There aren't any neighboring intervals.
Any two intervals have intervals between them.
On 03.11.2024 23:18, Jim Burns wrote:
On 11/3/2024 3:38 AM, WM wrote:
Further there are never
two irrational numbers
without a rational number between them.
(Even the existence of neighbouring intervals
is problematic.)
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong in geometry.
The measure outside of the intervals is infinite.
Hence there exists at least one point outside.
This point has two nearest intervals
On 11/4/2024 6:26 AM, WM wrote:
On 03.11.2024 23:18, Jim Burns wrote:
On 11/3/2024 3:38 AM, WM wrote:
Further there are never
two irrational numbers
without a rational number between them.
(Even the existence of neighbouring intervals
is problematic.)
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong in geometry.
The measure outside of the intervals is infinite.
Hence there exists at least one point outside.
This point has two nearest intervals
This point,
which is on the boundary of two intervals,
is not two irrational points.
Further there are never
two irrational numbers
without an interval between them.
On 11/4/2024 6:37 AM, Jim Burns wrote:
Further there are never
two irrational numbers
without an interval between them.
Unless they equal each other? ;^)
On 11/4/2024 2:02 PM, Jim Burns wrote:
On 11/4/2024 3:54 PM, Chris M. Thomasson wrote:Well, yeah in a sense.
On 11/4/2024 6:37 AM, Jim Burns wrote:
Further there are never
two irrational numbers
without an interval between them.
Unless they equal each other? ;^)
And thus are one, not two, points?
A = (-1, .5, 3, 7)
B = (-1, .5, 3, 7)
A = B = true
For example,
for any two real numbers x, y ... bla bla
USUALLY does not exclude x = y. [Might look strange, but is just the
usual math lingo.]
.
.
.
For example,
for any two real numbers x, y ... bla bla
USUALLY does not exclude x = y. [Might look strange, but is just the
usual math lingo.]
.
.
.
For example,
for any two real numbers x, y ... bla bla
USUALLY does not exclude x = y. [Might look strange, but is just the
usual math lingo.]
.
.
.
Hint: Archimedes Plutonium had some problems with that convention. Seems
that Jim wants to follow his lead.
Am 05.11.2024 um 01:20 schrieb Moebius:
Hint: Archimedes Plutonium had some problems with that convention.
Seems that Jim wants to follow his lead.
Actually, I'd recommend not to follow JB, FtA and/or RD blindly.
Sometimes they just talk nonsense.
On 11/8/24 11:43 AM, WM wrote:
The infinite of the real axis is a big supply but an as big drain.
What "drain", the numbers exist.
I take it as evident that intervals of the measure 1/5 of the positive
real axis will not, by any shuffling, cover the real axis completely,
let alone infinitely often. I think who believes this is a deplorable
fanatic if not a fool.
Since 1/5 of infinity isn't a finite measure, you can't use finite logic
to handle them.
You are just proving your use of broken logic.
On 11/8/2024 5:18 AM, WM wrote:
My understanding of mathematics and geometry
is that
reordering cannot increase the measure
(only reduce it by overlapping).
This is a basic axiom which
will certainly be agreed to by
everybody not conditioned by matheology.
By
"everybody not conditioned by matheology"
you mean
"everybody who hasn't thought much about infinity"
Am Fri, 08 Nov 2024 17:43:15 +0100 schrieb WM:
I take it as evident that intervals of the measure 1/5 of the positiveWhat is the measure you are using and what does it give for the real
real axis will not, by any shuffling, cover the real axis completely,
let alone infinitely often. I think who believes this is a deplorable
fanatic if not a fool.
axis?
On 11/9/2024 1:42 PM, WM wrote:
But note that Cantor's bijection between naturals and rationals does
not insert any non-natural number into ℕ.
It confirms only that [both] sets [have the same size, actually aleph_0].
Am 08.11.2024 um 21:35 schrieb Moebius:
What is the measure you are using and what does it give for the real
axis?
Ob Du es nochmal schaffst, auf diesen saudummen Scheißdreck NICHT zu
antworten?
Ich kann es jedenfalls nicht mehr sehen/lesen.
On 08.11.2024 19:01, Jim Burns wrote:
On 11/8/2024 5:18 AM, WM wrote:
My understanding of mathematics and geometry
is that
reordering cannot increase the measure
(only reduce it by overlapping).
This is a basic axiom which
will certainly be agreed to by
everybody not conditioned by matheology.
By
"everybody not conditioned by matheology"
you mean
"everybody who hasn't thought much about infinity"
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.
Everybody who believes that
the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.
On 11/9/2024 6:45 AM, WM wrote:
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.
Our sets do not change.
The set
{[n-⅒,n+⅒]: n∈ℕ⁺}
with the midpoints at
⟨ 1, 2, 3, 4, 5, ... ⟩
does not _change_ to the set
{[iₙ/jₙ-⅒,iₙ/jₙ+⅒]: n∈ℕ⁺}
with the midpoints at
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩
----
Either
all instances of a 𝗰𝗹𝗮𝗶𝗺 about a set
are _only_ true or _only_ false
or
a set changes.
In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which
has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.
In the second case, with the changing sets,
who knows?
Perhaps something else could be done,
but not that.
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".
Consider again the two sets of midpoints
⟨ 1, 2, 3, 4, 5, ... ⟩ and
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩
They both _are_
And their points correspond
by i/j ↦ n = (i+j-1)(i+j-2)/2+i
It confirms only that both sets have the same size, actually aleph_0.
On 10.11.2024 00:27, Jim Burns wrote:
On 11/9/2024 6:45 AM, WM wrote:
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.
Our sets do not change.
The set
{[n-⅒,n+⅒]: n∈ℕ⁺}
with the midpoints at
⟨ 1, 2, 3, 4, 5, ... ⟩
does not _change_ to the set
{[iₙ/jₙ-⅒,iₙ/jₙ+⅒]: n∈ℕ⁺}
with the midpoints at
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩
It cannot do so because
the reality of the rationals is much larger than
the reality of the naturals.
Either
all instances of a 𝗰𝗹𝗮𝗶𝗺 about a set
are _only_ true or _only_ false
or
a set changes.
In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which >> has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.
But it
But it will never complete
an infinite set of claims.
It will forever remain in the status nascendi.
Therefore
irrelevant for actual or completed infinity.
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".
But they cannot become such sets.
< into each other.Consider again the two sets of midpoints
⟨ 1, 2, 3, 4, 5, ... ⟩ and
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩
They both _are_
And their points correspond
by i/j ↦ n = (i+j-1)(i+j-2)/2+i
But they cannot be completely transformed
On 11/10/2024 4:35 AM, WM wrote:
The set
{3,4,5}
does not _change_ to the set
{6,7,8}
because
our sets do not change.
Our sets do not change.
In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which >>> has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.
But it
"It" refers to who or what?
But it will never complete
an infinite set of claims.
We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
It will forever remain in the status nascendi.
Therefore
irrelevant for actual or completed infinity.
A finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false,
will forever remain
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false.
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".
But they cannot become such sets.
Our sets do not change.
But they cannot be completely transformed< into each other.
Our sets do not change.
Consider again the two sets of midpoints
⟨ 1, 2, 3, 4, 5, ... ⟩ and
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩
They both _are_
And their points correspond
by i/j ↦ n = (i+j-1)(i+j-2)/2+i
On 11/10/2024 1:35 AM, WM wrote:
On 10.11.2024 00:27, Jim Burns wrote:
On 11/9/2024 6:45 AM, WM wrote:
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.
Our sets do not change.
The set
{[n-⅒,n+⅒]: n∈ℕ⁺}
with the midpoints at
⟨ 1, 2, 3, 4, 5, ... ⟩
does not _change_ to the set
{[iₙ/jₙ-⅒,iₙ/jₙ+⅒]: n∈ℕ⁺}
with the midpoints at
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩
It cannot do so because the reality of the rationals is much larger
than the reality of the naturals.[...]
Cantor pairing can create a unique pair of natural numbers from a single natural number. Why do think of rationals at all!?
On 10.11.2024 18:49, Jim Burns wrote:
On 11/10/2024 4:35 AM, WM wrote:
On 10.11.2024 00:27, Jim Burns wrote:
In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which
has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.
But it
"It" refers to who or what?
To that finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲.
But it will never complete
an infinite set of claims.
We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
But you claim that
_all_ fractions are in bijection with
all natural numbers,
don't you?
It will forever remain in the status nascendi.
Therefore
irrelevant for actual or completed infinity.
A finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false,
will forever remain
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false.
Therefore
such a sequence does not entitle you
to claim infinite mappings.
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".
But they cannot become such sets.
Our sets do not change.
My intervals I(n) = [n - 1/10, n + 1/10]
must be translated to all the midpoints
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5,
2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
if you want to contradict my claim.
On 10.11.2024 21:36, Chris M. Thomasson wrote:
On 11/10/2024 1:35 AM, WM wrote:
On 10.11.2024 00:27, Jim Burns wrote:
On 11/9/2024 6:45 AM, WM wrote:
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.
Our sets do not change.
But intervals on the real axis can be translated.
On 11/11/2024 4:00 AM, WM wrote:
But intervals on the real axis can be translated.
The interval [4-⅒,4+⅒] can be translated to
the interval [1/3-⅒,1/3+⅒].
[4-⅒,4+⅒] does not change to [1/3-⅒,1/3+⅒]
[4-⅒,4+⅒] will continue being after translation.
[1/3-⅒,1/3+⅒] has never been [4-⅒,4+⅒].
On 11/11/2024 3:41 AM, WM wrote:
But it will never complete
an infinite set of claims.
We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
But you claim that
_all_ fractions are in bijection with
all natural numbers,
don't you?
Yes, I claim that.
This is one 𝗰𝗹𝗮𝗶𝗺:
⎛ All fractions are in bijection with
⎝ all natural numbers.
My intervals I(n) = [n - 1/10, n + 1/10]
must be translated to all the midpoints
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5,
2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
if you want to contradict my claim.
Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".
On 04.11.2024 15:37, Jim Burns wrote:
On 11/4/2024 6:26 AM, WM wrote:
On 03.11.2024 23:18, Jim Burns wrote:
On 11/3/2024 3:38 AM, WM wrote:
Further there are never
two irrational numbers
without a rational number between them.
(Even the existence of neighbouring intervals
is problematic.)
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong in geometry.
The measure outside of the intervals is infinite.
Hence there exists at least one point outside.
This point has two nearest intervals
This point,
which is on the boundary of two intervals,
is not two irrational points.
You are wrong.
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.
Further there are never
two irrational numbers
without an interval between them.
Not in reality. But in the used model.
The rationals are dense
but the intervals are not.
This proves that
the rationals are not countable.
On 11/4/2024 12:32 PM, WM wrote:
[...]
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝ jₖ = k-iₖ
proves that
the rationals are countable.
On 11/4/2024 12:32 PM, WM wrote:
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.
⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.
I mean 'exterior' in the topological sense.
For a point x in the boundary ∂A of set A
each open set Oₓ which holds x
holds points in A and points not.in A
Each of {...,-3,-2,-1,0,1,2,3,...} is
the midpoint of an interval.
There can't be any exterior point
a distance 1 from any interval.
There can't be any exterior point
a distance ⅟2 from any interval.
Nor ⅟3. Nor ⅟4. Nor any positive distance.
An exterior point which is not
a positive distance from any interval
is not an exterior point.
Therefore,
in what is _almost_ your conclusion,
there are no exterior points.
Instead, there are boundary points.
For each x not.in the intervals,
each open set Oₓ which holds x
holds points in the intervals and
points not.in the intervals.
x is a boundary point.
The rationals are dense
Yes.
Each multi.point interval [x,x′] holds
rationals.
but the intervals are not.
No.
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝ jₖ = k-iₖ
proves that
the rationals are countable.
On 05.11.2024 18:25, Jim Burns wrote:
On 11/4/2024 12:32 PM, WM wrote:
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.
⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.
This plentiness does not change
when the intervals are translated.
I mean 'exterior' in the topological sense.
For a point x in the boundary ∂A of set A
each open set Oₓ which holds x
holds points in A and points not.in A
The intervals are closed with irrational endpoints.
Each of {...,-3,-2,-1,0,1,2,3,...} is
the midpoint of an interval.
There can't be any exterior point
a distance 1 from any interval.
There can't be any exterior point
a distance ⅟2 from any interval.
Nor ⅟3. Nor ⅟4. Nor any positive distance.
Nice try.
But there are points outside of intervals,
and they are closer to interval ends
than to the interior, independent of
the configuration of the intervals.
Note that
only 3/oo of the points are inside.
An exterior point which is not
a positive distance from any interval
is not an exterior point.
Positive is what you can define,
but there is much more in smaller distance.
Remember the infinitely many unit fractions
within every eps > 0 that you can define.
Therefore,
in what is _almost_ your conclusion,
there are no exterior points.
There are 3/oo of all points exterior.
Instead, there are boundary points.
For each x not.in the intervals,
each open set Oₓ which holds x
holds points in the intervals and
points not.in the intervals.
x is a boundary point.
The intervals are closed
The rationals are dense
Yes.
Each multi.point interval [x,x′] holds
rationals.
but the intervals are not.
No.
Each multi.point interval [x,x′] holds
ε.cover intervals.
Therefore not all rationals are enumerated.
proves that
the rationals are countable.
Contradiction.
Something of your theory is inconsistent.
On 11/6/2024 5:35 AM, WM wrote:
On 05.11.2024 18:25, Jim Burns wrote:
On 11/4/2024 12:32 PM, WM wrote:
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.
⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.
This plentiness does not change
when the intervals are translated.
⎛ When the intervals are end.to.end.to.end,
⎜ there are exterior points
⎝ a distance 10¹⁰⁰⁰⁰⁰ from any interval.
Are there points 10¹⁰⁰⁰⁰⁰ from any interval
when midpoints of intervals include
each of {...,-3,-2,-1,0,1,2,3,...} ?
Isn't that a plentiness which changes?
The intervals are closed with irrational endpoints.
'Exterior' seems like a good way to say
'not in contact'.
It seems to me that you have a better argument
with open intervals instead of closed,
but let them be closed, if you like.
Either way,
there are no points 10¹⁰⁰⁰⁰⁰ from any interval.
Each of {...,-3,-2,-1,0,1,2,3,...} is
the midpoint of an interval.
There can't be any exterior point
a distance 1 from any interval.
There can't be any exterior point
a distance ⅟2 from any interval.
Nor ⅟3. Nor ⅟4. Nor any positive distance.
Nice try.
But there are points outside of intervals,
Are any of these points.outside
⅟2 from any interval? ⅟3? ⅟4?
If there is no point with more.than.⅟2
between it and any midpoint,
There are 3/oo of all points exterior.
Did you intend to write "interior"?
An exterior point is in
an open interval holding no rational.
There are no
open intervals holding no rational.
There are no exterior points.
Therefore not all rationals are enumerated.
Explain why.
Contradiction.
It contradicts a non.empty exterior.
It doesn't contradict an almost.all boundary.
Something of your theory is inconsistent.
Your intuition is disturbed by
an almost.all boundary.
Disturbed intuitions and inconsistencies
are different.
Instead, there are boundary points.
For each x not.in the intervals,
each open set Oₓ which holds x
holds points in the intervals and
points not.in the intervals.
x is a boundary point.
On 06.11.2024 15:22, Jim Burns wrote:
On 11/6/2024 5:35 AM, WM wrote:
The intervals are closed with irrational endpoints.
'Exterior' seems like a good way to say
'not in contact'.
Every point outside is not an endpoint
and is not in contact.
Something of your theory is inconsistent.
Your intuition is disturbed by
an almost.all boundary.
No. Your boundary is nonsense.
On 11/6/2024 11:24 AM, WM wrote:
On 06.11.2024 15:22, Jim Burns wrote:
On 11/6/2024 5:35 AM, WM wrote:
The intervals are closed with irrational endpoints.
'Exterior' seems like a good way to say
'not in contact'.
Every point outside is not an endpoint
Yes.
and is not in contact.
You don't know that.
The _union_ of
arbitrarily.many _open_ sets
is an open set.
However,
the _union_ of
these infinitely.many _closed_ intervals
with irrational endpoints
is an open interval
with rational endpoints.
"My" boundary is a definition.
The term "boundary" helps clarify
what I admit is a confusing situation.
https://en.wikipedia.org/wiki/Boundary_(topology)
⎛ It is the set of points p ∈ X such that
⎜ every neighborhood of p contains
⎜ at least one point of S and
⎝ at least one point not of S :
On 06.11.2024 20:08, Jim Burns wrote:
On 11/6/2024 11:24 AM, WM wrote:
On 06.11.2024 15:22, Jim Burns wrote:
On 11/6/2024 5:35 AM, WM wrote:
The intervals are closed with irrational endpoints.
'Exterior' seems like a good way to say
'not in contact'.
Every point outside is not an endpoint
Yes.
and is not in contact.
You don't know that.
From every positive point we know that
it is not 0 and
not in contact with (-oo, 0].
Same for every point not in an interval.
The _union_ of
arbitrarily.many _open_ sets
is an open set.
However,
the _union_ of
these infinitely.many _closed_ intervals
with irrational endpoints
is an open interval
with rational endpoints.
We use only intervals, not limits.
A point is in an interval or it is not.
On 11/6/2024 8:36 AM, WM wrote:
On 05.11.2024 18:25, Jim Burns wrote:
Instead, there are boundary points.
For each x not.in the intervals,
each open set Oₓ which holds x
holds points in the intervals and
points not.in the intervals.
x is a boundary point.
For every definable x we can decide whether it is inside the interval
including its endpoints or outside. No open intervals are necesseary
or useful. Your trick is cunning bot not accepted.
For any z we can decide if its inside or outside of <whatever>
Hint: we do not have to "decide" if a
number z is inside or outside of a certain interval. It IS EITHER inside
of the interval OR NOT. (No "decision" necessary.)
On 11/6/2024 2:50 PM, WM wrote:
From every positive point we know that
it is not 0 and
not in contact with (-oo, 0].
Same for every point not in an interval.
Is 0 "not in contact with" [-1,0) ⊆ ℝ
A point can be not.in the closure of each
of infinitely.many sets
and also in the closure of their union.
On 06.11.2024 21:20, Jim Burns wrote:
On 11/6/2024 2:50 PM, WM wrote:
From every positive point we know that
it is not 0 and
not in contact with (-oo, 0].
Same for every point not in an interval.
Is 0 "not in contact with" [-1,0) ⊆ ℝ
0 is not a positive point.
On 11/7/2024 3:46 AM, WM wrote:
I want to find out from you (WM)
what "not in contact with" means.
For a point
in the boundary but not in the set,
is it "not in contact with" the set?
Is it "in contact with" the set?
0 is in the boundary of [-1,0) > Is 0 "in contact with" [-1,0) ?
On 07.11.2024 12:18, Jim Burns wrote:
I want to find out from you (WM)
what "not in contact with" means.
For a point
in the boundary but not in the set,
is it "not in contact with" the set?
Is it "in contact with" the set?
It is not in contact with the set.
"For every eps" is not a valid criterion
because eps depends on
what you can define, not on what exists.
The endpoint is in contact with the set.
0 is in the boundary of [-1,0)
Is 0 "in contact with" [-1,0) ?
I am not an expert on these things.
I would say
it is in contact with the set
because a point of the set is next to it.
The closure of a set is in contact with the set.
⎛ The boundary of a set S holds
⎜ those points x′ such that
⎜ each interval [x,x″] with
⎜ x′ in its interior, x < x′ < x″,
⎝ holds points in S and points not.in S
On 06.11.2024 21:20, Jim Burns wrote:
[...]
Use the intervals
I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n].
Since n and q_n can be in bijection,
these intervals are sufficient to cover all q_n.
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".
On 11/7/2024 3:46 AM, WM wrote:
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".
Yes.
In that clever re.ordering, not scrunched together,
the whole positive axis
is in the ε.cover or
in the boundary of the ε.cover.
On 07.11.2024 16:29, Jim Burns wrote:
⎛ The boundary of a set S holds
⎜ those points x′ such that
⎜ each interval [x,x″] with
⎜ x′ in its interior, x < x′ < x″,
⎝ holds points in S and points not.in S
Do you think you need the boundary in my last example?
When we cover the real axis by intervals --------_1_--------_2_--------_3_--------_4_--------_5_--------_...
J(n) = [n - √2/10, n + √2/10]
and shuffle them in a clever way,
then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.
Do you believe this???
On 11/7/2024 12:59 PM, WM wrote:
When we cover the real axis by intervals
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
J(n) = [n - √2/10, n + √2/10]
and shuffle them in a clever way,
then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.
No irrational is not in contact with
the union of intervals.
The first is true because,
for each irrational x,
each interval of which x is in its interior
holds rationals, and
rationals are points in the union of intervals.
There is an enumeration of ℚ⁺
The infinite sum of measures = 2³ᐟ²/9
d is _not in contact with_ each interval.
However,
each interval of which d is in its interior
holds rationals, which are points in ⋃(ε.cover)
Thus, d is _in contact with_ ⋃(ε.cover)
but not with any of its intervals.
Do you believe this???
Don't you believe this???
