Three proofs of dark numbers contd

[WM]

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

[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

[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

[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

[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

[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

[zelos...@gmail.com]

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

[WM]

> 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

[Gus Gassmann]

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.

[Gus Gassmann]

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.

[Sergi o]

agree. WMs lost in the woods.

[Sergi o]

agree.

[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." There is no limit.

Regards, WM

[Sergi o]

which O, the one covering 1/1 ?

[Sergi o]

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

[Sergi o]

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

[Sergi o]

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

[Sergi o]

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

[Sergi o]

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
>

[zelos...@gmail.com]

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

[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.
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

[zelos...@gmail.com]

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

[FromTheRafters]

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?

[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.

>> 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

[zelos...@gmail.com]

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

[Sergi o]

On 11/17/2022 4:15 AM, WM wrote:
> 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.

Cantors matrix of the rationals is relevant.

Your's is not.

<snip WM lies>

>
> Regards, WM

[Sergi o]

On 11/17/2022 6:05 AM, WM wrote:
> 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:
>

<snip crap>

you continue to use a "step by step" process in an infinite set, than try to blame Cantor for your failure.

You get an F, for Fail.

>
> Regards, WM

[WM]

Sergi o schrieb am Donnerstag, 17. November 2022 um 14:44:43 UTC+1:
> On 11/17/2022 6:05 AM, WM wrote:

> you continue to use a "step by step" process in an infinite set, than try to blame Cantor for your failure.

I use a step-by-step process only for a finite number of steps. But I know that, if Cantor is right, a first O has to leave the matrix after a finite number of steps because infinitely many O's have to leave and there is only one infinity available, namely the natural numbers. Therefore the first O has to leave at a finite step --- if all O's leave at different distinct steps.

Regards, WM

[WM]

zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 14:37:34 UTC+1:
> torsdag 17 november 2022 kl. 13:05:48 UTC+1 skrev WM:

> > 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

Find first term which fails.

Regards, WM

[Sergi o]

On 11/17/2022 10:42 AM, WM wrote:
> Sergi o schrieb am Donnerstag, 17. November 2022 um 14:44:43 UTC+1:
>> On 11/17/2022 6:05 AM, WM wrote:
>
>> you continue to use a "step by step" process in an infinite set, than try to blame Cantor for your failure.
>
> I use a step-by-step process only for a finite number of steps.

which will ALWAYS fail with infinite sets, and you know that.

> But I know that, if Cantor is right,

Cantor is right, he Proved it. You cannot un-prove it. Especially when specifically using finite math on infinite sets, the classic mistake of math pupas.

> a first O has to leave the matrix

wrong again, sets are fixed. Including the set of rationals.

no elements enter or leave a set. What you actually do is create a new set.

> after a finite number of steps because infinitely many O's have to leave and there is only one infinity available, namely the natural numbers.

you are not thinking in terms of Math.

> Therefore the first O has to leave at a finite step --- if all O's leave at different distinct steps.

when does the last O leave ?

>
> Regards, WM

[Sergi o]

1/1

[WM]

Sergi o schrieb am Donnerstag, 17. November 2022 um 18:01:25 UTC+1:
> On 11/17/2022 10:42 AM, WM wrote:

> > But I know that, if Cantor is right,

> > a first O has to leave the matrix
> wrong again, sets are fixed. Including the set of rationals.

Just this proves Cantor wrong.

Regards, WM

[FromTheRafters]

After serious thinking WM wrote :

> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... which can be

The bijection is not a sequence, it is a relation between sets.

[WM]

This bijection is a sequence.

Regards, WM

[Sergi o]

wrong. The bijection is not a sequence, it is a relation between sets.

[Sergi o]

how?

[Fritz Feldhase]

> The bijection is not a sequence, it is a ...

Formally any function from IN into a set X is a sequence.

Hence especially a bijection from IN onto Q is a sequence.

See: https://de.wikipedia.org/wiki/Folge_(Mathematik)#Formale_Definition

[Fritz Feldhase]

On Thursday, November 17, 2022 at 10:00:19 PM UTC+1, Sergi o wrote:
> On 11/17/2022 2:07 PM, WM wrote:
> >
> > This bijection is a sequence.
> >

> wrong. The bijection is not a sequence, it is a [...]

Nope.

[Sergi o]

On 11/17/2022 6:10 PM, Fritz Feldhase wrote:
> On Thursday, November 17, 2022 at 10:00:19 PM UTC+1, Sergi o wrote:
>> On 11/17/2022 2:07 PM, WM wrote:
>>>
>>> This bijection is a sequence.
>>>
>> wrong. The bijection is not a sequence, it is a [...]
>
> Nope.
>
> Formally any function from IN into a set X is a sequence.

hmm... true, restrictive though.

>
> Hence especially a bijection from IN onto Q is a sequence.

and that sequence, is the result of a bijection.

>
> See: https://de.wikipedia.org/wiki/Folge_(Mathematik)#Formale_Definition


WM's "This bijection is a sequence" underscores his (internal) belief that Cantor's enumeration of the rationals is correct, as a "rule" (Cantors) is
applied to determine the actual sequence, and WM calls that a sequence.

[Chris M. Thomasson]

https://youtu.be/6AcOv8D3-pM

;^)

[Jim Burns]

On 11/17/2022 11:42 AM, WM wrote:
> Sergi o schrieb am Donnerstag,
> 17. November 2022 um 14:44:43 UTC+1:
>> On 11/17/2022 6:05 AM, WM wrote:

>> you continue to use a "step by step" process
>> in an infinite set,
>> than try to blame Cantor for your failure.
>
> I use a step-by-step process only for
> a finite number of steps.

