Fritz Feldhase schrieb am Mittwoch, 16. November 2022 um 00:33:11 UTC+1:
> On Tuesday, November 15, 2022 at 3:40:36 PM UTC+1, Sergi o wrote:
>
> > you use the word "define" or "definable" however you mean "finite" which is wrong.
>
> Indeed! Whenever he talks about "definable XXX" he means _finitely many_ XXX.

Of course. Nobody can define infinitely many elements individually.
>
> And his darkies are "all the others" (which "makes" a set infinite).

Of course. Defined numbers will never make a set infinite.

Regards, WM

Fritz Feldhase schrieb am Mittwoch, 16. November 2022 um 00:38:51 UTC+1:
> On Tuesday, November 15, 2022 at 8:02:00 PM UTC+1, Gus Gassmann wrote:
>
> > Every fraction is indexed in a finite step. But there are infinitely many fractions, so you need infinitely many "steps" to index them all.
> Right.
> > This is where the limit enters.
> Nope. This is where Chuck Norris enters!

Whether or not a limit enters is irrelevant. As long as X and O are exchanged without losses, no O will get lost. If O's get lost, in the limit or by simply assuming it or by whatever mechanism, this cannot be checked. The indexing cannot happen in a definable way. It is clearly no mapping according to Cantor: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)] The first loss of an O cannot happen at a determined place.

Regards, WM

Gus Gassmann schrieb am Mittwoch, 16. November 2022 um 04:05:02 UTC+1:
> On Tuesday, 15 November 2022 at 17:23:14 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Dienstag, 15. November 2022 um 19:36:08 UTC+1:
> > > On Tuesday, 15 November 2022 at 13:28:09 UTC-4, WM wrote:
> >
> > > > "The infinite sequence thus defined has the peculiar property to contain all positive
> > > > rational numbers and each of them only once at a determined place."
> > > > Therefore every place can be checked step by step because it is a finite place.
> > > Of course. And whenever you have checked that finite place, you have infinitely many more finite places left to check.
> > Therefore I gave a general proof: For all finite places that could be checked (in principle) there is no loss of O's. Hence at all existing finite places the indexing fails to cover more fractions than were covered in the beginning.
> >
> > There are two alternatives: Either not all fractions are indexed, or they are indexed in the limit where no individual indexing is possible.
> I have said this for a while now: You have no fucking clue what a limit is and how to interpret it.

It is irrelevant for the present discussion. If, as it i claimed, every fractions is indexed, then all the O's will have to leave the matrix. That means there must be a first step where one or more O's leave the matrix. Such a step cannot be determined. Hence there is no mapping in Cantor's sense "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)] but the usual meaningless handwaving of matheology, filled with hollow words like quantifier swap or limit.

> You don't even get the simplest of all limits, the behaviour of the fractions 1/n as n -> oo. The limit *IS* zero, but there is *NO* n for which 1/n = 0.

The limit is not attained. But according to set theory the complete indexing of the positive fractions is attained.

> How could there be?

The limit in analysis is not attained but approached as closely as desired. Simple as that.

> Perhaps you'd be so kind and write down the definition of this limit for yourself? (Better yet, since you are clearly unable to do even that for yourself, go copy it from some authoritative source, perhaps the one you copied for your book.)

The source of my knowledge are lectures at the university of Göttingen, one of the top institutes of mathematics worldwide.

