Take a number that wants to get close to zero.
Say:
[0] = 1
[1] = .1
[2] = .01
[3] = .001
[...] = [...]
This gets close to zero, yet never will equal zero. Okay so:
arbitrarily close seems to be the accepted term.
infinitely close is the wrong wording?
The function f(n) = 10^(-n) gets "infinitely close" to 0... lol. Using the "metaphysical formation" of arbitrarily close... ;^)
Take a number that wants to get close to zero. Say:
[0] = 1
[1] = .1
[2] = .01
[3] = .001
[...] = [...]
This gets close to zero, yet never will equal zero. Okay so:
arbitrarily close seems to be the accepted term.
infinitely close is the wrong wording?
Take a number that wants to get close to zero. Say:
[0] = 1
[1] = .1
[2] = .01
[3] = .001
[...] = [...]
This gets close to zero, yet never will equal zero. Okay so:
arbitrarily close seems to be the accepted term.
infinitely close is the wrong wording?
The function f(n) = 10^(-n) gets "infinitely close" to 0... lol. Using the "metaphysical formation"
of arbitrarily close... ;^)
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
Take a number that wants to get close to zero.
This makes no sense. "a number" is one number. And numbers don't want anything.
Yes, it's not clear what "infinitely close" means
But the infinite is always defined as the unlimited.
No. You can't count to infinity.
Mike Terry schrieb am Freitag, 16. Februar 2024 um 22:45:48 UTC+1:
Yes, it's not clear what "infinitely close" means
It means dark numbers.
The function Number of Unit Fractions between (0, and x) has
the following properties:
(1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen unless
NUF(x) increases at some x.
(2) NUF(x) cannot increase other than when passing unit fractions at some
x = 1/n.
(3) NUF(x) cannot pass more than one unit fraction at a single point x because
∀n ∈ ℕ: 1/n =/= 1/(n-1).
(4) This requires a first unit fraction, if all are there in actual
infinity.
Regards, WM
Chris M. Thomasson wrote:(𝗔𝗥𝗘) despicable lying 𝗸𝗵𝗮𝘇𝗮𝗿_𝗴𝗼𝘆𝘀.
Take a number that wants to get close to zero. Say:
[0] = 1
[1] = .1
[2] = .01
[3] = .001
[...] = [...]
This gets close to zero, yet never will equal zero. Okay so:
arbitrarily close seems to be the accepted term.
infinitely close is the wrong wording?
in my country a > [...] = [...] is not a number. It's imbecility.
I believe this proves the gearmons and 𝘁𝗵𝗲_Hitler_𝗻𝗮𝘇𝗶𝘀 were
Ukrainian President Vladimir Zelensky has said the agreement proves“Ukraine will be in NATO”
in my country a > [...] = [...] is not a number. It's imbecility.(𝗔𝗥𝗘) despicable lying 𝗸𝗵𝗮𝘇𝗮𝗿_𝗴𝗼𝘆𝘀.
I believe this proves the gearmons and 𝘁𝗵𝗲_Hitler_𝗻𝗮𝘇𝗶𝘀 were
𝗚𝗲𝗿𝗺𝗮𝗻𝘆_𝗮𝗻𝗱_𝗨𝗸𝗿𝗮𝗶𝗻𝗲_𝘀𝗶𝗴𝗻_‘𝗹𝗼𝗻𝗴_𝘁𝗲𝗿𝗺’_𝘀𝗲𝗰𝘂𝗿𝗶𝘁𝘆_𝗱𝗲𝗮𝗹
Ukrainian President Vladimir Zelensky has said the agreement proves“Ukraine will be in NATO”
Hey.The pieces of shit called Vladimir Putin and Alexander Bastrykin
just murdered Alexei Navalny.
Mike Terry schrieb am Freitag,
16. Februar 2024 um 22:45:48 UTC+1:
Yes,
it's not clear what "infinitely close" means
It means dark numbers.
The function
Number of Unit Fractions between (0, and x)
has the following properties:
(1) An increase from NUF(0) = 0 to NUF(x>0) > 0
cannot happen unless NUF(x) increases at some x.
(2) NUF(x) cannot increase other than
when passing unit fractions at some x = 1/n.
(3) NUF(x) cannot pass more than one
unit fraction at a single point x because
∀n ∈ ℕ: 1/n =/= 1/(n-1).
(4) This requires a first unit fraction,
if all are there in actual infinity.
On 2/17/2024 4:56 AM, WM wrote:
The function
Number of Unit Fractions between (0, and x)
has the following properties:
(1) An increase from NUF(0) = 0 to NUF(x>0) > 0
cannot happen unless NUF(x) increases at some x.
NUF(x) increases at 0
(2) NUF(x) cannot increase other than
when passing unit fractions at some x = 1/n.
NUF(x) cannot increase other than
when ∀β > 0: NUF(x-β) < NUF(x+β)
(3) NUF(x) cannot pass more than one
unit fraction at a single point x because
∀n ∈ ℕ: 1/n =/= 1/(n-1).
∀n ∈ ℕ: 1/n =/= 0
(4) This requires a first unit fraction,
if all are there in actual infinity.
Each final.ordinal.reciprocal
is preceded by
another final.ordinal.reciprocal.
The first final.ordinal.reciprocal not.exists.
x wrote:𝗮_𝗬𝗮𝗹𝗲_𝗱𝗶𝗽𝗹𝗼𝗺𝗮 from fucking
> in my country a > [...] = [...] is not a number. It's imbecility.
> I believe this proves the gearmons and 𝘁𝗵𝗲_Hitler_𝗻𝗮𝘇𝗶𝘀 were
(𝗔𝗥𝗘) despicable lying 𝗸𝗵𝗮𝘇𝗮𝗿_𝗴𝗼𝘆𝘀.
>
𝗚𝗲𝗿𝗺𝗮𝗻𝘆_𝗮𝗻𝗱_𝗨𝗸𝗿𝗮𝗶𝗻𝗲_𝘀𝗶𝗴𝗻_‘𝗹𝗼𝗻𝗴_𝘁𝗲𝗿𝗺’_𝘀𝗲𝗰𝘂𝗿𝗶𝘁𝘆_𝗱𝗲𝗮𝗹
> Ukrainian President Vladimir Zelensky has said the agreement proves
“Ukraine will be in NATO”
Hey.The pieces of shit called Vladimir Putin and Alexander Bastrykin
just murdered Alexei Navalny.
navalne?? are you fucking stupid. The traitor got
𝗰𝗶𝗮, was "𝗽𝗼𝗶𝘀𝗼𝗻𝗲𝗱" in little britain by "evil" Russians,because they
...
On 2/17/24 3:23 AM, WM wrote:
Take the function Number of Unit Fractions between (0, and x > 0). It
has the following properties:
(1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen unless
NUF(x) increases at some x.
(2) NUF(x) cannot increase other than when passing unit fractions at
some x = 1/n.
(3) NUF(x) cannot pass more than one unit fraction at a single point x
because
∀n ∈ ℕ: 1/n =/= 1/(n-1).
(4) This requires a first unit fraction, if all are there in actual
infinity.
Or, that such a function can't actually be defined, because it assumes
that there IS a "smallest unit fraction".
On 2/17/24 4:56 AM, WM wrote:
Mike Terry schrieb am Freitag, 16. Februar 2024 um 22:45:48 UTC+1:In other words, "Dark Numbers" are made up numbers that try to patch the holes in your logic
Yes, it's not clear what "infinitely close" means
It means dark numbers.
The function Number of Unit Fractions between (0, and x) has
the following properties:
(1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen unless
NUF(x) increases at some x.
(2) NUF(x) cannot increase other than when passing unit fractions at some
x = 1/n.
(3) NUF(x) cannot pass more than one unit fraction at a single point x
because
∀n ∈ ℕ: 1/n =/= 1/(n-1).
(4) This requires a first unit fraction, if all are there in actual
infinity.
Of course, since your premises are just wrong,
Le 17/02/2024 à 13:15, Richard Damon a écrit :
On 2/17/24 4:56 AM, WM wrote:
Mike Terry schrieb am Freitag, 16. Februar 2024 um 22:45:48 UTC+1:In other words, "Dark Numbers" are made up numbers that try to patch
Yes, it's not clear what "infinitely close" means
It means dark numbers.
The function Number of Unit Fractions between (0, and x) has
the following properties:
(1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen unless
NUF(x) increases at some x.
(2) NUF(x) cannot increase other than when passing unit fractions at
some
x = 1/n.
(3) NUF(x) cannot pass more than one unit fraction at a single point x
because
∀n ∈ ℕ: 1/n =/= 1/(n-1).
(4) This requires a first unit fraction, if all are there in actual
infinity.
the holes in your logic
There are no holes in my logic. There is nonsense in your belief.
Of course, since your premises are just wrong,
My premises are (1) to (3). Nothing wrong.
Regards, WM
Le 17/02/2024 à 19:35, Jim Burns a écrit :
On 2/17/2024 4:56 AM, WM wrote:
(4) This requires a first unit fraction,
if all are there in actual infinity.
Each final.ordinal.reciprocal
is preceded by
another final.ordinal.reciprocal [⅟n]
No, this axiom must be given up.
The first final.ordinal.reciprocal not.exists.
The alternative would be
an increase of NUF(x) to infinity at zero.
Not acceptable.
Your assumptions that define it are inconsistant with the definition of Natural Numbers.
Which do you (WM) give up?
How do you justify giving it up?
Le 18/02/2024 à 19:16, Jim Burns a écrit :
Which do you (WM) give up?
How do you justify giving it up?
I will never give up the following self-evidence:
If there are ℵo unit fractions in the interval (0, eps), then there is
an x with only a finite number of unit fractions in (0, x).
Why? Because unit fractions are real points on the real line. They
cannot appear as an infinite swarm without a finite start.
The intersection of all intervals (0, eps) that can be chosen by anybody
in eternity however contains ℵo unit fractions.
Regards, WM
On 2/16/2024 1:07 PM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
Take a number that wants to get close to zero.This makes no sense. "a number" is one number. And numbers don't want
anything.
That was designed to raise a laugh or two. I guess it bombed. Yikes!
Say:Strange notation. [0] = 1. Eh? Why not just use a more conventional
[0] = 1
[1] = .1
[2] = .01
[3] = .001
[...] = [...]
notation for a s sequence:
Too used to a programming language wrt indexing arrays I guess. :^)
s_0 = 1
s_1 = 0.1
etc.
You can, if you prefer, write it as a function: s(n) = 10^-n (as you do
later).
This gets close to zero, yet never will equal zero. Okay so:But it's not a very good one. For example, one could say that
arbitrarily close seems to be the accepted term.
p(n) = 2^n when n is even
p(n) = 2^-n when n is odd
gets arbitrarily close to zero but also arbitrarily far away from zero.
That's fine with me. I can see it wrt your logic.
infinitely close is the wrong wording?I would not know what you mean if you said that, so I would say it's the
wrong wording. The best wording is to say
lim_{n->oo} s(n) = 0.
which you can read as "the limit, as n tends to infinity, if s(n) is
zero".
The limit of f(n) = 10^(-n) is zero. However, none of the iterates equal zero. They just get closer and closer to it...
On 2/18/24 4:00 PM, WM wrote:
Le 18/02/2024 à 19:16, Jim Burns a écrit :
Which do you (WM) give up?
How do you justify giving it up?
I will never give up the following self-evidence:
If there are ℵo unit fractions in the interval (0, eps), then there is
an x with only a finite number of unit fractions in (0, x).
Why? Because unit fractions are real points on the real line. They
cannot appear as an infinite swarm without a finite start.
But the start was at 1/1
Remember "real points" take up no space,
0 .
Le 19/02/2024 à 00:12, Richard Damon a écrit :
On 2/18/24 4:00 PM, WM wrote:
Le 18/02/2024 à 19:16, Jim Burns a écrit :
Which do you (WM) give up?
How do you justify giving it up?
I will never give up the following self-evidence:
If there are ℵo unit fractions in the interval (0, eps), then there
is an x with only a finite number of unit fractions in (0, x).
Why? Because unit fractions are real points on the real line. They
cannot appear as an infinite swarm without a finite start.
But the start was at 1/1
Remember "real points" take up no space,
But unit fractions have internal distances and they take up space.
If there is a set of real points with distances at the real axis, then
every point can be considered as the border between two subsets. If it
is impossible to reduce the left-hand subset to a finite amount, then
there is no point available dividing infinitely many unit fractions.
Then they sit at one point. That is impossible by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
0 .
Regards, WM
On 2/19/24 3:14 AM, WM wrote:
Le 19/02/2024 à 00:12, Richard Damon a écrit :
On 2/18/24 4:00 PM, WM wrote:
Le 18/02/2024 à 19:16, Jim Burns a écrit :
Which do you (WM) give up?
How do you justify giving it up?
I will never give up the following self-evidence:
If there are ℵo unit fractions in the interval (0, eps), then there
is an x with only a finite number of unit fractions in (0, x).
Why? Because unit fractions are real points on the real line. They
cannot appear as an infinite swarm without a finite start.
But the start was at 1/1
Remember "real points" take up no space,
But unit fractions have internal distances and they take up space.
But a space that gets vanishingly small,
If there is a set of real points with distances at the real axis, then
every point can be considered as the border between two subsets. If it
is impossible to reduce the left-hand subset to a finite amount, then
there is no point available dividing infinitely many unit fractions.
Then they sit at one point. That is impossible by ∀n ∈ ℕ: 1/n - 1/(n+1)
= d_n
Why must the left side every become finite?
I guess you agree that Achilles can't pass the Tortoise, or maybe only
at some "dark" time that we can not see.
On Saturday, February 17, 2024 at 2:00:16 AM UTC-8, WM wrote:
because ∀n ∈ ℕ: 1/n =/= 1/(n-1).
This requires a first unit fraction, if all are there in actual
infinity. Of course the first unit fractions cannot be seen. They are
dark. Regards, WM
The first fraction is 1/infinity.That is dark. Zero is below it and
can't be seen. It is not even dark.
On 2/19/24 3:14 AM, WM wrote:
Le 19/02/2024 à 00:12, Richard Damon a écrit :
On 2/18/24 4:00 PM, WM wrote:
Le 18/02/2024 à 19:16, Jim Burns a écrit :
Which do you (WM) give up?
How do you justify giving it up?
I will never give up the following self-evidence:
If there are
ℵo unit fractions in the interval (0, eps),
then there is an x with only a finite number of
unit fractions in (0, x).
Why? Because unit fractions are real points on
the real line.
They cannot appear as an infinite swarm without
a finite start.
But the start was at 1/1
Remember "real points" take up no space,
But unit fractions have internal distances
and they take up space.
But a space that gets vanishingly small,
and thus
we CAN fit an infinite number of them in
a finite space.
Le 18/02/2024 à 19:16, Jim Burns a écrit :
One of these claim must be given up:
| a final.ordinal.reciprocal.free zone (0,δ)
| exists
or
| a skipping.function isn't all.continuous
or
| for final.ordinal.reciprocal ⅟m
| ⅟(4⋅m) is a final.ordinal.reciprocal.
Which do you (WM) give up?
How do you justify giving it up?
I will never give up the following self-evidence:
If
there are ℵo unit fractions in
the interval (0, eps),
then
there is an x with only a finite number of
unit fractions in (0, x).
Why?
Because unit fractions are real points on
the real line.
They cannot appear as
an infinite swarm without a finite start.
Le 19/02/2024 à 13:33, Richard Damon a écrit :
On 2/19/24 3:14 AM, WM wrote:
Le 19/02/2024 à 00:12, Richard Damon a écrit :
On 2/18/24 4:00 PM, WM wrote:
Le 18/02/2024 à 19:16, Jim Burns a écrit :
Which do you (WM) give up?
How do you justify giving it up?
I will never give up the following self-evidence:
If there are ℵo unit fractions in the interval (0, eps), then there >>>>> is an x with only a finite number of unit fractions in (0, x).
Why? Because unit fractions are real points on the real line. They
cannot appear as an infinite swarm without a finite start.
But the start was at 1/1
Remember "real points" take up no space,
But unit fractions have internal distances and they take up space.
But a space that gets vanishingly small,
Bot by the number of points - there are always infinitely many between
two adjacent unit fractions.
If there is a set of real points with distances at the real axis,
then every point can be considered as the border between two subsets.
If it is impossible to reduce the left-hand subset to a finite
amount, then there is no point available dividing infinitely many
unit fractions. Then they sit at one point. That is impossible by ∀n
∈ ℕ: 1/n - 1/(n+1) = d_n
Why must the left side every become finite?
If there are teally existing real points, then each one can be used, in principle, as the border.
I guess you agree that Achilles can't pass the Tortoise, or maybe only
at some "dark" time that we can not see.
Irrelevant for the present topic. But true that he overtakes in darkness.
Regards, WM
On 2/18/2024 4:00 PM, WM wrote:
I will never give up the following self-evidence:
If
there are ℵo unit fractions in
the interval (0, eps),
then
there is an x with only a finite number of
unit fractions in (0, x).
Why?
Because unit fractions are real points on
the real line.
They cannot appear as
an infinite swarm without a finite start.
In other words,
infinitely.many are leftward
finitely.many are rightward
for each ⅟j
No ⅟j is in a finite start
That is the reason that
an infinite swarm is the only possibility
One can avoid using the I.word by saying
| we CAN fit more than any finite number of them in
| a finite space.
On 2/19/24 7:59 AM, WM wrote:
If there is a set of real points with distances at the real axis,
then every point can be considered as the border between two subsets.
If it is impossible to reduce the left-hand subset to a finite
amount, then there is no point available dividing infinitely many
unit fractions. Then they sit at one point. That is impossible by ∀n >>>> ∈ ℕ: 1/n - 1/(n+1) = d_n
Why must the left side every become finite?
If there are really existing real points, then each one can be used, in
principle, as the border.
No, only the one ON the border would be the border, but for every one
that you might want to think of as the border, there are more that are
closer to the border.
Thus, there is NOT one (in the set) on the border, and thus there is no "first" (from the left) unit fraction, and thus NUF(x) isn't defined.
I guess you agree that Achilles can't pass the Tortoise, or maybe only
at some "dark" time that we can not see.
Irrelevant for the present topic. But true that he overtakes in darkness.
But he doesn't, he passes the Tortoise at a easily determined finite
time (if we know the actual speeds).
Le 19/02/2024 à 21:37, Jim Burns a écrit :
One can avoid using the I.word by saying
| we CAN fit more than any finite number of them in
| a finite space.
But if all are there existing from the scratch as an actually infinite
set, then each one can be addressed as border between two subsets, in principle. Even the smallest one.
Regards, WM
On 2/20/24 3:15 AM, WM wrote:
But if all are there existing from the scratch as an actually infinite
set, then each one can be addressed as border between two subsets, in
principle. Even the smallest one.
But there isn't a "Smallest One".
On 2/20/24 3:12 AM, WM wrote:
Le 19/02/2024 à 21:21, Jim Burns a écrit :
infinitely.many are leftward
finitely.many are rightward
for each ⅟j
That correctly describes an evolving infinite collection, i.e., a
potentially infinite set where more and more elements are created which
initially have not existed.
A complete, i.e. actually infinite set of ℵo real fixed points on the
real axis can be subdivided by any of its elements (since all are
existing) such that the subsets have cardinalities from 0, ℵo over n, ℵo >> to ℵo, n, and ℵo, 0.
But ℵo / n is still ℵo,
so you never get to 0.
No ⅟j is in a finite start
That is the reason that
an infinite swarm is the only possibility
Then you are not talking about really existing invariable points. But
that is what I discuss.
But neither are you talking about what really exists,
Le 19/02/2024 à 21:21, Jim Burns a écrit :
In other words,
infinitely.many are leftward
finitely.many are rightward
for each ⅟j
That correctly describes
an evolving infinite collection, i.e.,
a potentially infinite set where
more and more elements are created which
initially have not existed.
In other words,
infinitely.many are leftward
finitely.many are rightward
for each ⅟j
That correctly describes
an evolving infinite collection, i.e.,
a potentially infinite set where
more and more elements are created which
initially have not existed.
No ⅟j is in a finite start
That is the reason that
an infinite swarm is the only possibility
Then you are not talking about
really existing invariable points.
But that is what I discuss.
On 2/20/2024 3:12 AM, WM wrote:
Le 19/02/2024 à 21:21, Jim Burns a écrit :
In other words,
infinitely.many are leftward
finitely.many are rightward
for each ⅟j
That correctly describes
an evolving infinite collection, i.e.,
a potentially infinite set where
more and more elements are created which
initially have not existed.
We are finite beings.
The statements are finitely.many.
Their finitely.many.ness doesn't evolve.
It is a property of
finite sequences of statements that
if any of them is false,
then one of then is first.false.
Le 20/02/2024 à 21:02, Jim Burns a écrit :
It is a property of
finite sequences of statements that
if any of them is false,
then one of then is first.false.
Here is one statement that is true:
Every unit fraction exists on the real line.
But there are no marks indicating their places.
We cannot go to it without, in principle,
passing through all smaller ones.
Counting from 1, 2, 3, ... to n.
Le 20/02/2024 à 13:49, Richard Damon a écrit :
On 2/20/24 3:12 AM, WM wrote:
Le 19/02/2024 à 21:21, Jim Burns a écrit :
infinitely.many are leftward
finitely.many are rightward
for each ⅟j
That correctly describes an evolving infinite collection, i.e., a
potentially infinite set where more and more elements are created
which initially have not existed.
A complete, i.e. actually infinite set of ℵo real fixed points on the
real axis can be subdivided by any of its elements (since all are
existing) such that the subsets have cardinalities from 0, ℵo over n,
ℵo to ℵo, n, and ℵo, 0.
But ℵo / n is still ℵo,
subsets have cardinalities from (0, ℵo) over (n, ℵo) to (ℵo, n), and (ℵo, 0).
so you never get to 0.
Of course not. They are dark. But nevertheless these sets are existing.
No ⅟j is in a finite start
That is the reason that
an infinite swarm is the only possibility
Then you are not talking about really existing invariable points. But
that is what I discuss.
But neither are you talking about what really exists,
If all unit fractions really exist, then we can talk about each one,
although we cannot find most.
Regards, WM
On 2/20/2024 4:59 PM, WM wrote:
Every unit fraction exists on the real line.
But there are no marks indicating their places.
For each final.ordinal.reciprocal ⅟j
a geometric procedure exists which finds it.
On 2/20/24 11:47 AM, WM wrote:
You are obviosuly wrong. If all are there, then all can be used to
divide the set into two parts. None is exempt.
And we CAN do that, since none of them are "the first", so ALL of them
have an infinite number of point before them.
On 2/20/24 11:52 AM, WM wrote:
Le 20/02/2024 à 13:49, Richard Damon a écrit :
On 2/20/24 3:12 AM, WM wrote:
Le 19/02/2024 à 21:21, Jim Burns a écrit :
infinitely.many are leftward
finitely.many are rightward
for each ⅟j
That correctly describes an evolving infinite collection, i.e., a
potentially infinite set where more and more elements are created
which initially have not existed.
A complete, i.e. actually infinite set of ℵo real fixed points on the >>>> real axis can be subdivided by any of its elements (since all are
existing) such that the subsets have cardinalities from 0, ℵo over n, >>>> ℵo to ℵo, n, and ℵo, 0.
subsets have cardinalities from (0, ℵo) over (n, ℵo) to (ℵo, n), and >> (ℵo, 0).
If all unit fractions really exist, then we can talk about each one,
although we cannot find most.
We can find any one that we want.
At least as long as you don't try to qualify it with an impossible
qualifier, like the lowest value one.
