Jim Burns schrieb am Dienstag, 24. Mai 2022 um 21:07:37 UTC+2:
> On 5/24/2022 12:26 PM, WM wrote:

> After all swaps,
> no O remains in a place with a swap out of it.

It appears so to you. But the share cannot change.
>
> > They are dark.
> They do not exist.

Then explain how the share of X can grow by simply moving them around.

Regards, WM

Fritz Feldhase schrieb am Dienstag, 24. Mai 2022 um 22:12:20 UTC+2:
> On Tuesday, May 24, 2022 at 9:07:37 PM UTC+2, Jim Burns wrote:
>
> > After all swaps,
>
> Are you performing a supertask? (Are you Chuck Norris?)
>
> WM is right: If you are considering _the sequence_ of matrices there is no "after all swaps".

And during all finite steps the share of X remains constant. That means by only natural numbers it is impossible to index all fractions.
>
> If you are considering the sequence (1, 1/2, 1/4, 1/8, ...) there is no "after all divisions" (by 2).

Not even all terms of the sequence are existing.
>
> Btw. Your chess examply is quite helpful for pointing out a fundamental error in WM's "argument". There are positions in chess which cannot be reached by any number of regular moves (starting with the initial position). Still this does not mean that these positions do not exist!

Cantor claims that all fractions can be indexed in a regular way. Why did you construct your matrix if you knew it was in vain? Note that all you can construct can be reached within a finite number of swaps.
>
> In the same way there *is* a matrix containing just the fractions 1/1, 2/1, 3/1, ... though it cannot be "reached" by any number of swaps.

Chuckle. Magic shall save matheology?

Note: At all finite swaps the number of X remain constant. And the beyond is not of interest.

Regards, WM

William schrieb am Dienstag, 24. Mai 2022 um 22:56:35 UTC+2:
> On Tuesday, May 24, 2022 at 1:26:55 PM UTC-3, WM wrote:
> > William schrieb am Dienstag, 24. Mai 2022 um 15:04:35 UTC+2:
> > > On Tuesday, May 24, 2022 at 7:55:26 AM UTC-3, WM wrote:
> > > > William schrieb am Montag, 23. Mai 2022 um 15:18:21 UTC+2:
> > > >> We are looking at the sets indexes if the 0's.
> > > > We are looking at the final .result
> > > Indeed, and the final result is that the final set of indexes of the O's (an intersection of a sequence of sets each of which has cardinality aleph_0) is empty.
> > By exchanging X and O never an O disappears.
> At each exchange the index set changes.

But it does not increase.

> The statement "never an O disappears" means that the index set changes to another set that has cardinality aleph_0.

The set of indices outside of the first column is always finite. Therefore it can never cover any column between second and oo.
> It is a simple consequence of the Axiom of infinity that the intersection of a sequence of sets of cardinality aleph_0 can be empty.

The intersection is irrelevant. Relevant is that never any other column can be completed because as soon as this would happen, infinitely many indices had been issued. And more are not available.

Regards, WM

Fritz Feldhase schrieb am Mittwoch, 25. Mai 2022 um 03:27:59 UTC+2:
> On Wednesday, May 25, 2022 at 2:44:17 AM UTC+2, Jim Burns wrote:

> Again, "there is no" state/time/whatever which we could (reasonably) refer to as "after all swaps".

Then there is no state where all fractions are enumerated.

"Zwei n-fach geordnete Mengen M und N werden 'ähnlich' genannt, wenn es möglich ist, sie gegenseitig eindeutig und vollständig, Element für Element, einander so zuzuordnen," [Cantor, p. 424]

This completeness can be checked at every index.
>
> It just doesn't make any sense. (Except you are willing to consider the process of swapping as a supertask. But supertasks aren't part of set-theoretic arguments. Moreover the are still open to debate.)

The counting of fractions is a supertask.

> To make a long story short: We SHOULD NOT refer to supertasks when trying to explain the correctness of Cantor's argument. This would be rather counterproductive (to say the least).

Relevant is that never (by supertask or by formula) any other column can be completed because as soon as this would happen, infinitely many indices had been issued. And more are not available.

Regards, WM

horand....@gmail.com schrieb am Mittwoch, 25. Mai 2022 um 13:12:45 UTC+2:

> It is a limiting position, though. WM tried to throw a monkey wrench into the discussion way back, but all it yielded was his self-delusion that he had actually accomplished something.

Before the "limit" there must be a first column completed (because the Indices are linearly ordered. Never two or more are issued simultaneously.) But never (neither by supertask nor by formula) any other column can be completed because as soon as this would happen, infinitely many indices had been issued. And more are not available.

Regards, WM

On Wednesday, 25 May 2022 at 11:21:20 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 25. Mai 2022 um 13:12:45 UTC+2:
>
> > It is a limiting position, though. WM tried to throw a monkey wrench into the discussion way back, but all it yielded was his self-delusion that he had actually accomplished something.
> Before the "limit" there must

.... be an understanding on your part what a limit actually is. Your drivel only shows your utter ignorance. (I removed it to spare others the stench.) And now go play elsewhere, Wolfi.

On Wednesday, May 25, 2022 at 11:08:13 AM UTC-3, WM wrote:
>> ,,, more are not available.

Piffle. Using infinitely many indexes from a set of cardinality alpeh_0 does not mean "more are not available". A set of cardnality aleph_0, contains infinitely many disjoint subsets of cardinalisty aleph_0. ( Example, consider subset S_q consisting of powers of the qth prime.).

--
William Hughes

On 5/25/2022 9:16 AM, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 25. Mai 2022 um 03:27:59 UTC+2:
>> On Wednesday, May 25, 2022 at 2:44:17 AM UTC+2, Jim Burns wrote:
>
>> Again, "there is no" state/time/whatever which we could (reasonably) refer to as "after all swaps".
>
> Then there is no state where all fractions are enumerated.

wrong.

>

>>
>> It just doesn't make any sense. (Except you are willing to consider the process of swapping as a supertask. But supertasks aren't part of set-theoretic arguments. Moreover the are still open to debate.)
>
> The counting of fractions is a supertask.

wrong

>
>> To make a long story short: We SHOULD NOT refer to supertasks when trying to explain the correctness of Cantor's argument. This would be rather counterproductive (to say the least).
>
> Relevant is that never (by supertask or by formula) any other column can be completed because as soon as this would happen, infinitely many indices had been issued. And more are not available.

wrong

>
> Regards, WM