On 10/30/2023 12:52 PM, WM wrote:
> Fritz Feldhase schrieb am Montag, 30. Oktober 2023 um 20:37:10 UTC+1:
>> On Monday, October 30, 2023 at 8:33:59 PM UTC+1, WM wrote:
>>> Fritz Feldhase schrieb am Montag, 30. Oktober 2023 um 20:30:20 UTC+1:
>>>>
>>>> B is NOT "smaller than" A. Both are IN x IN matrices (i.e. have the same "size").
>>>>
>>>> But the set of elements of B is a proper subset of the set of elements of A. I guess that's what you MEAN.
>>>>
>>> What I mean is that A contains B as a submatrix
>> Please define /submatrix/, then show/prove your claim. Than you.
>
> The submatrix covers the upper left part of the matrix.
>>
>> Hint: Both - A and B - are IN x IN matrices (i.e. have the same "size" or "format", etc.).
>
> For the visible part that is true. Both, A and B, contain as indices i and j the potentially infinite collection ℕ_def. But A contains more dark indices because A contains positions (m, n) inhabited by fractions which cannot be determined as individuals.

Huh? Fractions that cannot be determined as individuals? I have some
advise for you:

https://youtu.be/uoghPts44Ng

Listen to it.

On 10/30/2023 2:51 PM, WM wrote:
> Jim Burns schrieb am Montag,
> 30. Oktober 2023 um 19:25:38 UTC+1:

>> If _both_
>> for each element in A,
>> there is a unique element in B
>> _and_
>> for each element in B,
>> there is a unique element in A,
>> then
>> B is the same "size" as A
>> (Cantor-Schröder-Bernstein)

> B is smaller than A

....despite there being,
by arithmetic,
for each element in A,
a unique element in B.

> B is smaller than A because,
> according to the rules,

| Push the natnumbers of the first column
| (without queue) into the field of fractions
| and store the hit fraction always there
| where the natnumber has come from.
| Try to push the natnumbers such that
| all matrix positions are occupied by them.
| Date: Thu, 19 Oct 2023 06:04:10 -0700 (PDT)

> never a fraction is sitting at
> an indexed position.

If
each natnumber of the first column
is pushed into the field of fractions,
then
at each fraction p/q
there is a natnumber n
n = p+(s-1)(s-2)/2
s = p+q
and,
each natnumber n
is at a fraction p/q
p = n-(s-1)(s-2)/2
q = s-p
s = ⌈½+(2⋅n+¼)¹ᐟ²⌉

Because arithmetic.

Your claim is that,
if not all positions are indexed,
then not all positions are indexed.

And that, therefore, arithmetic fails.

https://www.researchgate.net/figure/Then-a-Miracle-Occurs-Copyrighted-artwork-by-Sydney-Harris-Inc-All-materials-used-with_fig2_302632920
"then a miracle occurs"

> But all the fractions remain within the matrix.
>
> Can't you understand the rules published in the OP?
> Or don't you accept them?
> Why?

Because arithmetic.
n = p+(p+q-1)(p+q-2)/2
p+q = ⌈½+(2⋅n+¼)¹ᐟ²⌉

Chris M. Thomasson schrieb am Montag, 30. Oktober 2023 um 21:36:31 UTC+1:
> On 10/30/2023 12:33 PM, WM wrote:

> > What I mean is that A contains B as a submatrix but A contains also fractions which sit at places not belonging to B because they are not indexed, contrary to all places of B.
> Cantor pairing indexes them.

Look into the rules. Fractions never sit at an indexed place.