You are insisting that finitely-many
be treated as infinitely-many.

You (WM) have this broken version of "all".
"All the gizmos produced in this factory"
includes gizmos made at other factories.

> But I know that, if Cantor is right,

If Cantor claimed what you claim for Cantor,
then Cantor would be wrong.

In mathematics,
when a highly respected mathematician
says something wrong,
they are wrong.

A recent, very public example:
Edward Nelson was writing up a proof of
the inconsistency of arithmetic,
Terence Tao spotted the error, and,
after some discussion, Edward Nelson
saw it too.

*Nelson immediately withdrew his claim*

https://golem.ph.utexas.edu/category/2011/09/the_inconsistency_of_arithmeti.html

| You are quite right,
| and my original response was wrong.
| Thank you for spotting my error.
|
| I withdraw my claim.
|
| Posted by: Edward Nelson
| on October 1, 2011 1:39 PM
| Permalink | Reply to this

I think the most important training one
receives in mathematics programs is
how to _fail_ well.

Edward Nelson made a mistake.
He dropped his claim and moved on.
I'm sure that was extremely unpleasant
for him. He did it anyway.

You (WM) made a mistake, once upon a time.
And, here you are, decades later,
still wallowing in it.

Set aside all your failures.
A major indication that you have
no more than epsilon of mathematical training
is how bad you are at failing.

[Fritz Feldhase]

On Friday, November 18, 2022 at 4:04:58 AM UTC+1, Sergi o wrote:
> On 11/17/2022 6:10 PM, Fritz Feldhase wrote:
> >
> > Formally any function from IN into a set X is a sequence.
> >

> > Hence especially a bijection from IN onto Q is a sequence.
> >
> and that sequence, is the result of a bijection.

No, it's not "the result of a bijection", it *is* the bijection.

See: https://de.wikipedia.org/wiki/Folge_(Mathematik)#Formale_Definition

[zelos...@gmail.com]

It is a redherring, I don't need to find "one that fails" because the entire argument is INVALID

[WM]

Jim Burns schrieb am Freitag, 18. November 2022 um 06:26:16 UTC+1:
> On 11/17/2022 11:42 AM, WM wrote:
> > Sergi o schrieb am Donnerstag,
> > 17. November 2022 um 14:44:43 UTC+1:
> >> On 11/17/2022 6:05 AM, WM wrote:
>
> >> you continue to use a "step by step" process
> >> in an infinite set,
> >> than try to blame Cantor for your failure.
> >
> > I use a step-by-step process only for
> > a finite number of steps.
> You are insisting that finitely-many
> be treated as infinitely-many.

Up to every finite index the number of steps is finite, isn't it?

>
> You (WM) have this broken version of "all".

That is Cantor's version.

> > But I know that, if Cantor is right,
> If Cantor claimed what you claim for Cantor,
> then Cantor would be wrong.

He claimed this: "every number p/q comes at an absolutely fixed place of a simple infinite sequence", "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."

That means all (cancelled) fractions.

>
> In mathematics,
> when a highly respected mathematician
> says something wrong,
> they are wrong.

That is what I have proved. Cantor is wrong.

>
> A recent, very public example:
> Edward Nelson was writing up a proof of
> the inconsistency of arithmetic,
> Terence Tao spotted the error, and,
> after some discussion, Edward Nelson
> saw it too.
>
> *Nelson immediately withdrew his claim*

But no one has spotted an error in my proof yet. Try it.

>
> You (WM) made a mistake, once upon a time.

Spot it.

> A major indication that you have
> no more than epsilon of mathematical training
> is how bad you are at failing.

This is not spotting an error but the usual insult uttered by helpless would be spotters.

> > But I know that, if Cantor is right,
> > a first O has to leave the matrix
> > after a finite number of steps
> > because infinitely many O's have to leave
> > and there is only one infinity available,
> > namely the natural numbers.
> > Therefore
> > the first O has to leave at a finite step ---
> > if all O's leave at different distinct steps.

Here you could add a mathematical argument. But obviously you cannot because there is none.

Regards, WM

[WM]

Fritz Feldhase schrieb am Freitag, 18. November 2022 um 06:57:59 UTC+1:
> On Friday, November 18, 2022 at 4:04:58 AM UTC+1, Sergi o wrote:
> > On 11/17/2022 6:10 PM, Fritz Feldhase wrote:
> > >
> > > Formally any function from IN into a set X is a sequence.
> > >
> > > Hence especially a bijection from IN onto Q is a sequence.
> > >
> > and that sequence, is the result of a bijection.
> No, it's not "the result of a bijection", it *is* the bijection.

Yes. Perhaps Sergio means the result of Cantor's process of setting up a bijection. Like in:

- die entstehenden, im allgemeinen zweifach unendlichen Reihen ebenso behandelt und diesen Prozeß so lange fortführt, bis man nur rationale Zahlen vor sich sieht.

- das Hemmungs- oder Beschränkungsprinzip nenne, entgegen, wodurch dem durchaus endlosen Bildungsprozeß sukzessive gewisse Schranken auferlegt werden,

- diesem endlosen Prozeß einen gewissen vorläufigen Abschluß zu geben,

- und es erfährt daher der aus unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand.

- Zeichen für die beim subjektiven Zählprozeß gezählten Einzeldinge sein sollen.

- Man überzeugt sich, daß dieser Bildungsprozeß der Alefs und der ihnen entsprechenden Zahlenklassen des Systems  absolut grenzenlos ist.

