doing the same. Why are you posting?I am not interested in what you find valid. I am satisfied if mySatz: Jede Menge, die mindestens drei Elemente enthält (nennen wirA quantifier shift is never valid.
diese Mengen DE), enthält drei Elemente gemeinsam mit allen DE-Mengen.
Es gibt also drei Elemente, die in allen DE-Mengen enthalten sind. Das
ist ein erlaubter Quantorentausch.
reasoning cannot be contradicted by counterexample.
Am Sat, 12 Oct 2024 20:55:44 +0200 schrieb WM:
2n > n. That holds for all finite numbers. It halves the density.How do you define that?
Am Sat, 12 Oct 2024 20:13:11 +0200 schrieb WM:
What is counting? Seriously.
But that's too easy.I am not interested in what you find valid. I am satisfied if mySatz: Jede Menge, die mindestens drei Elemente enthält (nennen wirA quantifier shift is never valid.
diese Mengen DE), enthält drei Elemente gemeinsam mit allen DE-Mengen. >>>> Es gibt also drei Elemente, die in allen DE-Mengen enthalten sind. Das >>>> ist ein erlaubter Quantorentausch.
reasoning cannot be contradicted by counterexample.
Why are you posting?
On 12.10.2024 20:26, Tom Bola wrote:
But the doubles are larger. Hence after doubling the set has a smaller
density
Not with the (always) Dedekind-infinite sets
2n > n. That holds for all finite numbers. It halves the density.
On 10.10.2024 21:54, joes wrote:
Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:
On 10.10.2024 20:45, Alan Mackenzie wrote:Exactly! There are furthermore no infinite doubles of naturals (2n).
WM <wolfgang.mueckenheim@tha.de> wrote:There are no infinite n = natural numbers.
If all natnumbers are there and if 2n is greater than n, then theFor any finite n greater than zero, 2n is greater than n. The same
doubled numbers do not fit into ℕ.
does not hold for infinite n.
But the doubles are larger. Hence after doubling the set has a smaller density and therefore a larger extension on the real line. Hence not all natural numbers have been doubled.
Additionally: if n is finite, so is 2n. It cannot go beyond w.Deplorable. But note that all natural numbers are finite and follow this >>> law: When doubled then 2n > n. If a set of natural numbers is doubled,No. Not even close.Either doublingNumbers multiplied by 2 do not remain unchanged.
creates new natural numbers. Then not all have been doubled. Or all
have been doubled, then some products fall outside of ℕ.
then the results cover a larger set than before..
Then there is no complete set. The doubling can be repeated and
repeated. Always new numbers are created. Potential infinity.
Regards, WM
On 10.10.2024 21:51, joes wrote:What? There's a segment for every number.
Am Wed, 09 Oct 2024 15:24:21 +0200 schrieb WM:
On 09.10.2024 14:38, joes wrote:
Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:
Of course. If infinitely many numbers are indices, then infinitely many cannot be contents.Only the intersection of finitely many segments is infinite.Intersection with what?There are no other end segments, none are finite.They all have an infinite intersection.
Then you need to explain why the shifted proposition should be true.The proof of the cake lies in the eating.You haven't proved your logical rule in general. In particular, youValid quantifier shift.Invalid quantifier shift.because 'infinite' DOES NOT mean 'very large'.Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
can't use it here to argue it is valid, that would be circular.
I do not intend to prove quantifier shift in general. I prove: If every endsegment is infinite, then infinitely many numbers are in all
endsegments of this subset.
Not valid for infinite sets.Can you explain what a quantifier shift is?The above arguing. If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.
Proof: Inclusion monotony.
On 09.10.2024 19:40, joes wrote:Yes, and what do you claim?
Am Wed, 09 Oct 2024 17:39:36 +0200 schrieb WM:It is claimed that all endsegments are infinite, hence have lost only finitely many numbers as indices, but have an empty intersection.
But those endsegments which have lost only finitely many numbers andNot what we are discussing.
yet contain infinitely many, have an infinite intersection.
If there were a largest n, all previous segments were finite.So, for an infinite number of infinite sets infinitely many numbersTrue. Therefore not all endsegments are infinity.
have been "lost", leaving nothing.
On 12.10.2024 21:23, joes wrote:How do you define "space"?
Am Sat, 12 Oct 2024 20:55:44 +0200 schrieb WM:
Same number of numbers occupy twice the space on the real line. Alternatively: Half number of numbers occupy same space.2n > n. That holds for all finite numbers. It halves the density.How do you define that?
On 10/11/2024 11:02 AM, Jim Burns wrote:
⎜ the typical “working mathematician” is
⎜ a Platonist on weekdays and a formalist on Sundays.
I'm familiar with that you tend to repost that,
yet platonism and formalism don't necessarily
contradict nor preclude each other, with regards
to neither necessarily being fictionalism.
Mathematics is not "pretend",
and it's in accords to
some ultimate ontological commitment
to the _truth_ of the mathematical _facts_,
not "make-believe".
"Aristotle
"Zeno
"Cantor
"Russell
Frege
Anaximander
Goedel
Cohen
some ultimate ontological commitment
to the _truth_ of the mathematical _facts_,
Am Thu, 10 Oct 2024 20:12:56 +0200 schrieb WM:Well, they are.
It is claimed that all endsegments are infinite [...]
[and that they] have an empty intersection.
Yes, and what do you claim?
[...] not all endsegments are infinit[e].
Am Sat, 12 Oct 2024 18:51:10 +0200 schrieb WM:
On 10.10.2024 21:51, joes wrote:
Am Wed, 09 Oct 2024 15:24:21 +0200 schrieb WM:
On 09.10.2024 14:38, joes wrote:
Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Invalid quantifier shift.
Valid quantifier shift.
You haven't proved your logical rule in general. In particular, you
can't use it here to argue it is valid, that would be circular.
I do not intend to prove quantifier shift in general. I prove: If every
endsegment is infinite, then infinitely many numbers are in all
endsegments [...]
Then you need to explain why the shifted proposition should be true.
[...] If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.
Proof: Inclusion monotony.
Not valid for [...]
Am Sat, 12 Oct 2024 18:51:10 +0200 schrieb WM:
On 10.10.2024 21:51, joes wrote:
Am Wed, 09 Oct 2024 15:24:21 +0200 schrieb WM:
On 09.10.2024 14:38, joes wrote:
Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:
Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Invalid quantifier shift.
Valid quantifier shift.
You haven't proved your logical rule in general. In particular, you
can't use it here to argue it is valid, that would be circular.
I do not intend to prove quantifier shift in general. I prove: If every
endsegment is infinite, then infinitely many numbers are in all
endsegments [...]
Then you need to explain why the shifted proposition should be true.
[...] If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.
Proof: Inclusion monotony.
Not valid for [...]
On 10/11/2024 3:15 PM, Richard Damon wrote:
Right, but it means we lack some of the ability to actually PERCEIVE
the reality of the infinite.
That's interesting to me because it kind of reminds me of 2d vs 3d vs 4d observers. Us 3d beings can look down at a 2d beings world and see all
of it. A 4d being can look at our 3d world and see all of it. Think of a
4d being at a location in n-ary space that has a non-zero 4d component
to its vectors. We would not be able to see it because is off axis in a higher dimension. However, it can see us, right through us, all the way down...
When one tries to do that, it is too easy to presume characteristics
of the infinite are just like the finite, when they aren't (like
having "ends").
On 11.10.2024 20:02, Jim Burns wrote:
We have a description of ℕ and its elements,
free of mentions of 'potential' and 'actual'.
From its description,
we can reason about it and its elements.
"Potential ℕ" vs. "actual ℕ" leaves unchanged
which claims we reason to.
Potential infinity:
All natural numbers when double yield
natural numbers
but larger numbers are among the result.
In actual infinity
no new naturals can be created
but since doubling
doubles the value of each number,
numbers of the second number class are created.
On 10.10.2024 22:47, Jim Burns wrote:
On 10/9/2024 11:39 AM, WM wrote:
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Each γ≠0 preceding ω is (our) finite.
There are two alternatives:
Either doubling creates natnumbers,
then they are not among those doubled,
then we have not doubles all.
Or we have doubled all
but then larger numbers have been created.
β < ω ∧ γ < ω ⇒ β+γ < ω
⎛ Assume a counterexample.
⎜ Assume
⎜ β < ω ∧ γ < ω ∧ β+γ ≥ ω
Dark numbers have no discernible individuality.
β < ω ∧ γ < ω ⇒ β×γ < ω
⎛ Assume a counterexample.
⎜ Assume
⎜ β < ω ∧ γ < ω ∧ β×γ ≥ ω
Dark numbers have no discernible individuality.
On 9/18/2024 7:31 AM, Jim Burns wrote:
On 9/18/2024 8:39 AM, WM wrote:
On 16.09.2024 19:30, Jim Burns wrote:
On 9/15/2024 3:47 PM, WM wrote:
I don't believe in gaps on the real line.
There aren't gaps and there aren't next.numbers
in numbers.situating.splits of rationals with
countable.to.numerators.and.denominators
So what is next instead?
What is between one and the next?
A gap.
With regard to a very _strict) line of thinking
(no mixing and matching), say 100% natural numbers...
There is NO "gap" between, say:
3 and 4
There is no gap in the real line.
There is no next in the real line.
If there were, there'd be a gap.
WM used his keyboard to write :
I use properties of finite numbers.
Wrong, you disbelieve (when it suits you) the idea of cardinality.
On 10/12/24 2:55 PM, WM wrote:
2n > n. That holds for all finite numbers. It halves the density.
And half of infinity is infinity.
On 10/12/24 1:49 PM, WM wrote:
On 10.10.2024 21:54, joes wrote:
Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:
On 10.10.2024 20:45, Alan Mackenzie wrote:Exactly! There are furthermore no infinite doubles of naturals (2n).
WM <wolfgang.mueckenheim@tha.de> wrote:There are no infinite n = natural numbers.
If all natnumbers are there and if 2n is greater than n, then theFor any finite n greater than zero, 2n is greater than n. The same >>>>> does not hold for infinite n.
doubled numbers do not fit into ℕ.
But the doubles are larger. Hence after doubling the set has a smaller
density and therefore a larger extension on the real line. Hence not
all natural numbers have been doubled.
The doubles are larger than the element they replace, but that value was always in the set to begin with, so it never creates a "new" term.
law: When doubled then 2n > n. If a set of natural numbers is doubled, >>>> then the results cover a larger set than before..Additionally: if n is finite, so is 2n. It cannot go beyond w.
Then there is no complete set. The doubling can be repeated and
repeated. Always new numbers are created. Potential infinity.
You got it! The complete set of infinity is not available to finite
beings to directly observe and handle, because it is just too big for
use to work on.
On 10/12/24 2:19 PM, WM wrote:
On 11.10.2024 03:38, Richard Damon wrote:
The SIZE of the set of natural numbers is infinite, and thus obeys
the laws of infinite numbers. An infinite number, which has a finite
number, added to, multiplied by, or used as a power, is still that
same infinite number. It may seem impossible, but that is the nature
of infinite numbers.
No natural number is infinite. They all obey the law of finite
numbers. That includes the law that 2n > n.
Right, but for any number n that is a natural number 2n is also a
natural number and in the set.
On 10/12/24 2:39 PM, WM wrote:
On 11.10.2024 20:02, Jim Burns wrote:
"Potential ℕ" vs. "actual ℕ" leaves unchanged
which claims we reason to.
Potential infinity: All natural numbers when double yield natural
numbers but larger numbers are among the result.
In actual infinity no new naturals can be created but since doubling
doubles the value of each number, numbers of the second number class
are created.
No, in actual infinity no new natural can be created,
but every natural
number can find the one that is twice itself in the set,
Am Sat, 12 Oct 2024 19:49:23 +0200 schrieb WM:
On 10.10.2024 21:54, joes wrote:Taking "density" here to mean cardinality
Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:But the doubles are larger. Hence after doubling the set has a smaller
On 10.10.2024 20:45, Alan Mackenzie wrote:Exactly! There are furthermore no infinite doubles of naturals (2n).
WM <wolfgang.mueckenheim@tha.de> wrote:There are no infinite n = natural numbers.
If all natnumbers are there and if 2n is greater than n, then theFor any finite n greater than zero, 2n is greater than n. The same
doubled numbers do not fit into ℕ.
does not hold for infinite n.
density and therefore a larger extension on the real line. Hence not all
natural numbers have been doubled.
The real line reaches until, but does not
include omega, no matter your step size.
What value do you suppose n^2
and n^n diverge to?
No! Actual infinity already includes all doubles of all numbers.Then there is no complete set. The doubling can be repeated andAdditionally: if n is finite, so is 2n. It cannot go beyond w.Deplorable. But note that all natural numbers are finite and followNo. Not even close.Either doubling creates new natural numbers. Then not all have been >>>>>> doubled. Or all have been doubled, then some products fall outside >>>>>> of ℕ.Numbers multiplied by 2 do not remain unchanged.
this law: When doubled then 2n > n. If a set of natural numbers is
doubled, then the results cover a larger set than before..
repeated. Always new numbers are created. Potential infinity.
On 12.10.2024 22:47, Richard Damon wrote:Of course, and if the infinity is actual, they must already be in there.
On 10/12/24 2:39 PM, WM wrote:True.
On 11.10.2024 20:02, Jim Burns wrote:No, in actual infinity no new natural can be created,
"Potential ℕ" vs. "actual ℕ" leaves unchanged which claims we reason >>>> to.Potential infinity: All natural numbers when double yield natural
numbers but larger numbers are among the result.
In actual infinity no new naturals can be created but since doubling
doubles the value of each number, numbers of the second number class
are created.
but every natural number can find the one that is twice itself in theFalse. n*2 > n for all numbers, even for infinite numbers.
set,
Am Sat, 12 Oct 2024 18:51:10 +0200 schrieb WM:
On 10.10.2024 21:51, joes wrote:
Am Wed, 09 Oct 2024 15:24:21 +0200 schrieb WM:
On 09.10.2024 14:38, joes wrote:
Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:
What? There's a segment for every number.Of course. If infinitely many numbers are indices, then infinitely manyOnly the intersection of finitely many segments is infinite.Intersection with what?There are no other end segments, none are finite.They all have an infinite intersection.
cannot be contents.
I do not intend to prove quantifier shift in general. I prove: If everyThen you need to explain why the shifted proposition should be true.
endsegment is infinite, then infinitely many numbers are in all
endsegments of this subset.
Not valid for infinite sets.Can you explain what a quantifier shift is?The above arguing. If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.
Proof: Inclusion monotony.
On 13.10.2024 00:50, joes wrote:Any number div. by an inf. number ought to return density zero. OTOH,
Am Sat, 12 Oct 2024 19:49:23 +0200 schrieb WM:Cardinality is nonsense. Density is numbers per places for every finite
On 10.10.2024 21:54, joes wrote:Taking "density" here to mean cardinality
Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:But the doubles are larger. Hence after doubling the set has a smaller
On 10.10.2024 20:45, Alan Mackenzie wrote:Exactly! There are furthermore no infinite doubles of naturals (2n).
WM <wolfgang.mueckenheim@tha.de> wrote:There are no infinite n = natural numbers.
If all natnumbers are there and if 2n is greater than n, then the >>>>>>> doubled numbers do not fit into ℕ.For any finite n greater than zero, 2n is greater than n. The same >>>>>> does not hold for infinite n.
density and therefore a larger extension on the real line. Hence not
all natural numbers have been doubled.
set.
That is 2N u {2w} = G u {w}. If you insist on multiplying the other wayThe real line reaches until, but does not include omega, no matter yourThen double ℕ U {ω}.
step size.
w^2 and w^w, which both still have cardinality Aleph_0.What value do you suppose n^2 and n^n diverge to?More difficult to determine than 2n.
w*2 != 2w = wThen you are outside of basic mathematics. The double number is largerNo! Actual infinity already includes all doubles of all numbers.Then there is no complete set. The doubling can be repeated andAdditionally: if n is finite, so is 2n. It cannot go beyond w.Deplorable. But note that all natural numbers are finite and followNo. Not even close.Either doubling creates new natural numbers. Then not all haveNumbers multiplied by 2 do not remain unchanged.
been doubled. Or all have been doubled, then some products fall
outside of ℕ.
this law: When doubled then 2n > n. If a set of natural numbers is
doubled, then the results cover a larger set than before..
repeated. Always new numbers are created. Potential infinity.
than the doubled - in every case, even for infinite numbers.
On 12.10.2024 22:47, Richard Damon wrote:And there being inf. many places, that makes both amounts infinite.
On 10/12/24 2:55 PM, WM wrote:
The density of the natural umbers is 1 number per place, the density of2n > n. That holds for all finite numbers. It halves the density.And half of infinity is infinity.
the doubled numbers is 1/2 number per place.
On 12.10.2024 22:07, FromTheRafters wrote:It is a property of all sets, be they finite or actually infinite.
WM used his keyboard to write :
Cardinality is a property of potentially infinite sets.I use properties of finite numbers.Wrong, you disbelieve (when it suits you) the idea of cardinality.
On 12.10.2024 22:47, Richard Damon wrote:The set N does not have an upper bound on even numbers, or at all.
On 10/12/24 2:19 PM, WM wrote:But not in the set of numbers to be doubled. Half of the doubled numbers
On 11.10.2024 03:38, Richard Damon wrote:Right, but for any number n that is a natural number 2n is also a
The SIZE of the set of natural numbers is infinite, and thus obeysNo natural number is infinite. They all obey the law of finite
the laws of infinite numbers. An infinite number, which has a finite
number, added to, multiplied by, or used as a power, is still that
same infinite number. It may seem impossible, but that is the nature
of infinite numbers.
numbers. That includes the law that 2n > n.
natural number and in the set.
are not in that set.
On 12.10.2024 22:47, Richard Damon wrote:Doubling does not create a second infinity.
On 10/12/24 1:49 PM, WM wrote:
On 10.10.2024 21:54, joes wrote:
Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:But the doubles are larger. Hence after doubling the set has a smaller
On 10.10.2024 20:45, Alan Mackenzie wrote:Exactly! There are furthermore no infinite doubles of naturals (2n).
WM <wolfgang.mueckenheim@tha.de> wrote:There are no infinite n = natural numbers.
If all natnumbers are there and if 2n is greater than n, then the >>>>>>> doubled numbers do not fit into ℕ.For any finite n greater than zero, 2n is greater than n. The same >>>>>> does not hold for infinite n.
density and therefore a larger extension on the real line. Hence not
all natural numbers have been doubled.
Not with even numbers.The doubles are larger than the element they replace, but that valueDoubling creates terms which were not in the doubled set.
was always in the set to begin with, so it never creates a "new" term.
No, we are taking the complete, actually infinite set which reachesThen there is no complete set. The doubling can be repeated andWhen doubled then 2n > n. If a set of natural numbers is doubled,Additionally: if n is finite, so is 2n. It cannot go beyond w.
then the results cover a larger set than before..
repeated. Always new numbers are created. Potential infinity.
Am 13.10.2024 um 00:44 schrieb joes:
Am Thu, 10 Oct 2024 20:12:56 +0200 schrieb WM:
It is claimed that all endsegments are infinite [...]
[and that they] have an empty intersection.
Yes, and what do you claim?Well, they are.
[...] not all endsegments are infinit[e].
Satz: ALLE Endsegmente sind unendlich.
Beweis: Ein /Endsegment/ ist definiert als das Komplement eines
endlichen Anfangsabschnitts (der natürlichen Zahlen) bezüglich IN.
D. h. E ist ein Endsegment genau dann, wenn es einen endlichen Anfangsabschnitt A gibt mit E = IN \ A. Daraus ergibt sich sofort, dass
ALLE Endsegmente unendlich sind, weil IN unendlich ist und jeder
endliche (sic!) Anfangsabschnitt endlich ist.
Moreover, Satz: Der Schnitt über ALLE Endsegmente ist leer.
Beweis: Angenommen es ist nicht so. Dann müsste es eine natürliche Zahl geben, die in allen Endsegmenten enthalten ist. Sei WM so eine Zahl. WM
ist aber nicht im Endsegment IN \ {n e IN : n <= WM} enthalten.
Am Mon, 14 Oct 2024 12:35:56 +0200 schrieb WM:
False. n*2 > n for all numbers, even for infinite numbers.Of course, and if the infinity is actual, they must already be in there.
Am Mon, 14 Oct 2024 12:21:39 +0200 schrieb WM:
Cardinality is a property of potentially infinite sets.
Am Mon, 14 Oct 2024 12:28:58 +0200 schrieb WM:
On 12.10.2024 22:47, Richard Damon wrote:And there being inf. many places, that makes both amounts infinite.
On 10/12/24 2:55 PM, WM wrote:The density of the natural umbers is 1 number per place, the density of
2n > n. That holds for all finite numbers. It halves the density.And half of infinity is infinity.
the doubled numbers is 1/2 number per place.
No, we are taking the complete, actually infinite set which reaches
to "before" w.
[...] If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.
[...] If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.
On 10/12/2024 2:06 PM, WM wrote:Correct.
On 10.10.2024 22:47, Jim Burns wrote:
On 10/9/2024 11:39 AM, WM wrote:
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Each γ≠0 preceding ω is (our) finite.
There are two alternatives:
Either doubling creates natnumbers,
then they are not among those doubled,
then we have not doubled all.
Or we have doubled all
but then larger numbers have been created.
There are two alternatives:
Either 𝔊 exists such that 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
Or {2,4,6,...} ᵉᵃᶜʰ< ω
⎛ Assume 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
⎜
⎜ 2⋅𝔊 < ω
⎜ For each j such that 0 < j ≤ 2⋅𝔊
⎜ j-1 exists.
⎜ 2⋅𝔊 = (2⋅𝔊+1)-1 exists
⎜ 2⋅𝔊+1 = (2⋅𝔊+2)-1 exists
⎜ 2⋅𝔊+2 = 2⋅(𝔊+1)
⎜ 2⋅(𝔊+1) < ωMistake.
⎜ However,No, your mistake.
⎜ ω ≤ 2⋅(𝔊+1)
⎝ Contradiction.
On 14.10.2024 14:00, joes wrote:It does not. Changing the elements does not change their number.
Am Mon, 14 Oct 2024 12:35:56 +0200 schrieb WM:
They cannot be there because doubling all elements of a set of naturals doubles the space covered on the number line.False. n*2 > n for all numbers, even for infinite numbers.Of course, and if the infinity is actual, they must already be in
there.
On 14.10.2024 14:02, joes wrote:The amount of numbers in G and N. You cannot make an infinite set
Am Mon, 14 Oct 2024 12:28:58 +0200 schrieb WM:Neither 1 nor 1/2 are infinite. These values are calculated by all
On 12.10.2024 22:47, Richard Damon wrote:And there being inf. many places, that makes both amounts infinite.
On 10/12/24 2:55 PM, WM wrote:The density of the natural umbers is 1 number per place, the density
2n > n. That holds for all finite numbers. It halves the density.And half of infinity is infinity.
of the doubled numbers is 1/2 number per place.
finite subsets of the set and in the limit.
On 14.10.2024 14:15, joes wrote:No, there is no consequent infinity. The even numbers do not go
No, we are taking the complete, actually infinite set which reaches toand fills the space between 0 and ω evenly. Same happens with the
"before" w.
doubled set between 0 and ω2.
Am Mon, 14 Oct 2024 15:40:01 +0200 schrieb WM:
On 14.10.2024 14:15, joes wrote:
No, we are taking the complete, actually infinite set
[of ordinal numbers which lie] "before" ω.
and fills the space between 0 and ω evenly. Same happens with the
doubled set between 0 and ω2.
No, there is no consequent infinity. The even numbers do not go
0, 2, 4, ..., ω, ω+2, ω+4, ..., ω*2
Am 14.10.2024 um 18:04 schrieb joes:
Am Mon, 14 Oct 2024 15:40:01 +0200 schrieb WM:
On 14.10.2024 14:15, joes wrote:
No, we are taking the complete, actually infinite set [of ordinal
numbers which lie] "before" ω.
and fills the space between 0 and ω evenly. Same happens with the
doubled set between 0 and ω2.
No, there is no consequent infinity. The even numbers do not go
0, 2, 4, ..., ω, ω+2, ω+4, ..., ω*2
Just let Mückenhirn know:
An e IN: n*2 < ω .
Proof: An e IN: n*2 e IN (easy) and An e IN: n < ω. qed
Am Mon, 14 Oct 2024 15:34:59 +0200 schrieb WM:
On 14.10.2024 14:00, joes wrote:It does not. Changing the elements does not change their number.
Am Mon, 14 Oct 2024 12:35:56 +0200 schrieb WM:They cannot be there because doubling all elements of a set of naturals
False. n*2 > n for all numbers, even for infinite numbers.Of course, and if the infinity is actual, they must already be in
there.
doubles the space covered on the number line.
Am 13.10.2024 um 10:52 schrieb Moebius:
WM faselt wieder einmal etwas daher:
[...] If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.
Eine falsche Behauptung.
1. Jedes Endsegment E besitzt eine unendliche Teilmenge, nämlich E selbst. 2. Es gibt keine unendliche Menge, die als Teilmenge in allen
Endsegmenten enthalten ist.
Beweis: Sei WM eine Menge, die als Teilmenge in allen Endsegmenten
enthalten ist. (Dass es solche Mengen gibt, ist klar, die leere Menge
ist z. B. so eine Menge.) Wir nehmen an, dass WM nicht leer ist, also
mind. ein Element enthält. Sei wm so ein Element, also wm e WM.
Am Mon, 14 Oct 2024 15:40:01 +0200 schrieb WM:
On 14.10.2024 14:15, joes wrote:No, there is no consequent infinity. The even numbers do not go
No, we are taking the complete, actually infinite set which reaches toand fills the space between 0 and ω evenly. Same happens with the
"before" w.
doubled set between 0 and ω2.
0, 2, 4, ..., w, w+2, w+4, ..., w*2
On 13.10.2024 18:28, Jim Burns wrote:
On 10/12/2024 2:06 PM, WM wrote:
On 10.10.2024 22:47, Jim Burns wrote:
On 10/9/2024 11:39 AM, WM wrote:
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Each γ≠0 preceding ω is (our) finite.
There are two alternatives:
Either doubling creates natnumbers,
then they are not among those doubled,
then we have not doubled all.
Or we have doubled all
but then larger numbers have been created.
There are two alternatives:
Either 𝔊 exists such that 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
Or {2,4,6,...} ᵉᵃᶜʰ< ω
⎛ Assume 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
⎜
⎜ 2⋅𝔊 < ω
⎜ For each j such that 0 < j ≤ 2⋅𝔊
⎜ j-1 exists.
⎜ 2⋅𝔊 = (2⋅𝔊+1)-1 exists
⎜ 2⋅𝔊+1 = (2⋅𝔊+2)-1 exists
⎜ 2⋅𝔊+2 = 2⋅(𝔊+1)
Correct.
⎜ For each j such that 0 < j ≤ 2⋅(𝔊+1)
⎜ j-1 exists.
⎜ 2⋅(𝔊+1) < ω
Mistake.
⎜ However,
⎜ ω ≤ 2⋅(𝔊+1)
⎝ Contradiction.
No, your mistake.
On 10/14/2024 10:43 AM, WM wrote:
On 13.10.2024 18:28, Jim Burns wrote:
On 10/12/2024 2:06 PM, WM wrote:
On 10.10.2024 22:47, Jim Burns wrote:
On 10/9/2024 11:39 AM, WM wrote:
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Each γ≠0 preceding ω is (our) finite.
There are two alternatives:
Either doubling creates natnumbers,
then they are not among those doubled,
then we have not doubled all.
Or we have doubled all
but then larger numbers have been created.
There are two alternatives:
Either 𝔊 exists such that 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
Or {2,4,6,...} ᵉᵃᶜʰ< ω
⎛ Assume 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
⎜
⎜ 2⋅𝔊 < ω
⎜ For each j such that 0 < j ≤ 2⋅𝔊
⎜ j-1 exists.
⎜ 2⋅𝔊 = (2⋅𝔊+1)-1 exists
⎜ 2⋅𝔊+1 = (2⋅𝔊+2)-1 exists
⎜ 2⋅𝔊+2 = 2⋅(𝔊+1)
Correct.
⎜ For each j such that 0 < j ≤ 2⋅(𝔊+1)
⎜ j-1 exists. ⎜ 2⋅(𝔊+1) < ω
Mistake.
ω is the first not.finite ordinal.
Each finite ordinal is before ω
If 2⋅𝔊 is countable.to from 0
then 2⋅(𝔊+1) is countable.to from 0
If 2⋅𝔊 is finite
then 2⋅(𝔊+1) is finite
If 2⋅𝔊 < ω
then 2⋅(𝔊+1) < ω
On 10/14/2024 2:07 PM, WM wrote:
Try again
considering the darkness of most numbers.
⎛ A set S of ordinals holds first.S or is empty.
Whether k or ξ or any ordinal is dark
is irrelevant to that description.
On 14.10.2024 19:13, Jim Burns wrote:
On 10/14/2024 10:43 AM, WM wrote:
On 13.10.2024 18:28, Jim Burns wrote:
On 10/12/2024 2:06 PM, WM wrote:
On 10.10.2024 22:47, Jim Burns wrote:
On 10/9/2024 11:39 AM, WM wrote:
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Each γ≠0 preceding ω is (our) finite.
