zelos...@gmail.com schrieb am Dienstag, 25. Januar 2022 um 06:19:57 UTC+1:
> måndag 24 januari 2022 kl. 18:00:52 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Montag, 24. Januar 2022 um 12:19:46 UTC+1:
> > > måndag 24 januari 2022 kl. 10:52:13 UTC+1 skrev WM:
> >
> > > > Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
> > > > 1,2,4,7,OOOOO...
> > > > 3,5,8,OOOOOO...
> > > > 6,OOOOOOOO...
> > > > OOOOOOOOO...
> > > > OOOOOOOOO...
> > > > OOOOOOOOO...
> > > > OOOOOOOOO...
> > > > OOOOOOOOO...
> > > > 9,OOOOOOOO...
> > > > ...
> > > >
> > > > reproducing the first eight terms of the sequence (k(m/n)) where
> > > >
> > > > k = (m + n - 1)(m + n - 2)/2 + m ?
> > > >
> > > Care to actually bother not fucking obfuscating it?
> > >
> > > We have a bijection, it is done. Get over it already
> > That is not the question here. The question has been defined above.
> >
> You are doing those things just to obfuscate,

No, I am doing these things in order to show you problems of set theory. No O can leave the matrix. The X act like Leonidas at the Thermopylae.

Regards, WM

sergio schrieb am Montag, 24. Januar 2022 um 18:17:31 UTC+1:
> On 1/24/2022 11:03 AM, WM wrote:
> > sergio schrieb am Montag, 24. Januar 2022 um 16:24:06 UTC+1:
> >> On 1/24/2022 3:51 AM, WM wrote:
> >
> >>> There is only one relevant question here: Is there any mistake in
> >>>
> >>> 1,2,4,7,OOOOO...
> >>> 3,5,8,OOOOOO...
> >>> 6,OOOOOOOO...
> >>> OOOOOOOOO...
> >>> OOOOOOOOO...
> >>> OOOOOOOOO...
> >>> OOOOOOOOO...
> >>> OOOOOOOOO...
> >>> 9,OOOOOOOO...
> >>> ...
> >>>
> >>> reproducing the first eight terms of the sequence (k(m/n)) where
> >>>
> >>> k = (m + n - 1)(m + n - 2)/2 + m ?
> >>>
> >> yes mistakes, and you are doing it wrong. your 9 is in the wrong place, 9 goes here;
> >
> > No, above you see the eighth term of the sequence of configurations. All indexes k from 9 on are attached to integer fractions n/1 with n = k.

> there is your problem.

Yes --- and nobody can solve it adhering to the framework of set theory. No O can leave the battlefield. The only exit, the first column is defended by the X's like the Thermopylae by Leonidas and his brave men.

Regards, WM

tisdag 25 januari 2022 kl. 11:18:55 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 25. Januar 2022 um 06:19:57 UTC+1:
> > måndag 24 januari 2022 kl. 18:00:52 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Montag, 24. Januar 2022 um 12:19:46 UTC+1:
> > > > måndag 24 januari 2022 kl. 10:52:13 UTC+1 skrev WM:
> > >
> > > > > Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
> > > > > 1,2,4,7,OOOOO...
> > > > > 3,5,8,OOOOOO...
> > > > > 6,OOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > 9,OOOOOOOO...
> > > > > ...
> > > > >
> > > > > reproducing the first eight terms of the sequence (k(m/n)) where
> > > > >
> > > > > k = (m + n - 1)(m + n - 2)/2 + m ?
> > > > >
> > > > Care to actually bother not fucking obfuscating it?
> > > >
> > > > We have a bijection, it is done. Get over it already
> > > That is not the question here. The question has been defined above.
> > >
> > You are doing those things just to obfuscate,
> No, I am doing these things in order to show you problems of set theory. No O can leave the matrix. The X act like Leonidas at the Thermopylae.
>
> Regards, WM
no, you are doing it to obfuscate because have proven that it works.

And even, EVEN, if we assume that your retarded shite was even REMOTELY correct, which it isn't, guess what? IT DOES NOT FUCKING MATTER!

There are INFINITELY many bijections, INFINITELY many surjections in both directions, all of which shows they must be of same cardinality.

