On Sunday, November 19, 2023 at 9:01:28 PM UTC+1, Chris M. Thomasson wrote:

> Fwiw, its fun to use these mapped Cantor pairs as notes for some music.
>
> https://youtu.be/XkwgJt5bxKI

I've already listened to it. Sounds great!

Keep up the good work, man!

On Sunday, November 19, 2023 at 7:24:11 PM UTC+1, Fritz Feldhase wrote:
> On Sunday, November 19, 2023 at 7:10:04 PM UTC+1, Fritz Feldhase wrote:
> > On Sunday, November 19, 2023 at 11:10:30 AM UTC+1, WM wrote:
> > >
> > > Cantor [...] define[s] his matrix B.
> > >
> > Right. Actually, he defined the function f: IN x IN --> IN, (n, m) |-> (m + n - 1)(m + n - 2)/2 + m, which can be interpreted as a matrix of size "IN x IN".
> >
> > This function is a bijection from IN x IN onto IN.
> >
> Um nun die unendliche Folge zu erhalten, von der Cantor im Folgenden spricht, braucht man nur die Umkehrfunktion f^-1 von f zu betrachten. Es gilt dann:
>
> "every [fraction] p/q comes at an absolutely fixed position of a simple infinite sequence"
>
> und:
>
> "The infinite sequence thus defined has the peculiar property to contain the positive [fractions] completely, and each of them only once at a determined place."
>
> This infinite sequence starts with the following elements (in the given order):
>
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

Actually, we can "construct" this sequence explicitely (not referring to f)..

We first obeserve that the set of fractions {a/b : a,b e IN} can be "decomposed" into (infinitely many) disjoined sets (the following way):

S_n := {a/b : a,b e IN & a+b = n+1} (n e IN).

Then S_1 = {1/1}, S_2 = {1/2, 2/1}, S_3 = {1/3, 2/2, 3/1}, ...

Actually: An,m e IN: n =/= m -> S_n n S_m = { } and U_(n e IN) S_n = {a/b : a,b e IN}.

Now we define (infinitely many) disjount sets of ordered pairs (the following way):

P_n := {(n*(n+1)/2 + a, a/b) : a,b e IN & a+b = n+1} (n e IN).

Then P_1 = {(1, 1/1)}, P_2 = {(2, 1/2), (3, 2/1)}, P_3 = {(4, 1/3), (5, 2/2), (6, 3/1)}, ...

And (obviously) U{P_n : n e IN} just is the infinite sequence Cantor was talking about (above):

U{P_n : n e IN} = {(1, 1/1), (2, 1/2), (3, 2/1), (4, 1/3), (5, 2/2), (6, 3/1), ...} = (1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ...).

"The infinite sequence thus defined has the peculiar property to contain the positive fractions completely, and each of them only once at a determined place."

Fritz Feldhase schrieb:
> On Sunday, November 19, 2023 at 7:24:11 PM UTC+1, Fritz Feldhase wrote:
>> On Sunday, November 19, 2023 at 7:10:04 PM UTC+1, Fritz Feldhase wrote:
>>> On Sunday, November 19, 2023 at 11:10:30 AM UTC+1, WM wrote:
>>>>
>>>> Cantor [...] define[s] his matrix B.
>>>>
>>> Right. Actually, he defined the function f: IN x IN --> IN, (n, m) |-> (m + n - 1)(m + n - 2)/2 + m, which can be interpreted as a matrix of size "IN x IN".
>>>
>>> This function is a bijection from IN x IN onto IN.
>>>
>> Um nun die unendliche Folge zu erhalten, von der Cantor im Folgenden spricht, braucht man nur die Umkehrfunktion f^-1 von f zu betrachten. Es gilt dann:
>>
>> "every [fraction] p/q comes at an absolutely fixed position of a simple infinite sequence"
>>
>> und:
>>
>> "The infinite sequence thus defined has the peculiar property to contain the positive [fractions] completely, and each of them only once at a determined place."
>>
>> This infinite sequence starts with the following elements (in the given order):
>>
>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
>
> Actually, we can "construct" this sequence explicitely (not referring to f).
>
> We first obeserve that the set of fractions {a/b : a,b e IN} can be "decomposed" into (infinitely many) disjoined sets (the following way):
>
> S_n := {a/b : a,b e IN & a+b = n+1} (n e IN).
>
> Then S_1 = {1/1}, S_2 = {1/2, 2/1}, S_3 = {1/3, 2/2, 3/1}, ...
>
> Actually: An,m e IN: n =/= m -> S_n n S_m = { } and U_(n e IN) S_n = {a/b : a,b e IN}.
>
> Now we define (infinitely many) disjount sets of ordered pairs (the following way):
>
> P_n := {(n*(n+1)/2 + a, a/b) : a,b e IN & a+b = n+1} (n e IN).
>
> Then P_1 = {(1, 1/1)}, P_2 = {(2, 1/2), (3, 2/1)}, P_3 = {(4, 1/3), (5, 2/2), (6, 3/1)}, ...
>
> And (obviously) U{P_n : n e IN} just is the infinite sequence Cantor was talking about (above):
>
> U{P_n : n e IN} = {(1, 1/1), (2, 1/2), (3, 2/1), (4, 1/3), (5, 2/2), (6, 3/1), ...} = (1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ...).
>
> "The infinite sequence thus defined has the peculiar property to contain the positive fractions completely, and each of them only once at a determined place."
>