There are two alternatives:
Either doubling creates natnumbers,
then they are not among those doubled,
then we have not doubled all.
Or we have doubled all
but then larger numbers have been created.
There are two alternatives:
Either 𝔊 exists such that 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
Or {2,4,6,...} ᵉᵃᶜʰ< ω
⎛ Assume 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
⎜
⎜ 2⋅𝔊 < ω
⎜ For each j such that 0 < j ≤ 2⋅𝔊
⎜ j-1 exists.
⎜ 2⋅𝔊 = (2⋅𝔊+1)-1 exists
⎜ 2⋅𝔊+1 = (2⋅𝔊+2)-1 exists
⎜ 2⋅𝔊+2 = 2⋅(𝔊+1)
Correct.
⎜ For each j such that 0 < j ≤ 2⋅(𝔊+1)
⎜ j-1 exists.
⎜ 2⋅(𝔊+1) < ω
Mistake.
ω is the first not.finite ordinal.
Each finite ordinal is before ω
Yes.
If 2⋅𝔊 is countable.to from 0
then 2⋅(𝔊+1) is countable.to from 0
Mistake.
If 2⋅𝔊 is finite
then 2⋅(𝔊+1) is finite
Mistake.
If 2⋅𝔊 < ω
then 2⋅(𝔊+1) < ω
Mistake.
Try again
considering the darkness of most numbers.
On 14.10.2024 21:47, Jim Burns wrote:
On 10/14/2024 2:07 PM, WM wrote:
Try again
considering the darkness of most numbers.
⎛ A set S of ordinals holds first.S or is empty.
That is true for visible ordinals only.
Whether k or ξ or any ordinal is dark
is irrelevant to that description.
That description is irrelevant to
the mathematics of actual infinity
which is what I pursue.
On 14.10.2024 18:04, joes wrote:Which "half"?
Am Mon, 14 Oct 2024 15:40:01 +0200 schrieb WM:Either the doubled numbers are natural, then half of them have not been
On 14.10.2024 14:15, joes wrote:No, there is no consequent infinity. The even numbers do not go 0, 2,
No, we are taking the complete, actually infinite set which reachesand fills the space between 0 and ω evenly. Same happens with the
to "before" w.
doubled set between 0 and ω2.
4, ..., w, w+2, w+4, ..., w*2
among the original set, or all natural numbers have been doubled, then
the result contains infinite numbers.
That doubling _all_ natural numbers only yields _all_ natural numbers is impossible.Of course. But all doubled naturals are themselves natural.
On 14.10.2024 17:05, Moebius wrote:Jedes Endsegment besitzt sogar seinen ganzen Inhalt mit seinem Vorgänger
Am 13.10.2024 um 10:52 schrieb Moebius:Endsegmente bilden eine abnehmende Mengenfolge, wobei jedes Endsegment
WM faselt wieder einmal etwas daher:Eine falsche Behauptung.
[...] If every endsegment has an infinite subset, then there exists
one and the same infinite subset of every endsegment.
1. Jedes Endsegment E besitzt eine unendliche Teilmenge, nämlich E
selbst.
2. Es gibt keine unendliche Menge, die als Teilmenge in allen
Endsegmenten enthalten ist.
Durch Wiederholung wird Deine extrem dumme Behauptung nicht besser.
außer dem ersten, nämlich ℕ, ein Element weniger als sein Vorgänger besitzt. Man spricht auch von einer inklusionsmonoton abnehmenden
Mengefolge. Jedes Endsegment besitzt also seinen gesamten Inhalt
gemeinsam mit allen Vorgängern. Das ist für alle Endsegmente richtig und zeigt insbesondere, dass unendliche Endsegmente eine unendliche Menge gemeinsam mit allen Vorgängern besitzen.
Da diese unendliche Menge nichtWarum ist die nicht verfügbar?
als Indizes für die unendlichen Endsegmente verfügbar ist, ist die Menge der unendlichen Endsegmente endlich, genauer: potentiell unendlich.
Doch, kann man, indem man die unendliche Menge aller Endsegmente aktual betrachtet. Oder einfach N\{0}=E(1). Oder E(n)->E(n+1).Beweis: Sei WM eine Menge, die als Teilmenge in allen EndsegmentenWenn alle Endsegmente unendlich sind, kann man ihre Inhalte nicht näher angeben oder auffinden. Deswegen ist auch wm ein nicht auffindbares
enthalten ist. (Dass es solche Mengen gibt, ist klar, die leere Menge
ist z. B. so eine Menge.) Wir nehmen an, dass WM nicht leer ist, also
mind. ein Element enthält. Sei wm so ein Element, also wm e WM.
Element.
Am Mon, 14 Oct 2024 18:25:50 +0200 schrieb WM:
On 14.10.2024 17:05, Moebius wrote:
Am 13.10.2024 um 10:52 schrieb Moebius:
WM faselt wieder einmal etwas daher:
[...] If every endsegment has an infinite subset, then there exists >>>>>> one and the same infinite subset of every endsegment.
Eine falsche Behauptung. Denn:
1. Jedes Endsegment E besitzt eine unendliche Teilmenge, nämlich E
selbst.
2. Es gibt keine unendliche Menge, die als Teilmenge in allen
Endsegmenten enthalten ist.
Endsegmente bilden eine abnehmende Mengenfolge, wobei jedes Endsegment
außer dem ersten, nämlich ℕ, ein Element weniger als sein Vorgänger
besitzt. Man spricht auch von einer inklusionsmonoton abnehmenden
Mengefolge. Jedes Endsegment besitzt also seinen gesamten Inhalt
gemeinsam mit allen Vorgängern. Das ist für alle Endsegmente richtig und >> zeigt insbesondere, dass [jedes] Endsegment eine unendliche Menge
gemeinsam mit allen [seinen] Vorgängern besitz[t].
Beweis: Sei WM eine Menge, die als Teilmenge in allen Endsegmenten
enthalten ist. (Dass es solche Mengen gibt, ist klar, die leere Menge
ist z. B. so eine Menge.) Wir nehmen an, dass WM nicht leer ist, also
mind. ein Element enthält. Sei wm so ein Element, also wm e WM.
Wenn alle Endsegmente unendlich sind, kann man ihre Inhalte nicht näher
angeben oder auffinden. Deswegen ist auch wm ein nicht auffindbares
Element.
On 12.10.2024 22:07, FromTheRafters wrote:
WM used his keyboard to write :
I use properties of finite numbers.
Wrong, you disbelieve (when it suits you) the idea of cardinality.
Cardinality is a property of potentially infinite sets.
Regards, WM
On 12.10.2024 22:47, Richard Damon wrote:
On 10/12/24 2:39 PM, WM wrote:
On 11.10.2024 20:02, Jim Burns wrote:
"Potential ℕ" vs. "actual ℕ" leaves unchanged
which claims we reason to.
Potential infinity: All natural numbers when double yield natural
numbers but larger numbers are among the result.
In actual infinity no new naturals can be created but since doubling
doubles the value of each number, numbers of the second number class
are created.
No, in actual infinity no new natural can be created,
True.
but every natural number can find the one that is twice itself in the
set,
False. n*2 > n for all numbers, even for infinite numbers.
Regards, WM
On 14.10.2024 18:04, joes wrote:
Am Mon, 14 Oct 2024 15:40:01 +0200 schrieb WM:
On 14.10.2024 14:15, joes wrote:No, there is no consequent infinity. The even numbers do not go
No, we are taking the complete, actually infinite set which reaches to >>>> "before" w.and fills the space between 0 and ω evenly. Same happens with the
doubled set between 0 and ω2.
0, 2, 4, ..., w, w+2, w+4, ..., w*2
Either the doubled numbers are natural, then half of them have not been
among the original set, or all natural numbers have been doubled, then
the result contains infinite numbers.
That doubling _all_ natural numbers only yields _all_ natural numbers is impossible.
Regards, WM
On 10/14/24 6:31 AM, WM wrote:
On 12.10.2024 22:47, Richard Damon wrote:
On 10/12/24 2:19 PM, WM wrote:
On 11.10.2024 03:38, Richard Damon wrote:
The SIZE of the set of natural numbers is infinite, and thus obeys
the laws of infinite numbers. An infinite number, which has a
finite number, added to, multiplied by, or used as a power, is
still that same infinite number. It may seem impossible, but that
is the nature of infinite numbers.
No natural number is infinite. They all obey the law of finite
numbers. That includes the law that 2n > n.
Right, but for any number n that is a natural number 2n is also a
natural number and in the set.
But not in the set of numbers to be doubled. Half of the doubled
numbers are not in that set.
Why not?
Remember, the set is infinite, so doesn't have an end to go past.
"countable cardinality" is so dirt-obvious
that it should go without saying,
Am 15.10.2024 um 17:48 schrieb joes:
Endsegmente bilden eine abnehmende Mengenfolge, wobei jedes Endsegment
außer dem ersten, nämlich ℕ, ein Element weniger als sein Vorgänger >>> besitzt. Man spricht auch von einer inklusionsmonoton abnehmenden
Mengefolge. Jedes Endsegment besitzt also seinen gesamten Inhalt
gemeinsam mit allen Vorgängern. Das ist für alle Endsegmente richtig und >>> zeigt insbesondere, dass [jedes] Endsegment eine unendliche Menge
gemeinsam mit allen [seinen] Vorgängern besitz[t].
Richtig.
Was es aber NICHT zeigt, ist, dass jedes Endsegment eine unendliche
Menge gemeinsam mit allen seinen Nachfolgern besitzt.
Wenn es zu jedem Endsegment so eine Menge gäbe, dann müsste insbesondere das erste Endsegment so eine Menge gemeinsam mit allen seinen
Nachfolgern besitzen. Es müsste also eine unendliche Menge geben, die in allen Endsegmenten (als Teilmenge) enthalten ist.
Tatsächlich gilt: Es gibt keine nicht-leere Menge, die als Teilmenge in allen Endsegmenten enthalten ist.
On 10/14/24 12:37 PM, WM wrote:
That doubling _all_ natural numbers only yields _all_ natural numbers
is impossible.
OF course it is impossible, it only yield half of the numbers, as no odd numbers are in the set.
Am Mon, 14 Oct 2024 18:37:31 +0200 schrieb WM:
On 14.10.2024 18:04, joes wrote:Which "half"?
Am Mon, 14 Oct 2024 15:40:01 +0200 schrieb WM:Either the doubled numbers are natural, then half of them have not been
On 14.10.2024 14:15, joes wrote:No, there is no consequent infinity. The even numbers do not go 0, 2,
No, we are taking the complete, actually infinite set which reachesand fills the space between 0 and ω evenly. Same happens with the
to "before" w.
doubled set between 0 and ω2.
4, ..., w, w+2, w+4, ..., w*2
among the original set, or all natural numbers have been doubled, then
the result contains infinite numbers.
So, what is that perfectly reasonable explanation?
Jedes Endsegment besitzt sogar seinen ganzen Inhalt mit seinem Vorgänger
und damit auch mit dem ersten gemeinsam. Und weil das erste unendlich ist, gibt es auch unendlich viele Segmente.
Take the index, using natural numbers
1: { 1 }
2: { 1, 2 }
3: { 1, 2, 3 }
4: { 1, 2, 3, 4 }
...
vs:
1: { 2 }
2: { 2, 4 }
3: { 2, 4, 6 }
4: { 2, 4, 6, 8 }
...
Am Mon, 14 Oct 2024 15:34:59 +0200 schrieb WM:
On 14.10.2024 14:00, joes wrote:It does not. Changing the elements does not change their number.
Am Mon, 14 Oct 2024 12:35:56 +0200 schrieb WM:They cannot be there because doubling all elements of a set of naturals
False. n*2 > n for all numbers, even for infinite numbers.Of course, and if the infinity is actual, they must already be in
there.
doubles the space covered on the number line.
On 15.10.2024 02:00, Jim Burns wrote:
So, what is that perfectly reasonable explanation?
There is a general rule not open to further discussion:
When doubling natural numbers we obtain natural numbers which have not
been doubled.
In potential infinity we obtain more even natural numbers than have been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
Regards, WM
On 15.10.2024 17:48, joes wrote:Was? Ein Segment hat einen Index und einen unendlichen Inhalt. Es gibt unendliche viele Segmente, einen für jede Zahl (die als Index dient).
Jedes Endsegment besitzt sogar seinen ganzen Inhalt mit seinemUnd trotzdem bleiben noch unendlich viele Zahlen nicht verindext?
Vorgänger und damit auch mit dem ersten gemeinsam. Und weil das erste
unendlich ist, gibt es auch unendlich viele Segmente.
Am Wed, 16 Oct 2024 10:32:46 +0200 schrieb WM:
On 15.10.2024 17:48, joes wrote:
Jedes Endsegment besitzt sogar seinen ganzen Inhalt mit seinem
Vorgänger und damit auch mit dem ersten gemeinsam. Und weil das erste
unendlich ist, gibt es auch unendlich viele Segmente.
Und trotzdem bleiben noch unendlich viele Zahlen nicht verindext?
Was? Ein Segment hat einen Index und einen unendlichen Inhalt. Es gibt unendliche viele Segmente, einen für jede Zahl (die als Index dient).
Jede Zahl ist Index eines unendlichen Segments, das alle Nachfolger
enthält. Die Menge aller Indizes bis zu einem n ist N\E(n). Wo hakt's?
On 15.10.2024 00:27, Chris M. Thomasson wrote:
Take the index, using natural numbersThere is a general rule not open to further discussion:
1: { 1 }
2: { 1, 2 }
3: { 1, 2, 3 }
4: { 1, 2, 3, 4 }
...
vs:
1: { 2 }
2: { 2, 4 }
3: { 2, 4, 6 }
4: { 2, 4, 6, 8 }
...
When doubling natural numbers we obtain natural numbers which have not
been doubled.
In potential infinity we obtain more even natural numbers than have been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
Regards, WM
On 10/16/24 4:30 AM, WM wrote:
There is a general rule not open to further discussion:But that isn't actually a general rule,
When doubling natural numbers we obtain even numbers which have not
been doubled.
In potential infinity we obtain more even natural numbers than have
been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ. >>
NO natural number when doubled results in a value that wasn't a natural number all the time.
Am Wed, 16 Oct 2024 10:32:46 +0200 schrieb WM:
On 15.10.2024 17:48, joes wrote:Was? Ein Segment hat einen Index und einen unendlichen Inhalt. Es gibt unendliche viele Segmente, einen für jede Zahl (die als Index dient).
Jedes Endsegment besitzt sogar seinen ganzen Inhalt mit seinemUnd trotzdem bleiben noch unendlich viele Zahlen nicht verindext?
Vorgänger und damit auch mit dem ersten gemeinsam. Und weil das erste
unendlich ist, gibt es auch unendlich viele Segmente.
Jede Zahl ist Index eines unendlichen Segments, das alle Nachfolger
enthält. Die Menge aller Indizes bis zu einem n ist N\E(n). Wo hakt's?
On 16.10.2024 11:20, joes wrote:So eins gibt's nicht. Jedes hat einen endlichen Index.
Am Wed, 16 Oct 2024 10:32:46 +0200 schrieb WM:Bei den unendlich vielen Endsegmenten, die unendlich viele Zahlen als
On 15.10.2024 17:48, joes wrote:Was? Ein Segment hat einen Index und einen unendlichen Inhalt. Es gibt
Jedes Endsegment besitzt sogar seinen ganzen Inhalt mit seinemUnd trotzdem bleiben noch unendlich viele Zahlen nicht verindext?
Vorgänger und damit auch mit dem ersten gemeinsam. Und weil das erste >>>> unendlich ist, gibt es auch unendlich viele Segmente.
unendliche viele Segmente, einen für jede Zahl (die als Index dient).
Jede Zahl ist Index eines unendlichen Segments, das alle Nachfolger
enthält. Die Menge aller Indizes bis zu einem n ist N\E(n). Wo hakt's?
Indizes vrbraucht haben, von denen also unendlich viele keinen
unendlichen Inhalt mehr haben können.
Am Wed, 16 Oct 2024 17:16:47 +0200 schrieb WM:
On 16.10.2024 11:20, joes wrote:
Wo hakt's?
Bei den unendlich vielen Endsegmenten, die unendlich viele Zahlen als
Indizes verbraucht haben, von denen also unendlich viele keinen
unendlichen Inhalt mehr haben können.
So eins gibt's nicht. Jedes hat einen endlichen Index [und einen unendlichen Inhalt --Moebius]
On 16.10.2024 10:27, WM wrote:
On 15.10.2024 02:00, Jim Burns wrote:
On 10/14/2024 4:03 PM, WM wrote:
On 14.10.2024 21:47, Jim Burns wrote:
On 10/14/2024 2:07 PM, WM wrote:
Try again
considering the darkness of most numbers.
⎛ A set S of ordinals holds first.S or is empty.
That is true for visible ordinals only.
That description is irrelevant to
the mathematics of actual infinity
which is what I pursue.
Perhaps
there is a perfectly reasonable explanation
why you (WM) can only pursue that
while sounding as though you don't know
what an ordinal is.
So, what is that perfectly reasonable explanation?
There is a general rule not open to further discussion:
There is a general rule not open to further discussion:
When doubling natural numbers
we obtain natural numbers which
have not been doubled.
CORRECTION:
When doubling natural numbers
we obtain even numbers which
have not been doubled.
In potential infinity
we obtain more even natural numbers
than have been doubled.
In actual infinity
we double ℕ and obtain
neither ℕ or a subset of ℕ.
On 10/16/2024 4:50 AM, WM wrote:
There is a general rule not open to further discussion:
When doubling natural numbers
we obtain natural numbers which
have not been doubled.
CORRECTION:
When doubling natural numbers
we obtain even numbers which have not been doubled.
The set S of ordinals which
are finite and for which
their double is not finite
is empty.
When doubling natural numbers (finite ordinals)
we obtain natural numbers.
When doubling all natural numbers
we obtain only natural numbers.
In potential infinity
we obtain more even natural numbers
than have been doubled.
In actual infinity
we double ℕ and obtain
neither ℕ or a subset of ℕ.
There is a general rule not open to further discussion:
The natural numbers ℕ equal the finite ordinals 𝕆ᶠⁱⁿ
Am Wed, 16 Oct 2024 17:16:47 +0200 schrieb WM:
Bei den unendlich vielen Endsegmenten, die unendlich viele Zahlen alsSo eins gibt's nicht. Jedes hat einen endlichen Index.
Indizes verbraucht haben, von denen also unendlich viele keinen
unendlichen Inhalt mehr haben können.
WM formulated on Wednesday :
On 16.10.2024 13:54, Richard Damon wrote:
On 10/16/24 4:30 AM, WM wrote:
There is a general rule not open to further discussion:But that isn't actually a general rule,
When doubling natural numbers we obtain even numbers which have not
been doubled.
In potential infinity we obtain more even numbers than have
been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
It is a general rule that doubling creates larger numbers.
It is a general rule that doubling finite cardinal numbers creates
larger finite cardinal numbers.
In transfinite cardinalities Aleph_Zero
times two is still Aleph_zero though.
On 10/16/24 4:50 AM, WM wrote:
On 16.10.2024 10:27, WM wrote:
On 15.10.2024 02:00, Jim Burns wrote:
So, what is that perfectly reasonable explanation?
There is a general rule not open to further discussion:
When doubling natural numbers we obtain natural numbers which have
not been doubled.
CORRECTION: When doubling natural numbers we obtain even numbers which
have not been doubled.
But that isn't actually a general rule,
On 16.10.2024 17:33, joes wrote:Nein? Das Nachfolgersegment "verliert" genau diesen Index.
Am Wed, 16 Oct 2024 17:16:47 +0200 schrieb WM:
Dann ist auch ein endlicher Index und die darauf folgenden unendlichBei den unendlich vielen Endsegmenten, die unendlich viele Zahlen alsSo eins gibt's nicht. Jedes hat einen endlichen Index.
Indizes verbraucht haben, von denen also unendlich viele keinen
unendlichen Inhalt mehr haben können.
vielen Zahlen im Schnitt aller.
On 16.10.2024 17:34, Jim Burns wrote:How do you mean?
On 10/16/2024 4:50 AM, WM wrote:
Maybe. Then not all natural numbers have been doubled.The set S of ordinals which are finite and for which their double isThere is a general rule not open to further discussion:CORRECTION:
When doubling natural numbers we obtain natural numbers which have
not been doubled.
When doubling natural numbers we obtain even numbers which have not
been doubled.
not finite is empty.
Maybe?!When doubling natural numbers (finite ordinals)Maybe.
we obtain natural numbers.
This is the same statement as above.When doubling all natural numbers we obtain only natural numbers.That is impossible.
wtf "possible" "worthless" it is actually the case.That is only possible in potential infinity. But there the result is worthless.There is a general rule not open to further discussion:In potential infinity we obtain more even natural numbers than have
been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
The natural numbers ℕ equal the finite ordinals 𝕆ᶠⁱⁿ
On 10/16/2024 8:49 AM, Moebius wrote:
"It is not guaranteed that n+1 exists for every n." (WM, sci.math)^^^^^^^^^^^^^^
One of WM's hyper moron quotes...
Am Wed, 16 Oct 2024 18:49:52 +0200 schrieb WM:
On 16.10.2024 17:33, joes wrote:Nein? Das Nachfolgersegment "verliert" genau diesen Index.
Am Wed, 16 Oct 2024 17:16:47 +0200 schrieb WM:Dann ist auch ein endlicher Index und die darauf folgenden unendlich
Bei den unendlich vielen Endsegmenten, die unendlich viele Zahlen alsSo eins gibt's nicht. Jedes hat einen endlichen Index.
Indizes verbraucht haben, von denen also unendlich viele keinen
unendlichen Inhalt mehr haben können.
vielen Zahlen im Schnitt aller.
On 16.10.2024 19:17, joes wrote:Na klar, es sind ja unendlich viele.
Am Wed, 16 Oct 2024 18:49:52 +0200 schrieb WM:Werden auf diese Weise alle Indizes in Endsegmenten verloren? Oder
On 16.10.2024 17:33, joes wrote:Nein? Das Nachfolgersegment "verliert" genau diesen Index.
Am Wed, 16 Oct 2024 17:16:47 +0200 schrieb WM:Dann ist auch ein endlicher Index und die darauf folgenden unendlich
Bei den unendlich vielen Endsegmenten, die unendlich viele ZahlenSo eins gibt's nicht. Jedes hat einen endlichen Index.
als Indizes verbraucht haben, von denen also unendlich viele keinen
unendlichen Inhalt mehr haben können.
vielen Zahlen im Schnitt aller.
nicht?
Am Wed, 16 Oct 2024 19:23:35 +0200 schrieb WM:
[Gehen] auf diese Weise alle Indizes [...] verloren? Oder nicht?
On 16.10.2024 17:34, Jim Burns wrote:
On 10/16/2024 4:50 AM, WM wrote:
On 16.10.2024 10:27, WM wrote:
There is a general rule
not open to further discussion:
When doubling natural numbers
we obtain natural numbers which
have not been doubled.
CORRECTION:
When doubling natural numbers we obtain
even numbers which have not been doubled.
The set S of ordinals which
are finite and for which
their double is not finite
doesn't hold first.S = 𝔊
⎛ Proof:
⎝ not( countable.to 2⋅(𝔊-1) ∧ not.countable.to 2⋅𝔊 )
The set S of ordinals which
are finite and for which
their double is not finite
is empty.
Maybe.
Then not all natural numbers have been doubled.
When doubling natural numbers (finite ordinals)
we obtain natural numbers.
Maybe.
When doubling all natural numbers
we obtain only natural numbers.
That is impossible.
In potential infinity
we obtain more even natural numbers
than have been doubled.
In actual infinity
we double ℕ and obtain
neither ℕ or a subset of ℕ.
There is a general rule not open to further discussion:
The natural numbers ℕ equal the finite ordinals 𝕆ᶠⁱⁿ
That is only possible in potential infinity.
But there the result is worthless.
On 16.10.2024 18:09, FromTheRafters wrote:
WM formulated on Wednesday :
On 16.10.2024 13:54, Richard Damon wrote:
On 10/16/24 4:30 AM, WM wrote:
There is a general rule not open to further discussion:But that isn't actually a general rule,
When doubling natural numbers we obtain even numbers which have not
been doubled.
In potential infinity we obtain more even numbers than have been
doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
It is a general rule that doubling creates larger numbers.
It is a general rule that doubling finite cardinal numbers creates
larger finite cardinal numbers.
Yes.
In transfinite cardinalities Aleph_Zero times two is still Aleph_zero
though.
Here we talk only about natural numbers to be doubled.
Regards, WM
On 10/16/2024 4:54 AM, Richard Damon wrote:
On 10/16/24 4:27 AM, WM wrote:
On 15.10.2024 02:00, Jim Burns wrote:
So, what is that perfectly reasonable explanation?
There is a general rule not open to further discussion:
When doubling natural numbers we obtain natural numbers which have
not been doubled.
In potential infinity we obtain more even natural numbers than have
been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ. >>>
Regards, WM
But that isn't actually a general rule, so you are just admitting that
you aren't talking about the same Natural Numbers that everyone else
is talking about, but some bastadization that is just fininte.
NO natural number when doubled results in a value that wasn't a
natural number all the time.
Humm. Don't tell me WM thinks that 2 is not a natural because 1*2=2
;^D
Your problem is that your "Acutal Infinity" isn't actually Infinity,
and like isn't actually "actually".
On 10/16/2024 11:06 AM, Chris M. Thomasson wrote:
[...]
and people who follow only one,
ignorant the other,
need to look up from their nose
because it's leading them.
there is closure in the set of naturals for addition.
Am Wed, 16 Oct 2024 19:23:35 +0200 schrieb WM:
On 16.10.2024 19:17, joes wrote:Na klar, es sind ja unendlich viele.
Am Wed, 16 Oct 2024 18:49:52 +0200 schrieb WM:Werden auf diese Weise alle Indizes in Endsegmenten verloren? Oder
On 16.10.2024 17:33, joes wrote:Nein? Das Nachfolgersegment "verliert" genau diesen Index.
Am Wed, 16 Oct 2024 17:16:47 +0200 schrieb WM:Dann ist auch ein endlicher Index und die darauf folgenden unendlich
Bei den unendlich vielen Endsegmenten, die unendlich viele ZahlenSo eins gibt's nicht. Jedes hat einen endlichen Index.
als Indizes verbraucht haben, von denen also unendlich viele keinen >>>>>> unendlichen Inhalt mehr haben können.
vielen Zahlen im Schnitt aller.
nicht?
No natural number is
the first to not.have a natural.number.double.
When doubling all natural numbers
we obtain only natural numbers.
That is impossible.
There is no first natural number from which we obtain
(by doubling) anything not.a.natural.number.
The only set of natural numbers with no first
is the empty set..
There is no ▒▒▒▒▒ natural number from which we obtain
(by doubling) anything not.a.natural.number.
On 17.10.2024 00:39, Jim Burns wrote:
No natural number is
the first to not.have a natural.number.double.
True.
When doubling all natural numbers
we obtain only natural numbers.
That is impossible.
There is no first natural number from which we obtain
(by doubling) anything not.a.natural.number.
True.
The only set of natural numbers with no first
is the empty set..
No, the set of dark numbers is another set without smallest element.
There is no ▒▒▒▒▒ natural number from which we obtain
(by doubling) anything not.a.natural.number.
Correct is: There is no such _definable_ natnumber.
There is a general rule not open to further discussion:
When doubling natural numbers we obtain even numbers which have not been doubled.
In potential infinity we obtain more even natural numbers than have been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
Regards WM
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
All of these "rules" are so loosely and ambiguously formulated, that they don't actually say anything at all - they are meaningless.
On 10/16/24 11:20 AM, WM wrote:
On 16.10.2024 13:54, Richard Damon wrote:
On 10/16/24 4:30 AM, WM wrote:
There is a general rule not open to further discussion:But that isn't actually a general rule,
When doubling natural numbers we obtain even numbers which have not
been doubled.
In potential infinity we obtain more even natural numbers than have
been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
It is a general rule that doubling creates larger numbers.
Yes, but none get larger than the domain of the Natural Numbers, the set
we started from.
This just shows the unusual behavior of infinite sets.
you assume that the actual
infinity is finite
WM <wolfgang.mueckenheim@tha.de> wrote:
There is a general rule not open to further discussion:
When doubling natural numbers we obtain even numbers which have not been
doubled.
In potential infinity we obtain more even natural numbers than have been
doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
All of these "rules" are so loosely and ambiguously formulated, that they don't actually say anything at all - they are meaningless.
[...] there is definitely _both_ the
"asymptotic density" AND the "countable cardinality".
if you double any natural number, you get another natural number.
On 17.10.2024 20:39, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
There is a general rule not open to further discussion:
When doubling natural numbers we obtain even numbers which have not been >>> doubled.
In potential infinity we obtain more even natural numbers than have been >>> doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
All of these "rules" are so loosely and ambiguously formulated, that they
don't actually say anything at all - they are meaningless.
These rules are basic.
You don't understand them. Perhaps too much at once.
Start with 2n > n for every natural number. (0 is not a natnumber.)
If you can't understand or don't believe, then there is no common basis
for discussion.
Regards WM
WM <wolfgang.mueckenheim@tha.de> wrote:
If you can't understand or don't believe, then there is no common basis
for discussion.