Get over it

On Tuesday, 25 January 2022 at 04:35:44 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Montag, 24. Januar 2022 um 22:39:24 UTC+1:
> > On Monday, 24 January 2022 at 13:03:53 UTC-4, WM wrote:
>
> > > > > There is only one relevant question here: Is there any mistake in
> > > > >
> > > > > 1,2,4,7,OOOOO...
> > > > > 3,5,8,OOOOOO...
> > > > > 6,OOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > 9,OOOOOOOO...
> > > > > ...
> > > All indexes k from 9 on are attached to integer fractions n/1 with n = k.
> > You are trying to mimic Cantor's counting of the positive rational numbers, and you think the numbers in the top row, starting from 2, represent integers?
> No, they represent the unit fractions 1/1, 1/2, 1/3, 1/4, ... and are indexed by the shown integers.

My bad. I misread the crucial line that you subsequently snipped:

> No, above you see the eighth term of the sequence of configurations. All indexes k from 9 on are attached to integer fractions n/1 with n = k.

OK. That and five bucks will get you a cup of coffee. Note that there still are aleph_0 X's in the first column, and there will be at every finite step. You still have no clue how infinity works.

On Tuesday, 25 January 2022 at 06:18:55 UTC-4, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 25. Januar 2022 um 06:19:57 UTC+1:
> > måndag 24 januari 2022 kl. 18:00:52 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Montag, 24. Januar 2022 um 12:19:46 UTC+1:
> > > > måndag 24 januari 2022 kl. 10:52:13 UTC+1 skrev WM:
> > >
> > > > > Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
> > > > > 1,2,4,7,OOOOO...
> > > > > 3,5,8,OOOOOO...
> > > > > 6,OOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > OOOOOOOOO...
> > > > > 9,OOOOOOOO...
> > > > > ...
> > > > >
> > > > > reproducing the first eight terms of the sequence (k(m/n)) where
> > > > >
> > > > > k = (m + n - 1)(m + n - 2)/2 + m ?
> > > > >
> > > > Care to actually bother not fucking obfuscating it?
> > > >
> > > > We have a bijection, it is done. Get over it already
> > > That is not the question here. The question has been defined above.
> > >
> > You are doing those things just to obfuscate,
> No, I am doing these things in order to show you problems of set theory. No O can leave the matrix. [...]

Not at any finite step! However, for every position that has an O there does exist a finite step (given by the Cantor function) at which the O gets exchanged for an X. You have no FUCKING CLUE what infinity actually means.

horand....@gmail.com schrieb am Dienstag, 25. Januar 2022 um 13:58:50 UTC+1:
> On Tuesday, 25 January 2022 at 06:18:55 UTC-4, WM wrote:

> > No, I am doing these things in order to show you problems of set theory. No O can leave the matrix. [...]
>
> Not at any finite step!

There are no other ways to leave.

> However, for every position that has an O there does exist a finite step (given by the Cantor function) at which the O gets exchanged for an X.

That does not help to leave the battlefield. Never any O can leave. That means all infinitely many O's remain, indicating not indexed fractions. But no such fraction can be found. This proges that there are not findable fractions.

> You have no FUCKING CLUE what infinity actually means.

I am not interested in those clues. I use logic.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 25. Januar 2022 um 12:41:26 UTC+1:
> tisdag 25 januari 2022 kl. 11:18:55 UTC+1 skrev WM:

> > > > > > Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
> > > > > > 1,2,4,7,OOOOO...
> > > > > > 3,5,8,OOOOOO...
> > > > > > 6,OOOOOOOO...
> > > > > > OOOOOOOOO...
> > > > > > OOOOOOOOO...
> > > > > > OOOOOOOOO...
> > > > > > OOOOOOOOO...
> > > > > > OOOOOOOOO...
> > > > > > 9,OOOOOOOO...
> > > > > > ...
> > > > > >
> > > > > > reproducing the first eight terms of the sequence (k(m/n)) where
> > > > > >
> > > > > > k = (m + n - 1)(m + n - 2)/2 + m ?

> And even, EVEN, if we assume that your retarded shite was even REMOTELY correct,

if not, you could find a mistake

> which it isn't, guess what? IT DOES NOT FUCKING MATTER!