On 07.11.2024 20:06, Jim Burns wrote:
On 11/7/2024 3:46 AM, WM wrote:
On 07.11.2024 16:29, Jim Burns wrote:
and shuffle them in a clever way,
then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".
Yes.
In that clever re.ordering, not scrunched together,
the whole positive axis
is in the ε.cover or
in the boundary of the ε.cover.
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].
⎛ ε.cover = {[x⁽ᵋₖ,xᵋ⁾ₖ]:k∈ℕ⁺}
⎜ x⁽ᵋₖ = iₖ/jₖ-2¹ᐟ²⋅10⁻ᵏ
⎝ xᵋ⁾ₖ = iₖ/jₖ+2¹ᐟ²⋅10⁻ᵏ
There is an enumeration of ℚ⁺
the set of ratios of ℕ⁺ countable.to from.1
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ/jₖ
⎜ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-((iₖ+jₖ)-1)((iₖ+jₖ)-2)/2
⎜ jₖ = (iₖ+jₖ)-iₖ
⎝ (iₖ+jₖ-1)(iₖ+jₖ-2)/2+iₖ = k
d is a Cantorian anti.diagonal of
the Cantorian rational.list ⟨iₖ/jₖ⟩
#d#ₖ = (#iₖ/jₖ#ₖ+5) mod 10
#x#ₖ is the -k.th digit of
the decimal representation of real number x
No boundaries are involved because
every interval of length 1/5 contains
infinitely many rationals and
therefore is essentially covered by
infinitely many intervals of length 1/5
- if Cantor is right.
On 11/7/2024 2:33 PM, WM wrote:They are the intervals that we start with.
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].
Those are not the cleverly.re.ordered intervals.
Then consider the two only alternatives: Either by reordering (one afterNo boundaries are involved because
every interval of length 1/5 contains infinitely many rationals and
therefore is essentially covered by infinitely many intervals of
length 1/5
- if Cantor is right.
I haven't claimed anything at all about
your all.1/5.length intervals.
On 08.11.2024 00:29, Jim Burns wrote:
On 11/7/2024 2:33 PM, WM wrote:
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].
Those are not the cleverly.re.ordered intervals.They are the intervals that we start with.
No boundaries are involved because
every interval of length 1/5 contains infinitely many rationals and
therefore is essentially covered by infinitely many intervals of
length 1/5
- if Cantor is right.
I haven't claimed anything at all aboutThen consider the two only alternatives: Either by reordering (one after
your all.1/5.length intervals.
the other or simultaneously) the measure of these intervals can grow
from 1/10 of the real axis to infinitely many times the real axis, or not.
My understanding of mathematics and geometry is that reordering cannot increase the measure (only reduce it by overlapping). This is a basic
axiom which will certainly be agreed to by everybody not conditioned by matheology. But there is also an analytical proof: Every reordering of
any finite set of intervals does not increase their measure. The limit
of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an interval.
Regards, WM
On 11/8/24 5:18 AM, WM wrote:
My understanding of mathematics and geometry is that reordering cannot
increase the measure (only reduce it by overlapping). This is a basic
axiom which will certainly be agreed to by everybody not conditioned
by matheology. But there is also an analytical proof: Every reordering
of any finite set of intervals does not increase their measure. The
limit of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an interval.
which makes the error that the properties of finite objects apply to the infinite objects, which isn't true, and what just breaks your logic.
You take it as a given, but that just means that your logic is unable to actually handle the infinite.
On 08.11.2024 13:28, Richard Damon wrote:
On 11/8/24 5:18 AM, WM wrote:
My understanding of mathematics and geometry is that reordering
cannot increase the measure (only reduce it by overlapping). This is
a basic axiom which will certainly be agreed to by everybody not
conditioned by matheology. But there is also an analytical proof:
Every reordering of any finite set of intervals does not increase
their measure. The limit of a constant sequence is this constant
however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an
interval.
which makes the error that the properties of finite objects apply to
the infinite objects, which isn't true, and what just breaks your logic.
The infinite of the real axis is a big supply but an as big drain.
You take it as a given, but that just means that your logic is unable
to actually handle the infinite.
I take it as evident that intervals of the measure 1/5 of the positive
real axis will not, by any shuffling, cover the real axis completely,
let alone infinitely often. I think who believes this is a deplorable
fanatic if not a fool.
Regards, WM
On 08.11.2024 00:29, Jim Burns wrote:
On 11/7/2024 2:33 PM, WM wrote:
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].
Those are not the cleverly.re.ordered intervals.
They are the intervals that we start with.
No boundaries are involved because
every interval of length 1/5 contains
infinitely many rationals and
therefore is essentially covered by
infinitely many intervals of length 1/5
- if Cantor is right.
I haven't claimed anything at all about
your all.1/5.length intervals.
Then consider the two only alternatives:
Either by reordering
(one after the other or simultaneously)
the measure of these intervals
the measure of these intervals can grow
from 1/10 of the real axis
to infinitely many times the real axis,
or not.
My understanding of mathematics and geometry
is that
reordering cannot increase the measure
(only reduce it by overlapping).
This is a basic axiom which
will certainly be agreed to by
everybody not conditioned by matheology.
But there is also an analytical proof:
Every reordering of
any finite set of intervals
does not increase their measure.
The limit of a constant sequence is
this constant however.
This geometrical consequence of Cantor's theory
has, to my knowledge, never been discussed.
By the way I got the idea after a posting of yours:
Each of {...,-3,-2,-1,0,1,2,3,...} is
the midpoint of an interval.
On 08.11.2024 13:28, Richard Damon wrote:What is the measure you are using and what does it give for the real
On 11/8/24 5:18 AM, WM wrote:
The infinite of the real axis is a big supply but an as big drain.My understanding of mathematics and geometry is that reordering cannotwhich makes the error that the properties of finite objects apply to
increase the measure (only reduce it by overlapping). This is a basic
axiom which will certainly be agreed to by everybody not conditioned
by matheology. But there is also an analytical proof: Every reordering
of any finite set of intervals does not increase their measure. The
limit of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an
interval.
the infinite objects, which isn't true, and what just breaks your
logic.
I take it as evident that intervals of the measure 1/5 of the positive
You take it as a given, but that just means that your logic is unable
to actually handle the infinite.
real axis will not, by any shuffling, cover the real axis completely,
let alone infinitely often. I think who believes this is a deplorable
fanatic if not a fool.
Am Fri, 08 Nov 2024 17:43:15 +0100 schrieb WM:
On 08.11.2024 13:28, Richard Damon wrote:
On 11/8/24 5:18 AM, WM wrote:The infinite of the real axis is a big supply but an as big drain.
My understanding of mathematics and geometry is that reordering cannot >>>> increase the measure (only reduce it by overlapping). This is a basicwhich makes the error that the properties of finite objects apply to
axiom which will certainly be agreed to by everybody not conditioned
by matheology. But there is also an analytical proof: Every reordering >>>> of any finite set of intervals does not increase their measure. The
limit of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an
interval.
the infinite objects, which isn't true, and what just breaks your
logic.
I take it as evident that intervals of the measure 1/5 of the positive
You take it as a given, but that just means that your logic is unable
to actually handle the infinite.
real axis will not, by any shuffling, cover the real axis completely,
let alone infinitely often. I think who believes this is a deplorable
fanatic if not a fool.
What is the measure you are using and what does it give for the real
axis?
Am 08.11.2024 um 20:05 schrieb joes:
Am Fri, 08 Nov 2024 17:43:15 +0100 schrieb WM:
On 08.11.2024 13:28, Richard Damon wrote:
On 11/8/24 5:18 AM, WM wrote:The infinite of the real axis is a big supply but an as big drain.
My understanding of mathematics and geometry is that reordering cannot >>>>> increase the measure (only reduce it by overlapping). This is a basic >>>>> axiom which will certainly be agreed to by everybody not conditioned >>>>> by matheology. But there is also an analytical proof: Every reordering >>>>> of any finite set of intervals does not increase their measure. Thewhich makes the error that the properties of finite objects apply to
limit of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge, >>>>> never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an
interval.
the infinite objects, which isn't true, and what just breaks your
logic.
I take it as evident that intervals of the measure 1/5 of the positive
You take it as a given, but that just means that your logic is unable
to actually handle the infinite.
real axis will not, by any shuffling, cover the real axis completely,
let alone infinitely often. I think who believes this is a deplorable
fanatic if not a fool.
What is the measure you are using and what does it give for the real
axis?
Ob Du es nochmal schaffst, auf diesen saudummen Scheißdreck NICHT zu antworten?
On 11.11.2024 19:23, Jim Burns wrote:
On 11/11/2024 3:41 AM, WM wrote:
But it will never complete
an infinite set of claims.
We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
But you claim that
_all_ fractions are in bijection with
all natural numbers,
don't you?
Yes, I claim that.
For that claim you need an infinite set of claims.
This is one 𝗰𝗹𝗮𝗶𝗺:
⎛ All fractions are in bijection with
⎝ all natural numbers.
It is wrong.
My intervals I(n) = [n - 1/10, n + 1/10]
must be translated to all the midpoints
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5,
2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
if you want to contradict my claim.
Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".
No, I claim that intervals can be translated.
(The set of intervals remains constant
in size and multitude.)
For every finite subset this is possible.
It happens that WM formulated :
For that claim you need an infinite set of claims.
Good thing we have such.
This is one ?????:
⎛ All fractions are in bijection with
⎝ all natural numbers.
It is wrong.
You 'think' it is wrong because you intuitively think it 'should' be
wrong. Saying it is wrong does not make it wrong no matter how many
times that you say it.
On 11.11.2024 22:50, FromTheRafters wrote:Yes, we do claim something for each, every one and all of the
It happens that WM formulated :
No.For that claim you need an infinite set of claims.Good thing we have such.
It absolutely does, for all fractions n.Therefore I prove it. The set of intervals I(n) = [n - 1/10, n + 1/10]You 'think' it is wrong because you intuitively think it 'should' beThis is one ?????:It is wrong.
⎛ All fractions are in bijection with ⎝ all natural numbers.
wrong. Saying it is wrong does not make it wrong no matter how many
times that you say it.
cannot cover all fractions on the real axis.
On 11/11/2024 1:00 AM, WM wrote:
I(n) = [n - 1/10, n + 1/10]
can be translated until all rational numbers
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
are midpoints.
Obviously that is impossible because the density 1/5 of the intervals
can never increase. It is possible however to shift an arbitrarily
large (a potentially infinite) number of intervals to rational midpoints.
I don't think you know how to take any natural number and turn it into a unique pair, and then back again via Cantor pairing.
Am Tue, 12 Nov 2024 09:58:12 +0100 schrieb WM:
Therefore I prove it. The set of intervals I(n) = [n - 1/10, n + 1/10]It absolutely does, for all fractions n.
cannot cover all fractions on the real axis.
On 11/11/2024 3:33 PM, WM wrote:
This is one 𝗰𝗹𝗮𝗶𝗺:
⎛ All fractions are in bijection with
⎝ all natural numbers.
It is wrong.
It is one claim.
Apparently, to you, it means that
sets change.
Our sets do not change.
On 11.11.2024 19:23, Jim Burns wrote:
On 11/11/2024 3:41 AM, WM wrote:
My intervals I(n) = [n - 1/10, n + 1/10]
must be translated to all the midpoints
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5,
2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
if you want to contradict my claim.
Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".
No, I claim that intervals can be translated.
(The set of intervals remains constant
in size and multitude.)
On 11/11/2024 3:33 PM, WM wrote:
On 11.11.2024 19:23, Jim Burns wrote:
On 11/11/2024 3:41 AM, WM wrote:
My intervals I(n) = [n - 1/10, n + 1/10]
must be translated to all the midpoints
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5,
2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
if you want to contradict my claim.
Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".
No, I claim that intervals can be translated.
By which, you mean that translation changes intervals.
(The set of intervals remains constant
in size and multitude.)
The set of intervals remains constant. Absolutely.
Sets do not change.
Intervals do not change.
Mathematical objects do not change.
On 12.11.2024 16:58, Jim Burns wrote:
On 11/11/2024 3:33 PM, WM wrote:
On 11.11.2024 19:23, Jim Burns wrote:
On 11/11/2024 3:41 AM, WM wrote:
My intervals I(n) = [n - 1/10, n + 1/10]
must be translated to all the midpoints
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5,
2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
if you want to contradict my claim.
Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".
No, I claim that intervals can be translated.
By which, you mean that translation changes intervals.
No, the intervals remain constant
in size and multitude.
(The set of intervals remains constant
in size and multitude.)
The set of intervals remains constant. Absolutely.
Sets do not change.
Intervals do not change.
Mathematical objects do not change.
Therefore
the intervals covering all naturals
cannot cover more.
But the rationals are more in the sense that
they include all naturals and 1/2.
By your argument
Cantor has been falsified.
On 11/12/2024 11:43 AM, WM wrote:
No, the intervals remain constant
in size and multitude.
Intervals which are constant _only_
in size and multitude
are not constant absolutely.
These intervals
{[n-⅒,n+⅒]: n∈ℕ⁺}
cover all naturals ℕ⁺ and
do not cover all fractions ℕ⁺/ℕ⁺
But the rationals are more in the sense that
they include all naturals and 1/2.
These intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
cover all fractions ℕ⁺/ℕ⁺
These intervals
{[n-⅒,n+⅒]: n∈ℕ⁺}
and these intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
are different intervals.
If a = b then they [a and b] are equal?
For instance 5 = 5, the self relation equality?
If a = b then they [a and b] are equal?
For instance 5 = 5, the self relation equality?
[...] different ways to represent zero?
0 = 3-4+1 = 4-4 = 2*0 = 0
ect.
On 12.11.2024 20:01, Jim Burns wrote:
On 11/12/2024 11:43 AM, WM wrote:
No, the intervals remain constant
in size and multitude.
Intervals which are constant _only_
in size and multitude
are not constant absolutely.
They would suffer [suffice] to cover
all rationals completely
if Cantor's bijection was complete.
On 11/12/2024 4:38 PM, WM wrote:
On 12.11.2024 20:01, Jim Burns wrote:
On 11/12/2024 11:43 AM, WM wrote:
No, the intervals remain constant
in size and multitude.
Intervals which are constant _only_
in size and multitude
are not constant absolutely.
They would suffer [suffice] to cover
all rationals completely
if Cantor's bijection was complete.
I think that you want to use 'suffice' here.
Your English is generally excellent,
but 'suffice' is not a very common word.
One instance I'm fond of:
⎛ Some say the world will end in fire,
⎜ Some say in ice.
⎜ From what I’ve tasted of desire
⎜ I hold with those who favor fire.
⎜ But if it had to perish twice,
⎜ I think I know enough of hate
⎜ To say that for destruction ice
⎜ Is also great
⎝ And would suffice.
-- Robert Frost, "Fire and Ice"
On 12.11.2024 20:01, Jim Burns wrote:
These intervals
{[n-⅒,n+⅒]: n∈ℕ⁺}
cover all naturals ℕ⁺ and
do not cover all fractions ℕ⁺/ℕ⁺
Right.
But the rationals are more in the sense that
they include all naturals and 1/2.
These intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
cover all fractions ℕ⁺/ℕ⁺
But these are more intervals.
On 11/12/2024 4:38 PM, WM wrote:
But the rationals are more in the sense that
they include all naturals and 1/2.
These intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
cover all fractions ℕ⁺/ℕ⁺
But these are more intervals.
Are there more, though?
Or are there fewer?
i/j ↦ (i+j-1)(i+j-1)+2⋅i
⟨ 1/1 1/2 2/1 1/3 2/2 3/1 1/4 2/3 ... ⟩
↦
⟨ 2 4 6 8 10 12 14 16 ... ⟩
⟨ 1/1 1/2 2/1 1/3 2/2 3/1 1/4 2/3 ... ⟩
↦
⟨ 2 3 5 7 11 13 17 19 ... ⟩
⟨ 1/1 1/2 2/1 ... ⟩
↦
⟨ 10^10 10^10^10 10^10^10^10 ... ⟩
Or do infinite sets have different rules
than finite sets do?
On 13.11.2024 10:08, Jim Burns wrote:
On 11/12/2024 4:38 PM, WM wrote:
On 12.11.2024 20:01, Jim Burns wrote:
On 11/12/2024 11:43 AM, WM wrote:
But the rationals are more in the sense that
they include all naturals and 1/2.
These intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
cover all fractions ℕ⁺/ℕ⁺
But these are more intervals.
Are there more, though?
Or are there fewer?
i/j ↦ (i+j-1)(i+j-1)+2⋅i
⟨ 1/1 1/2 2/1 1/3 2/2 3/1 1/4 2/3 ... ⟩
↦
⟨ 2 4 6 8 10 12 14 16 ... ⟩
or
⟨ 1/1 1/2 2/1 1/3 2/2 3/1 1/4 2/3 ... ⟩
↦
⟨ 2 3 5 7 11 13 17 19 ... ⟩
or
⟨ 1/1 1/2 2/1 ... ⟩
↦
⟨ 10^10 10^10^10 10^10^10^10 ... ⟩
Or do infinite sets have different rules
than finite sets do?
If infinite sets obey the rules sketched above,
then set theorists must discard geometry
because
by shifting intervals
the relative covering 1/5 of ℝ+ becomes oo*ℝ,
and analysis
because
the constant sequence 1/5, 1/5, 1/5, ...
has limit oo,
and logic
because of
Bob.
by shifting intervals
the relative covering 1/5 of ℝ+ becomes oo*ℝ,
the constant sequence 1/5, 1/5, 1/5, ...
has limit oo,
Bob.
On 11/13/2024 11:31 AM, WM wrote:
If infinite sets obey the rules sketched above,
... _and are finite_ ...
then set theorists must discard geometry
because
by shifting intervals
the relative covering 1/5 of ℝ+ becomes oo*ℝ,
and analysis
because
the constant sequence 1/5, 1/5, 1/5, ...
has limit oo,
and logic
because of
Bob.
----
by shifting intervals
the relative covering 1/5 of ℝ+ becomes oo*ℝ,
By definition,
the value of a measure is an extended real≥0
An extended real≥0 is either
Archimedean == having a countable.to bound, or
non.Archimedean == not.having a countable.to bound.
The extended reals≥0 have only
the standard reals≥0, which are Archimedean, and
a single non.Archimedean point≥0 +∞
No,
the measure doesn't _become_ +∞
It has the same value +∞ before and after shifting.
----
the constant sequence 1/5, 1/5, 1/5, ...
has limit oo,
----
Bob.
KING BOB!
https://www.youtube.com/watch?v=TjAg-8qqR3g
If,
in a set A which
can match one of its proper subsets B,
[...] different ways to represent zero?
0 = 3-4+1 = 4-4 = 2*0 = 0
ect.
Am 13.11.2024 um 03:44 schrieb Chris M. Thomasson:
[...] different ways to represent zero?
0 = 3-4+1 = 4-4 = 2*0 = 0
ect.
Right. "3-4+1", "4-4", "2*0", "0", are just different NAMEs (math:
TERMs) to denote (refer to) _one and the same_ number, namely zero
(usualy, denoted with "0").
Hint: Superman is Clark Kent.
On 13.11.2024 20:38, Jim Burns wrote:
----
Bob.
KING BOB!
https://www.youtube.com/watch?v=TjAg-8qqR3g
If,
in a set A which
can match one of its proper subsets B,
That is nonsense too.
Jim Burns formulated on Wednesday :
On 11/13/2024 4:29 PM, WM wrote:
On 13.11.2024 20:38, Jim Burns wrote:
----
Bob.
KING BOB!
https://www.youtube.com/watch?v=TjAg-8qqR3g
If,
in a set A which
can match one of its proper subsets B,
That is nonsense too.
In my source window:
[...]That is nonsense too.
A finite ð˜€ð—²ð—¾ð˜‚ð—²ð—»ð—°ð—² of ð—°ð—¹ð—®ð—¶ð—ºð˜€ in
which
each claim is true.or.not.first.false
is
a finite ð˜€ð—²ð—¾ð˜‚ð—²ð—»ð—°ð—² of ð—°ð—¹ð—®ð—¶ð—ºð˜€ in
which
each claim is true.
================================================
I follow some of this mostly from context. :)
A finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 in which
each claim is true.or.not.first.false
is
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 in which
each claim is true.
Some claims are true and we know it
because
they claim that
when we say this, we mean that,
and we, conscious of our own minds, know that
when we say this, we mean that.
Some 𝗰𝗹𝗮𝗶𝗺𝘀 are not.first.false and we know it
because
we can see that
no assignment of truth.values exists
in which 𝘁𝗵𝗲𝘆 are first.false.
𝗾 is not first.false in ⟨ 𝗽 𝗽⇒𝗾 𝗾 ⟩.
Some finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲𝘀 of 𝗰𝗹𝗮𝗶𝗺𝘀 are
each true.or.not.first.false
and we know it.
When we know that,
we know each claim is true.
We know each claim is true, even if
it is a claim physically impossible to check,
like it would be physically impossible
to check each one of infinitely.many.
On 14.11.2024 00:16, Jim Burns wrote:
Therefore
a geometric representation let alone proof of
most of Cantor's bijections is impossible.
Therefore
a geometric representation let alone proof of
most of Cantor's bijections is impossible.