It's not a matter of belief. It's a matter of correct and rigorous mathematics.
On 17.10.2024 00:39, Jim Burns wrote:
No natural number is
the first to not.have a natural.number.double.
True.
When doubling all natural numbers
we obtain only natural numbers.
That is impossible.
There is no first natural number from which we obtain
(by doubling) anything not.a.natural.number.
True.
The only set of natural numbers with no first
is the empty set..
No, the set of dark numbers is
another set without smallest element.
There is no ▒▒▒▒▒ natural number from which we obtain
(by doubling) anything not.a.natural.number.
Correct is:
There is no such _definable_ natnumber.
There is a general rule not open to further discussion:
When doubling natural numbers we obtain
even numbers which have not been doubled.
In potential infinity we obtain
more even natural numbers than have been doubled.
In actual infinity we double ℕ and obtain
neither ℕ or a subset of ℕ.
On 17.10.2024 02:18, Richard Damon wrote:
On 10/16/24 11:20 AM, WM wrote:
On 16.10.2024 13:54, Richard Damon wrote:
On 10/16/24 4:30 AM, WM wrote:
There is a general rule not open to further discussion:But that isn't actually a general rule,
When doubling natural numbers we obtain even numbers which have not
been doubled.
In potential infinity we obtain more even natural numbers than have
been doubled.
In actual infinity we double ℕ and obtain neither ℕ or a subset of ℕ.
It is a general rule that doubling creates larger numbers.
Yes, but none get larger than the domain of the Natural Numbers, the
set we started from.
When doubling natural numbers we obtain even numbers which have not been doubled.
This just shows the unusual behavior of infinite sets.
2n > n is always true, in finite and in infinite sets.
you assume that the actual infinity is finite
No, it is simply complete. Therefore not for all n 2n can be a natnumber.
Regards, WM
On 17.10.2024 00:39, Jim Burns wrote:Aha!
No natural number is the first to not.have a natural.number.double.True.
Contradiction to your agreement below.When doubling all natural numbers we obtain only natural numbers.That is impossible.
Thus, your "dark" numbers can't be natural.There is no first natural number from which we obtain (by doubling)True.
anything not.a.natural.number.
You mean, like the unit fractions?The only set of natural numbers with no first is the empty set..No, the set of dark numbers is another set without smallest element.
[spam snipped]There is no ▒▒▒▒▒ natural number from which we obtain (by doubling)
anything not.a.natural.number.
On 16.10.2024 21:48, FromTheRafters wrote:That is not what closure means: that you always stay in the same set
there is closure in the set of naturals for addition.In potential infinity we obtain more even natural numbers than have been doubled. (Closure)
On 10/17/2024 04:25 AM, Jim Burns wrote:
Loosen the restriction on the discussion,
lose the force.multiplier.
There isn't much useful to be said
about things which _might or might not_
be well.ordered. Etc.
It is a fact:
that in number theory,
according to number theorists,
that there's a descriptive aspect of numbers,
and it's "asymptotic density",
and according to "asymptotic density",
half of the integers are even.
Whether it's SETS of numbers
or the sets of NUMBERS, in
number theory, the set of integers,
has associated with the set of even integers,
a relative size relation, of: one half.
So, if you don't recognize that as a fact,
then, you're not talking about numbers.
Numbers the arithmetization, ...,
which is a most usual first thing in
all manners of descriptive set theory
as would-be relevant.
From an arithmetization:
all the properties
of integers so follow,
and one of them is "asymptotic density".
Half of the integers are even.
It seems as if his brain melts when he thinks of multiple any natural
number by two... Dark thought cross his mind... ;^D
It seems as if his brain melts when he thinks of multiple any natural
number by two... Dark thought cross his mind... ;^D
On 9/16/2024 2:38 PM, Jim Burns wrote:
On 9/16/2024 4:15 PM, Chris M. Thomasson wrote:
WM is a strange one.
If no point is between them,
then they are the same.
If no point is between different points,
then the points aren't in the complete line.
If p0 and p1 are the same point then p1 - p0 is zero.
Say:
p0 = (1, 2, 3)
p1 = (1, 2, 3)
pdif = p1 - p0
pdif would be (0, 0, 0)
See?
On 9/15/2024 3:47 PM, WM wrote:
On 14.09.2024 20:35, joes wrote:
Am Sat, 14 Sep 2024 16:01:02 +0200 schrieb WM:
Two points are next to each other
means that
no point is between them.
Which is the case for no two (different) reals.
For no two different visible numbers,
to be precise.
On 10/18/2024 11:58 AM, Moebius wrote:
Am 18.10.2024 um 20:36 schrieb Chris M. Thomasson:
It seems as if his brain melts when he thinks of multiple any natural
number by two... Dark thought cross his mind... ;^D
Well, WM "comprehends" the infinitely many natural numbers (and the
numbers beyond) the following way (I'd say):
1, 2, 3, ..., n (note: if n is very large ... dusk sets in), ... (the
dark realm)
So if you multiply a very large n e IN by 2 the result may belong to
the dark realm, and who can say if those darkies are natural numbers
or not?!
Hummm... I wonder if ((42424242^69696969) * 2) makes the following song
cross WM's mind:
(You Don't Know the Power of the Dark Side Music Video) https://youtu.be/HEYrRNMr2kg
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
42424242^69696969) * 2
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
(42424242^69696969) * 2
On 10/18/2024 02:05 AM, FromTheRafters wrote:
[...]
I know that
Burns is a platonist at least one day each week,
so, hope and a bottle of ketchup isn't a hamburger,
yet at least there's hope.
On 10/18/2024 3:26 PM, Moebius wrote:
Am 19.10.2024 um 00:16 schrieb Moebius:
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
(42424242^69696969) * 2
Btw. I'm quite sure that this number belongs to the dark side!
Or, numbers just for men... 69?
On 10/18/2024 3:16 PM, Moebius wrote:
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
Hummm... I wonder if ((42424242^69696969) * 2) makes the following
song cross WM's mind:
(You Don't Know the Power of the Dark Side Music Video)
https://youtu.be/HEYrRNMr2kg
Horrible.
Gorgio Moroder did that 45 jears ago. And better.
https://www.youtube.com/watch?v=2tMnde9fjBg
Iirc, he did this one as well:
https://youtu.be/x1afn71-0sI
On 10/18/2024 3:16 PM, Moebius wrote:
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
On 10/18/2024 11:58 AM, Moebius wrote:
Am 18.10.2024 um 20:36 schrieb Chris M. Thomasson:
It seems as if his brain melts when he thinks of multiple any
natural number by two... Dark thought cross his mind... ;^D
Well, WM "comprehends" the infinitely many natural numbers (and the
numbers beyond) the following way (I'd say):
1, 2, 3, ..., n (note: if n is very large ... dusk sets in), ...
(the dark realm)
So if you multiply a very large n e IN by 2 the result may belong to
the dark realm, and who can say if those darkies are natural numbers
or not?!
Hummm... I wonder if ((42424242^69696969) * 2) makes the following
song cross WM's mind:
(You Don't Know the Power of the Dark Side Music Video)
https://youtu.be/HEYrRNMr2kg
Horrible.
Gorgio Moroder did that 45 jears ago. And better.
https://www.youtube.com/watch?v=2tMnde9fjBg
WM needs to break free from the TrollShell 2000.0402 bot anomaly very pre-alpha .012 ago...
https://youtu.be/zuo0YOiPPlk?list=RDMM
On 10/18/2024 5:29 PM, Chris M. Thomasson wrote:
On 10/18/2024 5:12 PM, Moebius wrote:
Am 19.10.2024 um 02:09 schrieb Chris M. Thomasson:
On 10/18/2024 3:16 PM, Moebius wrote:
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
On 10/18/2024 11:58 AM, Moebius wrote:
Am 18.10.2024 um 20:36 schrieb Chris M. Thomasson:
It seems as if his brain melts when he thinks of multiple any
natural number by two... Dark thought cross his mind... ;^D
Well, WM "comprehends" the infinitely many natural numbers (and
the numbers beyond) the following way (I'd say):
1, 2, 3, ..., n (note: if n is very large ... dusk sets in), ... >>>>>>> (the dark realm)
So if you multiply a very large n e IN by 2 the result may belong >>>>>>> to the dark realm, and who can say if those darkies are natural
numbers or not?!
Hummm... I wonder if ((42424242^69696969) * 2) makes the following >>>>>> song cross WM's mind:
(You Don't Know the Power of the Dark Side Music Video)
https://youtu.be/HEYrRNMr2kg
Horrible.
Gorgio Moroder did that 45 jears ago. And better.
https://www.youtube.com/watch?v=2tMnde9fjBg
WM needs to break free from the TrollShell 2000.0402 bot anomaly
very pre-alpha .012 ago...
https://youtu.be/zuo0YOiPPlk?list=RDMM
I guess, I'm to old for that shit. Sorry.
:^D Shit happens Man.
I met the one woman at a casino who was around 54 years old. That's what
she told me for some reason. She was young at heart drinking and
partying and shit! Fun.
On 10/18/2024 5:29 PM, Chris M. Thomasson wrote:
On 10/18/2024 5:12 PM, Moebius wrote:
Am 19.10.2024 um 02:09 schrieb Chris M. Thomasson:
On 10/18/2024 3:16 PM, Moebius wrote:
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
On 10/18/2024 11:58 AM, Moebius wrote:
Am 18.10.2024 um 20:36 schrieb Chris M. Thomasson:
It seems as if his brain melts when he thinks of multiple any
natural number by two... Dark thought cross his mind... ;^D
Well, WM "comprehends" the infinitely many natural numbers (and
the numbers beyond) the following way (I'd say):
1, 2, 3, ..., n (note: if n is very large ... dusk sets in), ... >>>>>>> (the dark realm)
So if you multiply a very large n e IN by 2 the result may belong >>>>>>> to the dark realm, and who can say if those darkies are natural
numbers or not?!
Hummm... I wonder if ((42424242^69696969) * 2) makes the following >>>>>> song cross WM's mind:
(You Don't Know the Power of the Dark Side Music Video)
https://youtu.be/HEYrRNMr2kg
Horrible.
Gorgio Moroder did that 45 jears ago. And better.
https://www.youtube.com/watch?v=2tMnde9fjBg
WM needs to break free from the TrollShell 2000.0402 bot anomaly
very pre-alpha .012 ago...
https://youtu.be/zuo0YOiPPlk?list=RDMM
I guess, I'm to old for that shit. Sorry.
:^D Shit happens Man.
I met the one woman at a casino who was around 54 years old. That's what
she told me for some reason. She was young at heart drinking and
partying and shit! Fun.
On 10/18/2024 5:33 PM, Moebius wrote:
Am 19.10.2024 um 02:30 schrieb Chris M. Thomasson:
On 10/18/2024 5:29 PM, Chris M. Thomasson wrote:
On 10/18/2024 5:12 PM, Moebius wrote:
Am 19.10.2024 um 02:09 schrieb Chris M. Thomasson:
On 10/18/2024 3:16 PM, Moebius wrote:
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
On 10/18/2024 11:58 AM, Moebius wrote:
Am 18.10.2024 um 20:36 schrieb Chris M. Thomasson:
It seems as if his brain melts when he thinks of multiple any >>>>>>>>>> natural number by two... Dark thought cross his mind... ;^D >>>>>>>>>Well, WM "comprehends" the infinitely many natural numbers (and >>>>>>>>> the numbers beyond) the following way (I'd say):
1, 2, 3, ..., n (note: if n is very large ... dusk sets
in), ... (the dark realm)
So if you multiply a very large n e IN by 2 the result may
belong to the dark realm, and who can say if those darkies are >>>>>>>>> natural numbers or not?!
Hummm... I wonder if ((42424242^69696969) * 2) makes the
following song cross WM's mind:
(You Don't Know the Power of the Dark Side Music Video)
https://youtu.be/HEYrRNMr2kg
Horrible.
Gorgio Moroder did that 45 jears ago. And better.
https://www.youtube.com/watch?v=2tMnde9fjBg
WM needs to break free from the TrollShell 2000.0402 bot anomaly
very pre-alpha .012 ago...
https://youtu.be/zuo0YOiPPlk?list=RDMM
I guess, I'm to old for that shit. Sorry.
:^D Shit happens Man.
I met the one woman at a casino who was around 54 years old. That's
what she told me for some reason. She was young at heart drinking and
partying and shit! Fun.
Ha ha ha! You made my day, man! :-)
I will be 47 this year! December. Well, 54 and 47 are not that far off,
so to speak! wow. Wow, how time goes by... I wonder how old WM is?
On 10/18/2024 5:39 PM, Moebius wrote:
Am 19.10.2024 um 02:37 schrieb Chris M. Thomasson:Wow! I was born in 1977.
On 10/18/2024 5:33 PM, Moebius wrote:
Am 19.10.2024 um 02:30 schrieb Chris M. Thomasson:
On 10/18/2024 5:29 PM, Chris M. Thomasson wrote:
On 10/18/2024 5:12 PM, Moebius wrote:
Am 19.10.2024 um 02:09 schrieb Chris M. Thomasson:
On 10/18/2024 3:16 PM, Moebius wrote:
Am 18.10.2024 um 22:48 schrieb Chris M. Thomasson:
On 10/18/2024 11:58 AM, Moebius wrote:
Am 18.10.2024 um 20:36 schrieb Chris M. Thomasson:
It seems as if his brain melts when he thinks of multiple >>>>>>>>>>>> any natural number by two... Dark thought cross his mind... ;^D >>>>>>>>>>>Well, WM "comprehends" the infinitely many natural numbers >>>>>>>>>>> (and the numbers beyond) the following way (I'd say):
1, 2, 3, ..., n (note: if n is very large ... dusk sets
in), ... (the dark realm)
So if you multiply a very large n e IN by 2 the result may >>>>>>>>>>> belong to the dark realm, and who can say if those darkies >>>>>>>>>>> are natural numbers or not?!
Hummm... I wonder if ((42424242^69696969) * 2) makes the
following song cross WM's mind:
(You Don't Know the Power of the Dark Side Music Video)
https://youtu.be/HEYrRNMr2kg
Horrible.
Gorgio Moroder did that 45 jears ago. And better.
https://www.youtube.com/watch?v=2tMnde9fjBg
WM needs to break free from the TrollShell 2000.0402 bot anomaly >>>>>>>> very pre-alpha .012 ago...
https://youtu.be/zuo0YOiPPlk?list=RDMM
I guess, I'm to old for that shit. Sorry.
:^D Shit happens Man.
I met the one woman at a casino who was around 54 years old. That's
what she told me for some reason. She was young at heart drinking
and partying and shit! Fun.
Ha ha ha! You made my day, man! :-)
I will be 47 this year! December. Well, 54 and 47 are not that far
off, so to speak! wow. Wow, how time goes by... I wonder how old WM is?
See: https://de.wikipedia.org/wiki/Wolfgang_M%C3%BCckenheim
" Mückenheim writes in the foreword to his book "Mathematics for the
First Semesters": "With the finiteness of every set, the set of all
digits of a number is also finite."
Sounds like a troll a hyper-ultra finite mindfuck would think for sure!
Thanks for the info, Moebius.
I am right there, as time ticks by... Does WM will get darker as time
ticks? oh shit. UltraHyperFinite... ;^)
On 10/18/2024 5:51 PM, Moebius wrote:
WM is one of the rather rare [case of] "academic cranks".Strange to me, for I thought WM was a most certifiable a troll, indeed.
Of course (!), NOT a mathematician by profession.
On 10/18/2024 5:57 PM, Moebius wrote:
Am 19.10.2024 um 02:52 schrieb Chris M. Thomasson:
On 10/18/2024 5:51 PM, Moebius wrote:
WM is one of the rather rare [case of] "academic cranks".Strange to me, for I thought WM was a most certifiable a troll, indeed.
Of course (!), NOT a mathematician by profession.
No, he's a physicist "gone cracy".
Damn. At least he is way smarter than AP, for sure. Imvho...
On 10/18/2024 5:57 PM, Moebius wrote:
Am 19.10.2024 um 02:55 schrieb Chris M. Thomasson:
On 10/18/2024 5:54 PM, Moebius wrote:
Am 19.10.2024 um 02:49 schrieb Chris M. Thomasson:
I am right there, as time ticks by... Does WM will get darker as
time ticks? oh shit. UltraHyperFinite... ;^)
I guess, it's called /dementia/. :-/
(Poor man.)
DAMN!!!!!!!! That sucks. So far, I feel fine.
Same, same. But who knwow? : :-P
Well, my typos are grand. Treating usenet like a quick message medium
from time to time. That is a sign! ;^o
On 10/18/2024 5:58 PM, Chris M. Thomasson wrote:
On 10/18/2024 5:57 PM, Moebius wrote:
Am 19.10.2024 um 02:52 schrieb Chris M. Thomasson:
On 10/18/2024 5:51 PM, Moebius wrote:No, he's a physicist "gone cracy".
WM is one of the rather rare [case of] "academic cranks".Strange to me, for I thought WM was a most certifiable a troll, indeed. >>>
Of course (!), NOT a mathematician by profession.
Can a HyperUltra Finite <etc>
Iirc, AP was a dishwasher or something?
Still AP inspired me to create this: [...]
though I have conversed with AP quite a bit on this group. Damn.
On 10/18/2024 6:21 PM, Moebius wrote:
Am 19.10.2024 um 03:13 schrieb Chris M. Thomasson:
Still AP inspired me to create this: [...]
See?!
Please don't compare AP with WM.
Yeah. Sorry about that. Sorry to AP! Please try to forgive me.
I thought WM was a most certifiable a troll, indeed
No, he's a physicist "gone cracy".
WM <wolfgang.mueckenheim@tha.de> wrote:
If you can't understand or don't believe, then there is no common basis
for discussion.
It's not a matter of belief. It's a matter of correct and rigorous mathematics.
Anyway, the size of these infinite sets
are equal despite outpacing and asymtotic density considerations.
There is no ω
such that the numbers are
evenly.spaced between 0 and ω
because
that describes a finite ordinal, not ω
On 10/17/2024 2:22 PM, WM wrote:
A nonempty set without a first element
is not a set of only finite ordinals.
On 10/17/24 2:46 PM, WM wrote:
When doubling natural numbers we obtain even numbers which have not
been doubled.
Then your "Actual Infinity" wasn't actually infinte.
As it must contain *ALL* the Natural Numbers to be that set.
2n > n is always true, in finite and in infinite sets.
In FINITE or ORDINAL systems, and there 2n will be in the same actually infinite set as n.
In Infinte Cardinal spaces (like Aleph_0 is in) 2n == n
On 10/17/24 2:46 PM, WM wrote:
On 17.10.2024 02:18, Richard Damon wrote:
On 10/16/24 1:18 PM, WM wrote:
Of course it is a general rule. Doubling creates larger numbers.
But those numbers are still in the set, and all the members of the
set were doubled, so none of the doubled values got us to a value
that wasn't doubled.
Therefore not all doubles are natural.
So, what ones aren't?
It is proven that the set of Natural Numbers are closed under
multiplication.
WM submitted this idea :
On 18.10.2024 14:26, Jim Burns wrote:
There is no ω
such that the numbers are
evenly.spaced between 0 and ω
because
that describes a finite ordinal, not ω
What is immediately before ω if not finite numbers?
What is immediately before zero in the
naturals?
A number immediately before an infinite ordinal
is an infinite ordinal.
What is immediately before ωNo number exists immediately before ω
ω-1 can't be infinite and must be infinite.
On 18.10.2024 14:26, Jim Burns wrote:
There is no ω
such that the numbers are
evenly.spaced between 0 and ω
because
that describes a finite ordinal, not ω
What is immediately before ω
if not finite numbers?
What is immediately before ω
if not finite numbers?
On 18.10.2024 03:26, Richard Damon wrote:
On 10/17/24 2:46 PM, WM wrote:
When doubling natural numbers we obtain even numbers which have not
been doubled.
Then your "Actual Infinity" wasn't actually infinte.
As it must contain *ALL* the Natural Numbers to be that set.
Therefore the double numbers are not natural but infinite.
2n > n is always true, in finite and in infinite sets.
In FINITE or ORDINAL systems, and there 2n will be in the same
actually infinite set as n.
That is not possible if all natural numbers are doubled. The result
covers the interval (0, ω*2) twice as large as the original one (0, ω).
In Infinte Cardinal spaces (like Aleph_0 is in) 2n == n
No.
Regards, WM
On 18.10.2024 03:26, Richard Damon wrote:Same thing. The set of natural numbers is infinite, whether actual or potential. What is a "collection"?
On 10/17/24 2:46 PM, WM wrote:
On 17.10.2024 02:18, Richard Damon wrote:
On 10/16/24 1:18 PM, WM wrote:
It is proven that the set of Natural Numbers are closed underThat is not the set but the potentially infinite collection of Peano
multiplication.
numbers.
WM was thinking very hard :
On 19.10.2024 12:56, Jim Burns wrote:
What is immediately before ωNo number exists immediately before ω
Immediately or not: The number before ω is finite.
No number is before omega,
omega is the naturally ordered set of natural
numbers.
ω-1 can't be infinite and must be infinite.
ω-1 is a natural number
No it is not.
Am Sat, 19 Oct 2024 11:24:54 +0200 schrieb WM:
On 18.10.2024 03:26, Richard Damon wrote:
It is proven that the set of Natural Numbers [is] closed under
multiplication.
That is not the set but [bla bla bla]
[...] The set of natural numbers is infinite
On 10/19/2024 1:10 AM, WM wrote:
When doubling natural numbers we obtain even numbers which have not
been doubled.
This is a sentence that every mathematician can understand.
On 17.10.2024 23:22, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
If you can't understand or don't believe, then there is no common basis
for discussion.
It's not a matter of belief. It's a matter of correct and rigorous
mathematics.
When doubling natural numbers we obtain even numbers which have not been doubled.
This is a sentence that every mathematician can understand.
It is true because the interval covered by the doubled numbers is
twice as large as the interval covered by the numbers to be doubled.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 17.10.2024 23:22, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
If you can't understand or don't believe, then there is no common basis >>>> for discussion.
It's not a matter of belief. It's a matter of correct and rigorous
mathematics.
When doubling natural numbers we obtain even numbers which have not been
doubled.
This is a sentence that every mathematician can understand.
It is not - it is ill formed and ambiguous. It doesn't say which
natural numbers are being doubled.
It is unmathematical in that it
seems to posit a doubling being done one element at a time
rather than
the standard mathematical concept of a mapping from N -> N where n is
mapped to 2n. In this standard notion, all numbers are doubled, and we encounter no undoubled even natural numbers.
It is true because the interval covered by the doubled numbers is
twice as large as the interval covered by the numbers to be doubled.
The interval is infinite. "Doubling" an infinite set yields a set of
the same size as the original - there is a 1-1 correspondence between
them.
Am 19.10.2024 um 13:21 schrieb joes:
Am Sat, 19 Oct 2024 11:24:54 +0200 schrieb WM:
On 18.10.2024 03:26, Richard Damon wrote:
It is proven that the set of Natural Numbers [is] closed under
multiplication.
That is not the set but the potentially infinite collection of Peano numbers.
we are talking about THE SET OF NATURAL NUMBERS, IN,
and its elements -called NATURAL NUMBERS- here.
This set is infinite.
Hint: Damit dieses "Reden" sinnvoll ist, bedarf es natürlich eines entsprechenden mengentheoretischen Kontexts (->SET THEORY). (Das mag
viell. nicht jedem Gesprächsteilnehmer hier klar sein, ist aber so.)
[...] The set of natural numbers is infinite
Indeed.
On 19.10.2024 14:20, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 17.10.2024 23:22, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
If you can't understand or don't believe, then there is no common basis >>>>> for discussion.
It's not a matter of belief. It's a matter of correct and rigorous
mathematics.
When doubling natural numbers we obtain even numbers which have not been >>> doubled.
This is a sentence that every mathematician can understand.
It is not - it is ill formed and ambiguous. It doesn't say which
natural numbers are being doubled.
That is not said because it is true for all sets of natural numbers.
Your critique is therefore not justified but due to your lack of comprehension.
It is unmathematical in that it seems to posit a doubling being done
one element at a time
Wrong again. Even all natural numbers can be multiplied by 2.
rather than the standard mathematical concept of a mapping from N -> N
where n is mapped to 2n. In this standard notion, all numbers are
doubled, and we encounter no undoubled even natural numbers.
Therefore the standard notion is wrong, if the natural numbers are a set.
The interval occupied by the numbers is doubled when all numbers are multiplied by 2. If even the second half, which has not been multiplied, contains natural numbers numbers, then there are more after the
procedure than before. Hence the "set" has changed and therefore is not
a set. Note that sets do not change.
It is true because the interval covered by the doubled numbers is
twice as large as the interval covered by the numbers to be doubled.
The interval is infinite. "Doubling" an infinite set yields a set of
the same size as the original - there is a 1-1 correspondence between
them.
Multiplying n by 2 does not yield the same number.
Multiplying all n by 2 does not yield the same numbers.
Mathematics!
Regards, WM
On 19.10.2024 14:06, Moebius wrote:It is usually called N. Now the Mückenhirn unnaturals may not be closed
Am 19.10.2024 um 13:21 schrieb joes:
Am Sat, 19 Oct 2024 11:24:54 +0200 schrieb WM:
On 18.10.2024 03:26, Richard Damon wrote:
It is proven that the set of Natural Numbers [is] closed underThat is not the set but the potentially infinite collection of Peano
multiplication.
numbers.
We don't get more. "Doubling" is a total function: every natural iswe are talking about THE SET OF NATURAL NUMBERS, IN,If we process all natural numbers and get more than we have processed,
and its elements -called NATURAL NUMBERS- here.
This set is infinite.
then this is not a set but a collection, because sets do not change.
Why should it.Hint: Damit dieses "Reden" sinnvoll ist, bedarf es natürlich einesand does not change --- neither shrink nor grow.
entsprechenden mengentheoretischen Kontexts (->SET THEORY). (Das mag
viell. nicht jedem Gesprächsteilnehmer hier klar sein, ist aber so.)
[...] The set of natural numbers is infinite
On 19.10.2024 14:20, Alan Mackenzie wrote:It is not true for infinite sets.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 17.10.2024 23:22, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
That is not said because it is true for all sets of natural numbers.When doubling natural numbers we obtain even numbers which have notIt is not - it is ill formed and ambiguous. It doesn't say which
been doubled.
This is a sentence that every mathematician can understand.
natural numbers are being doubled.
Your critique is therefore not justified but due to your lack of comprehension.
You seem unable to imagine that.It is unmathematical in that itWrong again. Even all natural numbers can be multiplied by 2.
seems to posit a doubling being done one element at a time
"if"rather than the standard mathematical concept of a mapping from N -> NTherefore the standard notion is wrong, if the natural numbers are a
where n is mapped to 2n. In this standard notion, all numbers are
doubled, and we encounter no undoubled even natural numbers.
set.
The interval occupied by the numbers is doubled when all numbers are multiplied by 2. If even the second half, which has not been multiplied,You just said all numbers are multiplied. What is the "second half"?
contains natural numbers, then there are more after theTell that the potential infinity.
procedure than before. Hence the "set" has changed and therefore is not
a set. Note that sets do not change.
In part it does. All even numbers are also naturals.Multiplying n by 2 does not yield the same number.It is true because the interval covered by the doubled numbers isThe interval is infinite. "Doubling" an infinite set yields a set of
twice as large as the interval covered by the numbers to be doubled.
the same size as the original - there is a 1-1 correspondence between
them.
Multiplying all n by 2 does not yield the same numbers.
On 18.10.2024 00:34, Jim Burns wrote:
On 10/1v7/2024 2:22 PM, WM wrote:
On 17.10.2024 00:39, Jim Burns wrote:
The only set of natural numbers with no first
is the empty set..
No, the set of dark numbers is
another set without smallest element.
A nonempty set without a first element
is not a set of only finite ordinals.
The set of dark numbers contains
only natural numbers.
What you call a "set of finite ordinals" is
not a set
but a potentially infinite collection.
No, the set of dark numbers is
another set without smallest element.
A nonempty set without a first element
is not a set of only finite ordinals.
The set of dark numbers contains
only natural numbers.
Proof:
If you double all your finite ordinals
you obtain only finite ordinals again,
although the covered interval is
twice as large as the original interval
covered by "all" your finite ordinals.
WM <wolfgang.mueckenheim@tha.de> wrote:
rather than the standard mathematical concept of a mapping from N -> N
where n is mapped to 2n. In this standard notion, all numbers are
doubled, and we encounter no undoubled even natural numbers.
Therefore the standard notion is wrong, if the natural numbers are a set.
You mean it's wrong because it doesn't gel with your intuition?
If you think you can obtain an
"undoubled" number in that mapping, please feel free to give an example.
You can't, of course, you'll just say that all such are "dark numbers",
Note that I haven't talked about "sets which change" - that's entirely
your idea. I talked about a map from N -> N, where n maps to 2n.
WM expressed precisely :
On 19.10.2024 13:15, FromTheRafters wrote:
WM was thinking very hard :
On 19.10.2024 12:56, Jim Burns wrote:
What is immediately before ωNo number exists immediately before ω
Immediately or not: The number before ω is finite.
