On 1/26/2022 11:04 AM, WM wrote:
> 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

again, you stop on the way to infinity

When one does a 2D array to 1D array conversion, which this is ( matrix to natural numbers), one does not lose any elements at all. They are in one to
one mapping.

"As the result we find that the number of indexes X does never change" => it is ℵ0, the number of entries in the matrix is ℵ0

"The number of O's remains constant" => constant at ℵ0

"And beyond nothing can happen anymore." see ? you stopped.

What disproves you, and your conclusions, is the fact that you cannot name any fraction in the matrix of rationals, that has not been enumerated.

On 1/26/2022 11:04 AM, WM wrote:
> 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

no. you cannot name a fraction that is not enumerated.

On 1/26/2022 5:26 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.

The O that starts at 1/2 in the matrix
does not remain in ℕ⁺⨯ℕ⁺
I name it _Bob_
https://www.youtube.com/watch?v=TjAg-8qqR3g&t=45s

The O from 1/3 is named Kevin.
The O from 2/2 is named Stuart.
They also don't remain in ℕ⁺⨯ℕ⁺
There are more not remaining in ℕ⁺⨯ℕ⁺

>> 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.

Right.
Each O swaps with a different X.
So there can't be more O's than X's.

> That does not change.
> Since the X's origin is the matrix, the O's remain.

No.

A single swap or a two-ended-stepping-order-able
collection of swaps happen within a two-ended-
stepping-order-able collection.
In this case, what is swapped is conserved.
Consider sheep and pebbles.

However.

Consider Bob, the O that starts at 1/2.

Bob swaps with the X at 2/1
From 2/1, Bob swaps with the X at 3/1
within two-ended-stepping-order-able {1,...,3}⨯{1}
Then with the X at 6/1
within two-ended-stepping-order-able {1,...,6}⨯{1}
Then with the X at 21/1
within two-ended-stepping-order-able {1,...,21}⨯{1}
Then with the X at 231/1
within two-ended-stepping-order-able {1,...,231}⨯{1}
Then with the X at 26796/1
within two-ended-stepping-order-able {1,...,26796}⨯{1}
....

Each swap leaves Bob in some {1,...,k}⨯{1}
and thus in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}

But, after each swap, there is another swap,
a swap which leaves Bob out of some {1,...,j}⨯{1}

From 2/1, Bob swaps with the X at 3/1
left out of {1,...,2}⨯{1}
Then with the X at 6/1
left out of {1,...,3}⨯{1}
Then with the X at 21/1
left out of {1,...,6}⨯{1}
Then with the X at 231/1
left out of {1,...,21}⨯{1}
Then with the X at 26796/1
left out of {1,...,231}⨯{1}
Then with the X at 359026206/1
left out of {1,...,26796}⨯{1}
....

Where is Bob after _all_ the swaps?

Not in {1,...,2}⨯{1}
Not in {1,...,3}⨯{1}
Not in {1,...,6}⨯{1}
Not in {1,...,21}⨯{1}
Not in {1,...,231}⨯{1}
Not in {1,...,26796}⨯{1}
....

After _all_ the swaps,
Bob is out of _all_ the {1,...,k}⨯{1}
But ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
Bob is out of ℕ⁺⨯{1}
after _all_ the swaps.

| We have _local_ Bob-conservation in each {1,...,k}⨯{1}
| We do not have _global_ Bob-conservation in ℕ⁺⨯{1}

> We know every transposition does not change the number
> of O's.

Yes. Local Bob-conservation.

> You claim that in the limit this would change.

I claim that some claims are reliably true for collections
which are two-ended-stepping-order-able and are NOT reliably
true for collections which are NOT two-ended-stepping-order-
-able.

So, roughly speaking, yes.
The details are important. They're how I prove my claim.

> That is a silly claim.
> If it were true, everything could happen in the limit.
> Cantor's theore would break down nevertheless.

Not everything can happen "in the limit".[1]

We can describe one of infinitely-many (or one of
a non-two-ended-stepping-order-able collection) and
take only reliable statement-steps from the description.

In this way, we can learn what happens and does not happen
"in the limit". Not everything happens.

[1]
In my opinion,
thinking of infinite sets as having this extra
X-doubling O-disappearing property
gets the situation exactly backwards.

It's finite sets which have an extra property which
conserves X's and O's. And sheep.

Compare the finite/infinite distinction for sets
to the continuous/discontinuous distinction for functions.

Continuous functions are functions and also meet the
additional requirements to be continuous.
Finite sets are sets and also meet the additional
requirements to be finite.

Continuous functions are useful in many situations
we concern ourselves with. It makes perfect sense to
pay special attention to them. That does not make
discontinuous functions somehow illogical.