On 11/13/2024 3:28 PM, Moebius wrote:
Am 14.11.2024 um 00:27 schrieb Moebius:
Am 13.11.2024 um 03:44 schrieb Chris M. Thomasson:
[...] different ways to represent zero?
0 = 3-4+1 = 4-4 = 2*0 = 0
ect.
Right. "3-4+1", "4-4", "2*0", "0", are just different NAMEs (math:
TERMs) to denote (refer to) _one and the same_ number, namely zero
(usualy, denoted with "0").
Hint: Superman is Clark Kent.
Superman = Clark Kent
Superman = Clark Kent = Kal-El
On 11/13/2024 3:28 PM, Moebius wrote:
Am 14.11.2024 um 00:27 schrieb Moebius:
Am 13.11.2024 um 03:44 schrieb Chris M. Thomasson:
[...] different ways to represent zero?
0 = 3-4+1 = 4-4 = 2*0 = 0
ect.
Right. "3-4+1", "4-4", "2*0", "0", are just different NAMEs (math:
TERMs) to denote (refer to) _one and the same_ number, namely zero
(usualy, denoted with "0").
Hint: Superman is Clark Kent.
Superman = Clark Kent
Superman = Clark Kent = Kal-El
On 11/14/2024 2:19 PM, Moebius wrote:
Am 14.11.2024 um 23:09 schrieb Chris M. Thomasson:
On 11/13/2024 3:28 PM, Moebius wrote:
Am 14.11.2024 um 00:27 schrieb Moebius:
Am 13.11.2024 um 03:44 schrieb Chris M. Thomasson:
[...] different ways to represent zero?
0 = 3-4+1 = 4-4 = 2*0 = 0
ect.
Right. "3-4+1", "4-4", "2*0", "0", are just different NAMEs (math:
TERMs) to denote (refer to) _one and the same_ number, namely zero
(usualy, denoted with "0").
Hint: Superman is Clark Kent.
Superman = Clark Kent
Superman = Clark Kent = Kal-El
Right. The names "Superman", "Clark Kent" and "Kal-El" denote one and
the same person (being).
We might extend that list by, say, "the only child of Jor-El and Lara
Lor-Van", etc.
0 = the predecessor of 1 = 1 - 1 = (the x in IR such that 1 + x = 1).
For fun, define the "+" symbol to mean "having a child", so:
(Jor-El + Lara Lor-Van) = (Superman = Clark Kent = Kal-El)
? lol.
On 11/14/2024 2:19 PM, Moebius wrote:
Am 14.11.2024 um 23:09 schrieb Chris M. Thomasson:
On 11/13/2024 3:28 PM, Moebius wrote:
Am 14.11.2024 um 00:27 schrieb Moebius:
Am 13.11.2024 um 03:44 schrieb Chris M. Thomasson:
[...] different ways to represent zero?
0 = 3-4+1 = 4-4 = 2*0 = 0
ect.
Right. "3-4+1", "4-4", "2*0", "0", are just different NAMEs (math:
TERMs) to denote (refer to) _one and the same_ number, namely zero
(usualy, denoted with "0").
Hint: Superman is Clark Kent.
Superman = Clark Kent
Superman = Clark Kent = Kal-El
Right. The names "Superman", "Clark Kent" and "Kal-El" denote one and
the same person (being).
We might extend that list by, say, "the only child of Jor-El and Lara
Lor-Van", etc.
0 = the predecessor of 1 = 1 - 1 = (the x in IR such that 1 + x = 1).
For fun, define the "+" symbol to mean "having a child", so:
(Jor-El + Lara Lor-Van) = (Superman = Clark Kent = Kal-El)
? lol.
Am 14.11.2024 um 23:27 schrieb Chris M. Thomasson:
On 11/14/2024 2:19 PM, Moebius wrote:
Am 14.11.2024 um 23:09 schrieb Chris M. Thomasson:
On 11/13/2024 3:28 PM, Moebius wrote:
Am 14.11.2024 um 00:27 schrieb Moebius:
Am 13.11.2024 um 03:44 schrieb Chris M. Thomasson:
[...] different ways to represent zero?
0 = 3-4+1 = 4-4 = 2*0 = 0
ect.
Right. "3-4+1", "4-4", "2*0", "0", are just different NAMEs (math: >>>>>> TERMs) to denote (refer to) _one and the same_ number, namely zero >>>>>> (usualy, denoted with "0").
Hint: Superman is Clark Kent.
Superman = Clark Kent
Superman = Clark Kent = Kal-El
Right. The names "Superman", "Clark Kent" and "Kal-El" denote one and
the same person (being).
We might extend that list by, say, "the only child of Jor-El and Lara
Lor-Van", etc.
0 = the predecessor of 1 = 1 - 1 = (the x in IR such that 1 + x = 1).
For fun, define the "+" symbol to mean "having a child", so:
(Jor-El + Lara Lor-Van) = (Superman = Clark Kent = Kal-El)
? lol.
Nope, that won't work.
For this we might define + , say, the following way:
x + y = the child of x and y if x and y have exactly one child
= the set of the children of x and y otherwise.
Then
Jor-El + Lara Lor-Van = Superman = Clark Kent = Kal-El
and, say,
Superman + Donald Trump = {} .
"having a child (or children)" is a predicate, while "the child of ...
and ..." is a term.
On 11/14/2024 5:20 AM, WM wrote:
On 14.11.2024 00:16, Jim Burns wrote:
Therefore
a geometric representation let alone proof of
most of Cantor's bijections is impossible.
Consider geometry.
For two triangles △A′B′C′ and △A″B″C″
if
△A′B′C′ and △A″B″C″ are similar triangles
⎛ μ∠C′A′B′ = μ∠C″A″B″
⎜ μ∠A′B′C′ = μ∠A″B″C″
⎜ μ∠B′C′A′ = μ∠B″C″A″
⎝ △A′B′C′ ∼ △A″B″C″
then
corresponding sides are in the same ratio
( μA︫︭′︭B︫′/μA︫︭″︭B︫″ = μB︫︭′︭C︫′/μB︫︭″︭C︫″ = μA︫︭′︭C︫′/μA︫︭″︭C︫″
For similar △A′B′C′ ∼ △A″B″C″
if μA︫︭′︭B︫′ = 1
and μB︫︭″︭C︫″ = x
then 1/μA︫︭″︭B︫″ = μB︫︭′︭C︫′/x
and μA︫︭″︭B︫″ = μB︫︭′︭C︫′ = x¹ᐟ²
For similar △A′B′C′ ∼ △A″B″C″
if μA︫︭′︭B︫′ = 1
and μA︫︭″︭B︫″ = x
and μB︫︭′︭C︫′ = y
then 1/x = y/μB︫︭″︭C︫″
and μB︫︭″︭C︫″ = x⋅y
For similar △A′B′C′ ∼ △A″B″C″
if μA︫︭′︭B︫′ = 1
and μA︫︭″︭B︫″ = x
and μB︫︭″︭C︫″ = y
then 1/x = μB︫︭′︭C︫′/y
and μB︫︭′︭C︫′ = y/x
The ceiling ⌈x⌉ of x is the first integer ≥ x
Therefore
a geometric representation let alone proof of
most of Cantor's bijections is impossible.
Cantor's bijection is
k ↦ i/j
i︭+︭j := ⌈(2⋅k+¼)¹ᐟ²+½⌉
i := k-(i︭+︭j-1)⋅(i︭+︭j-2)/2
j := i︭+︭j-i
(i+j-1)⋅(i+j-2)/2+i = k
Setting aside for a moment
what you _think_ Cantor's bijection is,
what part of _that_
is impossible to represent geometrically?
On 11/14/2024 5:20 AM, WM wrote:
Here is a single claim which is true:
You don't say what reason you (WM) have
for knowing that that single claim is true.
Am Fri, 15 Nov 2024 11:04:33 +0100 schrieb WM:
It can be proven for every finite geometric figure that covering it byDuh. If some configuration doesn't cover it, shuffling the pieces changes nothing. But there may be other configurations that do cover the figure.
small pieces or intervals does not depend on the individuality and
therefore on the order of the pieces.
That means if there is a configuration where the figure is not covered
completely, every possible shuffling will also fail.
It can be proven for every finite geometric figure that covering it byDuh. If some configuration doesn't cover it, shuffling the pieces changes nothing. But there may be other configurations that do cover the figure.
small pieces or intervals does not depend on the individuality and
therefore on the order of the pieces.
That means if there is a configuration where the figure is not covered completely, every possible shuffling will also fail.
On 14.11.2024 19:31, Jim Burns wrote:
On 11/14/2024 5:20 AM, WM wrote:
Here is a single claim which is true:
You don't say what reason you (WM) have
for knowing that that single claim is true.
It can be proven
for every finite geometric figure
that covering it by small pieces or intervals
does not depend on the individuality and
therefore on the order of the pieces.
That means
if there is a configuration where
the figure is not covered completely,
every possible shuffling will also fail.
For infinite figures
we use the analytical limit
as is normal in mathematics.
On 14.11.2024 19:31, Jim Burns wrote:
On 11/14/2024 5:20 AM, WM wrote:
Therefore
a geometric representation let alone proof of
most of Cantor's bijections is impossible.
Consider geometry.
For two triangles △A′B′C′ and △A″B″C″
if
△A′B′C′ and △A″B″C″ are similar triangles
then
corresponding sides are in the same ratio
Therefore
a geometric representation let alone proof of
most of Cantor's bijections is impossible.
Your writing is unreadable
but that does not matter because
of course only a disproof is possible,
since there are no bijections.
Setting aside for a moment
what you _think_ Cantor's bijection is,
what part of _that_
is impossible to represent geometrically?
It is impossible to cover the matrix
XOOO...
XOOO...
XOOO...
XOOO...
...
by shuffling, shifting, reordering the X,
because they are not distinguishable.
On 11/15/2024 09:55 AM, Jim Burns wrote:
On 11/15/2024 5:10 AM, WM wrote:
On 14.11.2024 19:31, Jim Burns wrote:
Setting aside for a moment
what you _think_ Cantor's bijection is,
what part of _that_
is impossible to represent geometrically?
It is impossible to cover the matrix
XOOO...
XOOO...
XOOO...
XOOO...
...
by shuffling, shifting, reordering the X,
because they are not distinguishable.
⟨k,1⟩ ↦ ⟨i,j⟩ ↤ ⟨k,1⟩
⎛ i = k-⌈(2⋅k+¼)¹ᐟ²-1/2⌉⋅⌈(2⋅k+¼)¹ᐟ²-3/2⌉/2
⎜ j = ⌈(2⋅k+¼)¹ᐟ²+1/2⌉⋅⌈(2⋅k+¼)¹ᐟ²-1/2⌉/2-1-k
⎝ k = (i+j-1)⋅(i+j-2)/2+i
Each ⟨k,1⟩ sends X to ⟨i,j⟩
Each ⟨i,j⟩ receives X from ⟨k,1⟩
According to geometry.
Which I predict makes geometry wrong[WM], too.
Non-standard models of integers exist.
Non-standard models of integers exist.
On 11/15/2024 10:05 AM, Ross Finlayson wrote:
On 11/15/2024 09:55 AM, Jim Burns wrote:
[...]
Non-standard models of integers exist.
"A restriction of comprehension is not a truth."
Each ⟨k,1⟩ sends X to ⟨i,j⟩
Each ⟨i,j⟩ receives X from ⟨k,1⟩
On 11/15/2024 5:04 AM, WM wrote:
For infinite figures
we use the analytical limit
as is normal in mathematics.
The reason you (WM) give is that
☠⎛ whatever is true of all finite
☠⎝ is also true of the infinite.
You (WM) are misinterpreting the infinite as
☠( just like the finite, but bigger.
The finite are the countable.to from.nothing.
Anything countable.to from.nothing is finite.
It's what.we.mean.
And our sets do not change.
On 11/15/2024 12:11 PM, Jim Burns wrote:
On 11/15/2024 1:24 PM, Ross Finlayson wrote:
On 11/15/2024 10:05 AM, Ross Finlayson wrote:
Non-standard models of integers exist.
"A restriction of comprehension is not a truth."
A finite sequence of claims, each claim of which
is true.or.not.first.false, is
a finite sequence of claims, each claim of which
is true.
A fragment, then.
Ugg, an infinite set is never exhausted. Taking a gallon of water out of
an infinite pool of water means you are holding a gallon water, but the infinite pool is still infinite. The same. Taking an infinite amount of
water means the pool is still full of water.
On 11/15/2024 1:53 PM, WM wrote:
The sequence 1/5, 1/5, 1/5, ... has the limit 1/5 after all terms.
Ugg, an infinite set is never exhausted. Taking a gallon of water out of
an infinite pool of water means you are holding a gallon water, but the infinite pool is still infinite. The same. Taking an infinite amount of
water means the pool is still full of water.
Hint: Let's consider your claim: "an infinite set is never exhausted".
But IN \ {1} \ {2} \ {3} \ ... _should_ be {}, I'd say. After all, which natural number would "remain" (=be) in the set
IN \ {1} \ {2} \ {3} \ ...
? :-P
Yeah, slightly "paradoxical". IN \ {1} is infinite, IN \ {1} \ {2} is infinite, IN \ {1} \ {2} \ {3} is infinite, etc. Actually, for each and everey natural number n: IN \ {1} \ ... \ {n} is infinite (in THIS sense
your "never" is true). But what's about IN \ {1} \ {2} \ {3} \ ...?
WHICH natural number would be in this set? :-P
Be aware of the infinite!
Remember:
; You (WM) are misinterpreting the infinite as
; ☠( just like the finite, but bigger.
.
.
.
Am 15.11.2024 um 22:58 schrieb Chris M. Thomasson:
[...] Taking a gallon of water out
of an infinite pool of water means you are holding a gallon water, but
the infinite pool is still infinite. The same. Taking an infinite
amount of water means the pool is still full of water.
I already told you that this is not necessarily the case.
On 11/16/2024 1:07 PM, Chris M. Thomasson wrote:
(infinity - infinity) = infinity
(infinity - infinity) = undefined
Taking out the "infinite amount of H20 molecules" numerated by 1, 2, 3,
... would lead to an EMPTY pool.
On the other hand, taking out the "infinite amount of H20 molecules" numerated by 2, 4, 6, ... would not drain the pool. :-P It would still
be filled (sort of) with infinitely many H20 molecules. :-P
Be aware if the infinite, man!
On 11/16/2024 1:09 PM, Chris M. Thomasson wrote:
(infinity - infinity) = undefined
?
Say taking all of the infinite even numbers away... That leaves an
infinite number of odd numbers? This type of thought.
On 11/16/2024 1:29 PM, FromTheRafters wrote:
Chris M. Thomasson pretended:
(infinity - infinity) = undefined
?
https://en.wikipedia.org/wiki/
L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms
On 11/16/2024 1:29 PM, FromTheRafters wrote:
Chris M. Thomasson pretended :
On 11/16/2024 1:07 PM, Chris M. Thomasson wrote:
(infinity - infinity) = infinity
(infinity - infinity) = undefined
?
https://en.wikipedia.org/wiki/
L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms
For some strange reason reading that made me think of the following song:
https://youtu.be/Sdq4T3iRV80?list=RDMMy3hf0T4qpYg
lol!
Am 16.11.2024 um 22:48 schrieb Chris M. Thomasson:
On 11/16/2024 1:29 PM, FromTheRafters wrote:
Chris M. Thomasson pretended :
On 11/16/2024 1:07 PM, Chris M. Thomasson wrote:
(infinity - infinity) = infinity
(infinity - infinity) = undefined
?
https://en.wikipedia.org/wiki/
L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms
For some strange reason reading that made me think of the following song:
https://youtu.be/Sdq4T3iRV80?list=RDMMy3hf0T4qpYg
lol!
Yusuf / Cat Stevens --- good man! Damn!
On 11/16/2024 2:11 PM, Moebius wrote:
Am 16.11.2024 um 22:48 schrieb Chris M. Thomasson:
On 11/16/2024 1:29 PM, FromTheRafters wrote:
Chris M. Thomasson pretended:
(infinity - infinity) = undefined
?
https://en.wikipedia.org/wiki/
L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms
Or this one: https://en.wikipedia.org/wiki/Indeterminate_form
"oo - oo" is an "indeterminate form".
I must be missing something: [...]
[1] = a gallon of water out of an infinite pool
[2] = another gallon of water out of an infinite pool
[3] = on and on... taken to infinity...
The pool would always have infinite water for this process?
Am 16.11.2024 um 23:16 schrieb Chris M. Thomasson:
On 11/16/2024 2:11 PM, Moebius wrote:
Am 16.11.2024 um 22:48 schrieb Chris M. Thomasson:
On 11/16/2024 1:29 PM, FromTheRafters wrote:
Chris M. Thomasson pretended:
(infinity - infinity) = undefined
?
https://en.wikipedia.org/wiki/
L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms
Or this one: https://en.wikipedia.org/wiki/Indeterminate_form
"oo - oo" is an "indeterminate form".
I must be missing something: [...]
[1] = a gallon of water out of an infinite pool
[2] = another gallon of water out of an infinite pool
[3] = on and on... taken to infinity...
The pool would always have infinite water for this process?
Assume that the H2O molecules in the pool are "numerated" by 1, 2,
3, ... (i.e. that ALL H2O molecules in the pool are "numerated" by
natural numbers).
Taking out the "infinite amount of H2O molecules" numerated by 1, 2,
3, ... would lead to an EMPTY pool.
On the other hand, taking out the "infinite amount of H2O molecules" numerated by 2, 4, 6, ... would not drain the pool. 😛 It would still be filled (sort of) with infinitely many H20 molecules. 😛
Be aware if the infinite, man!
.
.
.
Am 16.11.2024 um 23:16 schrieb Chris M. Thomasson:
On 11/16/2024 2:11 PM, Moebius wrote:
Am 16.11.2024 um 22:48 schrieb Chris M. Thomasson:
On 11/16/2024 1:29 PM, FromTheRafters wrote:
Chris M. Thomasson pretended:
(infinity - infinity) = undefined
?
https://en.wikipedia.org/wiki/
L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms
Or this one: https://en.wikipedia.org/wiki/Indeterminate_form
"oo - oo" is an "indeterminate form".
I must be missing something: [...]
[1] = a gallon of water out of an infinite pool
[2] = another gallon of water out of an infinite pool
[3] = on and on... taken to infinity...
The pool would always have infinite water for this process?
Assume that the H2O molecules in the pool are "numerated" by 1, 2,
3, ... (i.e. that ALL H2O molecules in the pool are "numerated" by
natural numbers).
Taking out the "infinite amount of H2O molecules" numerated by 1, 2,
3, ... would lead to an EMPTY pool.
On the other hand, taking out the "infinite amount of H2O molecules" numerated by 2, 4, 6, ... would not drain the pool. 😛 It would still be filled (sort of) with infinitely many H20 molecules. 😛
Be aware if the infinite, man!
.
.
.
Am 16.11.2024 um 23:16 schrieb Chris M. Thomasson:
On 11/16/2024 2:11 PM, Moebius wrote:
Am 16.11.2024 um 22:48 schrieb Chris M. Thomasson:
On 11/16/2024 1:29 PM, FromTheRafters wrote:
Chris M. Thomasson pretended:
(infinity - infinity) = undefined
?
https://en.wikipedia.org/wiki/
L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms
Or this one: https://en.wikipedia.org/wiki/Indeterminate_form
"oo - oo" is an "indeterminate form".
I must be missing something: [...]
[1] = a gallon of water out of an infinite pool
[2] = another gallon of water out of an infinite pool
[3] = on and on... taken to infinity...
The pool would always have infinite water for this process?
Assume that the H2O molecules in the pool are "numerated" by 1, 2,
3, ... (i.e. that ALL H2O molecules in the pool are "numerated" by
natural numbers).
Taking out the "infinite amount of H2O molecules" numerated by 1, 2,
3, ... would lead to an EMPTY pool.
On the other hand, taking out the "infinite amount of H2O molecules" numerated by 2, 4, 6, ... would not drain the pool. 😛 It would still be filled (sort of) with infinitely many H20 molecules. 😛
Be aware if the infinite, man!
.
.
.
On 11/16/2024 2:22 PM, Moebius wrote:
Am 16.11.2024 um 23:16 schrieb Chris M. Thomasson:
On 11/16/2024 2:11 PM, Moebius wrote:
Am 16.11.2024 um 22:48 schrieb Chris M. Thomasson:
On 11/16/2024 1:29 PM, FromTheRafters wrote:
Chris M. Thomasson pretended:
(infinity - infinity) = undefined
?
https://en.wikipedia.org/wiki/
L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms
Or this one: https://en.wikipedia.org/wiki/Indeterminate_form
"oo - oo" is an "indeterminate form".
I must be missing something: [...]
[1] = a gallon of water out of an infinite pool
[2] = another gallon of water out of an infinite pool
[3] = on and on... taken to infinity...
The pool would always have infinite water for this process?
Assume that the H2O molecules in the pool are "numerated" by 1, 2,
3, ... (i.e. that ALL H2O molecules in the pool are "numerated" by
natural numbers).
Taking out the "infinite amount of H2O molecules" numerated by 1, 2,
3, ... would lead to an EMPTY pool.