That is not my aim. I will only show that the O's cannot leave, either in any finite step nor later.
>
> There are INFINITELY many bijections, INFINITELY many surjections in both directions, all of which shows they must be of same cardinality.

All identified elements are in bijections. The fractions covered by O's cannot be identified. There is only one explanation.

Regards, WM

On 1/25/2022 11:35 AM, WM wrote:
> horand....@gmail.com schrieb am Dienstag, 25. Januar 2022 um 13:58:50 UTC+1:
>> On Tuesday, 25 January 2022 at 06:18:55 UTC-4, WM wrote:
>
>>> No, I am doing these things in order to show you problems of set theory. No O can leave the matrix. [...]
>>
>> Not at any finite step!
>
> There are no other ways to leave.

as usual, your construct is flawed, you change the matrix set at every step

>
>> However, for every position that has an O there does exist a finite step (given by the Cantor function) at which the O gets exchanged for an X.
>
> That does not help to leave the battlefield. Never any O can leave. That means all infinitely many O's remain, indicating not indexed fractions. But no such fraction can be found. This proges that there are not findable fractions.

you have shown that your math fails. you change the matrix set in every step.

Cantor Ennumeration does not change the matrix at all.

>
>> You have no FUCKING CLUE what infinity actually means.
>
> I am not interested in those clues. I use logic.

no. you use deception.

>
> Regards, WM

On 1/25/2022 11:40 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 25. Januar 2022 um 12:41:26 UTC+1:
>> tisdag 25 januari 2022 kl. 11:18:55 UTC+1 skrev WM:
>
>>>>>>> Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
>>>>>>> 1,2,4,7,OOOOO...
>>>>>>> 3,5,8,OOOOOO...
>>>>>>> 6,OOOOOOOO...
>>>>>>> OOOOOOOOO...
>>>>>>> OOOOOOOOO...
>>>>>>> OOOOOOOOO...
>>>>>>> OOOOOOOOO...
>>>>>>> OOOOOOOOO...
>>>>>>> 9,OOOOOOOO...
>>>>>>> ...
>>>>>>>
>>>>>>> reproducing the first eight terms of the sequence (k(m/n)) where
>>>>>>>
>>>>>>> k = (m + n - 1)(m + n - 2)/2 + m ?
>
>> And even, EVEN, if we assume that your retarded shite was even REMOTELY correct,
>
> if not, you could find a mistake
>
>> which it isn't, guess what? IT DOES NOT FUCKING MATTER!
>
> That is not my aim. I will only show that the O's cannot leave, either in any finite step nor later.

all Os are covered by natural numbers. Name one that is not.

>>
>> There are INFINITELY many bijections, INFINITELY many surjections in both directions, all of which shows they must be of same cardinality.
>
> All identified elements are in bijections. The fractions covered by O's cannot be identified. There is only one explanation.

and that explanation is; you make mistakes

>
> Regards, WM

On 1/25/2022 3:42 AM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 25. Januar 2022 um 01:36:58 UTC+1:
>> On 1/23/2022 2:12 PM, WM wrote:

>>> Whatever you try, the O's will not disappear,
>>> neither during no after the aprocess of indexing.
>>
>> Whatever you want to call it, "disappearing", or
>> "explosion", or "matheology", the X's and O's on n/d
>> can be covered by the X's from k/1 alone,
>> where k = (n+d-1)*(n+d-2)/2 + n
>
> No.
>
>> Let X or O be on n/d.
>>
>> Is there an X on k/1 to cover it?
>>
>> Yes.
>> k = (n+d-1)*(n+d-2)/2 + n
>
> Nevertheless all O's remain.

We know that no O remains in ℕ⁺⨯ℕ⁺
For each O, we know where in ℕ⁺⨯{1}
the X comes from that replaces it.
k = (n+d-1)*(n+d-2)/2 + n

Yes,
ℕ⁺⨯{1} is a proper subset of ℕ⁺⨯ℕ⁺

However,
the rule that a set's elements cannot match
to a proper subset only applies to sets and subsets
which are two-ended-stepping-order-able.