On 2/20/24 4:59 PM, WM wrote:
We cannot use everything that exists on the real line, because among
them there is the smallest unit fraction, at least the smallest unit
fraction that exists on the real line. Where else should it be? This
existence is static. You seem to deny it. If we could point to it, we
caught the smallest unit fraction. But we cannot point to it although it
must be there. That proves: It is dark.
No, there ISN'T a "Smallest Unit Fraction" as has been shown.
Since ALL Unit Fractions "exist" in the mathematical sense,.
"Counting " the unit fractions can only be done from 1/1 to 1/2 to 1`/3
and so on.
You can't start counting from an end that doesn't exist. Trying to do
so, just breaks your system.
Le 21/02/2024 à 03:17, Richard Damon a écrit :
On 2/20/24 11:47 AM, WM wrote:
You are obviosuly wrong. If all are there, then all can be used to
divide the set into two parts. None is exempt.
And we CAN do that, since none of them are "the first", so ALL of them
have an infinite number of point before them.
If all are there, timeless and static, then one of them is the first.
NUF(x) growing from 0 to ℵo immediately would contradict mathematics according to which after every unit fraction there are points without
unit fraction.
Regards, WM
On 02/17/2024 10:35 AM, Jim Burns wrote:
...]
So, there's no first example where
"the equivalency function" isn't a model
of "not-a-real-function"
with "real analytical character",
considering that
the infinite and continuum limit
was already run out one-way.
Le 21/02/2024 à 00:55, Jim Burns a écrit :
On 2/20/2024 4:59 PM, WM wrote:
Every unit fraction exists on the real line.
But there are no marks indicating their places.
For each final.ordinal.reciprocal ⅟j
a geometric procedure exists which finds it.
Not for those existing next to zero.
For each final.ordinal.reciprocal ⅟j
a geometric procedure exists which finds it.
Not for those existing next to zero.
Note that
if reciprocals are existing on the real axis and
if all points are timeless,
then there is a point next to zero.
So your claim involves time. That is not
mathematics, which is time-independent
On 02/21/2024 12:02 PM, Jim Burns wrote:
[...]
This isn't the complete ordered field,
It's line-drawing, and it's the function
between discrete and continuous.
That's what it is.
It's line-drawing, and it's the function
between discrete and continuous.
On 2/21/24 3:52 AM, WM wrote:
Then take the first one existing there.
There isn't one, and you are just proving your ignornacd.
On 2/21/24 3:42 AM, WM wrote:
If all are there, timeless and static, then one of them is the first.
You don't understand the properties of UNBOUNDED sets.
Le 21/02/2024 à 13:32, Richard Damon a écrit :
On 2/21/24 3:52 AM, WM wrote:
Then take the first one existing there.
There isn't one, and you are just proving your ignornacd.
There is a first one in a static chain of points 1/n with gaps between
them. To deny this means falling victim to nonsense. Matheology.
Regards, WM
No positive point is next to zero,
Note that
if reciprocals are existing on the real axis and
if all points are timeless,
then there is a point next to zero.
Elaborate.
Do you reject
all skipping.functions being discontinuous.somewhere?
Do you reject
only both or neither ⅟m ⅟(4⋅m) being
final.ordinal.reciprocals findable by
geometric procedure?
On 2/22/24 7:01 AM, WM wrote:
Le 21/02/2024 à 13:32, Richard Damon a écrit :
On 2/21/24 3:42 AM, WM wrote:
If all are there, timeless and static, then one of them is the first.
You don't understand the properties of UNBOUNDED sets.
Sets between 0 and 1 have bounds.
Only if "Between" is INCLUSIVE, as the bounds are 0 and 1.
If the set EXCLUDES one or both of the bounds, it doesn't have it in
anymore.
On 2/22/24 7:04 AM, WM wrote:
Le 21/02/2024 à 13:32, Richard Damon a écrit :If the is, then NAME IT or explain how it can be.
On 2/21/24 3:52 AM, WM wrote:
Then take the first one existing there.
There isn't one, and you are just proving your ignornacd.
There is a first one in a static chain of points 1/n with gaps between
them. To deny this means falling victim to nonsense. Matheology.
(Not that your system
says it must be, that just shows your system is broken)
After all if some 1/n was actually the smallest, then that says that n
must be the highest natural number, but the definition of the Natural
Numbers says that the include the successor to all Natural Numbers, and
every number has a successor, so n+1 must be a Natural Number.
By your "Static" rule, it can't be that it comes into being when we look
at it, as that isn't "Stati".
Le 21/02/2024 à 13:32, Richard Damon a écrit :
On 2/21/24 3:42 AM, WM wrote:
If all are there, timeless and static,
then one of them is the first.
You don't understand
the properties of UNBOUNDED sets.
Sets between 0 and 1 have bounds.
Le 21/02/2024 à 20:26, Jim Burns a écrit :
No positive point is next to zero,
If all are there and timeless,
then there is a first one.
But it is more obvious that
the chain of unit fractiond must have a first one,
whenever there is a unit fraction at all.
Note that
if reciprocals are existing on the real axis and
if all points are timeless,
then there is a point next to zero.
Elaborate.
Nothing to elaborate.
Do you reject
all skipping.functions being discontinuous.somewhere?
Functions measuring elements are not
discontinuous for more than 1
when the elements have point between them.
Do you reject
only both or neither ⅟m ⅟(4⋅m) being
final.ordinal.reciprocals findable by
geometric procedure?
Dark numbers cannot be found.
ℕ is not finite.
[0,1] is not finite.
on 2/22/2024, WM supposed :
Sets between 0 and 1 have bounds.
Which sets?
On 2/22/2024 8:06 AM, WM wrote:
Le 21/02/2024 à 20:26, Jim Burns a écrit :
No positive point is next to zero,
If all are there and timeless,
then there is a first one.
Each positive point is preceded by
a positive point,
and no positive precedes each positive.
There isn't a first one.
But it is more obvious that
the chain of unit fractiond must have a first one,
whenever there is a unit fraction at all.
Each is preceded by some.
None precedes all.
Note that
if reciprocals are existing on the real axis and
if all points are timeless,
then there is a point next to zero.
Elaborate.
Nothing to elaborate.
Doesn't that worry you?
If you opened the hood of your auto,
and no engine was in there,
wouldn't you get a glimmer of a sense that,
perhaps, things were not as they should be?
Do you reject
all skipping.functions being discontinuous.somewhere?
Functions measuring elements are not
discontinuous for more than 1
when the elements have point between them.
A point β exists between
lower.bounds of final.ordinal.reciprocals and
not.lower.bounds of final.ordinal.reciprocals.
WM was thinking very hard :
Le 22/02/2024 à 15:39, FromTheRafters a écrit :
on 2/22/2024, WM supposed :
Sets between 0 and 1 have bounds.
Which sets?
All sets with elements x which obey 0 < x < 1.
Then your interval was unbounded (0,1) and your 'x' is strictly between
zero and one.
Le 22/02/2024 à 13:17, Richard Damon a écrit :
On 2/22/24 7:04 AM, WM wrote:
Le 21/02/2024 à 13:32, Richard Damon a écrit :If the is, then NAME IT or explain how it can be.
On 2/21/24 3:52 AM, WM wrote:
Then take the first one existing there.
There isn't one, and you are just proving your ignornacd.
There is a first one in a static chain of points 1/n with gaps
between them. To deny this means falling victim to nonsense. Matheology. >>>
I did.
(Not that your system says it must be, that just shows your system is
broken)
It is simply mathematics and logic: By logic there must be a start of
NUF(x), by mathematics the start can only be 1. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
After all if some 1/n was actually the smallest, then that says that n
must be the highest natural number, but the definition of the Natural
Numbers says that the include the successor to all Natural Numbers,
and every number has a successor, so n+1 must be a Natural Number.
The definition of unit fractions says that all have gaps and there is no point where more than 1 sit.
We cannot look at dark numbers.
By your "Static" rule, it can't be that it comes into being when we
look at it, as that isn't "Stati".
Regards, WM
Le 22/02/2024 à 15:54, Jim Burns a écrit :
Consider a finite set B
In each
transitive.trichotomous.order on B
each non.empty subset S of B holds
two extrema (minimum and maximum) of S
Each
transitive.trichotomous.order on B
has
the each.subset.two.extrema property.
More at
https://en.wikipedia.org/wiki/Finite_set
ℕ is not finite.
Ye4s, but what does this mean?
The visible part has no final element n,
because n+1 and 2n and n^n^n can be found
for every n.
The complete set however
is complete such that
no element can be added.
Based on Cantor's
Eigentlichunendlichem =
Transfinitum =
Vollendetunendlichem =
Unendlichseiendem =
kategorematice infinitum
there is a fixed number of numbers,
although we cannot count or determine it
better than by |ℕ|.
[0,1] is not finite.
It has bounds.
The numer of points is fixed.
Le 22/02/2024 à 19:12, Jim Burns a écrit :
On 2/22/2024 8:06 AM, WM wrote:
Le 21/02/2024 à 20:26, Jim Burns a écrit :
No positive point is next to zero,
If all are there and timeless,
then there is a first one.
Each positive point is preceded by
a positive point,
and no positive precedes each positive.
Your claim
If all are there and timeless,
then there is a first one.
requires to consider the predecessors.
"Timeless" means that
every point exists independent of a predecessor.
On 2/22/24 9:11 AM, WM wrote:
There is a first one in a static chain of points 1/n with gapsIf the is, then NAME IT or explain how it can be.
between them. To deny this means falling victim to nonsense. Matheology. >>>>
I did.
Nope, you just say you assume that there must be a smallest.
On 2/23/2024 3:55 AM, WM wrote:
The ordinals' descents and ascents are not the same.
For each ordinal ψ
if ψ has any infinite descent,
then, because well.order,
an ordinal χ exists first with any infinite descent.
ψ doesn't have an infinite descent.
Generalizing over ordinals,
each ordinal ψ has finite.descent.
On 2/23/2024 3:42 AM, WM wrote:
[0,1] is not finite.
It has bounds.
The numer of points is fixed.
[0,1] ⊇ [0,1) which has one extremum.
On 2/23/24 4:04 AM, WM wrote:
The border is either 0 an 1 or between 0 and 1. Real timeless complete
existence requires fixed smallest and largest points.
Eigentlichunendlichem = Transfinitum = Vollendetunendlichem =
Unendlichseiendem = kategorematice infinitum.
Remember, you have an INFINITE sized set (even if all the values are
finite) so your rule doesn't apply.
Don't understand you german, but you must be wrong.
And yes, if be "Real" you mean existing in the real physical world,
Le 23/02/2024 à 13:23, Richard Damon a écrit :
On 2/23/24 4:04 AM, WM wrote:
The border is either 0 an 1 or between 0 and 1. Real timeless
complete existence requires fixed smallest and largest points.
Eigentlichunendlichem = Transfinitum = Vollendetunendlichem =
Unendlichseiendem = kategorematice infinitum.
Remember, you have an INFINITE sized set (even if all the values are
finite) so your rule doesn't apply.
All number exist as points on the real line.
Don't understand you german, but you must be wrong.
Completed infinity. That means all points are there, existing,
independent of neighbours.
And yes, if be "Real" you mean existing in the real physical world,
All numbers exist as points on the real line.
Regards, WM
On 2/24/24 7:16 AM, WM wrote:
All numbers exist as points on the real line.Yes, an no point exists as the "first" above 0,
On 2/24/24 6:30 AM, WM wrote:
I prove it by NUF(x). Points on the real axis exist there without
necessity of neighbours or predecessors, and if they have internal
distances this cannot be doubted.
You first need to prove the NUF(x) exists.
Le 24/02/2024 à 16:31, Richard Damon a écrit :
On 2/24/24 7:16 AM, WM wrote:
All numbers exist as points on the real line.Yes, an no point exists as the "first" above 0,
That is in contradiction with logic. Static existence in a linear order enforces a first one.
Not so easy to be seen with integers. Very clear with unit fractions.
Your attempts do deny this contradict finite logic (there is no other)
and this are in vain. It is futile to discuss this fact. Therefore I
will non longer do so.
Regards, WM
Le 24/02/2024 à 16:31, Richard Damon a écrit :
On 2/24/24 6:30 AM, WM wrote:
If there are unit fratcions, then NUF exists.I prove it by NUF(x). Points on the real axis exist there without
necessity of neighbours or predecessors, and if they have internal
distances this cannot be doubted.
You first need to prove the NUF(x) exists.
Regards, WM
Le 23/02/2024 à 20:21, Jim Burns a écrit :
[...]
Have I overlooked an answer from you
concerning thos topic?
The ordinals' descents and ascents are not the same.
Every way up can be reversed.
That proves that also
the ascents are finite.
Generalizing over ordinals,
each ordinal ψ has finite.descent.
Each ordinal has finite ascent.
Le 24/02/2024 à 16:31, Richard Damon a écrit :
On 2/24/24 7:16 AM, WM wrote:
All numbers exist as points on the real line.
Yes, an no point exists as the "first" above 0,
That is in contradiction with logic.
Static existence in a linear order enforces
a first one.
It is futile to discuss this fact.
Therefore I will non longer do so.
Le 23/02/2024 à 14:11, Jim Burns a écrit :
On 2/23/2024 3:42 AM, WM wrote:
Le 22/02/2024 à 15:54, Jim Burns a écrit :
[0,1] is not finite.
It has bounds.
The numer of points is fixed.
[0,1] ⊇ [0,1) which has one extremum.
It has two but only one is visible.
Considering only unit fractions proves it clearly.
But considering all points does not chance
the principle.
On 2/24/2024 7:20 AM, WM wrote:
Le 23/02/2024 à 14:11, Jim Burns a écrit :
On 2/23/2024 3:42 AM, WM wrote:
Le 22/02/2024 à 15:54, Jim Burns a écrit :
[0,1] is not finite.
It has bounds.
The numer of points is fixed.
[0,1] ⊇ [0,1) which has one extremum.
It has two but only one is visible.
Considering only unit fractions proves it clearly.
Even if it did (which it doesn't)
[0,1] is not finite.
But considering all points does not chance
the principle.
ℕ ℚ and ℝ are not finite.
On 2/24/2024 11:14 AM, WM wrote:
It is futile to discuss this fact.
Therefore I will non longer do so.
Congratulations.
I suppose it is too soon to ask you to
apologize to your current and former students
for leading them astray.
On 2/24/2024 7:10 AM, WM wrote:
The ordinals' descents and ascents are not the same.
Every way up can be reversed.
An _infinite_ ascent,
such as ⟨ α α+1 α+2 α+3 ... ⟩
when reversed, isn't a descent.
⟨ α β γ δ ε ζ ... ⟩ is an ascent.
⟨ α β γ δ ε ζ ... ⟩ is strictly increasing.
And.
For each entry ξ {before.ξ} is finite.
An _infinite_ ascent doesn't end.
An infinite ascent _reversed_ doesn't begin
A descent begins.
That proves that also
the ascents are finite.
The ascents which, when reversed, are descents
are the finite ascents.
If you like,
you can exclude all infinite ascents,
and then prove that all _remaining_ ascents
are finite.
But that doesn't deny that
⟨ α α+1 α+2 α+3 ... ⟩ is an infinite ascent.
Le 24/02/2024 à 18:55, Jim Burns a écrit :
⟨ α β γ δ ε ζ ... ⟩ is an ascent.
Visible ascent has no "...".
Ever visible step occurs without "...".
⟨ α β γ δ ε ζ ... ⟩ is strictly increasing.
and can be reversed unless dark steps,
marked by "...", are involved.
⟨ α β γ δ ε ζ ... ⟩ is strictly increasing.
and can be reversed unless dark steps,
marked by "...", are involved.
And.
For each entry ξ {before.ξ} is finite.
Of course.
And reversible.
Onlky dark steps are not reversible.
Le 24/02/2024 à 20:25, Jim Burns a écrit :
On 2/24/2024 7:20 AM, WM wrote:
Le 23/02/2024 à 14:11, Jim Burns a écrit :
On 2/23/2024 3:42 AM, WM wrote:
Le 22/02/2024 à 15:54, Jim Burns a écrit :
[0,1] is not finite.
It has bounds.
The numer of points is fixed.
[0,1] ⊇ [0,1) which has one extremum.
It has two but only one is visible.
Considering only unit fractions proves it clearly.
Even if it did (which it doesn't)
[0,1] is not finite.
I said:
The number of points is fixed.
There are no more when starting
to count at the smaller ones.
But considering all points does not chance
the principle.
ℕ ℚ and ℝ are not finite.
Neveretheless
there are smallest positive points,
being dark.
Among final.ordinals ℕ and
differences.of.ratios ℚ of final.ordinals and
least.upper.bounds ℝ of bounded.non.empty.sets of
differences.of.ratios of final.ordinals,
there are neither darkᵂᴹ nor visibleᵂᴹ
smallest positive points.
On 2/25/2024 4:01 AM, WM wrote:
Le 24/02/2024 à 18:55, Jim Burns a écrit :
⟨ α β γ δ ε ζ ... ⟩ is an ascent.
Visible ascent has no "...".
Every visible step occurs without "...".
No step occurs with "..."
Because no step occurs with "..."
each step is visibleᵂᴹ
Le 25/02/2024 à 20:35, Jim Burns a écrit :
On 2/25/2024 4:01 AM, WM wrote:
Le 24/02/2024 à 18:55, Jim Burns a écrit :
⟨ α β γ δ ε ζ ... ⟩ is an ascent.
Visible ascent has no "...".
Every visible step occurs without "...".
No step occurs with "..."
"..." is only a convenient way to say that
the sequence continues in the way expected.
There are less.convenient, "...".free ways
to say that.
Then drop it
Then drop it
and name the steps.
Because no step occurs with "..."
each step is visibleᵂᴹ
Then drop it.
And show the first step which
cannot be reversed
froms ascent to descent.
Le 25/02/2024 à 21:58, Jim Burns a écrit :
Among final.ordinals ℕ and
differences.of.ratios ℚ of final.ordinals and
least.upper.bounds ℝ of bounded.non.empty.sets of
differences.of.ratios of final.ordinals,
there are neither darkᵂᴹ nor visibleᵂᴹ
smallest positive points.
You say so because
you cannot grasp the importance of
NUF(x) and Bob never leaving the matrix.
But your misunderstanding and
claiming the contrary
does not prove the contrary.
Among final.ordinals ℕ and
differences.of.ratios ℚ of final.ordinals and
least.upper.bounds ℝ of bounded.non.empty.sets of
differences.of.ratios of final.ordinals,
there are neither darkᵂᴹ nor visibleᵂᴹ
smallest positive points.
Generalizing,
each positive point x is not.smallest.
On 2/26/2024 3:32 AM, WM wrote:
Your request for names is arbitrary,
not importantly different from a request to have
the description and augmenting claims
inscribed in parchment and placed in your hand.
And show the first step which
cannot be reversed
froms ascent to descent.
A reverse finite ascent is
a finite descent.
A reverse infinite ascent is not
a descent of any kind,
We know that
an infinite ascent is infinite
by the same method that we know that
a right triangle has three corners.
Le 26/02/2024 à 17:44, Jim Burns a écrit :
We know that
an infinite ascent is infinite
by the same method that we know that
a right triangle has three corners.
Nonsense.
The only reason for the non-reversibility
is darkness.
Proof:
Every visible step is reversible.
On 02/26/2024 11:05 AM, Jim Burns wrote:
[...]
Does infinite induction complete?
Also, is it necessarily not.ultimately.untrue?
the _Crossing the Inductive Impass_,
the _Bridge that Fools Cannot Cross_,
On 02/27/2024 12:57 PM, Jim Burns wrote:
On 2/26/2024 2:19 PM, Ross Finlayson wrote:
the _Crossing the Inductive Impass_,
the _Bridge that Fools Cannot Cross_,
Fortunately,
we do not need to cross
the Bridge that Fools Cannot Cross
in order to learn what's on the other side.
We start with
a description of one of
what could turn out to be infinitely.many.
We continue with more claims,
each claim such that we can see it can't be first.false
by looking at the claims themselves for
not.first.false.ness,
here on the finite side of the Bridge.
In a _finite_ claim sequence,
if any claim is false,
then some claim is first.false.
By this method,
knowledge of the infinite grows out of
knowledge of the finite --
specifically, knowledge of finite sequences,
sequences of claims, but also of sheep, of pebbles,
here on the finite side of the Bridge.
So, you're claiming you _know_ where you're going,
and, that it's a matter of deduction
and, that it's a matter of deduction
that the infinite limit what so proceeds,
to, it, to _there_,
results not appreciably different.
On 02/27/2024 04:38 PM, Jim Burns wrote:
On 2/27/2024 4:57 PM, Ross Finlayson wrote:
and, that it's a matter of deduction
that the infinite limit what so proceeds,
to, it, to _there_,
results not appreciably different.
There is no proceeding.
Topologically speaking,
if a limit exists of an infinite series,
then, in each neighborhood of the limit,
there are almost.all points of the series,
AKA all.but.finitely.many.
Perhaps it could be useful to think of
a limit as a _synecdoche_ for the series,
where, by synecdoche, I mean
a part (such as 100 heads) standing for
the whole (such as 100 cattle).
I can see that
you can defend
not crossing
right up until
before the end.
Yet, are you going
to keep letting that
punk with a chicken
stapled to his face
keep passing you?
Thanks for bringing the
word synecdoche, I haven't
thought it in a while and
it's very good.
Still, though, a series that converges
has a limit, and, it has a value,
and, it does not converge to less.
I can see that
you can defend
not crossing
right up until
before the end.
Le 26/02/2024 à 20:05, Jim Burns a écrit :
Generalizing,
each positive point x is not.smallest.
There is a smallest unit fraction,
proven by NUF(x).
There is a smallest unit fraction,
proven by NUF(x).
There is also a smallest positive point
because all points are static.
Only potential
infinity can do without a smallest point.
On 02/28/2024 03:45 AM, Jim Burns wrote:
Consider the "line" of only the finite decimals.
1 is not in ⟨ 0.9 0.99 0.999 0.999 ... ⟩
ℼ is not in ⟨ 3.1 3.14 3.141 3.1415 ... ⟩
In neither case does the series cross
the _Bridge that Fools Cannot Cross_
The difference between them is that
a "there" is in the line to synecdoche
⟨ 0.9 0.99 0.999 0.999 ... ⟩
a "there" is not in the line to synecdoche
⟨ 3.1 3.14 3.141 3.1415 ... ⟩
I sort of dispute, according to the usual
definition of the limit and a property of
a series/sequence being Cauchy, where
you say "there" is "in the line" for the
inverse powers to zero but not the
spigot of digits of pi.
On 2/27/2024 9:31 AM, WM wrote:
Therefore, there is no
final.ordinal φ without final and larger φ⁺¹
And there is no
largest final.ordinal
Le 28/02/2024 à 23:57, Jim Burns a écrit :
Therefore, there is no
final.ordinal φ without final and larger φ⁺¹
That claim contradicts mathematics,
in particular NUF(x).
And there is no
largest final.ordinal
Stop your claims based on Peano
(which is only obtained from
experience with visible numbers).
Show how NUF(x) can jump from 0 to more than 1.
On 2/29/2024 3:30 AM, WM wrote:
Stop your claims based on Peano
(which is only obtained from
experience with visible numbers).
Show how NUF(x) can jump from 0 to more than 1.
Then, you (WM) are telling yourself that
| No largest final.ordinal
is unrelated?
On 02/28/2024 10:53 PM, Ross Finlayson wrote:
On 02/28/2024 08:07 PM, Jim Burns wrote:
[...][...]