Cantor himself gave the following construction:

----------

"We also have

(8) aleph_0*aleph_0 = aleph_0 .

Proof. By (6) of § 3, aleph_0*aleph_0 is the cardinal number of the
aggregate of bindings

{(my,ny)},

where my and ny are any finite cardinal numbers which are independent
of one another. If also lambda represents any finite cardinal number,
so that {lambda}, {my}, and {ny} are only different notations for the
same aggregate of all finite numbers, we have to show that

{(my,ny)} ~ {lambda}.

Let us denote my + ny by rho; then rho takes all the numerical values
2, 3, 4, ..., and there are in all rho - 1 elements (my,ny) for which
my + ny = rho, namely :

(1, rho - 1), (2, rho - 2), ... (rho - 1, 1).

In this sequence imagine first the elemental (1,1), for which rho = 2,
put, then the two elements for which rho = 3, then the three elements
for which rho = 4, and so on. Thus we get all the elements (my,ny) in
a simple series:

(1,1); (1,2), (2;1); (1,3), (2,2), (3;1); (1,4), (2,3), ...,

and here, as we easily see, the element (my,ny) comes at the lambda-th
place, where

(9) lambda = my + (my + ny - 1)(my + ny - 2)/2 .

The variable lambda takes every numerical value 1, 2, 3, ..., once.
Consequently, by means of (9), a reciprocally univocal relation
subsists between the aggregates {lambda} and {(my,ny)}."
(Georg Cantor: Contributions to the founding of the theory of
transfinite numbers. Dover Publications, 1915; p.106f.)

----------

Of special interest is the last sentence:

| By means of (9), a reciprocally univocal relation
| subsists between the aggregates {lambda} and {(my,ny)}.

(In the german original of this work:
"es besteht vermöge (9) eine gegenseitig eindeutige Beziehung
zwischen den Mengen {lambda} und {(my,ny)}")

No "billiard games", no "single steps", no "supertask",
no "not indexed fractions" - but a clear bijection from N to NxN :-)

DIeter Heidorn

On 20.11.2023 10:05, Dieter Heidorn wrote:
> Fritz Feldhase schrieb:
>> On Sunday, November 19, 2023 at 7:24:11 PM UTC+1, Fritz Feldhase wrote:
>>> On Sunday, November 19, 2023 at 7:10:04 PM UTC+1, Fritz Feldhase wrote:
>>>> On Sunday, November 19, 2023 at 11:10:30 AM UTC+1, WM wrote:

>>>>
>>>> This function is a bijection from IN x IN onto IN.
>>>>
>>> Um nun die unendliche Folge zu erhalten, von der Cantor im Folgenden
>>> spricht, braucht man nur die Umkehrfunktion f^-1 von f zu betrachten.
>>> Es gilt dann:
>>>
>>> "every [fraction] p/q comes at an absolutely fixed position of a
>>> simple infinite sequence"
>>>
>>> und:
>>>
>>> "The infinite sequence thus defined has the peculiar property to
>>> contain the positive [fractions] completely, and each of them only
>>> once at a determined place."

So Cantor infers "all" from "every".
>>>
>>> This infinite sequence starts with the following elements (in the
>>> given order):
>>>
>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
>>> 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
>>
>> Actually, we can "construct" this sequence explicitely (not referring
>> to f).

in the same way we do that with matrices.
>>

> (In the german original of this work:
>    "es besteht vermöge (9) eine gegenseitig eindeutige Beziehung
>     zwischen den Mengen {lambda} und {(my,ny)}")

Just that is utulized in my proof using matrices.>
> No "billiard games", no "single steps",

We can consider single steps. But we need not. My matrices are
completely defined by few sentencesin the OP.