Regards, WM

[WM]

zelos...@gmail.com schrieb am Freitag, 18. November 2022 um 07:24:59 UTC+1:
> torsdag 17 november 2022 kl. 17:45:15 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 14:37:34 UTC+1:
> > > torsdag 17 november 2022 kl. 13:05:48 UTC+1 skrev WM:
> >
> > > > 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
> > Find first term which fails.
> >

> It is a redherring, I don't need to find "one that fails" because the entire argument is INVALID

Why should the application of another language be invalid?

Fact is the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... can be checked by my matrices up to every finite term.

Your statement "It cannot" is wrong.

Regards, WM

[zelos...@gmail.com]

fredag 18 november 2022 kl. 10:55:35 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 18. November 2022 um 07:24:59 UTC+1:
> > torsdag 17 november 2022 kl. 17:45:15 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Donnerstag, 17. November 2022 um 14:37:34 UTC+1:
> > > > torsdag 17 november 2022 kl. 13:05:48 UTC+1 skrev WM:
> > >
> > > > > 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
> > > Find first term which fails.
> > >
> > It is a redherring, I don't need to find "one that fails" because the entire argument is INVALID
> Why should the application of another language be invalid?

It is not another language, it isapplying an irrelevant thing.

>
> Fact is the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... can be checked by my matrices up to every finite term.

No, it cannot because it is irrelevant. The checking is done through other means!

>
> Your statement "It cannot" is wrong.

everything you say is wrong

>
> Regards, WM

[Sergi o]

On 11/18/2022 3:43 AM, WM wrote:
> Jim Burns schrieb am Freitag, 18. November 2022 um 06:26:16 UTC+1:
>> On 11/17/2022 11:42 AM, WM wrote:
>>> Sergi o schrieb am Donnerstag,
>>> 17. November 2022 um 14:44:43 UTC+1:
>>>> On 11/17/2022 6:05 AM, WM wrote:
>>
>>>> you continue to use a "step by step" process
>>>> in an infinite set,
>>>> than try to blame Cantor for your failure.
>>>
>>> I use a step-by-step process only for
>>> a finite number of steps.
>> You are insisting that finitely-many
>> be treated as infinitely-many.
>
> Up to every finite index the number of steps is finite, isn't it?

WM wont acknowledge he is using finite math on infinite set which results in failure, so we get his misdirection.

>>
>> You (WM) have this broken version of "all".
>
> That is Cantor's version.

more misdirection.

>
>>> But I know that, if Cantor is right,
>> If Cantor claimed what you claim for Cantor,
>> then Cantor would be wrong.
>
> He claimed this: "every number p/q comes at an absolutely fixed place of a simple infinite sequence", "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."

Cantor is correct.

>
> That means all (cancelled) fractions.
>>
>> In mathematics,
>> when a highly respected mathematician
>> says something wrong,
>> they are wrong.
>
> That is what I have proved. Cantor is wrong.

WM's proof is a spoof, stickies and pasties replacing fractions! what a goof!

>>
>> A recent, very public example:
>> Edward Nelson was writing up a proof of
>> the inconsistency of arithmetic,
>> Terence Tao spotted the error, and,
>> after some discussion, Edward Nelson
>> saw it too.
>>
>> *Nelson immediately withdrew his claim*
>
> But no one has spotted an error in my proof yet. Try it.

wrong, it is full of errors you choose to ignore, which says you do not know algebra, or feeble.

>>
>> You (WM) made a mistake, once upon a time.
>
> Spot it.

your daffynition of "defined"

>
>> A major indication that you have
>> no more than epsilon of mathematical training
>> is how bad you are at failing.
>
> This is not spotting an error but the usual insult uttered by helpless would be spotters.

so far you have errors in every single post, you are a troll.

>
>>> But I know that, if Cantor is right,
>>> a first O has to leave the matrix
>>> after a finite number of steps
>>> because infinitely many O's have to leave
>>> and there is only one infinity available,
>>> namely the natural numbers.
>>> Therefore
>>> the first O has to leave at a finite step ---
>>> if all O's leave at different distinct steps.
>
> Here you could add a mathematical argument. But obviously you cannot because there is none.

you did not present any!

>
> Regards, WM

[Sergi o]

but you cannot "check" for the entire sequence, as Cantor does, so you have failed.

>
>
> Regards, WM

[WM]

zelos...@gmail.com schrieb am Freitag, 18. November 2022 um 13:55:19 UTC+1:
> fredag 18 november 2022 kl. 10:55:35 UTC+1 skrev WM:

> > Why should the application of another language be invalid?
> It is not another language, it isapplying an irrelevant thing.

The only difference between Cantor's process and mine is this: I remember where the indices have their origin.

> >
> > Fact is the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... can be checked by my matrices up to every finite term.
> No, it cannot because it is irrelevant.

Nevertheless it can be checked nd the sequence can be constructed until every finite term.

> The checking is done through other means!

No, it is not done at all. Cantor takes the indices from wherever. I take them from well-defined places.

Regards, WM

[Sergi o]

On 11/18/2022 11:29 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 18. November 2022 um 13:55:19 UTC+1:
>> fredag 18 november 2022 kl. 10:55:35 UTC+1 skrev WM:
>
>>> Why should the application of another language be invalid?
>> It is not another language, it isapplying an irrelevant thing.
>
> The only difference between Cantor's process and mine is this: I remember where the indices have their origin.

Burp! Why are you posting your crap in sci.math ?

