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

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

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

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

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

On 11/17/2022 10:45 AM, WM wrote:
> 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

1/1

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

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.

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.

Regards, WM

On 11/17/2022 2:07 PM, WM wrote:
> 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.
>
> Regards, WM

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

On 11/17/2022 11:53 AM, WM wrote:
> 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

how?

> 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

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.

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

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

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.

On 11/17/2022 2:15 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.
> 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.

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

;^)

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.

> 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.

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

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

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

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

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

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

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

On 11/18/2022 3:55 AM, WM wrote:
> 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.

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

>
>
> Regards, 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