sergio schrieb am Mittwoch, 19. Januar 2022 um 20:17:46 UTC+1:
> There is no need to show where indexes come from.

But it is interesting. And it proves that the O will never change their number.

Regards, WM

On Wednesday, January 19, 2022 at 6:24:18 PM UTC-4, WM wrote:

> An infinite number of finite lines.

Correct. Each line has n o's where n is an element of N_P. Each element of N_P is finite. N_P is infinite.

> That implies that the infinity is potential.

Nope, the number of lines does not change.

--
William Hughes

On 1/19/2022 1:56 PM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:

>> n/d ---> i/1
>> moves the X's and O's within the defined 𝔽⁺
>
> Yes.
>
> Every transposition where the X_i1 moves to O_nd and
> vice versa will preserve the numbers of X's and O's.

Every set {τ} of one transposition has a
two-ended stepping-order.
{τ} preserves X's and O's.
{τ} is of the first kind.

Preserving X's and O's is a property of
collections of the first kind.

>> You want the reason to be existence of dark numbers.
>> But i = (n+d-1)*(n+d-2)/2 + n
>> Whether or not dark numbers exist, no O moves to
>> a dark number.
>
> Right.
>
>> Dark numbers don't explain how all of
>> 𝔽⁺ = { n/d | n,d ∈ ⋃{{1,...,k}} }
>> fits in only its first column.
>
> They don't fit in the first column.

They fit in the first column.

Let i = (n+d-1)*(n+d-2)/2 + n

If {1,...,n} and {1,...,d} exist and so ∈ ℕ⁺
then {1,...,i} exists and so ∈ ℕ⁺
Each n/d has its own i/1

Also,
if n/d ≠ n'/d' then i ≠ i'
No i/1 has two n/d.

>> The reason that all of 𝔽⁺ can move to only its first column
>> is that 𝔽⁺ is NOT the first kind of collection,
>> the sheep-kind, the pebble-kind.
>
> Whatever.
> Even if all O's fit in the first column,
> they stay within the matrix.
>
>> You:
>> Look! 𝔽⁺ does not behave like the first kind of collection.
>
> I am not interested in the kind of collection.
> I am only interested in the fact that no O can leave,
> and hence not all fractions can be indeXed.

Some collections cannot match to a proper subset or a proper
superset. For those collections, it's meaningful to say
no O can leave. Those are collections of the first kind.

You don't realize it, but you _are_ interested in the
kind of collection, because not all collections are
of the first kind. For the second kind it, is not
meaningful to say no O can leave. By using that language,
you claim the collection of fractions is of the first
kind. It isn't.

On 1/19/2022 4:26 PM, WM wrote:
> sergio schrieb am Mittwoch, 19. Januar 2022 um 20:17:46 UTC+1:
>> There is no need to show where indexes come from.
>
> But it is interesting. And it proves that the O will never change their number.
>
> Regards, WM
>

nope, it only shows you intentionally constructed the matrix wrong, to show your pseudoisms.

On 1/19/2022 4:24 PM, WM wrote:
> William schrieb am Mittwoch, 19. Januar 2022 um 21:24:37 UTC+1:
>> On Wednesday, January 19, 2022 at 2:32:18 PM UTC-4, WM wrote:
>>> William schrieb am Mittwoch, 19. Januar 2022 um 16:06:05 UTC+1:
>>>> On Wednesday, January 19, 2022 at 5:25:44 AM UTC-4, WM wrote:
>>>>> William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
>>>>>> On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
>>>>>
>>>>>>>>> But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
>>>>>>>>>
>>>>>>>>> o
>>>>>>>>> oo
>>>>>>>>> ooo
>>>>>>>>> ...
>>>>>>>>>
>>>>>>>>> than are o's avaible.
>>>>>>>> Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
>>>>>>> FISONs have less than aleph_0 o's.
>>>>>> Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.
>>>>> OK, if you believe in infinite FISONs,
>>>> Nope. I do not believe in an infinite FISON. Each line is finite.
>>> Each line is the union of all preceding lines + o.
>> Correct. Does not, nothing you say can, change the fact that there are an infinite number of lines.
>
> An infinite number of finite lines. That implies that the infinity is potential.

nope. there are 2 dimensions to it, one is finite, the other is infinite.