Finite sets are useful in many situations
we concern ourselves with. It makes perfect sense to
pay special attention to them. That does not make
infinite sets somehow illogical.

On Wednesday, 26 January 2022 at 13:05:16 UTC-4, 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.

Not "after every finite step", you imbecile.*AT* every finite step. Your use of natural language is telling in its deception (mostly of yourself, since everybody else has figured out your dirty tricks). The whole point is that at every finite step there still are aleph_0 further steps to go. Your stupidity is infuriating!

On 1/26/2022 12:42 PM, Jim Burns wrote:
> On 1/26/2022 5:26 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.
>
> The O that starts at 1/2 in the matrix
> does not remain in ℕ⁺⨯ℕ⁺
> I name it _Bob_
> https://www.youtube.com/watch?v=TjAg-8qqR3g&t=45s
>
> The O from 1/3 is named Kevin.
> The O from 2/2 is named Stuart.
> They also don't remain in ℕ⁺⨯ℕ⁺
> There are more not remaining in ℕ⁺⨯ℕ⁺
>
>>> 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.
>
> Right.
> Each O swaps with a different X.
> So there can't be more O's than X's.
>
>> That does not change.
>> Since the X's origin is the matrix, the O's remain.
>
> No.
>
> A single swap or a two-ended-stepping-order-able
> collection of swaps happen within a two-ended-
> stepping-order-able collection.
> In this case, what is swapped is conserved.
> Consider sheep and pebbles.
>
> However.
>
> Consider Bob, the O that starts at 1/2.
>
> Bob swaps with the X at 2/1
> From 2/1, Bob swaps with the X at 3/1
> within two-ended-stepping-order-able {1,...,3}⨯{1}
> Then with the X at 6/1
> within two-ended-stepping-order-able {1,...,6}⨯{1}
> Then with the X at 21/1
> within two-ended-stepping-order-able {1,...,21}⨯{1}
> Then with the X at 231/1
> within two-ended-stepping-order-able {1,...,231}⨯{1}
> Then with the X at 26796/1
> within two-ended-stepping-order-able {1,...,26796}⨯{1}
> ...
>
> Each swap leaves Bob in some {1,...,k}⨯{1}
> and thus in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
>
> But, after each swap, there is another swap,
> a swap which leaves Bob out of some {1,...,j}⨯{1}
>
> From 2/1, Bob swaps with the X at 3/1
> left out of {1,...,2}⨯{1}
> Then with the X at 6/1
> left out of {1,...,3}⨯{1}
> Then with the X at 21/1
> left out of {1,...,6}⨯{1}
> Then with the X at 231/1
> left out of {1,...,21}⨯{1}
> Then with the X at 26796/1
> left out of {1,...,231}⨯{1}
> Then with the X at 359026206/1
> left out of {1,...,26796}⨯{1}
> ...
>
> Where is Bob after _all_ the swaps?
>
> Not in {1,...,2}⨯{1}
> Not in {1,...,3}⨯{1}
> Not in {1,...,6}⨯{1}
> Not in {1,...,21}⨯{1}
> Not in {1,...,231}⨯{1}
> Not in {1,...,26796}⨯{1}
> ...
>
> After _all_ the swaps,
> Bob is out of _all_ the {1,...,k}⨯{1}
> But ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
> Bob is out of ℕ⁺⨯{1}
> after _all_ the swaps.
>
> | We have _local_ Bob-conservation in each {1,...,k}⨯{1}
> | We do not have _global_ Bob-conservation in ℕ⁺⨯{1}
>
>> We know every transposition does not change the number
>> of O's.
>
> Yes. Local Bob-conservation.
>
>> You claim that in the limit this would change.
>
> I claim that some claims are reliably true for collections
> which are two-ended-stepping-order-able and are NOT reliably
> true for collections which are NOT two-ended-stepping-order-
> -able.
>
> So, roughly speaking, yes.
> The details are important. They're how I prove my claim.
>
>> That is a silly claim.
>> If it were true, everything could happen in the limit.
>> Cantor's theore would break down nevertheless.
>
> Not everything can happen "in the limit".[1]
>
> We can describe one of infinitely-many (or one of
> a non-two-ended-stepping-order-able collection) and
> take only reliable statement-steps from the description.
>
> In this way, we can learn what happens and does not happen
> "in the limit". Not everything happens.
>
> [1]
> In my opinion,
> thinking of infinite sets as having this extra
> X-doubling O-disappearing property
> gets the situation exactly backwards.
>
> It's finite sets which have an extra property which
> conserves X's and O's. And sheep.
>
> Compare the finite/infinite distinction for sets
> to the continuous/discontinuous distinction for functions.
>
> Continuous functions are functions and also meet the
> additional requirements to be continuous.
> Finite sets are sets and also meet the additional
> requirements to be finite.
>
> Continuous functions are useful in many situations
> we concern ourselves with. It makes perfect sense to
> pay special attention to them. That does not make
> discontinuous functions somehow illogical.
>
> Finite sets are useful in many situations
> we concern ourselves with. It makes perfect sense to
> pay special attention to them. That does not make
> infinite sets somehow illogical.
>