On the other hand, taking out the "infinite amount of H2O molecules"
numerated by 2, 4, 6, ... would not drain the pool. 😛 It would still
be filled (sort of) with infinitely many H20 molecules. 😛
Beware if the infinite, man!
Shit.... Humm.... For fun, what about an infinite waterfall dumping into
an already infinite pool, two separate objects. The pool will always
accepts more water since it is infinite. Now, taking an infinite number
of gallons of water out of the pool is interesting because of the
infinite waterfall?
Damn infinity! beware? ;^)
Hint: If the first gallon of water consists of the H2O molecules
numbered by 1, ..., n_1, the second gallon of water consists of the H2O molecules numbered by (n_1)+1, ..., n_2 (with n_2 > (n_1)+1), and so on,
the "outcome" would be an empty pool.
(Hint: try to name the number
attached to an H2O molecule which "remains" in the pool.)
For example "aleph_0 - aleph_0" is not defined.
WM was thinking very hard :
On 16.11.2024 22:33, Moebius wrote:
For example "aleph_0 - aleph_0" is not defined.
Small wonder. ℵo means only infinitely many: |ℕ|, |ℚ|, and many others.
|ℕ|-|ℕ| however is defined.
No, it is not.
WM presented the following explanation :
On 17.11.2024 12:01, FromTheRafters wrote:
WM was thinking very hard :
On 16.11.2024 22:33, Moebius wrote:
For example "aleph_0 - aleph_0" is not defined.
Small wonder. ℵo means only infinitely many: |ℕ|, |ℚ|, and many others.
|ℕ|-|ℕ| however is defined.
No, it is not.
If sets are invariable then ℕ \ ℕ is empty.
If |ℕ| concerns only the elements of ℕ, then |ℕ|-|ℕ|= 0.
So, you're saying that if I take aleph_zero natural numbers and I remove
the aleph_zero odd numbers from consideration in a new set, I will have
a new emptyset instead of E?
Your math is seriously flawed.
WM pretended :
On 17.11.2024 12:38, FromTheRafters wrote:
WM presented the following explanation :
On 17.11.2024 12:01, FromTheRafters wrote:
WM was thinking very hard :
On 16.11.2024 22:33, Moebius wrote:
For example "aleph_0 - aleph_0" is not defined.
Small wonder. ℵo means only infinitely many: |ℕ|, |ℚ|, and many >>>>>> others.
|ℕ|-|ℕ| however is defined.
No, it is not.
If sets are invariable then ℕ \ ℕ is empty.
If |ℕ| concerns only the elements of ℕ, then |ℕ|-|ℕ|= 0.
So, you're saying that if I take aleph_zero natural numbers and I
remove the aleph_zero odd numbers from consideration in a new set, I
will have a new emptyset instead of E?
Try to understand. "aleph_0 - aleph_0" is not defined.
Try to understand that |N| equals aleph_zero.
On 17.11.2024 17:59, FromTheRafters wrote:
WM pretended :
On 17.11.2024 12:38, FromTheRafters wrote:Try to understand that |N| equals aleph_zero.
WM presented the following explanation :
On 17.11.2024 12:01, FromTheRafters wrote:
WM was thinking very hard :
On 16.11.2024 22:33, Moebius wrote:
For example "aleph_0 - aleph_0" is not defined.
Small wonder. ℵo means only infinitely many: |ℕ|, |ℚ|, and many >>>>>>> others. |ℕ|-|ℕ| however is defined.
No, it is not.
If sets are invariable then ℕ \ ℕ is empty.
If |ℕ| concerns only the elements of ℕ, then |ℕ|-|ℕ|= 0.
So, you're saying that if I take aleph_zero natural numbers and I
remove the aleph_zero odd numbers from consideration in a new set, I
will have a new emptyset instead of E?
Try to understand. "aleph_0 - aleph_0" is not defined.
Of course. ℵo equals |ℕ|, equals |ℚ|, equals all countable sets. It is simply another name for infinitely "many". |ℕ| however is a fixed
infinite number. Note that sets are invariable.
Am Sun, 17 Nov 2024 18:38:00 +0100 schrieb WM:
Of course. ℵo equals |ℕ|, equals |ℚ|, equals all countable sets. It is >> simply another name for infinitely "many". |ℕ| however is a fixed
infinite number. Note that sets are invariable.
Yes, those are all the same number.
On 17.11.2024 12:01, FromTheRafters wrote:The number |N|=|Q|=Aleph_0 is the same regardless of the set.
WM was thinking very hard :
On 16.11.2024 22:33, Moebius wrote:
For example "aleph_0 - aleph_0" is not defined.
Small wonder. ℵo means only infinitely many: |ℕ|, |ℚ|, and many
others. |ℕ|-|ℕ| however is defined.
No, it is not.
If sets are invariable then ℕ \ ℕ is empty.
If |ℕ| concerns only the elements of ℕ, then |ℕ|-|ℕ|= 0.
It happens that WM formulated :
On 17.11.2024 17:59, FromTheRafters wrote:
WM pretended :
On 17.11.2024 12:38, FromTheRafters wrote:
WM presented the following explanation :
On 17.11.2024 12:01, FromTheRafters wrote:
WM was thinking very hard :
On 16.11.2024 22:33, Moebius wrote:
For example "aleph_0 - aleph_0" is not defined.
Small wonder. ℵo means only infinitely many: |ℕ|, |ℚ|, and many >>>>>>>> others.
|ℕ|-|ℕ| however is defined.
No, it is not.
If sets are invariable then ℕ \ ℕ is empty.
If |ℕ| concerns only the elements of ℕ, then |ℕ|-|ℕ|= 0.
So, you're saying that if I take aleph_zero natural numbers and I
remove the aleph_zero odd numbers from consideration in a new set,
I will have a new emptyset instead of E?
Try to understand. "aleph_0 - aleph_0" is not defined.
Try to understand that |N| equals aleph_zero.
Of course. ℵo equals |ℕ|, equals |ℚ|, equals all countable sets. It is >> simply another name for infinitely "many". |ℕ| however is a fixed
infinite number. Note that sets are invariable.
But you said both that it equals zero and that it is undefined. You
should pick one and be consistent.
It is sometimes better to think of cardinality as a number indicating a notion of 'set size' rather than a notion of 'number many'. For finite
sets these two notions merge.
Am Sun, 17 Nov 2024 12:19:32 +0100 schrieb WM:
On 17.11.2024 12:01, FromTheRafters wrote:
WM was thinking very hard :
On 16.11.2024 22:33, Moebius wrote:
For example "aleph_0 - aleph_0" is not defined.
Small wonder. ℵo means only infinitely many: |ℕ|, |ℚ|, and many
others. |ℕ|-|ℕ| however is defined.
No, it is not.
If sets are invariable then ℕ \ ℕ is empty.
If |ℕ| concerns only the elements of ℕ, then |ℕ|-|ℕ|= 0.
The number |N|=|Q|=Aleph_0 is the same regardless of the set.
If you define the difference to be zero, it doesn't matter.
WM pretended :
On 17.11.2024 21:59, FromTheRafters wrote:
It is sometimes better to think of cardinality as a number indicating
a notion of 'set size' rather than a notion of 'number many'. For
finite sets these two notions merge.
In the infinite they differ:
ℵo is not the same as |ℕ| although |ℕ| is in the set of numbers
described by ℵo.
aleph_zero is the same object as the cardinality of N.
WM brought next idea :
|ℕ| - |ℕ| = 0 because if you subtract one element from ℕ then you haveno longer ℕ and therefore no longer |ℕ| describing it.
Still wrong.
on 11/18/2024, WM supposed :
On 18.11.2024 18:11, FromTheRafters wrote:
WM pretended :
On 17.11.2024 21:59, FromTheRafters wrote:
It is sometimes better to think of cardinality as a number
indicating a notion of 'set size' rather than a notion of 'number
many'. For finite sets these two notions merge.
In the infinite they differ:
ℵo is not the same as |ℕ| although |ℕ| is in the set of numbers
described by ℵo.
aleph_zero is the same object as the cardinality of N.
Yes, but |ℕ| is the number of elements of ℕ.
Or, more generally, it is the "SIZE" of the set. Size equals the number
of elements in a *FINITE* set.
You said aleph_zero minus aleph_zero was different to the cardinality of
N
aleph_zero *IS* the cardinality of N.
on 11/18/2024, WM supposed :
On 18.11.2024 18:15, FromTheRafters wrote:
WM brought next idea :
|ℕ| - |ℕ| = 0 because if you subtract one element from ℕ then you haveno longer ℕ and therefore no longer |ℕ| describing it.
Still wrong.
If you remove one element from ℕ, then you have still ℵo but no longer >> all elements of ℕ.
But you do have now a proper subset of the naturals the same size as
before.
If |ℕ| describes the number of elements, then it has changed to |ℕ| - 1.
Minus one is not defined.
If you don't like |ℕ| then call this number the number of natural
numbers.
Why would I do that when it is the *SIZE* of the smallest infinite set.
On 17.11.2024 21:16, joes wrote:Duh.
Am Sun, 17 Nov 2024 12:19:32 +0100 schrieb WM:
On 17.11.2024 12:01, FromTheRafters wrote:
WM was thinking very hard :
On 16.11.2024 22:33, Moebius wrote:
For example "aleph_0 - aleph_0" is not defined.
Small wonder. ℵo means only infinitely many: |ℕ|, |ℚ|, and many >>>>> others. |ℕ|-|ℕ| however is defined.
No, it is not.
If sets are invariable then ℕ \ ℕ is empty.
If |ℕ| concerns only the elements of ℕ, then |ℕ|-|ℕ|= 0.
The number |N|=|Q|=Aleph_0 is the same regardless of the set.
Yes it is the expression that the set has infinitely many elements. But
it is not specific to any set.
If you subtract any natural number from ℕ, then you have another set, no longer ℕ. But the cardinality remains ℵo. Therefore it does not describe the set ℕ.I can't follow.
On 18.11.2024 22:58, FromTheRafters wrote:|N\{0}| = |N| = Aleph_0
on 11/18/2024, WM supposed :
On 18.11.2024 18:15, FromTheRafters wrote:
WM brought next idea :
|ℕ| - |ℕ| = 0 because if you subtract one element from ℕ then you >>>>>> have no longer ℕ and therefore no longer |ℕ| describing it.
You can still say it is a subset, like Cantor did with "reality".It has one element less, hence the "size" ℵo is a very unsharp measure.If you remove one element from ℕ, then you have still ℵo but no longer >>> all elements of ℕ.But you do have now a proper subset of the naturals the same size as
before.
And |N\{2}| = Aleph_0.Subtracting an element is defined. |ℕ| - 1 is defined as the number of elements minus 1.If |ℕ| describes the number of elements, then it has changed to |ℕ| - >>> 1.Minus one is not defined.
There are infinitely many of them.The set of prime numbers is smaller.If you don't like |ℕ| then call this number the number of naturalWhy would I do that when it is the *SIZE* of the smallest infinite set.
numbers.
On 18.11.2024 18:15, FromTheRafters wrote:Yes, you have a subset of the same cardinality.
WM brought next idea :
|ℕ| - |ℕ| = 0 because if you subtract one element from ℕ then you >>>> have no longer ℕ and therefore no longer |ℕ| describing it.
If you remove one element from ℕ, then you have still ℵo but no longer all elements of ℕ.
If |ℕ| describes the number of elements, then it has|N| - 1 = Aleph_0 - 1 = Aleph_0
changed to |ℕ| - 1.
WM used his keyboard to write :
N is the set, and |N| is the size of the set.
aleph_zero *IS* the cardinality of N.
Of course, but it is also the cardinality of all other countable sets
which have more or fewer elements.
No, only countably infinite sets. You have that all finite sets are
countable too and have fewer elements.
Am Mon, 18 Nov 2024 23:16:22 +0100 schrieb WM:
|N\{0}| = |N| = Aleph_0
And |N\{2}| = Aleph_0.
Am Mon, 18 Nov 2024 19:20:24 +0100 schrieb WM:
If |ℕ| describes the number of elements, then it has|N| - 1 = Aleph_0 - 1 = Aleph_0
changed to |ℕ| - 1.
WM wrote on 11/18/2024 :
On 18.11.2024 22:58, FromTheRafters wrote:
on 11/18/2024, WM supposed :
On 18.11.2024 18:15, FromTheRafters wrote:
WM brought next idea :
|ℕ| - |ℕ| = 0 because if you subtract one element from ℕ then you >>>>>>> haveno longer ℕ and therefore no longer |ℕ| describing it.
Still wrong.
If you remove one element from ℕ, then you have still ℵo but no
longer all elements of ℕ.
But you do have now a proper subset of the naturals the same size as
before.
It has one element less, hence the "size" ℵo is a very unsharp measure.
Comparing the size of sets by bijection. Bijection of finite sets give
you a same number of elements, bijection of infinite sets give you same
size of set.
If |ℕ| describes the number of elements, then it has changed to |ℕ| >>>> - 1.
Minus one is not defined.
Subtracting an element is defined. |ℕ| - 1 is defined as the number of
elements minus 1.
Nope!
If you don't like |ℕ| then call this number the number of natural
numbers.
Why would I do that when it is the *SIZE* of the smallest infinite set.
The set of prime numbers is smaller.
No, it is not.
There is a bijection.
On 18.11.2024 23:40, FromTheRafters wrote:
WM wrote on 11/18/2024 :Why? Because only potential infinity is involved. True bijections pr5ove equinumerosity.
On 18.11.2024 22:58, FromTheRafters wrote:Comparing the size of sets by bijection. Bijection of finite sets give
on 11/18/2024, WM supposed :It has one element less, hence the "size" ℵo is a very unsharp
On 18.11.2024 18:15, FromTheRafters wrote:But you do have now a proper subset of the naturals the same size as
WM brought next idea :If you remove one element from ℕ, then you have still ℵo but no
|ℕ| - |ℕ| = 0 because if you subtract one element from ℕ then you
have no longer ℕ and therefore no longer |ℕ| describing it.
longer all elements of ℕ.
before.
measure.
you a same number of elements, bijection of infinite sets give you same
size of set.
And what, pray tell, is Aleph_0 - 1 ?The number of ℕ \ {1} is 1 less than ℕ.Nope!Subtracting an element is defined. |ℕ| - 1 is defined as the number of >>> elements minus 1.If |ℕ| describes the number of elements, then it has changed to |ℕ| >>>>> - 1.Minus one is not defined.
It is a subset.It is, because 4 and 8 are missing.No, it is not.The set of prime numbers is smaller.If you don't like |ℕ| then call this number the number of naturalWhy would I do that when it is the *SIZE* of the smallest infinite
numbers.
set.
All of them do.There is a bijection.Only between numbers which have more successors than predecessors,
although it is claimed that no successors are remaining.--
Am Tue, 19 Nov 2024 16:24:52 +0100 schrieb WM:
Why? Because only potential infinity is involved. True bijections prove
equinumerosity.
What is a "true" bijection?
The number of ℕ \ {1} is 1 less than ℕ.And what, pray tell, is Aleph_0 - 1 ?
> There is a bijection.All of them do.
Only between numbers which have more successors than predecessors,
WM submitted this idea :
There is a bijection.
Only between numbers which have more successors than predecessors,
although it is claimed that no successors are remaining.
You are not making any sense.
WM was thinking very hard :
Only finite sets are countable.
Wrong, it is countable if there is a bijection.
But all "countably" infinite sets have the same size, which according
to the different numbers of elements is a very unsharp measure.
That you don't like cardinal arithmetic, doesn't make cardinal
arithmetic wrong.
WM explained on 11/20/2024 :
set theory claims that all natural
numbers can be counted to such that no successors remain.
No it doesn't.
WM formulated on Wednesday :
It does not make it wrong, but it unmasks it at imprecise. That's why
I don't like it. We can do better.
It works well enough.
On 20.11.2024 15:15, FromTheRafters wrote:
WM explained on 11/20/2024 :
set theory claims that all natural numbers can be counted to such that
no successors remain.
No it doesn't.
Even all rationals and algebraics.
"we get the epitome (ω) of all real algebraic numbers [...] and with
respect to this order we can talk about the nth algebraic number where
not a single one of this epitome has been forgotten"
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place"
On 19.11.2024 17:27, joes wrote:And a "false" one?
Am Tue, 19 Nov 2024 16:24:52 +0100 schrieb WM:
A true bijection between two sets proves that both sets have the sameWhy? Because only potential infinity is involved. True bijectionsWhat is a "true" bijection?
prove equinumerosity.
number of elements.
Great.It is "infinitely many" like Aleph_0.The number of ℕ \ {1} is 1 less than ℕ.And what, pray tell, is Aleph_0 - 1 ?
Exactly, even though I lost track of which two sets.Then the numbers of elements would be identical.> There is a bijection.All of them do.
Only between numbers which have more successors than predecessors,
On 19.11.2024 17:42, FromTheRafters wrote:It is not claimed.
WM submitted this idea :
You are not making any sense.There is a bijection.Only between numbers which have more successors than predecessors,
although it is claimed that no successors are remaining.
Try to count to a natural number that has fewer successors thanBecause there are no such numbers.
predecessors. Impossible.
But set theory claims that all natural numbersFuck no. Get your quantifiers in order: every single natural number
can be counted to such that no successors remain.
Am Wed, 20 Nov 2024 17:51:19 +0100 schrieb WM:
On 20.11.2024 15:15, FromTheRafters wrote:
WM explained on 11/20/2024 :
set theory claims that all natural numbers can be counted to such that >>>> no successors remain.
No it doesn't.
Even all rationals and algebraics.
"we get the epitome (ω) of all real algebraic numbers [...] and with
respect to this order we can talk about the nth algebraic number where
not a single one of this epitome has been forgotten"
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place"
You are once again lacking in precision:
every natural is finite and thus
countable.
Am Wed, 20 Nov 2024 13:04:04 +0100 schrieb WM:
Try to count to a natural number that has fewer successors thanBecause there are no such numbers.
predecessors. Impossible.
But set theory claims that all natural numberset your quantifiers in order:
can be counted to such that no successors remain.
every single natural number
is very clearly finite;
the cardinal number corresponding
to the set of all of them is countably infinite.
Am Wed, 20 Nov 2024 12:57:22 +0100 schrieb WM:
On 19.11.2024 17:27, joes wrote:And a "false" one?
Am Tue, 19 Nov 2024 16:24:52 +0100 schrieb WM:A true bijection between two sets proves that both sets have the same
Why? Because only potential infinity is involved. True bijectionsWhat is a "true" bijection?
prove equinumerosity.
number of elements.
Great.It is "infinitely many" like Aleph_0.The number of ℕ \ {1} is 1 less than ℕ.And what, pray tell, is Aleph_0 - 1 ?
WM was thinking very hard :
On 20.11.2024 15:15, FromTheRafters wrote:
WM explained on 11/20/2024 :
set theory claims that all natural numbers can be counted to such
that no successors remain.
No it doesn't.
Even all rationals and algebraics.
"we get the epitome (ω) of all real algebraic numbers [...] and with
respect to this order we can talk about the nth algebraic number where
not a single one of this epitome has been forgotten"
"The infinite sequence thus defined has the peculiar property to
contain the positive rational numbers completely, and each of them
only once at a determined place"
Which doesn't support your claim at all. No mention whatsoever of no successors remaining.
On 11/20/2024 12:05 PM, FromTheRafters wrote:
WM wrote on 11/20/2024 :
What colour has the real axis after you have solved both tasks?
Depending on the order of the tasks. I think half red or half black.
Well you have to reference academic reference and describe "supertask"
besides "asymptotics" about where "the asymptotic density of black or
red respectively is 1 in the limit",
WM wrote on 11/20/2024 :
On 20.11.2024 15:14, FromTheRafters wrote:
WM formulated on Wednesday :
It does not make it wrong, but it unmasks it at imprecise. That's
why I don't like it. We can do better.
It works well enough.
Really? Then you can answer the following questions:
Let every unit interval after a natural number on the real axis be
coloured white with exception of the powers of 2 which are coloured
black. Is it possible to shift the black intervals so that the whole
real axis becomes black?
No, of course not.
Or: Let every unit interval after a natural number on the real axis be
coloured as above with exception of the intervals after the odd prime
numbers which are coloured red. Is it possible to shift the red
intervals so that the whole real axis becomes red?
No, of course not.
What colour has the real axis after you have solved both tasks?
Depending on the order of the tasks. I think half red or half black.
On 20.11.2024 19:12, joes wrote:You misunderstood him. I don't see anything about successors.
Am Wed, 20 Nov 2024 17:51:19 +0100 schrieb WM:It was Cantor who said the above. There is no lack of precision.
On 20.11.2024 15:15, FromTheRafters wrote:You are once again lacking in precision:
WM explained on 11/20/2024 :Even all rationals and algebraics.
set theory claims that all natural numbers can be counted to suchNo it doesn't.
that no successors remain.
"we get the epitome (ω) of all real algebraic numbers [...] and with
respect to this order we can talk about the nth algebraic number where
not a single one of this epitome has been forgotten"
"The infinite sequence thus defined has the peculiar property to
contain the positive rational numbers completely, and each of them
only once at a determined place"
Numbers "missing" is meaningless. What did you mean to say here?every natural is finite and thus countable.According to Cantor there is no number missing, let alone infinitely
many.