>
> Regards, WM

onsdag 19 januari 2022 kl. 19:17:36 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 19. Januar 2022 um 12:31:53 UTC+1:
> > onsdag 19 januari 2022 kl. 10:38:59 UTC+1 skrev WM:
> > > Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
> > > > On 1/18/2022 2:26 PM, WM wrote:
> > > > > Jim Burns schrieb
> > > > > am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
> > > > >> Without ending somewhere, they're not the first kind.
> > > > >
> > > > > But without ending all O's will remain.
> > > > If the O on n/d remains,
> > > Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
> > > > the replacement has ended
> > > > before step i = (n+d-1)*(n+d-2)/2 + n
> > > No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
> > >
> > > Look, I did nothing else but ask where Cantor's indexes come from. Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!
> > >
> > Again, your re-arranging is entirely fuckign IRRELEVANT
> Why do you think so?
>
> Of course nobody need be interested in my model. But for those beein interested I can state that my model applies Cantor's first diagonal arguments meticulously. Every fraction is indexed as he did. The only difference is that I ask where the indexes come from.
>
> Since only transpositions are applied, always exchanging two items, an index and a void card at *finite places*, but never creating indices, it is clear that not all fractions can be indexed. And there is absolutely no mistake in my model because it is so simple. Everybody not being prejudiced by inertia and othodox belief, should easily be able to understand this most important correction of mathematics since Hippasos.
>
> Regards, WM

it is irrelevant because the goal is finding A bijection, not that ALL functions are bijections.

We have found one, ergo, it is done.

Your re-arrangement to find a non-bijection is entirely irrelevant then.

Jim Burns schrieb am Donnerstag, 20. Januar 2022 um 00:32:55 UTC+1:
> On 1/19/2022 1:56 PM, WM wrote:
> > Jim Burns schrieb
> > am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:
>
> >> n/d ---> i/1
> >> moves the X's and O's within the defined 𝔽⁺
> >
> > Yes.
> >
> > Every transposition where the X_i1 moves to O_nd and
> > vice versa will preserve the numbers of X's and O's.
>
> Every set {τ} of one transposition has a
> two-ended stepping-order.

Every actually infinite set either will become completed or not. If it has been completed, there is an end. If it does never become completed, we are always in the first order.

> Preserving X's and O's is a property of
> collections of the first kind.

As long as the drain is blocked, no O can leave. Afterwards no O can leave.

> > They don't fit in the first column.
>
> They fit in the first column.

Here is another picture of the same facts. Consider the infinite matrix

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
....

and clear all columns with exception of the first (mark and delete them):

1/1, _, _, _, ...
2/1, _, _, _, ...
3/1, _, _, _, ...
4/1, _, _, _, ...
....

Now try to fill the whole matrix with the remaining fractions. That is what Cantor promises to do.

If you rotate the first column by 90° into the first line, you will fail. But if you start to fill the uper left-hand corner, then you will succeed? No, it is only owed to your short-sightedness. Because you cannot see the end of the first line and of any line or column.
>
> Let i = (n+d-1)*(n+d-2)/2 + n
>
> If {1,...,n} and {1,...,d} exist and so ∈ ℕ⁺
> then {1,...,i} exists and so ∈ ℕ⁺
> Each n/d has its own i/1

As I have shown in severeal ways (blocked drain, or the above geometric proof) this claim is wrong. Therefore an intelligent person has to consider an explanation - not to bring up his destroyed hope over and over again.
>
> > Whatever.
> > Even if all O's fit in the first column,
> > they stay within the matrix.

That is already true by the fact that every transposition occurs between finite places - in infinity.

> Some collections cannot match to a proper subset or a proper
> superset. For those collections, it's meaningful to say
> no O can leave. Those are collections of the first kind.