agree, to me they are separate branches of math, with some crossover.
Such as continuous analog sound is sampled into discontinuous functions and into sets of numbers, an mp3 file etc...

Cantors enumeriation is one of the simplest 2D matrix to 1D array conversions there is.

those that say it is false, are stuck in 11th grade math and cannot pass.

WM has brought this to us :
> 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:

That's what happens when you try to unclog your matrix drain with the
garbage disposal on.

onsdag 26 januari 2022 kl. 18:05:00 UTC+1 skrev 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
ALL are in the bijection, why is this so fucking hard for you to get?

There is no FINITE steps, there are no steps PERIOD!

sergio schrieb am Mittwoch, 26. Januar 2022 um 18:15:52 UTC+1:
> On 1/26/2022 11:05 AM, WM wrote:

> > Every index is moved to the place which Cantor has prescribed:

> that is wrong.

Where is the first deviation?
>
> 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 O's covering fractions are what matters.

Regards, WM

sergio schrieb am Mittwoch, 26. Januar 2022 um 19:26:13 UTC+1:
> On 1/26/2022 11:04 AM, WM wrote:
> "As the result we find that the number of indexes X does never change" => it is ℵ0, the number of entries in the matrix is ℵ0
>
> "The number of O's remains constant" => constant at ℵ0
>
> "And beyond nothing can happen anymore." see ? you stopped.

No, I considered only the possible claim by matheologians that there was a beyond.
> What disproves you, and your conclusions, is the fact that you cannot name any fraction in the matrix of rationals, that has not been enumerated.

This does not disprove the persistence of all O's in case of actual infinity. Therefore either actual infinity is disproved or at least the existence of dark numbers is proved.

Regards, WM

Jim Burns schrieb am Mittwoch, 26. Januar 2022 um 19:42:39 UTC+1:
> On 1/26/2022 5:26 AM, WM wrote:

> > Name one that does not remain.
> The O that starts at 1/2 in the matrix
> does not remain in ℕ⁺⨯ℕ⁺

That is a claim.

> I name it _Bob_

Where and when does Bob leave? What is his last place in the matrix?

> https://www.youtube.com/watch?v=TjAg-8qqR3g&t=45s
>
> The O from 1/3 is named Kevin.
> The O from 2/2 is named Stuart.
> They also don't remain in ℕ⁺⨯ℕ⁺
> There are more not remaining in ℕ⁺⨯ℕ⁺

First answer my question.

> > Since the X's origin is the matrix, the O's remain.
> No.

Then name a transposition where an O disappears.
>
> Consider Bob, the O that starts at 1/2.
>
> Bob swaps with the X at 2/1
> From 2/1, Bob swaps with the X at 3/1
> within two-ended-stepping-order-able {1,...,3}⨯{1}
> Then with the X at 6/1
> within two-ended-stepping-order-able {1,...,6}⨯{1}
> Then with the X at 21/1
> within two-ended-stepping-order-able {1,...,21}⨯{1}
> Then with the X at 231/1
> within two-ended-stepping-order-able {1,...,231}⨯{1}
> Then with the X at 26796/1
> within two-ended-stepping-order-able {1,...,26796}⨯{1}
> ...
>
> Each swap leaves Bob in some {1,...,k}⨯{1}
> and thus in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
>
> But, after each swap, there is another swap,
> a swap which leaves Bob out of some {1,...,j}⨯{1}

No swap removes him from the matrix. But you must believe this counterfactually, countermathematically, counterlogically.
>
> Where is Bob after _all_ the swaps?
>
> Not in {1,...,2}⨯{1}
> Not in {1,...,3}⨯{1}
> Not in {1,...,6}⨯{1}
> Not in {1,...,21}⨯{1}
> Not in {1,...,231}⨯{1}
> Not in {1,...,26796}⨯{1}
> ...
>
Not at a definable place, but not outside of the battlefield. That is proved here.

> Bob is out of ℕ⁺⨯{1}
> after _all_ the swaps.