Regards, WM

Gus Gassmann schrieb am Mittwoch, 16. November 2022 um 04:17:15 UTC+1:
> On Tuesday, 15 November 2022 at 17:30:49 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Dienstag, 15. November 2022 um 20:02:00 UTC+1:
> > > On Tuesday, 15 November 2022 at 13:42:35 UTC-4, WM wrote:
> >
> > > > > > (3) Therefore it is impossible to index all fractions in a definable way. Indexing many fractions "in the limit" [...]
> > > > >
> > > > > Nobody is doing that,
> > > > You tried it until I showed that you are wrong.
> > > Every fraction is indexed in a finite step. But there are infinitely many fractions, so you need infinitely many "steps" to index them all.
> > No limit enters. In all infinitely many steps no O is removed from the matrix.
> > > This is where the limit enters.
> > This limit does not remove any O from the matrix.

in a determined way.
> > No. The lossless exchange of X and O does not remove any O from the matrix.
> Where did I say it does? I explained it to you,the 'O's move to the first column and are being pushed further and further down that column. Since eventually *EVERY* position in the matrix that you care to name will

Every position that can be named will carry an X. But no O has left the matrix. That's the point!

> > > What do you think the Cantor formula k = (m + n - 1)*(m + n - 2)/2 + m gives you???
> > It describes a tiny part of the upper left corner of the matrix. No O leaves the matrix.
> It describes the ENTIRE matrix.

You claim it. It is wrong. The O's have not left the matrix.

> For every k you *KNOW* that all positions in rows m' and columns n' with k' = (m' + n' - 1)*(m' + n' - 2)/2 + m' and k' < k contain only 'X's. And that's *ALL* you need to determine the limit. In fact, you could write out a proof by induction for this fact.

Of course. All determined positions of the matrix can be reached by induction and are finally carrying X's. Induction is valid for definable natnumbers, i.e., ends of FISONs. But that does not make the O's leave the matrix.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 07:19:22 UTC+1:
> tisdag 15 november 2022 kl. 18:28:09 UTC+1 skrev WM:

> > "The infinite sequence thus defined has the peculiar property to contain all positive
> > rational numbers and each of them only once at a determined place."
> > Therefore every place can be checked step by step because it is a finite place.

> You don't need to "check" it

But I can do it, if it is possibke, that means if a bijection exists as Cantor describes it: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

> We just check that for any rational number there is a natural number assigned to it, tada, ALL has been "checked" at once

As this does not explain the loss of the O's it is tad, as you say.

Regards, WM

Fritz Feldhase schrieb am Mittwoch, 16. November 2022 um 10:18:54 UTC+1:
> On Wednesday, November 16, 2022 at 6:34:20 AM UTC+1, Chris M. Thomasson wrote:
> > On 11/15/2022 3:33 PM, Fritz Feldhase wrote:

> Yes. He calls this idiotic idea "MathRealism". (Actually, a rather crude form of ultrafinitism.)
>
> But that's NOT what he's talkig about here.
>
> His stance here is: Either there is no infinty or there are (necessarily) dark numbers.

Very well observed.
>
> Hence those "elements" that "make" a set infinite are "dark" for him.
>
The places where the O's sit cannot be determined. But the O's cannot have left the matrix. I hope you will not defend your statement that in a lossless exchange losses can appear?

> Though WM uses these terms as if they would refer to a _property_ of a natural number.)

They refer to the large remainder of undefined numbers.

> > He has to be a really hard core worshiper at the Church of UltraFinitism...
> Well, that's just ONE of his multiple "points of view".

Very well observed.

Regards, WM

onsdag 16 november 2022 kl. 10:55:48 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 07:19:22 UTC+1:
> > tisdag 15 november 2022 kl. 18:28:09 UTC+1 skrev WM:
>
> > > "The infinite sequence thus defined has the peculiar property to contain all positive
> > > rational numbers and each of them only once at a determined place."
> > > Therefore every place can be checked step by step because it is a finite place.
> > You don't need to "check" it
>
> But I can do it, if it is possibke, that means if a bijection exists as Cantor describes it: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
>
> > We just check that for any rational number there is a natural number assigned to it, tada, ALL has been "checked" at once
>
> As this does not explain the loss of the O's it is tad, as you say.
>
> Regards, WM

You can find it in this instance but that is not a requirement.

zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 12:52:36 UTC+1:
> onsdag 16 november 2022 kl. 10:55:48 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 07:19:22 UTC+1:
> > > tisdag 15 november 2022 kl. 18:28:09 UTC+1 skrev WM:
> >
> > > > "The infinite sequence thus defined has the peculiar property to contain all positive
> > > > rational numbers and each of them only once at a determined place."
> > > > Therefore every place can be checked step by step because it is a finite place.
> > > You don't need to "check" it
> >
> > But I can do it, if it is possibke, that means if a bijection exists as Cantor describes it: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
> >
> > > We just check that for any rational number there is a natural number assigned to it, tada, ALL has been "checked" at once
> >
> > As this does not explain the loss of the O's it is tad, as you say.