If you really disproved Cantor, publish a paper in a reviewed Professional Journal.

but you only have crap, so you try to nail it on the walls of sci.math, but it slides off.


<snip crap>

> Regards, WM

[Fritz Feldhase]

On Friday, November 18, 2022 at 7:36:20 PM UTC+1, Sergi o wrote:
> >
> > The only difference between Cantor's process and mine is <bla>

> >
> Burp! Why are you posting your crap in sci.math ?
>
> If you really disproved Cantor, publish a paper in a reviewed Professional Journal.
>

> but you only have crap [...]

Yeah, it's crap!

[WM]

Sergi o schrieb am Freitag, 18. November 2022 um 19:36:20 UTC+1:
> On 11/18/2022 11:29 AM, WM wrote:
> > zelos...@gmail.com schrieb am Freitag, 18. November 2022 um 13:55:19 UTC+1:
> >> fredag 18 november 2022 kl. 10:55:35 UTC+1 skrev WM:
> >
> >>> Why should the application of another language be invalid?
> >> It is not another language, it isapplying an irrelevant thing.
> >
> > The only difference between Cantor's process and mine is this: I remember where the indices have their origin.
> Burp!

Can you see it?

> Why are you posting your crap in sci.math ?

Could be that some readers are intelligent enough to understand.

>
> If you really disproved Cantor, publish a paper in a reviewed Professional Journal.

Try to publish counter evidence of God's existence in L'Osservatore Romano?

Regards, WM

[WM]

Says Franz Fritsche who believes that lossless exchange can suffer from infinitely many losses.

Regards, WM

[Chris M. Thomasson]

Wrt are you talking about?

[Sergi o]

your "lossless" exchanges creates an infinite amount of litter;

1. each time a O stickie is put onto a fraction covering it, then pulled off later and discarded,
2. each time an X pastie is put onto a fraction hiding it, or on top of an O stickie and then discarded,

do you pick up after yourself, or just let it rot in the environment, like your O's ?

[Chris M. Thomasson]

I wonder if WM is familiar with lossless compression...

[Sergi o]

FTX said they were a lossless exchange, too.

lossless compression (the still FAT boys)=> PNG, GIF, ZIP, RAW, BMP, FLAC, ALAC, WMA

[zelos...@gmail.com]

fredag 18 november 2022 kl. 18:29:09 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 18. November 2022 um 13:55:19 UTC+1:
> > fredag 18 november 2022 kl. 10:55:35 UTC+1 skrev WM:
>
> > > Why should the application of another language be invalid?
> > It is not another language, it isapplying an irrelevant thing.
> The only difference between Cantor's process and mine is this: I remember where the indices have their origin.
> > >
> > > Fact is the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... can be checked by my matrices up to every finite term.
> > No, it cannot because it is irrelevant.
> Nevertheless it can be checked nd the sequence can be constructed until every finite term.

That is not how the fucking function is made you retard!

> > The checking is done through other means!
> No, it is not done at all. Cantor takes the indices from wherever. I take them from well-defined places.

No, you do not. You make up a shit thing and think it is valid, IT IS NOT!

All cantor and mathematics do is define a function from N to Q+, that is it!

>
> Regards, WM

[Chris M. Thomasson]

lossless compression is good, in the sense that the data is preserved.

[Chris M. Thomasson]

Total scam!

[Sergi o]

lossie compression, I tried a loss* compression on the alphabet, => delete the 10 least use letters, Z,Q,J,X,K,V,B,Y, and *ust put a "*" in its place. I
removed the *e*s from a t*pewriter, but that caused accidents. An*how It never caught on.

how about going from an 8 by 16 (or 8 by 8) pi*al/character to a 4 by 6 ?
We have too many letters an*wa*, and far to many words.

The ne*t generation should be taught words that are 8 letters or less, and do math in base 5. so instead of *ear 2022, it would be 31042

I may be o*er thinking this.

[Chris M. Thomasson]

"Lossy" compression is, well, "okay" for images... Fractal compression
is lossy. However, I created a fractal cipher that actually has a chance
to compress. I setup some terms that can make it lossless, but its a bit
touchy, floating point precision and all of that fun stuff, so to speak.
I think I showed you my code on it. Storing data in the roots of complex
numbers. I can show you my code if you have not seen it already...

[FromTheRafters]

WM formulated on Thursday :
> FromTheRafters schrieb am Donnerstag, 17. November 2022 um 20:46:44 UTC+1:
>> After serious thinking WM wrote :
>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... which can be
>> The bijection is not a sequence, it is a relation between sets.

>
> This bijection is a sequence.

Then why all the so-called matrices?

[FromTheRafters]

WM used his keyboard to write :

> zelos...@gmail.com schrieb am Freitag, 18. November 2022 um 13:55:19 UTC+1:
>> fredag 18 november 2022 kl. 10:55:35 UTC+1 skrev WM:
>
>>> Why should the application of another language be invalid?
>> It is not another language, it isapplying an irrelevant thing.
>
> The only difference between Cantor's process and mine is this: I remember
> where the indices have their origin.

Which matters not.

[FromTheRafters]

Sergi o formulated on Friday :

There is lossless JPEG too, but is RAW or BMP ever even compressed?

[FromTheRafters]

WM formulated the question :

> FromTheRafters schrieb am Donnerstag, 17. November 2022 um 20:46:44 UTC+1:
>> After serious thinking WM wrote :
>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... which can be
>> The bijection is not a sequence, it is a relation between sets.
>
> This bijection is a sequence.

This particular proof/demonstration of the bijection is.