> Therefore your argument shows that
> too less fractions can be indeXed.
>
>>> I can prove from the blocked drain that never
>>> an O will disappear.
>>
>> Keep the drain blocked.
>> No X's or O's are mysteriously located in the dark.
>
> When the indexing has been "completed",
> all O's remain and all are in the dark.

There is no last O. There is no last X.

Nevertheless, we can describe each O and each X,
and take reliable statement-steps from the description.
Our description and statement-steps are _complete_
when we make them.

No last O, no last X, but no O and no X is left out of
the description, none drops out at any reliable
statement-step. They are complete.

There is no dark in ℕ⁺⨯ℕ⁺ or in ℕ⁺⨯{1}

Define
ℕ⁺ = ⋃{{1,...,k}} union all {1,...,k}

Anything in ℕ⁺ ends some {1,...,k}
a collection beginning at 1
in which each cut is a step,
each step is a count,
and each successor has a successor.

>> Your "proof" refers to two-ended-stepping-orderable
>
> My proof refers to the fact that no X will let pass
> an O --- and nothing else.

No O in ℕ⁺⨯ℕ⁺ passes out of ℕ⁺⨯{1}

k = (n+d-1)*(n+d-2)/2 + n
takes the O at 1/2 and transfers it to 2/1 ∈ ℕ⁺⨯{1}
From 2/1 it's transferred to 3/1 ∈ ℕ⁺⨯{1}
From 3/1 it's transferred to 6/1 ∈ ℕ⁺⨯{1}
....

The O that starts at 1/2 follows this path
1/2, 2/1, 3/1, 6/1, 21/1, 231/1, 26796/1 ...

For each step in the path, the steps _up to_
have a two-ended stepping-order.

However,
none of the steps in the path is the last step
in the path == There is no last step.

Each step k/1 in the path can kick the O that started
at 1/2 further down the path to k'/1
where k' = (k+1-1)*(k+1-2)/2 + k
There is always further down the path to kick it to,
and it always lands inside ℕ⁺⨯{1}

> It is like Leonidas at the Thermopyles.

300 Spartans can be two-ended-ly stepping-ordered.
ℕ⁺⨯{1} can't be.

Imagine King Canute commanding the ocean ℕ⁺⨯{1}
to NOT rise and cover all of ℕ⁺⨯ℕ⁺
The ocean rises.

On 1/25/2022 2:42 AM, WM wrote:
> Jim Burns schrieb am Dienstag, 25. Januar 2022 um 01:36:58 UTC+1:
>> On 1/23/2022 2:12 PM, WM wrote:
>
>>> Whatever you try, the O's will not disappear,
>>> neither during no after the aprocess of indexing.
>> Whatever you want to call it, "disappearing", or
>> "explosion", or "matheology", the X's and O's on n/d
>> can be covered by the X's from k/1 alone,
>> where k = (n+d-1)*(n+d-2)/2 + n
>
> No.
>>
>>
>> Let X or O be on n/d.
>>
>> Is there an X on k/1 to cover it?
>>
>> Yes.
>> k = (n+d-1)*(n+d-2)/2 + n
>
> Nevertheless all O's remain. Therefore your argument shows that too less fractions can be indeXed.

no. All Os are indexed. This is simple 2D array to 1D array indexing.

>
>>> I can prove from the blocked drain that never
>>> an O will disappear.
>> Keep the drain blocked.
>> No X's or O's are mysteriously located in the dark.
>
> When the indexing has been "completed", all O's remain and all are in the dark.

nope.

>>
>> Your "proof" refers to two-ended-stepping-orderable
>
> My proof refers to the fact that no X will let pass an O --- and nothing else. It is like Leonidas at the Thermopyles.

your proof is a spoof. Leonidas was overwhelmed by the Xerxes

>
> Regards, WM

On Tuesday, 25 January 2022 at 13:35:30 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Dienstag, 25. Januar 2022 um 13:58:50 UTC+1:
> > On Tuesday, 25 January 2022 at 06:18:55 UTC-4, WM wrote:
>
> > > No, I am doing these things in order to show you problems of set theory. No O can leave the matrix. [...]
> >
> > Not at any finite step!
> There are no other ways to leave.
> > However, for every position that has an O there does exist a finite step (given by the Cantor function) at which the O gets exchanged for an X.
> That does not help to leave the battlefield. Never any O can leave. That means all infinitely many O's remain, indicating not indexed fractions. But no such fraction can be found. This proges that there are not findable fractions.
> > You have no FUCKING CLUE what infinity actually means.
> I am not interested in those clues. I use logic.