Set theory claims that all natural numbers can be counted to suchObviously. There is no end to the successors such that you are done
that no successors remain. That is false.
On 20.11.2024 19:18, joes wrote:So?
Am Wed, 20 Nov 2024 13:04:04 +0100 schrieb WM:
All successors are natural numbers.Try to count to a natural number that has fewer successors thanBecause there are no such numbers.
predecessors. Impossible.
If all can be counted, then no successors remain.All at once or every single one?
You have shown that you don't understand them.That is a foolish excuse.But set theory claims that all natural numbers can be counted to suchet your quantifiers in order:
that no successors remain.
There are countably infinite numbers, but ok.every single natural number is very clearly finite;Every number that can be counted to is finite.
But every number that canEvery number, period. There is no number without successors.
be counted to has more successors than predecessors.
Therefore not every number can be counted to.Well, the ordinal numbers less than epsilon_0 are called countably
WTF there is no largest number. How do you think counting changesthe cardinal number corresponding to the set of all of them isThe set of all numbers that can be counted to is finite, namely a
countably infinite.
number that is counted to. This cannot change by counting.
According to Cantor there is no number missing, let alone infinitely
many. [WM]
On 11/20/2024 3:57 PM, joes wrote:
Am Wed, 20 Nov 2024 19:37:20 +0100 schrieb WM:
According to Cantor there is no number missing, let alone infinitely
many.
Numbers "missing" is meaningless. [...]
On 11/20/2024 4:11 PM, Chris M. Thomasson wrote:
On 11/20/2024 4:10 PM, Chris M. Thomasson wrote:
On 11/20/2024 3:57 PM, joes wrote:
Am Wed, 20 Nov 2024 19:37:20 +0100 schrieb WM:
On 20.11.2024 19:12, joes wrote:You misunderstood him. I don't see anything about successors.
Am Wed, 20 Nov 2024 17:51:19 +0100 schrieb WM:It was Cantor who said the above. There is no lack of precision.
On 20.11.2024 15:15, FromTheRafters wrote:You are once again lacking in precision:
WM explained on 11/20/2024 :Even all rationals and algebraics.
set theory claims that all natural numbers can be counted to such >>>>>>>>> that no successors remain.No it doesn't.
"we get the epitome (ω) of all real algebraic numbers [...] and with >>>>>>> respect to this order we can talk about the nth algebraic number >>>>>>> where
not a single one of this epitome has been forgotten"
"The infinite sequence thus defined has the peculiar property to >>>>>>> contain the positive rational numbers completely, and each of them >>>>>>> only once at a determined place"
Numbers "missing" is meaningless. What did you mean to say here?every natural is finite and thus countable.According to Cantor there is no number missing, let alone infinitely >>>>> many.
Set theory claims that all natural numbers can be counted to suchObviously. There is no end to the successors such that you are done
that no successors remain. That is false.
counting them after some finite number.
A balanced scale with no weights on either side. Place a unit weight
on the right side and say this equal to 0 + 1. Wrt unit weight...
? ;^) lol.
say balanced condition is zero. So, with the single unit weight on the
right, place a unit weight on the left. We got a balanced scale
representing 1 - 1 - 0?
1-1=0 DAMN
TYPOS! GRRRR! Sorry Everybody! ;^o
WM explained on 11/20/2024 :
"not a single one of this epitome has been forgotten" means not a
single one remaining, let alone more than one.
No it doesn't, it means the system used to list them is inherently
unable to miss any.
Am Wed, 20 Nov 2024 19:42:41 +0100 schrieb WM:
On 20.11.2024 19:18, joes wrote:So?
Am Wed, 20 Nov 2024 13:04:04 +0100 schrieb WM:All successors are natural numbers.
Try to count to a natural number that has fewer successors thanBecause there are no such numbers.
predecessors. Impossible.
If all can be counted, then no successors remain.All at once or every single one?
But every number that canEvery number, period. There is no number without successors.
be counted to has more successors than predecessors.
After serious thinking WM wrote :
On 20.11.2024 21:05, FromTheRafters wrote:
WM wrote on 11/20/2024 :
On 20.11.2024 15:14, FromTheRafters wrote:
WM formulated on Wednesday :
It does not make it wrong, but it unmasks it at imprecise. That's
why I don't like it. We can do better.
It works well enough.
Really? Then you can answer the following questions:
Let every unit interval after a natural number on the real axis be
coloured white with exception of the powers of 2 which are coloured
black. Is it possible to shift the black intervals so that the whole
real axis becomes black?
No, of course not.
Thank you. You are the first person not bewitched by set theory.
The real line is infinitely long in both directions. Your shifted
intervals are all on the positive real axis.
Depending on the order of the tasks. I think half red or half black.
Recidivistic?
I recognize McDuck in disguise, or your notion that there are fewer
primes than there are natural numbers. You keep finding different ways
to say the same wrong thing.
According to Cantor there is no number missing, let alone infinitely
many. [WM]
Stelle aus Cantors Oevre zitieren, wo er sagt, that "there is no number missing"?
WM formulated the question :
On 21.11.2024 00:04, FromTheRafters wrote:
WM explained on 11/20/2024 :
"not a single one of this epitome has been forgotten" means not a
single one remaining, let alone more than one.
No it doesn't, it means the system used to list them is inherently
unable to miss any.
None is missed. That is no successors remain missed.
It says nothing about successors at all, just that there is a systematic method for listing them without missing any.
WM wrote :
Let every unit interval after a natural number on the real axis be >>>>>> coloured white with exception of the powers of 2 which are
coloured black. Is it possible to shift the black intervals so
that the whole real axis becomes black?
The analytical limit is wrong in your opinion?
Which limit?
Which limit?
On 21.11.2024 14:10, FromTheRafters wrote:What successors are you talking about?
WM formulated the question :
On 21.11.2024 00:04, FromTheRafters wrote:
WM explained on 11/20/2024 :None is missed. That is no successors remain missed.
"not a single one of this epitome has been forgotten" means not aNo it doesn't, it means the system used to list them is inherently
single one remaining, let alone more than one.
unable to miss any.
You are mistaken. The successors are also all listed - we are talkingIt says nothing about successors at all, just that there is aAnd that is wrong because for every listed natural number almost all are
systematic method for listing them without missing any.
not listed.
On 21.11.2024 14:05, FromTheRafters wrote:Yes, of the infinite real axis none are black.
WM wrote :
Let every unit interval after a natural number on the real axis be >>>>>>> coloured white with exception of the powers of 2 which are
coloured black. Is it possible to shift the black intervals so
that the whole real axis becomes black?
The limit of the sequence f(n) of relative coverings in the interval (0,The analytical limit is wrong in your opinion?Which limit?
n] is 0, not 1.
I mean, bijections are also complete for every finite subset.Which limit?That is where Cantor's bijections are believed to be complete.
On 21.11.2024 01:08, joes wrote:So?
Am Wed, 20 Nov 2024 19:42:41 +0100 schrieb WM:
On 20.11.2024 19:18, joes wrote:
Am Wed, 20 Nov 2024 13:04:04 +0100 schrieb WM:All successors are natural numbers.
Try to count to a natural number that has fewer successors thanBecause there are no such numbers.
predecessors. Impossible.
Every single natural can be counted to.Counting concerns every single number.If all can be counted, then no successors remain.All at once or every single one?
Not finitely.Then not all can be counted.But every number that can be counted to has more successors thanEvery number, period. There is no number without successors.
predecessors.
Am Thu, 21 Nov 2024 17:55:25 +0100 schrieb WM:
On 21.11.2024 14:05, FromTheRafters wrote:Yes, of the infinite real axis none are black.
WM wrote :The limit of the sequence f(n) of relative coverings in the interval (0,
Let every unit interval after a natural number on the real axis be >>>>>>>> coloured white with exception of the powers of 2 which are
coloured black. Is it possible to shift the black intervals so >>>>>>>> that the whole real axis becomes black?
The analytical limit is wrong in your opinion?Which limit?
n] is 0, not 1.
I mean, bijections are also complete for every finite subset.Which limit?That is where Cantor's bijections are believed to be complete.
Am Thu, 21 Nov 2024 12:50:19 +0100 schrieb WM:
On 21.11.2024 01:08, joes wrote:So?
Am Wed, 20 Nov 2024 19:42:41 +0100 schrieb WM:
On 20.11.2024 19:18, joes wrote:
Am Wed, 20 Nov 2024 13:04:04 +0100 schrieb WM:All successors are natural numbers.
Try to count to a natural number that has fewer successors thanBecause there are no such numbers.
predecessors. Impossible.
Every single natural can be counted to.Counting concerns every single number.If all can be counted, then no successors remain.All at once or every single one?
On 20.11.2024 19:20, joes wrote:Determined by the sets differing only in a finite number of elements?
Am Wed, 20 Nov 2024 12:57:22 +0100 schrieb WM:
On 19.11.2024 17:27, joes wrote:
Am Tue, 19 Nov 2024 16:24:52 +0100 schrieb WM:A true bijection between two sets proves that both sets have the same
Why? Because only potential infinity is involved. True bijectionsWhat is a "true" bijection?
prove equinumerosity.
number of elements.
Thanks for agreeing with |N| = |N\{0}|.And a "false" one?Does allow for very different numbers like prime numbers and algebraic numbers.
Infinitely great.Great.It is "infinitely many" like Aleph_0.The number of ℕ \ {1} is 1 less than ℕ.And what, pray tell, is Aleph_0 - 1 ?
On 11/21/2024 1:45 PM, WM wrote:
On 21.11.2024 22:05, joes wrote:
Am Thu, 21 Nov 2024 12:50:19 +0100 schrieb WM:
On 21.11.2024 01:08, joes wrote:So?
Am Wed, 20 Nov 2024 19:42:41 +0100 schrieb WM:
On 20.11.2024 19:18, joes wrote:
Am Wed, 20 Nov 2024 13:04:04 +0100 schrieb WM:All successors are natural numbers.
Try to count to a natural number that has fewer successors than >>>>>>>> predecessors. Impossible.Because there are no such numbers.
If all can be counted, then no successors remain.
Every single natural can be counted to.All at once or every single one?Counting concerns every single number.
Nonsense.
What one cannot be counted to?
Am Wed, 20 Nov 2024 19:44:39 +0100 schrieb WM:
Determined by the sets differing only in a finite number of elements?
A true bijection between two sets proves that both sets have the sameWhy? Because only potential infinity is involved. True bijectionsWhat is a "true" bijection?
prove equinumerosity.
number of elements.
Thanks for agreeing with |N| = |N\{0}|.And a "false" one?Does allow for very different numbers like prime numbers and algebraic
numbers.
Infinitely great.Great.It is "infinitely many" like Aleph_0.The number of ℕ \ {1} is 1 less than ℕ.And what, pray tell, is Aleph_0 - 1 ?
On 11/21/2024 1:45 PM, WM wrote:
Counting concerns every single number.Every single natural can be counted to.
Nonsense.
What one cannot be counted to?
Am 22.11.2024 um 03:58 schrieb Chris M. Thomasson:
On 11/21/2024 1:45 PM, WM wrote:
On 21.11.2024 22:05, joes wrote:
Counting concerns every single number.Every single natural can be counted to.
Nonsense.
Proof by induction:
1 can be counted to (obviously). If n (where n is a natural number) can
be counted to, then n+1 can be counted to (obviously). Hence for each
and every natural numbers n: n can be counted too. qed
What one cannot be counted to?
On 21.11.2024 21:59, joes wrote:Less than every positive number = zero. Also I was pulling your leg.
Am Thu, 21 Nov 2024 17:55:25 +0100 schrieb WM:Not none but less than any definable eps.
On 21.11.2024 14:05, FromTheRafters wrote:Yes, of the infinite real axis none are black.
WM wrote :The limit of the sequence f(n) of relative coverings in the interval
Which limit?The analytical limit is wrong in your opinion?Let every unit interval after a natural number on the real axis >>>>>>>>> be coloured white with exception of the powers of 2 which are >>>>>>>>> coloured black. Is it possible to shift the black intervals so >>>>>>>>> that the whole real axis becomes black?
(0, n] is 0, not 1.
Haha what I was talking about a function likeNo, for any finite subset (0, n] there is no complete bijection. Too few black unit intervals available to cover (0, n].I mean, bijections are also complete for every finite subset.Which limit?That is where Cantor's bijections are believed to be complete.
On 21.11.2024 23:45, joes wrote:Ah yes, the sets {-1} u N and {-2} u N have different sizes.
Am Wed, 20 Nov 2024 19:44:39 +0100 schrieb WM:Not differing at all.
Determined by the sets differing only in a finite number of elements?
A true bijection between two sets proves that both sets have theWhy? Because only potential infinity is involved. True bijections >>>>>>> prove equinumerosity.What is a "true" bijection?
same number of elements.
Good. Then we can consider those sets to have the same number.Of course. ℵo means nothing but infinitely many.Thanks for agreeing with |N| = |N\{0}|.It is "infinitely many" like Aleph_0.The number of ℕ \ {1} is 1 less than ℕ.And what, pray tell, is Aleph_0 - 1 ?
On 22.11.2024 08:49, Moebius wrote:True. But not the intersection of all of them, since that is the (ugh)
Am 22.11.2024 um 03:58 schrieb Chris M. Thomasson:
On 11/21/2024 1:45 PM, WM wrote:
On 21.11.2024 22:05, joes wrote:
Induction proves that every initial segment of endsegments has anProof by induction:Nonsense.Counting concerns every single number.Every single natural can be counted to.
infinite intersection.
Yes they do (with what?).1 can be counted to (obviously). If n (where n is a natural number) canBut not all endsegments have an infinite intersection.
be counted to, then n+1 can be counted to (obviously). Hence for each
and every natural numbers n: n can be counted too. qed
All endsegments have an empty intersection.No, all segments have an infinite intersection with each other,
Since every endsegment can lose only oneExactly. That is all of them, there are infinitely segments.
number, there must be infinitely many endsegments involved in reducing
the intersection from infinite to empty.
They don't exist. That is why they are disregarded.Just the indices involved in reducing the intersection from infinite to empty. They are dark.What one cannot be counted to?
Am Fri, 22 Nov 2024 13:00:52 +0100 schrieb WM:
Good. Then we can consider those sets to have the same number.Of course. ℵo means nothing but infinitely many.Thanks for agreeing with |N| = |N\{0}|.It is "infinitely many" like Aleph_0.The number of ℕ \ {1} is 1 less than ℕ.And what, pray tell, is Aleph_0 - 1 ?
Am Fri, 22 Nov 2024 13:08:28 +0100 schrieb WM:
All endsegments have an empty intersection.No, all segments have an infinite intersection with each other,
namely the one that comes "later", with a larger index.
Since every endsegment can lose only oneExactly. That is all of them, there are infinitely segments.
number, there must be infinitely many endsegments involved in reducing
the intersection from infinite to empty.
They don't exist.Just the indices involved in reducing the intersection from infinite toWhat one cannot be counted to?
empty. They are dark.
On 22.11.2024 13:32, joes wrote:As it should. You can give each prime an index.
Am Fri, 22 Nov 2024 13:00:52 +0100 schrieb WM:
That is the big mistake. It makes you think that the sets of naturalsGood. Then we can consider those sets to have the same number.Of course. ℵo means nothing but infinitely many.Thanks for agreeing with |N| = |N\{0}|.It is "infinitely many" like Aleph_0.The number of ℕ \ {1} is 1 less than ℕ.And what, pray tell, is Aleph_0 - 1 ?
and of prime numbers could cover each other.
Am Fri, 22 Nov 2024 15:51:11 +0100 schrieb WM:
On 22.11.2024 13:32, joes wrote:As it should. You can give each prime an index.
> Am Fri, 22 Nov 2024 13:00:52 +0100 schrieb WM:
>>>>>>>> The number of ℕ \ {1} is 1 less than ℕ.
>>>>>>> And what, pray tell, is Aleph_0 - 1 ?
>>>>>> It is "infinitely many" like Aleph_0.
>>> Thanks for agreeing with |N| = |N\{0}|.
>> Of course. ℵo means nothing but infinitely many.
> Good. Then we can consider those sets to have the same number.
That is the big mistake. It makes you think that the sets of naturals
and of prime numbers could cover each other.
On 22.11.2024 13:32, joes wrote:
Then we can consider those sets
to have the same number.
That is the big mistake.
It makes you think that
the sets of naturals and of prime numbers
could cover each other.
On 22.11.2024 16:11, joes wrote:
Am Fri, 22 Nov 2024 15:51:11 +0100 schrieb WM:
the sets of naturals and of prime numbers [can] cover each other.As it should. You can give each prime an index.
On 22.11.2024 16:11, joes wrote:
Am Fri, 22 Nov 2024 15:51:11 +0100 schrieb WM:
the sets of naturals and of prime numbers [can] cover each other.As it should. You can give each prime an index.
Indeed! The two formulas
| p(1) = min P
| p(n+1) = min {p e P : p > p(n)} (for all n e IN)
(recursively) define the function p: IN --> P. Where IN is the set of
all natural numbers and P is the set of all prime numbers.
Hint: p(1) = 2, p(2) = 3, p(3) = 5, ...
Actually, if p e P, then there is an (index) n e IN such that p(n) = p.
(It's easy to prove that p: IN --> P is a bijection.)
On 22.11.2024 16:11, joes wrote:
Am Fri, 22 Nov 2024 15:51:11 +0100 schrieb WM:
the sets of naturals and of prime numbers [can] cover each other.As it should. You can give each prime an index.
Indeed! The two formulas
| p(1) = min P
| p(n+1) = min {p e P : p > p(n)} (for all n e IN)
(recursively) define the function p: IN --> P. Where IN is the set of
all natural numbers and P is the set of all prime numbers.
Hint: p(1) = 2, p(2) = 3, p(3) = 5, ...
Actually, if p e P, then there is an (index) n e IN such that p(n) = p.
(It's easy to prove that p: IN --> P is a bijection.)
On 22.11.2024 16:11, joes wrote:
Am Fri, 22 Nov 2024 15:51:11 +0100 schrieb WM:
the sets of naturals and of prime numbers [can] cover each other.As it should. You can give each prime an index.
Indeed! The two formulas
| p(1) = min P
| p(n+1) = min {p e P : p > p(n)} (for all n e IN)
(recursively) define the function p: IN --> P. Where IN is the set of
all natural numbers and P is the set of all prime numbers.
Hint: p(1) = 2, p(2) = 3, p(3) = 5, ...
Actually, if p e P, then there is an (index) n e IN such that p(n) = p.
(It's easy to prove that p: IN --> P is a bijection.)
After serious thinking WM wrote :
On 22.11.2024 16:11, joes wrote:
Am Fri, 22 Nov 2024 15:51:11 +0100 schrieb WM:
On 22.11.2024 13:32, joes wrote:As it should. You can give each prime an index.
> Am Fri, 22 Nov 2024 13:00:52 +0100 schrieb WM:
>>>>>>>> The number of ℕ \ {1} is 1 less than ℕ.
>>>>>>> And what, pray tell, is Aleph_0 - 1 ?
>>>>>> It is "infinitely many" like Aleph_0.
>>> Thanks for agreeing with |N| = |N\{0}|.
>> Of course. ℵo means nothing but infinitely many.
> Good. Then we can consider those sets to have the same number.
That is the big mistake. It makes you think that the sets of naturals
and of prime numbers could cover each other.
But not every index a prime.
Why not? It looks like a bijection to me.
Am 23.11.2024 um 06:18 schrieb Moebius:
On 22.11.2024 16:11, joes wrote:
Am Fri, 22 Nov 2024 15:51:11 +0100 schrieb WM:
the sets of naturals and of prime numbers [can] cover each other.As it should. You can give each prime an index.
Indeed! The two formulas
| p(1) = min P
| p(n+1) = min {p e P : p > p(n)} (for all n e IN)
(recursively) define the function p: IN --> P. Where IN is the set of
all natural numbers and P is the set of all prime numbers.
Hint: p(1) = 2, p(2) = 3, p(3) = 5, ...
Actually, if p e P, then there is an (index) n e IN such that p(n) = p.
(It's easy to prove that p: IN --> P is a bijection.)
p(n) is the n-th prime number in the sequence of prime numbers ordered
by size.
ℙ covers ℕ, and ℕ covers ℙ
WM laid this down on his screen :
Let every unit interval after a natural number on the real axis be
coloured white with exception of the intervals after the prime numbers
which are coloured red. It is impossible to shift the red intervals so
that the whole real axis becomes red. Every interval (10n, 10 (n+1)]
is deficient - on the whole real axis.
So what? Your imaginings don't affect the fact that there is a bijection.
WM has brought this to us :
On 23.11.2024 13:20, FromTheRafters wrote:
WM laid this down on his screen :
Let every unit interval after a natural number on the real axis be
coloured white with exception of the intervals after the prime
numbers which are coloured red. It is impossible to shift the red
intervals so that the whole real axis becomes red. Every interval
(10n, 10 (n+1)] is deficient - on the whole real axis.
So what? Your imaginings don't affect the fact that there is a
bijection.
If there was a bijection,
There is.
then the whole axis could become red.
What makes you think that?
Do you believe that?
No, of course not.