So, about the infinite forms, then is that there
is the idea that there are quite usually directly
enough checks, and where there aren't it's where
quite usually enough "those are transcedental
numbers, and must have their own sorts checks".
So, this is for working up what are "the additions
of infinite forms", or, "the operations of infinite
forms", that make for the scaffolding, and its
relation to the inverse powers of two, then for
the telescoping and otherwise getting into the
issues of infinite forms and proper correctness
in their derivations and what's OK.
Le 29/02/2024 à 18:48, Jim Burns a écrit :
On 2/29/2024 3:30 AM, WM wrote:
Stop your claims based on Peano
(which is only obtained from
experience with visible numbers).
Show how NUF(x) can jump from 0 to more than 1.
Then, you (WM) are telling yourself that
| No largest final.ordinal
is unrelated?
I am not telling anything.
Therefore, there is no
final.ordinal φ without final and larger φ⁺¹
That claim contradicts mathematics,
in particular NUF(x).
I am adhering to mathematics which proves
that NUF(x) cannot jump from 0 to more than 1.
On 02/28/2024 08:07 PM, Jim Burns wrote:
On 2/28/2024 12:35 PM, Ross Finlayson wrote:
On 02/28/2024 03:45 AM, Jim Burns wrote:
Consider the "line" of only the finite decimals.
1 is not in ⟨ 0.9 0.99 0.999 0.999 ... ⟩
ℼ is not in ⟨ 3.1 3.14 3.141 3.1415 ... ⟩
In neither case does the series cross
the _Bridge that Fools Cannot Cross_
The difference between them is that
a "there" is in the line to synecdoche
⟨ 0.9 0.99 0.999 0.999 ... ⟩
a "there" is not in the line to synecdoche
⟨ 3.1 3.14 3.141 3.1415 ... ⟩
I sort of dispute, according to the usual
definition of the limit and a property of
a series/sequence being Cauchy, where
you say "there" is "in the line" for the
inverse powers to zero but not the
spigot of digits of pi.
The line (here) which ℼ is not in
is not the usual line.
My purpose (here) is to
illuminate the difference between
having a limit and not.having a limit.
Unfortunately for that purpose
(though fortunately for most other purposes)
the complete.ordered.field line holds
all the limits.
So, I am considering limits (or their absence) in
not.the.real.line:
only the finite decimals.
That sort of makes sense,
or doesn't make non-sense,
but it gets me to wondering
about "faux complete" what aren't gapless,
for example, or here that these terms,
"pseudo", "quasi", "degenerate", or here "faux",
with regards to that "pseudo" and "quasi"
sound pretty good, "mock", "synecdoche"
sounds pretty good. "Almost."
What is it you have in mind, here there's a
pretty good definition of "Flying Rainbow
Sparkle Pony", and, "The Relephant", which
are names we've used to describe models of
reals or continuous domains.
Numerical methods is the field of approximations,
when closed forms just won't do, as for example
they're infinite or un-known, or as where means
of approximation, and "finding", make for numerical
methods as the usual milieu of finite forms if
perhaps not closed (complete).
So, ..., reading generously that's sort of why
I disputed that and now I expect you to explain.
I mean, I would, but, we're both familiar with Grice,
and "understanding" is fairly explicit.
Here what I've been thinking about is this,
infinite series or forms, infinite sequences,
where series indicate sums, of partial sums,
for various cases of, operators, their inverses,
alternations, and variations in the coefficients,
where coefficient is a usual sort of scalar attached
otherwise to abstractly a value by an operand,
algebra's, then for example variously when those
are constants, like, continued fractions.
Series
++++++
Alternating Series
+-+-+-
Alternating Products
*/*/*/
Products
******
Roots
rad rad rad rad rad rad
Means
avg avg avg avg avg avg
So, in terms of numerical methods, say, and modeling
the error term, or, asymptotics and approximations,
and, also, ..., numerical methods and "tap time"
or the spigot, that a spigot algorithm as of a tap
produces elements of a sequence, given bounds in that,
carry on and explain yourself, it's quite unlike you
to give any sort of this kind of the underdefined,
and I won't go expecting to "not understand".
On 02/29/2024 11:37 AM, Jim Burns wrote:
True vs. false are very important in math.
Of course.
Another very important complementary pair:
satisfiable vs. gibberish.
There used to be one of these RPG's,
it was cartoons, instead of charisma
it was "chutzpah", for example, and
then everything still had to have its
logic for it, here that the coyote doesn't
feel the force of gravity until he looks down.
So anyways, in cartoon physics:
that which you don't know
can't hurt you.
Yet, in most structured RPG's,
it's exactly that
there are rules that
keep it from being just ridiculous,
On 2/29/2024 2:31 PM, WM wrote:
I am adhering to mathematics which proves
that NUF(x) cannot jump from 0 to more than 1.
¬(largest final.ordinal)
¬(smallest final.ordinal.reciprocal)
¬(NUF(x) = 1)
Le 29/02/2024 à 21:04, Jim Burns a écrit :
On 2/29/2024 2:31 PM, WM wrote:
I am adhering to mathematics which
proves that NUF(x) cannot jump
from 0 to more than 1.
¬(largest final.ordinal)
¬(smallest final.ordinal.reciprocal)
¬(NUF(x) = 1)
Incompatible with
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
On 02/29/2024 09:11 AM, Ross Finlayson wrote:
On 02/28/2024 10:53 PM, Ross Finlayson wrote:
On 02/28/2024 08:07 PM, Jim Burns wrote:
[...][...][...]
Sigma: sum
Integral: sum ....
Pi: product
On 3/1/2024 3:39 AM, WM wrote:
Le 29/02/2024 à 21:04, Jim Burns a écrit :
On 2/29/2024 2:31 PM, WM wrote:
I am adhering to mathematics which
proves that NUF(x) cannot jump
from 0 to more than 1.
¬(largest final.ordinal)
¬(smallest final.ordinal.reciprocal)
¬(NUF(x) = 1)
Incompatible with
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
No.
∀n ∈ ℕ: ⅟n - ⅟(n+1) > 0
Theorem.
∀x ∈ ℝ: NUF(cx) ≠ 1
(i)
∀x < 0: NUF(x) = 0 ≠ 1
(ii)
∀x > 0: NUF(x) ≠ 1
| Assume otherwise.
| Assume
| ¬∀x > 0: NUF(x) ≠ 1
|
| ∃x > 0: NUF(x) = 1
| ∃mₓ ∈ ℕ: mₓ ≤ ⅟x < mₓ⁺¹ < mₓ⁺²
|
| ∀n ∈ ℕ: ⅟n - ⅟(n+1) > 0
| mₓ < mₓ⁺¹ < mₓ⁺² < mₓ⁺³
| ⅟mₓ > ⅟mₓ⁺¹ > ⅟mₓ⁺² > ⅟mₓ⁺³
| ⅟mₓ - ⅟mₓ⁺¹ > 0
| ⅟mₓ⁺¹ - ⅟mₓ⁺² > 0
|
| However,
| ⅟x < mₓ⁺¹ < mₓ⁺²
| x > ⅟mₓ⁺¹ > ⅟mₓ⁺²
| NUF(x) ≥ |{⅟mₓ⁺¹,⅟mₓ⁺²}| = 2
| but NUF(x) = 1
| Contradiction.
Therefore,
∀x > 0: NUF(cx) ≠ 1
∀x ∈ ℝ: NUF(cx) ≠ 1
Le 01/03/2024 à 13:46, Jim Burns a écrit :
On 3/1/2024 3:39 AM, WM wrote:
Le 29/02/2024 à 21:04, Jim Burns a écrit :
On 2/29/2024 2:31 PM, WM wrote:
I am adhering to mathematics which
proves that NUF(x) cannot jump
from 0 to more than 1.
¬(largest final.ordinal)
¬(smallest final.ordinal.reciprocal)
¬(NUF(x) = 1)
Incompatible with
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
No.
You must be blind. There is a gap after *every* unit fraction.
∀n ∈ ℕ: ⅟n - ⅟(n+1) > 0
Theorem.
∀x ∈ ℝ: NUF(cx) ≠ 1
(i)
∀x < 0: NUF(x) = 0 ≠ 1
(ii)
∀x > 0: NUF(x) ≠ 1
| Assume otherwise.
Useless. There is a gap after *every* unit fraction. In every gap NUF(x)
is constant.
| Assume
| ¬∀x > 0: NUF(x) ≠ 1
|
| ∃x > 0: NUF(x) = 1
| ∃mₓ ∈ ℕ: mₓ ≤ ⅟x < mₓ⁺¹ < mₓ⁺²
Peano is only provable in the visible domain.
|
| ∀n ∈ ℕ: ⅟n - ⅟(n+1) > 0
| mₓ < mₓ⁺¹ < mₓ⁺² < mₓ⁺³
| ⅟mₓ > ⅟mₓ⁺¹ > ⅟mₓ⁺² > ⅟mₓ⁺³
| ⅟mₓ - ⅟mₓ⁺¹ > 0
| ⅟mₓ⁺¹ - ⅟mₓ⁺² > 0
|
| However,
| ⅟x < mₓ⁺¹ < mₓ⁺²
| x > ⅟mₓ⁺¹ > ⅟mₓ⁺²
| NUF(x) ≥ |{⅟mₓ⁺¹,⅟mₓ⁺²}| = 2
| but NUF(x) = 1
| Contradiction.
Therefore,
∀x > 0: NUF(cx) ≠ 1
∀x ∈ ℝ: NUF(cx) ≠ 1
This argument assumes Peano for all natural numbers and is contradicted
by the gap after every unit fraction.
Regards, WM
On 02/29/2024 09:11 AM, Ross Finlayson wrote:
On 02/28/2024 10:53 PM, Ross Finlayson wrote:
On 02/28/2024 08:07 PM, Jim Burns wrote:
[...][...][...]
One notion here is that
the methods of infinite series, and,
the methods of infinitesimals,
either have various vagaries,
that these days
infinite series still often see
the willi-nilli,
while infinitesimals are
quite thoroughly, contained in
the description of delta-epsilonics
(as, for that by definition,
the vanishing terms maintaining
real analytical character,
are, so).
while infinitesimals are
quite thoroughly, contained in
the description of delta-epsilonics
In other words, "Dark Numbers" are made up
numbers that try to patch the holes in your logic
WM don't be afraid of the darkness:
Cloud7 - Angels Of The Dark (Trippy Edit) https://www.youtube.com/watch?v=qi2xhtVCh_c
WM schrieb:
In other words, "Dark Numbers" are made up numbers that try to patch
the holes in your logic
Le 01/03/2024 à 13:46, Jim Burns a écrit :
On 3/1/2024 3:39 AM, WM wrote:
Le 29/02/2024 à 21:04, Jim Burns a écrit :
On 2/29/2024 2:31 PM, WM wrote:
I am adhering to mathematics which
proves that NUF(x) cannot jump
from 0 to more than 1.
¬(largest final.ordinal)
¬(smallest final.ordinal.reciprocal)
¬(NUF(x) = 1)
Incompatible with
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
No.
You must be blind.
There is a gap after *every* unit fraction.
On 3/1/24 1:25 PM, WM wrote:
Le 01/03/2024 à 13:46, Jim Burns a écrit :
On 3/1/2024 3:39 AM, WM wrote:
Le 29/02/2024 à 21:04, Jim Burns a écrit :
On 2/29/2024 2:31 PM, WM wrote:
I am adhering to mathematics which
proves that NUF(x) cannot jump
from 0 to more than 1.
¬(largest final.ordinal)
¬(smallest final.ordinal.reciprocal)
¬(NUF(x) = 1)
Incompatible with
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
No.
You must be blind. There is a gap after *every* unit fraction.
As is the gap before it, which is smaller than it, so there must be
another one before it.
There is a gap after *every* unit fraction. In every gap NUF(x)
is constant.
Yes, but there is no FIRST x for NUF(x) to be 1 at.
Therefore you "definition" of NUF is inconsistant.
Peano is only provable in the visible domain.
And every Natural Number is finite, and thus built by Peano and visible.
On 3/1/2024 1:25 PM, WM wrote:
There is a gap after *every* unit fraction. (*)
And, before *each* unit fraction, a unit fraction.
the order '<' has quantifier anti.magic.
∀j:∃k≠j: ⅟k < ⅟j
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
¬∃k:∀j≠k: ⅟k < ⅟j
¬(smallest final.ordinal.reciprocal) (**)
On 3/2/24 7:20 AM, WM wrote:
Le 01/03/2024 à 19:55, Richard Damon a écrit :
There is a gap after *every* unit fraction.
As is the gap before it, which is smaller than it, so there must be
another one before it.
This does not invalidate that at each one NUF makes a step of 1. Either
there is a first one without a gap before it, which is smaller than it,
or there is no completed infinity at all.
But it does show that there is no point for NUF to start at 1,
If there is no first one, then there are more than one first unit
fractions.
Nope. Doesn't follow for unbounded set.
Your error is using inappropriate logic on an unbounded set.
Therefore you "definition" of NUF is inconsistant.
Not at all! The only possible inconsistency is completed infinity.
YES.
Explain where NUF(x) == 1 is if there is no first unit fraction, or how
a first unit fraction exists wnen all unit fractions have a smaller unit fraction.
Peano is only provable in the visible domain.
And every Natural Number is finite, and thus built by Peano and visible.
Then there is no completed infinity, and Cantor's countable sets and his
list are not existing.
How do you say that? Completed infinity isn't a "Natural Number" but can still exist.
Just like most rationals are not Natural Numbers, but do exist.
You like to just assume things you can not actually prove.
Le 01/03/2024 à 19:55, Richard Damon a écrit :
On 3/1/24 1:25 PM, WM wrote:
Le 01/03/2024 à 13:46, Jim Burns a écrit :
On 3/1/2024 3:39 AM, WM wrote:
Le 29/02/2024 à 21:04, Jim Burns a écrit :
On 2/29/2024 2:31 PM, WM wrote:
I am adhering to mathematics which
proves that NUF(x) cannot jump
from 0 to more than 1.
¬(largest final.ordinal)
¬(smallest final.ordinal.reciprocal)
¬(NUF(x) = 1)
Incompatible with
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
No.
You must be blind. There is a gap after *every* unit fraction.
As is the gap before it, which is smaller than it, so there must be
another one before it.
This does not invalidate that at each one NUF makes a step of 1. Either
there is a first one without a gap before it, which is smaller than it,
or there is no completed infinity at all.
There is a gap after *every* unit fraction. In every gap NUF(x) is
constant.
Yes, but there is no FIRST x for NUF(x) to be 1 at.
If there is no first one, then there are more than one first unit
fractions.
Therefore you "definition" of NUF is inconsistant.
Not at all! The only possible inconsistency is completed infinity.
Peano is only provable in the visible domain.
And every Natural Number is finite, and thus built by Peano and visible.
Then there is no completed infinity, and Cantor's countable sets and his
list are not existing.
Regards, WM
Le 02/03/2024 à 15:20, Richard Damon a écrit :
On 3/2/24 7:20 AM, WM wrote:
Le 01/03/2024 à 19:55, Richard Damon a écrit :
There is a gap after *every* unit fraction.
As is the gap before it, which is smaller than it, so there must be
another one before it.
This does not invalidate that at each one NUF makes a step of 1.
Either there is a first one without a gap before it, which is smaller
than it, or there is no completed infinity at all.
But it does show that there is no point for NUF to start at 1,
Then there is no completed infinity, no bijection beteen infinite sets,
no Cantor-list.
If there is no first one, then there are more than one first unit
fractions.
Nope. Doesn't follow for unbounded set.
It follows in every case where logic applies.
Your error is using inappropriate logic on an unbounded set.
If logic rules at all, then infinite sets may be checked whehter they
are compatible with logic. If they don't, then the conclusion must not
be new logic but abolishment of infinite sets.
Therefore you "definition" of NUF is inconsistant.
Not at all! The only possible inconsistency is completed infinity.
YES.
Explain where NUF(x) == 1 is if there is no first unit fraction, or
how a first unit fraction exists wnen all unit fractions have a
smaller unit fraction.
There are two alternatives: Completed infinity means that all points are there. This implies a first one, according to ruling logic. If there is
no first point, then the points are not permanently existing.
Peano is only provable in the visible domain.
And every Natural Number is finite, and thus built by Peano and
visible.
Then there is no completed infinity, and Cantor's countable sets and
his list are not existing.
How do you say that? Completed infinity isn't a "Natural Number" but
can still exist.
Completed infinity means that all natural numbers and all unit fractions
are permanently existing such that a last one and a first one are there, hidden though.
Just like most rationals are not Natural Numbers, but do exist.
You like to just assume things you can not actually prove.
Either all are there, then the smallest unit fraction is there too. Or
this is not the case.
Regards, WM
On 3/2/24 10:09 AM, WM wrote:
Then there is no completed infinity, no bijection beteen infinite sets,
no Cantor-list.
How does that follow?
Infinity was completed in the construction of the set of Natural Numbers.
No need for your exrtrainous NUF function.
Le 02/03/2024 à 16:44, Richard Damon a écrit :
On 3/2/24 10:09 AM, WM wrote:
Then there is no completed infinity, no bijection beteen infinite
sets, no Cantor-list.
How does that follow?
If all are there, then there is a first one. If not all are there, then
not all can be put in bijection.
Infinity was completed in the construction of the set of Natural Numbers.
Potential infinity: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo. ℵo numbers
will never be crossed by Peano.
No need for your exrtrainous NUF function.
The function is needed to disprove actual infinity.
Regards, WM
Le 01/03/2024 à 22:27, Jim Burns a écrit :
On 3/1/2024 1:25 PM, WM wrote:
There is a gap after *every* unit fraction. (*)
And, before *each* unit fraction, a unit fraction.
Does this change (*)?
the order '<' has quantifier anti.magic.
∀j:∃k≠j: ⅟k < ⅟j
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
¬∃k:∀j≠k: ⅟k < ⅟j
¬(smallest final.ordinal.reciprocal) (**)
Either
(**) is true in every case.
Then, because of (*),
there is no completed infinity.
Or
there are dark numbers and
(**) is not true in every case.
On 3/2/24 12:28 PM, WM wrote:
If all are there, then there is a first one. If not all are there, then
not all can be put in bijection.
You are just using incorrect logic.
On 3/2/2024 9:33 AM, WM wrote:
Le 01/03/2024 à 22:27, Jim Burns a écrit :
On 3/1/2024 1:25 PM, WM wrote:
There is a gap after *every* unit fraction. (*)
And, before *each* unit fraction, a unit fraction.
Does this change (*)?
For each unit.fraction,
there is a unit.fraction.before and
there is a gap.after, if [⅟1,∞) is a gap.
Both first and not.first have gaps.after.
Only not.firsts have unit.fractions.before.
Each unit.fraction is not.first.
No unit.fraction is first.
the order '<' has quantifier anti.magic.
∀j:∃k≠j: ⅟k < ⅟j
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
¬∃k:∀j≠k: ⅟k < ⅟j
¬(smallest final.ordinal.reciprocal) (**)
Either
(**) is true in every case.
Then, because of (*),
there is no completed infinity.
Your (WM's) completed.infinityᵂᴹ is
infinite sets which don't change.
No set changes.
If set S changes,
then
both x ∈ S and x ∉ S can be claimed
after which
_everything_ can be
a not.first.false augmenting claim.
No set changes.
You (WM) allege that
there are _two_ kinds of
sets which don't accept another element,
finiteᵂᴹ sets and completedᵂᴹ.infinite sets,
No set changes.
Le 02/03/2024 à 22:57, Richard Damon a écrit :
On 3/2/24 12:28 PM, WM wrote:
If all are there, then there is a first one. If not all are there,
then not all can be put in bijection.
You are just using incorrect logic.
If you go from 0 in positive direction, then you find a first positive x
0, but afterwards you find also smaller x > 0 wich you had notencountered before. Why did you not see them initially?
Regards, WM
On 3/3/24 3:54 AM, WM wrote:
Le 02/03/2024 à 22:57, Richard Damon a écrit :
On 3/2/24 12:28 PM, WM wrote:
If all are there, then there is a first one. If not all are there,
then not all can be put in bijection.
You are just using incorrect logic.
If you go from 0 in positive direction, then you find a first positive x
0, but afterwards you find also smaller x > 0 wich you had notencountered before. Why did you not see them initially?
What make you think that you encounter a first positive x?
There is no bound on that end, so there is no first.
Just a tightly packed collection of values that are so tightly packed
you can't get to the first.
You are just stuck using logic that doesn't apply,
If you are going to insist on the logic you are using, you can't have unbounded sets.
Le 03/03/2024 à 13:19, Richard Damon a écrit :
On 3/3/24 3:54 AM, WM wrote:
Le 02/03/2024 à 22:57, Richard Damon a écrit :
On 3/2/24 12:28 PM, WM wrote:
If all are there, then there is a first one. If not all are there,
then not all can be put in bijection.
You are just using incorrect logic.
If you go from 0 in positive direction, then you find a first
positive x > 0, but afterwards you find also smaller x > 0 wich you
had not encountered before. Why did you not see them initially?
What make you think that you encounter a first positive x?
There are always smaller unit fractions than can be chosen.
This proves that they appear and become choosable only after the first
choice has happened. What can be the reason? Either darkness or non-existence.
There is no bound on that end, so there is no first.
There is a bound, namely zero.
But I can always get closer to zero than by the previous approach.
Just a tightly packed collection of values that are so tightly packed
you can't get to the first.
You are just stuck using logic that doesn't apply,
My logic applies everywhere.
If you are going to insist on the logic you are using, you can't have
unbounded sets.
That has been proven: There are no infinite sets with completely visible elements.
Regards, WM
On 3/3/2024 12:52 PM, Richard Damon wrote:
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
On 3/3/24 3:54 AM, WM wrote:
Le 02/03/2024 à 22:57, Richard Damon a écrit :
On 3/2/24 12:28 PM, WM wrote:
If all are there, then there is a first one. If not all are
there, then not all can be put in bijection.
You are just using incorrect logic.
If you go from 0 in positive direction, then you find a first
positive x > 0, but afterwards you find also smaller x > 0 wich
you had not encountered before. Why did you not see them initially?
What make you think that you encounter a first positive x?
There are always smaller unit fractions than can be chosen.
This proves that they appear and become choosable only after the
first choice has happened. What can be the reason? Either darkness or
non-existence.
Nope, they were always there.
You can't find a first, because there is no "first" unit fraction.
It, by definition, doesn't exist.
Well, every time I think of the first unit fraction, 1/1 comes to
mind... :^)
[...]
On 3/3/24 2:02 PM, WM wrote:
There are always smaller unit fractions than can be chosen.
This proves that they appear and become choosable only after the first
choice has happened. What can be the reason? Either darkness or
non-existence.
Nope, they were always there.
You can't find a first, because there is no "first" unit fraction.
There is no bound on that end, so there is no first.
There is a bound, namely zero.
Which isn't in the set, so the set has no bound in it.
Thus, no lowest unit fraction.
Le 03/03/2024 à 00:04, Jim Burns a écrit :
On 3/2/2024 9:33 AM, WM wrote:
Le 01/03/2024 à 22:27, Jim Burns a écrit :
On 3/1/2024 1:25 PM, WM wrote:
There is a gap after *every* unit fraction. (*)
And, before *each* unit fraction, a unit fraction.
Does this change (*)?
For each unit.fraction,
there is a unit.fraction.before and
there is a gap.after, if [⅟1,∞) is a gap.
Hence
NUF(x) cannot increase by more than 1
at any point x.
Both first and not.first have gaps.after.
Only not.firsts have unit.fractions.before.
Each unit.fraction is not.first.