Yes, an iterative function from such as this produces a sequence. But
consider two sets of elements, one on the left and one on the right. We
don't care what the elements are. If the relation is left total (all
elements of left set get 'used' in the pairing relation, nobody left
behind) and right unique (no double-dealing two-timers between sets)
and the right and left can be exchanged and the same still holds true
(injective and surjective) it is a bijection. No sequence needed. They
are equinumerous (same size) and in finite sets it also means they have
the same number of elements.

This particular Cantor pairing is or can be a sequence to help you
(well, everybody else) to see the bijection (the stripe pattern) but
does not restrict anyone to a step by step completion algorithm.

https://home.cs.colorado.edu/~srirams/courses/csci2824-spr14/functionTypes-18.html

[Chris M. Thomasson]

On 11/21/2022 8:47 PM, FromTheRafters wrote:
> WM formulated on Thursday :
>> FromTheRafters schrieb am Donnerstag, 17. November 2022 um 20:46:44
>> UTC+1:
>>> After serious thinking WM wrote :
>>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... which

[...]

You are missing all a lot of Cantor parings. What about 0/1 and 1/0? Humm...

[Chris M. Thomasson]

Oops! I responded to FromTheRafters instead of WM.

1/1, 1/2, 2/1, 1/3, 2/2, 3/1 misses a lot of Cantor pairings. There is
0/3, 2/0, ect...

Why does WM seem to think that a Cantor pair must be a proper fraction?
Well, it sure seems that way.

[Chris M. Thomasson]

I don't think so. A BMP can store lossless compressed data. Also, a PNG
is nice because it is lossless.

[FromTheRafters]

Chris M. Thomasson used his keyboard to write :

In this specific case, we are dealing with rationals whose numerator
and denominator are positive, not non-negative.

[Sergi o]

lossless JPEG is different from normal JPEG, and uses predictive coding, so bits are changed

RAW is not quite standerdized, and RAW can have compression

PNG is for transfering images on internet, but color space is different from other color pallets, can have lossless compression.

WM's Lossless Exchange (WMLE) suffers from loss of stickies and pasties on Xs and Os, and know one knows what the full picture is.

[Chris M. Thomasson]

Signed zero for fun:

1/+0 vs 1/-0

;^)

[Sergi o]

like a plot of 1/x^2, or 1/x^(even)

[Chris M. Thomasson]

Fwiw, I forgot one of the states of zero.

-0, 0, +0

We can define it as ternary. :^)

[Jim Burns]

On 11/18/2022 4:43 AM, WM wrote:
> Jim Burns schrieb am Freitag,
> 18. November 2022 um 06:26:16 UTC+1:
>> On 11/17/2022 11:42 AM, WM wrote:
>>> Sergi o schrieb am Donnerstag,
>>> 17. November 2022 um 14:44:43 UTC+1:

>>>> you continue to use a "step by step" process
>>>> in an infinite set,
>>>> than try to blame Cantor for your failure.
>>>
>>> I use a step-by-step process only for
>>> a finite number of steps.
>>
>> You are insisting that finitely-many
>> be treated as infinitely-many.
>
> Up to every finite index
> the number of steps is finite, isn't it?

A step for which
there is no split before that step
which doesn't have a one-by-one move
across that split
is a _finite step_

A move across a split is _one-by-one_ if
there is a last step before
and a first step after
and nothing between them.

Yes,
each finite step
is a finite step.


Also,
for each finite step,
there is a step next after it.

For each step next after a finite step,
there is no split before that next step
without a one-by-one move across that split

Thus,
each step next after a finite step
is also a finite step.


However,
any step after all finite steps
is NOT a finite step.

To be a finite step,
there would need to be,
for each split before it,
a one-by-one move across that split.

A one-by-one move requires
a last before
and a first after
and nothing between them.

However,
each step next after a finite step
is a finite step.

There is no
last finite before all finites.
So
there is no
one-by-one move from before to after
all finites.
So,
a step after all finite steps
is not a finite step.

Note that this arises from
what we mean by "finite".
not from your dark numbers.

>> You (WM) have this broken version of "all".
>
> That is Cantor's version.

If Cantor said
"all gizmos made in this factory"
includes gizmos made in different factories,
then Cantor would be wrong.

Cantor didn't say anything even vaguely like that.
Set aside the issue of who said what.
_Whoever says that_ is wrong.

Cantor is not our Pope.
Things are not correct *because*
Cantor or Frege or Leibnitz or Euler
said them.

In mathematics,
sola scriptura speaks ex cathedra.

>>> But I know that, if Cantor is right,
>>
>> If Cantor claimed what you claim for Cantor,
>> then Cantor would be wrong.
>
> He claimed this:
> "every number p/q comes at an absolutely fixed
> place of a simple infinite sequence",
> "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."

Cantor's fraction-sequence 𝐶 begins at 1/1
For each split of 𝐶
some fraction p/q is last before that split and
some fraction p'/q' such that
| if q ≠ 1
| then p'/q' = p⁺⁺/q⁻⁻
| else p'/q' = 1/p⁺⁺
is first after that split.

For each fraction p/q
there is a unique index k
s = p+q
k = (s-1)(s-2)/2+p

For each index k
there is a unique fraction p/q
s = ⌈(8k+1)¹ᐟ²+1)/2)⌉
p = k-(s-1)(s-2)/2
q = s-p

> That means all (cancelled) fractions.

We describe one of some things (fractions, indexes).
The descriptive claims are true of
each of those things,
in the same why that
each gizmo made in this factory
is made in this factory, not elsewhere.