No number is before omega,
In analysis we have to deal only with the infinitely small and the
infinitely large as a limit-notion, as something becoming, emerging,
produced, i.e., as we put it, with the potential infinite. [Hilbert].
Is this from a 1925 lecture?#
All these becoming, emerging, produced numbers
What does that even mean?
omega is the naturally ordered set of natural numbers.
That is another meaning of the word but not relevant, in particular
not if there is no set of natural numbers because new numbers can be
produced and added.
???
WM <wolfgang.mueckenheim@tha.de> wrote:
When multiplying all natural numbers by 2, then the number of numbers
remains the same but the density is reduced and therefore the interval
is doubled.
That's not mathematics.
It's merely your intuition, derived from finite
sets and misapplied to infinite sets.
In mathematics, there is no meaningful distinction between what you think
of as two different forms of infinity.
On 19.10.2024 16:24, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
rather than the standard mathematical concept of a mapping from N -> N >>>> where n is mapped to 2n. In this standard notion, all numbers are
doubled, and we encounter no undoubled even natural numbers.
Therefore the standard notion is wrong, if the natural numbers are a set.
You mean it's wrong because it doesn't gel with your intuition?
No, it does not comply with mathematics.
When multiplying all natural numbers by 2, then the number of numbers
remains the same but the density is reduced and therefore the interval
is doubled.
2 > n. Hence either natural numbers are created which have not been multiplied, then ℕ is not a set, or other numbers are created, then ℕ
is a set.
If you think you can obtain an "undoubled" number in that mapping,
please feel free to give an example.
I can prove it by 2n > n.
You can't, of course, you'll just say that all such are "dark
numbers",
Either dark numbers or natnumbers which have not been processed. There
is no other way because 2n > n.
Note that I haven't talked about "sets which change" - that's entirely
your idea. I talked about a map from N -> N, where n maps to 2n.
This Bourbaki-notion can be applied to potentially infinite sets only.
Try to understand the correct mathematics.
Regards, WM
Am Sat, 19 Oct 2024 15:38:53 +0200 schrieb WM:
On 19.10.2024 14:20, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 17.10.2024 23:22, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
It is not true for infinite sets.That is not said because it is true for all sets of natural numbers.When doubling natural numbers we obtain even numbers which have notIt is not - it is ill formed and ambiguous. It doesn't say which
been doubled.
This is a sentence that every mathematician can understand.
natural numbers are being doubled.
Your critique is therefore not justified but due to your lack of
comprehension.
You seem unable to imagine that.It is unmathematical in that itWrong again. Even all natural numbers can be multiplied by 2.
seems to posit a doubling being done one element at a time
"if"rather than the standard mathematical concept of a mapping from N -> NTherefore the standard notion is wrong, if the natural numbers are a
where n is mapped to 2n. In this standard notion, all numbers are
doubled, and we encounter no undoubled even natural numbers.
set.
The interval occupied by the numbers is doubled when all numbers areYou just said all numbers are multiplied. What is the "second half"?
multiplied by 2. If even the second half, which has not been multiplied,
Multiplying all n by 2 does not yield the same numbers.In part it does. All even numbers are also naturals.
Am Sat, 19 Oct 2024 15:18:56 +0200 schrieb WM:
If we process all natural numbers and get more than we have processed,We don't get more.
then this is not a set but a collection, because sets do not change.
"Doubling" is a total function: every natural is
mapped to exactly one other.
Why should it.Hint: Damit dieses "Reden" sinnvoll ist, bedarf es natürlich einesand does not change --- neither shrink nor grow.
entsprechenden mengentheoretischen Kontexts (->SET THEORY). (Das mag
viell. nicht jedem Gesprächsteilnehmer hier klar sein, ist aber so.)
[...] The set of natural numbers is infinite
On 19.10.2024 12:56, Jim Burns wrote:
A number immediately before an infinite ordinal
is an infinite ordinal.
That is the traditional opinion.
It has lead to internal contradictions
(vanishing Bob).
On 19.10.2024 19:32, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
When multiplying all natural numbers by 2, then the number of numbers
remains the same but the density is reduced and therefore the interval
is doubled.
That's not mathematics.
It is mathematics.
It's merely your intuition, derived from finite sets and misapplied to
infinite sets.
It is the basic mathematics of natural numbers: 2n > n.
If you believe in a different version of mathematics, try to find
people who are interested in that version. I am not.
In mathematics, there is no meaningful distinction between what you think
of as two different forms of infinity.
You have not the faintest idea of infinity.
Regards, WM
WM has brought this to us :
Either dark numbers or natnumbers which have not been processed. There
is no other way because 2n > n.
Processed? Created? As I told you before, these are objects not projects.
On 10/19/2024 7:10 AM, WM wrote:
It has lead to internal contradictions (vanishing Bob).
The contradictions are with
_what you think_ 'infinite' means.
On 10/19/2024 4:16 AM, WM wrote:
What you call a "set of finite ordinals" is
not a set
but a potentially infinite collection.
There is a general rule not open to further discussion:
Finite sets aren't potentially infinite collections.
Proof:
If you double all your finite ordinals
you obtain only finite ordinals again,
Yes.
although the covered interval is
twice as large as the original interval
covered by "all" your finite ordinals.
No.
The least.upper.bound of finites is ω
The least.upper.bound of doubled finites is ω
WM <wolfgang.mueckenheim@tha.de> wrote:
An infinite set is one which has a proper subset which can be
put into 1-1 correspondence with the original set. That is the
definition.
On 19.10.2024 20:22, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
An infinite set is one which has a proper subset which can be
put into 1-1 correspondence with the original set. That is the
definition.
According to Dedekind every set {1, 2, 3, ..., n} is in correspondence
with the set {2, 4, 6, ..., 2n} which covers twice the interval,
containing numbers not in the original set.
This does not change when the whole set ℕ is multiplied by 2.
The result covers twice the interval, ....
.... containing numbers not in the original set ℕ.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
If you believe in a different version of mathematics, try to find
people who are interested in that version. I am not.
The other posters on this newsgroups, they are not hard to find. We all understand standard mathematics. You do not.
Gloria? [...]
https://youtu.be/nNEb2k_EmMg?list=RDy3hf0T4qpYg
[...] Creatures of the night type lyrics:
https://youtu.be/RP0_8J7uxhs
On 19.10.2024 20:12, Jim Burns wrote:
On 10/19/2024 7:10 AM, WM wrote:
It has lead to internal contradictions
(vanishing Bob).
The contradictions are with
_what you think_ 'infinite' means.
The contradiction is independent of infinity.
It is your claim that
exchanging two objects
can result in the loss of one of them.
On 19.10.2024 18:04, Jim Burns wrote:
On 10/19/2024 4:16 AM, WM wrote:
What you call a "set of finite ordinals" is
not a set
but a potentially infinite collection.
There is a general rule not open to further discussion:
Finite sets aren't potentially infinite collections.
Potentially infinite collections are
finite sets open to change.
Proof:
If you double all your finite ordinals
you obtain only finite ordinals again,
Yes.
although the covered interval is
twice as large as the original interval
covered by "all" your finite ordinals.
No.
The least.upper.bound of finites is ω
The least.upper.bound of doubled finites is ω
Doubling halves the density and doubles the interval,
creating numbers which had not been doubled.
2n > n does not fail for any natural number.
The least.upper.bound of finites is ω
The least.upper.bound of doubled finites is ω
On 18.10.2024 14:26, Jim Burns wrote:
There is no ω
such that the numbers are
evenly.spaced between 0 and ω
because
that describes a finite ordinal, not ω
What is immediately before ω if not finite numbers?
Regards, WM
On 10/19/2024 2:28 PM, WM wrote:
The contradiction is independent of infinity.
It is your claim that
infinitely.many exchanges in an infinite set
(vanishing Bob)
exchanging two objects
can result in the loss of one of them.
I fixed that for you.
WM <wolfgang.mueckenheim@tha.de> wrote:
The result covers twice the interval, ....
It does not, except in the sense that twice infinite = infinite.
.... containing numbers not in the original set ℕ.
No. If you think that, then give an example of a 2n which "isn't in the original set N".
You won't and you can't. But you'll likely come back
to your standard get-out clause about (non existent) "dark numbers".
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
No k exists such that
2⋅k is a finite and 2⋅k+2 > 2⋅k is not a finite.
On 10/19/24 4:28 AM, WM wrote:
On 18.10.2024 14:26, Jim Burns wrote:
There is no ω
such that the numbers are
evenly.spaced between 0 and ω
because
that describes a finite ordinal, not ω
What is immediately before ω if not finite numbers?
The SET of Natural Numbers, not any one in particular.
Which is what builds the actual infinite set. "Actual Infinity" is what
you get when you conplete that construction process.
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
If all finites are doubled, then not all results can be in that set.
Either more finites appear, or the results are infinite.
No k exists such that
2⋅k is a finite and 2⋅k+2 > 2⋅k is not a finite.
All doubled numbers result in larger numbers. That cannot be avoided.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
If all finites are doubled, then not all results can be in that set.
Either more finites appear, or the results are infinite.
That's your intuition getting the better of you again.
When "all
finites" (by which I assume you mean natural numbers) are doubled, all
the doubled numbers are finite, too.
We're talking about a mapping
between infinite sets, not a process. Nothing "appears".
If you think some of the doubled numbers are infinite, please give an
example of a natural number which when doubled becomes infinite.
On 19.10.2024 22:19, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The result covers twice the interval, ....
It does not, except in the sense that twice infinite = infinite.
Twice the infinite contains numbers not in the first infinite.
What else could twice mean?
.... containing numbers not in the original set ℕ.
No. If you think that, then give an example of a 2n which "isn't in the
original set N".
These numbers are dark. Their existence is proven by the fact that
∀n ∈ ℕ, n < 2n.
You won't and you can't. But you'll likely come back
to your standard get-out clause about (non existent) "dark numbers".
The only alternative is that doubling numbers creates only doubled
numbers.
It is not acceptable.
It cannot be apologized by the possibility to map all numbers on even numbers.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 19.10.2024 22:19, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The result covers twice the interval, ....
It does not, except in the sense that twice infinite = infinite.
Twice the infinite contains numbers not in the first infinite.
It does not.
What else could twice mean?
It doesn't really mean much at all.
These numbers are dark. Their existence is proven by the fact that
∀n ∈ ℕ, n < 2n.
Their non-existence I proved in this thread, many posts ago.
The only alternative is that doubling numbers creates only doubled
numbers.
That's more or less what happens, except that numbers aren't "created".
They just are.
It is not acceptable.
It is accepted by mathematicians.
It cannot be apologized by the possibility to map all numbers on even
numbers.
There exists a 1-1 correspondence between all natural numbers and all
even natural numbers, a proper subset of all natural numbers. The set of
all natural numbers is thus infinite.
On 20.10.2024 00:08, Jim Burns wrote:
On 10/19/2024 2:28 PM, WM wrote:
The contradiction is independent of infinity.
It is your claim that
infinitely.many exchanges in an infinite set
(vanishing Bob)
Every exchange is _one_ lossless exchange.
exchanging two objects
can result in the loss of one of them.
I fixed that for you.
It is nonsense like:
∀n ∈ ℕ: |{2, 4, 6, ..., 2n}|/|{1, 2, 3, 4, 5, 6, ..., 2n} = 1/2
but |{2, 4, 6, ...}|/|{1, 2, 3, 4, 5, 6, ...} = 1.
Regards, WM
On 20.10.2024 04:48, Richard Damon wrote:
On 10/19/24 4:28 AM, WM wrote:
On 18.10.2024 14:26, Jim Burns wrote:
There is no ω
such that the numbers are
evenly.spaced between 0 and ω
because
that describes a finite ordinal, not ω
What is immediately before ω if not finite numbers?
The SET of Natural Numbers, not any one in particular.
In this linearly ordered set there can only be one number before ω.
Regards, WM
On 20.10.2024 04:48, Richard Damon wrote:
Which is what builds the actual infinite set. "Actual Infinity" is
what you get when you conplete that construction process.
If it is completed and then all numbers are doubled, what do you get then?
Regards, WM
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
If all finites are doubled, then not all results can be in that set.
Either more finites appear, or the results are infinite.
No k exists such that
2⋅k is a finite and 2⋅k+2 > 2⋅k is not a finite.
All doubled numbers result in larger numbers. That cannot be avoided.
Regards, WM
On 20.10.2024 11:39, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 19.10.2024 22:19, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The result covers twice the interval, ....
It does not, except in the sense that twice infinite = infinite.
Twice the infinite contains numbers not in the first infinite.
It does not.
2n > n holds for all natural numbers.
What else could twice mean?
It doesn't really mean much at all.
It means that after doubling more numbers are there than have been doubled.
These numbers are dark. Their existence is proven by the fact that
∀n ∈ ℕ, n < 2n.
Their non-existence I proved in this thread, many posts ago.
Based on wrong assumptions.
The only alternative is that doubling numbers creates only doubled
numbers.
That's more or less what happens, except that numbers aren't "created".
They just are.
That's not what happens in mathematics.
It is not acceptable.
It is accepted by mathematicians.
Only by those who never thought about that topic.
It cannot be apologized by the possibility to map all numbers on even
numbers.
There exists a 1-1 correspondence between all natural numbers and all
even natural numbers, a proper subset of all natural numbers. The set of
all natural numbers is thus infinite.
The mapping needs larger numbers than have been mapped.
That is not a matter of infinite sets but basic mathematics of all
natural numbers. The correspondence between all natural numbers and
all even natural numbers requires a variable "all". Potential infinity.
Regards, WM
On 20.10.2024 11:21, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
If all finites are doubled, then not all results can be in that set.
Either more finites appear, or the results are infinite.
That's your intuition getting the better of you again.
It is not intuition but fact that 2n > n.
When "all finites" (by which I assume you mean natural numbers) are
doubled, all the doubled numbers are finite, too.
Maybe, but they are not all in the original set. Hence more natural
numbers are necessary than have been doubled or mapped.
We're talking about a mapping between infinite sets, not a process.
Nothing "appears".
Say mapping or multiplying, it is a process. But that is not important. Important is only that in the image there are numbers which have not
been mapped because 2n > n.
If you think some of the doubled numbers are infinite, please give an
example of a natural number which when doubled becomes infinite.
In the image there are numbers which are not in the original set.
If all natural numbers have been mapped, then there are larger numbers
in the image. That is an unavoidable consequence.
These numbers cannot be seen, but that does not negate the consequence.
If this is not accepted, then not all natural numbers of the image have
been in the original set.
Regards, WM
On 10/19/2024 03:54 PM, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
On 19.10.2024 18:04, Jim Burns wrote:
The least.upper.bound of finites is ω
What ω is
is such that
k < ω ⇔ k is a finite ordinal.
No k exists such that
k is a finite and k+1 > k is not a finite.
No k exists such that
k is an upper.bound of the finites.
ω is but anything prior to ω isn't
an upper.bound of the finites.
ω is the least.upper.bound of the finites.
The least.upper.bound of doubled finites is ω
A doubled finite is finite.
No k exists such that
2⋅k is a finite and 2⋅k+2 > 2⋅k is not a finite.
No k exists such that
2⋅k is an upper.bound of the doubled finites.
ω is but anything prior to ω isn't
an upper.bound of the doubled finites.
ω is the least.upper.bound of the doubled finites.
Isn't mathematics true?
The omega is usually called
a fixed-point besides being
a limit ordinal, also it's called
a compactification
or one-point compactification of the integers
for the most usual sort of idea of
a non-standard countable model of integers
with exactly one infinite member.
On 10/20/24 3:48 AM, WM wrote:´
All doubled numbers result in larger numbers. That cannot be avoided.
But since there isn't a "largest" number,
WM <wolfgang.mueckenheim@tha.de> wrote:
Say mapping or multiplying, it is a process. But that is not important.
Important is only that in the image there are numbers which have not
been mapped because 2n > n.
That "because" attempts to connect two unrelated statements there are no "numbers which have not been mapped". If you think there are, give an example.
If this is not accepted, then not all natural numbers of the image have
been in the original set.
<Sigh>. The sets are infinite.
On 20.10.2024 14:56, Alan Mackenzie wrote:"created".
WM <wolfgang.mueckenheim@tha.de> wrote:
The only alternative is that doubling numbers creates only doubled
numbers.
That's more or less what happens, except that numbers aren't
They just are.
That's not what happens in mathematics.
How would you know?
I know that 2n > n.
The mapping needs larger numbers than have been mapped.
It does not. Each number in the image is in the range.
Impossible. The density is halved in the image, the number of elements
is the same, therefore the image is twice as large and half of its
numbers are not in the original set.
Regards, WM
On 10/20/2024 06:36 AM, Jim Burns wrote:
Is there some sort of protocol
which you (RF) recognize
in order to talk about a thing
and not.talk about things not.that?
Here there's freedom of speech it one of
what we call constitutional liberties,
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
If all finites are doubled,
then not all results can be in that set.
On 20.10.2024 00:08, Jim Burns wrote:
On 10/19/2024 2:28 PM, WM wrote:
The contradiction is independent of infinity.
It is your claim that
infinitely.many exchanges in an infinite set
(vanishing Bob)
Every exchange is _one_ lossless exchange.
exchanging two objects
can result in the loss of one of them.
I fixed that for you.
It is nonsense like:
∀n ∈ ℕ:
|{2, 4, 6, ..., 2n}|/|{1, 2, 3, 4, 5, 6, ..., 2n} = 1/2
but
|{2, 4, 6, ...}|/|{1, 2, 3, 4, 5, 6, ...} = 1.
On 10/20/2024 3:48 AM, WM wrote:
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
If all finites are doubled,
then not all results can be in that set.
If all finites are doubled,
then all results are in the set of finites.
We know this by
On 10/20/24 3:54 AM, WM wrote:
On 20.10.2024 04:48, Richard Damon wrote:A set of exactly the same size using only some of the numbers of the
Which is what builds the actual infinite set. "Actual Infinity" is
what you get when you conplete that construction process.
If it is completed and then all numbers are doubled, what do you get
then?
original set.
That is part of the property of the infinite.
On 20.10.2024 14:20, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Say mapping or multiplying, it is a process. But that is not important.
Important is only that in the image there are numbers which have not
been mapped because 2n > n.
That "because" attempts to connect two unrelated statements there are no "numbers which have not been mapped". If you think there are, give an example.
They can be proven by mathematics: Density reduced, fixed number,
interval enlarged.
If this is not accepted, then not all natural numbers of the image have
been in the original set.
<Sigh>. The sets are infinite.
Infinite does not mean inaccessible to logic.
Regards, WM
On 20.10.2024 13:56, Richard Damon wrote:
On 10/20/24 3:48 AM, WM wrote:
´All doubled numbers result in larger numbers. That cannot be avoided.
But since there isn't a "largest" number,
There is completeness.
Regards, WM
On 10/20/2024 3:40 AM, WM wrote:
On 20.10.2024 00:08, Jim Burns wrote:
On 10/19/2024 2:28 PM, WM wrote:
The contradiction is independent of infinity.
It is your claim that
infinitely.many exchanges in an infinite set
(vanishing Bob)
Every exchange is _one_ lossless exchange.
It is my claim that
infinitely.many exchanges in an infinite set
can result in the loss of one of them.
1 is not infinite.
It is nonsense like:
∀n ∈ ℕ:
|{2, 4, 6, ..., 2n}|/|{1, 2, 3, 4, 5, 6, ..., 2n} = 1/2
but
|{2, 4, 6, ...}|/|{1, 2, 3, 4, 5, 6, ...} = 1.
A finite set can be ordered such that
each subset holds its top and bottom or is empty.
No, the alternative is that infinite sets work like infinite sets
without an end, and that this means that some of the properties of
finite sets don't hold for infinite sets.
On 20.10.2024 20:20, Jim Burns wrote:
On 10/20/2024 3:48 AM, WM wrote:
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
If all finites are doubled,
then not all results can be in that set.
If all finites are doubled,
then all results are in the set of finites.
But not in the mapped or multiplied range.
If all finites are doubled,
then all results are in the set of finites.
But not in the mapped or multiplied range.
On 10/20/24 3:15 PM, WM wrote:
When a set of numbers is mapped then the number of numbers remains fixed.Right, so the "ALeph_0" Natural Numbers were mapped to the Even Natural Numbers, a set with exactly that same size, which is also a proper
subset of it.
On 10/20/24 10:26 AM, WM wrote:
On 20.10.2024 13:56, Richard Damon wrote:
On 10/20/24 3:48 AM, WM wrote:´
All doubled numbers result in larger numbers. That cannot be avoided.
But since there isn't a "largest" number,
There is completeness.
Which meaning of "Completeness" do you mean?
For set theory, the "Completeness" of the Natural Numbers says there is
a suprema of the set
On 10/20/2024 3:24 PM, WM wrote:
On 20.10.2024 20:20, Jim Burns wrote:
On 10/20/2024 3:48 AM, WM wrote:
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
A doubled finite is finite.
If all finites are doubled,
then not all results can be in that set.
If all finites are doubled,
then all results are in the set of finites.
But not in the mapped or multiplied range.
If n is countable.to from 0
then ⟨0,1,...,n-1,n⟩ exists
If ⟨0,1,...,n-1,n⟩ exists
then ⟨n,n+1,...,2⋅n-1,2⋅n⟩ exists
If ⟨0,1,...,n-1,n⟩ and ⟨n,n+1,...,2⋅n-1,2⋅n⟩ exist
then ⟨0,1,...,n-1,n,n+1,...,2⋅n-1,2⋅n⟩ exists
If ⟨0,1,...,n-1,n,n+1,...,2⋅n-1,2⋅n⟩ exists
then 2⋅n is countable.to from 0
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
If n is countable.to from 0
then 2⋅n is countable.to from 0
If all finites are doubled,
then all results are in the set of finites.
But not in the mapped or multiplied range.
If n is countable.to from 0
then 2⋅n is countable.to from 0
On 20.10.2024 21:42, Jim Burns wrote:
On 10/20/2024 3:24 PM, WM wrote:
On 20.10.2024 20:20, Jim Burns wrote:
On 10/20/2024 3:48 AM, WM wrote:
On 20.10.2024 00:54, Jim Burns wrote:
A doubled finite is finite.
If all finites are doubled,
then not all results can be in that set.
If all finites are doubled,
then all results are in the set of finites.
But not in the mapped or multiplied range.
If n is countable.to from 0
then ⟨0,1,...,n-1,n⟩ exists
If ⟨0,1,...,n-1,n⟩ exists
then ⟨n,n+1,...,2⋅n-1,2⋅n⟩ exists
If ⟨0,1,...,n-1,n⟩ and ⟨n,n+1,...,2⋅n-1,2⋅n⟩ exist
then ⟨0,1,...,n-1,n,n+1,...,2⋅n-1,2⋅n⟩ exists
If ⟨0,1,...,n-1,n,n+1,...,2⋅n-1,2⋅n⟩ exists
then 2⋅n is countable.to from 0
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
If n is countable.to from 0
then 2⋅n is countable.to from 0
If all finites are doubled,
then all results are in the set of finites.
But not in the mapped or multiplied range.
If n is countable.to from 0
then 2⋅n is countable.to from 0
Nevertheless
2n is not in the set {1, ..., n}.
A doubled finite is finite.
On 10/20/2024 12:54 PM, WM wrote:
2n is not in the set {1, ..., n}.
On the other hand, for all n in {1, 2, 3, ...} 2n is in the set {1, 2,
3, ...}.
If n is in the set ℕ≠{1,...,n} of finites,
then 2⋅n is in the set ℕ≠{1,...,n} of finites.
On 20.10.2024 21:31, Richard Damon wrote:
On 10/20/24 3:15 PM, WM wrote:
When a set of numbers is mapped then the number of numbers remainsRight, so the "ALeph_0" Natural Numbers were mapped to the Even
fixed.
Natural Numbers, a set with exactly that same size, which is also a
proper subset of it.
I did not talk about cardinality which is nonsense but about the real
number of elements, Cantor called it reality.
Regards, WM
On 20.10.2024 22:23, Jim Burns wrote:
If n is in the set ℕ≠{1,...,n} of finites,
then 2⋅n is in the set ℕ≠{1,...,n} of finites.
When the elements of any set of naturals are multiplied by 2,
their density is halved,
their reality remains the same,
their extension is doubled.
Therefore
the image contains numbers which are not in the range.
On 20.10.2024 21:42, Jim Burns wrote:We are not doubling a finite set. And 2n is in N for every n e N.
On 10/20/2024 3:24 PM, WM wrote:
On 20.10.2024 20:20, Jim Burns wrote:
On 10/20/2024 3:48 AM, WM wrote:
On 20.10.2024 00:54, Jim Burns wrote:
On 10/19/2024 2:19 PM, WM wrote:
Nevertheless 2n is not in the set {1, ..., n}.If n is countable.to from 0 then 2⋅n is countable.to from 0If all finites are doubled,But not in the mapped or multiplied range.
then all results are in the set of finites.
On 20.10.2024 04:48, Richard Damon wrote:The set N u {w} is not "linearly ordered".
On 10/19/24 4:28 AM, WM wrote:In this linearly ordered set there can only be one number before ω.
On 18.10.2024 14:26, Jim Burns wrote:The SET of Natural Numbers, not any one in particular.
There is no ω such that the numbers are evenly.spaced between 0 and ω >>>> because that describes a finite ordinal, not ωWhat is immediately before ω if not finite numbers?
On 19.10.2024 17:21, joes wrote:What else would it be.
Am Sat, 19 Oct 2024 15:38:53 +0200 schrieb WM:
On 19.10.2024 14:20, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 17.10.2024 23:22, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
rather than the standard mathematical concept of a mapping from N ->Therefore the standard notion is wrong, if the natural numbers are a
N where n is mapped to 2n. In this standard notion, all numbers are
doubled, and we encounter no undoubled even natural numbers.
set.
What is the second half of N? What do we get when we double that.The interval occupied by the numbers is doubled when all numbers areYou just said all numbers are multiplied. What is the "second half"?
multiplied by 2. If even the second half, which has not been
multiplied,
Look, if we construct the set of powers of 2, by starting with 1 and multiplying (a finite number of times) by 2, this set is infiniteAll even numbers which have been multiplied are natural numbers by definition. And by the same definition there were not more even natural numbers. But as the result there are more.Multiplying all n by 2 does not yield the same numbers.In part it does. All even numbers are also naturals.
On 18.10.2024 03:26, Richard Damon wrote:Then half of an infinite number is finite?
On 10/17/24 2:46 PM, WM wrote:
Therefore the double numbers are not natural but infinite.When doubling natural numbers we obtain even numbers which have notThen your "Actual Infinity" wasn't actually infinte.
been doubled.
As it must contain *ALL* the Natural Numbers to be that set.
This destroys the ordering. If m>n, then 2m>2n.That is not possible if all natural numbers are doubled. The result2n > n is always true, in finite and in infinite sets.In FINITE or ORDINAL systems, and there 2n will be in the same actually
infinite set as n.
covers the interval (0, ω*2) twice as large as the original one (0, ω).
--In Infinte Cardinal spaces (like Aleph_0 is in) 2n == n
On 19.10.2024 12:56, Jim Burns wrote:There is no single number before w. Exactly all naturals are smaller
Immediately or not: The number before ω is finite.What is immediately before ωNo number exists immediately before ω
Almost like w itself!ω-1 can't be infinite and must be infinite.ω-1 is a natural number, but the dark numbers occupying almost all space between 0 and ω act like an infinity. We cannot look or count through
them although all are finite, alas too large.
Am Sat, 19 Oct 2024 13:10:20 +0200 schrieb WM:
ω-1 is a natural number
Am Sun, 20 Oct 2024 09:51:37 +0200 schrieb WM:
On 20.10.2024 04:48, Richard Damon wrote:
On 10/19/24 4:28 AM, WM wrote:
What is [...] before ω [...]?
The [elements of the] SET of Natural Numbers, not any one in particular.
In this linearly ordered set there can only be one number before ω.
The set N u {w} is not "linearly ordered".
2 * any_natural_number = a_natural_number
Got it?
Am 21.10.2024 um 01:06 schrieb joes:
Am Sat, 19 Oct 2024 13:10:20 +0200 schrieb WM:
ω-1 is a natural number
Oh really, which one? :-)
Too much brain coolants? Wine perhaps?
Am Sat, 19 Oct 2024 10:23:05 +0200 schrieb WM:
That is not possible if all natural numbers are doubled. The resultThis destroys the ordering. If m>n, then 2m>2n.
covers the interval (0, ω*2) twice as large as the original one (0, ω).
Am Sun, 20 Oct 2024 09:51:37 +0200 schrieb WM:
On 20.10.2024 04:48, Richard Damon wrote:The set N u {w} is not "linearly ordered".
On 10/19/24 4:28 AM, WM wrote:In this linearly ordered set there can only be one number before ω.
On 18.10.2024 14:26, Jim Burns wrote:The SET of Natural Numbers, not any one in particular.
There is no ω such that the numbers are evenly.spaced between 0 and ω >>>>> because that describes a finite ordinal, not ωWhat is immediately before ω if not finite numbers?
On 10/21/2024 12:14 AM, WM wrote:
also the transfinite numbers are ordered. But for dark numbers the
ordering is inaccessible anyhow.