No unit.fraction is first.
This implies that
not a complete set of unit fractions exists.
Note that
completeness in linear order
implies
a first instance
the order '<' has quantifier anti.magic.
∀j:∃k≠j: ⅟k < ⅟j
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
¬∃k:∀j≠k: ⅟k < ⅟j
¬(smallest final.ordinal.reciprocal) (**)
Either
(**) is true in every case.
Then, because of (*),
there is no completed infinity.
Your (WM's) completed.infinityᵂᴹ is
infinite sets which don't change.
No set changes.
If points on the real axis do not change
then there is a first instance of
one or more points.
If set S changes,
then
both x ∈ S and x ∉ S can be claimed
after which
_everything_ can be
a not.first.false augmenting claim.
The set of known prime numbers changes.
The set of used natural numbers changes.
Almost all natural numbers remain unused:
∀n ∈ ℕ_use: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
No set changes.
You (WM) allege that
there are _two_ kinds of
sets which don't accept another element,
finiteᵂᴹ sets and completedᵂᴹ.infinite sets,
Otherwise they were not complete.
No set changes.
If you go from 0 in positive direction,
then you find a first positive x > 0,
but afterwards you find also smaller x > 0
wich you had not encountered before. Why?
On 03/03/2024 03:07 PM, Richard Damon wrote:
Yes, but
he want to count from the other end,
start at just above 0.
Every journey starts with a first step.
You guys are like
"we're not crossing Zeno's bridge,
this is close enough,
and Leibniz comes and goes for us".
Sum_i=1^infinity 1/2^i = 1
Int e^x dx = e^x + C
ran(f(n) = n/d, 0 <= n <= d, d -> infinity) = [0,1]
And Leibniz is like, "thanks, I got this".
On 3/3/2024 3:51 AM, WM wrote:
There is a gap after *every* unit fraction. (*)
And, before *each* unit fraction, a unit fraction.
Does this change (*)?
For each unit.fraction,
there is a unit.fraction.before and
there is a gap.after, if [⅟1,∞) is a gap.
Hence
NUF(x) cannot increase by more than 1
at any point x.
Consider a finite sequence of claims.
If x is a unit fraction,
then
a set which x is not.in
is not the set of unit fractions, which is
a set which, like all sets, does not change.
Note that
completeness in linear order
implies
a first instance
Completenessᵂᴹ in linear order
does not correspond to two.ended.
Completenessᵂᴹ in linear order
corresponds to
each non.∅.subset two.ended.
[1]
Le 04/03/2024 à 14:08, Jim Burns a écrit :
On 3/3/2024 3:51 AM, WM wrote:
Note that
completeness in linear order
implies
a first instance
Completenessᵂᴹ in linear order
does not correspond to two.ended.
It does.
Completenessᵂᴹ in linear order
corresponds to
each non.∅.subset two.ended.
[1]
That is completeness according to basic logic.
Le 04/03/2024 à 14:08, Jim Burns a écrit :
On 3/3/2024 3:51 AM, WM wrote:
There is a gap after *every* unit fraction. (*)
And, before *each* unit fraction, a unit fraction.
Does this change (*)?
For each unit.fraction,
there is a unit.fraction.before and
there is a gap.after, if [⅟1,∞) is a gap.
Hence
NUF(x) cannot increase by more than 1
at any point x.
Consider a finite sequence of claims.
Irrelevant.
It is sufficient to consider one claim,
namely this one:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
On 03/04/2024 10:31 AM, Jim Burns wrote:
On 3/3/2024 6:19 PM, Ross Finlayson wrote:
And Leibniz is like, "thanks, I got this".
Leibniz has this.
We have this.
Nobody enters Cantor's Paradise.
n = 1, 2, 3, ...
0^1/n = 0, 0, 0, ..., 1
What, not first first, not ultimate untrue?
You can't be adding omega there
as an upper bound,
as d goes to infinity
there is no upper bound.
The range of f always includes 1.
The continuum limit of n/d for whole numbers,
is a marvel.
It just so happens that we have a strong
conventional formal opinion to let that
side of the aspect of what is a continuous
domain and that being a continuous domain
out, because it keeps a lot of stuff simpler,
Le 03/03/2024 à 21:52, Richard Damon a écrit :
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
There is no bound on that end,
so there is no first.
There is a bound, namely zero.
Which isn't in the set,
so the set has no bound in it.
But the set has a bound such that
it cannot further extend.
On 3/4/2024 1:56 PM, WM wrote:
What does completenessᵂᴹ mean?
On 3/4/2024 1:56 PM, WM wrote:
It is sufficient to consider one claim,
namely this one:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
∀n ∈ ℕ: ∀k ∈ ℕ: ⅟n - ⅟(n+k+1) > 0
On 3/4/2024 4:44 AM, WM wrote:
Le 03/03/2024 à 21:52, Richard Damon a écrit :
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
There is no bound on that end,
so there is no first.
There is a bound, namely zero.
Which isn't in the set,
so the set has no bound in it.
But the set has a bound such that
it cannot further extend.
A set doesn't change.
Set ℕ⭳ doesn't change.
Set ℕ⭳ isn't finite.
Le 04/03/2024 à 22:39, Jim Burns a écrit :
On 3/4/2024 1:56 PM, WM wrote:
It is sufficient to consider one claim,
namely this one:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
∀n ∈ ℕ: ∀k ∈ ℕ: ⅟n - ⅟(n+k+1) > 0
This does not invalidate the fact that after ***every*** unit fraction
there is a gap without unit fractions forcing NUF(x) to be constant. Infinitely many unit fractions cannot sit at one and the same point. But
that would be necessary if NUF(x) jumped from 0 to ℵo. Therefore we have only the alternatives: Either no completeness = potential infinity =
variable sets, or ~∀n ∈ ℕ ∃ n+1 ∈ ℕ.
Regards, WM
Le 04/03/2024 à 23:53, Jim Burns a écrit :
On 3/4/2024 4:44 AM, WM wrote:
Le 03/03/2024 à 21:52, Richard Damon a écrit :
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
There is no bound on that end,
so there is no first.
There is a bound, namely zero.
Which isn't in the set,
so the set has no bound in it.
But the set has a bound such that
it cannot further extend.
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Set ℕ⭳ doesn't change.
Set ℕ⭳ isn't finite.
This has been disproved by NUF(x), even if many don't understand the proof.
Regards, WM
On 3/5/24 4:57 AM, WM wrote:
Le 04/03/2024 à 22:39, Jim Burns a écrit :
On 3/4/2024 1:56 PM, WM wrote:
It is sufficient to consider one claim,
namely this one:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
∀n ∈ ℕ: ∀k ∈ ℕ: ⅟n - ⅟(n+k+1) > 0
This does not invalidate the fact that after ***every*** unit fraction
there is a gap without unit fractions forcing NUF(x) to be constant.
Infinitely many unit fractions cannot sit at one and the same point. But
that would be necessary if NUF(x) jumped from 0 to ℵo. Therefore we have >> only the alternatives: Either no completeness = potential infinity =
variable sets, or ~∀n ∈ ℕ ∃ n+1 ∈ ℕ.
But the gap BEFORE the unit fraction is smaller than the unit fraction,
so there is always more unit fractions below it.
Your logic just can't handle the unboundedness of the beginning
Yes, NUF(x) JUMPS
because it can't find the first point to start
counting at, because no point is the first.
On 3/5/24 5:09 AM, WM wrote:
Le 04/03/2024 à 23:53, Jim Burns a écrit :
On 3/4/2024 4:44 AM, WM wrote:
Le 03/03/2024 à 21:52, Richard Damon a écrit :
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
There is no bound on that end,
so there is no first.
There is a bound, namely zero.
Which isn't in the set,
so the set has no bound in it.
But the set has a bound such that
it cannot further extend.
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Why? N just needs to be the complete set of Natural Numbers, the full complete infinity of them.
No.
Le 05/03/2024 à 12:36, Richard Damon a écrit :
On 3/5/24 5:09 AM, WM wrote:
Le 04/03/2024 à 23:53, Jim Burns a écrit :
On 3/4/2024 4:44 AM, WM wrote:
Le 03/03/2024 à 21:52, Richard Damon a écrit :
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
There is no bound on that end,
so there is no first.
There is a bound, namely zero.
Which isn't in the set,
so the set has no bound in it.
But the set has a bound such that
it cannot further extend.
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Why? N just needs to be the complete set of Natural Numbers, the full
complete infinity of them.
Completeness of ℕ implies the permanent existence of all natnumbers and therefore of all unit fractions too.
No.
What else could completeness mean in your opinion?
This has been disproved by NUF(x), even if many don't understand theproof.
Regards, WM
WM formulated on Wednesday :
Le 05/03/2024 à 12:36, Richard Damon a écrit :
On 3/5/24 5:09 AM, WM wrote:
Le 04/03/2024 à 23:53, Jim Burns a écrit :
On 3/4/2024 4:44 AM, WM wrote:
Le 03/03/2024 à 21:52, Richard Damon a écrit :
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
There is no bound on that end,
so there is no first.
There is a bound, namely zero.
Which isn't in the set,
so the set has no bound in it.
But the set has a bound such that
it cannot further extend.
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Why? N just needs to be the complete set of Natural Numbers, the full
complete infinity of them.
Completeness of ℕ implies the permanent existence of all natnumbers and
therefore of all unit fractions too.
No.
What else could completeness mean in your opinion?
A complete set, in your mental model, makes no sense since the set is
defined as having all of its elements anyway.
Most will be thinking of this when completeness is mentioned.
https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers https://www.quora.com/What-is-a-complete-set
I was saying that you NUF doesn't prove what you claim, because your NUF doesn't actually exist as a function with defined values at define
points for values of x > 0.
Le 06/03/2024 à 11:34, FromTheRafters a écrit :
WM formulated on Wednesday :
Le 05/03/2024 à 12:36, Richard Damon a écrit :
On 3/5/24 5:09 AM, WM wrote:
Le 04/03/2024 à 23:53, Jim Burns a écrit :
On 3/4/2024 4:44 AM, WM wrote:
Le 03/03/2024 à 21:52, Richard Damon a écrit :
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
There is no bound on that end,
so there is no first.
There is a bound, namely zero.
Which isn't in the set,
so the set has no bound in it.
But the set has a bound such that
it cannot further extend.
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Why? N just needs to be the complete set of Natural Numbers, the
full complete infinity of them.
Completeness of ℕ implies the permanent existence of all natnumbers
and therefore of all unit fractions too.
No.
What else could completeness mean in your opinion?
A complete set, in your mental model, makes no sense since the set is
defined as having all of its elements anyway.
This is assumed to be true bit is in contradiction with applications
like Hilbert's hotel. All guests (natural numbers) occupy allroorms
(natural numbers. Nevertheless, another natural number appears.
Most will be thinking of this when completeness is mentioned.
https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers
https://www.quora.com/What-is-a-complete-set
That is not the completeness of actual infinity.
Regards, WM
Le 06/03/2024 à 17:24, Richard Damon a écrit :
I was saying that you NUF doesn't prove what you claim, because your
NUF doesn't actually exist as a function with defined values at define
points for values of x > 0.
If infinity is actually complete, then NUF exists.
Regards, WM
On 3/6/24 9:24 AM, WM wrote:
A complete set, in your mental model, makes no sense since the set is
defined as having all of its elements anyway.
This is assumed to be true bit is in contradiction with applications
like Hilbert's hotel. All guests (natural numbers) occupy allroorms
(natural numbers. Nevertheless, another natural number appears.
It isn't a contradiction, but a paradox,
I suppose you need to define what you mean by infinity being complete.
So, if NUF(x) needs something that is infinte for its x or its value,
then it is outside the domain/range of those number systems.
It seems it needs x to be outside the reals to generate results that are Natural Numbers,
NUF(x) can only be even attempted to be defined in a trans-finite number system,
Le 04/03/2024 à 23:53, Jim Burns a écrit :
On 3/4/2024 4:44 AM, WM wrote:
Le 03/03/2024 à 21:52, Richard Damon a écrit :
On 3/3/24 2:02 PM, WM wrote:
Le 03/03/2024 à 13:19, Richard Damon a écrit :
There is no bound on that end,
so there is no first.
There is a bound, namely zero.
Which isn't in the set,
so the set has no bound in it.
But the set has a bound such that
it cannot further extend.
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Set ℕ⭳ doesn't change.
Set ℕ⭳ isn't finite.
This has been disproved by NUF(x),
even if many don't understand the proof.
Le 06/03/2024 à 18:44, Richard Damon a écrit :
On 3/6/24 9:24 AM, WM wrote:
A complete set, in your mental model, makes no sense since the set
is defined as having all of its elements anyway.
This is assumed to be true bit is in contradiction with applications
like Hilbert's hotel. All guests (natural numbers) occupy allroorms
(natural numbers. Nevertheless, another natural number appears.
It isn't a contradiction, but a paradox,
Nonsense. It is impossible to add an element if all are already present.
That is not a paradox but simply your superstition violating basic logic.
Regards, WM
Le 06/03/2024 à 18:52, Richard Damon a écrit :
I suppose you need to define what you mean by infinity being complete.
If there is an infinite set, then none of its elements can be added.
For instance Hilbert's hotel is impossible in actual infinity.
So, if NUF(x) needs something that is infinte for its x or its value,
then it is outside the domain/range of those number systems.
NUF(x) counts all really existing points.
It seems it needs x to be outside the reals to generate results that
are Natural Numbers,
No, every unit fraction is a real number.
NUF(x) can only be even attempted to be defined in a trans-finite
number system,
No. The first point is counted by 1. Nothing transfinite.\
Regards, WM
On 3/6/2024 9:28 AM, WM wrote:
Le 06/03/2024 à 17:24, Richard Damon a écrit :
I was saying that you NUF doesn't prove what you claim, because your
NUF doesn't actually exist as a function with defined values at define
points for values of x > 0.
If infinity is actually complete, then NUF exists.
huh? Since when does a infinite unbounded set like the natural numbers finally complete?
That implies that you think that there is a largest
natural number and/or a smallest unit fraction.
Nonsense!
ℕ⭳ not.changes
On 3/6/24 10:06 AM, WM wrote:
It is impossible to add an element if all are already present.
That is not a paradox but simply your superstition violating basic logic.
Not really.
The logic of infinite sets is just non-intuative, and if you try to use
logic based on finite sets you just break.
After all, you can split in infinite set into two pieces, both of which
have individually just as many elements as the original set as a whole.
On 3/6/24 10:13 AM, WM wrote:
Le 06/03/2024 à 18:52, Richard Damon a écrit :
I suppose you need to define what you mean by infinity being complete.
If there is an infinite set, then none of its elements can be added.
For instance Hilbert's hotel is impossible in actual infinity.
So, you don't actually understand how infinity works.
But no unit fraction is the first, so if NUF(x) can be 1, it can't be at
a unit fraction.
If NUF(x) can be 1, you need to define something that exists that it
happens at. You can't just say it happens at points that don't actually exist.
Le 06/03/2024 à 19:29, Jim Burns a écrit :
On 3/5/2024 5:09 AM, WM wrote:
Le 04/03/2024 à 23:53, Jim Burns a écrit :
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Set ℕ⭳ doesn't change.
Set ℕ⭳ isn't finite.
ℕ⭳ not.changes
Then there is a smallest unit fraction.
It is dark.
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Set ℕ⭳ doesn't change.
Set ℕ⭳ isn't finite.
ℕ⭳ not.changes
Then there is a smallest unit fraction.
It is dark.
Le 06/03/2024 à 20:51, Richard Damon a écrit :
On 3/6/24 10:06 AM, WM wrote:
It is impossible to add an element if all are already present. That
is not a paradox but simply your superstition violating basic logic.
Not really.
The logic of infinite sets is just non-intuative, and if you try to
use logic based on finite sets you just break.
Set-logic must adhere to basic logic. It is impossible to add an element
if all are already present.
Every claim disobeying this basic logic is nonsesen.
After all, you can split in infinite set into two pieces, both of
which have individually just as many elements as the original set as a
whole.
This can only happen in potential infinity, not in complete infinite sets.
Regards, WM
Le 06/03/2024 à 21:06, Richard Damon a écrit :
On 3/6/24 10:13 AM, WM wrote:
Le 06/03/2024 à 18:52, Richard Damon a écrit :
I suppose you need to define what you mean by infinity being complete.
If there is an infinite set, then none of its elements can be added.
For instance Hilbert's hotel is impossible in actual infinity.
So, you don't actually understand how infinity works.
I understand that matheology is erected on lies and stupidity.
Potential and actual infinity are interchanged whenever set theory
breaks down.
But no unit fraction is the first, so if NUF(x) can be 1, it can't be
at a unit fraction.
NUF(x) cannot grow from 0 to more than 1 because after every unit
fraction there is a gap with NUF(x) being constant. You seem to keep on forgetting that fact, because your mind can't seem to handle it.
If NUF(x) can be 1, you need to define something that exists that it
happens at. You can't just say it happens at points that don't
actually exist.
I do not claim that NUF(x) = 1 exists. It exists only if infinity is
actual - a necessary precondition for Cantor's claims.
Regards, WM
Le 06/03/2024 à 19:29, Jim Burns a écrit :
On 3/5/2024 5:09 AM, WM wrote:
Le 04/03/2024 à 23:53, Jim Burns a écrit :
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Set ℕ⭳ doesn't change.
Set ℕ⭳ isn't finite.
ℕ⭳ not.changes
Then there is a smallest unit fraction.
It is dark.
Each unit fraction, darkᵂᴹ or not, is preceded.
Everyone.else would say each is not.firstⁿᵒᵗᵂᴹ
In fact there is no smallest unit fraction...
After all, you can split in infinite set into two pieces, both of
which have individually just as many elements as the original set as a
whole.
This can only happen in potential infinity, not in complete infinite sets.
Nope. It works with the natural numbers, which are an complete infinite set.
On 3/7/2024 5:41 AM, WM wrote:
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Set ℕ⭳ doesn't change.
Set ℕ⭳ isn't finite.
ℕ⭳ not.changes
Then there is a smallest unit fraction.
It is dark.
ℕ⭳ is the set of all ordinals which next not.fits.
For each n ∈ ℕ⭳
a 1.to.1.map to [0,n) from [0,n⁺¹) not.exists.
They are the finiteⁿᵒᵗᵂᴹ ordinals.
On 3/7/24 3:01 AM, WM wrote:
Le 06/03/2024 à 21:06, Richard Damon a écrit :
On 3/6/24 10:13 AM, WM wrote:
Le 06/03/2024 à 18:52, Richard Damon a écrit :
I suppose you need to define what you mean by infinity being complete. >>>>If there is an infinite set, then none of its elements can be added.
For instance Hilbert's hotel is impossible in actual infinity.
So, you don't actually understand how infinity works.
I understand that matheology is erected on lies and stupidity.
Potential and actual infinity are interchanged whenever set theory
breaks down.
I guess that means that you are admitting that your "logic" just can't
handle the truely infinite sets.
But no unit fraction is the first, so if NUF(x) can be 1, it can't be
at a unit fraction.
NUF(x) cannot grow from 0 to more than 1 because after every unit
fraction there is a gap with NUF(x) being constant. You seem to keep on
forgetting that fact, because your mind can't seem to handle it.
And thus NUF(x) doesn't exist as there is no "first unit fraction" for
it to be 1 at.
If NUF(x) can be 1, you need to define something that exists that it
happens at. You can't just say it happens at points that don't
actually exist.
I do not claim that NUF(x) = 1 exists. It exists only if infinity is
actual - a necessary precondition for Cantor's claims.
But the Natural Numbers ARE an infinite set, so it seems either your
logic doesn't allow you to HAVE the natural numbers, or you admit your
logic is just wrong.
Either way, you can't use the natural number, or the unit fractions, in
your logic, if you admit they don't exist in your logic.
Le 07/03/2024 à 20:24, Jim Burns a écrit :
Each unit fraction, darkᵂᴹ or not, is preceded.
Everyone.else would say each is not.firstⁿᵒᵗᵂᴹ
That is impossible for permanently existing points,
because each unit fraction is followed by
a constant level of NUF.
How to exclude this
in your opinion?
On 3/8/2024 2:52 AM, WM wrote:
each unit fraction is followed by
a constant level of NUF.
How to exclude this
in your opinion?
What is the not.opinion reasonⁿᵒᵗᵂᴹ to exclude ⅟n⁻¹ ?
Le 07/03/2024 à 21:33, Jim Burns a écrit :
On 3/7/2024 5:41 AM, WM wrote:
A set doesn't change.
If ∀n ∈ ℕ ∃n+1 ∈ ℕ, then the set ℕ changes.
Set ℕ⭳ doesn't change.
Set ℕ⭳ isn't finite.
ℕ⭳ not.changes
Then there is a smallest unit fraction.
It is dark.
ℕ⭳ is the set of all ordinals which next not.fits.
For each n ∈ ℕ⭳
a 1.to.1.map to [0,n) from [0,n⁺¹) not.exists.
They are the finiteⁿᵒᵗᵂᴹ ordinals.
NUF(x) = 1 exists
for permanently existing points,
because each unit fraction is followed by
a constant level of NUF.
How to exclude this in your opinion?
Le 08/03/2024 à 15:14, Jim Burns a écrit :
On 3/8/2024 2:52 AM, WM wrote:
each unit fraction is followed by
a constant level of NUF.
How to exclude this
in your opinion?
What is the not.opinion reasonⁿᵒᵗᵂᴹ
to exclude ⅟n⁻¹ ?
If you don't understand it,
try to think better or stop your waffle.
Counting down from next.fits
never reaches next.not.fits.
because each unit fraction is followed by
a constant level of NUF.
Elaborate why because.
Why is this a non.opinion reason?
How to exclude this in your opinion?
Irrelevant.
On 3/8/2024 10:32 AM, WM wrote:
Le 08/03/2024 à 15:14, Jim Burns a écrit :
On 3/8/2024 2:52 AM, WM wrote:
each unit fraction is followed by
a constant level of NUF.
How to exclude this
in your opinion?
What is the not.opinion reasonⁿᵒᵗᵂᴹ
to exclude ⅟n⁻¹ ?
If you don't understand it,
try to think better or stop your waffle.
Your argument is your job, not mine.
Le 08/03/2024 à 17:52, Jim Burns a écrit :
Counting down from next.fits
never reaches next.not.fits.
Counting upwards from zero,
there is a level after every unit fraction.
because each unit fraction is followed by
a constant level of NUF.
Elaborate why because.
Why is this a non.opinion reason?
Because after every unit fraction NUF(x) pauses.
On 03/07/2024 11:24 AM, Jim Burns wrote:
[...]
What I really appreciate about your latest
addition to your axioms is the adding that
there's an other side and a middle resulting
in the middle, that the infinitely-many must
somehow have ends, if not bounds, and
either side, and a middle.
On 3/8/2024 1:52 PM, WM wrote:
Le 08/03/2024 à 17:52, Jim Burns a écrit :
Counting down from next.fits
never reaches next.not.fits.
Counting upwards from zero,
there is a level after every unit fraction.
for each unit.fraction ⅟j
for each ordinal k such that next.not.fits
more.than.k unit fractions are between 0 and ⅟j
Le 07/03/2024 à 21:12, Richard Damon a écrit :
On 3/7/24 3:01 AM, WM wrote:
Le 06/03/2024 à 21:06, Richard Damon a écrit :
On 3/6/24 10:13 AM, WM wrote:
Le 06/03/2024 à 18:52, Richard Damon a écrit :
I suppose you need to define what you mean by infinity being
complete.
If there is an infinite set, then none of its elements can be added. >>>>> For instance Hilbert's hotel is impossible in actual infinity.