Regards, WM

Jim Burns schrieb am Montag, 30. Oktober 2023 um 21:59:21 UTC+1:
> On 10/30/2023 2:51 PM, WM wrote:
> > Jim Burns schrieb am Montag,
> > 30. Oktober 2023 um 19:25:38 UTC+1:
> >> If _both_
> >> for each element in A,
> >> there is a unique element in B
> >> _and_
> >> for each element in B,
> >> there is a unique element in A,
> >> then
> >> B is the same "size" as A
> >> (Cantor-Schröder-Bernstein)
> > B is smaller than A
> ...despite there being,
> by arithmetic,
> for each element in A,
> a unique element in B.
> > B is smaller than A because,
> > according to the rules,
> | Push the natnumbers of the first column
> | (without queue) into the field of fractions
> | and store the hit fraction always there
> | where the natnumber has come from.
> | Try to push the natnumbers such that
> | all matrix positions are occupied by them.
> |
> Date: Thu, 19 Oct 2023 06:04:10 -0700 (PDT)
> > never a fraction is sitting at
> > an indexed position.
> If
> each natnumber of the first column
> is pushed into the field of fractions,
> then
> at each fraction p/q
> there is a natnumber n

It indexes the place (i, j) where p/q was sitting until it wandered to the place the natnumber came from.

> Your claim is that,
> if not all positions are indexed,
> then not all positions are indexed.

Since not all positions are indexed, not all positions are indexed.

> > Can't you understand the rules published in the OP?
> > Or don't you accept them?
> > Why?
> Because arithmetic.
> n = p+(p+q-1)(p+q-2)/2

n = i+(i+j-1)(i+j-2)/2.

p/q escapes. Look into the rules. Fractions never sit at an indexed place.

Regards, WM

On 10/30/2023 2:36 PM, WM wrote:
> Jim Burns schrieb am Montag, 30. Oktober 2023 um 21:59:21 UTC+1:
>> On 10/30/2023 2:51 PM, WM wrote:
>>> Jim Burns schrieb am Montag,
>>> 30. Oktober 2023 um 19:25:38 UTC+1:
>>>> If _both_
>>>> for each element in A,
>>>> there is a unique element in B
>>>> _and_
>>>> for each element in B,
>>>> there is a unique element in A,
>>>> then
>>>> B is the same "size" as A
>>>> (Cantor-Schröder-Bernstein)
>>> B is smaller than A
>> ...despite there being,
>> by arithmetic,
>> for each element in A,
>> a unique element in B.
>>> B is smaller than A because,
>>> according to the rules,
>> | Push the natnumbers of the first column
>> | (without queue) into the field of fractions
>> | and store the hit fraction always there
>> | where the natnumber has come from.
>> | Try to push the natnumbers such that
>> | all matrix positions are occupied by them.
>> |
>> Date: Thu, 19 Oct 2023 06:04:10 -0700 (PDT)
>>> never a fraction is sitting at
>>> an indexed position.
>> If
>> each natnumber of the first column
>> is pushed into the field of fractions,
>> then
>> at each fraction p/q
>> there is a natnumber n
>
> It indexes the place (i, j) where p/q was sitting until it wandered to the place the natnumber came from.

Huh? You can map back and forth from natural number index to pair, and
from pair back to natural index. Nothing is lost. You just need to study up.

;^o

>
>> Your claim is that,
>> if not all positions are indexed,
>> then not all positions are indexed.
>
> Since not all positions are indexed, not all positions are indexed.
>
>>> Can't you understand the rules published in the OP?
>>> Or don't you accept them?
>>> Why?
>> Because arithmetic.
>> n = p+(p+q-1)(p+q-2)/2
>
> n = i+(i+j-1)(i+j-2)/2.
>
> p/q escapes. Look into the rules. Fractions never sit at an indexed place.
>
> Regards, WM

On Monday, October 30, 2023 at 3:48:17 PM UTC-3, WM wrote:
> William schrieb am Montag, 30. Oktober 2023 um 18:10:00 UTC+1:
> > On Monday, October 30, 2023 at 8:35:17 AM UTC-3, WM wrote:
> > > William schrieb am Montag, 30. Oktober 2023 um 03:28:33 UTC+1:
> > > > On Sunday, October 29, 2023 at 5:52:08 PM UTC-3, WM wrote:
> > > > > A ordered pair is "indexed" at every element of the matrix B
> > > > > Many ordered pairs (m, n) defining positions of the matrix are indexed.
> > > > Actually, all ordered pairs.
> > > Not those occupied by fractions
> > ??? All positions in B are occupied by natural numbers

So there is a bijection between the positions of B (the elements of |Nx|N) and the natural numbers |N.