Why are the dark numbers unordered?
half of an infinite set
can be the same size as the whole,
On 10/20/2024 4:42 PM, WM wrote:
On 20.10.2024 22:23, Jim Burns wrote:
If n is in the set ℕ≠{1,...,n} of finites,
then 2⋅n is in the set ℕ≠{1,...,n} of finites.
When the elements of any set of naturals are multiplied by 2,
∃{1,2,...,n-1,n} ⇔
∃{1,2,...,n-1,n,n+1,...,n+n-1,n+n}
their density is halved,
their reality remains the same,
their extension is doubled.
Therefore
the image contains numbers which are not in the range.
n ∈ ℕ ⇔ ∃{1,2,...,n-1,n}
∃{1,2,...,n-1,n} ⇔ ∃{1,2,...,n-1,n,n+1,...,n+n-1,n+1}
∃{1,2,...,n-1,n,n+1,...,n+n-1,n+1} ⇔ n+n ∈ ℕ
n ∈ ℕ ⇔ n+n ∈ ℕ
WM formulated the question :
Nevertheless 2n is not in the set {1, ..., n}.
Neither is n+1, what's your point?
On 21.10.2024 01:04, joes wrote:Yes, but the ordinals up to w*2 are two consecutive infinities.
Am Sat, 19 Oct 2024 10:23:05 +0200 schrieb WM:
No, also the transfinite numbers are ordered. But for dark numbers the ordering is inaccessible anyhow.That is not possible if all natural numbers are doubled. The resultThis destroys the ordering. If m>n, then 2m>2n.
covers the interval (0, ω*2) twice as large as the original one (0,
ω).
On 20.10.2024 23:00, FromTheRafters wrote:
WM formulated the question :
If the range really is complete, it needs to be infinite, so it canWhen doubling numbers, their distance increases in positive direction,Nevertheless 2n is not in the set {1, ..., n}.Neither is n+1, what's your point?
hence larger numbers are in the image than in the range.
If the range was complete, the image shows that the range was not
complete.
Am Mon, 21 Oct 2024 10:14:07 +0200 schrieb WM:
If the range was complete, the image shows that the range was notIf the range really is complete, it needs to be infinite, so it can
complete.
include the larger numbers.
WM brought next idea :
If the range was complete, the image shows that the range was not
complete.
What's the preimage?
On 21.10.2024 11:21, FromTheRafters wrote:
WM brought next idea :
If the range was complete,
the image shows that
the range was not complete.
What's the preimage?
The range.
It is a complete set by assumption.
All its numbers can be mapped,
some of them not to the preimage.
On 21.10.2024 10:21, joes wrote:Dude. An infinite set can contain an m>n for every n in it.
Am Mon, 21 Oct 2024 10:14:07 +0200 schrieb WM:
No. No set of numbers can include larger numbers.If the range was complete, the image shows that the range was notIf the range really is complete, it needs to be infinite, so it can
complete.
include the larger numbers.
On 20.10.2024 23:41, Jim Burns wrote:
On 10/20/2024 4:42 PM, WM wrote:
On 20.10.2024 22:23, Jim Burns wrote:
If n is in the set ℕ≠{1,...,n} of finites,
then 2⋅n is in the set ℕ≠{1,...,n} of finites.
When the elements of any set of naturals are multiplied by 2,
∃{1,2,...,n-1,n} ⇔
∃{1,2,...,n-1,n,n+1,...,n+n-1,n+n}
their density is halved,
their reality remains the same,
their extension is doubled.
Therefore the image contains
numbers which are not in the range.
n ∈ ℕ ⇔ ∃{1,2,...,n-1,n}
n is not among the dark numbers.
∃{1,2,...,n-1,n} ⇔ ∃{1,2,...,n-1,n,n+1,...,n+n-1,n+1}
∃{1,2,...,n-1,n,n+1,...,n+n-1,n+1} ⇔ n+n ∈ ℕ
n ∈ ℕ ⇔ n+n ∈ ℕ
That and more holds for all definable numbers.
But:
When doubling numbers,
their distance increases in positive direction,
hence larger numbers are in the image
than in the range.
If the range was complete,
the image shows that the range was not complete.
On 10/20/2024 11:33 AM, Jim Burns wrote:
On 10/20/2024 12:30 PM, Ross Finlayson wrote:
On 10/20/2024 06:36 AM, Jim Burns wrote:
Is there some sort of protocol
which you (RF) recognize
in order to talk about a thing
and not.talk about things not.that?
Here there's freedom of speech it one of
what we call constitutional liberties,
I take your answer to mean
"no, there is no such protocol".
Surely,
with your degree in mathematics,
you understand that
you are depriving yourself of
the use of a powerful tool.
Au contraire, to the contrary,
I proffer that
extending the relevant domain while keeping it
a completely connected relevant domain
_is_ a most proper and surmounting
improvement of the discourse,
to include the wider considerations of
a topic in the _foundations_ of the theory,
not merely a single theorem
under a microscope.
Anyways you still haven't picked "anti and only".
On 10/21/2024 6:01 AM, WM wrote:
How can you (WM) not.know what 'preimage' means?
Am Mon, 21 Oct 2024 11:59:00 +0200 schrieb WM:
On 21.10.2024 10:21, joes wrote:An infinite set can contain an m>n for every n in it.
Am Mon, 21 Oct 2024 10:14:07 +0200 schrieb WM:No. No set of numbers can include larger numbers.
If the range was complete, the image shows that the range was notIf the range really is complete, it needs to be infinite, so it can
complete.
include the larger numbers.
Dude. An infinite set can contain an m > n for every n in it.
.(0)1 says its the closest to zero.
.(0)01 says that little shit is all hype!
On 10/21/2024 12:14 AM, WM wrote:
for dark numbers the ordering is inaccessible
Why are the dark numbers unordered?
On 10/20/2024 4:37 PM, Moebius wrote:
Am 20.10.2024 um 21:57 schrieb Chris M. Thomasson:
Too much brain coolants? Wine perhaps?
Psychosis perhaps?
"Psychosis is a condition of the mind or psyche that results in
difficulties determining what is real and what is not real. Symptoms
may include delusions [...] among other features." (Wikipedia)
I hope not! I hope WM is just a troll, sharp as a fiddle.
With a view to a kill:
https://youtu.be/gkuUnnlAPZk
Makes me think of another song [...]
https://youtu.be/y3hf0T4qpYg?list=RDy3hf0T4qpYg
On 10/19/2024 2:46 PM, Moebius wrote:
"Branigan died in her sleep at her lodge in East Quogue, New York, on
August 26, 2004, aged 52. The cause was attributed to a previously
undiagnosed cerebral aneurysm."
Ahhhh shit! God damn it. I know she passed away, but not the details!
Grrrr! Damn it.
On 21.10.2024 14:30, Jim Burns wrote:
On 10/21/2024 6:01 AM, WM wrote:
How can you (WM) not.know what 'preimage' means?
I have lectured analysis in German only.
I assumed that preimage is the set mapped to the image.
On 10/21/2024 6:01 AM, WM wrote:
On 21.10.2024 11:21, FromTheRafters wrote:
What's the preimage?
The range.
It is a complete set by assumption.
All its numbers can be mapped,
some of them not to the preimage.
On 20.10.2024 23:54, Richard Damon wrote:
half of an infinite set can be the same size as the whole,
That is nonsense but it is not the point.
The point is this: When doubling numbers, their distance increases in positive direction, hence larger numbers are in the image than in the
range.
If the range was complete, the image shows that the range was not complete.
Regards, WM
On 10/21/2024 2:25 PM, WM wrote:
On 21.10.2024 14:30, Jim Burns wrote:
On 10/21/2024 6:01 AM, WM wrote:
How can you (WM) not.know what 'preimage' means?
I have lectured analysis in German only.
I assumed that preimage is the set mapped to the image.
It's not an improvement that you allegedly think that
odd numbers are the result of doubling.
The range.
It is a complete set by assumption.
All its numbers can be mapped,
some of them not to the preimage.
On 10/21/2024 2:59 AM, WM wrote:
No set of numbers can include larger numbers.
What do you mean? Define a "large" number, say a natural one?
On 10/21/24 4:09 AM, WM wrote:
The point is this: When doubling numbers, their distance increases in
positive direction, hence larger numbers are in the image than in the
range.
Nope, because the input range is infinite.
If the range was complete, the image shows that the range was not
complete.
Nope, just shows that you don't understand what "complete" means for an infinite set.
On 21.10.2024 21:38, Chris M. Thomasson wrote:Namely, infinite numbers.
On 10/21/2024 2:59 AM, WM wrote:
No set of natural numbers can include *larger* numbers than it consistsNo set of numbers can include larger numbers.What do you mean? Define a "large" number, say a natural one?
of.
If the set is actually infinite, then all its numbers are there andContradiction to above.
can be doubled. Because of 2n > n larger numbers are created.
If theyIt did.
are natural numbers they prove that the set originally was not complete because it did not contain the larger numbers.
(In potential infinity this is not a problem because there is no--
completeness demanded).
Am Tue, 22 Oct 2024 10:05:38 +0200 schrieb WM:
Because of 2n > n larger numbers [bla bla bla]
On 21.10.2024 22:34, Jim Burns wrote:
On 10/21/2024 2:25 PM, WM wrote:
On 21.10.2024 14:30, Jim Burns wrote:
How can you (WM)
not.know what 'preimage' means?
I have lectured analysis in German only.
I assumed that
preimage is the set mapped to the image.
It's not an improvement that
you allegedly think that
odd numbers are the result of doubling.
I never did.
Are your comprehension problems new?
Am 22.10.2024 um 11:39 schrieb joes:
die Menge IN enthält ALLE natürlichen Zahlen.
Insbesondere enthält IN also zu jeder Zahl n auch die Zahl 2n = n + n.
Diese Zahl ist genau n Schritte von n "entfernt",
Das alles spielt sich für jede natürliche Zahl n "im Endlichen" ab.
On 21.10.2024 22:34, Jim Burns wrote:
On 10/21/2024 2:25 PM, WM wrote:
On 21.10.2024 14:30, Jim Burns wrote:
On 10/21/2024 6:01 AM, WM wrote:
The range.
A complete set of natural numbers.
It is a complete set by assumption.
All its numbers can be mapped,
some of them not to the preimage.
Not to the mapped set.
All its numbers can be mapped,
some of them not to the preimage.
On 10/22/2024 4:13 AM, WM wrote:
ℕ is defined such that
n ∈ ℕ ⇔ ∃{0,1,...,n-1,n}
∀n ∈ ℕ: 2×n ∈ ℕ
On 22.10.2024 18:03, Jim Burns wrote:
ℕ is defined such that
n ∈ ℕ ⇔ ∃{0,1,...,n-1,n}
Most of all
it is an invariable set
with all its elements existing
and subject to doubling.
∀n ∈ ℕ: 2×n ∈ ℕ
Not if all elements are existing
before multiplication already.
The description of each element in ℕ
requires its double to also be in ℕ
...not so different from the way in which
the description of a right triangle
requires
the square of its longest side to equal
the sum of the squares of the two other sides.
On 20.10.2024 20:50, Jim Burns wrote:Uncontroversial.
On 10/20/2024 3:40 AM, WM wrote:Every exchange is one and has to obey logic.
On 20.10.2024 00:08, Jim Burns wrote:It is my claim that infinitely.many exchanges in an infinite set can
On 10/19/2024 2:28 PM, WM wrote:
Every exchange is _one_ lossless exchange.The contradiction is independent of infinity.infinitely.many exchanges in an infinite set (vanishing Bob)
It is your claim that
result in the loss of one of them.
1 is not infinite.
If your claim was acceptable, then every enumeration of an infinite setAn enumeration is not an exchange.
could lose elements. Therefore it is rejected as detrimental to set
theory.
On 22.10.2024 18:03, Jim Burns wrote:
On 10/22/2024 4:13 AM, WM wrote:
ℕ is defined such that
n ∈ ℕ ⇔ ∃{0,1,...,n-1,n}
Most of all it is an invariable set with all its elements existing and subject to doubling.
∀n ∈ ℕ: 2×n ∈ ℕ
Not if all elements are existing before multiplication already.
Regards, WM
On 10/22/24 12:12 PM, WM wrote:
On 22.10.2024 18:03, Jim Burns wrote:
∀n ∈ ℕ: 2×n ∈ ℕ
Not if all elements are existing before multiplication already.
IF not, then your actual infinity wasn't actually infinite
Your "actual infinity" seems to be just an unimaginably large value, not infinite, as your actual infinity has an end, it has an element without
a successor, so it isn't the set it claims to be.
On 10/21/2024 11:09 AM, Ross Finlayson wrote:
On 10/21/2024 09:21 AM, Jim Burns wrote:
On 10/20/2024 8:20 PM, Ross Finlayson wrote:
Anyways you still haven't picked "anti and only".
I vaguely recall that
you (RF) made some incorrect claims about
Cantor's argument from anti.diagonals,
and you asked for my participation in some way.
Could you refresh my memory? TIA.
Then I suggested that I would put anti-diagonal
in one fist, only-diagonal in the other, then
hide them behind my back and perhaps exchange
them, then that you get to pick.
You get to pick, was the idea, then I laughed
and said that I had put them together, so,
you get both or none.
"You" here meaning anybody, ...,
because it's a mathematical statement
so is the same for anyone.
On 22.10.2024 19:38, Jim Burns wrote:
The description of each element in ℕ
requires its double to also be in ℕ
All that is in ℕ, according to your opinion,
is accepted.
If you find that the set ic complete,
then it is doubled.
...not so different from the way in which
the description of a right triangle
requires
the square of its longest side to equal
the sum of the squares of the two other sides.
Your above expression is in fact
a very closely related one.
On 23.10.2024 13:37, Richard Damon wrote:
On 10/22/24 12:12 PM, WM wrote:
On 22.10.2024 18:03, Jim Burns wrote:
∀n ∈ ℕ: 2×n ∈ ℕ
Not if all elements are existing before multiplication already.
IF not, then your actual infinity wasn't actually infinite
It is infinite like the fractions between 0 and 1. When doubling we get even-numerator fractions, some of which greater the 1.
Your "actual infinity" seems to be just an unimaginably large value,
not infinite, as your actual infinity has an end, it has an element
without a successor, so it isn't the set it claims to be.
The completed infinite cannot avoid to be complete. But it is infinite because the end cannot be determined because of the dark domain.
Regards, WM
On 10/23/24 10:46 AM, WM wrote:
On 23.10.2024 13:37, Richard Damon wrote:
On 10/22/24 12:12 PM, WM wrote:
On 22.10.2024 18:03, Jim Burns wrote:
∀n ∈ ℕ: 2×n ∈ ℕ
Not if all elements are existing before multiplication already.
IF not, then your actual infinity wasn't actually infinite
It is infinite like the fractions between 0 and 1. When doubling we
get even-numerator fractions, some of which greater the 1.
But from 0 to 1 isn't an infinite distance
At 1/2, we double to hit the boundry.
But, with the Natual Numbers, there isn't a value that is 1/2 of the
"highest value" since there isn't a highest number that is a Natural
Number, since every Natural Number has a successor that is higher.
Your "actual infinity" seems to be just an unimaginably large value,
not infinite, as your actual infinity has an end, it has an element
without a successor, so it isn't the set it claims to be.
The completed infinite cannot avoid to be complete. But it is infinite
because the end cannot be determined because of the dark domain.
No, the end cannot be determined, because it isn't there.
There is nothing about being complete that means it needs to have an
"end"
On 10/22/2024 2:34 PM, WM wrote:
On 22.10.2024 19:38, Jim Burns wrote:
The description of each element in ℕ
requires its double to also be in ℕ
All that is in ℕ, according to your opinion,
is accepted.
In the ℕ which is our ℕ,
n ∈ ℕ ⇔ ∃⟨0,1,...,n-1,n⟩
In the ℕ which is our ℕ,
for each j ∈ ℕ, ∃k ∈ ℕ: k = j+1
for each j ∈ ℕ, j=0 ∨ ∃i ∈ ℕ: i+1=j
for each S s ℕ, S = {} ∨ ∃m ∈ S: m=min.S
Bob is in room 0 of our ℕ.Hotel
Swap guests in 0⇄1, 1⇄2, 2⇄3, 3⇄4, ...
After all swaps,
there is no first room Bob is in.
There is no ▒▒▒▒▒ room Bob is in.
'Bye, Bob.
If you find that the set is complete,
then it is doubled.
In a WM.complete ℕ.Hotel,
dark rooms are added for Bob to disappear to
when he isn't in the visible rooms,
repairing 'bye.Bob,
leaving can't.see.Bob.
However,
none of these swaps 0⇄1, 1⇄2, 2⇄3, 3⇄4, ...
move Bob to a dark room.
Bob disappeared
without going to a dark room.
Adding dark rooms without Bob in them
does not repair 'bye.Bob.
There is no
'bye.Bob.repairing WM.complete Hotel.
On 10/24/24 7:07 AM, WM wrote:
On 24.10.2024 05:04, Richard Damon wrote:
On 10/23/24 10:46 AM, WM wrote:
On 23.10.2024 13:37, Richard Damon wrote:
On 10/22/24 12:12 PM, WM wrote:
On 22.10.2024 18:03, Jim Burns wrote:
∀n ∈ ℕ: 2×n ∈ ℕ
Not if all elements are existing before multiplication already.
IF not, then your actual infinity wasn't actually infinite
It is infinite like the fractions between 0 and 1. When doubling we
get even-numerator fractions, some of which greater the 1.
But from 0 to 1 isn't an infinite distance
Measured in rational points it is infinite.
But values are not measured in "rational points".
It is claimed that there are all numbers. "That we have for instance
when we consider the entirety of the numbers 1, 2, 3, 4, ... itself as
a completed unit, or the points of a line as an entirety of things
which is completely available. That sort of infinity is named actual
infinite." [D. Hilbert: "Über das Unendliche", Mathematische Annalen
95 (1925) p. 167]
So?
If you HAVE all the numberes, 1, 2, 3, 4, ... that set goes on FOREVER
and doesn't have a upper end.
If you have the COMPLETE unit, it doesn't have a highest number.
Your operation of doubling the values on the line from 0 to 1 isn't
operating the property that that set is infinite on, so doesn't follow
the law of the infinite.
There is nothing about being complete that means it needs to have an
"end"
Whatever, it is complete and all its numbers can be doubled. Some are
resulting in larger numbers than have been doubled.
Nope, as every number (A Natural Number) doubles to another number in
that set (The Natural Numbers) so you never left the set.
On 24.10.2024 05:04, Richard Damon wrote:
On 10/23/24 10:46 AM, WM wrote:
On 23.10.2024 13:37, Richard Damon wrote:
On 10/22/24 12:12 PM, WM wrote:
On 22.10.2024 18:03, Jim Burns wrote:
∀n ∈ ℕ: 2×n ∈ ℕ
Not if all elements are existing before multiplication already.
IF not, then your actual infinity wasn't actually infinite
It is infinite like the fractions between 0 and 1. When doubling we
get even-numerator fractions, some of which greater the 1.
But from 0 to 1 isn't an infinite distance
Measured in rational points it is infinite.
At 1/2, we double to hit the boundry.
There we have crossed many dark fractions already.
Same at ω/2.
It is claimed that there are all numbers. "That we have for instance
But, with the Natual Numbers, there isn't a value that is 1/2 of the
"highest value" since there isn't a highest number that is a Natural
Number, since every Natural Number has a successor that is higher.
when we consider the entirety of the numbers 1, 2, 3, 4, ... itself as a completed unit, or the points of a line as an entirety of things which
is completely available. That sort of infinity is named actual
infinite." [D. Hilbert: "Über das Unendliche", Mathematische Annalen 95 (1925) p. 167]
Your "actual infinity" seems to be just an unimaginably large value,
not infinite, as your actual infinity has an end, it has an element
without a successor, so it isn't the set it claims to be.
The completed infinite cannot avoid to be complete. But it is
infinite because the end cannot be determined because of the dark
domain.
No, the end cannot be determined, because it isn't there.
There is nothing about being complete that means it needs to have an
"end"
Whatever, it is complete and all its numbers can be doubled. Some are resulting in larger numbers tha have been doubled.
Regards, WM
On 23.10.2024 20:39, Jim Burns wrote:
On 10/22/2024 2:34 PM, WM wrote:
On 22.10.2024 19:38, Jim Burns wrote:
The description of each element in ℕ
requires its double to also be in ℕ
All that is in ℕ, according to your opinion,
is accepted.
In the ℕ which is our ℕ,
n ∈ ℕ ⇔ ∃⟨0,1,...,n-1,n⟩
In the ℕ which is our ℕ,
for each j ∈ ℕ, ∃k ∈ ℕ: k = j+1
for each j ∈ ℕ, j=0 ∨ ∃i ∈ ℕ: i+1=j
for each S s ℕ, S = {} ∨ ∃m ∈ S: m=min.S
All that is accepted and doubled.
Bob is in room 0 of our ℕ.Hotel
Swap guests in 0⇄1, 1⇄2, 2⇄3, 3⇄4, ...
After all swaps,
there is no first room Bob is in.
There is no ▒▒▒▒▒ room Bob is in.
'Bye, Bob.
Obviously he has occupied a dark room.
If you find that the set is complete,
then it is doubled.
In a WM.complete ℕ.Hotel,
dark rooms are added for Bob to disappear to
when he isn't in the visible rooms,
repairing 'bye.Bob,
leaving can't.see.Bob.
However,
none of these swaps
move Bob to a dark room.
No definable swap.
But in case of completeness of definable rooms,
Bob could pass all rooms
(because in case of completeness all rooms exist)
and occupy the last room
(because in case of all rooms
there is a last room necessary to establish completeness).
This can only be prevented by dark rooms.
Bob disappeared
without going to a dark room.
Impossible with an indestructible Bob.
Adding dark rooms without Bob in them
does not repair 'bye.Bob.
There is no
'bye.Bob.repairing WM.complete Hotel.
There is no chance to repair mathematics
when Bob disappears.
On 10/24/2024 6:58 AM, WM wrote:
All that is accepted and doubled.
The subset of those with a double not.in ℕ
holds no minimum.
After all swaps,
Bob is not in any room he was ever in.
'Bye, Bob.
Note well:
After all _infinitely many_ swaps,
Bob is not in any of _infinitely many_ rooms he was ever in.
Infinite is different from finite.
The problem with Bob disappearing is that
Bob disappearing is not a problem.
It's not repairable because it's not broken.
On 10/24/2024 11:58 AM, WM wrote:
[...] I can understand that
you have to apply Freudian repression. Eyes wide shut.
On 24.10.2024 19:34, Jim Burns wrote:
On 10/24/2024 6:58 AM, WM wrote:
All that is accepted and doubled.
The subset of those with a double not.in ℕ
holds no minimum.
The whole set with all its numbers exists
and can be mapped to the double numbers.
That halves the density
and doubles the covered interval.
After all swaps,
Bob is not in any room he was ever in.
'Bye, Bob.
After all swaps Bob is in a room,
because there is nowhere an outlet.
Note well:
After all _infinitely many_ swaps,
Bob is not in any of _infinitely many_ rooms
he was ever in.
Infinite is different from finite.
Bobs cannot dissolve into nothing.
The problem with Bob disappearing is that
Bob disappearing is not a problem.
It's not repairable because it's not broken.
Lossless exchanges remain lossless forever.
On 10/24/2024 2:58 PM, WM wrote:
The whole set with all its numbers exists
and can be mapped to the double numbers.
Yes.
∃⟨0,1,...,n-1,n⟩ ⇒
∃⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩
That halves the density
and doubles the covered interval.
No.
⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩ is finite.
After all swaps,
Bob is not in any room he was ever in.
'Bye, Bob.
After all swaps Bob is in a room,
because there is nowhere an outlet.
That Bob is not in any room he was ever in
and there is nowhere an outlet
proves the set is infinite.
Bobs cannot dissolve into nothing.
Bob can be in an infinite set.
The problem with Bob disappearing is that
Bob disappearing is not a problem.
It's not repairable because it's not broken.
Lossless exchanges remain lossless forever.
After all swaps,
Bob is not in any room he was ever in.
Am 25.10.2024 um 06:33 schrieb Jim Burns:
On 10/24/2024 2:58 PM, WM wrote:
The whole set with all its numbers exists
and can be mapped to the double numbers.
Yes.
∃⟨0,1,...,n-1,n⟩ ⇒
∃⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩
That halves the density
and doubles the covered interval.
No.
⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩ is finite.
The whole interval (0, ω) is not finite,
let alone the doubled interval.
If you dislike this result,
then you oppose to complete infinity.
On 10/25/2024 7:42 AM, WM wrote:
The whole interval (0, ω) is not finite,
let alone the doubled interval.
⟦0,ω⦆ is the set of finite ordinals.
That is the definition of finite ordinal.
That is the definition of ω,
the first ordinal after all finite ordinals.
γ before ω: γ is finite.
γ ∈ ⟦0,ω⦆ ⇒
∀β ∈ ⦅0,γ⟧: ∃α: α+1=β
ω before ξ: ξ is not finite.
ω ∈ ⦅0,ξ⟧ ⇒
¬∀β ∈ ⦅0,ξ⟧: ∃α: α+1=β
(Keep in mind that ¬∃α: α+1=ω )
A better question is:
why do you (WM) support it?
On 25.10.2024 16:53, Jim Burns wrote:
On 10/25/2024 7:42 AM, WM wrote:
The whole interval (0, ω) is not finite,
let alone the doubled interval.
⟦0,ω⦆ is the set of finite ordinals.
That is the definition of finite ordinal.
That is the definition of ω,
the first ordinal after all finite ordinals.
Correct so far.
γ before ω: γ is finite.
γ ∈ ⟦0,ω⦆ ⇒
∀β ∈ ⦅0,γ⟧: ∃α: α+1=β
ω before ξ: ξ is not finite.
ω ∈ ⦅0,ξ⟧ ⇒
¬∀β ∈ ⦅0,ξ⟧: ∃α: α+1=β
(Keep in mind that ¬∃α: α+1=ω )
That is wrong in complete infinity.
A better question is:
why do you (WM) support it?
I support it in order to show that
your infinity is inconsistent.
Example:
Almost all unit fractions
cannot be discerned by definable real numbers.
If they are existing,
they are indiscernible, i.e. dark.
On 10/25/2024 12:57 PM, WM wrote:
A better question is:
why do you (WM) support it?
I support it in order to show that
your infinity is inconsistent.
My infinity? You mean
γ ∈ ⟦0,ω⦆ ⇔ ∀β ∈ ⦅0,γ⟧: ∃α: α+1=β
?
Almost all unit fractions
cannot be discerned by definable real numbers.
All unit fractions are reciprocals of
positive countable.to.from.0 numbers.
If they are existing,
they are indiscernible, i.e. dark.
If it is existing,
it is a reciprocal of
a positive countable.to.from.0 number.
I haven't found a reason to discern.
On 10/23/2024 08:05 AM, Jim Burns wrote:
Put d and anything else behind your back.
Swap swap swap, pull them out.
I pick both.
One of those picked is in [0,1]ᴿ\f(ℕ)
f(ℕ) ≠ [0,1]ᴿ
If you think I'm wrong, say why.
Ah, the main thing I want you to notice,
is, that the usual definition of "function",
in the usual descriptive milieu of functions
according to ZF set theory, is: a sub-set
of the Cartesian product of left-hand-side
and right-hand-side, or "Cartesian functions".
So, then this "only diagonal", has that
this function EF the n/d n->d d-> oo
is NOT a Cartesian function:
it simply does NOT have
the same definition of function as
being a Cartesian function, so
all that results from Cantor-Schroeder-Bernstein
about the transitivity of functions,
that's built upon Cartesian functions:
does not hold.
[...] that the most direct mapping between
discrete domain and continuous range is
this totally simple continuum limit of n/d
for natural integers as only d is not finite
and furthermore
is constant monotone strictly increasing
with a bounded range in [a,b], an infinite domain.
On 25.10.2024 21:05, Jim Burns wrote:
On 10/25/2024 12:57 PM, WM wrote:
A better question is:
why do you (WM) support it?
I support it in order to show that
your infinity is inconsistent.
My infinity? You mean
γ ∈ ⟦0,ω⦆ ⇔ ∀β ∈ ⦅0,γ⟧: ∃α: α+1=β
?
Mainly, among other points, the claim that
all unit fractions can be defined and he claim that
a Bob ca disappear in lossless exchanges.
Almost all unit fractions
cannot be discerned by definable real numbers.
All unit fractions are reciprocals of
positive countable.to.from.0 numbers.
That does not change the facts.
If they are existing,
they are indiscernible, i.e. dark.
If it is existing,
it is a reciprocal of
a positive countable.to.from.0 number.
Most of them are indiscernible too.
I haven't found a reason to discern.
Don't you claim that
every unit fraction can be discerned, i.e.
separated from the smaller ones by a real number?
On 10/25/2024 3:15 PM, WM wrote:
Mainly, among other points, the claim that
all unit fractions can be defined and the claim that
a Bob can disappear in lossless exchanges.
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
Almost all unit fractions
cannot be discerned by definable real numbers.
All unit fractions are reciprocals of
positive countable.to.from.0 numbers.
That does not change the facts.
Clearly, the disagreement of proofs
does not change which claims you call facts.
Don't you claim that
every unit fraction can be discerned, i.e. separated from the smaller
ones by a real number?