We reason from those descriptive claims
to further claims,
but only by claim-to-claim steps which we know
cannot lose the true-of-each-one property.

[WM]

Jim Burns schrieb am Mittwoch, 23. November 2022 um 01:49:14 UTC+1:
> On 11/18/2022 4:43 AM, WM wrote:
> > Jim Burns schrieb am Freitag,

> > Up to every finite index
> > the number of steps is finite, isn't it?
>
> A step for which
> there is no split before that step
> which doesn't have a one-by-one move
> across that split
> is a _finite step_

A definable finite step is the last step of a FISON.
An undefinable finite step belongs to the ℵo-infinit set of dark numbers which cannot be discerned and hence cannot be put in order. No first no last, but only darkness.

>
> A move across a split is _one-by-one_ if
> there is a last step before
> and a first step after
> and nothing between them.

Definable steps.

>
> Yes,
> each finite step
> is a finite step.
>

Yes. In the darkness there are no steps.

> each step next after a finite step
> is also a finite step.

Yes.

>
> However,
> any step after all finite steps
> is NOT a finite step.

There are not "all" finite steps. Potential infinity.

>
> A one-by-one move requires
> a last before
> and a first after
> and nothing between them.

Yes.

>
> However,
> each step next after a finite step
> is a finite step.

Yes.

> So
> there is no
> one-by-one move from before to after
> all finites.
> So,
> a step after all finite steps
> is not a finite step.
>
> Note that this arises from
> what we mean by "finite".
> not from your dark numbers.

All that is true. But to believe that an inclusion monotonic sequence of infinite sets like endsegments has an empty intersection is wrong. To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.

> Cantor is not our Pope.
> Things are not correct *because*
> Cantor or Frege or Leibnitz or Euler
> said them.

So it is. Therefore I am entitled to show where he went astray.

>
> We reason from those descriptive claims
> to further claims,
> but only by claim-to-claim steps which we know
> cannot lose the true-of-each-one property.

To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.

Regards, WM

[FromTheRafters]

After serious thinking Chris M. Thomasson wrote :

Ones' Complement binary representations have a zero and a negative
zero.

What makes your plus zero 'different' from your zero such that it would
be needed? Two ways to say nothing is somehow not enough? :)

[FromTheRafters]

WM expressed precisely :

> Jim Burns schrieb am Mittwoch, 23. November 2022 um 01:49:14 UTC+1:
>> On 11/18/2022 4:43 AM, WM wrote:
>>> Jim Burns schrieb am Freitag,
>
>>> Up to every finite index
>>> the number of steps is finite, isn't it?
>>
>> A step for which
>> there is no split before that step
>> which doesn't have a one-by-one move
>> across that split
>> is a _finite step_
>
> A definable finite step is the last step of a FISON.
> An undefinable finite step belongs to the ℵo-infinit set of dark numbers
> which cannot be discerned and hence cannot be put in order. No first no last,
> but only darkness.

Please don't abuse the steps, they are for demonstration purposes only.
A bijective mapping takes 'very little' time and should work for sets
that aren't ordinal numbers too. It is convenient to *show* a bijection
of naturals by matching two known ordinal numbers at a time 'checking'
each 'step' of the 'way' that they conform to a specific generation
algorithm but these stepwise checks are not definitive.

[zelos...@gmail.com]

All of this boils down to bullshit!



For fuck sake we don't even need to use the fucking function given!

we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
so |N^2| = |N|
We also have |c-n|=c
we then also have |Q+|=|Nx(N/{0})|=|NxN|=|N^2|=|N|

and by definition of how cardinal arithmetic works, we have a bijection from N to Q+

Now shut the fuck up with your bullshit.

[WM]

FromTheRafters schrieb am Mittwoch, 23. November 2022 um 13:53:33 UTC+1:
> WM expressed precisely :

>
> Please don't abuse the steps, they are for demonstration purposes only.

They are defined by the formula
k = (m + n - 1)(m + n - 2)/2 + m
and be the 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, 2/5, 3/4, 4/3, 5/2, 6/1, ...

> A bijective mapping takes 'very little' time and should work for sets
> that aren't ordinal numbers too. It is convenient to *show* a bijection
> of naturals by matching two known ordinal numbers at a time 'checking'
> each 'step' of the 'way' that they conform to a specific generation
> algorithm but these stepwise checks are not definitive.

This however is definitive and does not require checking single steps but holds for every k in above formula: To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.

Regards, WM

[WM]

and I have proved that that is nonsense.

[Sergi o]

On 11/23/2022 7:02 AM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 23. November 2022 um 13:53:33 UTC+1:
>> WM expressed precisely :
>>
>> Please don't abuse the steps, they are for demonstration purposes only.
>
> They are defined by the formula
> k = (m + n - 1)(m + n - 2)/2 + m
> and be the 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, 2/5, 3/4, 4/3, 5/2, 6/1, ...

However, your three dots ... , to you, indicate the start of your dark numbers.

>> A bijective mapping takes 'very little' time and should work for sets
>> that aren't ordinal numbers too. It is convenient to *show* a bijection
>> of naturals by matching two known ordinal numbers at a time 'checking'
>> each 'step' of the 'way' that they conform to a specific generation
>> algorithm but these stepwise checks are not definitive.
>
> This however is definitive and does not require checking single steps but holds for every k in above formula:

so, you now believe in "variables" ? how can that be ?