In addition to not having a FUCKING clue about infinity, you have no FUCKING clue about logic. Every spot that originally is occupied by an 'O' is eventually covered with an 'X'. EVERY FUCKING spot! Provably. It is an infinite process, and even though you cannot zip along it step by step an expect to finish it, it has a fucking *LIMIT*, and that limit is all X's. But limits is another thing you have no FUCKING CLUE about.

tisdag 25 januari 2022 kl. 18:40:37 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 25. Januar 2022 um 12:41:26 UTC+1:
> > tisdag 25 januari 2022 kl. 11:18:55 UTC+1 skrev WM:
>
> > > > > > > Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
> > > > > > > 1,2,4,7,OOOOO...
> > > > > > > 3,5,8,OOOOOO...
> > > > > > > 6,OOOOOOOO...
> > > > > > > OOOOOOOOO...
> > > > > > > OOOOOOOOO...
> > > > > > > OOOOOOOOO...
> > > > > > > OOOOOOOOO...
> > > > > > > OOOOOOOOO...
> > > > > > > 9,OOOOOOOO...
> > > > > > > ...
> > > > > > >
> > > > > > > reproducing the first eight terms of the sequence (k(m/n)) where
> > > > > > >
> > > > > > > k = (m + n - 1)(m + n - 2)/2 + m ?
> > And even, EVEN, if we assume that your retarded shite was even REMOTELY correct,
> if not, you could find a mistake
> > which it isn't, guess what? IT DOES NOT FUCKING MATTER!
> That is not my aim. I will only show that the O's cannot leave, either in any finite step nor later.
> >
> > There are INFINITELY many bijections, INFINITELY many surjections in both directions, all of which shows they must be of same cardinality.
> All identified elements are in bijections. The fractions covered by O's cannot be identified. There is only one explanation.
>
> Regards, WM

>if not, you could find a mistake

I have pointed out several times, I am no longer playing your game because you are too retarded adn dishonest.

>All identified elements are in bijections. The fractions covered by O's cannot be identified. There is only one explanation.

This is an empty assertion, the bijection is for ALL of N onto ALL of Q, nothing is excluded. Get over it.

your "identified" means NOTHING in mathematics.

sergio schrieb am Dienstag, 25. Januar 2022 um 19:00:53 UTC+1:
> On 1/25/2022 11:40 AM, WM wrote:

> > That is not my aim. I will only show that the O's cannot leave, either in any finite step nor later.
> all Os are covered by natural numbers. Name one that is not.

All O's remain on the matrix. Name one that does not.

Regards, WM

Jim Burns schrieb am Dienstag, 25. Januar 2022 um 19:13:29 UTC+1:
> On 1/25/2022 3:42 AM, WM wrote:
> >> k = (n+d-1)*(n+d-2)/2 + n
> >
> > Nevertheless all O's remain.
> We know that no O remains in ℕ⁺⨯ℕ⁺

You are wrong.Name one that does not remain.

> For each O, we know where in ℕ⁺⨯{1}
> the X comes from that replaces it.

And we know that every X is replaced by an O. That does not change. Since the X's origin is the matrix, the O's remain.

We know every transposition does not change the number of O's. You claim that in the limit this would change. That is a silly claim. If it were true, everything could happen in the limit. Cantor's theore would break down nevertheless.

> k = (n+d-1)*(n+d-2)/2 + n
> Yes,
> ℕ⁺⨯{1} is a proper subset of ℕ⁺⨯ℕ⁺
>
> However,
> the rule that a set's elements cannot match
> to a proper subset only applies to sets and subsets
> which are two-ended-stepping-order-able.

Irrelevant. The rule that the first column belongs to ℕ⁺⨯ℕ⁺ is true in all cases.

> > When the indexing has been "completed",
> > all O's remain and all are in the dark.
> There is no last O. There is no last X.