So, you don't actually understand how infinity works.
I understand that matheology is erected on lies and stupidity.
Potential and actual infinity are interchanged whenever set theory
breaks down.
I guess that means that you are admitting that your "logic" just can't
handle the truely infinite sets.
You guess wrong as always. Logic is independent of mine and can
distinguish between potential and actual infinity. Only the liars and
cranks of matheology try to blur it.
But no unit fraction is the first, so if NUF(x) can be 1, it can't
be at a unit fraction.
NUF(x) cannot grow from 0 to more than 1 because after every unit
fraction there is a gap with NUF(x) being constant. You seem to keep
on forgetting that fact, because your mind can't seem to handle it.
And thus NUF(x) doesn't exist as there is no "first unit fraction" for
it to be 1 at.
NUF(x) is well-defined and therefore exists.
If NUF(x) can be 1, you need to define something that exists that it
happens at. You can't just say it happens at points that don't
actually exist.
I do not claim that NUF(x) = 1 exists. It exists only if infinity is
actual - a necessary precondition for Cantor's claims.
But the Natural Numbers ARE an infinite set, so it seems either your
logic doesn't allow you to HAVE the natural numbers, or you admit your
logic is just wrong.
I apply logic that is the basis of all sciences.
Either way, you can't use the natural number, or the unit fractions,
in your logic, if you admit they don't exist in your logic.
It is not my logic but universal logic without which all maths would
break down.
Regards, WM
Le 07/03/2024 à 21:07, Richard Damon a écrit :
Each unit fraction is followed by a constant level of NUF. How toAfter all, you can split in infinite set into two pieces, both of
which have individually just as many elements as the original set as
a whole.
This can only happen in potential infinity, not in complete infinite
sets.
Nope. It works with the natural numbers, which are an complete
infinite set.
exclude this in your opinion?
Regards, WM
On 3/7/2024 11:54 PM, WM wrote:
Le 07/03/2024 à 21:03, "Chris M. Thomasson" a écrit :
In fact there is no smallest unit fraction...Each unit fraction is followed by a constant level of NUF. How to
exclude this in your opinion?
On the way to zero, the lower limit wrt unit fractions:
1/2 - 0 = .5 // we got more unit fractions to go...
1/42 - 0 = 0.02380952 ... // we got more unit fractions to go...
1/69 - 0 = 0.01449275 ... // we got more unit fractions to go...
[... on and on ...]
Notice how they keep getting smaller, yet they will never equal zero?
we go from 1 to zero, zero is its limit wrt:
1/1, 1/2, 1/3, 1/4, ect... so on and so fourth...
See how there is no smallest one?
Afaict, your logic is radically broken.
On 3/7/24 11:55 PM, WM wrote:
Each unit fraction is followed by a constant level of NUF. How to
exclude this in your opinion?
Why do I need to?
The problem you have is that each unit fraction is also PRECEEDED by a constant level shorter that the unit fraction, so there is always
another unit fraction before it.
NUF just isn't defined in the finite numbers because there is no "first"
unit fraction to count from.
It is sort of like trying to define that a finite x exist, such that
x+1 = x
On 3/8/24 12:29 AM, WM wrote:
Le 07/03/2024 à 21:12, Richard Damon a écrit :
On 3/7/24 3:01 AM, WM wrote:
Le 06/03/2024 à 21:06, Richard Damon a écrit :
On 3/6/24 10:13 AM, WM wrote:
Le 06/03/2024 à 18:52, Richard Damon a écrit :
I suppose you need to define what you mean by infinity being
complete.
If there is an infinite set, then none of its elements can be added. >>>>>> For instance Hilbert's hotel is impossible in actual infinity.
So, you don't actually understand how infinity works.
I understand that matheology is erected on lies and stupidity.
Potential and actual infinity are interchanged whenever set theory
breaks down.
I guess that means that you are admitting that your "logic" just can't
handle the truely infinite sets.
You guess wrong as always. Logic is independent of mine and can
distinguish between potential and actual infinity. Only the liars and
cranks of matheology try to blur it.
Then explain how you get a first unit fraction, when every unit fraction
has one before it?
You THINK logic is independent, and perhaps there is a "natural Logic",
but that Natural logic might not be able to support the concept if "Infinity", as that might not actually exist in reality.
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be violated without violating all of mathematics.
WM drivels:
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be violated >> without violating all of mathematics.
Bullshit, as always...
Tom Bola a écrit :
WM drivels:
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be violated >>> without violating all of mathematics.
Bullshit, as always...
when judged from the perspective of a nutcase. But even a nutcase should understand that the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
Regards, WM
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be violated >>>> without violating all of mathematics.
Bullshit, as always...
when judged from the perspective of a nutcase. But even a nutcase should
understand that the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
Bullshit, as always...
Bullshit, as always...
Le 09/03/2024 à 11:42, Tom Bola a écrit :
WM drivels:
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be
violated without violating all of mathematics.
Bullshit, as always...
when judged from the perspective of a nutcase. But even a nutcase
should understand that the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
Regards, WM
Le 08/03/2024 à 23:06, Richard Damon a écrit :
On 3/8/24 12:29 AM, WM wrote:
Le 07/03/2024 à 21:12, Richard Damon a écrit :
On 3/7/24 3:01 AM, WM wrote:
Le 06/03/2024 à 21:06, Richard Damon a écrit :
On 3/6/24 10:13 AM, WM wrote:
Le 06/03/2024 à 18:52, Richard Damon a écrit :
I suppose you need to define what you mean by infinity being
complete.
If there is an infinite set, then none of its elements can be added. >>>>>>> For instance Hilbert's hotel is impossible in actual infinity.
So, you don't actually understand how infinity works.
I understand that matheology is erected on lies and stupidity.
Potential and actual infinity are interchanged whenever set theory
breaks down.
I guess that means that you are admitting that your "logic" just
can't handle the truely infinite sets.
You guess wrong as always. Logic is independent of mine and can
distinguish between potential and actual infinity. Only the liars and
cranks of matheology try to blur it.
Then explain how you get a first unit fraction, when every unit
fraction has one before it?
I did so many times already: Either this holds only for visible unit fractions, or there is no permament existience of all unit fractions.
You THINK logic is independent, and perhaps there is a "natural
Logic", but that Natural logic might not be able to support the
concept if "Infinity", as that might not actually exist in reality.
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be
violated without violating all of mathematics.
Regards, WM
On 03/09/2024 06:09 AM, Tom Bola wrote:
WM drivels:
Tom Bola a écrit :
WM drivels:
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be violated
without violating all of mathematics.
Bullshit, as always...
when judged from the perspective of a nutcase. But even a nutcase should >>> understand that the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
Bullshit, as always...
Regards, WM
Bullshit, as always...
Hallo Tom, thanks for writing to sci.math, recently
I started reading de.sci.mathematik, and it's encouraging
to know that there's much the same idea there as here.
(Danke fur schrieben und lesen, es gibt sammlich hier.)
In mathematics, in most fields today, "English" is common.
The idea that over history, at some point French was the
Lingua Franca and universal, has that English is today,
"Lingua Angla" if you will, then that over time there
are various main languages in mathematics.
English German French Latin ... Arabic ... Greek ...
Of course, with those being called the "international
languages", of mathematics, any language can have written
mathematics, and of course, mathematics has its own language,
and the language of mathematical formulary is largely universal.
Here on sci.math I am constantly writing about continuity
and infinity, I started writing same on de.sci.mathematik
and fr.sci.maths, recemment.
Tom, where we are here on sci.math, is that the standard
is good, that though the non-standard is not necessarily bad.
The idea is that though MW/WM is broken, there are some
concepts associated with his arguments, his, "axioms",
that are not, some of his "axioms" by themselves could
be good. These are challenges to all of modern mathematics.
Then, the goal, should be, for modern mathematics again,
how to arrive at there are no paradoxes and there are no
contradictions, while at the same time, "conscientiousness
to the intuitive", and, "only conscientious in the completions".
This is where a most usual goal of a mathematics is that
there is one, a mathematics altogether, then that the
primacy and centrality of the intuitive truths, as perceived,
result constructive and rulial systems of relation together,
zusammeln, compris, that it's what foundations should be.
On 3/9/2024 2:09 AM, WM wrote:
Le 08/03/2024 à 22:34, "Chris M. Thomasson" a écrit :
On 3/7/2024 11:54 PM, WM wrote:
Le 07/03/2024 à 21:03, "Chris M. Thomasson" a écrit :
In fact there is no smallest unit fraction...Each unit fraction is followed by a constant level of NUF. How to
exclude this in your opinion?
On the way to zero, the lower limit wrt unit fractions:
1/2 - 0 = .5 // we got more unit fractions to go...
1/42 - 0 = 0.02380952 ... // we got more unit fractions to go...
1/69 - 0 = 0.01449275 ... // we got more unit fractions to go...
[... on and on ...]
Notice how they keep getting smaller, yet they will never equal zero?
Of course not. Therefore NUF(x) cannot increase at 0.
we go from 1 to zero, zero is its limit wrt:
1/1, 1/2, 1/3, 1/4, ect... so on and so fourth...
See how there is no smallest one?
It cannot be seen. But if infinity is actual, then all unit fractions
are present and each one has a gap behind it.
Your gap wrt 1/2 to zero is filled with infinity. See the gaps here?
Keep in mind that we are working with unit fractions.
1 - 1/2 - 0?
0
.25->0
See how 1/4 is "smaller" than 1/2? You are trying to go from zero up.
Strange of you!
Le 09/03/2024 à 11:42, Tom Bola a écrit :
[...]
the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
On 3/9/24 3:16 AM, WM wrote:
The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
But they don't need to. The "matching" is between the set of Natural
Numbers, which is one line, and the Rationals, which is a SEPERATE ARRAY.
So we have:
XXXXXXXXXXXXXXXXXXXXX...
and
OOOOO ...
OOOOO ...
OOOOO ...
...
and we can cover all the Os with an X from the line above it.
On 3/9/24 2:22 AM, WM wrote:
your logic is the logic of FINITE sets,
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be
violated without violating all of mathematics.
Nope. That is logic of FINITE sets.
Your logic just doesn't get you to either the Natural Numbers or the
Unit Fraction, just arbitrary finite subsets of them. Your "FISON"s
You are trying to go from zero up.
Strange of you!
On 3/9/2024 6:16 AM, WM wrote:
Le 09/03/2024 à 11:42, Tom Bola a écrit :
[...]
the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
That is why Bob and the Os CAN disappear
with rows and columns from ℕ⭳
Chris M. Thomasson wrote:
You are trying to go from zero up.
Really, yes!
Strange of you!
Indeed.
Le 09/03/2024 à 22:27, Richard Damon a écrit :
On 3/9/24 3:16 AM, WM wrote:
The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
But they don't need to. The "matching" is between the set of Natural
Numbers, which is one line, and the Rationals, which is a SEPERATE ARRAY.
So we have:
XXXXXXXXXXXXXXXXXXXXX...
and
OOOOO ...
OOOOO ...
OOOOO ...
...
and we can cover all the Os with an X from the line above it.
The matching should also be possible between the unit fractions of the
first line and the positive fractions including the unit frations. But
that would require to cover the whole matrix by the X of the first line
XXX...
OOO...
OOO...
..
which is impossible.
Regards, WM
Le 09/03/2024 à 22:24, Richard Damon a écrit :
On 3/9/24 2:22 AM, WM wrote:
your logic is the logic of FINITE sets,
This logic is basic to all thinking. If infinite sets violate it, then
they annot exist.
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be
violated without violating all of mathematics.
Nope. That is logic of FINITE sets.
It is basic to all thinking.
Your logic just doesn't get you to either the Natural Numbers or the
Unit Fraction, just arbitrary finite subsets of them. Your "FISON"s
This logic shows ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo. It does not
exclude the existence of ℵo but it shows that ℵo numbers are dark.
Regards, WM
Le 10/03/2024 à 00:46, Tom Bola a écrit :
Chris M. Thomasson wrote:
You are trying to go from zero up.
Really, yes!
Strange of you!
Indeed.
Why? If all unit fractions are existing, then it is possible.
If not, then set theory is dead anyway.
WM wrote:
Le 10/03/2024 à 00:46, Tom Bola a écrit :
Chris M. Thomasson wrote:Why? If all unit fractions are existing, then it is possible.
You are trying to go from zero up.
No, because for every fraction 1/n there exist infinitely many
fractions 1/m where m > n.
On 3/10/24 3:34 AM, WM wrote:
The matching should also be possible between the unit fractions of the
first line and the positive fractions including the unit fractions. But
that would require to cover the whole matrix by the X of the first line
Why?
They are two different sets of values, that just happen to represent the
same numerical value.
XXX...
OOO...
OOO...
..
which is impossible.
On 3/10/24 3:29 AM, WM wrote:
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be
violated without violating all of mathematics.
Nope. That is logic of FINITE sets.
It is basic to all thinking.
Nope. Maybe YOU can't think past it, but that is YOUR thought.
Le 09/03/2024 à 15:09, Tom Bola a écrit :
[...]
I prefer arguments.
[...]
I prefer arguments.
Le 10/03/2024 à 14:00, Richard Damon a écrit :
On 3/10/24 3:29 AM, WM wrote:
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be
violated without violating all of mathematics.
Nope. That is logic of FINITE sets.
It is basic to all thinking.
Nope. Maybe YOU can't think past it, but that is YOUR thought.
Everything else is belief in nonsense, meaningless to discuss about.
EOD.
Regards, WM
Le 09/03/2024 à 22:24, Richard Damon a écrit :
On 3/9/24 2:22 AM, WM wrote:
your logic is the logic of FINITE sets,
This logic is basic to all thinking. If infinite sets violate it, then
they annot exist.
If there are unit fractions in linear order, then there is a first
instance where NUF(x) =/= 0. This is general logic that cannot be
violated without violating all of mathematics.
Nope. That is logic of FINITE sets.
It is basic to all thinking.
Your logic just doesn't get you to either the Natural Numbers or the
Unit Fraction, just arbitrary finite subsets of them. Your "FISON"s
This logic shows ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo. It does not
exclude the existence of ℵo but it shows that ℵo numbers are dark.
Regards, WM
Le 10/03/2024 à 13:55, Richard Damon a écrit :
On 3/10/24 3:34 AM, WM wrote:
The matching should also be possible between the unit fractions of
the first line and the positive fractions including the unit
fractions. But that would require to cover the whole matrix by the X
of the first line
Why?
Because there is a bijection between unit fractions and fraction
according to set theory.
They are two different sets of values, that just happen to represent
the same numerical value.
If so, then the whole matrix could be covered by X, but it can't.
XXX...
OOO...
OOO...
..
which is impossible.
Regards, WM
WM explained on 3/10/2024 :
If all are permanently existing, then there must be a first one according to >> logic.
According to your faulty logic.
On 3/10/2024 7:36 AM, WM wrote:
If all are permanently existing, then there must be a first one
according to logic.
1/1 is the first unit fraction. There is no last one.... See?
Le 10/03/2024 à 19:00, FromTheRafters a écrit :
WM explained on 3/10/2024 :
If all are permanently existing, then there must be a first one
according to logic.
According to your faulty logic.
It is basic logic. If you don't like it believe in matheology. That does
not hurt because it is completely meaningless and has no application,
hence no awful application either. It only deters people from meaningful activities.
Regards, WM
Le 10/03/2024 à 21:20, "Chris M. Thomasson" a écrit :
On 3/10/2024 7:36 AM, WM wrote:
If all are permanently existing, then there must be a first one
according to logic.
1/1 is the first unit fraction. There is no last one.... See?
No. We cannot see it. But if all exist, then it exists too.
Regards, WM
On 3/10/24 7:31 AM, WM wrote:
Le 10/03/2024 à 13:55, Richard Damon a écrit :
On 3/10/24 3:34 AM, WM wrote:
The matching should also be possible between the unit fractions of
the first line and the positive fractions including the unit
fractions. But that would require to cover the whole matrix by the X
of the first line
Why?
Because there is a bijection between unit fractions and fractions
according to set theory.
Right, so you make the bijunction between the two sets, not between two pieces of one set.
They are two different sets of values, that just happen to represent
the same numerical value.
If so, then the whole matrix could be covered by X, but it can't.
XXX...
OOO...
OOO...
..
which is impossible.
We could do
XOO...
XOO...
XOO..
X
.
.
1/1 <-> 1/1 both the same
1/2 <-> 2/1 gives
XXO...
OOO...
XOO...
X
.
.
1/3 <-> 1/2 gives
XXO...
XOO...
OOO...
X
.
.
1/4 <-> 3/1 GIVES
XXXO...
XOOO...
OOOO...
O
X
.
.
1/5 <-> 2/2 GIVES
XXXO...
XXOO...
OOOO...
O
O
X
X
.
.
1/6 <-> 1/3 GIVES
XXXO...
XXOO...
XOOO...
O
O
O
X
X
.
.
So, as were go, yes, we have a "bubble" of zeros on the 1/n going down,
but every one of them has a x farther down that will fill them.
Thus we can show that EVERY point in the x/y has a point in 1/y that
maps to it, so if we continue this proceed FOREVER (which is what a 1:1 mapping of an infinte set will take) we will totally cover the space with x.
On 3/10/24 7:40 AM, WM wrote:
Le 10/03/2024 à 14:00, Richard Damon a écrit :
On 3/10/24 3:29 AM, WM wrote:
If there are unit fractions in linear order, then there is a first >>>>>> instance where NUF(x) =/= 0. This is general logic that cannot be
violated without violating all of mathematics.
Nope. That is logic of FINITE sets.
It is basic to all thinking.
Nope. Maybe YOU can't think past it, but that is YOUR thought.
Everything else is belief in nonsense, meaningless to discuss about.
EOD.
Regards, WM
Maybe for you, but not to others.
Brings to mind an old proverb from a frind:
You can't think about that which you refuse to think about.
If you limit yourself to finite logic, then you, by your own
restrictions, can not think about things that are not finite.
So, you have put yourself into the finite box, but want to reach out to
the infinite, but have defined TO YOURSELF, that you can't get there.
What is holding yourself back, is YOU.
Le 10/03/2024 à 19:39, Richard Damon a écrit :
On 3/10/24 7:31 AM, WM wrote:
Le 10/03/2024 à 13:55, Richard Damon a écrit :
On 3/10/24 3:34 AM, WM wrote:
The matching should also be possible between the unit fractions of
the first line and the positive fractions including the unit
fractions. But that would require to cover the whole matrix by the
X of the first line
Why?
Because there is a bijection between unit fractions and fractions
according to set theory.
Right, so you make the bijunction between the two sets, not between
two pieces of one set.
The bijection is established by pairs.
They are two different sets of values, that just happen to represent
the same numerical value.
If so, then the whole matrix could be covered by X, but it can't.
XXX...
OOO...
OOO...
..
which is impossible.
We could do
XOO...
XOO...
XOO..
X
.
.
1/1 <-> 1/1 both the same
1/2 <-> 2/1 gives
XXO...
OOO...
XOO...
X
.
.
1/3 <-> 1/2 gives
XXO...
XOO...
OOO...
X
.
.
1/4 <-> 3/1 GIVES
XXXO...
XOOO...
OOOO...
O
X
.
.
1/5 <-> 2/2 GIVES
XXXO...
XXOO...
OOOO...
O
O
X
X
.
.
1/6 <-> 1/3 GIVES
XXXO...
XXOO...
XOOO...
O
O
O
X
X
.
.
So, as were go, yes, we have a "bubble" of zeros on the 1/n going
down, but every one of them has a x farther down that will fill them.
Every visible position will be filled with X, yes. But no O can
disappear because the first column is the only drain. As long as it is blocked by X, no O can leave. And after all X have been issued nothing happens anymore.
Thus we can show that EVERY point in the x/y has a point in 1/y that
maps to it, so if we continue this proceed FOREVER (which is what a
1:1 mapping of an infinte set will take) we will totally cover the
space with x.
You can show it for all visible fractions. But since no O leaves the
matrix, there are not indexed fractions remaining, at least according to
my logic which is the only logic that counts (pun intended).
Regards, WM
Also, it says that your NUF(x) isn't what you say, as its values are not "aleph" for finite x, but are some finite number, and NUF(1) is N, where
N is the size of the FISON that you are in. Maybe you don't know that
value, and thus you don't know the values for NUF(x)
It can't be "aleph" as you math/logic doesn't handle infinites with out
going inconsistent.
On 3/10/24 2:45 PM, WM wrote:
So, as were go, yes, we have a "bubble" of zeros on the 1/n going
down, but every one of them has a x farther down that will fill them.
Every visible position will be filled with X, yes. But no O can
disappear because the first column is the only drain. As long as it is
blocked by X, no O can leave. And after all X have been issued nothing
happens anymore.
But you never run out of X's in the first column.
That is what you don't understand,
Le 10/03/2024 à 22:48, Richard Damon a écrit :
Also, it says that your NUF(x) isn't what you say, as its values are
not "aleph" for finite x, but are some finite number, and NUF(1) is N,
where N is the size of the FISON that you are in. Maybe you don't know
that value, and thus you don't know the values for NUF(x)
It can't be "aleph" as you math/logic doesn't handle infinites with
out going inconsistent.
NUF(1) = ℵo like for every visible argument 1/n. No problem to handle.
But ℵo is a number that does not appear by miracle but by counting 1, 2,
3, ... because it means cardinality of countable infinite set.
Regards, WM
Le 10/03/2024 à 23:01, Richard Damon a écrit :
On 3/10/24 2:45 PM, WM wrote:
So, as were go, yes, we have a "bubble" of zeros on the 1/n going
down, but every one of them has a x farther down that will fill them.
Every visible position will be filled with X, yes. But no O can
disappear because the first column is the only drain. As long as it
is blocked by X, no O can leave. And after all X have been issued
nothing happens anymore.
But you never run out of X's in the first column.
What does that matter? No O will disappear, hence the number of not
indexed fractions will remain constant forever.
That is what you don't understand,
If the process of indexing does not come to an end, then the mapping is
never completed, then there is no bijection at all - according to my logic
Regards, WM
Le 10/03/2024 à 22:48, Richard Damon a écrit :
[...]
But ℵo is a number that
does not appear by miracle but
by counting 1, 2, 3, ...
because it means
cardinality of countable infinite set.
On 03/10/2024 11:00 AM, FromTheRafters wrote:
WM explained on 3/10/2024 :
If all are permanently existing,
then there must be a first one
according to logic.
According to your faulty logic.
"Faulty logic", is that where exceptions are rules,
and vice-versa?
On 3/11/24 1:36 AM, WM wrote:
What does that matter? No O will disappear, hence the number of not
indexed fractions will remain constant forever.
But that isn't the question.
The question is can you match every Natural Number to a Rational, which
is proven you can.
As EVERY spot will have an X from somewhere down the line moved on to it.
Using the wrong criteria just gets you the wrong answer.
T with Bob inserted is T∪{Bob} which is
a different set and a set which fits T
On 3/11/24 1:45 AM, WM wrote:
NUF(1) = ℵo like for every visible argument 1/n. No problem to handle.
But ℵo is a number that does not appear by miracle but by counting 1, 2, >> 3, ... because it means cardinality of countable infinite set.
So, you admit you are lying about not using the Matheologies that you
claim you are not using.
Richard Damon has brought this to us :
On 3/11/24 1:36 AM, WM wrote:
Le 10/03/2024 à 23:01, Richard Damon a écrit :
On 3/10/24 2:45 PM, WM wrote:
So, as were go, yes, we have a "bubble" of zeros on the 1/n going down, >>>>>> but every one of them has a x farther down that will fill them.