>
> Regards, WM

[Sergi o]

On 11/23/2022 7:04 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 23. November 2022 um 13:59:43 UTC+1:
>> onsdag 23 november 2022 kl. 13:06:13 UTC+1 skrev WM:
>
>> For fuck sake we don't even need to use the fucking function given!
>>
>> we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
>> so |N^2| = |N|
>
> and I have proved that that is nonsense.

you have stated previously you do not believe in proofs, even simple ones, and you have never provided a proof.

why should we believe you now?


>

>
> Regards, WM

[zelos...@gmail.com]

onsdag 23 november 2022 kl. 14:04:36 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 23. November 2022 um 13:59:43 UTC+1:
> > onsdag 23 november 2022 kl. 13:06:13 UTC+1 skrev WM:
>
> > For fuck sake we don't even need to use the fucking function given!
> >
> > we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
> > so |N^2| = |N|
> and I have proved that that is nonsense.

no you haven't, all you have proven is that your fucking matrix is bullshit. The theorem I state is more powerful than your shit

[Fritz Feldhase]

On Wednesday, November 23, 2022 at 3:33:19 PM UTC+1, Sergi o wrote:

> why should we believe you now?

Because he's a psychotic crank full of shit?

[Fritz Feldhase]

On Wednesday, November 23, 2022 at 2:04:36 PM UTC+1, WM wrote:

> and I have proved that that is nonsense.

All you have proved is that you are a full blown crank full of shit.

[Jim Burns]

On 11/23/2022 7:06 AM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 23. November 2022 um 01:49:14 UTC+1:
>> On 11/18/2022 4:43 AM, WM wrote:

>>> Up to every finite index
>>> the number of steps is finite, isn't it?
>>
>> A step for which
>> there is no split before that step
>> which doesn't have a one-by-one move
>> across that split
>> is a _finite step_
>
> A definable finite step is
> the last step of a FISON.

Yes but
that description doesn't make us
any better off than we were before.
𝐅𝐈𝐒𝐎𝐍 ==
𝐅inite 𝐈nitial 𝐒egment 𝐎f 𝐍aturals.

"Finite is finite." So?

Our over-arching strategy is to _start_ with
claims true of each one of whatever we're
reasoning about, and then _keep_ that
being-true-of-each as we advance to further
claims.

The more true-of-each detail we start with,
the more we have to work with, when we try to
find claims we care about to which we can
advance in true-of-each-keeping ways.

> An undefinable finite step belongs to the
> ℵo-infinit set of dark numbers which cannot be
> discerned and hence cannot be put in order.
> No first no last, but only darkness.

Your undefinable finites are not in FISONs.

We can start with being in a FISON as
a description of one of what we're reasoning
about, and then none of the claims we start with
or advance to are about undefinable finites.

If I say
| _All_ the gizmos made in this factory
| have new-style thingummies.
then
pointing to a gizmo made elsewhere
_is not a counter-example_

If I say
| _All_ the FISON-ends >= 1 have
| unique prime factorizations.
then
pointing to a dark number
_is not a counter-example_

>> However,
>> any step after all finite steps
>> is NOT a finite step.
>
> There are not "all" finite steps.
> Potential infinity.

Because you decided to call "finite steps"
things which I am NOT talking about.

However,
I am STILL NOT talking about your other things.

An ordered collection
which begins at 0 and ends somewhere and
which, for each split, there is a
predecessor-successor pair across it
is
an ordered collection
which begins at 0 and ends somewhere and
which, for each split, there is a
predecessor-successor pair across it,
no matter what you or I or Georg Cantor
call it.

We reason _from the description_
You call something else a blablablah
finite step, but you haven't changed
_the description_ from which I'm
reasoning.

[Chris M. Thomasson]

On 11/23/2022 5:02 AM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 23. November 2022 um 13:53:33 UTC+1:
>> WM expressed precisely :
>>
>> Please don't abuse the steps, they are for demonstration purposes only.
>
> They are defined by the formula
> k = (m + n - 1)(m + n - 2)/2 + m
> and be the 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, 2/5, 3/4, 4/3, 5/2, 6/1, ...

[...]

You are missing a lot of pairs here...

[Chris M. Thomasson]

Ditto.

[Chris M. Thomasson]

:^) Well, humm... perhaps... We can say a step-by-step process whose
limit is zero, say:

.1, .01, .001, .0001, ...

Gets to "positive zero", while:

-1, -.01, -.001, -.0001, ...

Gets to "negative zero".

And zero, is just "zero with no sign"?


.1, .01, .001, .0001, ... = +0
-1, -.01, -.001, -.0001, ... = -0
0 = 0

Make any sense to you at all? Or, is that just stepping deep into the
heart of kookville? Humm...

[WM]

Jim Burns schrieb am Mittwoch, 23. November 2022 um 19:04:17 UTC+1:

> Our over-arching strategy is to _start_ with
> claims true of each one of whatever we're
> reasoning about, and then _keep_ that
> being-true-of-each as we advance to further
> claims.

Exchanging the X's and O's will delete all O's in a definable way but only after all definable terms of the sequence have kept all and every O.

>
> Your undefinable finites are not in FISONs.

So it is.

>
> We reason _from the description_

Then you should understand this description: Exchanging the X's and O's will never delete any O.

Regards, WM

[Sergi o]

On 11/23/2022 4:09 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 23. November 2022 um 19:04:17 UTC+1:
>
>> Our over-arching strategy is to _start_ with
>> claims true of each one of whatever we're
>> reasoning about, and then _keep_ that
>> being-true-of-each as we advance to further
>> claims.
>
> Exchanging the X's and O's will delete all O's in a definable way but only after all definable terms of the sequence have kept all and every O.

as WM for each fraction, will remove AND delete the O stickie on the fraction, and replace it with a X pastie.