On 10/30/2023 2:30 PM, WM wrote:
> Chris M. Thomasson schrieb am Montag, 30. Oktober 2023 um 21:36:31 UTC+1:
>> On 10/30/2023 12:33 PM, WM wrote:
>
>>> What I mean is that A contains B as a submatrix but A contains also fractions which sit at places not belonging to B because they are not indexed, contrary to all places of B.
>> Cantor pairing indexes them.
>
> Look into the rules. Fractions never sit at an indexed place.

You are not familiar with Cantor pairing. That is your fault, not mine.

On Monday, October 30, 2023 at 4:30:20 PM UTC-3, Fritz Feldhase wrote:
> On Monday, October 30, 2023 at 7:48:17 PM UTC+1, WM wrote:
>
> > but B is smaller than A

There is no such thing as matrix A. At every step n in |N we have matrix A(n). B is not a submatrix of A(n) for any n.

On 10/30/2023 2:30 PM, WM wrote:
> Chris M. Thomasson schrieb am Montag, 30. Oktober 2023 um 21:36:31 UTC+1:
>> On 10/30/2023 12:33 PM, WM wrote:
>
>>> What I mean is that A contains B as a submatrix but A contains also fractions which sit at places not belonging to B because they are not indexed, contrary to all places of B.
>> Cantor pairing indexes them.
>
> Look into the rules. Fractions never sit at an indexed place.

Are you thick or something?

https://groups.google.com/g/sci.math/c/TujbFEXtH2w/m/R4Vyt5rICQAJ

WM expressed precisely :
> Fritz Feldhase schrieb am Montag, 30. Oktober 2023 um 20:37:10 UTC+1:
>> On Monday, October 30, 2023 at 8:33:59 PM UTC+1, WM wrote:
>>> Fritz Feldhase schrieb am Montag, 30. Oktober 2023 um 20:30:20 UTC+1:
>>>>
>>>> B is NOT "smaller than" A. Both are IN x IN matrices (i.e. have the same
>>>> "size").
>>>>
>>>> But the set of elements of B is a proper subset of the set of elements of
>>>> A. I guess that's what you MEAN.
>>>>
>>> What I mean is that A contains B as a submatrix
>> Please define /submatrix/, then show/prove your claim. Than you.
>
> The submatrix covers the upper left part of the matrix.

So it is finite then?

On 10/30/2023 5:36 PM, WM wrote:
> Jim Burns schrieb am Montag,
> 30. Oktober 2023 um 21:59:21 UTC+1:
>> On 10/30/2023 2:51 PM, WM wrote:

>>> never a fraction is sitting at
>>> an indexed position.
>>
>> If
>> each natnumber of the first column
>> is pushed into the field of fractions,
>> then
>> at each fraction p/q
>> there is a natnumber n
>
> It indexes the place (i, j) where
> p/q was sitting until it wandered to
> the place the natnumber came from.

The place the natnumber n came from
has rows i and columns j in
minimally-each-accessible sets.
Also p/q wanders _from_ a place which
has rows i and columns j in
the same minimally-each-accessible sets.

There is closure of the arithmetic operations
within a minimally-each-accessible set.

It is NOT closure within any FINITE set.

>> Your claim is that,
>> if not all positions are indexed,
>> then not all positions are indexed.
>
> Since not all positions are indexed,
> not all positions are indexed.

Why did you snip this?
>> And that, therefore, arithmetic fails.

Aren't you trying to argue that
arithmetic fails?

>>> Can't you understand
>>> the rules published in the OP?
>>> Or don't you accept them?
>>> Why?
>>
>> Because arithmetic.
>> n = p+(p+q-1)(p+q-2)/2
>
> n = i+(i+j-1)(i+j-2)/2.
>
> p/q escapes.

p/q only wanders within
the minimally-each-accessible rows and columns.