No. That would imply a transposition that eliminates Bob. Where could he go? And how could he defeat Leonidas?
>
> > You claim that in the limit this would change.
> I claim that some claims are reliably true for collections
> which are two-ended-stepping-order-able and are NOT reliably
> true for collections which are NOT two-ended-stepping-order-
> -able.

The permanent existence of Bob is true in every case. In order to believe in Cantor, you have to invent countermathematical claims.
>
> So, roughly speaking, yes.
> The details are important. They're how I prove my claim.

You do not prove this claim because applying mathematics and logic proves the contrary.

> > That is a silly claim.
> > If it were true, everything could happen in the limit.
> > Cantor's theore would break down nevertheless.
> Not everything can happen "in the limit".[1]

In particular it is impossible that aleph_0 O's disappear in the limit.
>
> We can describe one of infinitely-many (or one of
> a non-two-ended-stepping-order-able collection) and
> take only reliable statement-steps from the description.
>
> In this way, we can learn what happens and does not happen
> "in the limit". Not everything happens.

Nothng happens. There is no limit of definable elements. Between them and the limit there are dark elements.
>
> [1]
> In my opinion,
> thinking of infinite sets as having this extra
> X-doubling O-disappearing property
> gets the situation exactly backwards.

No, it gets it from Cantor's results which have started from finite sets. According to Cantor,
there is never an obstacle or halt in this process of assignment. That is quite different from a limit.
>
> It's finite sets which have an extra property which
> conserves X's and O's. And sheep.

It's logic and mathematics which conserves X's and O's, because there is no exist and no first leave.
>
> Compare the finite/infinite distinction for sets
> to the continuous/discontinuous distinction for functions.
>
> Continuous functions are functions and also meet the
> additional requirements to be continuous.
> Finite sets are sets and also meet the additional
> requirements to be finite.

Mathematics sets meet the additional requirement to obey mathematical constraints.
>
> Finite sets are useful in many situations
> we concern ourselves with. It makes perfect sense to
> pay special attention to them. That does not make
> infinite sets somehow illogical.

No. But what you require them to accomplish makes them illogical. No O can leave, neither after a finite nor after an infinite number of steps.

Regards, WM

horand....@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 21:56:05 UTC+1:
> On Wednesday, 26 January 2022 at 13:05:16 UTC-4, WM wrote:

> > 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.

> Not "after every finite step",*AT* every finite step.

At every finite step no O exits.

> The whole point is that at every finite step there still are aleph_0 further steps to go.

Of course. And at no finite step an O leaves. Therefore never an O leaves. Never all fractions are indexed.

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 09:37:51 UTC+1:
> onsdag 26 januari 2022 kl. 18:05:00 UTC+1 skrev WM:

> > 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.
> >
> ALL are in the bijection, why is this so fucking hard for you to get?

What is the frist term when an O exists? Where does it go then?
>
> There is no FINITE steps, there are no steps PERIOD!

But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exist?

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 09:37:51 UTC+1:
> onsdag 26 januari 2022 kl. 18:05:00 UTC+1 skrev WM:

> > 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.
> >
> ALL are in the bijection, why is this so fucking hard for you to get?

What is the frist term when an O exits? Where does it go then?
>
> There is no FINITE steps, there are no steps PERIOD!

But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?

Regards, WM

On Thursday, 27 January 2022 at 07:11:03 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 21:56:05 UTC+1:
> > On Wednesday, 26 January 2022 at 13:05:16 UTC-4, WM wrote:
>
> > > 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.
> > Not "after every finite step",*AT* every finite step.
>
> At every finite step no O exits.

And nobody ever claimed that they do. You play a funny version of tic-tac-toe in that you shift the symbols around instead of starting from an empty square, but it does not matter. The board is filled in with X's one step at a time. This process has infinitely many steps, and it is no big surprise that after finitely many steps infinitely many O's remain. I state for the Kth time: You have no clue what infinity actually means.

> > The whole point is that at every finite step there still are aleph_0 further steps to go.
> Of course. And at no finite step an O leaves. Therefore never an O leaves.. Never all fractions are indexed.

Ah, no. No reciprocal 1/n is ever zero. In fact, to every reciprocal there exist aleph_0 further reciprocals in the interval (1/n), and yet the limit of the sequence of (1/n) as n -> oo *IS* zero. *WHEN* do you think *THAT* happens?

I repeat myself for the K+1st time: You have no clue what infinity actually means.

With your little game of tic-tac-toe you have a limit, too, and that limit leaves the entire infinite square covered in X's, even though at every finite point there are still countably many O's on the board. That last point is not overly surprising, since at every finite point there are also countably many X's left to move. The O's do not leave through the drain, because that, like your brain, is clogged with your shit.