If you claim that other collections can transpose between places with infinite coordinates, then you are outside of mathematics. If not, your second kind does not help.
>
> You don't realize it, but you _are_ interested in the
> kind of collection, because not all collections are
> of the first kind. For the second kind it, is not
> meaningful to say no O can leave. By using that language,
> you claim the collection of fractions is of the first
> kind. It isn't.

If you claim that other collections can transpose between places with infinite coordinates, then you are outside of mathematics. Do you really claim so? Or do you claim that no O ever will land at a place with infinite coordinates, but all O's will land outside of the matrix?

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 06:36:35 UTC+1:
> onsdag 19 januari 2022 kl. 19:17:36 UTC+1 skrev WM:

> > Since only transpositions are applied, always exchanging two items, an index and a void card at *finite places*, but never creating indices, it is clear that not all fractions can be indexed. And there is absolutely no mistake in my model because it is so simple. Everybody not being prejudiced by inertia and othodox belief, should easily be able to understand this most important correction of mathematics since Hippasos.
> >
> it is irrelevant because the goal is finding A bijection,

But you have not. This is proved by the blocked drain for unmatched fractions O.

> not that ALL functions are bijections.

If there is a bihection, then no transposition can destroy it.
>
> We have found one, ergo, it is done.

You have not.
>
> Your re-arrangement to find a non-bijection is entirely irrelevant then.

A contradiction is very relevant.

Here is another picture of the same facts. Consider the infinite matrix

1/1, _, _, _, ...
2/1, _, _, _, ...
3/1, _, _, _, ...
4/1, _, _, _, ...
....

and try to fill the whole matrix *including the first column* with the fractions. It is obviously foolish, only to consider this attempt.

Regards, WM

WM presented the following explanation :
> zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 06:36:35 UTC+1:
>> onsdag 19 januari 2022 kl. 19:17:36 UTC+1 skrev WM:
>
>>> Since only transpositions are applied, always exchanging two items, an
>>> index and a void card at *finite places*, but never creating indices, it is
>>> clear that not all fractions can be indexed. And there is absolutely no
>>> mistake in my model because it is so simple. Everybody not being prejudiced
>>> by inertia and othodox belief, should easily be able to understand this
>>> most important correction of mathematics since Hippasos.
>>>
>> it is irrelevant because the goal is finding A bijection,
>
> But you have not. This is proved by the blocked drain for unmatched fractions
> O.
>
>> not that ALL functions are bijections.
>
> If there is a bihection, then no transposition can destroy it.
>>
>> We have found one, ergo, it is done.
>
> You have not.
>>
>> Your re-arrangement to find a non-bijection is entirely irrelevant then.
>
> A contradiction is very relevant.
>
> Here is another picture of the same facts. Consider the infinite matrix
>
> 1/1, _, _, _, ...
> 2/1, _, _, _, ...
> 3/1, _, _, _, ...
> 4/1, _, _, _, ...
> ...
>
> and try to fill the whole matrix *including the first column* with the
> fractions. It is obviously foolish, only to consider this attempt.

Indeed, so why do you insist upon doing it this wrong way. Stsrt at the
corner and serpentine. There is nothing to stop that 'process' so every
element will 'eventually' be 'hit'.

WM presented the following explanation :
> Jim Burns schrieb am Donnerstag, 20. Januar 2022 um 00:32:55 UTC+1:
>> On 1/19/2022 1:56 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:
>>>> n/d ---> i/1
>>>> moves the X's and O's within the defined ?⁺
>>>
>>> Yes.
>>>
>>> Every transposition where the X_i1 moves to O_nd and
>>> vice versa will preserve the numbers of X's and O's.
>>
>> Every set {τ} of one transposition has a
>> two-ended stepping-order.
>
> Every actually infinite set either will become completed or not.