> no "supertask",

And what about the algebraic numbers?

> no "not indexed fractions"

All matrices A(0), A(1), A(2), ... have the same extension. If B is
produced, finally, then the fractions have not left A. They sit at not
indexed positions within A. If B is not produced but only all
approximations B(n), then the same is valid for the A(n).

Regards, WM
>

On 19.11.2023 18:20, William wrote:
> On Sunday, November 19, 2023 at 1:09:58 PM UTC-4, WM wrote:
>> it tells me that all matrices A(i) are larger than B
>
> Nope. It tells you that all the A(i) are the same "size". It does not tell you that A(0) is "larger" than B.

(0) and every A(i) is larger than the elements of B, because every A(i)
contains fractions which sit at not indexed positions, by definition of
the game.

If B is produced from its elements, then A is produced, containing all
fractions at not indexed positions. And A is as large as every A(i)
because nothing leaves the matrix.

Further this is proved by the formula valid for all indices of columns
and lines: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every index that can be defined has ℵo successors which cannot be
defined because they are there and remain there in every definable case.
They can only be covered collectively: |ℕ \ {1, 2, 3, ...}| = 0.

Regards, WM

On 19.11.2023 19:09, Fritz Feldhase wrote:
> On Sunday, November 19, 2023 at 11:10:30 AM UTC+1, WM wrote:
>
>> Cantor [...] define[s] his matrix B.

If it is constructed according to my prescription, then almost all
matrix positions are not indexed because they are inhabited by
fractions, but not by natural numbers.
>
> Right. Actually, he defined the function f: IN x IN --> IN, (n, m) |-> (m + n - 1)(m + n - 2)/2 + m), which can be interpreted as a matrix of size "IN x IN".
>
> This function is a bijection from IN x IN onto IN.

And what about the algebraic numbers? How do they get enumerated?

Regards, WM

On Monday, November 20, 2023 at 12:02:38 PM UTC+1, WM wrote:

> And what about the algebraic numbers? How do they get enumerated?

Huh?! Was haben denn die algebraischen Zahlen mit Cantors Beweis der Abzählbarkeit der (Menge der) rationalen Zahlen zu tun, Mückenheim?

Weißt Du eigentlich noch wo vorne und wo hinten ist?

On Monday, November 20, 2023 at 11:40:39 AM UTC+1, WM wrote:

> And what about the algebraic numbers?

Huh?! Was haben denn die algebraischen Zahlen mit Cantors Beweis der Abzählbarkeit der (Menge der) rationalen Zahlen zu tun, Mückenheim?

Kannst Du mal beim Thema bleiben?

On 20.11.2023 12:06, Fritz Feldhase wrote:
> On Monday, November 20, 2023 at 12:02:38 PM UTC+1, WM wrote:
>
>> And what about the algebraic numbers? How do they get enumerated?
>
> Huh?! Was haben denn die algebraischen Zahlen mit Cantors Beweis der Abzählbarkeit der (Menge der) rationalen Zahlen zu tun,

Here it is clearly visible that only a supertask can accomplish the
complete enumeration.

Regards, WM

On Monday, November 20, 2023 at 11:40:39 AM UTC+1, WM wrote:

> So Cantor infers "all" from "every".

"Infers" ist zuviel gesagt, Mückenheim.

Hinweis: In der Mathematik kann man sagen: "Für jedes x in X gilt: ...x..." oder "Für alle x in X gilt: ...x...".

There's no difference _concerning the truth or falsity_ between: "Each and every x in X is ..." and "All x in X are ...".

<facepalm>

See: https://en.wikipedia.org/wiki/Universal_quantification
and: https://de.wikipedia.org/wiki/Quantor#Schreib-_und_Sprechweise

Note: „Für alle/jedes x gilt …“

The following two sources might help too:
https://www.sgipt.org/wisms/analogik/alle/Segeth108.jpg
https://www.sgipt.org/wisms/analogik/alle/hermes41.GIF

You are dumb like shit Mückenheim AND uneducated.