But there are all O's.
>
> Nevertheless, we can describe each O and each X,
> and take reliable statement-steps from the description.
> Our description and statement-steps are _complete_
> when we make them.
>
> No last O, no last X, but no O and no X is left out of
> the description, none drops out at any reliable
> statement-step. They are complete.

There are all O's within the matrix. Otherwise a first one had to pass away..

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 06:47:53 UTC+1:

d a mistake
> I have pointed out several times, I am no longer playing your game because

you cannot solve the dilemma.

> your "identified" means NOTHING in mathematics.

It means all element which you can prove to exist in the bijection.

Regards, WM

horand....@gmail.com schrieb am Dienstag, 25. Januar 2022 um 21:40:51 UTC+1:
> Every spot that originally is occupied by an 'O' is eventually covered with an 'X'. EVERY FUCKING spot!

Wherever the X is taken from, we get an O. And that is always a matrix element. So no O can disappear. Logiciansunfortunately are too short-sighted to see this.

It is an infinite process, and even though you cannot zip along it step by step an expect to finish it

Of course not. Therefore I prove a general feature for all cases.

>, it has a fucking *LIMIT*,

It has a very impenetrable limit, namely aways an X blocking the drain.

> and that limit is all X's. But limits is another thing you have no FUCKING CLUE about.

All X's remain there, and all O's remain there too, since they cannot exit. You can only believe so counterfactually in order to avoid the insight that you have been stultified lifelong. I understand that you will not dare to see the obvious.

Regards, WM

On Wednesday, 26 January 2022 at 06:48:52 UTC-4, WM wrote:

> All X's remain there, and all O's remain there too, since they cannot exit. You can only believe so counterfactually in order to avoid the insight that you have been stultified lifelong. I understand that you will not dare to see the obvious.

Oh, I see the obvious and have pointed it out to you many times. You are too stupid to do any kind of mathematics, you are a dishonest, pompous and moronic arsehole, and you think that your bluster can make up for your fucking stupidity. You may be able to browbeat your students into regurgitating your shit on exams in exchange for grades, but don't delude yourself into thinking that that makes you a decent human being.

onsdag 26 januari 2022 kl. 11:28:20 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 06:47:53 UTC+1:
>
> d a mistake
> > I have pointed out several times, I am no longer playing your game because
> you cannot solve the dilemma.
> > your "identified" means NOTHING in mathematics.
> It means all element which you can prove to exist in the bijection.
>
> Regards, WM

>you cannot solve the dilemma.

There is nothing to solve because you present no dilemma, you make up shit, claim dilemma, but it isn't.

>It means all element which you can prove to exist in the bijection.

Which means NOTHING because in a bijection ALL elements of a set is project on ALL elements on the target set.

On 1/26/2022 4:20 AM, WM wrote:
> sergio schrieb am Dienstag, 25. Januar 2022 um 19:00:53 UTC+1:
>> On 1/25/2022 11:40 AM, WM wrote:
>
>>> That is not my aim. I will only show that the O's cannot leave, either in any finite step nor later.
>> all Os are covered by natural numbers. Name one that is not.
>
> All O's remain on the matrix. Name one that does not.
>
> Regards, WM

As you stated, the O where 1/2 was, was rejected and replaced by an X,

in accordance with

k = (m + n - 1)(m + n - 2)/2 + m

now you say it does not

which is it ?

On 1/26/2022 4:26 AM, WM wrote:
> Jim Burns schrieb am Dienstag, 25. Januar 2022 um 19:13:29 UTC+1:
>> On 1/25/2022 3:42 AM, WM wrote:
>
>>>> k = (n+d-1)*(n+d-2)/2 + n
>>>
>>> Nevertheless all O's remain.
>> We know that no O remains in ℕ⁺⨯ℕ⁺
>
> You are wrong.Name one that does not remain.

1/2, or any other fraction.

>
>> For each O, we know where in ℕ⁺⨯{1}
>> the X comes from that replaces it.
>
> And we know that every X is replaced by an O.

nope, you changed your construct ?