> You can find it in this instance but that is not a requirement.

Where does the first O leave the matrix? Since many are to follow, this must happen at a finite step which can be checked.

Regards, WM

On Wednesday, 16 November 2022 at 05:43:12 UTC-4, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 16. November 2022 um 04:05:02 UTC+1:
> > On Tuesday, 15 November 2022 at 17:23:14 UTC-4, WM wrote:
> > > Gus Gassmann schrieb am Dienstag, 15. November 2022 um 19:36:08 UTC+1:
> > > > On Tuesday, 15 November 2022 at 13:28:09 UTC-4, WM wrote:
> > >
> > > > > "The infinite sequence thus defined has the peculiar property to contain all positive
> > > > > rational numbers and each of them only once at a determined place."
> > > > > Therefore every place can be checked step by step because it is a finite place.
> > > > Of course. And whenever you have checked that finite place, you have infinitely many more finite places left to check.
> > > Therefore I gave a general proof: For all finite places that could be checked (in principle) there is no loss of O's. Hence at all existing finite places the indexing fails to cover more fractions than were covered in the beginning.
> > >
> > > There are two alternatives: Either not all fractions are indexed, or they are indexed in the limit where no individual indexing is possible.
> > I have said this for a while now: You have no fucking clue what a limit is and how to interpret it.
>
> It is irrelevant for the present discussion.

You don't even get how limits are the *KEY* to your stepwise matrix process. You truly are a clueless wanker.

On Wednesday, 16 November 2022 at 05:43:12 UTC-4, WM wrote:
[...]
> The source of my knowledge are lectures at the university of Göttingen, one of the top institutes of mathematics worldwide.

Yes, and the source of the proofs in your book are other books written by reputable mathematicians.

All that proves is that you can copy, and occasionally even without transcription errors.

On 11/16/2022 1:06 PM, Gus Gassmann wrote:
> On Wednesday, 16 November 2022 at 05:43:12 UTC-4, WM wrote:
>> Gus Gassmann schrieb am Mittwoch, 16. November 2022 um 04:05:02 UTC+1:
>>> On Tuesday, 15 November 2022 at 17:23:14 UTC-4, WM wrote:
>>>> Gus Gassmann schrieb am Dienstag, 15. November 2022 um 19:36:08 UTC+1:
>>>>> On Tuesday, 15 November 2022 at 13:28:09 UTC-4, WM wrote:
>>>>
>>>>>> "The infinite sequence thus defined has the peculiar property to contain all positive
>>>>>> rational numbers and each of them only once at a determined place."
>>>>>> Therefore every place can be checked step by step because it is a finite place.
>>>>> Of course. And whenever you have checked that finite place, you have infinitely many more finite places left to check.
>>>> Therefore I gave a general proof: For all finite places that could be checked (in principle) there is no loss of O's. Hence at all existing finite places the indexing fails to cover more fractions than were covered in the beginning.
>>>>
>>>> There are two alternatives: Either not all fractions are indexed, or they are indexed in the limit where no individual indexing is possible.
>>> I have said this for a while now: You have no fucking clue what a limit is and how to interpret it.
>>
>> It is irrelevant for the present discussion.
>
> You don't even get how limits are the *KEY* to your stepwise matrix process. You truly are a clueless wanker.

agree. WMs lost in the woods.