½⋅(⅟n+⅟(n+1)) existing is discerning?
I did not know that.
On 10/26/2024 9:04 AM, WM wrote:
On 26.10.2024 05:21, Jim Burns wrote:
On 10/25/2024 3:15 PM, WM wrote:
Mainly, among other points, the claim that
all unit fractions can be defined and the claim that
a Bob can disappear in lossless exchanges.
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that all natnumbers can be defined.
Huh? Confusing to me. Humm... Are you trying to suggest that a natural
number can _not_ be a natural number?
Am 26.10.2024 um 21:35 schrieb Chris M. Thomasson:
On 10/26/2024 9:04 AM, WM wrote:
On 26.10.2024 05:21, Jim Burns wrote:
On 10/25/2024 3:15 PM, WM wrote:
Mainly, among other points, the claim that
all unit fractions can be defined and the claim that
a Bob can disappear in lossless exchanges.
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that all natnumbers can be defined.
Huh? Confusing to me. Humm... Are you trying to suggest that a natural
number can _not_ be a natural number?
No. But most natnumbers cannot be defined. This can best be understood
by the unit fractions.
Try to understand the function NUF(x) = Number of Unit Fractions between
0 and x, which starts with 0 at 0 or less and after NUF(x') = 1 cannot
change to 2 without pausing for an interval consisting of uncountably
many real points. The reason is this: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Regards, WM
and after NUF(x') = 1
WM <wolfgang.mueckenheim@tha.de> wrote:
On 26.10.2024 05:21, Jim Burns wrote:
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that all natnumbers can be defined.
Wrong. Natural numbers are defined as those defined by the Peano
axioms. That is what we mean by "natural number". Anything which
"can't be defined" this way isn't a natural number.
On 26.10.2024 05:21, Jim Burns wrote:
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that all natnumbers can be defined.
Regards, WM
Am 27.10.2024 um 08:38 schrieb WM:
and after NUF(x') = 1
There is no x' e IR such that NUF(x') = 1.
Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.
If 'not defined' could be a proper subset of the naturals, then there
would be a first such 'not defined' in that subset. Of course WM can't substantiate any of his wild claims.
On 10/26/24 12:04 PM, WM wrote:
On 26.10.2024 05:21, Jim Burns wrote:
On 10/25/2024 3:15 PM, WM wrote:
Mainly, among other points, the claim that
all unit fractions can be defined and the claim that
a Bob can disappear in lossless exchanges.
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that all natnumbers can be defined.
But they have been.
0. Of course x' and x'' are dark rational numbers, the smallest unitfractions.
Try to define more unit fractions than remain undefined.
But none are.
Try to understand the function NUF(x) which starts with 0 at 0 and
after 1 cannot change to 2 without pausing for an interval consisting
of uncountably many real points.
Bucause NUF(x) has an illogical definition, it assumes something that
doesn't exist.
After NUF(x') = 1
On 10/27/24 3:38 AM, WM wrote:
Am 26.10.2024 um 21:35 schrieb Chris M. Thomasson:
On 10/26/2024 9:04 AM, WM wrote:
On 26.10.2024 05:21, Jim Burns wrote:
On 10/25/2024 3:15 PM, WM wrote:
Mainly, among other points, the claim that
all unit fractions can be defined and the claim that
a Bob can disappear in lossless exchanges.
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that all natnumbers can be defined.
Huh? Confusing to me. Humm... Are you trying to suggest that a
natural number can _not_ be a natural number?
No. But most natnumbers cannot be defined. This can best be understood
by the unit fractions.
So, what is the line between the DEFINED natural numbers and the "not defined"?
Am 27.10.2024 um 14:54 schrieb Moebius:
Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.
That [...] would require that [for each and every x > 0]
[there] exist [ℵo unit fractions] between 0 and x
After NUF(x') = 1
course x' and x'' are [...] rational numbers, the smallest unit fractions.
Am 27.10.2024 um 14:54 schrieb Moebius:
Am 27.10.2024 um 08:38 schrieb WM:
and after NUF(x') = 1
There is no x' e IR such that NUF(x') = 1.
If all unit fractions are existing, then there is each one. Never two or
more are at the same point. Hence there is a single one first.
Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.
That is blatantly wrong because it would require that ℵo unit fractions exist between 0 and each and every x > 0, i.e., the open interval (0, 1].
Regards, WM
Am 27.10.2024 um 14:54 schrieb Moebius:Which is obviously the case. If there were a real x with finitely many
Am 27.10.2024 um 08:38 schrieb WM:
and after NUF(x') = 1There is no x' e IR such that NUF(x') = 1.
Hint: For each and every x e IR, x <= 0: NUF(x) = 0 and for each andThat is blatantly wrong because it would require that ℵo unit fractions exist between 0 and each and every x > 0,
every x e IR, x > 0: NUF(x) = aleph_0.
i.e., the open interval (0, 1].No, you shifted the quantifiers again.
Am 27.10.2024 um 12:38 schrieb Richard Damon:By the Peano axioms.
On 10/26/24 12:04 PM, WM wrote:Impossible.
On 26.10.2024 05:21, Jim Burns wrote:But they have been.
On 10/25/2024 3:15 PM, WM wrote:
Mainly, among other points, the claim that all unit fractions can be >>>>> defined and the claim that a Bob can disappear in losslessThe proof that all unit fractions can be defined is to define them as
exchanges.
reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that all natnumbers can be defined.
Proof: If infinity is actual, then all elements of the setWhat does it mean for an element of a set to not exist?
of unit fractions exist.
WM pretended :
Am 27.10.2024 um 14:39 schrieb FromTheRafters:
If 'not defined' could be a proper subset of the naturals, then there
would be a first such 'not defined' in that subset. Of course WM
can't substantiate any of his wild claims.
Proof:
Blah blah blah.
Am Sun, 27 Oct 2024 17:12:23 +0100 schrieb WM:
Am 27.10.2024 um 14:54 schrieb Moebius:
Am 27.10.2024 um 08:38 schrieb WM:
and after NUF(x') = 1There is no x' e IR such that NUF(x') = 1.
Which is obviously the case.Hint: For each and every x e IR, x <= 0: NUF(x) = 0 and for each andThat is blatantly wrong because it would require that ℵo unit fractions
every x e IR, x > 0: NUF(x) = aleph_0.
exist between 0 and each and every x > 0,
If there were a real x with finitely many
UFs less than it, the finitely many larger UFs... couldn't have
infinitely many lesser UFs. Unless you claim finitely many UFs.
i.e., the open interval (0, 1].No, you shifted the quantifiers again.
Am Sun, 27 Oct 2024 17:18:32 +0100 schrieb WM:
Proof: If infinity is actual, then all elements of the setWhat does it mean for an element of a set to not exist?
of unit fractions exist.
Am 27.10.2024 um 17:07 schrieb WM:
After NUF(x') = 1
There is no x' e IR such that NUF(x') = 1.
Am 27.10.2024 um 17:12 schrieb WM:
Am 27.10.2024 um 14:54 schrieb Moebius:
Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.
That [...] would require that [for each and every x > 0] [there] exist
[ℵo unit fractions] between 0 and x
Exactly!
On 10/27/24 12:12 PM, WM wrote:
Am 27.10.2024 um 14:54 schrieb Moebius:
Am 27.10.2024 um 08:38 schrieb WM:
and after NUF(x') = 1
There is no x' e IR such that NUF(x') = 1.
If all unit fractions are existing, then there is each one. Never two
or more are at the same point. Hence there is a single one first.
No two together, but no need for one to be first at the large 'n' end of them.
Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.
That is blatantly wrong because it would require that ℵo unit
fractions exist between 0 and each and every x > 0, i.e., the open
interval (0, 1].
Right, like is what happens.
It may seem strange to a person stuck in finite logic, but is true when
you understand how infinity works.
So, you agree that there are no undefined Natural Numbers to be dark, as
the set of that defined Natural Numbers grow without end.
On 10/27/24 12:07 PM, WM wrote:
Am 27.10.2024 um 16:55 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 26.10.2024 05:21, Jim Burns wrote:
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that all natnumbers can be defined.
Wrong. Natural numbers are defined as those defined by the Peano
axioms. That is what we mean by "natural number". Anything which
"can't be defined" this way isn't a natural number.
Then there is no actual infinity.
Then why are all your arguements based on it?
Proof: If infinity is actual, then all elements of the set of unit
fractions exist. The function NUF(x) = Number of Unit Fractions
between 0 and x starts with 0 at 0. After NUF(x') = 1 it cannot change
to NUF(x'') = 2 without pausing for an interval consisting of
uncountably many real points. The reason is this: ∀n ∈ ℕ: 1/n -
1/(n+1) > 0. Of course x' and x'' are dark rational numbers, the
smallest unit fractions.
So, all you have done is shown that YOUR CONCEPT of Acutual Infinity
can't be correct.
On 27.10.2024 19:00, Richard Damon wrote:
So, you agree that there are no undefined Natural Numbers to be dark,
as the set of that defined Natural Numbers grow without end.
Nevertheless it remains finite and far less than the actual infinity.
Regards, WM
On 27.10.2024 18:08, Moebius wrote:
Am 27.10.2024 um 17:12 schrieb WM:This requirement is impossible to satisfy because every unit fraction is
Am 27.10.2024 um 14:54 schrieb Moebius:
Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.
That [...] would require that [for each and every x > 0] [there]
exist [ℵo unit fractions] between 0 and x
Exactly!
an x > 0.
Regards, WM
On 27.10.2024 19:00, Richard Damon wrote:lolno. N is infinite.
So, you agree that there are no undefined Natural Numbers to be dark,Nevertheless it remains finite and far less than the actual infinity.
as the set of that defined Natural Numbers grow without end.
On 27.10.2024 18:08, Moebius wrote:Huh? This produces an infinity of UFs.
Am 27.10.2024 um 17:12 schrieb WM:This requirement is impossible to satisfy because every unit fraction is
Am 27.10.2024 um 14:54 schrieb Moebius:Exactly!
Hint: For each and every x e IR, x <= 0: NUF(x) = 0 and for each andThat [...] would require that [for each and every x > 0] [there] exist
every x e IR, x > 0: NUF(x) = aleph_0.
[ℵo unit fractions] between 0 and x
an x > 0.
On 27.10.2024 19:00, Richard Damon wrote:No, there is nothing.
On 10/27/24 12:12 PM, WM wrote:NUF increases by 1 or more, but more would violate mathematics.
Am 27.10.2024 um 14:54 schrieb Moebius:No two together, but no need for one to be first at the large 'n' end
Am 27.10.2024 um 08:38 schrieb WM:If all unit fractions are existing, then there is each one. Never two
and after NUF(x') = 1There is no x' e IR such that NUF(x') = 1.
or more are at the same point. Hence there is a single one first.
of them.
Right, like is what happens.Hint: For each and every x e IR, x <= 0: NUF(x) = 0 and for each andThat is blatantly wrong because it would require that ℵo unit
every x e IR, x > 0: NUF(x) = aleph_0.
fractions exist between 0 and each and every x > 0,
It may seem strange to a person stuck in finite logic, but is true when
you understand how infinity works.
This infinity between 0 and (0, 1] is not what I can accept.i.e., the open interval (0, 1].
On 27.10.2024 19:04, joes wrote:I can't see how that follows?
Am Sun, 27 Oct 2024 17:12:23 +0100 schrieb WM:Between two unit fractions there are uncountably many x > 0. Therefore
Am 27.10.2024 um 14:54 schrieb Moebius:Which is obviously the case.
Am 27.10.2024 um 08:38 schrieb WM:That is blatantly wrong because it would require that ℵo unit
and after NUF(x') = 1There is no x' e IR such that NUF(x') = 1.
Hint: For each and every x e IR, x <= 0: NUF(x) = 0 and for each and
every x e IR, x > 0: NUF(x) = aleph_0.
fractions exist between 0 and each and every x > 0,
your claim is wrong.
No, there are inf. many.If there were a real x with finitely many UFs less than it, theThere is a smallest UF.
finitely many larger UFs... couldn't have infinitely many lesser UFs.
Unless you claim finitely many UFs.
That IS the case.This is the definition of NUF(x): There are NUF(x) UFs between 0 and x.i.e., the open interval (0, 1].No, you shifted the quantifiers again.
Not: For given x, there are infinitely many UFs.
There is no UFCorrect, there is no smallest UF >0. But the quantifier-shifted claim
smaller than all x > 0 because every UF is an x > 0 itself.
On 27.10.2024 19:01, joes wrote:Ah, but then the element does exist after all, but also somehow wasn't
Am Sun, 27 Oct 2024 17:18:32 +0100 schrieb WM:
The set can be extended this element - like Hilbert's Hotel, when everyProof: If infinity is actual, then all elements of the set of unitWhat does it mean for an element of a set to not exist?
fractions exist.
guest multiplies his room number with 10^10^10000.
Am 27.10.2024 um 12:38 schrieb Richard Damon:Thus the set of undefined numbers shrinks to size zero like the
On 10/27/24 3:38 AM, WM wrote:There is no line. The defined natural numbers are a potentially infinite
Am 26.10.2024 um 21:35 schrieb Chris M. Thomasson:So, what is the line between the DEFINED natural numbers and the "not
On 10/26/2024 9:04 AM, WM wrote:No. But most natnumbers cannot be defined. This can best be understood
On 26.10.2024 05:21, Jim Burns wrote:Huh? Confusing to me. Humm... Are you trying to suggest that a
On 10/25/2024 3:15 PM, WM wrote:No, you falsely assume that all natnumbers can be defined.
Mainly, among other points, the claim that all unit fractions can >>>>>>> be defined and the claim that a Bob can disappear in losslessThe proof that all unit fractions can be defined is to define them >>>>>> as reciprocals of positive countable.to.from.0 numbers.
exchanges.
That describes all of them and only them.
natural number can _not_ be a natural number?
by the unit fractions.
defined"?
set, i.e. it can grow without end.
On 26.10.2024 05:21, Jim Burns wrote:
On 10/25/2024 3:15 PM, WM wrote:
Mainly, among other points, the claim that
all unit fractions can be defined and the claim that
a Bob can disappear in lossless exchanges.
The proof that all unit fractions can be defined
is to define them
as reciprocals of positive countable.to.from.0 numbers.
That describes all of them and only them.
No, you falsely assume that
all natnumbers can be defined.
Almost all unit fractions
cannot be discerned by definable real numbers.
All unit fractions are reciprocals of
positive countable.to.from.0 numbers.
That does not change the facts.
Clearly, the disagreement of proofs
does not change which claims you call facts.
Try to define more unit fractions
than remain undefined.
Try to understand the function NUF(x)
which starts with 0 at 0 and
after 1 cannot change to 2 without
pausing for an interval consisting of
uncountably many real points.
Don't you claim that
every unit fraction can be discerned,
i.e. separated from the smaller ones
by a real number?
½⋅(⅟n+⅟(n+1)) existing is discerning?
I did not know that.
Giving a real number
_by its digits or as a fraction_
which is between two unit fractions discerns
the larger one and its larger predecessors.
On 10/28/24 5:49 AM, WM wrote:
Your proof fails because it assumes properties that are not actually required.
If you really mean that there must be a lowest unit fraction,
On 10/26/2024 12:04 PM, WM wrote:
No, you falsely assume that
all natnumbers can be defined.
I assume that
natural numbers are finite ordinals, and
that a finite ordinal is predecessored or is 0 and
each of its priors is predecessored or is 0, and
that an ordinal is successored and
a set of ordinals is minimummed or is {}
do not answer the question.Try to define more unit fractions
than remain undefined.
𝗖𝗹𝗮𝗶𝗺𝘀 exist which
Each rational can be given as a fraction.
On 10/28/24 6:36 AM, WM wrote:
NUF increases by 1 or more, but more would violate mathematics.
No, NUF(x) jumps from 0 to Aleph_0 in the domain of finite numbers,
because there is no finite x where it has the value of 0.
It is just a "undefined" function.
Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.
That is blatantly wrong because it would require that ℵo unit
fractions exist between 0 and each and every x > 0, i.e., the open
interval (0, 1].
Right, like is what happens.
It may seem strange to a person stuck in finite logic, but is true
when you understand how infinity works.
This infinity between 0 and (0, 1] is not what I can accept.
Note, it isn't an "infinity between" it is that the "bottom" of (0, 1] doesn't exist as a definable point.
On 10/28/24 6:32 AM, WM wrote:
On 27.10.2024 18:08, Moebius wrote:
Am 27.10.2024 um 17:12 schrieb WM:This requirement is impossible to satisfy because every unit fraction
Am 27.10.2024 um 14:54 schrieb Moebius:
Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.
That [...] would require that [for each and every x > 0] [there]
exist [ℵo unit fractions] between 0 and x
Exactly!
is an x > 0.
Of course it isn't.
You just don't understand how infinity works. Things without ends just
behave differently then things with finite ends,
On 10/28/24 6:30 AM, WM wrote:
On 27.10.2024 18:05, Moebius wrote:
Am 27.10.2024 um 17:07 schrieb WM:
After NUF(x') = 1
There is no x' e IR such that NUF(x') = 1.
There is NUF(0) = 0. NUF(1) > 0. Hence NUF increases. This cannot
happen by more than single unitfractions with interruptions, according
to mathematics:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
It can not happen AT FINITE VALUES, but can "between" them,
like just
above 0, where it jumps from 0 to Aleph_0
WM was thinking very hard :
The function NUF(x) = Number of Unit Fractions between 0 and x starts
with 0 at 0.
Followed by a discontinuity.
After NUF(x') = 1 it cannot change to NUF(x'') = 2 without pausing
for an interval consisting of uncountably many real points.
Your "Axiom of because I say so" is overworked.
Non sequitur. This is just the second part of your stepwise function.
It
doesn't have to happen step by step as you envision it.
Am Sun, 27 Oct 2024 17:20:17 +0100 schrieb WM:
There is no line. The defined natural numbers are a potentially infiniteThus the set of undefined numbers shrinks to size zero
set, i.e. it can grow without end.
like the
endsegments.
On 10/28/2024 12:47 PM, WM wrote:
On 28.10.2024 12:21, Richard Damon wrote:
On 10/28/24 6:32 AM, WM wrote:
On 27.10.2024 18:08, Moebius wrote:
Am 27.10.2024 um 17:12 schrieb WM:This requirement is impossible to satisfy because every unit
Am 27.10.2024 um 14:54 schrieb Moebius:
Hint: For each and every x e IR, x <= 0: NUF(x) = 0That [...] would require that [for each and every x > 0] [there]
and for each and every x e IR, x > 0: NUF(x) = aleph_0. [Moebius] >>>>>>
exist [ℵo unit fractions] between 0 and x [WM, corrected by me]
Exactly! [Moebius]
fraction is an x > 0. [WM]
Of course it isn't. [Richard Damon]
In order to maintain your nonsense, you have even to deny that all
unit fractions are positive or to believe that many sit between 0 and
(0, oo).
Huh? WTF?
On 10/28/2024 12:42 PM, WM wrote:
On 28.10.2024 12:17, Richard Damon wrote:
Your proof fails because it assumes properties that are not actually
required.
If you really mean that there must be a lowest unit fraction, [Richard Damon]
I mean that there are unit fractions
None is below zero. [WM]
Mathematics proves that never more than one is at any point.
You sure are trying to prove that you are an, idiot...?
On 10/28/2024 3:25 PM, Moebius wrote:
I told ye. Didn't I? Mad, gaga, [insane,] loco loco.No shit! Wow. Pretty fucked up man. ;^o
WM's mind is the brain insane? Well, perhaps not this bad:
https://youtu.be/5lvQx7dYY94
Shit happens... ;^o
On 10/28/2024 3:25 PM, Moebius wrote:
I told ye. Didn't I? Mad, gaga, [insane,] loco loco.No shit! Wow. Pretty fucked up man. ;^o
WM's mind is the brain insane? Well, perhaps not this bad:
https://youtu.be/5lvQx7dYY94
Shit happens... ;^o
Am 29.10.2024 um 00:02 schrieb Chris M. Thomasson:
On 10/28/2024 3:55 PM, Moebius wrote:
Btw. try this as an antidote: https://www.youtube.com/watch?
v=2EdLasOrG6c
:^D Heck, WM might even need a new religion. Not sure what one. Humm:
https://youtu.be/usADINi17cI
Na, perhaps another one? Humm... ;^D
Nice, but also good (maybe even better):
https://www.youtube.com/watch?v=4B7mKCsGSUc
( Actually, clearly much better... :-P )
On 10/28/2024 3:55 PM, Moebius wrote:
Btw. try this as an antidote: https://www.youtube.com/watch?v=2EdLasOrG6c
:^D Heck, WM might even need a new religion. Not sure what one. Humm:
https://youtu.be/usADINi17cI
Na, perhaps another one? Humm... ;^D
Am 29.10.2024 um 00:45 schrieb Moebius:
Am 29.10.2024 um 00:02 schrieb Chris M. Thomasson:
On 10/28/2024 3:55 PM, Moebius wrote:
Btw. try this as an antidote: https://www.youtube.com/watch?
v=2EdLasOrG6c
:^D Heck, WM might even need a new religion. Not sure what one. Humm:
https://youtu.be/usADINi17cI
Na, perhaps another one? Humm... ;^D
Nice, but also good (maybe even better):
https://www.youtube.com/watch?v=4B7mKCsGSUc
( Actually, clearly much better... :-P )
Hint: I generally prefer women voices (don't ask!), but in this case ...
(you see).
On 28.10.2024 12:17, Richard Damon wrote:
On 10/28/24 5:49 AM, WM wrote:
Your proof fails because it assumes properties that are not actually
required.
If you really mean that there must be a lowest unit fraction,
I mean that there are unit fractions.
None is below zero.
Mathematics proves that never more than one is at any point.
Regards, WM
On 28.10.2024 15:07, FromTheRafters wrote:
WM was thinking very hard :
The function NUF(x) = Number of Unit Fractions between 0 and x starts
with 0 at 0.
Followed by a discontinuity.
No. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds everywhere.
After NUF(x') = 1 it cannot change to NUF(x'') = 2 without pausing
for an interval consisting of uncountably many real points.
Your "Axiom of because I say so" is overworked.
It is mathematics:>> The reason is this: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Non sequitur. This is just the second part of your stepwise function.
It holds everywhere if mathematics holds everywhere.
It doesn't have to happen step by step as you envision it.
As mathematics dictates it.
You have confessed: In order to maintain actual infinity without dark numbers, you have to violate mathematics.
Regards, WM
WM explained :
On 28.10.2024 15:07, FromTheRafters wrote:
WM was thinking very hard :
The function NUF(x) = Number of Unit Fractions between 0 and x
starts with 0 at 0.
Followed by a discontinuity.
No. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds everywhere.
Not for n equal to zero.
After NUF(x') = 1 it cannot change to NUF(x'') = 2 without pausing
for an interval consisting of uncountably many real points.
Your "Axiom of because I say so" is overworked.
It is mathematics:>> The reason is this: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. >>>
Non sequitur. This is just the second part of your stepwise function.
It holds everywhere if mathematics holds everywhere.
It doesn't have to happen step by step as you envision it.
As mathematics dictates it.
You have confessed: In order to maintain actual infinity without dark
numbers, you have to violate mathematics.
Bull! Just another of your non sequiturs.
On 28.10.2024 12:21, Richard Damon wrote:Yes, they are infinitesimal.
On 10/28/24 6:30 AM, WM wrote:Yes, at undefinable values.
On 27.10.2024 18:05, Moebius wrote:It can not happen AT FINITE VALUES, but can "between" them,
Am 27.10.2024 um 17:07 schrieb WM:There is NUF(0) = 0. NUF(1) > 0. Hence NUF increases. This cannot
After NUF(x') = 1There is no x' e IR such that NUF(x') = 1.
happen by more than single unitfractions with interruptions, according
to mathematics:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
--like just above 0, where it jumps from 0 to Aleph_0Not at one point, definable or not.
On 10/28/24 3:42 PM, WM wrote:
I mean that there are unit fractions.Which doesn't mean there must be a first, as they aproach an
None is below zero.
Mathematics proves that never more than one is at any point.
accumulation point where the density becomes infinite.
Something which can't happen your world of finite logic, but does when
the logic can handle infinities.
On 10/28/24 3:52 PM, WM wrote:
On 28.10.2024 12:21, Richard Damon wrote:
On 10/28/24 6:36 AM, WM wrote:
NUF increases by 1 or more, but more would violate mathematics.
No, NUF(x) jumps from 0 to Aleph_0 in the domain of finite numbers,
because there is no finite x where it has the value of 0.
It has the value 0 for all x =< 0. And it cannot jump by more than 1
at any point.
Of course it can. just not at any finite value.
No, but the first steps happen at undefinable x.
No, it happens at an x that isn't a finite number.
If you allow your NUF to accept infintesimal numbers
This infinity between 0 and (0, 1] is not what I can accept.
Note, it isn't an "infinity between" it is that the "bottom" of (0,
1] doesn't exist as a definable point.
That is true. The bottom is dark.
No, the bottom is outside the set.
Your "Darkness" is just your attempt to hide the problems with your logic.
Am Mon, 28 Oct 2024 20:55:19 +0100 schrieb WM:
On 28.10.2024 12:21, Richard Damon wrote:Yes, they are infinitesimal.
On 10/28/24 6:30 AM, WM wrote:Yes, at undefinable values.
On 27.10.2024 18:05, Moebius wrote:It can not happen AT FINITE VALUES, but can "between" them,
Am 27.10.2024 um 17:07 schrieb WM:There is NUF(0) = 0. NUF(1) > 0. Hence NUF increases. This cannot
After NUF(x') = 1There is no x' e IR such that NUF(x') = 1.
happen by more than single unitfractions with interruptions, according >>>> to mathematics:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
like just above 0, where it jumps from 0 to Aleph_0Not at one point, definable or not.
On 29.10.2024 00:58, Richard Damon wrote:
On 10/28/24 3:42 PM, WM wrote:
I mean that there are unit fractions.Which doesn't mean there must be a first, as they aproach an
None is below zero.
Mathematics proves that never more than one is at any point.
accumulation point where the density becomes infinite.
Their density is bounded by uncountably many points between every pair
of consecutive unit fractions:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
The density is one point over uncountably many points, that is rather precisely 0.
Something which can't happen your world of finite logic, but does when
the logic can handle infinities.
Where does the density surpass 1/10? Can you find this point? If not it
is another proof of dark numbers.
Regards, WM
On 29.10.2024 01:21, Richard Damon wrote:
On 10/28/24 3:52 PM, WM wrote:
On 28.10.2024 12:21, Richard Damon wrote:
On 10/28/24 6:36 AM, WM wrote:
NUF increases by 1 or more, but more would violate mathematics.
No, NUF(x) jumps from 0 to Aleph_0 in the domain of finite numbers,
because there is no finite x where it has the value of 0.
It has the value 0 for all x =< 0. And it cannot jump by more than 1
at any point.
Of course it can. just not at any finite value.
There are no others on the real line.
No, but the first steps happen at undefinable x.
No, it happens at an x that isn't a finite number.
The it is dark.
If you allow your NUF to accept infintesimal numbers
No, that is strongly forbidden.
This infinity between 0 and (0, 1] is not what I can accept.
Note, it isn't an "infinity between" it is that the "bottom" of (0,
1] doesn't exist as a definable point.
That is true. The bottom is dark.
No, the bottom is outside the set.
Then it is dark bottom.
Your "Darkness" is just your attempt to hide the problems with your
logic.
And your bottom outside the set is what?
Regards, WM
On 29.10.2024 00:58, Richard Damon wrote:So not bounded at all.
On 10/28/24 3:42 PM, WM wrote:
Their density is bounded by uncountably many points between every pairI mean that there are unit fractions. None is below zero.Which doesn't mean there must be a first, as they aproach an
Mathematics proves that never more than one is at any point.
accumulation point where the density becomes infinite.
of consecutive unit fractions:
The density is one point over uncountably many points, that is rather precisely 0.
How do you define the density at a point?Something which can't happen your world of finite logic, but does whenWhere does the density surpass 1/10? Can you find this point?
the logic can handle infinities.
On 29.10.2024 01:21, Richard Damon wrote:That's the point (pun unintended).
On 10/28/24 3:52 PM, WM wrote:There are no others on the real line.
On 28.10.2024 12:21, Richard Damon wrote:Of course it can. just not at any finite value.
On 10/28/24 6:36 AM, WM wrote:It has the value 0 for all x =< 0. And it cannot jump by more than 1
NUF increases by 1 or more, but more would violate mathematics.No, NUF(x) jumps from 0 to Aleph_0 in the domain of finite numbers,
because there is no finite x where it has the value of 0.
at any point.
I.e. infinitesimal.The nit is dark.No, but the first steps happen at undefinable x.No, it happens at an x that isn't a finite number.
Then there is no x with NUF(x)=1.If you allow your NUF to accept infintesimal numbersNo, that is strongly forbidden.
The infimum is 0.Then it is dark bottom.No, the bottom is outside the set.That is true. The bottom is dark.This infinity between 0 and (0, 1] is not what I can accept.Note, it isn't an "infinity between" it is that the "bottom" of (0,
1] doesn't exist as a definable point.
Your "Darkness" is just your attempt to hide the problems with yourAnd your bottom outside the set is what?
logic.