Stop lying. Sets don't "become" anything.

torsdag 20 januari 2022 kl. 11:21:31 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 06:36:35 UTC+1:
> > onsdag 19 januari 2022 kl. 19:17:36 UTC+1 skrev WM:
>
> > > Since only transpositions are applied, always exchanging two items, an index and a void card at *finite places*, but never creating indices, it is clear that not all fractions can be indexed. And there is absolutely no mistake in my model because it is so simple. Everybody not being prejudiced by inertia and othodox belief, should easily be able to understand this most important correction of mathematics since Hippasos.
> > >
> > it is irrelevant because the goal is finding A bijection,
> But you have not. This is proved by the blocked drain for unmatched fractions O.
> > not that ALL functions are bijections.
> If there is a bihection, then no transposition can destroy it.
> >
> > We have found one, ergo, it is done.
> You have not.
> >
> > Your re-arrangement to find a non-bijection is entirely irrelevant then..
> A contradiction is very relevant.
> Here is another picture of the same facts. Consider the infinite matrix
> 1/1, _, _, _, ...
> 2/1, _, _, _, ...
> 3/1, _, _, _, ...
> 4/1, _, _, _, ...
> ...
> and try to fill the whole matrix *including the first column* with the fractions. It is obviously foolish, only to consider this attempt.
>
> Regards, WM

>But you have not. This is proved by the blocked drain for unmatched fractions O.

It proves nothing of the sort. You changing the bijection proves nothing about it. The formula is such that it will take every rational to a natural number.

>If there is a bihection, then no transposition can destroy it.

but yours isn't a transpoition, it is infinitely many and that can.

>You have not.

We have.

>A contradiction is very relevant.

Except you are doing nothing to show a contradiction

"This entirely different function doesn't work as a bijection" is not a contradiction to my function functioning.

To show it is not a bijection, you have to show that there is at least one I missed. I can show that the function hits all of them. You have failed to do this.

Fortunately for us I wouldn't even need to do that, if I can show a surjection from A to B, and a surjection from B to A, then there exists a bijection between both sets. It is trivial to make these surjections here.

On Thursday, 20 January 2022 at 00:23:05 UTC-4, sergio wrote:
> On 1/19/2022 4:26 PM, WM wrote:
> > sergio schrieb am Mittwoch, 19. Januar 2022 um 20:17:46 UTC+1:
> >> There is no need to show where indexes come from.
> >
> > But it is interesting. And it proves that the O will never change their number.
> >
> > Regards, WM
> >
> nope, it only shows you intentionally constructed the matrix wrong, to show your pseudoisms.

It doesn't even show that. All it shows is that he has no clue what it means for the left column to have an infinite number (i.e., an inexhaustible source) of X's. By going to the tic-tac-toe approach and reducing everything to X's and O's he has dumbed things down to third-grade level, and he *STILL* doesn't see his mistake. But I am sure his next move is going to mumble about how some X's are lit and others are dark and that he alone can tell the difference.

FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:28:18 UTC+1:
> WM presented the following explanation :

> > Here is another picture of the same facts. Consider the infinite matrix
> >
> > 1/1, _, _, _, ...
> > 2/1, _, _, _, ...
> > 3/1, _, _, _, ...
> > 4/1, _, _, _, ...
> > ...
> >
> > and try to fill the whole matrix *including the first column* with the
> > fractions. It is obviously foolish, only to consider this attempt.
> Indeed, so why do you insist upon doing it this wrong way. Start at the
> corner and serpentine. There is nothing to stop that 'process' so every
> element will 'eventually' be 'hit'.

There is no drain for O's either. So they remain on the matrix.

Regards, WM

FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:29:53 UTC+1:
> WM presented the following explanation :

> > Every actually infinite set either will become completed or not.
> Sets don't "become" anything.

Counting is a process which, according to matheology, can become completed. We find the terms Zuordnungsgesetz, Zuordnungsmodus, Zuordnungsprozeß, Zuordnungsverfahren, Zuordnungsverhältnis. [Cantor, pp. 239, 283f, 286f, 291, 293, 305f, 413]

Regards, WM

WM wrote on 1/20/2022 :
> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:28:18 UTC+1:
>> WM presented the following explanation :
>
>>> Here is another picture of the same facts. Consider the infinite matrix
>>>
>>> 1/1, _, _, _, ...
>>> 2/1, _, _, _, ...
>>> 3/1, _, _, _, ...
>>> 4/1, _, _, _, ...
>>> ...
>>>
>>> and try to fill the whole matrix *including the first column* with the
>>> fractions. It is obviously foolish, only to consider this attempt.
>> Indeed, so why do you insist upon doing it this wrong way. Start at the
>> corner and serpentine. There is nothing to stop that 'process' so every
>> element will 'eventually' be 'hit'.
>
> There is no drain for O's either. So they remain on the matrix.