On 11/16/2022 1:09 PM, Gus Gassmann wrote:
> On Wednesday, 16 November 2022 at 05:43:12 UTC-4, WM wrote:
> [...]
>> The source of my knowledge are lectures at the university of Göttingen, one of the top institutes of mathematics worldwide.
>
> Yes, and the source of the proofs in your book are other books written by reputable mathematicians.
>
> All that proves is that you can copy, and occasionally even without transcription errors

agree.

On 11/16/2022 12:05 PM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 12:52:36 UTC+1:
>> onsdag 16 november 2022 kl. 10:55:48 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 07:19:22 UTC+1:
>>>> tisdag 15 november 2022 kl. 18:28:09 UTC+1 skrev WM:
>>>
>>>>> "The infinite sequence thus defined has the peculiar property to contain all positive
>>>>> rational numbers and each of them only once at a determined place."
>>>>> Therefore every place can be checked step by step because it is a finite place.
>>>> You don't need to "check" it
>>>
>>> But I can do it, if it is possibke, that means if a bijection exists as Cantor describes it: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
>>>
>>>> We just check that for any rational number there is a natural number assigned to it, tada, ALL has been "checked" at once
>>>
>>> As this does not explain the loss of the O's it is tad, as you say.
>
>> You can find it in this instance but that is not a requirement.
>
> Where does the first O leave the matrix? Since many are to follow, this must happen at a finite step which can be checked.
>
> Regards, WM

which O, the one covering 1/1 ?

Gus Gassmann schrieb am Mittwoch, 16. November 2022 um 20:06:06 UTC+1:
> On Wednesday, 16 November 2022 at 05:43:12 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Mittwoch, 16. November 2022 um 04:05:02 UTC+1:
> > > On Tuesday, 15 November 2022 at 17:23:14 UTC-4, WM wrote:
> > > > Gus Gassmann schrieb am Dienstag, 15. November 2022 um 19:36:08 UTC+1:
> > > > > On Tuesday, 15 November 2022 at 13:28:09 UTC-4, WM wrote:
> > > >
> > > > > > "The infinite sequence thus defined has the peculiar property to contain all positive
> > > > > > rational numbers and each of them only once at a determined place."
> > > > > > Therefore every place can be checked step by step because it is a finite place.
> > > > > Of course. And whenever you have checked that finite place, you have infinitely many more finite places left to check.
> > > > Therefore I gave a general proof: For all finite places that could be checked (in principle) there is no loss of O's. Hence at all existing finite places the indexing fails to cover more fractions than were covered in the beginning.
> > > >
> > > > There are two alternatives: Either not all fractions are indexed, or they are indexed in the limit where no individual indexing is possible.
> > > I have said this for a while now: You have no fucking clue what a limit is and how to interpret it.
> >
> > It is irrelevant for the present discussion.
> You don't even get how limits are the *KEY* to your stepwise matrix process.

"The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." There is no limit.

Regards, WM

On 11/16/2022 3:26 AM, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 16. November 2022 um 00:33:11 UTC+1:
>> On Tuesday, November 15, 2022 at 3:40:36 PM UTC+1, Sergi o wrote:
>>
>>> you use the word "define" or "definable" however you mean "finite" which is wrong.
>>
>> Indeed! Whenever he talks about "definable XXX" he means _finitely many_ XXX.
>
> Of course. Nobody can define infinitely many elements individually.

wrong. N is an infinite set of the natural numbers.

>>
>> And his darkies are "all the others" (which "makes" a set infinite).
>
> Of course. Defined numbers will never make a set infinite.

you mean "finite" numbers will never make a set infinite.

>
> Regards, WM

On 11/16/2022 3:33 AM, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 16. November 2022 um 00:38:51 UTC+1:
>> On Tuesday, November 15, 2022 at 8:02:00 PM UTC+1, Gus Gassmann wrote:
>>
>>> Every fraction is indexed in a finite step. But there are infinitely many fractions, so you need infinitely many "steps" to index them all.
>> Right.
>>> This is where the limit enters.
>> Nope. This is where Chuck Norris enters!
>
> Whether or not a limit

there is no limit.