On 10/28/2024 11:31 AM, Jim Burns wrote:
[...]
Of course you would mean
"finite ordinals as
a brief summary model of discernibles the members" of
"the closure of all relations that make things numbers,
all those sets, too, all the related things",
"the" "numbers".
On 28.10.2024 19:31, Jim Burns wrote:
On 10/26/2024 12:04 PM, WM wrote:
No, you falsely assume that
all natnumbers can be defined.
I assume that
natural numbers are finite ordinals, and
that a finite ordinal is predecessored or is 0 and
each of its priors is predecessored or is 0, and
that an ordinal is successored and
a set of ordinals is minimummed or is {}
That is irrelevant
with respect to the fact that
between every chosen unit fraction and zero
there are ℵo unit fractions which
cannot be chosen.
between every chosen unit fraction and zero
there are ℵo unit fractions which
cannot be chosen.
On 29.10.2024 09:36, joes wrote:
Am Mon, 28 Oct 2024 20:55:19 +0100 schrieb WM:
Yes, at undefinable values.
Yes, they are infinitesimal.
There are no infinitesimal x > 0 on the real axis.
But your claim may be interpreted as dark numbers x
because they cannot be determined.
On 10/28/2024 4:55 PM, Moebius wrote:
***NO ONE*** can perform this in a better way!Actually, I always liked that song. Mellow 80's chill soft rock for
https://www.youtube.com/watch?v=VdQY7BusJNU
sure. Indeed.
On 10/28/2024 4:45 PM, Moebius wrote:
Am 29.10.2024 um 00:02 schrieb Chris M. Thomasson:
On 10/28/2024 3:55 PM, Moebius wrote:
Btw. try this as an antidote: https://www.youtube.com/watch?
v=2EdLasOrG6c
:^D Heck, WM might even need a new religion. Not sure what one. Humm:
https://youtu.be/usADINi17cI
Na, perhaps another one? Humm... ;^D
Nice, but also good (maybe even better):
https://www.youtube.com/watch?v=4B7mKCsGSUc
( Actually, clearly much better... :-P )
:^D Not bad at all.
On 29.10.2024 09:36, joes wrote:Your claimed "dark numbers" may be interpreted as infinitesimal.
Am Mon, 28 Oct 2024 20:55:19 +0100 schrieb WM:There are no infinitesimal x > 0 on the real axis. But your claim may be interpreted as dark numbers x because they cannot be determined.
On 28.10.2024 12:21, Richard Damon wrote:Yes, they are infinitesimal.
On 10/28/24 6:30 AM, WM wrote:Yes, at undefinable values.
On 27.10.2024 18:05, Moebius wrote:It can not happen AT FINITE VALUES, but can "between" them,
Am 27.10.2024 um 17:07 schrieb WM:There is NUF(0) = 0. NUF(1) > 0. Hence NUF increases. This cannot
After NUF(x') = 1There is no x' e IR such that NUF(x') = 1.
happen by more than single unitfractions with interruptions,
according to mathematics:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
--like just above 0, where it jumps from 0 to Aleph_0Not at one point, definable or not.
https://youtu.be/1Cw1ng75KP0?list=PLRYjIdaAzRVmmaJEG-elAlp91clvYXnws
WM formulated the question :
On 28.10.2024 21:51, FromTheRafters wrote:
WM explained :
On 28.10.2024 15:07, FromTheRafters wrote:
WM was thinking very hard :
The function NUF(x) = Number of Unit Fractions between 0 and x
starts with 0 at 0.
Followed by a discontinuity.
No. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds everywhere.
Not for n equal to zero.
I said ∀n ∈ ℕ. 0 is no natural number. NUF(0) = 0.
That is why your function is a step function with a discontinuity.
On 9/2/2024 10:07 AM, WM wrote:
How many [...] unit fractions are lessorequal than all unit
fractions? The correct answer is:
WM explained :
NUF(0) = 0, NUF(1) = ℵo. Therefore NUF must grow but cannot grow by
more than 1 at any point x of the real axis.
The number of unit fractions less than x is always aleph_zero for
positive x.
On 10/29/2024 4:48 AM, WM wrote:
On 29.10.2024 09:36, joes wrote:
Am Mon, 28 Oct 2024 20:55:19 +0100 schrieb WM:
Yes, at undefinable values.
Yes, they are infinitesimal.
There are no infinitesimal x > 0 on the real axis.
But your claim may be interpreted as dark numbers x
because they cannot be determined.
Not.existing is not.existing, dark or visible.
We can reason correctly about uncountably.many
points.between.splits.of.ℚ
On 10/29/24 4:42 AM, WM wrote:
NUF(x) MUST jump from 0 to Aleph_0 at all real values x, as below ANY
real number x, there are Aleph_0 unit fractions.
No, but the first steps happen at undefinable x.
No, it happens at an x that isn't a finite number.
Then it is dark.
And, as you said above, NOT REAL.
If you allow your NUF to accept infintesimal numbers
No, that is strongly forbidden.
Then it just jumps.
On 10/29/2024 6:13 PM, Moebius wrote:
Am 30.10.2024 um 00:00 schrieb Chris M. Thomasson:Is hot pop better than warm pop? ;^D
https://youtu.be/1Cw1ng75KP0?list=PLRYjIdaAzRVmmaJEG-elAlp91clvYXnws
Hot, but (actually) (musically) not THAT good (->pop). :-P
Let's say ... "nice" :-P
On 29.10.2024 18:34, Jim Burns wrote:
On 10/29/2024 4:48 AM, WM wrote:
On 29.10.2024 09:36, joes wrote:
Am Mon, 28 Oct 2024 20:55:19 +0100 schrieb WM:
Yes, at undefinable values.
Yes, they are infinitesimal.
There are no infinitesimal x > 0 on the real axis.
But your claim may be interpreted as dark numbers x
because they cannot be determined.
Not.existing is not.existing, dark or visible.
We can reason correctly about uncountably.many
points.between.splits.of.ℚ
NUF(0) = 0, NUF(1) = ℵo.
Therefore NUF must grow but cannot grow by
more than 1 at any point x of the real axis.
On 10/29/2024 08:15 AM, Jim Burns wrote:
I find the term "number" problematic until
the description of _which_ numbers is given.
Most people observe an extensionality in
the "real-valued", then for example somebody like
Hardy also relates real-valued values to points
on a line.
When you say "finite ordinals"
what's intended is
"all the body of structural relation
with regards to the ordinal and
with regards to the finite",
and it's doens't necessarily say so much about
"the numbers" or even "the natural numbers",
as with regards to that Peano's numbers aren't
merely zero and a closure to successor,
there's constant monotone increase and modularity,
and some have it's only where the operations of
the addition and division, semi-groups, come together.
On 10/30/2024 11:28 AM, Jim Burns wrote:
On 10/30/2024 1:36 AM, Ross Finlayson wrote:
On 10/29/2024 08:15 AM, Jim Burns wrote:
I find the term "number" problematic until
the description of _which_ numbers is given.
Most people observe an extensionality in
the "real-valued", then for example somebody like
Hardy also relates real-valued values to points
on a line.
If you want "number" to refer to one of
the smallest superset of ℚ with the LUB property (ie, to ℝ)
then I'm okay with that.
I'm only asking that it be clear that's what it refers to.
Nah, I have "number" as mostly reflecting "scalar",
though there's usual wider usage
on 10/30/2024, WM supposed :
Believe what you like without foundation.
If ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is true, the NUF(x) grows in steps of not >> more than 1.
Wrong.
∀ᴿx > 0: NUF(x) = |ℕ|
On 10/30/2024 12:48 PM, WM wrote:
On 30.10.2024 16:43, FromTheRafters wrote:
on 10/30/2024, WM supposed :
Believe what you like without foundation.
If ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is true, the NUF(x) grows in steps of >>>> not more than 1.
Wrong.
What? ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
1/1 - 1/2 is greater than zero.
Wow, what a discovery! lol. ;^/
1/2 - 1/3 is greater than zero.
1/3 - 1/4 is greater than zero.
... On and on ...
They never hit zero. Got it?
No unit [fraction] equals zero.
On 10/30/2024 12:48 PM, WM wrote:
On 30.10.2024 16:43, FromTheRafters wrote:
on 10/30/2024, WM supposed :
Believe what you like without foundation.
If ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is true, the NUF(x) grows in steps of >>>> not more than 1.
Wrong.
What? ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
1/1 - 1/2 is greater than zero.
Wow, what a discovery! lol. ;^/
1/2 - 1/3 is greater than zero.
1/3 - 1/4 is greater than zero.
... On and on ...
They never hit zero. Got it?
No unit [fraction] equals zero.
On 10/30/2024 1:19 PM, Moebius wrote:
Mückenheim's great discovery is that the DISTANCE between twoWe know that the unit fractions tend to zero but never equal it. So, WM thinks well, the limit means finite? Or some shit like that? Humm...
"adjacent" unit fractions is always > 0. Incredible!
For absolutely no reason he now "thinks" that this fact implies a
smallest unit fraction. :-)
Am 30.10.2024 um 21:34 schrieb Chris M. Thomasson:
On 10/30/2024 1:19 PM, Moebius wrote:
Mückenheim's great discovery is that the DISTANCE between twoWe know that the unit fractions tend to zero but never equal it. So,
"adjacent" unit fractions is always > 0. Incredible!
For absolutely no reason he now "thinks" that this fact implies a
smallest unit fraction. :-)
WM thinks well, the limit means finite? Or some shit like that? Humm...
Yeah. He just can't comprehend the idea of infinitely many <whatever>.
On 30.10.2024 18:05, Jim Burns wrote:
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
∀ᴿx > 0: NUF(x) = |ℕ|
Is ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 wrong?
On 10/30/2024 3:55 PM, WM wrote:
On 30.10.2024 18:05, Jim Burns wrote:
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
∀ᴿx > 0: NUF(x) = |ℕ|
Is ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 wrong?
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
WM explained :
On 30.10.2024 16:43, FromTheRafters wrote:
on 10/30/2024, WM supposed :
Believe what you like without foundation.
If ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is true, the NUF(x) grows in steps of >>>> not more than 1.
Wrong.
What? ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
No, the other part. Your 'conclusion' is a non sequitur.
Mückenheim's great discovery is that the DISTANCE between two "adjacent" unit fractions is always > 0. Incredible!
For absolutely no reason he now "thinks" that this fact implies a
smallest unit fraction.
WM submitted this idea :
On 30.10.2024 21:24, FromTheRafters wrote:
WM explained :
On 30.10.2024 16:43, FromTheRafters wrote:
on 10/30/2024, WM supposed :
Believe what you like without foundation.
If ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is true, the NUF(x) grows in steps of >>>>>> not more than 1.
Wrong.
What? ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
No, the other part. Your 'conclusion' is a non sequitur.
My conclusion is that all unit fractions are separated by large sets
of real points from each other. Never two or more unit fractions are
at the same point. Is that what you doubt? Hardly.
Then you must doubt that NUF(x) can grow only by 1 at any point x? But
why?
Because NUF() doesn't "grow" it just *is*.
WM wrote :
On 31.10.2024 09:29, FromTheRafters wrote:
WM submitted this idea :
On 30.10.2024 21:24, FromTheRafters wrote:
WM explained :
On 30.10.2024 16:43, FromTheRafters wrote:
on 10/30/2024, WM supposed :
Believe what you like without foundation.
If ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is true, the NUF(x) grows in steps >>>>>>>> of not more than 1.
Wrong.
What? ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ?
No, the other part. Your 'conclusion' is a non sequitur.
My conclusion is that all unit fractions are separated by large sets
of real points from each other. Never two or more unit fractions are
at the same point. Is that what you doubt? Hardly.
Then you must doubt that NUF(x) can grow only by 1 at any point x?
But why?
Because NUF() doesn't "grow" it just *is*.
According to set theory every function just "is". But we analyze or
describe its behaviour with increasing argument x as increasing,
constant or decreasing. Should that be forbidden in case of NUF in
order to avoid problems?
Well, it would cure your discontinuity dyslexia problems.
On 29.10.2024 11:02, FromTheRafters wrote:
WM formulated the question :
On 28.10.2024 21:51, FromTheRafters wrote:
WM explained :
On 28.10.2024 15:07, FromTheRafters wrote:
WM was thinking very hard :
The function NUF(x) = Number of Unit Fractions between 0 and x
starts with 0 at 0.
Followed by a discontinuity.
No. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds everywhere.
Not for n equal to zero.
I said ∀n ∈ ℕ. 0 is no natural number. NUF(0) = 0.
That is why your function is a step function with a discontinuity.
NUF(0) = 0, NUF(1) = ℵo. Therefore NUF must grow but cannot grow by more than 1 at any point x of the real axis.
Regards, WM
On 31.10.2024 01:14, Jim Burns wrote:
On 10/30/2024 3:55 PM, WM wrote:
On 30.10.2024 18:05, Jim Burns wrote:
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
∀ᴿx > 0: NUF(x) = |ℕ|
Is ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 wrong?
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?
Failure to recognize that discontinuity is mathematical noted.
WM explained :
On 31.10.2024 12:00, FromTheRafters wrote:
Failure to recognize that discontinuity is mathematical noted.
False. NUF(x) can only increase at unit fractions. They cause no
larger discontinuity than 1.
Wrong, your function, if swept as you seem to require, jumps from zero
to countable infinity and stays there.
Why do you call 1 a discontinuity?
On 10/30/24 11:38 AM, WM wrote:
NUF(x) MUST jump from 0 to Aleph_0 at all real values x, as below ANY
real number x, there are Aleph_0 unit fractions.
On 10/31/24 5:31 AM, WM wrote:
The problem is functions don't "grow" at a point
NUF(x) has an infinite slope at x = 0, as the unit fractions have an accumulaton point there.
On 10/30/24 7:53 AM, WM wrote:
NUF(0) = 0, NUF(1) = ℵo. Therefore NUF must grow but cannot grow byBecause it grows BETWEEN points on the real axis from 0 to ℵo.
more than 1 at any point x of the real axis.
Your problem is there isn't a "first" x > 0 above 0 to see it increase
at,
so it can just jump to ℵo.But it cannot do so at a single point.
On 10/31/2024 4:13 AM, WM wrote:
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?
Is
⎛ ∀ᴿx > 0:
⎜ ∀n ∈ ℕ:
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
wrong?
On 31.10.2024 13:22, Jim Burns wrote:
On 10/31/2024 4:13 AM, WM wrote:
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?
Is
⎛ ∀ᴿx > 0:
⎜ ∀n ∈ ℕ:
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
wrong?
One of two contradicting formulas must be dropped.
On 31.10.2024 12:36, Richard Damon wrote:At single points a function has a single value, not a jump. In the
On 10/30/24 7:53 AM, WM wrote:
I investigate its behaviour at single points.NUF(0) = 0, NUF(1) = ℵo. Therefore NUF must grow but cannot grow byBecause it grows BETWEEN points on the real axis from 0 to ℵo.
more than 1 at any point x of the real axis.
Like all your non-questions, this is not about UFs but rather theYour problem is there isn't a "first" x > 0 above 0 to see it increaseWe do not see dark numbers but I can prove their existence
See above.so it can just jump to ℵo.But it cannot do so at a single point.
On 10/31/2024 1:41 PM, WM wrote:
On 31.10.2024 13:22, Jim Burns wrote:
On 10/31/2024 4:13 AM, WM wrote:
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?
Is
⎛ ∀ᴿx > 0:
⎜ ∀n ∈ ℕ:
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
wrong?
One of two contradicting formulas must be dropped.
Which (inside.quantifiers) formula is
the last which you accept with all prior formulas?
Am Thu, 31 Oct 2024 18:28:15 +0100 schrieb WM:Ab in die Klapsmühle mit dem Kerl!
We do not see dark numbers but I can prove their existence
On 31.10.2024 17:02, FromTheRafters wrote:
WM explained :
On 31.10.2024 12:00, FromTheRafters wrote:
Failure to recognize that discontinuity is mathematical noted.
False. NUF(x) can only increase at unit fractions. They cause no
larger discontinuity than 1.
Wrong, your function, if swept as you seem to require, jumps from zero
to countable infinity and stays there.
It cannot jump to countable infinity at a single point because in a
single point only one unit fraction exists. I investigate its behaviour
at single points.
Why do you call 1 a discontinuity?
The increase from n to n+1 at one point is a discontinuity.
Regards, WM
On 10/31/2024 4:36 AM, Richard Damon wrote:
On 10/30/24 10:04 AM, WM wrote:
On 30.10.2024 13:57, FromTheRafters wrote:
WM explained :
NUF(0) = 0, NUF(1) = ℵo. Therefore NUF must grow but cannot grow by >>>>> more than 1 at any point x of the real axis.
The number of unit fractions less than x is always aleph_zero for
positive x.
Believe what you like without foundation.
If ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is true, the NUF(x) grows in steps of >>> not more than 1.
Regards, WM
No, what that expression shows is that for every n, the unit fraction
1/ n has another unit fraction 1/(n+1) that is smaller than it, so
there is no "first" unit fraction in that sense for NUF(x) to get to 1.
You are just proving you don't undetstand even your own mathematics.
Does WM think that there is a unit fraction that is "closest" to 0? If
so, he is moronic yet again... Wow.
On 31.10.2024 13:22, Jim Burns wrote:
On 10/31/2024 4:13 AM, WM wrote:
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?
Is
⎛ ∀ᴿx > 0:
⎜ ∀n ∈ ℕ:
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
wrong?
One of two contradicting formulas must be dropped.
Regards, WM
At single points a function has a single value, not a jump.
In the
case of NUF, that value is infinite everywhere except at 0.
On 10/31/24 1:41 PM, WM wrote:
On 31.10.2024 13:22, Jim Burns wrote:
On 10/31/2024 4:13 AM, WM wrote:
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?
Is
⎛ ∀ᴿx > 0:
⎜ ∀n ∈ ℕ:
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
wrong?
One of two contradicting formulas must be dropped.
Or, just admit that your NUF(x) is where the contradiction is and drop it.
On 10/31/24 1:35 PM, WM wrote:
On 31.10.2024 12:36, Richard Damon wrote:
On 10/30/24 11:38 AM, WM wrote:
NUF(x) MUST jump from 0 to Aleph_0 at all real values x, as below
ANY real number x, there are Aleph_0 unit fractions.
You cannot distinguish them by any real number? That proves that they
are dark.
They are not finite values.
It happens that WM formulated :
On 01.11.2024 00:43, Richard Damon wrote:
On 10/31/24 1:35 PM, WM wrote:
On 31.10.2024 12:36, Richard Damon wrote:
On 10/30/24 11:38 AM, WM wrote:
NUF(x) MUST jump from 0 to Aleph_0 at all real values x, as below >>>>>>> ANY real number x, there are Aleph_0 unit fractions.
You cannot distinguish them by any real number? That proves that
they are dark.
They are not finite values.
All unit fractions are finite values.
Each unit fraction is finite, the set of all unit fractions is not
finite. Not finite is 'infinite' and there is no potential or actual
anymore -- just finite and not finite.
On 31.10.2024 19:34, Jim Burns wrote:
On 10/31/2024 1:41 PM, WM wrote:
On 31.10.2024 13:22, Jim Burns wrote:
On 10/31/2024 4:13 AM, WM wrote:
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?
Is
⎛ ∀ᴿx > 0:
⎜ ∀n ∈ ℕ:
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
wrong?
One of two contradicting formulas must be dropped.
Which (inside.quantifiers) formula is
the last which you accept with all prior formulas?
Which one requires that
NUF(x) can grow at an x ∈ ℝ by more than 1?
Regards, WM
On 01.11.2024 11:57, FromTheRafters wrote:
[...]
Actual means that all are there,
including the smallest.
Why?
It is a point on the real line,
well separated from its neighbour.
After serious thinking WM wrote :
On 01.11.2024 11:57, FromTheRafters wrote:
It happens that WM formulated :
On 01.11.2024 00:43, Richard Damon wrote:
On 10/31/24 1:35 PM, WM wrote:
On 31.10.2024 12:36, Richard Damon wrote:
On 10/30/24 11:38 AM, WM wrote:
NUF(x) MUST jump from 0 to Aleph_0 at all real values x, as
below ANY real number x, there are Aleph_0 unit fractions.
You cannot distinguish them by any real number? That proves that
they are dark.
They are not finite values.
All unit fractions are finite values.
Each unit fraction is finite, the set of all unit fractions is not
finite. Not finite is 'infinite' and there is no potential or actual
anymore -- just finite and not finite.
Actual means that all are there,
They are all there anyway, by definition.
including the smallest.
No smallest, since you simply inverted the set of naturals which has no largest.
Infinite subsets
don't do that for you, even if you wish really hard.
It is a point on the real line, well separated from its neighbour.
No point on the real line is separated, that is for discrete sets.
Will you (WM) also be ignoring
the work which shows
NUF(x) grows at 0 to ℵ₀ ?
Here,
you say: the smallest exists.
That does not pop the smallest into existence.
It contracts our discourse to
lines which have a smallest.
Why?
It is a point on the real line,
well separated from its neighbour.
You can contract our discourse to
real lines with smallest points,
but they will also be
real lines without smallest points,
WM formulated the question :
Infinite subsets don't do that for you, even if you wish really hard.
They cannot evade if they are invariable.
Sets don't change.
It is a point on the real line, well separated from its neighbour.
No point on the real line is separated, that is for discrete sets.
All unit fractions belong to points which are separated by non-unit
fractions.
Okay, but you previously said the reals are separated.
On 31.10.2024 21:46, joes wrote:
At single points a function has a single value, not a jump.
It jumps in case of NUF by 1 at a unit fraction with respect to the
foregoing unit fraction and the many points between both.
In the
case of NUF, that value is infinite everywhere except at 0.
That is impossible because there are not infinitely many unit fractions between 0 and everywhere.
Regards, WM
On 01.11.2024 00:44, Richard Damon wrote:
On 10/31/24 1:41 PM, WM wrote:
On 31.10.2024 13:22, Jim Burns wrote:
On 10/31/2024 4:13 AM, WM wrote:
Neither
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?
Is
⎛ ∀ᴿx > 0:
⎜ ∀n ∈ ℕ:
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
wrong?
One of two contradicting formulas must be dropped.
Or, just admit that your NUF(x) is where the contradiction is and drop
it.
Why? If alleged sets of real numbers really consist of real numbers,
then we can treat them as real points.
Regards, WM
WM explained on 11/1/2024 :
On 01.11.2024 19:39, FromTheRafters wrote:
WM formulated the question :
Infinite subsets don't do that for you, even if you wish really hard. >>>>They cannot evade if they are invariable.
Sets don't change.
Therefore the elements do not depend on us and our knowledge. "If I
find x, then I can find x + 1" is not relevant. "For every x (that I
find) there is x + 1" is no relevant. All elements are there,
independent of what we know or do. Therefore the first and the last
are also there independent of us. If they weren't, their existence
would depend on some circumstances and could change.
Circumstances like "there is no last element"?
the set
of denominators have no largest element to 'start' with.
On 11/1/24 6:37 AM, WM wrote:
If alleged sets of real numbers really consist of real numbers,
then we can treat them as real points.
But there is no "Real Point" where NUF(x) could be 1, since there will
be multiple unit fractions below x for any x.
Infinite sets do not need to have both ends in them.
On 11/1/24 6:07 AM, WM wrote:
On 31.10.2024 21:46, joes wrote:
At single points a function has a single value, not a jump.
It jumps in case of NUF by 1 at a unit fraction with respect to the
foregoing unit fraction and the many points between both.
If it jumps a *A* point, it has two values at that point,
Why not?In the
case of NUF, that value is infinite everywhere except at 0.
That is impossible because there are not infinitely many unit
fractions between 0 and everywhere.
On 11/1/24 6:45 AM, WM wrote:
On 01.11.2024 00:43, Richard Damon wrote:So?
On 10/31/24 1:35 PM, WM wrote:
On 31.10.2024 12:36, Richard Damon wrote:
On 10/30/24 11:38 AM, WM wrote:
NUF(x) MUST jump from 0 to Aleph_0 at all real values x, as below >>>>>>> ANY real number x, there are Aleph_0 unit fractions.
You cannot distinguish them by any real number? That proves that
they are dark.
They are not finite values.
All unit fractions are finite values.
There is no Unit Fraction x where NUF(x) can be 1,
as there will always
be other unit fractions at x/2 and x/3
On 01.11.2024 22:53, FromTheRafters wrote:Independently nonexistent.
WM explained on 11/1/2024 :
On 01.11.2024 19:39, FromTheRafters wrote:
WM formulated the question :Therefore the elements do not depend on us and our knowledge. "If I
Sets don't change.Infinite subsets don't do that for you, even if you wish reallyThey cannot evade if they are invariable.
hard.
find x, then I can find x + 1" is not relevant. "For every x (that I
find) there is x + 1" is no relevant. All elements are there,
independent of what we know or do. Therefore the first and the last
are also there independent of us.
No, it's just a property of the infinite, especially the actual.That means, there is always another element. Potential infinity.If they weren't, their existenceCircumstances like "there is no last element"?
would depend on some circumstances and could change.
Invalid deduction.the set of denominators have no largest element to 'start' with.If all unit fractions are existing, then a smallest unit fraction is existing.
If NUF(x) has grown to ℵ₀ at x₀, then ℵ₀ unit fractions mustWell AT zero the value is 0. No UF is zero or less...
be between 0 and x₀. Hence at least ℵ₀ points with ℵ₀ intervals of uncountably many points must be between 0 and x₀. That cannot happen at x₀ = 0.
On 01.11.2024 16:04, Jim Burns wrote:If one exists. That does not mean that anything is "missing",
Here, you say: the smallest exists.I say: All exist. That implies the smallest.
That does not pop the smallest into existence.
Or infinite ones.It contracts our discourse to lines which have a smallest.Yes. The other alternative would consist of variable sets.
Am Sat, 02 Nov 2024 10:59:04 +0100 schrieb WM:
If all unit fractions are existing, then a smallest unit fraction is
existing.
Invalid deduction.
On 31.10.2024 21:46, joes wrote:Whether you define your pathological function with < or <=, at every
At single points a function has a single value, not a jump.It jumps in case of NUF by 1 at a unit fraction with respect to the
foregoing unit fraction and the many points between both.
--In the case of NUF, that value is infinite everywhere except at 0.
Am Fri, 01 Nov 2024 18:33:34 +0100 schrieb WM:
I say: All exist. That implies [a] smallest.
Am Fri, 01 Nov 2024 18:33:34 +0100 schrieb WM:
I say: All exist. That implies the smallest.
In the case of NUF, that value is infinite everywhere except at 0.
Am Fri, 01 Nov 2024 11:07:27 +0100 schrieb WM:
On 31.10.2024 21:46, joes wrote:
At single points a function has a single value, not a jump.
It jumps in case of NUF by 1 at a unit fraction with respect to the
foregoing unit fraction and [bla]
Am Fri, 01 Nov 2024 11:07:27 +0100 schrieb WM:
On 31.10.2024 21:46, joes wrote:Whether you define your pathological function with < or <=, at every
At single points a function has a single value, not a jump.It jumps in case of NUF by 1 at a unit fraction with respect to the
foregoing unit fraction and the many points between both.
point it has one value (note it doesn't have the SAME value at every
point [...]).
In the case of NUF, [we have:]
Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:
If an invariable set of numbers is there, then there is a smallest and a
largest number of those which are existing.
That's just wrong.
On 01.11.2024 13:33, FromTheRafters wrote:That's just wrong.
After serious thinking WM wrote :If an invariable set of numbers is there, then there is a smallest and a largest number of those which are existing.
On 01.11.2024 11:57, FromTheRafters wrote:They are all there anyway, by definition.
It happens that WM formulated :Actual means that all are there,
On 01.11.2024 00:43, Richard Damon wrote:Each unit fraction is finite, the set of all unit fractions is not
On 10/31/24 1:35 PM, WM wrote:All unit fractions are finite values.
On 31.10.2024 12:36, Richard Damon wrote:They are not finite values.
On 10/30/24 11:38 AM, WM wrote:You cannot distinguish them by any real number? That proves that >>>>>>> they are dark.
NUF(x) MUST jump from 0 to Aleph_0 at all real values x, as >>>>>>>>>> below ANY real number x, there are Aleph_0 unit fractions.
finite. Not finite is 'infinite' and there is no potential or actual
anymore -- just finite and not finite.
including the smallest.No smallest, since you simply inverted the set of naturals which has no
largest.
On 01.11.2024 14:21, Jim Burns wrote:
⎛ ∀ᴿx > 0:
⎜ ∀n ∈ ℕ:
⎜ x > 0
⎜ ⅟x > 0
⎜ n+⅟x ≥ ⅟x > 0
⎜ ⌈n+⅟x⌉ ≥ ⅟x > 0
⎜ ⌈n+⅟x⌉ ∈ ℕ
⎜ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ
⎜ x⋅⌈n+⅟x⌉ ≥ x⋅⅟x > 0
⎜ x⋅⌈n+⅟x⌉⋅⅟⌈n+⅟x⌉ ≥ 1⋅⅟⌈n+⅟x⌉ > 0
⎜ x ≥ ⅟⌈n+⅟x⌉ > 0
⎜ ⅟⌈n+⅟x⌉ ∈ (0,x] ∧ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
⎛ ∀n ∈ ℕ⁺:
⎜ n > 0
⎜ n+1 > n > 0
⎜ ⅟n⋅(n+1)⋅⅟(n+1) > ⅟n⋅n⋅⅟(n+1)
⎜ ⅟n > ⅟(n+1)
⎜ ⅟n - ⅟(n+1) > ⅟(n+1) - ⅟(n+1)
⎝ ⅟n - ⅟(n+1) > 0
Will you (WM) also be ignoring
the work which shows
NUF(x) grows at 0 to ℵ₀ ?