And.

After all indexing,
p/q is not among
the minimally-each-accessible rows and columns.

How?
By the first column being the same "size" as
the whole matrix superset.

This is the point of
Cantor's fraction-sequence,
Disappearing Bob,
Hilbert's Hotel,
etc etc etc:
Infinite is different from finite,
not merely tremendously larger.

> Look into the rules.

n = p+(p+q-1)(p+q-2)/2
p+q = ⌈½+(2⋅n+¼)¹ᐟ²⌉

> Fractions never sit at an indexed place.

If
any fraction sits at a place among
the minimally-each-accessible rows and columns,
then
not all places among
the minimally-each-accessible rows and columns
are indexed.
F ⇒ ¬I

I ⇒ ¬F is
true wherever F ⇒ ¬I is true and
false wherever F ⇒ ¬I is false.

Therefore,
I ⇒ ¬F
If
all places among
the minimally-each-accessible rows and columns
are indexed,
then
no fraction sits at a place among
the minimally-each-accessible rows and columns.

Jim Burns schrieb am Dienstag, 31. Oktober 2023 um 04:29:43 UTC+1:
> On 10/30/2023 5:36 PM, WM wrote:

> This is the point of
> Cantor's fraction-sequence,
> Disappearing Bob,

and related nonsense. Not of interest in mathematics.

But in the OP you see a game which requires disappearing fractions. Do you claim it too?

Regards, WM

William schrieb am Montag, 30. Oktober 2023 um 22:51:21 UTC+1:

> So there is a bijection between the positions of B (the elements of |Nx|N) and the natural numbers |N.

There is no bijection between the positions of A
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
....
and ℕ. But that is claimed by Cantor.

Regards, WM

William schrieb am Montag, 30. Oktober 2023 um 22:56:41 UTC+1:
> On Monday, October 30, 2023 at 4:30:20 PM UTC-3, Fritz Feldhase wrote:
> > On Monday, October 30, 2023 at 7:48:17 PM UTC+1, WM wrote:
> >
> > > but B is smaller than A
> There is no such thing as matrix A. At every step n in |N we have matrix A(n).

At every step we have no fraction indeXed, but every fraction is remaining in A(n) for every n.
Indexing can only happen such that one index is applied at every step. There are no steps reaching B.

>B is not a submatrix of A(n) for any n.

B is a submatrix of A.
If there is no matrix A, then not all indeXes can be applied. Then Cantor also fails.

Regards, WM

FromTheRafters schrieb am Dienstag, 31. Oktober 2023 um 02:15:51 UTC+1:
> WM expressed precisely :

> > The submatrix covers the upper left part of the matrix.
> So it is finite then?

It is "always finite". It is not fixed finite. That means it is potentially infinite.

Regards, WM

William schrieb am Montag, 30. Oktober 2023 um 22:51:21 UTC+1:

> So there is a bijection between the positions of B (the elements of |Nx|N) and the natural numbers |N.

There is no bijection between the positions of any A(n) or A
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
....
and ℕ. But that is claimed by Cantor.

Regards, WM

On 10/31/2023 5:20 AM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 31. Oktober 2023 um 04:29:43 UTC+1:

>> This is the point of
>> Cantor's fraction-sequence,
>> Disappearing Bob,
>
> and related nonsense.

From ℕ×ℕ to ℕ and back again:
n = p+(p+q-1)(p+q-2)/2
p+q = ⌈½+(2⋅n+¼)¹ᐟ²⌉

If you're claiming arithmetic is nonsense,
why do you snip my referring to that claim?

If you're embarrassed by that claim,
consider no longer claiming it.

> Not of interest in mathematics.

Opinions differ (disturbingly) on whether
arithmetic is of interest in mathematics.

> But in the OP you see a game which
> requires disappearing fractions.
> Do you claim it too?

I claim that arithmetic is correct.