<snip x and o poop>

>It is clearly no mapping

Wrong!

according to Cantor: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once
at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

Ha!! you are reading that wrong, it SAYS it is a one to one mapping

<snip x and o poop>

>
> Regards, WM

On 11/16/2022 4:02 AM, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 16. November 2022 um 10:18:54 UTC+1:
>> On Wednesday, November 16, 2022 at 6:34:20 AM UTC+1, Chris M. Thomasson wrote:
>>> On 11/15/2022 3:33 PM, Fritz Feldhase wrote:
>
>> Yes. He calls this idiotic idea "MathRealism". (Actually, a rather crude form of ultrafinitism.)
>>
>> But that's NOT what he's talkig about here.
>>
>> His stance here is: Either there is no infinty or there are (necessarily) dark numbers.
>
> Very well observed.

He observed your stance of pseudo-math.

>>
>> Hence those "elements" that "make" a set infinite are "dark" for him.
>>

<snip failed x and o desaster>

>
>> Though WM uses these terms as if they would refer to a _property_ of a natural number.)
>
> They refer to the large remainder of undefined numbers.

that is your self imposed limitation, and why your darkies never ever exist anywhere.

>
>>> He has to be a really hard core worshiper at the Church of UltraFinitism...
>> Well, that's just ONE of his multiple "points of view".
>
> Very well observed.

He observed your stance(s) of pseudo-math.

>
> Regards, WM

On 11/16/2022 3:55 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 07:19:22 UTC+1:
>> tisdag 15 november 2022 kl. 18:28:09 UTC+1 skrev WM:
>
>>> "The infinite sequence thus defined has the peculiar property to contain all positive
>>> rational numbers and each of them only once at a determined place."
>>> Therefore every place can be checked step by step because it is a finite place.
>
>> You don't need to "check" it
>
> CORRECTED=> a bijection exists as Cantor describes it: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
>
>> We just check that for any rational number there is a natural number assigned to it, tada, ALL has been "checked" at once
>
> As this does not explain the loss of the O's it is tad, as you say.

wrong. the only reason you use x and o is because you do not understand Algebra.

your x and o are a farce, a non-silly joke, you have failed.

>
> Regards, WM

On 11/16/2022 3:43 AM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 16. November 2022 um 04:05:02 UTC+1:
>> On Tuesday, 15 November 2022 at 17:23:14 UTC-4, WM wrote:
>>> Gus Gassmann schrieb am Dienstag, 15. November 2022 um 19:36:08 UTC+1:
>>>> On Tuesday, 15 November 2022 at 13:28:09 UTC-4, WM wrote:
>>>
>>>>> "The infinite sequence thus defined has the peculiar property to contain all positive
>>>>> rational numbers and each of them only once at a determined place."
>>>>> Therefore every place can be checked step by step because it is a finite place.
>>>> Of course. And whenever you have checked that finite place, you have infinitely many more finite places left to check.
>>> Therefore I gave a general proof: For all finite places that could be checked (in principle) there is no loss of O's. Hence at all existing finite places the indexing fails to cover more fractions than were covered in the beginning.
>>>
>>> There are two alternatives: Either not all fractions are indexed, or they are indexed in the limit where no individual indexing is possible.
>> I have said this for a while now: You have no fucking clue what a limit is and how to interpret it.
>
> It is irrelevant for the present discussion. If, as it i claimed, every fractions is indexed, then all the O's will have to leave the matrix. That means there must be a first step where one or more O's leave the matrix. Such a step cannot be determined. Hence there is no mapping in Cantor's sense "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)] but the usual meaningless handwaving of matheology, filled with hollow words like quantifier swap or limit.

and you are wrong.

>
>> You don't even get the simplest of all limits, the behaviour of the fractions 1/n as n -> oo. The limit *IS* zero, but there is *NO* n for which 1/n = 0.
>
> The limit is not attained. But according to set theory the complete indexing of the positive fractions is attained.

no, according to Cantor.