On Monday, November 20, 2023 at 12:13:14 PM UTC+1, WM wrote:
> On 20.11.2023 12:06, Fritz Feldhase wrote:
> > On Monday, November 20, 2023 at 12:02:38 PM UTC+1, WM wrote:
> > >
> > > And what about the algebraic numbers? How do they get enumerated?
> > >
> > Huh?! Was haben denn die algebraischen Zahlen mit Cantors Beweis der Abzählbarkeit der (Menge der) rationalen Zahlen zu tun,
> >
> Here it is clearly visible that only a supertask can accomplish the complete enumeration.

Du hast einen gewaltigen Sprung in der Schüssel, Mückenheim.

On 20.11.2023 12:22, Fritz Feldhase wrote:
> On Monday, November 20, 2023 at 11:40:39 AM UTC+1, WM wrote:
>
>> So Cantor infers "all" from "every".
>
> "Infers" ist zuviel gesagt,

No, he needs this wrong conclusion.
>
> Hinweis: In der Mathematik kann man sagen: "Für jedes x in X gilt: ...x..." oder "Für alle x in X gilt: ...x...".
>
> There's no difference _concerning the truth or falsity_ between: "Each and every x in X is ..." and "All x in X are ...".

Each term of the sequence has successors. All terms have no successors.
There is completeness.
Each matrix A(n) has successors. All matrices have no successors. There
is completeness A.

Regards, WM

On Monday, November 20, 2023 at 12:39:06 PM UTC+1, WM wrote:
> On 20.11.2023 12:22, Fritz Feldhase wrote:
> > On Monday, November 20, 2023 at 11:40:39 AM UTC+1, WM wrote:
> > >
> > > So Cantor infers "all" from "every".
> > >
> > No, he doesn't "infer" anything (in this connection), Mückenheim.
> >
> > Hinweis: In der Mathematik kann man sagen: "Für jedes x in X gilt: ...x..." oder "Für alle x in X gilt: ...x...".
> >
> > There's no difference _concerning the truth or falsity_ between: "Each and every x in X is ..." and "All x in X are ...".
> >
> Each <bla>

Please see a psychiatrist, soon!

Fritz Feldhase schrieb:
> On Monday, November 20, 2023 at 11:40:39 AM UTC+1, WM wrote:
>
>> And what about the algebraic numbers?
>
> Huh?! Was haben denn die algebraischen Zahlen mit Cantors Beweis der Abzählbarkeit der (Menge der) rationalen Zahlen zu tun, Mückenheim?

His problem is always the same: He doesn't understand bijections and
thinks that Cantors descriptions of methods

"to set [two sets] to correspond one to the other according to some
specific rule, in such a way that for every element of M there is an
element of N and vice versa, for every element of M there is an
element of N"
(Cantor: On infinite linear point-sets; Nr.1 (1879))

are meant as "step-wise supertasks".

In combination with his lack of knowledge concerning the properties of
countable sets, he then can't understand simple proofs of countability
of the set of algebraic numbers, e.g. these ones:

https://www.aleph1.info/?call=Puc&permalink=mengenlehre1_1_7_Z6

https://en.wikipedia.org/wiki/Countable_set#cite_note-27

(Nebenbei bemerkt: Heute hat er angefangen, mir seinen Mist per privater
mail ins Postfach zu kübeln. Hat er das bei dir auch schon einmal
getan?)

Dieter Heidorn

On Monday, November 20, 2023 at 4:23:23 PM UTC+1, Dieter Heidorn wrote:

> (Nebenbei bemerkt: Heute hat er angefangen, mir seinen Mist per privater
> mail ins Postfach zu kübeln. Hat er das bei dir auch schon einmal
> getan?)

Nein, Gott sei Dank, noch nicht. Ich habe ihn mit meiner oftmals etwas rupiggen Art offenbar zu sehr vergräzt.

On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
>... every A(i) contains fractions which sit at not indexed positions,

Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.

On 20.11.2023 16:23, Dieter Heidorn wrote:
> Fritz Feldhase schrieb:
>> On Monday, November 20, 2023 at 11:40:39 AM UTC+1, WM wrote:
>>
>>> And what about the algebraic numbers?
>>
>> Huh?! Was haben denn die algebraischen Zahlen mit Cantors Beweis der
>> Abzählbarkeit der (Menge der) rationalen Zahlen zu tun

Sie zeigen noch deutlicher, dass keine geschlossene Formel für eine
Bijektion vorliegt.
>
> His problem is always the same: He doesn't understand bijections

I recognize that you believe in supertasks but deny to do so.
>   "to set [two sets] to correspond one to the other according to some
>    specific rule, in such a way that for every element of M there is an
>    element of N and vice versa, for every element of M there is an
>    element of N"
>    (Cantor: On infinite linear point-sets; Nr.1 (1879))

Such a specific rule is applied in my Game and it shows that the matrix A(0)

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, ...
....

in no step A(n) changes its size, and the submatrix covered by of
naturals is significantly smaller for all matrices A(n).