WM wrote on 11/23/2024 :
On 23.11.2024 13:35, FromTheRafters wrote:
WM has brought this to us :
On 23.11.2024 13:20, FromTheRafters wrote:
WM laid this down on his screen :
Let every unit interval after a natural number on the real axis be >>>>>> coloured white with exception of the intervals after the prime
numbers which are coloured red. It is impossible to shift the red
intervals so that the whole real axis becomes red. Every interval
(10n, 10 (n+1)] is deficient - on the whole real axis.
So what? Your imaginings don't affect the fact that there is a
bijection.
If there was a bijection,
There is.
then the whole axis could become red.
What makes you think that?
A bijection proves that every prime number (and its colour) can be put
to a natural number (and colour it).
???
On 23.11.2024 19:43, FromTheRafters wrote:You do know the proof that there are infinitely many primes?
WM wrote on 11/23/2024 :A bijection between natural numbers and prime numbers proves that for
On 23.11.2024 13:35, FromTheRafters wrote:
WM has brought this to us :
On 23.11.2024 13:20, FromTheRafters wrote:There is.
WM laid this down on his screen :
Let every unit interval after a natural number on the real axis be >>>>>>> coloured white with exception of the intervals after the prime
numbers which are coloured red. It is impossible to shift the red >>>>>>> intervals so that the whole real axis becomes red. Every interval >>>>>>> (10n, 10 (n+1)] is deficient - on the whole real axis.
So what? Your imaginings don't affect the fact that there is a
bijection.
If there was a bijection,
then the whole axis could become red.What makes you think that?
every prime number there is a natural number: p_1, p_2, p_3, ...
If that is correct, then there are as many natural numbers as primeThat is the world we live in.
numbers and as many prime numbers as natural numbers. Then the following scenario is possible:
Cover the unit intervals of prime numbers by red hats. Then shift the
red hats so that all unit intervals of the positive real axis get red
hats.
Am Sat, 23 Nov 2024 20:40:40 +0100 schrieb WM:
A bijection between natural numbers and prime numbers proves that forYou do know the proof that there are infinitely many primes?
every prime number there is a natural number: p_1, p_2, p_3, ...
If that is correct, then there are as many natural numbers as primeThat is the world we live in.
numbers and as many prime numbers as natural numbers. Then the following
scenario is possible:
Cover the unit intervals of prime numbers by red hats. Then shift the
red hats so that all unit intervals of the positive real axis get red
hats.
On 11/23/2024 5:30 AM, WM wrote:
On 22.11.2024 22:50, Jim Burns wrote:
ℙ covers ℕ, and ℕ covers ℙ
Let every unit interval on
the infinite real axis
be coloured white. Cover the unit intervals of prime numbers
by red hats.
It is impossible to shift the red hats
Yes, because we are finite beings,
and there are infinitely.many red hats.
It is impossible to shift the red hats
so that all unit intervals of
the whole real axis get red hats.
There are too few prime numbers.
No.
⎛ Assume that that is so.
⎜ Assume that there are
⎜ enough red hats for the first 𝔊 numbers
⎜ but not enough for the 𝔊+1ᵗʰ
There are too few prime numbers.
No,
there being too few primes
leads to contradiction.
On 22.11.2024 22:50, Jim Burns wrote:
ℙ covers ℕ, and ℕ covers ℙ
Let every unit interval on
the infinite real axis
be coloured white.
Cover the unit intervals of prime numbers
by red hats.
It is impossible to shift the red hats
It is impossible to shift the red hats
so that all unit intervals of
the whole real axis
get red hats.
There are too few prime numbers.
There are too few prime numbers.
On 23.11.2024 19:43, FromTheRafters wrote:
WM wrote on 11/23/2024 :
On 23.11.2024 13:35, FromTheRafters wrote:
WM has brought this to us :
On 23.11.2024 13:20, FromTheRafters wrote:
WM laid this down on his screen :
Let every unit interval after a natural number on the real axis
be coloured white with exception of the intervals after the prime >>>>>>> numbers which are coloured red. It is impossible to shift the red >>>>>>> intervals so that the whole real axis becomes red. Every interval >>>>>>> (10n, 10 (n+1)] is deficient - on the whole real axis.
So what? Your imaginings don't affect the fact that there is a
bijection.
If there was a bijection,
There is.
then the whole axis could become red.
What makes you think that?
A bijection proves that every prime number (and its colour) can be
put to a natural number (and colour it).
???
A bijection between natural numbers and prime numbers proves that for
every prime number there is a natural number: p_1, p_2, p_3, ...
If that is correct, then there are as many natural numbers as prime
numbers and as many prime numbers as natural numbers. Then the following scenario is possible:
Cover the unit intervals of prime numbers by red hats. Then shift the
red hats so that all unit intervals of the positive real axis get red hats.
Regards, WM
On 11/23/24 2:40 PM, WM wrote:
On 23.11.2024 19:43, FromTheRafters wrote:
WM wrote on 11/23/2024 :
On 23.11.2024 13:35, FromTheRafters wrote:
WM has brought this to us :
On 23.11.2024 13:20, FromTheRafters wrote:
WM laid this down on his screen :
Let every unit interval after a natural number on the real axis >>>>>>>> be coloured white with exception of the intervals after theSo what? Your imaginings don't affect the fact that there is a
prime numbers which are coloured red. It is impossible to shift >>>>>>>> the red intervals so that the whole real axis becomes red. Every >>>>>>>> interval (10n, 10 (n+1)] is deficient - on the whole real axis. >>>>>>>
bijection.
If there was a bijection,
There is.
then the whole axis could become red.
What makes you think that?
A bijection proves that every prime number (and its colour) can be
put to a natural number (and colour it).
???
A bijection between natural numbers and prime numbers proves that for
every prime number there is a natural number: p_1, p_2, p_3, ...
If that is correct, then there are as many natural numbers as prime
numbers and as many prime numbers as natural numbers. Then the
following scenario is possible:
Cover the unit intervals of prime numbers by red hats. Then shift the
red hats so that all unit intervals of the positive real axis get red
hats.
Regards, WM
And you can, as
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
and in general, the red hat on the nth prime number can be moved to the number n
Since there are a countable infinite number of prime numbers, there
exist an nth prime number for every n,
so all the numbers get covered.
We have a 1:1 relationship (bijection) established between the set of
prime numbers and the set of Natural Numbers.
On 23.11.2024 21:18, Jim Burns wrote:No. That other place where we take the hats from, larger primes, are of
On 11/23/2024 5:30 AM, WM wrote:
On 22.11.2024 22:50, Jim Burns wrote:
ℙ covers ℕ, and ℕ covers ℙ
Let every unit interval on the infinite real axis be coloured white.
Cover the unit intervals of prime numbers by red hats.
It is impossible to shift the red hats
Yes, because we are finite beings,
and there are infinitely.many red hats.
No, the reason is that every shift removes the hat from its place and requires an other hat, taken from wherever, but with certainty leaving
an uncovered interval. That does never change.
Not true. The first n prime numbers are obviously in bijection with theThat need not be assumed but that is obviously so for every part of theIt is impossible to shift the red hats so that all unit intervals ofNo. Assume that that is so.
the whole real axis get red hats.
There are too few prime numbers.
real axis.
Right, that is your mistake. There are still hats left over.Assume that there are enough red hats for the first 𝔊 numbers butThat is a mistake. If there are enough hats for G natnumbers, then there
not enough for the 𝔊+1ᵗʰ
are also enough for G^G^G natnumbers. Alas they leave G^G^G unit
intervals without hats. That is the catch!
Like who?Shall unit intervals disappear like Bob?There are too few prime numbers.No, there being too few primes leads to contradiction.
WM wrote :
A bijection between natural numbers and prime numbers proves that for
every prime number there is a natural number: p_1, p_2, p_3, ...
If that is correct, then there are as many natural numbers as prime
numbers and as many prime numbers as natural numbers. Then the
following scenario is possible:
Cover the unit intervals of prime numbers by red hats. Then shift the
red hats so that all unit intervals of the positive real axis get red
hats.
What makes you think that that follows?
Am Sat, 23 Nov 2024 21:45:06 +0100 schrieb WM:
On 23.11.2024 21:18, Jim Burns wrote:No. That other place where we take the hats from, larger primes, are of course covered by even larger primes. We don't stop at some arbitrary
On 11/23/2024 5:30 AM, WM wrote:
On 22.11.2024 22:50, Jim Burns wrote:
ℙ covers ℕ, and ℕ covers ℙ
Let every unit interval on the infinite real axis be coloured white.
Cover the unit intervals of prime numbers by red hats.
It is impossible to shift the red hats
Yes, because we are finite beings,
and there are infinitely.many red hats.
No, the reason is that every shift removes the hat from its place and
requires an other hat, taken from wherever, but with certainty leaving
an uncovered interval. That does never change.
finite number, but continue forever.
Not true. The first n prime numbers are obviously in bijection with the numbers 1 to n.That need not be assumed but that is obviously so for every part of theIt is impossible to shift the red hats so that all unit intervals ofNo. Assume that that is so.
the whole real axis get red hats.
There are too few prime numbers.
real axis.
Right, that is your mistake. There are still hats left over.Assume that there are enough red hats for the first 𝔊 numbers butThat is a mistake. If there are enough hats for G natnumbers, then there
not enough for the 𝔊+1ᵗʰ
are also enough for G^G^G natnumbers. Alas they leave G^G^G unit
intervals without hats. That is the catch!
On 23.11.2024 21:18, Jim Burns wrote:
On 11/23/2024 5:30 AM, WM wrote:
It is impossible to shift the red hats
so that all unit intervals of
the whole real axis get red hats.
There are too few prime numbers.
No.
⎛ Assume that that is so.
That need not be assumed but
that is obviously so for every part of the real axis.
⎜ Assume that there are
⎜ enough red hats for the first 𝔊 numbers
⎜ but not enough for the 𝔊+1ᵗʰ
That is a mistake.
If there are enough hats for G natnumbers,
then there are also enough for G^G^G natnumbers.
Alas they leave G^G^G unit intervals without hats.
That is the catch!
There are too few prime numbers.
No,
there being too few primes
leads to contradiction.
Shall unit intervals disappear like Bob?
On 11/23/2024 3:45 PM, WM wrote:
⎜ Assume that there are
⎜ enough red hats for the first 𝔊 numbers
⎜ but not enough for the 𝔊+1ᵗʰ
That is a mistake.
If there are enough hats for G natnumbers,
then there are also enough for G^G^G natnumbers.
Thank you.
Alas they leave G^G^G unit intervals without hats.
That is the catch!
After all hat.shifts,
there is no first number without a hat.
After all hat.shifts,
the set of numbers without hats is empty.
There are too few prime numbers.
No,
there being too few primes
leads to contradiction.
Shall unit intervals disappear like Bob?
After all swaps,
Bob is not in any room (visible or dark)
which Bob has ever been in.
On 23.11.2024 22:48, Jim Burns wrote:No. "Almost all" means everything except for a finite number.
On 11/23/2024 3:45 PM, WM wrote:
But almost all numbers are without hat because the number of hats hasThank you.⎜ Assume that there are ⎜ enough red hats for the first 𝔊 numbers ⎜That is a mistake.
but not enough for the 𝔊+1ᵗʰ
If there are enough hats for G natnumbers,
then there are also enough for G^G^G natnumbers.
Alas they leave G^G^G unit intervals without hats.After all hat.shifts,
That is the catch!
there is no first number without a hat.
not increased.
It is still valid for every finite interval.After all hat.shifts,Then the number of hats must have increased.
the set of numbers without hats is empty.
Then number theory can be wasted, because its formula for n/logn primes
in (0, n] is no longer valid after Cantor-reordering.
Bob is in none of them.He is in a dark room.After all swaps, Bob is not in any room (visible or dark)Shall unit intervals disappear like Bob?There are too few prime numbers.No, there being too few primes leads to contradiction.
which Bob has ever been in.But the unit intervals remain steadfast on the real line.
On 23.11.2024 21:46, Richard Damon wrote:
On 11/23/24 2:40 PM, WM wrote:
On 23.11.2024 19:43, FromTheRafters wrote:
WM wrote on 11/23/2024 :
On 23.11.2024 13:35, FromTheRafters wrote:
WM has brought this to us :
On 23.11.2024 13:20, FromTheRafters wrote:
WM laid this down on his screen :
Let every unit interval after a natural number on the real axis >>>>>>>>> be coloured white with exception of the intervals after the
prime numbers which are coloured red. It is impossible to shift >>>>>>>>> the red intervals so that the whole real axis becomes red.
Every interval (10n, 10 (n+1)] is deficient - on the whole real >>>>>>>>> axis.
So what? Your imaginings don't affect the fact that there is a >>>>>>>> bijection.
If there was a bijection,
There is.
then the whole axis could become red.
What makes you think that?
A bijection proves that every prime number (and its colour) can be
put to a natural number (and colour it).
???
A bijection between natural numbers and prime numbers proves that for
every prime number there is a natural number: p_1, p_2, p_3, ...
If that is correct, then there are as many natural numbers as prime
numbers and as many prime numbers as natural numbers. Then the
following scenario is possible:
Cover the unit intervals of prime numbers by red hats. Then shift the
red hats so that all unit intervals of the positive real axis get red
hats.
Regards, WM
And you can, as
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
A very naive recipe.
and in general, the red hat on the nth prime number can be moved to
the number n
Since there are a countable infinite number of prime numbers, there
exist an nth prime number for every n,
Yes, for every n that belongs to a tiny initial segment.
so all the numbers get covered.
No.
We have a 1:1 relationship (bijection) established between the set of
prime numbers and the set of Natural Numbers.
No. Every hat taken from wherever leaves there a naked unit interval. Therefore for every interval (0, n] inside which hats are moved, the
relative covering is about n/logn.
Regards, WM
On 23.11.2024 22:48, Jim Burns wrote:
On 11/23/2024 3:45 PM, WM wrote:
⎜ Assume that there are
⎜ enough red hats for the first 𝔊 numbers
⎜ but not enough for the 𝔊+1ᵗʰ
That is a mistake.
If there are enough hats for G natnumbers,
then there are also enough for G^G^G natnumbers.
Thank you.
Alas they leave G^G^G unit intervals without hats.
That is the catch!
After all hat.shifts,
there is no first number without a hat.
But almost all numbers are without hat
because the number of hats has not increased.
On 11/23/24 4:11 PM, WM wrote:
Cover the unit intervals of prime numbers by red hats. Then shiftAnd you can, as
the red hats so that all unit intervals of the positive real axis
get red hats.
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
A very naive recipe.
But it works.
Yes, for every n that belongs to a tiny initial segment.
No, for EVERY n.
Show one that it doesn't work for!
so all the numbers get covered.
No.
WHich one doesn't.
On 11/23/2024 5:01 PM, WM wrote:
If there are enough hats for G natural numbers,
then there are also enough for G^G^G natural numbers.
If there are NOT enough for G^G^G natural numbers,
then there are also NOT enough for G natural numbers.
G precedes G^G^G.
If, for both G and G^G^G, there are NOT enough hats,
G^G^G is not first for which there are not enough.
That generalizes to
each natural number is not.first for which
there are NOT enough hats.
----
Consider the set of natural numbers for which
there are NOT enough hats.
Since it is a set of natural numbers,
there are two possibilities:
-- It could be the empty set.
-- It could be non.empty and hold a first number.
Its first number, if it existed, would be
the first natural number for which
there are NOT enough hats.
However,
the FIRST natural number for which
there are NOT enough hats
does not exist.
Therefore,
for each natural number,
there are enough hats.
on 11/23/2024, WM supposed :
On 23.11.2024 22:07, FromTheRafters wrote:
WM wrote :
A bijection between natural numbers and prime numbers proves that
for every prime number there is a natural number: p_1, p_2, p_3, ...
If that is correct, then there are as many natural numbers as prime
numbers and as many prime numbers as natural numbers. Then the
following scenario is possible:
Cover the unit intervals of prime numbers by red hats. Then shift
the red hats so that all unit intervals of the positive real axis
get red hats.
What makes you think that that follows?
Cantor.
Show me Cantor shifting interval hats!
Am Sat, 23 Nov 2024 23:01:00 +0100 schrieb WM:
But almost all numbers are without hat because the number of hats hasNo. "Almost all" means everything except for a finite number.
not increased.
Then number theory can be wasted, because its formula for n/logn primesIt is still valid for every finite interval.
in (0, n] is no longer valid after Cantor-reordering.
On 24.11.2024 03:22, Richard Damon wrote:
On 11/23/24 4:11 PM, WM wrote:
Cover the unit intervals of prime numbers by red hats. Then shiftAnd you can, as
the red hats so that all unit intervals of the positive real axis
get red hats.
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
A very naive recipe.
But it works.
It fails in every step to cover the interval (0, n] with hats taken from
this interval.
Yes, for every n that belongs to a tiny initial segment.
No, for EVERY n.
Show one that it doesn't work for!
The complete covering fails in every interval (0, n] with hats taken
from this interval.
so all the numbers get covered.
No.
WHich one doesn't.
Almost all. The reason is simple mathematics. For every interval (0, n]
the relative covering is 1/10, independent of how the hats are shifted.
This cannot be remedied in the infinite limit because outside of all
finite intervals (0, n] there are no further hats available.
On the other hand, we cannot find a first n that cannot be covered by a
hat. This dilemma cannot be resolved by negating one of the two facts.
It can only be solved by dark numbers.
Regards, WM
On 24.11.2024 00:44, joes wrote:
Am Sat, 23 Nov 2024 23:01:00 +0100 schrieb WM:
But almost all numbers are without hat because the number of hats hasNo. "Almost all" means everything except for a finite number.
not increased.
Here it means infinitely many more.
Then number theory can be wasted, because its formula for n/logn primesIt is still valid for every finite interval.
in (0, n] is no longer valid after Cantor-reordering.
It is also valid for the infinite interval because outside of all
intervals (0, n] there are no more hats available.
Further in number theory it is accepted for the whole set.
Regards, WM
WM formulated the question :
On 24.11.2024 12:07, FromTheRafters wrote:
on 11/23/2024, WM supposed :
On 23.11.2024 22:07, FromTheRafters wrote:
WM wrote :
A bijection between natural numbers and prime numbers proves that
for every prime number there is a natural number: p_1, p_2, p_3, ... >>>>>> If that is correct, then there are as many natural numbers as
prime numbers and as many prime numbers as natural numbers. Then
the following scenario is possible:
Cover the unit intervals of prime numbers by red hats. Then shift
the red hats so that all unit intervals of the positive real axis
get red hats.
What makes you think that that follows?
Cantor.
Show me Cantor shifting interval hats!
Constructing a bijection between the two sets is visualized by hats.
Where?
On 11/24/24 5:31 AM, WM wrote:
On 24.11.2024 03:22, Richard Damon wrote:
On 11/23/24 4:11 PM, WM wrote:
Cover the unit intervals of prime numbers by red hats. Then shiftAnd you can, as
the red hats so that all unit intervals of the positive real axis
get red hats.
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
A very naive recipe.
But it works.
It fails in every step to cover the interval (0, n] with hats taken
from this interval.
But that isn't the requirement.
The requirement it to map from ALL Prime Natural Numbers to ALL Natural Numbers
Yes, for every n that belongs to a tiny initial segment.
No, for EVERY n.
Show one that it doesn't work for!
The complete covering fails in every interval (0, n] with hats taken
from this interval.
Which isn't the interval in question.
Your funny-mental fallacy is that you think an infinite set can be
thought of as just some finite set allowed to keep growning until it
reaches infinity,
That is just the wrong model.
so all the numbers get covered.
No.
WHich one doesn't.
Almost all. The reason is simple mathematics. For every interval (0,
n] the relative covering is 1/10, independent of how the hats are
shifted. This cannot be remedied in the infinite limit because outside
of all finite intervals (0, n] there are no further hats available.
But finite sets aren't infinite sets, and don't act the same as them.
You can not just
use finite mathematics on infinite sets.
On 24.11.2024 03:22, Richard Damon wrote:Nobody cares about a nonbijection from N to N. We are interested in a
On 11/23/24 4:11 PM, WM wrote:
It fails in every step to cover the interval (0, n] with hats taken fromCover the unit intervals of prime numbers by red hats. Then shiftAnd you can, as the red hat on the number 2, can be moved to the
the red hats so that all unit intervals of the positive real axis
get red hats.
number 1 the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
this interval.
Think about it the other way around: when we take the primes andThe complete covering fails in every interval (0, n] with hats takenYes, for every n that belongs to a tiny initial segment.No, for EVERY n. Show one that it doesn't work for!
from this interval.
As above, nobody is arguing that the primes are somehow more infinite.Almost all. For every interval (0, n]WHich one doesn't.so all the numbers get covered.No.
the relative covering is 1/10, independent of how the hats are shifted.
This cannot be remedied in the infinite limit because outside of allYes it can, if you start out with a proper finite bijection.
finite intervals (0, n] there are no further hats available.
On the other hand, we cannot find a first n that cannot be covered by aOr getting over your wrong intuition.
hat. This dilemma cannot be resolved by negating one of the two facts.
It can only be solved by dark numbers.