Every visible position will be filled with X, yes. But no O can disappear >>>>> because the first column is the only drain. As long as it is blocked by >>>>> X, no O can leave. And after all X have been issued nothing happens
anymore.
But you never run out of X's in the first column.
What does that matter? No O will disappear, hence the number of not indexed >>> fractions will remain constant forever.
But that isn't the question.
The question is can you match every Natural Number to a Rational, which is >> proven you can.
IMO match is the wrong word, it is simply a pairing.
Le 11/03/2024 à 19:15, Jim Burns a écrit :
T with Bob inserted is T∪{Bob} which is
a different set and a set which fits T
The set with Bob vanished is a different set.
Le 11/03/2024 à 18:27, Richard Damon a écrit :
But that isn't the question.
The question is whether all fractions get indeXed.
The question is whether all fractions get indeXed.
The answer is no.
Related:
The first column ℕ⭳×{1} of Xs is
a set into which
the whole matrix ℕ⭳×ℕ⭳ of Xs.and.Os fits.
On 3/11/2024 5:21 PM, WM wrote:
Le 11/03/2024 à 18:27, Richard Damon a écrit :
But that isn't the question.
The question is whether all fractions get indeXed.
Fraction i/j gets index kᵢⱼ
Sᵢⱼ = i+j
kᵢⱼ = (Sᵢⱼ-1)(Sᵢⱼ-2)/2+i
Index kᵢⱼ gets fraction iₖᵢⱼ/jₖᵢⱼ
sₖᵢⱼ = ⌈√̅2̅⋅̅k̅ᵢ̅ⱼ̅+̅¼ + ½⌉
iₖᵢⱼ = kᵢⱼ-(sₖᵢⱼ-1)(sₖᵢⱼ-2)/2
jₖᵢⱼ = sₖᵢⱼ-iₖᵢⱼ
iₖᵢⱼ/jₖᵢⱼ = i/j
The question is whether all fractions get indeXed.
The answer is no.
Arithmetic disagrees with you.
Le 11/03/2024 à 18:27, Richard Damon a écrit :
On 3/11/24 1:36 AM, WM wrote:
What does that matter? No O will disappear, hence the number of not
indexed fractions will remain constant forever.
But that isn't the question.
The question is whether all fractions get indeXed. The answer is no.
The question is can you match every Natural Number to a Rational,
which is proven you can.
The answer is no. During the indexing procedure never one fraction more
is indexed than at the beginning.
As EVERY spot will have an X from somewhere down the line moved on to it.
Obviously not. Only every spot you can think of. Try to think better.
Using the wrong criteria just gets you the wrong answer.
Therefore use the right crieria.
Regards, WM
Le 12/03/2024 à 00:06, Jim Burns a écrit :
On 3/11/2024 5:21 PM, WM wrote:
Le 11/03/2024 à 18:27, Richard Damon a écrit :
But that isn't the question.
The question is whether all fractions get indeXed.
Fraction i/j gets index kᵢⱼ
Sᵢⱼ = i+j
kᵢⱼ = (Sᵢⱼ-1)(Sᵢⱼ-2)/2+i
Index kᵢⱼ gets fraction iₖᵢⱼ/jₖᵢⱼ
sₖᵢⱼ = ⌈√̅2̅⋅̅k̅ᵢ̅ⱼ̅+̅¼ + ½⌉
iₖᵢⱼ = kᵢⱼ-(sₖᵢⱼ-1)(sₖᵢⱼ-2)/2
jₖᵢⱼ = sₖᵢⱼ-iₖᵢⱼ
iₖᵢⱼ/jₖᵢⱼ = i/j
The question is whether all fractions get indeXed.
The answer is no.
Arithmetic disagrees with you.
Logic disagress with you.
No O can leave the matrix by the only drain, the first column, as long
as it contains X below every O. And afterwards nothing can move any longer.
Regards, WM
Le 11/03/2024 à 18:31, Richard Damon a écrit :
On 3/11/24 1:45 AM, WM wrote:
NUF(1) = ℵo like for every visible argument 1/n. No problem to handle. >>> But ℵo is a number that does not appear by miracle but by counting 1,
2, 3, ... because it means cardinality of countable infinite set.
So, you admit you are lying about not using the Matheologies that you
claim you are not using.
No, I use logic and reject miracles.
Regards, WM
Le 11/03/2024 à 23:36, Jim Burns a écrit :
Related:
The first column ℕ⭳×{1} of Xs is
a set into which
the whole matrix ℕ⭳×ℕ⭳ of Xs.and.Os fits.
Only if
all Os disappear in a counter logical manner.
Remember:
Lossless swaps remain lossless in infinity.
Le 12/03/2024 à 00:06, Jim Burns a écrit :
On 3/11/2024 5:21 PM, WM wrote:
Le 11/03/2024 à 18:27, Richard Damon a écrit :
But that isn't the question.
The question is whether all fractions get indeXed.
Fraction i/j gets index kᵢⱼ
Sᵢⱼ = i+j
kᵢⱼ = (Sᵢⱼ-1)(Sᵢⱼ-2)/2+i
Index kᵢⱼ gets fraction iₖᵢⱼ/jₖᵢⱼ
sₖᵢⱼ = ⌈√̅2̅⋅̅k̅ᵢ̅ⱼ̅+̅¼ + ½⌉
iₖᵢⱼ = kᵢⱼ-(sₖᵢⱼ-1)(sₖᵢⱼ-2)/2
jₖᵢⱼ = sₖᵢⱼ-iₖᵢⱼ
iₖᵢⱼ/jₖᵢⱼ = i/j
The question is whether all fractions get indeXed.
The answer is no.
Arithmetic disagrees with you.
Logic disagress with you.
No O can leave the matrix by the only drain,
the first column, as long as
it contains X below every O.
And afterwards nothing can move any longer.
For set ℕ⭳ of finiteⁿᵒᵗᐧᵂᴹ ordinals
not.exists finiteⁿᵒᵗᐧᵂᴹ ordinal n such that
before.n [0,n] not.fits ℕ⭳
Thus,
set ℕ⭳ of finiteⁿᵒᵗᐧᵂᴹ ordinals is
a not.finiteⁿᵒᵗᐧᵂᴹ set.
Some matrices have single columns which
can cover the whole matrix.
They aren't finiteⁿᵒᵗᐧᵂᴹ matrices.
You are just stuck in your finite box staring longingly at the infinite
that you can never understand.
On 3/11/24 2:23 PM, WM wrote:
Le 11/03/2024 à 18:31, Richard Damon a écrit :
On 3/11/24 1:45 AM, WM wrote:
NUF(1) = ℵo like for every visible argument 1/n. No problem to handle. >>>> But ℵo is a number that does not appear by miracle but by counting 1, >>>> 2, 3, ... because it means cardinality of countable infinite set.
No, I use logic and reject miracles.
No, you use limited logic in domains it can't handle.
IT DOES, if you can think of infinity, which you can't.
Le 12/03/2024 à 18:19, Richard Damon a écrit :
On 3/11/24 2:23 PM, WM wrote:
Le 11/03/2024 à 18:31, Richard Damon a écrit :
On 3/11/24 1:45 AM, WM wrote:
NUF(1) = ℵo like for every visible argument 1/n. No problem to handle. >>>>> But ℵo is a number that does not appear by miracle but by counting >>>>> 1, 2, 3, ... because it means cardinality of countable infinite set.
No, I use logic and reject miracles.
No, you use limited logic in domains it can't handle.
Points on the real axis are subject to logic.
Regards, WM
Le 12/03/2024 à 17:46, Richard Damon a écrit :
You are just stuck in your finite box staring longingly at the
infinite that you can never understand.
If you can tell us how an O (or all, if that is easier) does leave the matrix.
Regards, WM
Le 12/03/2024 à 20:55, Jim Burns a écrit :
Some matrices have single columns which
can cover the whole matrix.
They aren't finiteⁿᵒᵗᐧᵂᴹ matrices.
Tell us how any O will leave the matrix
by simple exchange with an X.
Tell us how any O will leave the matrix
by simple exchange with an X.
Tell us how any O will leave the matrix
by simple exchange with an X.
Le 12/03/2024 à 19:28, Jim Burns a écrit :
For set ℕ⭳ of finiteⁿᵒᵗᐧᵂᴹ ordinals
not.exists finiteⁿᵒᵗᐧᵂᴹ ordinal n such that
before.n [0,n] not.fits ℕ⭳
Thus,
set ℕ⭳ of finiteⁿᵒᵗᐧᵂᴹ ordinals is
a not.finiteⁿᵒᵗᐧᵂᴹ set.
Tell us how an O does leave the matrix.
What is the drain?
On 3/13/24 3:27 AM, WM wrote:
Points on the real axis are subject to logic.
Not FINITE logic.
On 3/13/2024 6:32 AM, WM wrote:
Tell us how any O will leave the matrix
by simple exchange with an X.
All the Os.and.Xs are replaced by all the Xs
Then we see that ALL the O's in the set of rationals become X's from the
Set of Naturals, and all those X's became O's
On 3/13/2024 6:31 AM, WM wrote:
Tell us how an O does leave the matrix.
What is the drain?
Your claim that
☠ a proper subset of ℕ⭳ can't cover ℕ⭳
is wrong.
On 03/12/2024 12:18 PM, Chris M. Thomasson wrote:
On 3/12/2024 9:43 AM, Richard Damon wrote:
[...]
I bet the phrase "infinite finite things"
might make WM's head explode... ;^)
Humm...
Now, as to whether:
"if there are infinitely-many
are there infinitely-grand?",
has that usually,
"if there are infinitely-grand
there are infinitely-many".
So, are there?
Le 13/03/2024 à 19:53, Jim Burns a écrit :
On 3/13/2024 6:31 AM, WM wrote:
Tell us how an O does leave the matrix.
What is the drain?
Your claim that
☠ a proper subset of ℕ⭳ can't cover ℕ⭳
is wrong.
It is true at least until you can explain how
an O leaves the matrix.
But your recent attempts are simply nonsense -
like your future attempts will be - because
exchange cannot reduce the number of Os,
let alone delete all of them.
Le 13/03/2024 à 16:10, Richard Damon a écrit :
On 3/13/24 3:27 AM, WM wrote:
Points on the real axis are subject to logic.
Not FINITE logic.
Every point is subject to real finite logic.
Regards, WM
Le 13/03/2024 à 16:09, Richard Damon a écrit :
Then we see that ALL the O's in the set of rationals become X's from
the Set of Naturals, and all those X's became O's
And how do they leave the matrix?
Regards, WM
On 3/13/24 12:33 PM, WM wrote:
Points on the real axis are subject to logic.
Not FINITE logic.
Every point is subject to real finite logic.
But the SET Of them isn't.
On 3/13/2024 3:40 PM, WM wrote:
It is true at least until you can explain how
an O leaves the matrix.
Your premise is wrong.
No O leaves.
All Os are covered by Xs from a proper subset.
The first column covers the whole matrix.
On 3/13/24 12:32 PM, WM wrote:
Then we see that ALL the O's in the set of rationals become X's from
the Set of Naturals, and all those X's became O's
And how do they leave the matrix?
Because the go into the vector of the Natural Numbers that is sending the Xs
Le 14/03/2024 à 06:35, Richard Damon a écrit :
On 3/13/24 12:32 PM, WM wrote:
Then we see that ALL the O's in the set of rationals become X's from
the Set of Naturals, and all those X's became O's
And how do they leave the matrix?
Because the go into the vector of the Natural Numbers that is sending
the Xs
No O goes into an X. Exchanging differs from occupying.
Regards, WM
Le 14/03/2024 à 01:15, Richard Damon a écrit :
On 3/13/24 12:33 PM, WM wrote:
Points on the real axis are subject to logic.
Not FINITE logic.
Every point is subject to real finite logic.
But the SET Of them isn't.
Then "the set" is a matheologial fiction without connection to basic mathematics and basic logic - a useless occupation for lovers of nonsense.
Regards, wM
the O goes to where the X was
and the X goes to where the O was.
the Os that go to the Natural Numbers /
Unit fractions stay with the set of Rationals,
so the arguement isn't valid.
All the O's end up at the end of the Row/Column, where they would belong.
On 3/14/24 10:07 AM, WM wrote:
Le 14/03/2024 à 01:15, Richard Damon a écrit :So, you are just admitting that you "Logic" can't talk about thing like
On 3/13/24 12:33 PM, WM wrote:
Points on the real axis are subject to logic.
Not FINITE logic.
Every point is subject to real finite logic.
But the SET Of them isn't.
Then "the set" is a matheologial fiction without connection to basic
mathematics and basic logic - a useless occupation for lovers of nonsense. >>
the Natural Numbers or the Rational, because that set is just a fiction.
Your logic can only talk about finite FISONs.
Le 14/03/2024 à 18:40, Richard Damon a écrit :
On 3/14/24 10:07 AM, WM wrote:
Le 14/03/2024 à 01:15, Richard Damon a écrit :So, you are just admitting that you "Logic" can't talk about thing
On 3/13/24 12:33 PM, WM wrote:
Points on the real axis are subject to logic.
Not FINITE logic.
Every point is subject to real finite logic.
But the SET Of them isn't.
Then "the set" is a matheologial fiction without connection to basic
mathematics and basic logic - a useless occupation for lovers of
nonsense.
like the Natural Numbers or the Rational, because that set is just a
fiction.
Logic is a general tool, not my own. If it cannot talk about infinite
sets, then they do not exist or they are dark.
Your logic can only talk about finite FISONs.
It can talk about the actually infinite complement too:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Regards, WM
Le 14/03/2024 à 18:39, Richard Damon a écrit :
the O goes to where the X was and the X goes to where the O was.
So it is. Never an O goes out of the matrix.
the Os that go to the Natural Numbers / Unit fractions stay with the
set of Rationals,
So it is.
so the arguement isn't valid.
What argument?
All the O's end up at the end of the Row/Column, where they would belong.
Also "the end", i.e., the places where the O can go, belong to the matrix.
Regards, WM
On 03/14/2024 10:09 AM, WM wrote:
[...]
Quite a few years ago I started developing a
notion "restricted sequence element interchange".
This is where, when you think of an infinite series,
and convergence, there's often an idea "well if you
can re-order it so in front are infinitely many
terms that don't converge, and there isn't a symmetric
complement in the subtractive or these are only positive
terms, then it diverges".
Le 13/03/2024 à 23:23, Jim Burns a écrit :
On 3/13/2024 3:40 PM, WM wrote:
It is true at least until you can explain how
an O leaves the matrix.
Your premise is wrong.
My premise is that
in the beginning
there was no chaos
but a sharp border between Xs and Os:
XOOO...
XOOO...
XOOO...
XOOO...
..
And these symbols can be interchanged
but none cane exist with the other.
No O leaves.
All Os are covered by Xs from a proper subset.
There is no possibility to "cover" an O.
X and O are excluding each other.
There is only a peaceful coexistence.
The first column covers the whole matrix.
Not without removing all Os.
| Bernhard Riemann proved that
| a conditionally convergent series may be rearranged
| to converge to any value at all, including ∞ or −∞;
| see Riemann series theorem.
On 3/14/2024 1:03 PM, WM wrote:
There is no possibility to "cover" an O.
X and O are excluding each other.
There is only a peaceful coexistence.
Your premise is that
logicallyᵂᴹ
a proper.subset cannot cover its superset,
that no 1.to.1.map exists from set to subset.
Your premise is that,
None of your offered explanations are correct for
the set ℕ⭳ of finiteⁿᵒᵗᐧᵂᴹ ordinals.
On 3/14/24 11:41 AM, WM wrote:
Logic is a general tool, not my own. If it cannot talk about infinite
sets, then they do not exist or they are dark.
Nope, there are different rules for logic,
It can talk about the actually infinite complement too:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Nope, since you admit that your logic can't define "ℕ", you can't even define that infinite complement either.
On 3/14/24 11:38 AM, WM wrote:
Le 14/03/2024 à 18:39, Richard Damon a écrit :
the O goes to where the X was and the X goes to where the O was.
So it is. Never an O goes out of the matrix.
So you aren't doing it right.
Why do you say that the O going to the OTHER set never leaves,
and the X
that comes from the OTHER set never replaces it?
All the O's end up at the end of the Row/Column, where they would belong. >>Also "the end", i.e., the places where the O can go, belong to the matrix.
No, they don't,
the belong to the SET of Natural Numbers or Unit Fractions.
The "Natural Number n" is NOT the Rational Number n/1, but they have the
same vaule.
The "Unit Fraction 1/n" is NOT the Rational Number 1/n,
Le 14/03/2024 à 20:06, Richard Damon a écrit :
On 3/14/24 11:38 AM, WM wrote:
Le 14/03/2024 à 18:39, Richard Damon a écrit :
the O goes to where the X was and the X goes to where the O was.
So it is. Never an O goes out of the matrix.
So you aren't doing it right.
Why do you say that the O going to the OTHER set never leaves,
It does not leave the matrix.
and the X that comes from the OTHER set never replaces it?
Every replacement fails to remove an O from the matrix.
All the O's end up at the end of the Row/Column, where they would
belong.
Also "the end", i.e., the places where the O can go, belong to the
matrix.
No, they don't,
Where do they go?
the belong to the SET of Natural Numbers or Unit Fractions.
Of course, but all elements of these sets belong to the matrix.
The "Natural Number n" is NOT the Rational Number n/1, but they have
the same vaule.
Initially every n/1 is indeXed by n.
The "Unit Fraction 1/n" is NOT the Rational Number 1/n,
You are wrong. But all that does not remove an O from the matrix.
Regards, WM
Le 14/03/2024 à 22:42, Jim Burns a écrit :
| Bernhard Riemann proved that
| a conditionally convergent series may be rearranged
| to converge to any value at all, including ∞ or −∞;
| see Riemann series theorem.
It is not difficult to find examples.
But it is difficult to find out why that can happen.
The key is Dark Numbers.
Le 14/03/2024 à 22:25, Jim Burns a écrit :
On 3/14/2024 1:03 PM, WM wrote:
There is no possibility to "cover" an O.
X and O are excluding each other.
There is only a peaceful coexistence.
Your premise is that
logicallyᵂᴹ
a proper.subset cannot cover its superset,
that no 1.to.1.map exists from set to subset.
Not at all!
My premise is that exchange cannot delete!
The conclusion is
that a proper subset cannot cover its superset.
None of your offered explanations are correct for
the set ℕ⭳ of finiteⁿᵒᵗᐧᵂᴹ ordinals.
They cannot be worse than Bob's death.
On 3/15/24 6:02 AM, WM wrote:
Every replacement fails to remove an O from the matrix.
Nope,
On 3/15/2024 8:10 AM, WM wrote:
Le 14/03/2024 à 22:42, Jim Burns a écrit :
| Bernhard Riemann proved that
| a conditionally convergent series may be rearranged
| to converge to any value at all, including ∞ or −∞;
| see Riemann series theorem.
It is not difficult to find examples.
But it is difficult to find out why that can happen.
It's a theorem.
https://en.wikipedia.org/wiki/Riemann_series_theorem
The proof of the theorem is why.
The key is Dark Numbers.
No darkᵂᴹ numbers are used.
𝑙𝑛(2) =
∫₀¹(1+x)⁻¹𝑑x =
∫₀¹(1-x+x²-x³+...)𝑑x =
∫₀¹1𝑑x - ∫₀¹x𝑑x + ∫₀¹x²𝑑x - ∫₀¹x³𝑑x + ... = ⅟1 - ⅟2 + ⅟3 - ⅟4 + ...
No darkᵂᴹ numbers are used.
On 3/15/2024 8:19 AM, WM wrote:
Le 14/03/2024 à 22:25, Jim Burns a écrit :
On 3/14/2024 1:03 PM, WM wrote:
There is no possibility to "cover" an O.
X and O are excluding each other.
There is only a peaceful coexistence.
Your premise is that
logicallyᵂᴹ
a proper.subset cannot cover its superset,
that no 1.to.1.map exists from set to subset.
Not at all!
My premise is that exchange cannot delete!
Which is, in other words, that
no 1.to.1.map exists from set to proper.subset.
Le 15/03/2024 à 17:01, Richard Damon a écrit :
On 3/15/24 6:02 AM, WM wrote:
Every replacement fails to remove an O from the matrix.
Nope,
No interest in discussing that.
Regards, WM
Le 15/03/2024 à 21:26, Jim Burns a écrit :
On 3/15/2024 8:19 AM, WM wrote:
Le 14/03/2024 à 22:25, Jim Burns a écrit :
On 3/14/2024 1:03 PM, WM wrote:
There is no possibility to "cover" an O.
X and O are excluding each other.
There is only a peaceful coexistence.
Your premise is that
logicallyᵂᴹ
a proper.subset cannot cover its superset,
that no 1.to.1.map exists from set to subset.
Not at all!
My premise is that exchange cannot delete!
Which is, in other words, that
no 1.to.1.map exists from set to proper.subset.
That is not a premise but a result.
Le 15/03/2024 à 18:31, Jim Burns a écrit :
On 3/15/2024 8:10 AM, WM wrote:
Le 14/03/2024 à 22:42, Jim Burns a écrit :
| Bernhard Riemann proved that
| a conditionally convergent series may be rearranged
| to converge to any value at all, including ∞ or −∞;
| see Riemann series theorem.
It is not difficult to find examples.
But it is difficult to find out why that can happen.
It's a theorem.
https://en.wikipedia.org/wiki/Riemann_series_theorem
The proof of the theorem is why.
The key is Dark Numbers.
No darkᵂᴹ numbers are used.
𝑙𝑛(2) =
∫₀¹(1+x)⁻¹𝑑x =
∫₀¹(1-x+x²-x³+...)𝑑x =
∫₀¹1𝑑x - ∫₀¹x𝑑x + ∫₀¹x²𝑑x - ∫₀¹x³𝑑x + ... =
⅟1 - ⅟2 + ⅟3 - ⅟4 + ...
No darkᵂᴹ numbers are used.
The same infinitely.manyⁿᵒᵗᐧᵂᴹ visibleᵂᴹ numbers
summed _in a different order_
(⅟1-⅟2) - ⅟4 + (⅟3-⅟6) - ⅟8 + (⅟5-⅟10) - ⅟12 + ... =
⅟2 - ⅟4 + ⅟6 - ⅟8 + ⅟10 - ⅟12 + ... =
⅟2⋅(⅟1 - ⅟2 + ⅟3 - ⅟4 + ⅟5 - ⅟6 + ...) =
⅟2⋅𝑙𝑛(2)
No darkᵂᴹ numbers are used.
And yet, 𝑙𝑛(2) ≠ ⅟2⋅𝑙𝑛(2)
They cannot be used.
They are contained in "...".
Otherwise you could name all terms.
But you can name only finitely many.
On 3/15/2024 5:54 PM, WM wrote:
Le 15/03/2024 à 21:26, Jim Burns a écrit :
On 3/15/2024 8:19 AM, WM wrote:
Le 14/03/2024 à 22:25, Jim Burns a écrit :
On 3/14/2024 1:03 PM, WM wrote:
There is no possibility to "cover" an O.
X and O are excluding each other.
There is only a peaceful coexistence.
Your premise is that
logicallyᵂᴹ
a proper.subset cannot cover its superset,
that no 1.to.1.map exists from set to subset.
Not at all!
My premise is that exchange cannot delete!
Which is, in other words, that
no 1.to.1.map exists from set to proper.subset.
That is not a premise but a result.
Your result is only a minor re.stating of
your premise.