>
> are meant as "step-wise supertasks".

The algebraic numbers are handled in this way. Grundsätzlich könntest Du
Deinen Fehler einsehen, wenn Du die Frage zu beantworten suchst, die ich
in dsm gestellt habe: Was unterscheidet die Abzählung der algebraischen
Zahlen von einer Supertask. Was unterscheidet sie von meinem Matrizenspiel?
>
> In combination with his lack of knowledge

You confuse knowledge with your belief in nonsense. The techniques
applied in set theory are presented and taught in my
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
> (Nebenbei bemerkt: Heute hat er angefangen, mir seinen Mist per privater
> mail ins Postfach zu kübeln.

Das habe ich nicht beabsichtigt, und ich entschuldige mich. Es muss wohl
ein Fehler im News-Server gewesen sein.

Regards, WM

On 20.11.2023 10:05, Dieter Heidorn wrote:

>
>      (9)     lambda = my + (my + ny - 1)(my + ny - 2)/2 .
>
>   | By means of (9), a reciprocally univocal relation
>   | subsists between the aggregates {lambda} and {(my,ny)}.

exists
>
> (In the german original of this work:
>    "es besteht vermöge (9) eine gegenseitig eindeutige Beziehung
>     zwischen den Mengen {lambda} und {(my,ny)}")

That is true for the set of fractions. The set of rationals numbers,
i.e., the set of different-value fractions requires sorting by hand.
>
> No "billiard games", no "single steps", no "supertask",
> no "not indexed fractions" - but a clear bijection from N to NxN :-)

Not for the enumeration of the rationals and for the enumeration of the
algebraics. Or better: The same enumeration as the enumeration of my
matrices. Not the slightest difference?

Regards, WM

On 20.11.2023 19:28, William wrote:
> On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
>> ... every A(i) contains fractions which sit at not indexed positions,
>
> Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.

Yes. The n finally indexed positions are the matrix B(n), a submatrix of
A(n).

> Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.

My game tells us (1) that the number of positions of A(n) is constant
for every n. Further it tells us that B(n) is a part of A(n). And (3) it
tells that all A(n) contain all fractions at not indexed positions. Ony
a very strong belief can claim that after all positions have been
indexed that can be indexed, the fractions have vanished.

Regards, WM>

On Tuesday, November 21, 2023 at 1:51:50 PM UTC-4, WM wrote:
> On 20.11.2023 19:28, William wrote:
> > On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
> >> ... every A(i) contains fractions which sit at not indexed positions,
> >
> > Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
> Yes. The n finally indexed positions are the matrix B(n), a submatrix of
> A(n).
> > Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.
> My game tells us (1) that the number of positions of A(n) is constant
> for every n. Further it tells us that B(n) is a part of A(n). And (3) it
> tells that all A(n) contain all fractions at not indexed positions. Ony
> a very strong belief can claim that after all positions have been
> indexed

At no step in your game have all positions been indexed.

On 21.11.2023 23:02, William wrote:
> On Tuesday, November 21, 2023 at 1:51:50 PM UTC-4, WM wrote:
>> On 20.11.2023 19:28, William wrote:
>>> On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
>>>> ... every A(i) contains fractions which sit at not indexed positions,
>>>
>>> Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
>> Yes. The n finally indexed positions are the matrix B(n), a submatrix of
>> A(n).
>>> Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.
>> My game tells us (1) that the number of positions of A(n) is constant
>> for every n. Further it tells us that B(n) is a part of A(n). And (3) it
>> tells that all A(n) contain all fractions at not indexed positions. Ony
>> a very strong belief can claim that after all positions have been
>> indexed
>
> At no step in your game have all positions been indexed.

Of course not. But either all natnumbers have found their finite
destination or at no step in applying his formula k = (m + n - 1)(m + n
- 2)/2 + m Cantor has applied all indices. But for all steps that are
existing (and can be checked) B(n) is smaller than A(n). Note that
already *one* not indexed fraction (like one missing real in Cantor's
list) would prove that the matrix positions are not denumerable. Such a
not indexed fraction is 1/2. Does it disappear "in the limit" B? Why
does the diagonal numbver not "disappear"?