Poor maintenance on the drains is probably the reason for your leaky
sets. Ensure the drains are properly seated before 'filling' your sets
with elements.

WM has brought this to us :
> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:29:53 UTC+1:
>> WM presented the following explanation :
>
>>> Every actually infinite set either will become completed or not.
>> Sets don't "become" anything.
>
> Counting is a process which, according to matheology, can become completed.
> We find the terms Zuordnungsgesetz, Zuordnungsmodus, Zuordnungsprozeß,
> Zuordnungsverfahren, Zuordnungsverhältnis. [Cantor, pp. 239, 283f, 286f, 291,
> 293, 305f, 413]

Who said anything about counting?

zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31 UTC+1:
> torsdag 20 januari 2022 kl. 11:21:31 UTC+1 skrev WM:

> > > Your re-arrangement to find a non-bijection is entirely irrelevant then.
> > A contradiction is very relevant.
> > Here is another picture of the same facts. Consider the infinite matrix
> > 1/1, _, _, _, ...
> > 2/1, _, _, _, ...
> > 3/1, _, _, _, ...
> > 4/1, _, _, _, ...
> > ...
> > and try to fill the whole matrix *including the first column* with the fractions. It is obviously foolish, only to consider this attempt.
> >
>
> >But you have not. This is proved by the blocked drain for unmatched fractions O.
> It proves nothing of the sort. You changing the bijection proves nothing about it.
I did not change it. I used Cantor's enumeration
k = (m + n - 1)(m + n - 2)/2 + m

> The formula is such that it will take every rational to a natural number.

The sequence of configurations is such that by transpositions the number of X's cannot increase and the number of O's cannot decrease. That is basic to all transpositions.

> >If there is a bijection, then no transposition can destroy it.

> but yours isn't a transpoition, it is infinitely many and that can.

No. Infinitely many black elements do not contain a white element.
>
> >A contradiction is very relevant.
> Except you are doing nothing to show a contradiction
> "This entirely different function doesn't work as a bijection" is not a contradiction to my function functioning.

You are wrong. I am using no entirely different function but the well known

k = (m + n - 1)(m + n - 2)/2 + m.
>
> To show it is not a bijection, you have to show that there is at least one I missed. I can show that the function hits all of them. You have failed to do this.

Dark numbers cannot be shown. But the indirect existence proof is done by the invariable number of O's.
>
> Fortunately for us I wouldn't even need to do that, if I can show a surjection from A to B, and a surjection from B to A, then there exists a bijection between both sets. It is trivial to make these surjections here.

Between the visible elements this is trivial. Nevertheless the invariable number of X's and O's shows the dark side.

You must believe:
1) For all k ∈ ℕ the number of X's and O's does not change by more than plus minus zero.
Or do you deny this?
2) The enumeration becomes ready. When this happens the number of O's is zero although there is no way where they could have gone.

Regards, WM

FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 16:09:02 UTC+1:
> WM has brought this to us :
> > FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:29:53 UTC+1:
> >> WM presented the following explanation :
> >
> >>> Every actually infinite set either will become completed or not.
> >> Sets don't "become" anything.
> >
> > Counting is a process which, according to matheology, can become completed.
> > We find the terms Zuordnungsgesetz, Zuordnungsmodus, Zuordnungsprozeß,
> > Zuordnungsverfahren, Zuordnungsverhältnis. [Cantor, pp. 239, 283f, 286f, 291,
> > 293, 305f, 413]
> Who said anything about counting?

That is our topic here. Going through the set 1, 2, 3, ... is called counting. Issuing natural numbers to fractions is counting them.