In conclusion, and for the K+2nd time: You have no clue what infinity actually means.

torsdag 27 januari 2022 kl. 11:45:54 UTC+1 skrev WM:
> sergio schrieb am Mittwoch, 26. Januar 2022 um 19:26:13 UTC+1:
> > On 1/26/2022 11:04 AM, WM wrote:
> > "As the result we find that the number of indexes X does never change" => it is ℵ0, the number of entries in the matrix is ℵ0
> >
> > "The number of O's remains constant" => constant at ℵ0
> >
> > "And beyond nothing can happen anymore." see ? you stopped.
> No, I considered only the possible claim by matheologians that there was a beyond.
> > What disproves you, and your conclusions, is the fact that you cannot name any fraction in the matrix of rationals, that has not been enumerated.
> This does not disprove the persistence of all O's in case of actual infinity. Therefore either actual infinity is disproved or at least the existence of dark numbers is proved.
>
> Regards, WM

You mean mathematicians, you're a crank.

torsdag 27 januari 2022 kl. 12:32:48 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 09:37:51 UTC+1:
> > onsdag 26 januari 2022 kl. 18:05:00 UTC+1 skrev WM:
>
> > > 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.
> > >
> > ALL are in the bijection, why is this so fucking hard for you to get?
> What is the frist term when an O exits? Where does it go then?
> >
> > There is no FINITE steps, there are no steps PERIOD!
> But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
>
> Regards, WM

>What is the frist term when an O exits? Where does it go then?

Irrelevant, you are obfuscating with this and trying to confuse things.

>But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?

Same here

You are asking "Why are all natural numbers finite but not N?"

Because N contains all of them, of course it is not finite then!

horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 12:40:44 UTC+1:
> On Thursday, 27 January 2022 at 07:11:03 UTC-4, WM wrote:
> > horand....@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 21:56:05 UTC+1:
> > > On Wednesday, 26 January 2022 at 13:05:16 UTC-4, WM wrote:
> >
> > > > 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.
> > > Not "after every finite step",*AT* every finite step.
> >
> > At every finite step no O exits.
> And nobody ever claimed that they do.

You claim that all fractions lose their O.

> The board is filled in with X's one step at a time.

No, all X are present from the beginning. They are only re-ordered.

> This process has infinitely many steps, and it is no big surprise that after finitely many steps infinitely many O's remain.

There is no "after infinitely many steps". Hence there are infinitely many O's remaining always, in fact precisely those from the start.

> I state for the Kth time: You have no clue what infinity actually means.

Whatever you may understand: If any O is leaving, where does it go?

> > > The whole point is that at every finite step there still are aleph_0 further steps to go.
> > Of course. And at no finite step an O leaves. Therefore never an O leaves. Never all fractions are indexed.
> Ah, no. No reciprocal 1/n is ever zero. In fact, to every reciprocal there exist aleph_0 further reciprocals in the interval (1/n), and yet the limit of the sequence of (1/n) as n -> oo *IS* zero. *WHEN* do you think *THAT* happens?

This example shows that you have no clue about infinity. Here we have no need to get rid of infinitely many elements.
>
> I repeat myself for the K+1st time: You have no clue what infinity actually means.
>
> With your little game of tic-tac-toe you have a limit, too, and that limit leaves the entire infinite square covered in X's, even though at every finite point there are still countably many O's on the board.

There are *all* on the board. And their only exit is blocked.

> That last point is not overly surprising, since at every finite point there are also countably many X's left to move. The O's do not leave through the drain,

How do they leave? And where do they go?

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 14:24:15 UTC+1:
> torsdag 27 januari 2022 kl. 12:32:48 UTC+1 skrev WM:

> > > There is no FINITE steps, there are no steps PERIOD!
> > But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
>
> >What is the frist term when an O exits? Where does it go then?
> Irrelevant, you are obfuscating with this and trying to confuse things.

No, I am using mathematics and logic.

> >But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?

> You are asking "Why are all natural numbers finite but not N?"

No, I am asking where do the O's go?
>
> Because N contains all of them, of course it is not finite then!

That has nothing to do with my question.