Result: It is simply impossible that the X in

XOOO...
XOOO...
XOOO...
XOOO...
....

are rearranged to cover all positions.

Regards, WM

On Wednesday, November 22, 2023 at 6:30:35 AM UTC-4, WM wrote:
> On 21.11.2023 23:02, William wrote:
> > On Tuesday, November 21, 2023 at 1:51:50 PM UTC-4, WM wrote:
> >> On 20.11.2023 19:28, William wrote:
> >>> On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
> >>>> ... every A(i) contains fractions which sit at not indexed positions,
> >>>
> >>> Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
> >> Yes. The n finally indexed positions are the matrix B(n), a submatrix of
> >> A(n).
> >>> Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.
> >> My game tells us (1) that the number of positions of A(n) is constant
> >> for every n. Further it tells us that B(n) is a part of A(n). And (3) it
> >> tells that all A(n) contain all fractions at not indexed positions. Ony
> >> a very strong belief can claim that after all positions have been
> >> indexed
> >
> > At no step in your game have all positions been indexed.
> Of course not.

Hence your game cannot tell you anything "after all positions have been indexed.

On 22.11.2023 11:40, William wrote:
> On Wednesday, November 22, 2023 at 6:30:35 AM UTC-4, WM wrote:
>> On 21.11.2023 23:02, William wrote:
>>> On Tuesday, November 21, 2023 at 1:51:50 PM UTC-4, WM wrote:
>>>> On 20.11.2023 19:28, William wrote:
>>>>> On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
>>>>>> ... every A(i) contains fractions which sit at not indexed positions,
>>>>>
>>>>> Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
>>>> Yes. The n finally indexed positions are the matrix B(n), a submatrix of
>>>> A(n).
>>>>> Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.
>>>> My game tells us (1) that the number of positions of A(n) is constant
>>>> for every n. Further it tells us that B(n) is a part of A(n). And (3) it
>>>> tells that all A(n) contain all fractions at not indexed positions. Ony
>>>> a very strong belief can claim that after all positions have been
>>>> indexed
>>>
>>> At no step in your game have all positions been indexed.
>> Of course not.
>
> Hence your game cannot tell you anything "after all positions have been indexed.

Of course not! The game proves that it is impossible to index all
positions, whether or not all natural numbers are applied as indices.
The latter is accomplished in the game if it is accomplished by Cantor's
procedure, his mapping which he erroneously calls a bijection. If the
former could be accomplished, then it would be possible that the X in

XOOO...
XOOO...
XOOO...
XOOO...
....

are rearranged to cover all positions. Obvious nonsense.

Regards, WM

On Wednesday, November 22, 2023 at 8:02:10 AM UTC-4, WM wrote:
> On 22.11.2023 11:40, William wrote:
> > On Wednesday, November 22, 2023 at 6:30:35 AM UTC-4, WM wrote:
> >> On 21.11.2023 23:02, William wrote:
> >>> On Tuesday, November 21, 2023 at 1:51:50 PM UTC-4, WM wrote:
> >>>> On 20.11.2023 19:28, William wrote:
> >>>>> On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
> >>>>>> ... every A(i) contains fractions which sit at not indexed positions,
> >>>>>
> >>>>> Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
> >>>> Yes. The n finally indexed positions are the matrix B(n), a submatrix of
> >>>> A(n).
> >>>>> Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.
> >>>> My game tells us (1) that the number of positions of A(n) is constant
> >>>> for every n. Further it tells us that B(n) is a part of A(n). And (3) it
> >>>> tells that all A(n) contain all fractions at not indexed positions. Ony
> >>>> a very strong belief can claim that after all positions have been
> >>>> indexed
> >>>
> >>> At no step in your game have all positions been indexed.
> >> Of course not.
> >
> > Hence your game cannot tell you anything "after all positions have been indexed.
> Of course not! The game proves that it is impossible to index all

Nope the game says nothing about "all"

B is not "accomplished by a procedure" As you note: B is defined, while your putative larger matrix A is not defined.

On 11/19/2023 12:23 PM, Fritz Feldhase wrote:
> On Sunday, November 19, 2023 at 8:58:47 PM UTC+1, Chris M. Thomasson wrote:
>
>> WM is hyper ultra finite.
>
> You mean his brain? Hyper ultra finite and tiny.

Perhaps... Humm, food for thought for sure. :^)