>>
>> Your undefinable finites are not in FISONs.
>
> So it is.

It is so.

>>
>> We reason _from the description_
>
> Then you should understand this description: Exchanging the X's and O's will never delete any O.

then you failed to delete the O pastie as you said you would above. Fail

>
> Regards, WM

[Sergi o]

makes a difference on the approach, as with 1/x^2.

[zelos...@gmail.com]

we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
so |N^2| = |N|

[Jim Burns]

On 11/23/2022 5:09 PM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 23. November 2022 um 19:04:17 UTC+1:

>> Your undefinable finites are not in FISONs.
>
> So it is.
>
>> We reason _from the description_
>
> Then you should understand this description:
> Exchanging the X's and O's will never delete any O.

Exchanging X's and O's (a bijection)
*within any FISON* will never delete any O

Each exchange is within some FISON.
It does not delete O's

Each FISON does not have all exchanges within it.
O's might be deleted by all the exchanges.
For example,
by all of this sequence of exchanges:
⟨1/1 1/1⟩
⟨1/2 2/1⟩
⟨2/1 3/1⟩
⟨1/3 4/1⟩
⟨2/2 5/1⟩
⟨3/1 6/1⟩
...

It is a _sequence_ of exchanges.
It begins at ⟨1/1 1/1⟩
For each split of the sequence,
there is an exchange ⟨p/q k/1⟩ last before
and an exchange ⟨p'/q' k'/1⟩ first after
such that
k' = k⁺⁺

if q ≠ 1
then p'/q' = p⁺⁺/q⁻⁻
else p'/q' = 1/p⁺⁺

All the O's are deleted.

There is no dark p q or k in any exchange
in the sequence.

| Assume otherwise.
| Assume some ⟨p/q k/1⟩ in the sequence
| contains a dark p q or k
|
| There is a _first_ ⟨p₁/q₁ k₁/1⟩
| in which p₁ q₁ or k₁ is dark
| and, in its predecessor ⟨p₀/q₀ k₀/1⟩
| p₀ q₀ and k₀ are NOT dark
|
| However,
| p₀ q₀ and k₀ are NOT dark
| and
| k₁ = k₀⁺⁺

| if q₀ ≠ 1
| then p₁/q₁ = p₀⁺⁺/q₀⁻⁻

| else p₁/q₁ = 1/p₀⁺⁺
| So
| p₁ q₁ and k₁ are NOT dark
| Contradiction.

Therefore,
there is no dark p q or k in any exchange
in the sequence.
Nonetheless, all O's are deleted by
all the exchanges.

[WM]

Jim Burns schrieb am Donnerstag, 24. November 2022 um 17:11:23 UTC+1:
> On 11/23/2022 5:09 PM, WM wrote:

> > Then you should understand this description:
> > Exchanging the X's and O's will never delete any O.
> Exchanging X's and O's (a bijection)
> *within any FISON* will never delete any O
>
> Each exchange is within some FISON.
> It does not delete O's

Fine.

>
> Each FISON does not have all exchanges within it.

But all definable exchanges occur within FISONs.

> O's might be deleted by all the exchanges.

Maybe, but not in a definable way.

> For example,
> by all of this sequence of exchanges:
> ⟨1/1 1/1⟩
> ⟨1/2 2/1⟩
> ⟨2/1 3/1⟩
> ⟨1/3 4/1⟩
> ⟨2/2 5/1⟩
> ⟨3/1 6/1⟩
> ...
>

Nice to see that you at least have understood what others claim to be incomprehensible for average mathematicians.

> It is a _sequence_ of exchanges.
> It begins at ⟨1/1 1/1⟩
> For each split of the sequence,
> there is an exchange ⟨p/q k/1⟩ last before
> and an exchange ⟨p'/q' k'/1⟩ first after
> such that
> k' = k⁺⁺
> if q ≠ 1
> then p'/q' = p⁺⁺/q⁻⁻
> else p'/q' = 1/p⁺⁺
> All the O's are deleted.

Not in a definable way. That is proved by the fact, that never anybody will be able to find a furst O deleted.

> Therefore,
> there is no dark p q or k in any exchange
> in the sequence.
> Nonetheless, all O's are deleted by
> all the exchanges.

That is wrong. Find the first O that is deleted in the visible domain.

Regards, WM

[zelos...@gmail.com]

We can get an injection from Q to N, and from N to Q, this shows the cardinality is once again, THE SAME!

[WM]

zelos...@gmail.com schrieb am Freitag, 25. November 2022 um 12:38:39 UTC+1:
> fredag 25 november 2022 kl. 10:55:25 UTC+1 skrev WM:
> > Jim Burns schrieb am Donnerstag, 24. November 2022 um 17:11:23 UTC+1:

> > > For example,
> > > by all of this sequence of exchanges:
> > > ⟨1/1 1/1⟩
> > > ⟨1/2 2/1⟩
> > > ⟨2/1 3/1⟩
> > > ⟨1/3 4/1⟩
> > > ⟨2/2 5/1⟩
> > > ⟨3/1 6/1⟩
> > > ...
> > >
> > Nice to see that you at least have understood what others claim to be incomprehensible for average mathematicians.

> We can get an injection from Q to N, and from N to Q, this shows the cardinality is once again, THE SAME!

Find the first O that is deleted. Every remianing O proves the existence of a fraction that is not indexed such that the index can be known.

Regards, WM

[zelos...@gmail.com]

Your O and X does not change the fact that I can find those two injections and thus proves you wrong.