Regards, WM

On 1/27/2022 5:07 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 26. Januar 2022 um 19:42:39 UTC+1:
>> On 1/26/2022 5:26 AM, WM wrote:
>
>>> Name one that does not remain.
>> The O that starts at 1/2 in the matrix
>> does not remain in ℕ⁺⨯ℕ⁺
>
> That is a claim.
>
>> I name it _Bob_
>
> Where and when does Bob leave? What is his last place in the matrix?
>
>> https://www.youtube.com/watch?v=TjAg-8qqR3g&t=45s
>>
>> The O from 1/3 is named Kevin.
>> The O from 2/2 is named Stuart.
>> They also don't remain in ℕ⁺⨯ℕ⁺
>> There are more not remaining in ℕ⁺⨯ℕ⁺
>
> First answer my question.
>
>>> Since the X's origin is the matrix, the O's remain.
>> No.
>
> Then name a transposition where an O disappears.
>>
>> Consider Bob, the O that starts at 1/2.
>>
>> Bob swaps with the X at 2/1
>> From 2/1, Bob swaps with the X at 3/1
>> within two-ended-stepping-order-able {1,...,3}⨯{1}
>> Then with the X at 6/1
>> within two-ended-stepping-order-able {1,...,6}⨯{1}
>> Then with the X at 21/1
>> within two-ended-stepping-order-able {1,...,21}⨯{1}
>> Then with the X at 231/1
>> within two-ended-stepping-order-able {1,...,231}⨯{1}
>> Then with the X at 26796/1
>> within two-ended-stepping-order-able {1,...,26796}⨯{1}
>> ...
>>
>> Each swap leaves Bob in some {1,...,k}⨯{1}
>> and thus in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
>>
>> But, after each swap, there is another swap,
>> a swap which leaves Bob out of some {1,...,j}⨯{1}
>
> No swap removes him from the matrix. But you must believe this counterfactually, countermathematically, counterlogically.
>>
>> Where is Bob after _all_ the swaps?
>>
>> Not in {1,...,2}⨯{1}
>> Not in {1,...,3}⨯{1}
>> Not in {1,...,6}⨯{1}
>> Not in {1,...,21}⨯{1}
>> Not in {1,...,231}⨯{1}
>> Not in {1,...,26796}⨯{1}
>> ...
>>
> Not at a definable place, but not outside of the battlefield. That is proved here.
>
>> Bob is out of ℕ⁺⨯{1}
>> after _all_ the swaps.
>
> No. That would imply a transposition that eliminates Bob. Where could he go? And how could he defeat Leonidas?
>>
>>> You claim that in the limit this would change.
>> I claim that some claims are reliably true for collections
>> which are two-ended-stepping-order-able and are NOT reliably
>> true for collections which are NOT two-ended-stepping-order-
>> -able.
>
> The permanent existence of Bob is true in every case. In order to believe in Cantor, you have to invent countermathematical claims.
>>
>> So, roughly speaking, yes.
>> The details are important. They're how I prove my claim.
>
> You do not prove this claim because applying mathematics and logic proves the contrary.
>
>>> That is a silly claim.
>>> If it were true, everything could happen in the limit.
>>> Cantor's theore would break down nevertheless.
>> Not everything can happen "in the limit".[1]
>
> In particular it is impossible that aleph_0 O's disappear in the limit.
>>
>> We can describe one of infinitely-many (or one of
>> a non-two-ended-stepping-order-able collection) and
>> take only reliable statement-steps from the description.
>>
>> In this way, we can learn what happens and does not happen
>> "in the limit". Not everything happens.
>
> Nothng happens. There is no limit of definable elements. Between them and the limit there are dark elements.
>>
>> [1]
>> In my opinion,
>> thinking of infinite sets as having this extra
>> X-doubling O-disappearing property
>> gets the situation exactly backwards.
>
> No, it gets it from Cantor's results which have started from finite sets. According to Cantor,
> there is never an obstacle or halt in this process of assignment. That is quite different from a limit.
>>
>> It's finite sets which have an extra property which
>> conserves X's and O's. And sheep.
>
> It's logic and mathematics which conserves X's and O's, because there is no exist and no first leave.
>>
>> Compare the finite/infinite distinction for sets
>> to the continuous/discontinuous distinction for functions.
>>
>> Continuous functions are functions and also meet the
>> additional requirements to be continuous.
>> Finite sets are sets and also meet the additional
>> requirements to be finite.
>
> Mathematics sets meet the additional requirement to obey mathematical constraints.
>>
>> Finite sets are useful in many situations
>> we concern ourselves with. It makes perfect sense to
>> pay special attention to them. That does not make
>> infinite sets somehow illogical.
>
> No. But what you require them to accomplish makes them illogical. No O can leave, neither after a finite nor after an infinite number of steps.
>
> Regards, WM

diversion, and failure.

Name a fraction in your matrix that is not included in the Enumeration, you cannot.

the rest of your smoke an mirrors cannot hide this fact.