>
>> How could there be?
>
> The limit in analysis is not attained but approached as closely as desired. Simple as that.

there is no limit in enumeration of the rationals.

>
>> Perhaps you'd be so kind and write down the definition of this limit for yourself? (Better yet, since you are clearly unable to do even that for yourself, go copy it from some authoritative source, perhaps the one you copied for your book.)
>
> The source of my knowledge are lectures at the university of Göttingen, one of the top institutes of mathematics worldwide.

and your math presented here is so deeply flawed, its not math at all.

you are a troll, who does not know algebra.

you do not represent any University or school.

>
> Regards, WM
>

onsdag 16 november 2022 kl. 19:05:38 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 12:52:36 UTC+1:
> > onsdag 16 november 2022 kl. 10:55:48 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 07:19:22 UTC+1:
> > > > tisdag 15 november 2022 kl. 18:28:09 UTC+1 skrev WM:
> > >
> > > > > "The infinite sequence thus defined has the peculiar property to contain all positive
> > > > > rational numbers and each of them only once at a determined place.."
> > > > > Therefore every place can be checked step by step because it is a finite place.
> > > > You don't need to "check" it
> > >
> > > But I can do it, if it is possibke, that means if a bijection exists as Cantor describes it: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
> > >
> > > > We just check that for any rational number there is a natural number assigned to it, tada, ALL has been "checked" at once
> > >
> > > As this does not explain the loss of the O's it is tad, as you say.
> > You can find it in this instance but that is not a requirement.
> Where does the first O leave the matrix? Since many are to follow, this must happen at a finite step which can be checked.
>
> Regards, WM

Your matrix is irrelevant! FOR FUCK SAKE! How stupid are you? THe matrix does not add anythign! it is a red herring!

zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 06:52:08 UTC+1:
> onsdag 16 november 2022 kl. 19:05:38 UTC+1 skrev WM:

> > Where does the first O leave the matrix? Since many are to follow, this must happen at a finite step which can be checked.
> >
> Your matrix is irrelevant!

The matrix allows to represent every finite term of Cantor's sequence.
There are infinitely many O's = not yet indexed fractions. If the are indexed in finite steps, then there must be a first O leaving at a finite step.. But this can be excluded for every finite step. Therefore the O's do not leave the matrix or they don't leave at finite steps. In both cases there is no bijection. So my matrix shows that the set theory of the last 150 years is wrong. Therefore it is highly relevant.

Regards, WM

torsdag 17 november 2022 kl. 11:15:34 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 06:52:08 UTC+1:
> > onsdag 16 november 2022 kl. 19:05:38 UTC+1 skrev WM:
>
> > > Where does the first O leave the matrix? Since many are to follow, this must happen at a finite step which can be checked.
> > >
> > Your matrix is irrelevant!
>
> The matrix allows to represent every finite term of Cantor's sequence.

No it doesn't, it represent your finite swapping which is not relevant to the fucking bijection as it does no finite swapping.

> There are infinitely many O's = not yet indexed fractions. If the are indexed in finite steps, then there must be a first O leaving at a finite step. But this can be excluded for every finite step. Therefore the O's do not leave the matrix or they don't leave at finite steps. In both cases there is no bijection. So my matrix shows that the set theory of the last 150 years is wrong. Therefore it is highly relevant.

It shows nothing of the sort, all it shows is that in finite steps, you cannot do something which is irrelevant given IT IS NOT MADE IN FUCKING FINITE STEPS!

>
> Regards, WM

WM explained on 11/17/2022 :
> zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 06:52:08
> UTC+1:
>> onsdag 16 november 2022 kl. 19:05:38 UTC+1 skrev WM:
>
>>> Where does the first O leave the matrix? Since many are to follow, this
>>> must happen at a finite step which can be checked.
>>>
>> Your matrix is irrelevant!
>
> The matrix allows to represent every finite term of Cantor's sequence.
> There are infinitely many O's = not yet indexed fractions.