On 3/15/24 2:48 PM, WM wrote:
Le 15/03/2024 à 17:01, Richard Damon a écrit :
On 3/15/24 6:02 AM, WM wrote:
Every replacement fails to remove an O from the matrix.
Nope,
No interest in discussing that.
Regards, WM
So, just closing your eyes to the truth again.
On 3/15/2024 5:52 PM, WM wrote:
Le 15/03/2024 à 18:31, Jim Burns a écrit :
On 3/15/2024 8:10 AM, WM wrote:
Le 14/03/2024 à 22:42, Jim Burns a écrit :
| Bernhard Riemann proved that
| a conditionally convergent series may be rearranged
| to converge to any value at all, including ∞ or −∞;
| see Riemann series theorem.
It is not difficult to find examples.
But it is difficult to find out why that can happen.
It's a theorem.
https://en.wikipedia.org/wiki/Riemann_series_theorem
The proof of the theorem is why.
The key is Dark Numbers.
No darkᵂᴹ numbers are used.
𝑙𝑛(2) =
∫₀¹(1+x)⁻¹𝑑x =
∫₀¹(1-x+x²-x³+...)𝑑x =
∫₀¹1𝑑x - ∫₀¹x𝑑x + ∫₀¹x²𝑑x - ∫₀¹x³𝑑x + ... =
⅟1 - ⅟2 + ⅟3 - ⅟4 + ...
No darkᵂᴹ numbers are used.
The same infinitely.manyⁿᵒᵗᐧᵂᴹ visibleᵂᴹ numbers
summed _in a different order_
(⅟1-⅟2) - ⅟4 + (⅟3-⅟6) - ⅟8 + (⅟5-⅟10) - ⅟12 + ... =
⅟2 - ⅟4 + ⅟6 - ⅟8 + ⅟10 - ⅟12 + ... =
⅟2⋅(⅟1 - ⅟2 + ⅟3 - ⅟4 + ⅟5 - ⅟6 + ...) =
⅟2⋅𝑙𝑛(2)
No darkᵂᴹ numbers are used.
And yet, 𝑙𝑛(2) ≠ ⅟2⋅𝑙𝑛(2)
They cannot be used.
Any yet, _without using darkᵂᴹ numbers_
your assumption leads to 𝑙𝑛(2) = ⅟2⋅𝑙𝑛(2)
They are contained in "...".
"..." isn't a container of any kind.
Suppose I repeat all that without '...'
which I have done previously.
Do your darkᵂᴹ numbers stop existing?
Otherwise you could name all terms.
But you can name only finitely many.
I can make a claim true of each term, and
follow it with claims not.first.false of each term.
On 3/15/2024 5:52 PM, WM wrote:
Le 15/03/2024 à 18:31, Jim Burns a écrit :
On 3/15/2024 8:10 AM, WM wrote:
Le 14/03/2024 à 22:42, Jim Burns a écrit :
| Bernhard Riemann proved that
| a conditionally convergent series may be rearranged
| to converge to any value at all, including ∞ or −∞;
| see Riemann series theorem.
It is not difficult to find examples.
But it is difficult to find out why that can happen.
It's a theorem.
https://en.wikipedia.org/wiki/Riemann_series_theorem
The proof of the theorem is why.
The key is Dark Numbers.
No darkᵂᴹ numbers are used.
𝑙𝑛(2) =
∫₀¹(1+x)⁻¹𝑑x =
∫₀¹(1-x+x²-x³+...)𝑑x =
∫₀¹1𝑑x - ∫₀¹x𝑑x + ∫₀¹x²𝑑x - ∫₀¹x³𝑑x + ... =
⅟1 - ⅟2 + ⅟3 - ⅟4 + ...
No darkᵂᴹ numbers are used.
The same infinitely.manyⁿᵒᵗᐧᵂᴹ visibleᵂᴹ numbers
summed _in a different order_
(⅟1-⅟2) - ⅟4 + (⅟3-⅟6) - ⅟8 + (⅟5-⅟10) - ⅟12 + ... =
⅟2 - ⅟4 + ⅟6 - ⅟8 + ⅟10 - ⅟12 + ... =
⅟2⋅(⅟1 - ⅟2 + ⅟3 - ⅟4 + ⅟5 - ⅟6 + ...) =
⅟2⋅𝑙𝑛(2)
No darkᵂᴹ numbers are used.
And yet, 𝑙𝑛(2) ≠ ⅟2⋅𝑙𝑛(2)
They cannot be used.
Any yet, _without using darkᵂᴹ numbers_
your assumption leads to 𝑙𝑛(2) = ⅟2⋅𝑙𝑛(2)
They are contained in "...".
"..." isn't a container of any kind.
Suppose I repeat all that without '...'
which I have done previously.
Do your darkᵂᴹ numbers stop existing?
Otherwise you could name all terms.
But you can name only finitely many.
I can make a claim true of each term, and
follow it with claims not.first.false of each term.
Le 15/03/2024 à 23:35, Jim Burns a écrit :
On 3/15/2024 5:54 PM, WM wrote:
Le 15/03/2024 à 21:26, Jim Burns a écrit :
On 3/15/2024 8:19 AM, WM wrote:
Le 14/03/2024 à 22:25, Jim Burns a écrit :
On 3/14/2024 1:03 PM, WM wrote:
There is no possibility to "cover" an O.
X and O are excluding each other.
There is only a peaceful coexistence.
Your premise is that
logicallyᵂᴹ
a proper.subset cannot cover its superset,
that no 1.to.1.map exists from set to subset.
Not at all!
My premise is that exchange cannot delete!
Which is, in other words, that
no 1.to.1.map exists from set to proper.subset.
That is not a premise but a result.
Your result is only a minor re.stating of
your premise.
The premise is this:
When an exchange between X and O is defined,
then none of the partners can disappear.
That is so simple and self-evident,
that it cannot reasonable be denied.
Nice that you agree that
this leads with no doubt to
the mathematical truth that
the (set of positive) fractions cannot be enumerated.
I did never claim more.
Le 15/03/2024 à 23:19, Richard Damon a écrit :
On 3/15/24 2:48 PM, WM wrote:
Le 15/03/2024 à 17:01, Richard Damon a écrit :
On 3/15/24 6:02 AM, WM wrote:
Every replacement fails to remove an O from the matrix.
Nope,
No interest in discussing that.
Regards, WM
So, just closing your eyes to the truth again.
If by definition only exchange of X and O is allowed, then the truth is
that none of the partners can disappear. A "logic" which violates this
truth is simply no logic.
Regards, WM
Le 15/03/2024 à 23:38, Jim Burns a écrit :
On 3/15/2024 5:52 PM, WM wrote:
They are contained in "...".
"..." isn't a container of any kind.
You are wrong.
In all explicit representations of infinite sequences
"..." contains almost all terms.
Suppose I repeat all that without '...'
which I have done previously.
Do your darkᵂᴹ numbers stop existing?
You cannot do an explicit representation
without "...".
You can give the formula.
But since the terms
⅟1 - ⅟2 + ⅟3 - ⅟4 + ...
do not change in the least by
reordering the visible part,
the sum remains the same as before.
I can make a claim true of each term, and
follow it with claims not.first.false of each term.
For each *visible* term where you can check it.
On 3/16/24 1:39 AM, WM wrote:
If by definition only exchange of X and O is allowed, then the truth is
that none of the partners can disappear. A "logic" which violates this
truth is simply no logic.
But the X's move from their INDEPENDENT SET into the MATRIX, and the Os
move from their MATRIX into the INDEPENDENT SET.
Thus ALL the Os leave the MATRIX and end up in the SET, and all the Xs
leave the SET and go into the MATRIX.
So, NO O reamin in the MATRIX and no X reamins in the SET.
So OF COURSE moving from the matrix to a part of the matrix can't remove
them from the matrix.
What were you thinking?
On 3/16/2024 5:17 AM, WM wrote:
Le 15/03/2024 à 23:38, Jim Burns a écrit :
On 3/15/2024 5:52 PM, WM wrote:
They are contained in "...".
"..." isn't a container of any kind.
You are wrong.
https://en.wikipedia.org/wiki/Ellipsis#In_mathematical_notation
| An ellipsis is also often used in mathematics
| to mean "and so forth".
| "I don't know what you mean by 'glory,'"
| Alice said.
In all explicit representations of infinite sequences
"..." contains almost all terms.
" ⅟1 - ⅟2 + ⅟3 - ⅟4 + ... "
is not explicit.
You cannot do an explicit representation
without "...".
You can give the formula.
The formula is very specific, clear, or detailed.
The formula is explicit.
But since the terms
⅟1 - ⅟2 + ⅟3 - ⅟4 + ...
do not change in the least by
reordering the visible part,
the sum remains the same as before.
...except for the sum NOT remaining
the same as before.
Le 17/03/2024 à 06:55, Jim Burns a écrit :
On 3/16/2024 5:17 AM, WM wrote:
Le 15/03/2024 à 23:38, Jim Burns a écrit :
On 3/15/2024 5:52 PM, WM wrote:
They are contained in "...".
"..." isn't a container of any kind.
You are wrong.
https://en.wikipedia.org/wiki/Ellipsis#In_mathematical_notation
| An ellipsis is also often used in mathematics
| to mean "and so forth".
Of course.
But while for infinite sequences
the "and so forth" is absolutely correct
(1, 2, 3, ... means all natural numbers)
most of these elements cannot be explicitly named.
Lossless exchange of X and O will result in
losses of Os <==>
All positive fractions can be enumerated.
This is a glorious result
sufficient for every mathematician to
determine his position.
In all explicit representations of
infinite sequences
"..." contains almost all terms.
" ⅟1 - ⅟2 + ⅟3 - ⅟4 + ... "
is not explicit.
It is what I call explicit.
If you know a more suitable English word
we can use it.
But I am sure you know what I mean.
You cannot do an explicit representation
without "...".
You can give the formula.
The formula is very specific, clear, or detailed.
The formula is explicit.
Some formulas are called explicit.
But what I mean is naming all elements
one by one.
But since the terms
⅟1 - ⅟2 + ⅟3 - ⅟4 + ...
do not change in the least by
reordering the visible part,
the sum remains the same as before.
...except for the sum NOT remaining
the same as before.
*In case of actual infinity*,
all terms and hence their sum remains
the same as before.
That is different in potential infinity
as applied by Riemann.
Le 16/03/2024 à 21:22, Richard Damon a écrit :
On 3/16/24 1:39 AM, WM wrote:
If by definition only exchange of X and O is allowed, then the truth
is that none of the partners can disappear. A "logic" which violates
this truth is simply no logic.
But the X's move from their INDEPENDENT SET into the MATRIX, and the
Os move from their MATRIX into the INDEPENDENT SET.
No. First all interger fractions are enumerated. Then the X moves from
the matrix (its first column) into the matrix (including its first
column). The O moves from the matrix (including its first column) into
the matrix (its first column). And every line of the matrix where the respective last O goes is full of Os.
Thus ALL the Os leave the MATRIX and end up in the SET, and all the Xs
leave the SET and go into the MATRIX.
So, NO O reamin in the MATRIX and no X reamins in the SET.
Absolutely wrong. All O remain within the matrix. The first column
belongs to the matrix.
So OF COURSE moving from the matrix to a part of the matrix can't
remove them from the matrix.
So it is. And never an O leavs the matrix.
What were you thinking?
I am glad that we have come to the agreement that the following is an equivalence:
Lossless exchange of X and O will result in losses of Os <==> All
positive fractions can be enumerated.
This is a nice result sufficient for every mathematician to determine
his position.
Regards, WM
On 3/17/2024 7:14 AM, WM wrote:
Le 17/03/2024 à 06:55, Jim Burns a écrit :
On 3/16/2024 5:17 AM, WM wrote:
Le 15/03/2024 à 23:38, Jim Burns a écrit :
On 3/15/2024 5:52 PM, WM wrote:
They are contained in "...".
"..." isn't a container of any kind.
You are wrong.
https://en.wikipedia.org/wiki/Ellipsis#In_mathematical_notation
| An ellipsis is also often used in mathematics
| to mean "and so forth".
Of course.
But while for infinite sequences
the "and so forth" is absolutely correct
Please remember that later,
when the description of the terms
have been augmented by claims which are
checkably not.first.false about the terms.
Lossless exchange of X and O will result in
losses of Os <==>
All positive fractions can be enumerated.
If any O is anywhere, not all swaps are swapped.
If all swaps are swapped, no O is anywhere.
This is a glorious result
sufficient for every mathematician to
determine his position.
In all explicit representations of
infinite sequences
"..." contains almost all terms.
" ⅟1 - ⅟2 + ⅟3 - ⅟4 + ... "
is not explicit.
It is what I call explicit.
If you know a more suitable English word
we can use it.
But what I mean is naming all elements
one by one.
*In case of actual infinity*,
all terms and hence their sum remains
the same as before.
There is no term not in.both.sums.
Their order is different.
The terms are the same.
Their sums are different.
Your "hence" is incorrect.
On 3/17/24 3:59 AM, WM wrote:
Le 16/03/2024 à 21:22, Richard Damon a écrit :
On 3/16/24 1:39 AM, WM wrote:
If by definition only exchange of X and O is allowed, then the truth
is that none of the partners can disappear. A "logic" which violates
this truth is simply no logic.
But the X's move from their INDEPENDENT SET into the MATRIX, and the
Os move from their MATRIX into the INDEPENDENT SET.
No. First all interger fractions are enumerated. Then the X moves from
the matrix (its first column) into the matrix (including its first
column). The O moves from the matrix (including its first column) into
the matrix (its first column). And every line of the matrix where the
respective last O goes is full of Os.
In other words, you are just making up your own rules.
Since you already admitted that you logic can't handle infinte sets, who
let you play with these matches?
Absolutely wrong. All O remain within the matrix. The first column
belongs to the matrix.
Of course, how could it otherwise?
So it is. And never an O leavs the matrix.
Of course they don't, since you aren't even trying.
If your even thought they could,
I am glad that we have come to the agreement that the following is an
equivalence:
Lossless exchange of X and O will result in losses of Os <==> All
positive fractions can be enumerated.
Nope.
The first is just your non-sense.
The second is a proven fact.
To do the proof you seem to be seeing dimly, you need to have TWO
DIFFERENT sets with the Xs in one and the Os in the other.
Le 17/03/2024 à 21:11, Richard Damon a écrit :
On 3/17/24 3:59 AM, WM wrote:
Le 16/03/2024 à 21:22, Richard Damon a écrit :
On 3/16/24 1:39 AM, WM wrote:
If by definition only exchange of X and O is allowed, then the
truth is that none of the partners can disappear. A "logic" which
violates this truth is simply no logic.
But the X's move from their INDEPENDENT SET into the MATRIX, and the
Os move from their MATRIX into the INDEPENDENT SET.
No. First all interger fractions are enumerated. Then the X moves
from the matrix (its first column) into the matrix (including its
first column). The O moves from the matrix (including its first
column) into the matrix (its first column). And every line of the
matrix where the respective last O goes is full of Os.
In other words, you are just making up your own rules.
No, I apply Cantor's rules precisely:
k = (m + n - 1)(m + n - 2)/2 + m.
The indexing follows his sequence:
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, ...
Since you already admitted that you logic can't handle infinte sets,
who let you play with these matches?
My logic handles infinite sets, for instance the Os and Xs.
Absolutely wrong. All O remain within the matrix. The first column
belongs to the matrix.
Of course, how could it otherwise?
So it is. And never an O leavs the matrix.
Of course they don't, since you aren't even trying.
The drain is blocked. No try could succeed.
If your even thought they could,
I never did.
I am glad that we have come to the agreement that the following is an
equivalence:
Lossless exchange of X and O will result in losses of Os <==> All
positive fractions can be enumerated.
Nope.
The first is just your non-sense.
You claimed that the O leave.
The second is a proven fact.
The second is nonsense, proven to every natural number which has lmost
all numbers as successors.
To do the proof you seem to be seeing dimly, you need to have TWO
DIFFERENT sets with the Xs in one and the Os in the other.
Yes. And then the first set is bijected with the first column. Is that forbidden?
Regards, WM
On 3/17/2024 7:14 AM, WM wrote:
The same terms added in a different order
yield different finite initial sums.
In an absolutely convergent infinite sum,
in every order, almost all the finite sums
are near the same point.
That is similar to how, in a finite sum,
the terms added in every order give the same sum.
An infinite sum is not a finite sum, though.
Do _absolutely_ convergent infinite sums
not have darkᵂᴹ terms?
Are the same terms visibleᵂᴹ in some series
and darkᵂᴹ in other series?
That is different in potential infinity
as applied by Riemann.
Potential.infinityᵂᴹ is infinityⁿᵒᵗᐧᵂᴹ.
Actual.infinityᵂᴹ is intended to sound impressive.
On 3/17/24 2:27 PM, WM wrote:
No, I apply Cantor's rules precisely:
k = (m + n - 1)(m + n - 2)/2 + m.
The indexing follows his sequence:
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, ...
But the k is from one independent set, and the m and n from the matrix.
He never is mapping m and n to the first row or column of that same matrix
My logic handles infinite sets, for instance the Os and Xs.
You admitted earlier that you logic was finite.
So, what made you think you should combine the two seperate sets of the matching into one?
You claimed that the O leave.
Yes, when you do it by the instructions, the O's leave the matrix into
the Set that is seperate.
Yes. And then the first set is bijected with the first column. Is that
forbidden?
Bijection is between TWO sets, not one and part of itself.
you need to
promote that part to be a whole independent if you want to count, so otherwise you have elements part of both.
Le 17/03/2024 à 22:54, Richard Damon a écrit :
On 3/17/24 2:27 PM, WM wrote:
No, I apply Cantor's rules precisely:
k = (m + n - 1)(m + n - 2)/2 + m.
The indexing follows his sequence:
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, ...
But the k is from one independent set, and the m and n from the matrix.
After the bijection k <--> k/1 the k is attached to the matrix place k/1.
He never is mapping m and n to the first row or column of that same
matrix
I have inserted the intermediate step k <--> k/1. Is it forbidden?
My logic handles infinite sets, for instance the Os and Xs.
You admitted earlier that you logic was finite.
All logic is finite. All ZFC-formulas are finite.
So, what made you think you should combine the two seperate sets of
the matching into one?
It is possible according to set theory and it shows that set theory is nonsense. That made me think to try it.
You claimed that the O leave.
Yes, when you do it by the instructions, the O's leave the matrix into
the Set that is seperate.
Now it is in the first column. What is wrong with that?
Yes. And then the first set is bijected with the first column. Is
that forbidden?
Bijection is between TWO sets, not one and part of itself.
Bijections are allowed between set and proper subset. That is even a definition of infinity. Look up Dedekind infinity.
you need to promote that part to be a whole independent if you want to
count, so otherwise you have elements part of both.
That defines Dedekind-infinity.
Regards, WM
Le 17/03/2024 à 18:26, Jim Burns a écrit :
On 3/17/2024 7:14 AM, WM wrote:
In an absolutely convergent infinite sum,
in every order, almost all the finite sums
are near the same point.
That is similar to how, in a finite sum,
the terms added in every order give the same sum.
An infinite sum is not a finite sum, though.
Up to every definable term it is finite.
Do _absolutely_ convergent infinite sums
not have darkᵂᴹ terms?
Of course they have dark terms too.
But you cannot sort them in a way
that would yield another limit.
Are the same terms visibleᵂᴹ in some series
and darkᵂᴹ in other series?
Visisbiliyt depends on time and system.
That is different in potential infinity
as applied by Riemann.
Potential.infinityᵂᴹ is infinityⁿᵒᵗᐧᵂᴹ.
It is without end.
Actual.infinityᵂᴹ is intended to sound impressive.
No it is a significant difference.
Hilbert's hotel for instance is not possible
in actual infinity because already
all natnumbers are present.
The same terms added in a different order
yield different finite initial sums.
For instance
the alternating harmonic series can be reordered,
such that
every positive term is followed by 10000 negative terms.
Do you think that
by this *reordering*
the set of negative terms gets
10000 times as big as the set of positive terms?
If all terms are there,
then this assumption is simply foolish.
So do I. Have you not yet understood my argument? The indices 1, 2, 3, ... required for enumeration are a set ℕ. The fractions to be enumerated are another set, represented by the matrix of not indexed fractions:That defines Dedekind-infinity.
But the subset has been made into its own set, not just some selected elements in the original set.
A subset is made from a set by copying the selected elements out of the original set.As we can see here, this Dedekind-infinity is self-contradictory concept.
On 3/18/2024 8:04 AM, WM wrote:
Le 17/03/2024 à 18:26, Jim Burns a écrit :
On 3/17/2024 7:14 AM, WM wrote:
In an absolutely convergent infinite sum,
in every order, almost all the finite sums
are near the same point.
That is similar to how, in a finite sum,
the terms added in every order give the same sum.
An infinite sum is not a finite sum, though.
Up to every definable term it is finite.
For each term, up to that term
isn't an infinite.sumⁿᵒᵗᐧᵂᴹ.
Do _absolutely_ convergent infinite sums
not have darkᵂᴹ terms?
Of course they have dark terms too.
But you cannot sort them in a way
that would yield another limit.
Are the same terms visibleᵂᴹ in some series
and darkᵂᴹ in other series?
Visibility depends on time and system.
Actual.infinityᵂᴹ is intended to sound impressive.
No it is a significant difference.
Hilbert's hotel for instance is not possible
in actual infinity because already
all natnumbers are present.
Your darkᵂᴹ numbers are inserted with
the intention of changing a set from having
behavior which you think is in error to
behavior which you decided a priori is correct.
The concept of changing which set about which
a question is being asked
strikes me as fairly Zen (from my non.Zen POV).
(I am told) a fair number of koans involve
unaskingᶻᵉⁿ the question,
which is what changing which set does.
That's a conceivably.interesting non.answer.
However, non.answers to mathematical questions
can have very.undesirable consequences.
Crops rotted in the field. Bridges collapsed.
Mathematics isn't _only_ a game.
One issue with your conceivably.interesting answer
is that you intend to _augment_ a misbehavingᵂᴹ set
until it behavesᵂᴹ.
Sadly, the misbehaviorᵂᴹ is because the set is infiniteⁿᵒᵗᐧᵂᴹ. Correcting it would require
the replacement of the set with a finiteⁿᵒᵗᐧᵂᴹ set.
_Augmenting_ will not do that.
The same terms added in a different order
yield different finite initial sums.
For instance
the alternating harmonic series can be reordered,
such that
every positive term is followed by 10000 negative terms.
Do you think that
by this *reordering*
the set of negative terms gets
10000 times as big as the set of positive terms?
⟨ 1 2 3 4 ... ⟩ is the same size as
⟨ 10000 20000 30000 40000 ... ⟩
On 03/18/2024 05:04 AM, WM wrote:
Le 17/03/2024 à 18:26, Jim Burns a écrit :
In an absolutely convergent infinite sum,
in every order, almost all the finite sums
are near the same point.
What are you trying to show, anyways?
Convergence of infinite series essentially
depends on prototypes of
infinite series that converge.
Le 18/03/2024 à 20:35, Jim Burns a écrit :
On 3/18/2024 8:04 AM, WM wrote:
Le 17/03/2024 à 18:26, Jim Burns a écrit :
On 3/17/2024 7:14 AM, WM wrote:
In an absolutely convergent infinite sum,
in every order, almost all the finite sums
are near the same point.
That is similar to how, in a finite sum,
the terms added in every order give the same sum.
An infinite sum is not a finite sum, though.
Up to every definable term it is finite.
For each term, up to that term
isn't an infinite.sumⁿᵒᵗᐧᵂᴹ.
Right.
Potentially infinite series have no sums
but only limits.
Often this is confused.