> That does not change. Since the X's origin is the matrix, the O's remain.
>
> We know every transposition does not change the number of O's. You claim that in the limit this would change. That is a silly claim. If it were true, everything could happen in the limit. Cantor's theore would break down nevertheless.

not at all, take Cantor's enumeration matrix, add a columb at the beginning of all zeros, proceed with Cantors diagonal threading not using your added
first columb and we arrive at Your Construct. OF course its full of zeros.

but it is far from what Cantor did, and you know it.

sergio schrieb am Mittwoch, 26. Januar 2022 um 15:04:47 UTC+1:
> On 1/26/2022 4:20 AM, WM wrote:
> > sergio schrieb am Dienstag, 25. Januar 2022 um 19:00:53 UTC+1:
> >> On 1/25/2022 11:40 AM, WM wrote:
> >
> >>> That is not my aim. I will only show that the O's cannot leave, either in any finite step nor later.
> >> all Os are covered by natural numbers. Name one that is not.
> >
> > All O's remain on the matrix. Name one that does not.
> >
> As you stated, the O where 1/2 was, was rejected and replaced by an X,
>
> in accordance with
> k = (m + n - 1)(m + n - 2)/2 + m
> now you say it does not
>
>
No, I say it remain in the matrix, in the end not at any definable place however. Here you can see whete it has gone after the eighth step. We see the following picture: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 have lost their indexes:

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

As the result we find that the number of indexes X does never change, i.e., in every finite step (where it only could change) it does not. The number of O's remains constant and much larger than the number of X's. After every finite step! And beyond nothing can happen anymore.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 14:54:35 UTC+1:
> onsdag 26 januari 2022 kl. 11:28:20 UTC+1 skrev WM:

> There is nothing to solve because you present no dilemma, you make up shit, claim dilemma, but it isn't.
> >It means all element which you can prove to exist in the bijection.
> Which means NOTHING because in a bijection ALL elements of a set is project on ALL elements on the target set.

Nevertheless all O's remain there. After eight steps for instance we see the following picture: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 have lost their indexes:

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

As the result we find that the number of indexes X does never change, i.e., in every finite step (where it only could change) it does not. The number of O's remains constant and much larger than the number of X's. After every finite step! And beyond nothing can happen anymore.

Regards, WM

horand....@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 12:28:11 UTC+1:
> On Wednesday, 26 January 2022 at 06:48:52 UTC-4, WM wrote:
>
> > All X's remain there, and all O's remain there too, since they cannot exit. You can only believe so counterfactually in order to avoid the insight that you have been stultified lifelong. I understand that you will not dare to see the obvious.
> Oh, I see the obvious and have pointed it out to you many times.

Do you see this?

Every index is moved to the place which Cantor has prescribed: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, ... . At this place there appears an X and where the index has been taken from there appears an O. After eight steps for instance we see the following picture: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 have lost their indexes:

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

As the result we find that the number of indexes X does never change, i.e., in every finite step (where it only could change) it does not. The number of O's remains constant and much larger than the number of X's. After every finite step! And beyond nothing can happen anymore.

Regards, WM

On 1/26/2022 11:05 AM, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 12:28:11 UTC+1:
>> On Wednesday, 26 January 2022 at 06:48:52 UTC-4, WM wrote:
>>
>>> All X's remain there, and all O's remain there too, since they cannot exit. You can only believe so counterfactually in order to avoid the insight that you have been stultified lifelong. I understand that you will not dare to see the obvious.
>> Oh, I see the obvious and have pointed it out to you many times.
>
> Do you see this?
>
> Every index is moved to the place which Cantor has prescribed: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, ... . At this place there appears an X and where the index has been taken from there appears an O. After eight steps for instance we see the following picture: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 have lost their indexes:
>
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
>
> As the result we find that the number of indexes X does never change, i.e., in every finite step (where it only could change) it does not. The number of O's remains constant and much larger than the number of X's. After every finite step! And beyond nothing can happen anymore.
>
> Regards, WM
>

that is wrong.

Add a row to the top, like I did below, and continue doing it your way, and you can see My added row is never lost, replaced, covered, changed, and
always has OOOOO... The number of O's remains huge. After every finite step! And beyond nothing can happen anymore.that will never go away...

OOOOOOOOO...
XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

The Great Sergio has show his supreamnessism in modifying matrix and misapplying Cantor Enumeration just like WM