Am Sun, 24 Nov 2024 11:31:00 +0100 schrieb WM:
On 24.11.2024 03:22, Richard Damon wrote:Nobody cares about a nonbijection from N to N. We are interested in a bijection from N to P, or {0, 1, ..., n} to {p0, p1, ..., p_n}
On 11/23/24 4:11 PM, WM wrote:It fails in every step to cover the interval (0, n] with hats taken from
Cover the unit intervals of prime numbers by red hats. Then shiftAnd you can, as the red hat on the number 2, can be moved to the
the red hats so that all unit intervals of the positive real axis
get red hats.
number 1 the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
this interval.
Think about it the other way around: when we take the primes andThe complete covering fails in every interval (0, n] with hats takenYes, for every n that belongs to a tiny initial segment.No, for EVERY n. Show one that it doesn't work for!
from this interval.
number them, we are never done, because there are inf.many primes.
Therefore we need the set of all naturals, N, to count them.
As above, nobody is arguing that the primes are somehow more infinite.Almost all. For every interval (0, n]WHich one doesn't.so all the numbers get covered.No.
the relative covering is 1/10, independent of how the hats are shifted.
This cannot be remedied in the infinite limit because outside of allYes it can, if you start out with a proper finite bijection.
finite intervals (0, n] there are no further hats available.
On the other hand, we cannot find a first n that cannot be covered by aOr getting over your wrong intuition.
hat. This dilemma cannot be resolved by negating one of the two facts.
It can only be solved by dark numbers.
On 24.11.2024 13:12, Richard Damon wrote:
On 11/24/24 5:31 AM, WM wrote:All finite sets are the infinite set.
On 24.11.2024 03:22, Richard Damon wrote:
On 11/23/24 4:11 PM, WM wrote:
Cover the unit intervals of prime numbers by red hats. Then shift >>>>>>> the red hats so that all unit intervals of the positive real axis >>>>>>> get red hats.And you can, as
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
A very naive recipe.
But it works.
It fails in every step to cover the interval (0, n] with hats taken
from this interval.
But that isn't the requirement.
The requirement it to map from ALL Prime Natural Numbers to ALL
Natural Numbers
Yes, for every n that belongs to a tiny initial segment.
No, for EVERY n.
Show one that it doesn't work for!
The complete covering fails in every interval (0, n] with hats taken
from this interval.
Which isn't the interval in question.
Your funny-mental fallacy is that you think an infinite set can be
thought of as just some finite set allowed to keep growning until it
reaches infinity,
That is just the wrong model.
so all the numbers get covered.
No.
WHich one doesn't.
Almost all. The reason is simple mathematics. For every interval (0,
n] the relative covering is 1/10, independent of how the hats are
shifted. This cannot be remedied in the infinite limit because
outside of all finite intervals (0, n] there are no further hats
available.
But finite sets aren't infinite sets, and don't act the same as them.
You can not just use finite mathematics on infinite sets.
But I can use the analytical limit of the constant sequence.
Regards, WM
On 11/24/24 7:26 AM, WM wrote:
On 24.11.2024 13:12, Richard Damon wrote:Nope, since the finite sets are not the same as the infinite set, the property you are looking at just doesn't exist in the infinite set.
On 11/24/24 5:31 AM, WM wrote:All finite sets are the infinite set.
On 24.11.2024 03:22, Richard Damon wrote:
On 11/23/24 4:11 PM, WM wrote:
Cover the unit intervals of prime numbers by red hats. ThenAnd you can, as
shift the red hats so that all unit intervals of the positive
real axis get red hats.
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
A very naive recipe.
But it works.
It fails in every step to cover the interval (0, n] with hats taken
from this interval.
But that isn't the requirement.
The requirement it to map from ALL Prime Natural Numbers to ALL
Natural Numbers
Yes, for every n that belongs to a tiny initial segment.
No, for EVERY n.
Show one that it doesn't work for!
The complete covering fails in every interval (0, n] with hats taken
from this interval.
Which isn't the interval in question.
Your funny-mental fallacy is that you think an infinite set can be
thought of as just some finite set allowed to keep growning until it
reaches infinity,
That is just the wrong model.
so all the numbers get covered.
No.
WHich one doesn't.
Almost all. The reason is simple mathematics. For every interval (0,
n] the relative covering is 1/10, independent of how the hats are
shifted. This cannot be remedied in the infinite limit because
outside of all finite intervals (0, n] there are no further hats
available.
But finite sets aren't infinite sets, and don't act the same as them.
You can not just use finite mathematics on infinite sets.
But I can use the analytical limit of the constant sequence.
Limit theory only works if the limit actually exists
You can get things that APPEAR to reach a limit, but actually don't.
On 24.11.2024 15:40, joes wrote:
Am Sun, 24 Nov 2024 11:31:00 +0100 schrieb WM:
On 24.11.2024 03:22, Richard Damon wrote:Nobody cares about a nonbijection from N to N. We are interested in a
On 11/23/24 4:11 PM, WM wrote:It fails in every step to cover the interval (0, n] with hats taken from >>> this interval.
Cover the unit intervals of prime numbers by red hats. Then shift >>>>>>> the red hats so that all unit intervals of the positive real axis >>>>>>> get red hats.And you can, as the red hat on the number 2, can be moved to the
number 1 the red hat on the number 3, can be moved to the number 2 >>>>>> the red hat on the number 5, can be moved to the number 3
bijection from N to P, or {0, 1, ..., n} to {p0, p1, ..., p_n}
That can be excluded because there are too few hats in all intervals.
Think about it the other way around: when we take the primes andThe complete covering fails in every interval (0, n] with hats takenYes, for every n that belongs to a tiny initial segment.No, for EVERY n. Show one that it doesn't work for!
from this interval.
number them, we are never done, because there are inf.many primes.
Therefore we need the set of all naturals, N, to count them.
Even the naturals divisible by 10000000 would suffice if Cantor was
right. But he is not.
As above, nobody is arguing that the primes are somehow more infinite.Almost all. For every interval (0, n]WHich one doesn't.so all the numbers get covered.No.
the relative covering is 1/10, independent of how the hats are shifted.
They are less than the naturals.
This cannot be remedied in the infinite limit because outside of allYes it can, if you start out with a proper finite bijection.
finite intervals (0, n] there are no further hats available.
If it could exist, then all hats could cover all naturals. But that is excluded by the relative covering of 1/10 in all intervals (0, n].
On the other hand, we cannot find a first n that cannot be covered by aOr getting over your wrong intuition.
hat. This dilemma cannot be resolved by negating one of the two facts.
It can only be solved by dark numbers.
The limit of a constant sequence is its constant. No intuition.
Regards, WM
On 11/23/2024 3:45 PM, WM wrote:
If there are enough hats for G natnumbers,
then there are also enough for G^G^G natnumbers.
On 24.11.2024 06:37, Jim Burns wrote:
If there are enough hats for G natural numbers,
then there are also enough for G^G^G natural numbers.
So it is.
But that does not negate the fact that
for every interval (0,n]
the relative covering is 1/10,
independent of how the hats are shifted.
This cannot be remedied in the infinite limit
because
outside of all finite intervals (0, n]
there are no further hats available.
If there are NOT enough for G^G^G natural numbers,
then there are also NOT enough for G natural numbers.
G precedes G^G^G.
If, for both G and G^G^G, there are NOT enough hats,
G^G^G is not first for which there are not enough.
That generalizes to
each natural number is not.first for which
there are NOT enough hats.
It seems so but
the sequence 1/10, 1/10, 1/10, ... has limit 1/10
with no doubt.
This dilemma is the reason
why dark numbers are required.
----
Consider the set of natural numbers for which
there are NOT enough hats.
It is dark.
Since it is a set of natural numbers,
there are two possibilities:
-- It could be the empty set.
-- It could be non.empty and hold a first number.
Both attempts fail.
On 24.11.2024 18:22, Richard Damon wrote:
On 11/24/24 7:26 AM, WM wrote:
On 24.11.2024 13:12, Richard Damon wrote:Nope, since the finite sets are not the same as the infinite set, the
On 11/24/24 5:31 AM, WM wrote:All finite sets are the infinite set.
On 24.11.2024 03:22, Richard Damon wrote:
On 11/23/24 4:11 PM, WM wrote:
Cover the unit intervals of prime numbers by red hats. ThenAnd you can, as
shift the red hats so that all unit intervals of the positive >>>>>>>>> real axis get red hats.
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3
A very naive recipe.
But it works.
It fails in every step to cover the interval (0, n] with hats taken
from this interval.
But that isn't the requirement.
The requirement it to map from ALL Prime Natural Numbers to ALL
Natural Numbers
Yes, for every n that belongs to a tiny initial segment.
No, for EVERY n.
Show one that it doesn't work for!
The complete covering fails in every interval (0, n] with hats
taken from this interval.
Which isn't the interval in question.
Your funny-mental fallacy is that you think an infinite set can be
thought of as just some finite set allowed to keep growning until it
reaches infinity,
That is just the wrong model.
so all the numbers get covered.
No.
WHich one doesn't.
Almost all. The reason is simple mathematics. For every interval
(0, n] the relative covering is 1/10, independent of how the hats
are shifted. This cannot be remedied in the infinite limit because
outside of all finite intervals (0, n] there are no further hats
available.
But finite sets aren't infinite sets, and don't act the same as them.
You can not just use finite mathematics on infinite sets.
But I can use the analytical limit of the constant sequence.
property you are looking at just doesn't exist in the infinite set.
The finite sets contains all hats because all natural numbers and all
10n are in finite sets. No hat is outside.>
Limit theory only works if the limit actually exists
If limits exist at all, then the limit of the sequence 1/10, 1/10,
1/10, ... does exist.
You can get things that APPEAR to reach a limit, but actually don't.
But if infinite sets do exist, then the set ℕ does exist, and all its elements are members of finite intervals (0, n].
Regards, WM
On 11/24/2024 4:24 AM, WM wrote:
On 24.11.2024 12:55, FromTheRafters wrote:
WM wrote:
Constructing a bijection between the two sets is visualized by hats.
Where?
Here!
A dunce hat? Humm, were you wearing such a hat when you were born? I
feel sorry for your Mom, wow! Ouch.
Let me guess... WM thinks [...]
On 11/23/2024 3:45 PM, WM wrote:
for every interval (0,n]
the relative covering is 1/10,
independent of how the hats are shifted.
This cannot be remedied in the infinite limit
because
outside of all finite intervals (0, n]
there are no further hats available.
All the hats for which
if there are G.many then there are G^G^G.many
are enough to "remedy" the 1/10.relative.covering
If each G can match G^G^G
Matching.a.proper.subset is
the sort of behavior which permits Bob to disappear
with enough room.swapping _inside_ the Hotel.
You (WM) treat that behavior as proof that
we are wrong and you (WM) are right.
What it is is proof that
not all sets behave like finite sets.
If there are enough hats for G natural numbers,
then there are also enough for G^G^G natural numbers.
The number G^G^G is not.first for which
there are NOT enough hats.
Therefore,
the number G^G^G is not.first for which
there are NOT enough hats.
A similar argument can be made for
each natural number.
Consider the set of natural numbers for which
there are NOT enough hats.
It is dark.
It is not dark what we mean by 'natural number'.
A natural number is countable.to from.0
The natural numbers "fail" at
being finitely.many.
It is nothing more than that.
On 11/24/24 12:32 PM, WM wrote:
The finite sets contain all hats because all natural numbers and all
10n are in finite sets. No hat is outside.>
But there is no finite set with ALL natural numbers.
Like usual, you mess up with your qualifiers.
Limit theory only works if the limit actually exists
If limits exist at all, then the limit of the sequence 1/10, 1/10,
1/10, ... does exist.
But the concept of 1/10th of an infinte set does not exist..
You can get things that APPEAR to reach a limit, but actually don't.
But if infinite sets do exist, then the set ℕ does exist, and all its
elements are members of finite intervals (0, n].
No, any given element is a member of a finite set, but you can't then
say that ALL are in such a set.
WIth infinity, Any and All are different qualifiers.
On 24.11.2024 20:14, Richard Damon wrote:
On 11/24/24 12:32 PM, WM wrote:
The finite sets contain all hats because all natural numbers and all
10n are in finite sets. No hat is outside.>
But there is no finite set with ALL natural numbers.
ℕ is fixed, that means |ℕ| is fixed.
Like usual, you mess up with your qualifiers.
Limit theory only works if the limit actually exists
If limits exist at all, then the limit of the sequence 1/10, 1/10,
1/10, ... does exist.
But the concept of 1/10th of an infinte set does not exist..
It does.
You can get things that APPEAR to reach a limit, but actually don't.
But if infinite sets do exist, then the set ℕ does exist, and all its
elements are members of finite intervals (0, n].
No, any given element is a member of a finite set, but you can't then
say that ALL are in such a set.
All are in the union of all finite sets.
WIth infinity, Any and All are different qualifiers.
Use them to increase your qualification.
Regards, WM
On 24.11.2024 19:16, Jim Burns wrote:
On 11/24/2024 6:06 AM, WM wrote:
On 24.11.2024 06:37, Jim Burns wrote:
Consider the set of natural numbers for which
there are NOT enough hats.
It is dark.
It is not dark what we mean by 'natural number'.
A natural number is countable.to from.0
That is a definable number.
Therefore,
the number G^G^G is not.first for which
there are NOT enough hats.
We do not disagree.
Therefore you need not
prove a difference for G and G^G^G.
A similar argument can be made for
each natural number.
No,
it can be made for each definable natural number,
it can be made for each definable natural number,
i.e., for a number belonging to
a tiny finite initial segment which is followed bay
almost all numbers.
a tiny finite initial segment which is followed bay
almost all numbers.
On 24.11.2024 20:14, Richard Damon wrote:Not all, but every.
On 11/24/24 12:32 PM, WM wrote:
The finite sets contain all hats because all natural numbers and all
10n are in finite sets. No hat is outside.>
What does that have to do with it?But there is no finite set with ALL natural numbers.ℕ is fixed, that means |ℕ| is fixed.
Like usual, you mess up with your qualifiers.
It has the same cardinality.It does.But the concept of 1/10th of an infinte set does not exist..Limit theory only works if the limit actually existsIf limits exist at all, then the limit of the sequence 1/10, 1/10,
1/10, ... does exist.
Why not just directly take N, made up of finite numbers?All are in the union of all finite sets.No, any given element is a member of a finite set, but you can't thenYou can get things that APPEAR to reach a limit, but actually don't.But if infinite sets do exist, then the set ℕ does exist, and all its
elements are members of finite intervals (0, n].
say that ALL are in such a set.
*quantifiersWIth infinity, Any and All are different qualifiers.Use them to increase your qualification.
On 24.11.2024 19:16, Jim Burns wrote:That's still true in the infinite case.
On 11/23/2024 3:45 PM, WM wrote:
for every interval (0,n]
the relative covering is 1/10,
independent of how the hats are shifted. This cannot be remedied in
the infinite limit because outside of all finite intervals (0, n]
there are no further hats available.
All the hats for which
if there are G.many then there are G^G^G.many
are enough to "remedy" the 1/10.relative.covering
Not at all!
If each G can match G^G^GThat is a false assumption.
If there are enough hats to cover G then there are enough hats to cover G^G^G. G does not cover G^G^G.
Matching a proper subset is the sort of behavior which permits Bob toExchange of two elements never leads to loss of one of them.
disappear with enough room swapping _inside_ the Hotel.
You (WM) treat that behavior as proof that we are wrong and you (WM)
are right.
Infinite sets are not finite.What it is is proof that not all sets behave like finite sets.Nonsense. Logic is also prevailing in infinite sets.
Right!If there are enough hats for G natural numbers,There is no such number because the set of definable hats is potentially infinite.
then there are also enough for G^G^G natural numbers.
The number G^G^G is not.first for which there are NOT enough hats.
Nor for G and 10G.Therefore,We do not disagree. Therefore you need not prove a difference for G and G^G^G.
the number G^G^G is not.first for which there are NOT enough hats.
Yes, like he said: for every natural.A similar argument can be made for each natural number.No, it can be made for each definable natural number, i.e., for a number belonging to a tiny finite initial segment which is followed bay almost
all numbers.
Yes, the naturals are all "definable". That is all mathematicians talkThat is a definable number.It is not dark what we mean by 'natural number'.Consider the set of natural numbers for which there are NOT enoughIt is dark.
hats.
A natural number is countable to from 0.
What would it even mean for the naturals to not be "complete"? EitherThe natural numbers "fail" at being finitely many.If they are infinitely many but complete, then they and their number
It is nothing more than that.
don't vary. |ℕ| - 1 < |ℕ| < |ℕ| + 1.
On 11/25/2024 7:11 AM, WM wrote:
On 24.11.2024 19:16, Jim Burns wrote:
On 11/24/2024 6:06 AM, WM wrote:
On 24.11.2024 06:37, Jim Burns wrote:
Consider the set of natural numbers for which
there are NOT enough hats.
It is dark.
It is not dark what we mean by 'natural number'.
A natural number is countable.to from.0
That is a definable number.
Thus
it is not dark that
what we mean by 'natural number' is
countable.to from.0 and is
what you call a definable number.
On 11/25/24 7:18 AM, WM wrote:
All are in the union of all finite sets.
But your logic can't handle an infinite union, as it can never complete
the process.
Am Mon, 25 Nov 2024 13:18:28 +0100 schrieb WM:
What does that have to do with it?But there is no finite set with ALL natural numbers.ℕ is fixed, that means |ℕ| is fixed.
Like usual, you mess up with your qualifiers.
It has the same cardinality.It does.But the concept of 1/10th of an infinte set does not exist..Limit theory only works if the limit actually existsIf limits exist at all, then the limit of the sequence 1/10, 1/10,
1/10, ... does exist.
Why not just directly take N, made up of finite numbers?All are in the union of all finite sets.No, any given element is a member of a finite set, but you can't thenYou can get things that APPEAR to reach a limit, but actually don't.But if infinite sets do exist, then the set ℕ does exist, and all its >>>> elements are members of finite intervals (0, n].
say that ALL are in such a set.
On 25.11.2024 22:05, joes wrote:It is possible to change N to N\0.
Am Mon, 25 Nov 2024 13:18:28 +0100 schrieb WM:
It is impossible to add or to delete an element.What does that have to do with it?But there is no finite set with ALL natural numbers.ℕ is fixed, that means |ℕ| is fixed.
Like usual, you mess up with your qualifiers.
It is impossible to change |ℕ| by 1 or more.
Countably infinite.Yes, it is much.It has the same cardinality.It does.But the concept of 1/10th of an infinte set does not exist..Limit theory only works if the limit actually existsIf limits exist at all, then the limit of the sequence 1/10, 1/10,
1/10, ... does exist.
Cantor proved nothing more.Why not? Do it. Consider the black hats at every 10 n and white hats atWhy not just directly take N, made up of finite numbers?All are in the union of all finite sets.No, any given element is a member of a finite set, but you can't thenYou can get things that APPEAR to reach a limit, but actuallyBut if infinite sets do exist, then the set ℕ does exist, and all
don't.
its elements are members of finite intervals (0, n].
say that ALL are in such a set.
all other numbers n. It is possible to shift the black hats such that
every interval (0, n] is completely covered by black hats. There is no
first n discernible that cannot be covered by black hat.
But the originNot if you really coloured ALL n.
of each used black hat larger than n is now covered by a white hat.
Without deleting all white hats it is not possible to cover all n byAn infinite exchange can.
black hats. But deleting white hats is prohibited by logic. Exchanging
can never delete one of the exchanged elements.
Therefore we have here,Thanks for shutting up.
like in all Cantor-pairings, the same impediment and further disussion
is futile.
Am Tue, 26 Nov 2024 10:11:22 +0100 schrieb WM:
ExchangingAn infinite exchange can.
can never delete one of the exchanged elements.
How much paint would be needed to paint the entire cartesian plane?
WM laid this down on his screen :
Consider the black hats at every 10 n and white hats at all other
numbers n. It is possible to shift the black hats such that every
interval (0, n] is completely covered by black hats. There is no first
n discernible that cannot be covered by black hat. But the origin of
each used black hat larger than n is now covered by a white hat.
Without deleting all white hats it is not possible to cover all n by
black hats. But deleting white hats is prohibited by logic. Exchanging
can never delete one of the exchanged elements. Therefore we have
here, like in all Cantor-pairings, the same impediment and further
disussion is futile: You must deny logic. I do not.
How much paint would be needed to paint the entire cartesian plane?
Too
much huh? then just paint the first quadrant with one quarter the amount
of paint. Your 'logic' is too easily denied.
WM pretended :
It is impossible to change |ℕ| by 1 or more.
Right, sets don't change. The set {2,3,4,...} does not equal the set of natural numbers, but |{2,3,4,...}| does equal |N|.
On 26.11.2024 12:15, FromTheRafters wrote:
WM pretended :
Then your |N| is an imprecise measure. My |N| is precise.It is impossible to change |ℕ| by 1 or more.
Right, sets don't change. The set {2,3,4,...} does not equal the set
of natural numbers, but |{2,3,4,...}| does equal |N|.
|{2,3,4,...}| = |N| - 1 =/= |N| .
Regards, WM
On 26.11.2024 16:06, Richard Damon wrote:
On 11/26/24 6:24 AM, WM wrote:
On 26.11.2024 12:15, FromTheRafters wrote:
WM pretended :Then your |N| is an imprecise measure. My |N| is precise.