Regards, WM

FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 16:07:04 UTC+1:
> WM wrote on 1/20/2022 :
> > FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:28:18 UTC+1:
> >> WM presented the following explanation :
> >
> >>> Here is another picture of the same facts. Consider the infinite matrix
> >>>
> >>> 1/1, _, _, _, ...
> >>> 2/1, _, _, _, ...
> >>> 3/1, _, _, _, ...
> >>> 4/1, _, _, _, ...
> >>> ...
> >>>
> >>> and try to fill the whole matrix *including the first column* with the
> >>> fractions. It is obviously foolish, only to consider this attempt.
> >> Indeed, so why do you insist upon doing it this wrong way. Start at the
> >> corner and serpentine. There is nothing to stop that 'process' so every
> >> element will 'eventually' be 'hit'.
> >
> > There is no drain for O's either. So they remain on the matrix.
> Poor maintenance on the drains is probably the reason for your leaky
> sets. Ensure the drains are properly seated before 'filling' your sets
> with elements.

Here we have no chance to remove the block of X's in the first column until all X's have been issued. But then we have no further use for the drain.

That is the surprising and basic feature of matheology: No O can leave as long as they are moved. No O can leave after all moves have ceased. But matheologians are convinced, in the zero slot between these two states all O's go down the drain.

Regardes, WM

On Thursday, January 20, 2022 at 4:02:30 PM UTC+1, WM wrote:
> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:29:53 UTC+1:
> > WM presented the following explanation :
> > > Every actually infinite set either will become completed or not.
> > Sets don't "become" anything.
> Counting is a process which, according to matheology, can become completed. We find the terms Zuordnungsgesetz, Zuordnungsmodus, Zuordnungsprozeß, Zuordnungsverfahren, Zuordnungsverhältnis. [Cantor, pp. 239, 283f, 286f, 291, 293, 305f, 413]
>
> Regards, WM

Finally - that's your problem - you don't know what 'Zuordnung' means.
Wie can fix that easily enough:

Literally it means 'assignment'.
In mathematics it means 'function' or 'relation' or 'mapping'.
It doesn't mean eenie-meenie-mynie-mo.

On 1/20/2022 9:19 AM, WM wrote:
> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 16:07:04 UTC+1:
>> WM wrote on 1/20/2022 :
>>> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:28:18 UTC+1:
>>>> WM presented the following explanation :
>>>
>>>>> Here is another picture of the same facts. Consider the infinite matrix
>>>>>
>>>>> 1/1, _, _, _, ...
>>>>> 2/1, _, _, _, ...
>>>>> 3/1, _, _, _, ...
>>>>> 4/1, _, _, _, ...
>>>>> ...
>>>>>
>>>>> and try to fill the whole matrix *including the first column* with the
>>>>> fractions. It is obviously foolish, only to consider this attempt.
>>>> Indeed, so why do you insist upon doing it this wrong way. Start at the
>>>> corner and serpentine. There is nothing to stop that 'process' so every
>>>> element will 'eventually' be 'hit'.
>>>
>>> There is no drain for O's either. So they remain on the matrix.
>> Poor maintenance on the drains is probably the reason for your leaky
>> sets. Ensure the drains are properly seated before 'filling' your sets
>> with elements.
>

<snip crap >

> Regardes, WM

On 1/20/2022 9:12 AM, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31 UTC+1:
>> torsdag 20 januari 2022 kl. 11:21:31 UTC+1 skrev WM:
>
>>>> Your re-arrangement to find a non-bijection is entirely irrelevant then.
>>> A contradiction is very relevant.
>>> Here is another picture of the same facts. Consider the infinite matrix
>>> 1/1, _, _, _, ...
>>> 2/1, _, _, _, ...
>>> 3/1, _, _, _, ...
>>> 4/1, _, _, _, ...
>>> ...
>>> and try to fill the whole matrix *including the first column* with the fractions. It is obviously foolish, only to consider this attempt.
>>>
>>
>>> But you have not. This is proved by the blocked drain for unmatched fractions O.
>> It proves nothing of the sort. You changing the bijection proves nothing about it.
>
> I did not change it. I used Cantor's enumeration
> k = (m + n - 1)(m + n - 2)/2 + m

Liar.