On 1/27/2022 4:41 AM, WM wrote:
> sergio schrieb am Mittwoch, 26. Januar 2022 um 18:15:52 UTC+1:
>> On 1/26/2022 11:05 AM, WM wrote:
>
>>> Every index is moved to the place which Cantor has prescribed:
>
>> that is wrong.
>
> Where is the first deviation?

an X at the bottom of the first column.

you have double defined X , one as covering a fraction and second as covering the first column. Fail

>>
>> 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 O's covering fractions are what matters.

no, you modified the matrix with the first column, I modify it with the first row.

>
> Regards, WM

On 1/27/2022 4:45 AM, WM wrote:
> sergio schrieb am Mittwoch, 26. Januar 2022 um 19:26:13 UTC+1:
>> On 1/26/2022 11:04 AM, WM wrote:
>> "As the result we find that the number of indexes X does never change" => it is ℵ0, the number of entries in the matrix is ℵ0
>>
>> "The number of O's remains constant" => constant at ℵ0
>>
>> "And beyond nothing can happen anymore." see ? you stopped.
>
> No, I considered only the possible claim by matheologians that there was a beyond.

red herring, you stopped.

>
>> What disproves you, and your conclusions, is the fact that you cannot name any fraction in the matrix of rationals, that has not been enumerated.
>
> This does not disprove the persistence of all O's in case of actual infinity. Therefore either actual infinity is disproved or at least the existence of dark numbers is proved.

that is diversion.

Your tic-tac-toe shell game shows your negative bias for Cantor's simple proof. You go to extremes by modifying the matrix in every step, which Cantor
never does. Fail

>
> Regards, WM

On Thursday, 27 January 2022 at 09:35:50 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 12:40:44 UTC+1:
> > On Thursday, 27 January 2022 at 07:11:03 UTC-4, WM wrote:
> > > horand....@gmail.com schrieb am Mittwoch, 26. Januar 2022 um 21:56:05 UTC+1:
> > > > On Wednesday, 26 January 2022 at 13:05:16 UTC-4, WM wrote:
> > >
> > > > > 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.
> > > > Not "after every finite step",*AT* every finite step.
> > >
> > > At every finite step no O exits.
> > And nobody ever claimed that they do.
> You claim that all fractions lose their O.

Only your delusional mind would think that those two statements are the same.

> > The board is filled in with X's one step at a time.
> No, all X are present from the beginning. They are only re-ordered.

That's *NOT* true. At every step you move one X from the left column to a position somewhere else in the map. Guess what? There are still the *SAME* number of X's in the left column as when you started, namely aleph_0. Your utter stupidity and ignorance regarding infinite objects makes not the least bit of difference.

> > This process has infinitely many steps, and it is no big surprise that after finitely many steps infinitely many O's remain.
> There is no "after infinitely many steps". Hence there are infinitely many O's remaining always, in fact precisely those from the start.

Of *COURSE* there is "after infinitely many steps". That's where the limit comes in, you fucking moron.

> > I state for the Kth time: You have no clue what infinity actually means..
> Whatever you may understand: If any O is leaving, where does it go?

I invite you to stick it where the sun don't shine. With the O's it's the same as with the X's: At every finite step, there are as many of them when you started, namely aleph_0. But the *LIMIT* matrix is one of only X's.

> > > > The whole point is that at every finite step there still are aleph_0 further steps to go.
> > > Of course. And at no finite step an O leaves. Therefore never an O leaves. Never all fractions are indexed.
> > Ah, no. No reciprocal 1/n is ever zero. In fact, to every reciprocal there exist aleph_0 further reciprocals in the interval (1/n), and yet the limit of the sequence of (1/n) as n -> oo *IS* zero. *WHEN* do you think *THAT* happens?
> This example shows that you have no clue about infinity. Here we have no need to get rid of infinitely many elements.

This remark shows several things: First you are too stupid to understand analogies. Second, you are too stupid to understand infinity. Third, you are too stupid to understand limits, and fourth, you are too stupid to create or follow a mathematical argument.
> > With your little game of tic-tac-toe you have a limit, too, and that limit leaves the entire infinite square covered in X's, even though at every finite point there are still countably many O's on the board.
> There are *all* on the board. And their only exit is blocked.

You only *THINK* the exit is blocked, because that's where you dump your *SHIT*, you fucking moron.

> > That last point is not overly surprising, since at every finite point there are also countably many X's left to move. The O's do not leave through the drain,
> How do they leave? And where do they go?

They do not "leave". They are replaced by X's. Just as at every finite step there are aleph_0 X's in the matrix, there also are aleph_0 O's. That you are too fucking stupid to grasp that is really not my fault.