From the correctness of arithmetic,
the correctness of
one "size" for both ℕ×ℕ and ℕ
becomes necessary.
n = p+(p+q-1)(p+q-2)/2
p+q = ⌈½+(2⋅n+¼)¹ᐟ²⌉

I solved the slope and area problem rigorously, not Newton, not Leibniz, not Cauchy or any other mainstream mathematics tosser.

I know the answers. They do not!

More here:

https://www.academia.edu/105576431/The_Holy_Grail_of_Calculus

On Tuesday, October 31, 2023 at 6:30:11 AM UTC-3, WM wrote:

> B is a submatrix of A.
There is no matrix A.

On Tuesday, October 31, 2023 at 6:31:39 AM UTC-3, WM wrote:

> It is "always finite". It is not fixed finite. That means it is potentially infinite.

B does not change. A "potentially infinite matrix" is nonsense.

On Tuesday, October 31, 2023 at 6:40:09 AM UTC-3, WM wrote:
> William schrieb am Montag, 30. Oktober 2023 um 22:51:21 UTC+1:
>
> > So there is a bijection between the positions of B (the elements of |Nx|N) and the natural numbers |N.
> There is no bijection between the positions of any A(n) [and |N]

Note A(n) has the same positions for each n. The fact the A(n) does not represent the bijection between the positions and |N
does not mean the bijection does not exist

William schrieb am Dienstag, 31. Oktober 2023 um 15:41:01 UTC+1:
> On Tuesday, October 31, 2023 at 6:30:11 AM UTC-3, WM wrote:
>
> > B is a submatrix of A.
> There is no matrix A.

Possible, but then not all indexes are applied, contrary to Cantor's affirmation.

Regards, WM

William schrieb am Dienstag, 31. Oktober 2023 um 15:56:03 UTC+1:
> On Tuesday, October 31, 2023 at 6:40:09 AM UTC-3, WM wrote:
> > William schrieb am Montag, 30. Oktober 2023 um 22:51:21 UTC+1:
> >
> > > So there is a bijection between the positions of B (the elements of |Nx|N) and the natural numbers |N.
> > There is no bijection between the positions of any A(n) [and |N]
>
> Note A(n) has the same positions for each n.

A(n) is the matrix after the nth push. It has all positions which A(1), A(2), ..., and A have. Of course. But in A no fractions are visible. That implies that also in A(1) not all positions are visible.

> The fact the A(n) does not represent the bijection between the positions and |N
> does not mean the bijection does not exist.

But it means that the bijection is not created by Cantor's approach.

Regards, WM

On Tuesday, October 31, 2023 at 12:37:32 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 31. Oktober 2023 um 15:41:01 UTC+1:
> > On Tuesday, October 31, 2023 at 6:30:11 AM UTC-3, WM wrote:
> >
> > > B is a submatrix of A.
> > There is no matrix A.
> Possible,

certain

>but then not all indexes are applied, contrary to Cantor's affirmation.

Silly. The fact that matrix A does not exist, does not mean a matrix which represents the bijection does not exist.

Fritz Feldhase schrieb am Dienstag, 31. Oktober 2023 um 14:02:34 UTC+1:
> On Tuesday, October 31, 2023 at 10:40:09 AM UTC+1, WM wrote:
> >
> > There is no bijection between the positions [elements] of [the 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, ...
> > 5/1, 5/2, 5/3, 5/4, ...
> > ...
> > and ℕ.

> It is {(n/m, m + (m + n - 1)(m + n - 2)/2) : n,m e IN} = {(1/1, 1), (1/2, 2), (2/1, 3), ...}.

Stay with the indexing the positions (m, n). After every push the number of natnumbers k and the number fractions p/q are constant and they all are within the matrix. After all pushes they remain in the matrix A too. But the fractions are invisible and their positions are not indexed (they are never indexed according to the rules). Your claim and
> See: https://de.wikipedia.org/wiki/Cantorsche_Paarungsfunktion

are refuted.

Regards, WM