It is impossible to change |ℕ| by 1 or more.
Right, sets don't change. The set {2,3,4,...} does not equal the set
of natural numbers, but |{2,3,4,...}| does equal |N|.
|{2,3,4,...}| = |N| - 1 =/= |N| .
Then your measure is incorrect, as by the DEFINITION of measures of
infinite sets, all countably infinite sets have the same "measure".
That is one special definition of a very imprecise measure. We can do
better.
Regards, WM
On 11/26/24 6:24 AM, WM wrote:
On 26.11.2024 12:15, FromTheRafters wrote:
WM pretended :Then your |N| is an imprecise measure. My |N| is precise.
It is impossible to change |ℕ| by 1 or more.
Right, sets don't change. The set {2,3,4,...} does not equal the set
of natural numbers, but |{2,3,4,...}| does equal |N|.
|{2,3,4,...}| = |N| - 1 =/= |N| .
Then your measure is incorrect, as by the DEFINITION of measures of
infinite sets, all countably infinite sets have the same "measure".
On 11/26/24 11:59 AM, WM wrote:
On 26.11.2024 16:06, Richard Damon wrote:
On 11/26/24 6:24 AM, WM wrote:
On 26.11.2024 12:15, FromTheRafters wrote:
WM pretended :Then your |N| is an imprecise measure. My |N| is precise.
It is impossible to change |ℕ| by 1 or more.
Right, sets don't change. The set {2,3,4,...} does not equal the
set of natural numbers, but |{2,3,4,...}| does equal |N|.
|{2,3,4,...}| = |N| - 1 =/= |N| .
Then your measure is incorrect, as by the DEFINITION of measures of
infinite sets, all countably infinite sets have the same "measure".
That is one special definition of a very imprecise measure. We can do
better.
But maybe you can't and get something consistant.
On 26.11.2024 18:50, Richard Damon wrote:
On 11/26/24 11:59 AM, WM wrote:
On 26.11.2024 16:06, Richard Damon wrote:
On 11/26/24 6:24 AM, WM wrote:
On 26.11.2024 12:15, FromTheRafters wrote:
WM pretended :Then your |N| is an imprecise measure. My |N| is precise.
It is impossible to change |ℕ| by 1 or more.
Right, sets don't change. The set {2,3,4,...} does not equal the
set of natural numbers, but |{2,3,4,...}| does equal |N|.
|{2,3,4,...}| = |N| - 1 =/= |N| .
Then your measure is incorrect, as by the DEFINITION of measures of
infinite sets, all countably infinite sets have the same "measure".
That is one special definition of a very imprecise measure. We can do
better.
But maybe you can't and get something consistant.
Of course. |{1, 2, 3, 4, ...}| = |ℕ| and |{2, 3, 4, ...}| = |ℕ| - 1 is consistent.
Regards, WM
On 11/27/24 5:12 AM, WM wrote:
Of course. |{1, 2, 3, 4, ...}| = |ℕ| and |{2, 3, 4, ...}| = |ℕ| - 1 is >> consistent.
So you think, but that is because you brain has been exploded by the contradiction.
We can get to your second set two ways, and the set itself can't know
which.
We could have built the set by the operation of removing 1 like your
math implies, or we can get to it by the operation of increasing each
element by its successor, which must have the same number of elements,
On 27.11.2024 13:32, Richard Damon wrote:What is "decreasing by a point"?
On 11/27/24 5:12 AM, WM wrote:
Of course. |{1, 2, 3, 4, ...}| = |ℕ| and |{2, 3, 4, ...}| = |ℕ| - 1 is >>> consistent.
So you think, but that is because you brain has been exploded by the
contradiction.
We can get to your second set two ways, and the set itself can't know
which.
We could have built the set by the operation of removing 1 like your
math implies, or we can get to it by the operation of increasing each
element by its successor, which must have the same number of elements,
Yes, the same number of elements, but not the same number of natural
numbers.
Decreasing every element in the real interval (0, 1] by one point
yields the real interval [0, 1). The set of points remains the same, the
set of positive points decreases by 1.
Replacing every element of the set {0, 1, 2, 3, ...} by its successorThere is no natural whose successor is omega.
yields {1, 2, 3, ..., ω}. The number of ordinals remains the same, the number of finite ordinals decreases.
On 27.11.2024 13:32, Richard Damon wrote:
On 11/27/24 5:12 AM, WM wrote:
Of course. |{1, 2, 3, 4, ...}| = |ℕ| and |{2, 3, 4, ...}| = |ℕ| - 1
is consistent.
So you think, but that is because you brain has been exploded by the
contradiction.
We can get to your second set two ways, and the set itself can't know
which.
We could have built the set by the operation of removing 1 like your
math implies, or we can get to it by the operation of increasing each
element by its successor, which must have the same number of elements,
Yes, the same number of elements, but not the same number of natural
numbers.
Hint: Decreasing every element in the real interval (0, 1] by one point yields the real interval [0, 1). The set of points remains the same, the
set of positive points decreases by 1.
Replacing every element of the set {0, 1, 2, 3, ...} by its successor
yields {1, 2, 3, ..., ω}. The number of ordinals remains the same, the number of finite ordinals decreases.
Regards, WM
On 11/27/24 12:13 PM, WM wrote:
Replacing every element of the set {0, 1, 2, 3, ...} by its successor
yields {1, 2, 3, ..., ω}. The number of ordinals remains the same, the
number of finite ordinals decreases.
Nope, because omega is NOT the successor for any natural number, the successor of EVERY Natural Number is a Natural Number.
It it the successor for the SET of natural numbers.
Am Wed, 27 Nov 2024 18:13:06 +0100 schrieb WM:
Decreasing every element in the real interval (0, 1] by one pointWhat is "decreasing by a point"?
yields the real interval [0, 1). The set of points remains the same, the
set of positive points decreases by 1.
Replacing every element of the set {0, 1, 2, 3, ...} by its successorThere is no natural whose successor is omega.
yields {1, 2, 3, ..., ω}. The number of ordinals remains the same, the
number of finite ordinals decreases.
WM explained :
On 27.11.2024 13:32, Richard Damon wrote:
On 11/27/24 5:12 AM, WM wrote:
Of course. |{1, 2, 3, 4, ...}| = |ℕ| and |{2, 3, 4, ...}| = |ℕ| - 1 >>>> is consistent.
So you think, but that is because you brain has been exploded by the
contradiction.
We can get to your second set two ways, and the set itself can't know
which.
We could have built the set by the operation of removing 1 like your
math implies, or we can get to it by the operation of increasing each
element by its successor, which must have the same number of elements,
Yes, the same number of elements, but not the same number of natural
numbers.
Hint: Decreasing every element in the real interval (0, 1] by one
point yields the real interval [0, 1). The set of points remains the
same, the set of positive points decreases by 1.
If you have a successor function for the real numbers, why don't you
share it with the rest of the world?
On 27.11.2024 13:32, Richard Damon wrote:
[...]
Yes, the same number of elements,
but not the same number of natural numbers.
On 11/27/2024 11:15 AM, WM wrote:
On 27.11.2024 18:58, Richard Damon wrote:
On 11/27/24 12:13 PM, WM wrote:
Replacing every element of the set {0, 1, 2, 3, ...} by its
successor yields {1, 2, 3, ..., ω}.
Nope, [...] the successor of EVERY natural number is a natural number
[hence] omega is NOT the successor for any natural number
The successor of the last natural number is ω.
There is no last natural number you fucking dipshit!
On 27.11.2024 18:58, Richard Damon wrote:
On 11/27/24 12:13 PM, WM wrote:
Replacing every element of the set {0, 1, 2, 3, ...} by its successor
yields {1, 2, 3, ..., ω}. The number of ordinals remains the same,
the number of finite ordinals decreases.
Nope, because omega is NOT the successor for any natural number, the
successor of EVERY Natural Number is a Natural Number.
The successor of the last natural number is ω.
It it the successor for the SET of natural numbers.
And that is nothing else but all natural numbers.
Regards, WM
On 27.11.2024 22:14, Richard Damon wrote:By itself.
On 11/27/24 2:15 PM, WM wrote:
By what?No, the set is different from its members.It it the successor for the SET of natural numbers.And that is nothing else but all natural numbers.
Finite cardinals can change by 1
The cardinal |ℕᶠⁱⁿ| cannot change by 1
Yes,
the sets can change membership by 1
However,
the cardinalities of those sets cannot change by 1
On 11/27/24 2:15 PM, WM wrote:
It it the successor for the SET of natural numbers.
And that is nothing else but all natural numbers.
No, the set is different from its members.
On 27.11.2024 20:47, Jim Burns wrote:
Finite cardinals can change by 1
The cardinal |ℕᶠⁱⁿ| cannot change by 1
Small wonder. Fuzzy properties like "many" cannot change by 1.
Yes,
the sets can change membership by 1
However,
the cardinalities of those sets cannot change by 1
This proves that cardinality is a fuzzy property.
Regards, WM
On 27.11.2024 22:14, Richard Damon wrote:
On 11/27/24 2:15 PM, WM wrote:
It it the successor for the SET of natural numbers.
And that is nothing else but all natural numbers.
No, the set is different from its members.
By what?
Regards, WM
On 11/27/24 4:33 PM, WM wrote:
On 27.11.2024 20:47, Jim Burns wrote:
Finite cardinals can change by 1
The cardinal |ℕᶠⁱⁿ| cannot change by 1
Small wonder. Fuzzy properties like "many" cannot change by 1.
Yes,
the sets can change membership by 1
However,
the cardinalities of those sets cannot change by 1
This proves that cardinality is a fuzzy property.
No, it shows that infinite values act differently than finite numbers.
On 11/27/24 4:43 PM, WM wrote:
On 27.11.2024 22:14, Richard Damon wrote:
On 11/27/24 2:15 PM, WM wrote:
It it the successor for the SET of natural numbers.
And that is nothing else but all natural numbers.
No, the set is different from its members.
By what?
It is a set, with set type properties, and its members are Natural
Numbers, with Natural Number like propertis.
The set is equal to N, which is the name of the set, but none of the
members are 'equal' to N, they are MEMBERS of it.
On 28.11.2024 01:31, Richard Damon wrote:
On 11/27/24 4:33 PM, WM wrote:
On 27.11.2024 20:47, Jim Burns wrote:
Finite cardinals can change by 1
The cardinal |ℕᶠⁱⁿ| cannot change by 1
Small wonder. Fuzzy properties like "many" cannot change by 1.
Yes,
the sets can change membership by 1
However,
the cardinalities of those sets cannot change by 1
This proves that cardinality is a fuzzy property.
No, it shows that infinite values act differently than finite numbers.
They do not act but are described. Blind men use fuzzy measured, seeing
men use precise measures.
Regards, WM
On 28.11.2024 01:30, Richard Damon wrote:
On 11/27/24 4:43 PM, WM wrote:
On 27.11.2024 22:14, Richard Damon wrote:
On 11/27/24 2:15 PM, WM wrote:
It it the successor for the SET of natural numbers.
And that is nothing else but all natural numbers.
No, the set is different from its members.
By what?
It is a set, with set type properties, and its members are Natural
Numbers, with Natural Number like propertis.
The set is equal to N, which is the name of the set, but none of the
members are 'equal' to N, they are MEMBERS of it.
The members together are ℕ.
The collection of definable natnumbers is ℕ_def.
The collection of all natnumbers is ℕ.
Regards, WM
On 11/27/2024 4:33 PM, WM wrote:
On 27.11.2024 20:47, Jim Burns wrote:
Finite cardinals can change by 1
The cardinal |ℕᶠⁱⁿ| cannot change by 1
Small wonder.
Fuzzy properties like "many" cannot change by 1.
⎛ ℕᶠⁱⁿ is the set of finite cardinals.
⎜ Bob is not a cardinal.
⎜
⎜ ∀ᶜᵃʳᵈξ: ξ ∈ ℕᶠⁱⁿ ⇔
⎜ ⟦0,ξ⦆∪{Bob} ≠ ⟦0,ξ⟧ ∧ |⟦0,ξ⦆∪{Bob}| ≠ |⟦0,ξ⟧|
⎜
⎝ ℕᶠⁱⁿ∪{Bob} ≠ ℕᶠⁱⁿ ∧ |ℕᶠⁱⁿ∪{Bob}| = |ℕᶠⁱⁿ|
On 27.11.2024 20:47, Jim Burns wrote:
Finite cardinals can change by 1
The cardinal |ℕᶠⁱⁿ| cannot change by 1
Small wonder.
Fuzzy properties like "many" cannot change by 1.
Yes,
the sets can change membership by 1
However,
the cardinalities of those sets cannot change by 1
This proves that cardinality is a fuzzy property.
This proves that cardinality is a fuzzy property.
After all the swaps
(of which no swap is a change in cardinality)
what remains is a proper subset
(which is not a change in cardinality).
On 29.11.2024 19:08, Jim Burns wrote:
After all the swaps
(of which no swap is a change in cardinality)
what remains is a proper subset
(which is not a change in cardinality).
Not proper.
No element can leave by swaps.
(which is not a change in cardinality).
Irrelevant. Cardinality is a fuzzy measure.
On 11/29/2024 2:37 PM, WM wrote:
On 29.11.2024 19:08, Jim Burns wrote:
No elements leave.After all the swaps
After all the swaps
what remains is a proper subset.
On 29.11.2024 21:54, Jim Burns wrote:
On 11/29/2024 2:37 PM, WM wrote:
On 29.11.2024 19:08, Jim Burns wrote:
JB contradicting himself:
After all the swaps
(of which no swap is a change in cardinality)
what remains is a proper subset
(which is not a change in cardinality).
The sets do not have any of
the cardinalities which would change.
The sets have a different cardinality,
one which does not change when the set changes.
After all the swaps
(of which no swap is a change in cardinality)
what remains is a proper subset
(which is not a change in cardinality).
Because infinite.
On 30.11.2024 02:44, Jim Burns wrote:
The sets do not have any of
the cardinalities which would change.
The sets have a different cardinality,
one which does not change when the set changes.
Then the intersection which is infinite
too remain infinite.
After all the swaps
(of which no swap is a change in cardinality)
what remains is a proper subset
(which is not a change in cardinality).
Because infinite.
Like the intersection.
Like the intersection.
Apparently, what you (WM) call "the intersection"
is each of infinitely.many intersections,
Our sets do not change.
On 30.11.2024 18:32, Jim Burns wrote:
Apparently, what you (WM) call "the intersection"
is each of infinitely.many intersections,
and their limit.
some infinite, some empty,
"changing" from one to another,
in a manner you accept or you do not accept.
Our sets do not change.
But there is a sequence of endsegments
E(1), E(2), E(3), ...
and a sequence of their intersections
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ... .
Both are identical -
from the first endsegment
on until every existing endsegment.
On 11/30/2024 2:07 PM, WM wrote:
But there is a sequence of endsegments
E(1), E(2), E(3), ...
With an empty set of common finite.cardinals.
and a sequence of their intersections
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ... .
With an empty set of common finite.cardinals.
Both are identical -
from the first endsegment
on until every existing endsegment.
Yes, they are identical. And identically empty.
Each infinite collection of
end.segments of
finite.cardinals
has an empty intersection.
On 30.11.2024 22:45, Jim Burns wrote:
On 11/30/2024 2:07 PM, WM wrote:
But there is a sequence of endsegments
E(1), E(2), E(3), ...
With an empty set of common finite.cardinals.
Inclusion monotony prevents
an empty set of common finite cardinals
without an empty endsegment.
Inclusion monotony prevents
an empty set of common finite cardinals
without an empty endsegment.
Yes, they are identical. And identically empty.
You said that the endsegments are never empty.
On 11/30/2024 4:56 PM, WM wrote:
Inclusion monotony prevents
an empty set of common finite cardinals without an empty endsegment.
No
inclusion.monotony does not prevent
an empty set of common finite.ordinals
without an empty end segment.
On 01.12.2024 00:34, Jim Burns wrote:
On 11/30/2024 4:56 PM, WM wrote:
Inclusion monotony prevents
an empty set of common finite cardinals without an empty endsegment.
No
It does.
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because
E(1)∩E(2)∩...∩E(n) = E(n).
inclusion.monotony does not prevent
an empty set of common finite.ordinals
without an empty end segment.
Stupid or impudent. No reason to continue.
Regards, WM
On 01.12.2024 00:34, Jim Burns wrote:
On 11/30/2024 4:56 PM, WM wrote:
Inclusion monotony prevents
an empty set of common finite cardinals
without an empty endsegment.
No
It does.
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit
because
E(1)∩E(2)∩...∩E(n) = E(n).
inclusion.monotony does not prevent
an empty set of common finite.ordinals
without an empty end segment.
Stupid or impudent. No reason to continue.
On 12/1/2024 5:02 AM, WM wrote:
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit
because
E(1)∩E(2)∩...∩E(n) = E(n).
Identical sequences without an empty end.segment.
Identical empty set of common finite.cardinals.
Inclusion monotony does not prevent
an empty set of common finite.cardinals
without an empty end.segment.
On 01.12.2024 21:02, Jim Burns wrote:
On 12/1/2024 5:02 AM, WM wrote:
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit
because
E(1)∩E(2)∩...∩E(n) = E(n).
Identical sequences without an empty end.segment.
Identical empty set of common finite.cardinals.
Inclusion monotony does not prevent
an empty set of common finite.cardinals
without an empty end.segment.
Stupid or impudent. No reason to continue.
Infinite[ly many] finite sets: <<< [!]
{ { 1 }, { 1, 2 }, { 1, 2, 3 }, { 1, 2, 3, 4 }, ... }
?
On 30.11.2024 22:45, Jim Burns wrote:Wrong.
On 11/30/2024 2:07 PM, WM wrote:
Inclusion monotony prevents an empty set of common finite cardinalsBut there is a sequence of endsegments E(1), E(2), E(3), ...With an empty set of common finite cardinals.
without an empty endsegment.
In non-empty endsegments there are common finite cardinals.Only for finite intersections.
The limit is empty.You said that the endsegments are never empty.and a sequence of their intersections E(1), E(1)∩E(2), E(1)∩E(2)∩E(3),Yes, they are identical. And identically empty.
Both are identical -
from the first endsegment on until every existing endsegment.
And as long as you only consider a finite number of segments, you don'tEach infinite collection of end segments ofBecause an infinite collection of endsegments requires infinitely many indices, that is all indices. As long as some contents n, n+1, n+2, ...
finite cardinals has an empty intersection.
is in endsegments, the indices reach only from 1 to n-1.
Am 02.12.2024 um 00:19 schrieb Chris M. Thomasson:
Infinite[ly many] finite sets: <<< [!]
{ { 1 }, { 1, 2 }, { 1, 2, 3 }, { 1, 2, 3, 4 }, ... }
?
In fact, in axiomatic set theory (due to von Neumann):
IN = {{}, {0}, {0, 1}, {0, 1, 2}, ...}
where 0 = {}, 1 = {0}, 2 = {0, 1}, ... etc. In general: n+1 = {0, ..., n}.
Am Sat, 30 Nov 2024 22:56:00 +0100 schrieb WM:
On 30.11.2024 22:45, Jim Burns wrote:Wrong.
On 11/30/2024 2:07 PM, WM wrote:Inclusion monotony prevents an empty set of common finite cardinals
But there is a sequence of endsegments E(1), E(2), E(3), ...With an empty set of common finite cardinals.
without an empty endsegment.
In non-empty endsegments there are common finite cardinals.Only for finite intersections.
The limit is empty.You said that the endsegments are never empty.and a sequence of their intersections E(1), E(1)∩E(2), E(1)∩E(2)∩E(3),Yes, they are identical. And identically empty.
Both are identical -
from the first endsegment on until every existing endsegment.
And as long as you only consider a finite number of segments,Each infinite collection of end segments ofBecause an infinite collection of endsegments requires infinitely many
finite cardinals has an empty intersection.
indices, that is all indices. As long as some contents n, n+1, n+2, ...
is in endsegments, the indices reach only from 1 to n-1.
you don't
have infinitely many of them.
On 15.12.2024 11:51, Mikko wrote:On the contrary. The density of the points prevents passing them.
On 2024-12-14 21:40:48 +0000, WM said:That contradicts the actual existence of all.
In a geometry where all points exist, all points can be passed.Yes but none of them can be passed before passing other points.
When other points are passed, the former has been passed before.When. This goes for all points, so none can actually be passed.
Otherwise it would not be the former.
Am Sun, 15 Dec 2024 12:12:13 +0100 schrieb WM:
On 15.12.2024 11:51, Mikko wrote:On the contrary. The density of the points prevents passing them.
On 2024-12-14 21:40:48 +0000, WM said:That contradicts the actual existence of all.
In a geometry where all points exist, all points can be passed.Yes but none of them can be passed before passing other points.
On 11/9/2024 3:49 AM, WM wrote:
Huh? Keep in mind that
On 12/26/2024 2:49 PM, Moebius wrote:
Am 26.12.2024 um 05:52 schrieb Chris M. Thomasson:
On 11/9/2024 3:49 AM, WM wrote:
idiotic nonsense.
Huh? Keep in mind that
Huh?! "Mind"?! Is that a joke?
WM has to have some kind of mind, right? ;^)
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