On 1/27/2022 6:07 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 26. Januar 2022 um 19:42:39 UTC+1:

>> Consider Bob, the O that starts at 1/2.
>>
>> Bob swaps with the X at 2/1
>> From 2/1, Bob swaps with the X at 3/1
>> within two-ended-stepping-order-able {1,...,3}⨯{1}
>> Then with the X at 6/1
>> within two-ended-stepping-order-able {1,...,6}⨯{1}
>> Then with the X at 21/1
>> within two-ended-stepping-order-able {1,...,21}⨯{1}
>> Then with the X at 231/1
>> within two-ended-stepping-order-able {1,...,231}⨯{1}
>> Then with the X at 26796/1
>> within two-ended-stepping-order-able {1,...,26796}⨯{1}
>> ...
>>
>> Each swap leaves Bob in some {1,...,k}⨯{1}
>> and thus in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
>>
>> But, after each swap, there is another swap,
>> a swap which leaves Bob out of some {1,...,j}⨯{1}
>
> No swap removes him from the matrix.

Right.
One swap moves Bob within the matrix.
Also {one swap} is a two-ended-stepping-order-able
collection of swaps.

I'll strengthen that claim.
_Each_ two-ended-stepping-order-able collection of
swaps within the matrix leaves Bob in the matrix
after its last swap.

What if the collection of swaps was NOT two-ended-
-stepping-order-able?

> But you must believe this counterfactually,
> countermathematically, counterlogically.

Sigh.
Go back and read my previous post
(for the first time, it seems).

>> Where is Bob after _all_ the swaps?
>>
>> Not in {1,...,2}⨯{1}
>> Not in {1,...,3}⨯{1}
>> Not in {1,...,6}⨯{1}
>> Not in {1,...,21}⨯{1}
>> Not in {1,...,231}⨯{1}
>> Not in {1,...,26796}⨯{1}
>> ...
>
> Not at a definable place,
> but not outside of the battlefield.

After _all_ the swaps,
Bob is out of _all_ the {1,...,k}⨯{1}
But ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
Bob is out of ℕ⁺⨯{1}
after _all_ the swaps.

> That is proved here.
>
>> Bob is out of ℕ⁺⨯{1}
>> after _all_ the swaps.
>
> No.
> That would imply a transposition that eliminates Bob.
> Where could he go? And how could he defeat Leonidas?

Each one of Bob's transpositions sets him down somewhere
in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}

But
each one of Bob's transpositions is followed by another
by another transposition which picks him up again.

_Because Bob has no last transposition_
each spot at which Bob is set down is _not_
a spot at which _all_ Bob's transpositions leave him.

Bob is not where he's been set down.
Bob is not where he's not been set down.
Bob is not in ℕ⁺⨯{1}

If your dark numbers are not in ℕ⁺⨯{1}
they are irrelevant, since my claim is that
Bob is not in ℕ⁺⨯{1}

( Although, it seems necessary to use strong magic to get
( Bob to a dark number on which he is never set.
( Whatever. Your dark numbers are about something else.

If your dark numbers are in ℕ⁺⨯{1}
we have already seen that Bob is not on any of them.

_Because Bob has no last transposition_
he can be set down on infinitely-many spots in ℕ⁺⨯{1}
and be left on none of them.

The key property is two-ended-stepping-order-ability
or its absence.

horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 17:06:10 UTC+1:
> On Thursday, 27 January 2022 at 09:35:50 UTC-4, WM wrote:

> > There are *all* on the board. And their only exit is blocked.
> You only *THINK* the exit is blocked

It is blockd by X as long as there are X. And even if not, where should the O's end?

> > How do they leave? And where do they go?
> They do not "leave". They are replaced by X's.

Pure replacing is not posible by transpositions. In my model the O appears wehre the X comes from.

> Just as at every finite step there are aleph_0 X's in the matrix, there also are aleph_0 O's.

The O's indicate not indexed fractions.

Regards, WM

horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 17:06:10 UTC+1:
> On Thursday, 27 January 2022 at 09:35:50 UTC-4, WM wrote:

> > > The board is filled in with X's one step at a time.
> > No, all X are present from the beginning. They are only re-ordered.
> That's *NOT* true. At every step you move one X from the left column to a position somewhere else in the map.

The left column belongs to the matrix. All X are present from the beginning and remain within the matrix, like the O's..

> > There is no "after infinitely many steps". Hence there are infinitely many O's remaining always, in fact precisely those from the start.
> Of *COURSE* there is "after infinitely many steps".

If you like, we can model it. Nothing changes after infinitely many steps. And before there have been infinitely many replacements, tra nspositions without any loss of X or O.

> They do not "leave". They are replaced by X's.

No O is replaced by an X. Only their positions are interchanged.

Regards, WM