What's taking so long?

zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 12:32:57 UTC+1:
> torsdag 17 november 2022 kl. 11:15:34 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 06:52:08 UTC+1:
> > > onsdag 16 november 2022 kl. 19:05:38 UTC+1 skrev WM:
> >
> > > > Where does the first O leave the matrix? Since many are to follow, this must happen at a finite step which can be checked.
> > > >
> > > Your matrix is irrelevant!
> >
> > The matrix allows to represent every finite term of Cantor's sequence.
> No it doesn't,

Here are the first terms:

Index 1 remains at the first term 1/1. The next term 1/2 is indexed with 2 which is taken from position 2/1.

XXOO...
OOOO...
XOOO...
XOOO...
XOOO...
    

Then index 3 it taken from 3/1 and attached to 2/1:

XXOO...
OOOO...
XOOO...
XOOO...
XOOO...
    

Then index 4 it taken from 4/1 and attached to 1/3:

XXXO...
XOOO...
OOOO...
OOOO...
XOOO...
    

Then index 5 it taken from 5/1 and attached to 2/2:

XXXO...
XXOO...
OOOO...
OOOO...
OOOO...
    

> it represent your finite swapping which is not relevant to the fucking bijection

The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... which can be checked by my matrices up to every finite term.

>> So my matrix shows that the set theory of the last 150 years is wrong. Therefore it is highly relevant.
> It shows nothing of the sort, all it shows is that in finite steps, you cannot do something which is irrelevant given IT IS NOT MADE IN FUCKING FINITE STEPS!

You are wrong. In finite steps I would reach the term which first omits an O - because infinitely many O's have to be omitted in infinitely many terms..

Regards, WM

torsdag 17 november 2022 kl. 13:05:48 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 12:32:57 UTC+1:
> > torsdag 17 november 2022 kl. 11:15:34 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 06:52:08 UTC+1:
> > > > onsdag 16 november 2022 kl. 19:05:38 UTC+1 skrev WM:
> > >
> > > > > Where does the first O leave the matrix? Since many are to follow, this must happen at a finite step which can be checked.
> > > > >
> > > > Your matrix is irrelevant!
> > >
> > > The matrix allows to represent every finite term of Cantor's sequence..
> > No it doesn't,
> Here are the first terms:
>
> Index 1 remains at the first term 1/1. The next term 1/2 is indexed with 2 which is taken from position 2/1.
>
> XXOO...
> OOOO...
> XOOO...
> XOOO...
> XOOO...
>     
>
> Then index 3 it taken from 3/1 and attached to 2/1:
>
> XXOO...
> OOOO...
> XOOO...
> XOOO...
> XOOO...
>     
>
> Then index 4 it taken from 4/1 and attached to 1/3:
>
> XXXO...
> XOOO...
> OOOO...
> OOOO...
> XOOO...
>     
>
> Then index 5 it taken from 5/1 and attached to 2/2:
>
> XXXO...
> XXOO...
> OOOO...
> OOOO...
> OOOO...
>     
> > it represent your finite swapping which is not relevant to the fucking bijection
> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... which can be checked by my matrices up to every finite term.

It cannot because again, your matrix is a fucking RED HERRING!

> >> So my matrix shows that the set theory of the last 150 years is wrong. Therefore it is highly relevant.
> > It shows nothing of the sort, all it shows is that in finite steps, you cannot do something which is irrelevant given IT IS NOT MADE IN FUCKING FINITE STEPS!
> You are wrong. In finite steps I would reach the term which first omits an O - because infinitely many O's have to be omitted in infinitely many terms.

I am right, you are the one wrong, stop with your fucking redherring!

You keep going on this because it obfuscate that you are a fucking idiot. Basic proofs in cardinal arithmetic alone shows you wrong!

>
> Regards, WM