Of course.
If NUF(x) has grown to ℵ₀ at x₀,
then ℵ₀ unit fractions must be between 0 and x₀.
Hence
at least ℵ₀ points with
ℵ₀ intervals of uncountably many points
must be between 0 and x₀.
That cannot happen at x₀ = 0.
Therefore
not even a fool could support your claim.
On 02.11.2024 02:04, Richard Damon wrote:
On 11/1/24 6:37 AM, WM wrote:
If alleged sets of real numbers really consist of real numbers, then
we can treat them as real points.
But there is no "Real Point" where NUF(x) could be 1, since there will
be multiple unit fractions below x for any x.
If actual infinity is assumed, then all points are there including a
first one, but its position cannot be determined. It is dark.
Regards, WM
Am Fri, 01 Nov 2024 11:07:27 +0100 schrieb WM:
On 31.10.2024 21:46, joes wrote:
In the case of NUF, that value is infinite everywhere except at 0.
Am 02.11.2024 um 13:17 schrieb joes:
Am Fri, 01 Nov 2024 18:33:34 +0100 schrieb WM:
I say: All exist. That implies [a] smallest.
No, it doesn't.
NUF(x) = 0 for x = 0 and
NUF(x) = aleph_0 for all x e IR, x > 0.
WM was thinking very hard :
That means, there is always another element. Potential infinity.
Sets don't change. Forget about amplifying 'not finite' with such as
'actual' and potential' -- infinite simply means not finite and 'actual/potential' is a distinction without a difference. A useless
concept outside of math philosophy.
the set of denominators have no largest element to 'start' with.
If all unit fractions are existing, then a smallest unit fraction is
existing. If NUF(x) has grown to ℵ₀ at x₀, then ℵ₀ unit fractions must
be between 0 and x₀. Hence at least ℵ₀ points with ℵ₀ intervals of >> uncountably many points must be between 0 and x₀. That cannot happen
at x₀ = 0.
Is that too hard to understand?
Apparently, for you.
Am 02.11.2024 um 14:21 schrieb joes:
Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:
If an invariable set of numbers is there, then there is a smallest and a >>> largest number of those which are existing.
or each and every n e IN there is an n' e IN (say n' = n+1)
On 11/1/2024 1:24 PM, WM wrote:
Hence
at least ℵ₀ points with
ℵ₀ intervals of uncountably many points
must be between 0 and x₀.
That cannot happen at x₀ = 0.
No.
There is no x₀ > 0 which,
between it and 0,
that does not happen.
On 11/2/24 6:23 AM, WM wrote:
If NUF(x) has grown to ℵ₀ at x₀, then ℵ₀ unit fractions must be
between 0 and x₀. Hence at least ℵ₀ points with ℵ₀ intervals of
uncountably many points must be between 0 and x₀. That cannot happen
at x₀ = 0.
Right, NUF(x)is 0 at x - 0, and Aleph_0 at any x > 0, since as you said,
for any positive finite x, there are Aleph_0 unit fractions below it.
On 11/2/24 6:18 AM, WM wrote:
There is no Unit Fraction x where NUF(x) can be 1,
No visible unit fraction.
But all unit fractions are visible.
On 11/2/24 6:15 AM, WM wrote:
On 02.11.2024 01:55, Richard Damon wrote:
Infinite sets do not need to have both ends in them.
Between 0 and 1 there are unit fractions. Hence there must be a first
one. The interval without unit fractions and the interval containing
unit fractions are separated by the smallest unit fraction.
Why?
That is just finite logic.
On 02.11.2024 01:55, Richard Damon wrote:
Infinite sets do not need to have both ends in them.
Between 0 and 1 there are unit fractions.
Hence there must be a first one.
The interval without unit fractions and
the interval containing unit fractions
are separated by the smallest unit fraction.
On 02.11.2024 15:10, Jim Burns wrote:
On 11/1/2024 1:24 PM, WM wrote:
Hence
at least ℵ₀ points with
ℵ₀ intervals of uncountably many points
must be between 0 and x₀.
That cannot happen at x₀ = 0.
No.
There is no x₀ > 0 which,
between it and 0,
that does not happen.
Then there is an x₀ where it happens.
x₀ =/= 0
On 11/02/2024 07:10 AM, Jim Burns wrote:
[...]
The delta-epsilonics of course,
or some put it "delta-epsilontics",
with little d and smaller e,
of often for induction arbitrary m and larger n,
is well-known to all students of calculus.
"The infinitesimal analysis", ....
I.e., in a manner of speaking,
the infinite transfinite cardinals
don't exist in delta-epsilonics
any more than plain manner-of-speaking "infinity".
On 02.11.2024 17:02, Richard Damon wrote:
On 11/2/24 6:23 AM, WM wrote:
If NUF(x) has grown to ℵ₀ at x₀, then ℵ₀ unit fractions must be >>> between 0 and x₀. Hence at least ℵ₀ points with ℵ₀ intervals of >>> uncountably many points must be between 0 and x₀. That cannot happen
at x₀ = 0.
Right, NUF(x)is 0 at x - 0, and Aleph_0 at any x > 0, since as you
said, for any positive finite x, there are Aleph_0 unit fractions
below it.
No, neither did I say so nor is it correct. Only for x > x₀ are ℵ₀ smaller unit fractions existing.
Regards, WM
On 02.11.2024 17:02, Richard Damon wrote:
On 11/2/24 6:15 AM, WM wrote:
On 02.11.2024 01:55, Richard Damon wrote:
Infinite sets do not need to have both ends in them.
Between 0 and 1 there are unit fractions. Hence there must be a first
one. The interval without unit fractions and the interval containing
unit fractions are separated by the smallest unit fraction.
Why?
That is just finite logic.
There is no other logic.
Regards, WM
On 02.11.2024 14:50, Moebius wrote:
Am 02.11.2024 um 14:21 schrieb joes:
Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:
If an invariable set of numbers is there, then there is a smallest
and a
largest number of those which are existing.
or each and every n e IN there is an n' e IN (say n' = n+1)
Actual infinity is not based on claims for each and every, but concerns
all.
Regards, WM
On 02.11.2024 17:02, Richard Damon wrote:
On 11/2/24 6:18 AM, WM wrote:
There is no Unit Fraction x where NUF(x) can be 1,
No visible unit fraction.
But all unit fractions are visible.
No.
Regards, WM
On 11/2/2024 6:15 AM, WM wrote:
On 02.11.2024 01:55, Richard Damon wrote:
Infinite sets do not need to have both ends in them.
Between 0 and 1 there are unit fractions.
Hence there must be a first one.
A.
⎛ S can be given a total ordering which is
⎜ well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S has
⎝ both a least and a greatest element in the subset.
[1]
A. isn't true of all sets.
If you augment ⅟ℕ with more elements
that is a different set.
A. is still not true of ⅟ℕ
If you, nonetheless, assert A.
A. is true of _all the sets in the discourse_
Because A. is not true of ⅟ℕ,
⅟ℕ is not in the discourse.
After A., claims about "sets" are not
claims about ⅟ℕ
On 11/2/2024 1:50 PM, WM wrote:
On 02.11.2024 15:10, Jim Burns wrote:
On 11/1/2024 1:24 PM, WM wrote:
Hence
at least ℵ₀ points with
ℵ₀ intervals of uncountably many points
must be between 0 and x₀.
That cannot happen at x₀ = 0.
Each x₀ > 0 is an x₀ where it happens.
On 11/2/24 1:56 PM, WM wrote:
On 02.11.2024 17:02, Richard Damon wrote:
On 11/2/24 6:23 AM, WM wrote:
If NUF(x) has grown to ℵ₀ at x₀, then ℵ₀ unit fractions must be >>>> between 0 and x₀. Hence at least ℵ₀ points with ℵ₀ intervals of >>>> uncountably many points must be between 0 and x₀. That cannot happen >>>> at x₀ = 0.
Right, NUF(x)is 0 at x - 0, and Aleph_0 at any x > 0, since as you
said, for any positive finite x, there are Aleph_0 unit fractions
below it.
No, neither did I say so nor is it correct. Only for x > x₀ are ℵ₀
smaller unit fractions existing.
So, for what finite number x > 0 is there less than that?
There isn't one, as we CAN PROVE that for all x > 0, there exist and
infinite number of unit fractions less that x
On 11/2/24 1:42 PM, WM wrote:
On 02.11.2024 14:50, Moebius wrote:
Am 02.11.2024 um 14:21 schrieb joes:
Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:
If an invariable set of numbers is there, then there is a smallest
and a
largest number of those which are existing.
or each and every n e IN there is an n' e IN (say n' = n+1)
Actual infinity is not based on claims for each and every, but
concerns all.
But if it applies to ALL, it must apply to ANY, so a property of ANY
must apply to each on of the ALL.
So, for ALL the Natural Numbers, there can't be a highest, because for
ANY Natural Number there is a following one
On 02.11.2024 21:34, Richard Damon wrote:
On 11/2/24 1:42 PM, WM wrote:
On 02.11.2024 14:50, Moebius wrote:
Am 02.11.2024 um 14:21 schrieb joes:
Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:
If an invariable set of numbers is there, then there is a smallest >>>>>> and a
largest number of those which are existing.
or each and every n e IN there is an n' e IN (say n' = n+1)
Actual infinity is not based on claims for each and every, but
concerns all.
But if it applies to ALL, it must apply to ANY, so a property of ANY
must apply to each on of the ALL.
So, for ALL the Natural Numbers, there can't be a highest, because for
ANY Natural Number there is a following one
That cannot be true for all dark numbers.
Regards, WM
On 02.11.2024 14:50, Moebius wrote:Lol. That actually sheds some light on your thought process:
Am 02.11.2024 um 14:21 schrieb joes:Actual infinity is not based on claims for each and every, but concerns
Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:or each and every n e IN there is an n' e IN (say n' = n+1)
If an invariable set of numbers is there, then there is a smallest
and a largest number of those which are existing.
all.
On 02.11.2024 17:02, Richard Damon wrote:There is the logic of the infinite, unless you claim that finite sets
On 11/2/24 6:15 AM, WM wrote:
On 02.11.2024 01:55, Richard Damon wrote:
Infinite sets do not need to have both ends in them.
Between 0 and 1 there are unit fractions. Hence there must be a first
one. The interval without unit fractions and the interval containing
unit fractions are separated by the smallest unit fraction.
Why? That is just finite logic.There is no other logic.
On 02.11.2024 19:45, Jim Burns wrote:Have you heard of the terms supremum and infimum?
On 11/2/2024 6:15 AM, WM wrote:It is true for all sets of separated points on the real line.
On 02.11.2024 01:55, Richard Damon wrote:
Infinite sets do not need to have both ends in them.
Between 0 and 1 there are unit fractions.
Hence there must be a first one.
A.
⎛ S can be given a total ordering which is ⎜ well-ordered both forwards >> and backwards.
⎜ That is, every non-empty subset of S has ⎝ both a least and a
greatest element in the subset.
[1]
A. isn't true of all sets.
No such points.If you augment ⅟ℕ with more elements that is a different set.
A. is still not true of ⅟ℕ
If you, nonetheless, assert A.
A. is true of _all the sets in the discourse_
Because A. is not true of ⅟ℕ,
⅟ℕ is not in the discourse.
After A., claims about "sets" are not claims about ⅟ℕ
cannot happen at points in smaller distance than uncountable many points
from zero.
On 11/2/24 5:37 PM, WM wrote:
There isn't one, as we CAN PROVE that for all x > 0, there exist and
infinite number of unit fractions less that x
For every visible number x larger, much larger than x₀ you can prove
that.
Nope, I can prove it for EVERY x.
On 03.11.2024 01:17, Richard Damon wrote:
On 11/2/24 5:37 PM, WM wrote:
There isn't one, as we CAN PROVE that for all x > 0, there exist and
infinite number of unit fractions less that x
For every visible number x larger, much larger than x₀ you can prove
that.
Nope, I can prove it for EVERY x.
You can prove that ℵo unit fractions fit between 0 and (0, 1]?
That is based on a inconsistent theory.
Regards, WM
Am Sat, 02 Nov 2024 18:42:15 +0100 schrieb WM:
On 02.11.2024 14:50, Moebius wrote:Lol. That actually sheds some light on your thought process:
Am 02.11.2024 um 14:21 schrieb joes:Actual infinity is not based on claims for each and every, but concerns
Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:or each and every n e IN there is an n' e IN (say n' = n+1)
If an invariable set of numbers is there, then there is a smallest
and a largest number of those which are existing.
all.
how do you suppose some property holds for all x, but not
for every?
Am Sat, 02 Nov 2024 22:23:08 +0100 schrieb WM:
It is true for all sets of separated points on the real line.Have you heard of the terms supremum and infimum?
cannot happen at points in smaller distance than uncountable many pointsNo such points.
from zero.
I *CAN* prove for every x in (0, 1], which is what I claimed.
On 02.11.2024 21:34, Richard Damon wrote:And that is why "dark" numbers are not natural (or the naturals are all
On 11/2/24 1:42 PM, WM wrote:That cannot be true for all dark numbers.
On 02.11.2024 14:50, Moebius wrote:But if it applies to ALL, it must apply to ANY, so a property of ANY
Am 02.11.2024 um 14:21 schrieb joes:Actual infinity is not based on claims for each and every, but
Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:or each and every n e IN there is an n' e IN (say n' = n+1)
If an invariable set of numbers is there, then there is a smallest >>>>>> and a largest number of those which are existing.
concerns all.
must apply to each on of the ALL.
So, for ALL the Natural Numbers, there can't be a highest, because for
ANY Natural Number there is a following one
On 03.11.2024 09:54, joes wrote:Dude. The logic itself is finite, but not what it is talking about.
Am Sat, 02 Nov 2024 18:57:51 +0100 schrieb WM:
On 02.11.2024 17:02, Richard Damon wrote:
Logic is always finite. Except that of some modern fools like Henkin.There is the logic of the infinite, unless you claim that finite setsWhy? That is just finite logic.There is no other logic.
are in fact infinite. Now that I say it, it seems remarkably close to
your beliefs.
On 03.11.2024 01:17, Richard Damon wrote:Quit twisting the quantifiers around you motherfucker.
On 11/2/24 5:37 PM, WM wrote:
You can prove that ℵo unit fractions fit between 0 and (0, 1]?Nope, I can prove it for EVERY x.There isn't one, as we CAN PROVE that for all x > 0, there exist andFor every visible number x larger, much larger than x₀ you can prove
infinite number of unit fractions less that x
that.
That is based on a inconsistent theory.
Am Sat, 02 Nov 2024 22:39:36 +0100 schrieb WM:
On 02.11.2024 21:34, Richard Damon wrote:
And that is why "dark" numbers are not natural (or the naturals are allSo, for ALL the Natural Numbers, there can't be a highest, because forThat cannot be true for all dark numbers.
ANY Natural Number there is a following one
not dark).
Quit twisting the quantifiers around
On 03.11.2024 13:13, Richard Damon wrote:
I *CAN* prove for every x in (0, 1], which is what I claimed.You cannot prove that ℵo unit fractions occupy at least ℵo points?
Regards, WM
On 11/3/24 9:09 AM, WM wrote:
On 03.11.2024 13:13, Richard Damon wrote:
I *CAN* prove for every x in (0, 1], which is what I claimed.You cannot prove that ℵo unit fractions occupy at least ℵo points?
Since there are Aleph_0 rational points in ANY finite length of the
number line
But Induction *IS* valid for all Natural Numbers.
On 03.11.2024 22:21, Richard Damon wrote:
On 11/3/24 9:09 AM, WM wrote:
On 03.11.2024 13:13, Richard Damon wrote:
I *CAN* prove for every x in (0, 1], which is what I claimed.You cannot prove that ℵo unit fractions occupy at least ℵo points?
Since there are Aleph_0 rational points in ANY finite length of the
number line
But not for every point x > 0. ℵo unit fractions occupy at least ℵo*2^ℵo
points, hence more than ℵo points. Your assertion is true only for every point that you can define. But you cannot define points between the
first ℵo unit fractions.
Regards, WM
On 11/02/2024 12:37 PM, Jim Burns wrote:
On 11/2/2024 2:02 PM, Ross Finlayson wrote:
The delta-epsilonics of course,
or some put it "delta-epsilontics",
with little d and smaller e,
of often for induction arbitrary m and larger n,
is well-known to all students of calculus.
"The infinitesimal analysis", ....
The delta.epsilonics well.known to students of calculus
is not infinitesimal analysis.
For δ > 0 and ε > 0
there are _finite_ j and k such that
δ > ⅟j > 0 and ε > ⅟k > 0
The delta-epsilonics is
a perfectly suitable approach to defining
infinite limit in theories of infinitesimal analysis.
How about atomism, is there a theory there with
truly abstractly uncuttable objects like the
theoretical atom?
On 03.11.2024 22:21, Richard Damon wrote:
But Induction *IS* valid for all Natural Numbers.
Only for all natnumbers which can be defined and which belong to a
finite initial segment which is followed by ℵo natnumbers most of which
you cannot define.
By induction you can prove the sum n(n+1)/2 for every initial segment 1+2+3+...+n. But not for all natural numbers.
Regards, WM
Yes, for every point x > 0, there are Aleph_0 unit fractions below it.
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial segment
1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
On 03.11.2024 22:21, Richard Damon wrote:
But Induction *IS* valid for all Natural Numbercs.
Only for all natnumbers which
can be defined and which
belong to a finite initial segment which
is followed by ℵo natnumbers most of which
you cannot define.
By induction you can prove the sum n(n+1)/2
for every initial segment 1+2+3+...+n.
But not for all natural numbers.
On 02.11.2024 20:52, Jim Burns wrote:
On 11/2/2024 1:50 PM, WM wrote:
On 02.11.2024 15:10, Jim Burns wrote:
On 11/1/2024 1:24 PM, WM wrote:
Hence
at least ℵ₀ points with
ℵ₀ intervals of uncountably many points
must be between 0 and x₀.
That cannot happen at x₀ = 0.
Each x₀ > 0 is an x₀ where it happens.
No.
At least ℵ₀ points
with ℵ₀ intervals of uncountably many points
must be between 0 and x₀.
This criterion is not satisfied by every point x > 0,
but it is satisfied by
every definable or visible point x > 0.
Why do you not see this requirement?
Ross, that post, like so many of yours, displays only local coherence.
Each sentence looks like it would make sense if only one were to read the surrounding context. On doing so, no overall sense is to be found, just
a vague drift from allusions to topics, never quite settling on anything definite.
On 11/04/2024 03:52 AM, Jim Burns wrote:
The delta-epsilonics of course,
or some put it "delta-epsilontics",
with little d and smaller e,
That one theory is
the theory well.known to students of calculus.
I mention this
(and mention it and mention it)
because
I am declining your invitation to
be the mark in a bait.and.switch.
Sure it is, the delta-epsilonics is well known,
and where often proofs in geometry will have been
introduced vis-a-vis usual sorts of plain "checks"
as "proofs", then delta-epsilonics is often the
first sort of deductive account, introduced to
what's usually upper-class secondary students,
then that the concept of the infinite limit
is made and then it's pointed out how that fits
with regards to classical expositions like Zeno's,
of what yet doesn't.
What I'm saying is that since antiquity,
it is known,
that there are at least two models of continuity,
and you may call it Archimedean and Democritan,
about the field of rationals versus atomism,
and that infinitesimal analysis includes both.
So, no, I'm not interested nor was it proffered
"bait-and-switch", though here there's still
the "pick one, get both".
The "clock arithmetic" is usually fmailiar to
students as with regards to the sweep of
the secondae hand around the clock, or as
with regards to the rollover of the odometer,
that a course-of-passage in _time_ is most
simply as in accords with the classical expositions
of constant and uniform motion.
Indeed the dialectic I should hope you know as
something akin to "thesis, anti-thesis, synthesis",
as with regards to that being "pick one, get both".
I.e., the complementary duals, these competing concerns,
meet in the middle, even if: "the middle of nowhere".
So, infinitesimal analysis includes delta-epsilonics,
if not the other way around.
Then, some "continuum infinitesimal analysis",
makes for "Standard Infinitesimals" like
these "iota-values" of "line-reals".
constant monotone strictly increasing
extent density completeness measure
On 11/04/2024 03:52 AM, Jim Burns wrote:
On 11/2/2024 6:01 PM, Ross Finlayson wrote:
On 11/02/2024 12:37 PM, Jim Burns wrote:
On 11/2/2024 2:02 PM, Ross Finlayson wrote:
The delta-epsilonics of course,
or some put it "delta-epsilontics",
with little d and smaller e,
of often for induction arbitrary m and larger n,
is well-known to all students of calculus.
"The infinitesimal analysis", ....
The delta.epsilonics well.known to students of calculus
is not infinitesimal analysis.
For δ > 0 and ε > 0
there are _finite_ j and k such that
δ > ⅟j > 0 and ε > ⅟k > 0
Sure it is,
the delta-epsilonics is well known,
What I'm saying is that since antiquity,
it is known,
that there are at least two models of continuity,
and you may call it Archimedean and Democritan,
about the field of rationals versus atomism,
and that infinitesimal analysis includes both.
So, infinitesimal analysis includes delta-epsilonics,
if not the other way around.
Am 04.11.2024 um 23:36 schrieb Alan Mackenzie:
Ross, that post, like so many of yours, displays only local coherence.
Each sentence looks like it would make sense if only one were to read the
surrounding context. On doing so, no overall sense is to be found, just
a vague drift from allusions to topics, never quite settling on anything
definite.
A floating stream of thoughts (sort of).
On 11/4/2024 3:18 PM, Jim Burns wrote:
On 11/4/2024 5:41 PM, Moebius wrote:
Am 04.11.2024 um 23:36 schrieb Alan Mackenzie:
Ross, that post, like so many of yours, displays only local coherence. >>>> Each sentence looks like it would make sense if only one were to
read the
surrounding context. On doing so, no overall sense is to be found,
just
a vague drift from allusions to topics, never quite settling on
anything
definite.
A floating stream of thoughts (sort of).
I have a theory, not confirmed by RF,
that Ross is conducting a brainstorming session,
in which credit is given for _creativity_ or
of the sheer number of ideas thrown out,
and not for any sort of thread tying them together.
As I say, not confirmed.
But my theory comforts me.
I don't try to connect Ross's posts to my posts.
He presents a buffet.
Take what you like, leave the rest.
Agreed! :^)
On 04.11.2024 13:14, Richard Damon wrote:
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial segment
1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
All defined numbers can be summed. Not all natural numbers can be summed.
Regards, WM
On 11/4/24 12:11 PM, WM wrote:
On 04.11.2024 13:14, Richard Damon wrote:
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial
segment 1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
All defined numbers can be summed. Not all natural numbers can be summed.
Why not?
On 11/2/2024 5:28 PM, WM wrote:
At least ℵ₀ points
with ℵ₀ intervals of uncountably many points
must be between 0 and x₀.
This criterion is not satisfied by every point x > 0,
No.
x₀ ≥ ⅟⌈n+⅟x₀⌉ > ⅟⌈n+1+⅟x₀⌉ > 0
but it is satisfied by
every definable or visible point x > 0.
every point.between.splits > 0 of
differences of ratios of countable.to numbers.
On 04.11.2024 22:46, Jim Burns wrote:
On 11/2/2024 5:28 PM, WM wrote:
At least ℵ₀ points
with ℵ₀ intervals of uncountably many points
must be between 0 and x₀.
This criterion is not satisfied by every point x > 0,
No.
x₀ ≥ ⅟⌈n+⅟x₀⌉ > ⅟⌈n+1+⅟x₀⌉ > 0
That requires that x₀ is definable.
But it is dark.
but it is satisfied by
every definable or visible point x > 0.
every point.between.splits > 0 of
differences of ratios of countable.to numbers.
Of definable splits.
On 02.11.2024 19:45, Jim Burns wrote:
On 11/2/2024 6:15 AM, WM wrote:
On 02.11.2024 01:55, Richard Damon wrote:
Infinite sets do not need to have both ends in them.
Between 0 and 1 there are unit fractions.
Hence there must be a first one.
A.
⎛ S can be given a total ordering which is
⎜ well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S has
⎝ both a least and a greatest element in the subset.
[1]
A. isn't true of all sets.
It is true for all sets of
separated points on the real line.
On 05.11.2024 04:08, Richard Damon wrote:
On 11/4/24 12:11 PM, WM wrote:
On 04.11.2024 13:14, Richard Damon wrote:
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial
segment 1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
All defined numbers can be summed. Not all natural numbers can be
summed.
Why not?
because most cannot be defined.
Regards, WM
On 11/5/24 11:45 AM, WM wrote:
On 05.11.2024 04:08, Richard Damon wrote:
On 11/4/24 12:11 PM, WM wrote:
On 04.11.2024 13:14, Richard Damon wrote:
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial
segment 1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
All defined numbers can be summed. Not all natural numbers can be
summed.
Why not?
because most cannot be defined.
But they ARE defined.
WM presented the following explanation :
On 06.11.2024 12:46, Richard Damon wrote:
On 11/5/24 11:45 AM, WM wrote:
On 05.11.2024 04:08, Richard Damon wrote:
On 11/4/24 12:11 PM, WM wrote:
On 04.11.2024 13:14, Richard Damon wrote:
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial
segment 1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
All defined numbers can be summed. Not all natural numbers can be
summed.
Why not?
because most cannot be defined.
But they ARE defined.
Then sum all of them.
-1/12,
really a sum as we have come to think of sums.
On 06.11.2024 12:46, Richard Damon wrote:
On 11/5/24 11:45 AM, WM wrote:
On 05.11.2024 04:08, Richard Damon wrote:
On 11/4/24 12:11 PM, WM wrote:
On 04.11.2024 13:14, Richard Damon wrote:
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial
segment 1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
All defined numbers can be summed. Not all natural numbers can be
summed.
Why not?
because most cannot be defined.
But they ARE defined.
Then sum all of them.
Regards, WM
On 11/6/24 10:15 AM, WM wrote:
On 06.11.2024 12:46, Richard Damon wrote:
On 11/5/24 11:45 AM, WM wrote:
On 05.11.2024 04:08, Richard Damon wrote:
On 11/4/24 12:11 PM, WM wrote:
On 04.11.2024 13:14, Richard Damon wrote:
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial
segment 1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
All defined numbers can be summed. Not all natural numbers can be
summed.
Why not?
because most cannot be defined.
But they ARE defined.
Then sum all of them.
They will sum to Aleph_0
Note, Addition on Natural Numbers is closed for FINITE sums (the sum of
a finite number of numbers), not necessarily for infinite series of them.
On 07.11.2024 03:27, Richard Damon wrote:
On 11/6/24 10:15 AM, WM wrote:
On 06.11.2024 12:46, Richard Damon wrote:
On 11/5/24 11:45 AM, WM wrote:
On 05.11.2024 04:08, Richard Damon wrote:
On 11/4/24 12:11 PM, WM wrote:
On 04.11.2024 13:14, Richard Damon wrote:
On 11/4/24 6:07 AM, WM wrote:
By induction you can prove the sum n(n+1)/2 for every initial >>>>>>>>> segment 1+2+3+...+n. But not for all natural numbers.
But all Natural Numbers can be defined.
All defined numbers can be summed. Not all natural numbers can be >>>>>>> summed.
Why not?
because most cannot be defined.
But they ARE defined.
Then sum all of them.
They will sum to Aleph_0
That is not a sum. It is just another name for infinity.
Note, Addition on Natural Numbers is closed for FINITE sums (the sum
of a finite number of numbers), not necessarily for infinite series of
them.
That is because infinitely many always contain dark numbers. All not
dark numbers can be summed.
Regards, WM
On 07.11.2024 03:27, Richard Damon wrote:Lol. You can also say the series of partial sums diverges. What did you
On 11/6/24 10:15 AM, WM wrote:That is not a sum. It is just another name for infinity.
On 06.11.2024 12:46, Richard Damon wrote:They will sum to Aleph_0
On 11/5/24 11:45 AM, WM wrote:Then sum all of them.
On 05.11.2024 04:08, Richard Damon wrote:But they ARE defined.
On 11/4/24 12:11 PM, WM wrote:because most cannot be defined.
On 04.11.2024 13:14, Richard Damon wrote:Why not?
On 11/4/24 6:07 AM, WM wrote:All defined numbers can be summed. Not all natural numbers can be >>>>>>> summed.
By induction you can prove the sum n(n+1)/2 for every initial >>>>>>>>> segment 1+2+3+...+n. But not for all natural numbers.But all Natural Numbers can be defined.
Am Thu, 07 Nov 2024 10:13:14 +0100 schrieb WM:
On 07.11.2024 03:27, Richard Damon wrote:
On 11/6/24 10:15 AM, WM wrote:
On 06.11.2024 12:46, Richard Damon wrote:
They will sum to Aleph_0
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