JVR schrieb am Donnerstag, 20. Januar 2022 um 16:41:52 UTC+1:
> On Thursday, January 20, 2022 at 4:02:30 PM UTC+1, WM wrote:
> > FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:29:53 UTC+1:
> > > WM presented the following explanation :
> > > > Every actually infinite set either will become completed or not.
> > > Sets don't "become" anything.
> > Counting is a process which, according to matheology, can become completed. We find the terms Zuordnungsgesetz, Zuordnungsmodus, Zuordnungsprozeß, Zuordnungsverfahren, Zuordnungsverhältnis. [Cantor, pp. 239, 283f, 286f, 291, 293, 305f, 413]
> >
> Finally - that's your problem - you don't know what 'Zuordnung' means.
> Wie can fix that easily enough:
>
> Literally it means 'assignment'.
> In mathematics it means 'function' or 'relation' or 'mapping'.
> It doesn't mean eenie-meenie-mynie-mo.

Either you are stupid or you are a liar, but probably you are both. When Cantor said "element by element", then this means precisely one by one and one after the other. Of course the frauds of matheology have tried to pervert this simple Zuordnung into an "all at once" in order to prevent inspection because they must have known for a long time already that there is no complete mapping between infinite sets - unless all are as stupid as you.

Regards, WM

On 1/20/2022 5:15 AM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 20. Januar 2022 um 00:32:55 UTC+1:
>> On 1/19/2022 1:56 PM, WM wrote:

>>> They don't fit in the first column.
>>
>> They fit in the first column.
>
> Here is another picture of the same facts.
> Consider the infinite matrix
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> and clear all columns with exception of the first
> (mark and delete them):
>
> 1/1, _, _, _, ...
> 2/1, _, _, _, ...
> 3/1, _, _, _, ...
> 4/1, _, _, _, ...
> ...
>
> Now try to fill the whole matrix with the remaining
> fractions.

1/1 2/1 4/1 7/1 11/1 ...
3/1 5/1 8/1 12/1 ...
6/1 9/1 13/1 ...
10/1 14/1 ...
15/1 ...
....

> That is what Cantor promises to do.
>
> If you rotate the first column by 90° into the first line,
> you will fail.

Cantor did not promise that it would succeed.
The promise was for the sequence of finite strips,
as above.

> But if you start to fill the uper left-hand corner,
> then you will succeed? No,

Yes.
Eppur si muove.

> then you will succeed? No, it is only owed to your
> short-sightedness. Because you cannot see the end of
> the first line and of any line or column.

No, your argument fails.
We do not need to see the ends (which do not exist)
in order to know each location formerly holding n/d
has i/1 now.

We describe and we reason from the description.
Our conclusion is true of _whatever is described_
If we cannot see what we describe, it is true
of what we cannot see.

It seems to me that our ability to reason about
what we cannot see is the source of what Eugene Wigner
called "the unreasonable effectiveness of mathematics
in the natural sciences". Consider the vastness of
today's scientific knowledge of the universe. It is
almost entirely unseen, unseeable. It has become our
knowledge because it is agreement with the teeny, tiny
part which we can see -- agreement according to the
rules of logic and mathematics.

No, we do not need to see in order to reason.
For a demonstration, I offer you the universe.

>> Let i = (n+d-1)*(n+d-2)/2 + n
>>
>> If {1,...,n} and {1,...,d} exist and so ∈ ℕ⁺
>> then {1,...,i} exists and so ∈ ℕ⁺
>> Each n/d has its own i/1
>
> As I have shown in severeal ways (blocked drain,

Pour a glass of water into another glass.
The first glass will be empty.

Walk along a beach at low tide. Look at the ocean.
Return six hours later.
_The same ocean_ now covers the beach.

Some things are like a glass of water.
Some things are like the ocean.
Describing them incorrectly is an error.

> or the above geometric proof)

i = (n+d-1)*(n+d-2)/2 + n

> this claim is wrong.
> Therefore an intelligent person has to consider
> an explanation

Ocean.
Diagonal.
Second kind.