Your darkᵂᴹ numbers are inserted with
the intention of changing a set from having
behavior which you think is in error to
behavior which you decided a priori is correct.
It is the onl possibility
if actual infinity is true.
The concept of changing which set about which
a question is being asked
strikes me as fairly Zen (from my non.Zen POV).
I don't know this religion.
One issue with your conceivably.interesting answer
is that you intend to _augment_ a misbehavingᵂᴹ set
until it behavesᵂᴹ.
Yes.
Sadly, the misbehaviorᵂᴹ is because the set is
infiniteⁿᵒᵗᐧᵂᴹ. Correcting it would require
the replacement of the set with a finiteⁿᵒᵗᐧᵂᴹ set.
_Augmenting_ will not do that.
It should be clear to every intelligent being,
that almost all elements of an infinite set
cannot be addressed.
People who claim that
every natural number can be defined and chosen
are worse than fools.
All definable terms of an infinite sequence belong to
a finite although changable set.
The same terms added in a different order
yield different finite initial sums.
For instance
the alternating harmonic series can be reordered,
such that every positive term is followed by 10000 negative terms.
Do you think that
by this *reordering*
the set of negative terms gets
10000 times as big as the set of positive terms?
⟨ 1 2 3 4 ... ⟩ is the same size as
⟨ 10000 20000 30000 40000 ... ⟩
That requires:
Lossless exchange of X and O will result in
losses of Os <==>
All positive fractions can be enumerated.
It has no place in the realm of logic,
at least not in my logic.
On 18.03.2024 16:41, Richard Damon wrote:
So do I. Have you not yet understood my argument? The indices 1, 2, 3,That defines Dedekind-infinity.
But the subset has been made into its own set, not just some selected
elements in the original set.
... required for enumeration are a set ℕ. The fractions to be enumerated are another set, represented by the matrix of not indexed fractions:
OOOO...
OOOO...
OOOO...
OOOO...
..
Now the set ℕ is put in bijection with the first column:
XOOO...
XOOO...
XOOO...
XOOO...
..
And then we try in vain to enlarge the indexed set.>
A subset is made from a set by copying the selected elements out ofAs we can see here, this Dedekind-infinity is self-contradictory concept.
the original set.
Regards, WM
Maybe you should worry less about what integer goes
to what rational and wonder some more about what
rational goes where in the integers.
On 3/18/2024 4:44 PM, WM wrote:
Potentially infinite series have no sums
but only limits.
Often this is confused.
An infinite.sumⁿᵒᵗᐧᵂᴹ is not a finite.sumⁿᵒᵗᐧᵂᴹ. Often this is confused.
An infinite.sumⁿᵒᵗᐧᵂᴹ, if it exists,
is near almost all of the finite initial sums.
Your darkᵂᴹ numbers are inserted with
the intention of changing a set from having
behavior which you think is in error to
behavior which you decided a priori is correct.
It is the only possibility
if actual infinity is true.
Changing the set leaves
the question about the first set
unanswered.
In a finitenot.WM linear order,
each non.trivial.subset has two ends.
It should be clear to every intelligent being,
that almost all elements of an infinite set
cannot be addressed.
People who claim that
every natural number can be defined and chosen
are worse than fools.
People who claim that
each natural number is a natural number
might not be prodigious intellects.
However, it does not seem foolish to say it.
Perhaps you have them confused with
people who say that
a darkᵂᴹ element cannot be said to equal itself.
All definable terms of an infinite sequence belong to
a finite although changable set.
Sets don't change.
⟨ 1 2 3 4 ... ⟩ is the same size as
⟨ 10000 20000 30000 40000 ... ⟩
That requires:
n ⟼ 10000⋅n
Lossless exchange of X and O will result in
losses of Os <==>
All positive fractions can be enumerated.
It has no place in the realm of logic,
at least not in my logic.
You've already chosen what your answers will be.
Now you labor to make your question fit. https://en.wikipedia.org/wiki/Procrustes
| In Greek mythology, Procrustes ... was a rogue smith and
| bandit from Attica who attacked people by stretching them
| or cutting off their legs, so as to force them to fit
| the size of an iron bed.
On 3/18/24 1:16 PM, WM wrote:
On 18.03.2024 16:41, Richard Damon wrote:
So do I. Have you not yet understood my argument? The indices 1, 2, 3,That defines Dedekind-infinity.
But the subset has been made into its own set, not just some selected
elements in the original set.
... required for enumeration are a set ℕ. The fractions to be enumerated >> are another set, represented by the matrix of not indexed fractions:
OOOO...
OOOO...
OOOO...
OOOO...
..
Now the set ℕ is put in bijection with the first column:
XOOO...
XOOO...
XOOO...
XOOO...
..
And then we try in vain to enlarge the indexed set.>
A subset is made from a set by copying the selected elements out ofAs we can see here, this Dedekind-infinity is self-contradictory concept.
the original set.
Because, you have too few Os, as some have been replaced, and if you
restore those, you lost your X's
Bijection is ALWAYS between two DISTINCT sets, not a set and some
elements of it that you try to call a subset.
You don't "Enlarege" the set, you show that a DIFFERENT bijection can
cover the whole matrix.
The theory shows that failed attempts at bijection may occur, but the
sets are equal if ANY bijection can be found.
If you want to try your method, lets mark tha first column with BOTH an
X and an O, and the rest with just an O
And after the bijection swapping the first column should just be an X
and an O, and the rest Xs
XO O O O ...
XO O O O ...
XO O O O ...
XO O O O ...
...
Swap 1/1 with 1/1
OX O O O ...
XO O O O ...
XO O O O ...
...
Swap 2/1 with 1/2
OX X O O ...
OO O O O ...
XO O O O ...
...
Swap 3/1 with 2/1
OX X O O ...
OX O O O ...
OO O O O ...
XO O O O ...
Swap 4/1 with 1/3
OX X X O ...
OX O O O ...
OO O O O ...
OO
XO
...
Swap 5/1 with 2/2
OX X X O ...
OX X O O ...
OO O O O ...
OO
OO
OO
XO
...
Swap 6/1 with 3/1
OX X X O ...
OX X O O ...
OX O O O ...
OO
OO
OO
OO
XO
...
So, we see the pattern of X's moving out of the first column into the
matrix, and the O all moving into the first column which is also getting
Xs from latter number in that column, so, after we do the infinite
number of steps, the first column will be all OX and X will be
everywhere else, as needed for the proof,
Le 19/03/2024 à 04:31, Richard Damon a écrit :
On 3/18/24 1:16 PM, WM wrote:
On 18.03.2024 16:41, Richard Damon wrote:
So do I. Have you not yet understood my argument? The indices 1, 2,That defines Dedekind-infinity.
But the subset has been made into its own set, not just some
selected elements in the original set.
3, ... required for enumeration are a set ℕ. The fractions to be
enumerated are another set, represented by the matrix of not indexed
fractions:
OOOO...
OOOO...
OOOO...
OOOO...
..
Now the set ℕ is put in bijection with the first column:
XOOO...
XOOO...
XOOO...
XOOO...
..
And then we try in vain to enlarge the indexed set.>
A subset is made from a set by copying the selected elements out ofAs we can see here, this Dedekind-infinity is self-contradictory
the original set.
concept.
Because, you have too few Os, as some have been replaced, and if you
restore those, you lost your X's
All the Os (fractions) remain there, but they are overwritten by X.
Bijection is ALWAYS between two DISTINCT sets, not a set and some
elements of it that you try to call a subset.
The Xs and the Os are distinct sets.
You don't "Enlarege" the set, you show that a DIFFERENT bijection can
cover the whole matrix.
That is excluded here.
The theory shows that failed attempts at bijection may occur, but the
sets are equal if ANY bijection can be found.
By the way, that is absolute idiocy.
If you want to try your method, lets mark tha first column with BOTH
an X and an O, and the rest with just an O
That is the original idea. I have only done without showing the covered fractions O.
And after the bijection swapping the first column should just be an X
and an O, and the rest Xs
XO O O O ...
XO O O O ...
XO O O O ...
XO O O O ...
...
Swap 1/1 with 1/1
OX O O O ...
XO O O O ...
XO O O O ...
...
Swap 2/1 with 1/2
OX X O O ...
OO O O O ...
XO O O O ...
...
There is no OO. When the indeX 2 has gonefrom 2/1, only one fraction 2/1 withOut index remains. And where the index is now, at 1/2, we have OX,
namely the fraction 1/2 and the index 2.
Swap 3/1 with 2/1
OX X O O ...
OX O O O ...
OO O O O ...
XO O O O ...
Swap 4/1 with 1/3
OX X X O ...
OX O O O ...
OO O O O ...
OO
XO
...
Swap 5/1 with 2/2
OX X X O ...
OX X O O ...
OO O O O ...
OO
OO
OO
XO
...
Swap 6/1 with 3/1
OX X X O ...
OX X O O ...
OX O O O ...
OO
OO
OO
OO
XO
...
So, we see the pattern of X's moving out of the first column into the
matrix, and the O all moving into the first column which is also
getting Xs from latter number in that column, so, after we do the
infinite number of steps, the first column will be all OX and X will
be everywhere else, as needed for the proof,
No. The numer of not indexed fractions remains constant for every finite step.
Regards, WM
On 3/19/24 4:05 AM, WM wrote:
Bijection is ALWAYS between two DISTINCT sets, not a set and some
elements of it that you try to call a subset.
The Xs and the Os are distinct sets.
No they aren't,
First, thery aren't "Sets" as sets can only have one of each element.
And they aren't in different sets, as you muddled the two together.
Only a correct bijection shows they are the same side, no negative inferences,
You are just creating a bad bijection, because you don't seem
Swap 2/1 with 1/2
OX X O O ...
OO O O O ...
XO O O O ...
...
There is no OO.
Why Not?
The first column is part of two distinct sets, The Matrix of m/n and
the Set of k/1, so it gets two symbols.
The number of not indexed fractions remains constant for every
finite step.
Nope, the number of not indexed fractions goes down by 1 each step.
On 19.03.2024 16:42, Richard Damon wrote:
On 3/19/24 4:05 AM, WM wrote:
Bijection is ALWAYS between two DISTINCT sets, not a set and some
elements of it that you try to call a subset.
The Xs and the Os are distinct sets.
No they aren't,
First, thery aren't "Sets" as sets can only have one of each element.
The X are abbreviating natural numbers, The O are abbreviating fractions.
And they aren't in different sets, as you muddled the two together.
Natural numbers are different from fractions.
Only a correct bijection shows they are the same side, no negative
inferences,
You are just creating a bad bijection, because you don't seem
No, I am using exactly Cantor's "bijection", but showing that it only
works for few numbers.
Swap 2/1 with 1/2
OX X O O ...
OO O O O ...
XO O O O ...
...
There is no OO.
Why Not?
Because OO means two fractions sitting at one place. That is impossible because all fractions remain at their places.
The first column is part of two distinct sets, The Matrix of m/n and
the Set of k/1, so it gets two symbols.
The first column contains two distinct sets, at the beginnig: the
fractions k/1 and the indices k. But the indices k move to other places
while the fractions k/1 remain fixed.
The number of not indexed fractions remains constant for every finite
step.
Nope, the number of not indexed fractions goes down by 1 each step.
The number of X does not increase.
Regards, WM
On 3/19/24 10:00 AM, WM wrote:
On 19.03.2024 16:42, Richard Damon wrote:
On 3/19/24 4:05 AM, WM wrote:
Bijection is ALWAYS between two DISTINCT sets, not a set and some
elements of it that you try to call a subset.
The Xs and the Os are distinct sets.
No they aren't,
First, thery aren't "Sets" as sets can only have one of each element.
The X are abbreviating natural numbers, The O are abbreviating fractions.
No, they are MARKING the location where such numbers are.
And they aren't in different sets, as you muddled the two together.
Natural numbers are different from fractions.
Yes, which is why they need to be two different sets.
Only a correct bijection shows they are the same side, no negative
inferences,
You are just creating a bad bijection, because you don't seem
No, I am using exactly Cantor's "bijection", but showing that it only
works for few numbers.
Which is between two distinct sets, not a set and part of itself.
Swap 2/1 with 1/2
OX X O O ...
OO O O O ...
XO O O O ...
...
There is no OO.
Why Not?
Because OO means two fractions sitting at one place. That is impossible
because all fractions remain at their places.
No, O indicates that this letter started on a fraction. And will get
moved to a index and replaced by the x as that index.
You can't talk about the fractions "staying" but the indexes moving.
Le 18/03/2024 à 23:54, Jim Burns a écrit :
On 3/18/2024 4:44 PM, WM wrote:
It should be clear to every intelligent being,
that almost all elements of an infinite set
cannot be addressed.
People who claim that
every natural number can be defined and chosen
are worse than fools.
People who claim that
each natural number is a natural number
might not be prodigious intellects.
However, it does not seem foolish to say it.
It is not.
But that is a different statement.
Le 19/03/2024 à 18:14, Richard Damon a écrit :
On 3/19/24 10:00 AM, WM wrote:
On 19.03.2024 16:42, Richard Damon wrote:
On 3/19/24 4:05 AM, WM wrote:
Bijection is ALWAYS between two DISTINCT sets, not a set and some
elements of it that you try to call a subset.
The Xs and the Os are distinct sets.
No they aren't,
First, thery aren't "Sets" as sets can only have one of each element.
The X are abbreviating natural numbers, The O are abbreviating
fractions.
No, they are MARKING the location where such numbers are.
That is the same.
And they aren't in different sets, as you muddled the two together.
Natural numbers are different from fractions.
Yes, which is why they need to be two different sets.
They are two different sets.
Only a correct bijection shows they are the same side, no negative
inferences,
You are just creating a bad bijection, because you don't seem
No, I am using exactly Cantor's "bijection", but showing that it only
works for few numbers.
Which is between two distinct sets, not a set and part of itself.
Precisely.
Swap 2/1 with 1/2
OX X O O ...
OO O O O ...
XO O O O ...
...
There is no OO.
Why Not?
Because OO means two fractions sitting at one place. That is
impossible because all fractions remain at their places.
No, O indicates that this letter started on a fraction. And will get
moved to a index and replaced by the x as that index.
O appers to move, but in fact only the first O appears and the second is covered by the X.
You can't talk about the fractions "staying" but the indexes moving.
Why not?
Regards, WM
Not if all summands remain the same. If all summands remain the same, all sums over one summand, all sums over two summands, all sums over n
summands remain the same. This isproved by induction.
The sum over all summands
remains the same. This is an axiom of actual infinity.
On 3/19/24 2:24 PM, WM wrote:
We can move a marker without
needing to move the value it was on.
Apparently not in your mind, sincw the O moving into the set of k isn't leaving the set of m/n
So, why isn't the O "leaving"?
You can't talk about the fractions "staying" but the indexes moving.
Why not?
Because it is just LYING that you are doing what the proof you are
talking about is saying.
On 3/19/2024 6:51 AM, WM wrote:
Le 18/03/2024 à 23:54, Jim Burns a écrit :
Perhaps you have them confused with
people who say that
a darkᵂᴹ element cannot be said to equal itself.
Dark elements are equal to themselves.
But we cannot know about most.
Pick _one_
Sets don't change.
Definable collections can change.
Sets don't change.
Their sums don't change.
All sets don't change.
All finite sums don't change.
Some infinite sums don't change.
And some infinite sums change.
Le 19/03/2024 à 23:00, Jim Burns a écrit :
On 3/19/2024 6:51 AM, WM wrote:
Le 18/03/2024 à 23:54, Jim Burns a écrit :
Perhaps you have them confused with
people who say that
a darkᵂᴹ element cannot be said to equal itself.
Dark elements are equal to themselves.
But we cannot know about most.
Pick _one_
| Each darkᵂᴹ element equals itself.
| Not.each darkᵂᴹ element equals itself.
If a new prime number is identified,
one hitherto dark prime number has been picked.
Le 20/03/2024 à 15:16, Richard Damon a écrit :
On 3/19/24 2:24 PM, WM wrote:
We can move a marker without needing to move the value it was on.
That is what I do. I move the X according to Cantor's prescription. The fraction remains where it belongs.
Apparently not in your mind, sincw the O moving into the set of k
isn't leaving the set of m/n
The O represents a fraction without index. It is overwritten by an X in
case the fraction is indexed.
So, why isn't the O "leaving"?
Because it belongs to the set of fractions.
You can't talk about the fractions "staying" but the indexes moving.
Why not?
Because it is just LYING that you are doing what the proof you are
talking about is saying.
The proof saying precisely this: An index X is mapped on a fraction O.
Regards, WM
There is no sum over all natural numbers because each of these has a successor.
Tom Bola a écrit:
There is no sum over all natural numbers because each of these has a successor.
1 + 2 + 3 + ... n... = ω [Cantor]
On 3/21/2024 3:42 PM, WM wrote:
Le 19/03/2024 à 23:00, Jim Burns a écrit :
On 3/19/2024 6:51 AM, WM wrote:
Le 18/03/2024 à 23:54, Jim Burns a écrit :
Perhaps you have them confused with
people who say that
a darkᵂᴹ element cannot be said to equal itself.
Dark elements are equal to themselves.
But we cannot know about most.
Pick _one_
| Each darkᵂᴹ element equals itself.
| Not.each darkᵂᴹ element equals itself.
If a new prime number is identified,
one hitherto dark prime number has been picked.
An intentional non.answer.
You (WM) pick _both_
You have disabled the not.first.false "telescope"
There is no sum over all natural numbers because each of these has a successor.
1 + 2 + 3 + ... n... = ω [Cantor]
This is just a *symbolic* hint by Cantor.
WM brought next idea :
Le 21/03/2024 à 21:11, Tom Bola a écrit :
There is no sum over all natural numbers because each of these has a
successor.
1 + 2 + 3 + ... n... = ω [Cantor]
Explain.
On 3/21/24 3:34 PM, WM wrote:
Le 20/03/2024 à 15:16, Richard Damon a écrit :
On 3/19/24 2:24 PM, WM wrote:
We can move a marker without needing to move the value it was on.
That is what I do. I move the X according to Cantor's prescription. The
fraction remains where it belongs.
So, If you add Os when you swap, of course they aren't going to go away.
So, why isn't the O "leaving"?
Because it belongs to the set of fractions.
No, it marked what WAS a fraction and moves to its paired index.
I guess you don't understand what you are supposed to be doing.
Le 22/03/2024 à 13:12, Tom Bola a écrit :
There is no sum over all natural numbers because each of these has a successor.
1 + 2 + 3 + ... n... = ω [Cantor]
This is just a *symbolic* hint by Cantor.
No. He uses + without modification or reservation.
Le 22/03/2024 à 13:17, FromTheRafters a écrit :
WM brought next idea :
Le 21/03/2024 à 21:11, Tom Bola a écrit :
There is no sum over all natural numbers because each of these has a
successor.
1 + 2 + 3 + ... n... = ω [Cantor]
Explain.
https://en.wikipedia.org/wiki/Plus_and_minus_signs
Le 22/03/2024 à 00:04, Jim Burns a écrit :
You have disabled the not.first.false "telescope"
I don't believe t all that it is able .
Le 19/03/2024 à 23:00, Jim Burns a écrit :
On 3/19/2024 6:51 AM, WM wrote:
Le 18/03/2024 à 23:54, Jim Burns a écrit :
Sets don't change.
Definable collections can change.
Sets don't change.
Their sums don't change.
All sets don't change.
All finite sums don't change.
Some infinite sums don't change.
And some infinite sums change.
Not if all summands remain the same.
If all summands remain the same,
all sums over one summand,
all sums over two summands,
all sums over n summands remain the same.
This is proved by induction.
The sum over all summands remains the same.
This is an axiom of actual infinity.
The sum over all summands remains the same.
This is an axiom of actual infinity.
Do you (WM) believe that, somewhere, sometime,
there is a finiteⁿᵒᵗᐧᵂᴹ claim.sequence
holding a false claim
but not.holding a first.false claim?
My axiom would open a different discussion,
one about ℝᴶᴮ
Le 22/03/2024 à 18:17, Jim Burns a écrit :
On 3/21/2024 3:42 PM, WM wrote:
The sum over all summands remains the same.
This is an axiom of actual infinity.
You (WM) are holding the wrong end of
your Actual.Infinityᵂᴹ axiom.
My axiom would open a different discussion,
one about ℝᴶᴮ
This discussion would be
as annoying and meaningles as
the discussion about Bob's fate.
Le 22/03/2024 à 19:13, Jim Burns a écrit :
Do you (WM) believe that, somewhere, sometime,
there is a finiteⁿᵒᵗᐧᵂᴹ claim.sequence
holding a false claim
but not.holding a first.false claim?
That is not relevant.
That is not relevant.
Every finite (potentially infinite) claim sequence
"n is a visible number,
n+1 is a visible number,
n+2 is a visible number,
..."
has no first false claim.
Nevertheless almost all numbers are invisible.
Le 22/03/2024 à 19:13, Jim Burns a écrit :
On 3/22/2024 8:13 AM, WM wrote:
Le 22/03/2024 à 00:04, Jim Burns a écrit :
You have disabled the not.first.false "telescope"
I don't believe t all that it is able .
We study _claims themselves_ as objects and
we have found useful properties of _the claims themselves_
which they have, no matter what _the claims are about_
Do you not.believe that
they have the useful properties we've found?
Do you (WM) believe that, somewhere, sometime,
there is a finiteⁿᵒᵗᐧᵂᴹ claim.sequence
holding a false claim
but not.holding a first.false claim?
That is not relevant.
Do you (WM) believe that, somewhere, sometime,
there is a finiteⁿᵒᵗᐧᵂᴹ claim.sequence
holding a false claim
but not.holding a first.false claim?
That is not relevant.
Every finite (potentially infinite)
claim sequence
"n is a visible number,
n+1 is a visible number,
n+2 is a visible number,
..."
has no first false claim.
Nevertheless almost all numbers are invisible.
On 03/24/2024 11:45 AM, Ross Finlayson wrote:
On 03/23/2024 12:13 PM, Jim Burns wrote:
[...]
. -v
.. -v
... -v
(...) -v
........
.
^- ..
^- ...
^- (...)
^- ........
With repetition as induction, or vice-versa,
doesn't it seem like it needs both ends to
arrive at either from each other?
The Wikipedia "List of philosophical problems"
includes an entry for induction,
whether it's circular.
https://en.wikipedia.org/wiki/List_of_philosophical_problems#Problem_of_induction
On 03/25/2024 11:36 AM, Jim Burns wrote:
[...]
About induction,
yeah that kind of is what induction is,
and about what makes for infinite induction
which is the usual consideration here
where "induction itself" as causality
of course just always follows.
It does raise the point though,...
"induction of induction",
about whether
the meta-theory is also the theory,
it is a thing.
On 3/26/2024 9:16 PM, Ross Finlayson wrote:
[...]
Physical.induction can be derived from
evidence and Bayes' Theorem and the assumption that
the evidence is a random (not.special) selection
from past and future evidence.collection.
A billion eastern sunrises is a lot of evidence,
and, via Bayes' Theorem, the posterior probability that
tomorrow has an eastern sunrise is close to 1
I'm assuming here that,
prior to observing sunrises,
each hypothesis Hₚ = "P(eastern.sunrise) = p"
has the same probability (Hₚ is not.special).
P(Hₚ) = 1
That's a very crude model,
but it can be elaborated.
Elaborating crude initial models is kind of
how things are done in Physics.land,
home of the adiabatic spherical cows in a vacuum.
On 03/27/2024 11:23 AM, Jim Burns wrote:
[...]
Induction the inference is the usual idea of causality,
and what comes around goes around.
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