> Or do you claim that no O ever will land at a place
> with infinite coordinates,

That one.

If {1,...,n} and {1,...,d} exist and so ∈ ℕ⁺
then {1,...,i} exists and so ∈ ℕ⁺
Each n/d has its own i/1

Remember that I define ℕ⁺ = ⋃{{1,...,k}}

> but all O's will land outside of the matrix?

No O starting inside the matrix ℕ⁺⨯ℕ⁺ lands outside
its first column ℕ⁺⨯{1}. We can prove that.

I think the best way to say this is that all O's
do not land at all. Anywhere. Not permanently.

I think that the best way to think about the O's
not-landing and ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1} is that
there are these special collections
( ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1} are not among them),
the ones with two-ended stepping-orders, for which
the O's not-landing would need explaining.
The first column _up to_ some entry is one of these
special collections. _All_ the first column ℕ⁺⨯{1} is
provably _not_ one of these special collections.
And so on, with other collections, other non-problem
problems.

Jim Burns schrieb am Donnerstag, 20. Januar 2022 um 19:16:18 UTC+1:
> On 1/20/2022 5:15 AM, WM wrote:
> > Jim Burns schrieb
> > am Donnerstag, 20. Januar 2022 um 00:32:55 UTC+1:
> >> On 1/19/2022 1:56 PM, WM wrote:
>
> >>> They don't fit in the first column.
> >>
> >> They fit in the first column.
> >
> > Here is another picture of the same facts.
> > Consider the infinite matrix
> >
> > 1/1, 1/2, 1/3, 1/4, ...
> > 2/1, 2/2, 2/3, 2/4, ...
> > 3/1, 3/2, 3/3, 3/4, ...
> > 4/1, 4/2, 4/3, 4/4, ...
> > ...
> >
> > and clear all columns with exception of the first
> > (mark and delete them):
> >
> > 1/1, _, _, _, ...
> > 2/1, _, _, _, ...
> > 3/1, _, _, _, ...
> > 4/1, _, _, _, ...
> > ...
> >
> > Now try to fill the whole matrix with the remaining
> > fractions.
> 1/1 2/1 4/1 7/1 11/1 ...
> 3/1 5/1 8/1 12/1 ...
> 6/1 9/1 13/1 ...
> 10/1 14/1 ...
> 15/1 ...
> ...
> > That is what Cantor promises to do.
> >
> > If you rotate the first column by 90° into the first line,
> > you will fail.
> Cantor did not promise that it would succeed.
> The promise was for the sequence of finite strips,
> as above.

But the number of fractions in the first column is in no way sufficient to solve the geometric problem which does not depend on the order of issued X's. Cantor and also his followers have never considered the geometrical side of the problem.

> > But if you start to fill the uper left-hand corner,
> > then you will succeed? No,
> Yes.
> Eppur si muove.

No. Note that the other columns to be filled are precisely as long as the first column.

> > then you will succeed? No, it is only owed to your
> > short-sightedness. Because you cannot see the end of
> > the first line and of any line or column.
> No, your argument fails.
> We do not need to see the ends (which do not exist)

But you claim that you get finished.
> in order to know each location formerly holding n/d
> has i/1 now.

That is irrelevant in geometry. Another question: Do you believe that you could succeed with not marked X's? Co you want to divorce geometry and set theory?
>
> I think the best way to say this is that all O's
> do not land at all. Anywhere. Not permanently.

Fine. Then we can be sure that they remain in the matrix.
>
> I think that the best way to think about the O's
> not-landing and ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1} is that
> there are these special collections
> ( ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1} are not among them),
> the ones with two-ended stepping-orders, for which
> the O's not-landing would need explaining.

Sorry, I have not got what you mean, but it appears interesting.

> The first column _up to_ some entry is one of these
> special collections. _All_ the first column ℕ⁺⨯{1} is
> provably _not_ one of these special collections.

But you do not use all the first column: "We do not need to see the ends (which do not exist)". Therefore it is irrelevant.
Regards, WM