On 12.11.2023 10:03, William wrote:
> On Sunday, November 12, 2023 at 4:33:41 AM UTC-4, WM wrote:

>> The function is only a shortcut describing the steps.
>
> Nope.

What else? What else does f(m, n) = k = (m + n - 1)(m + n - 2)/2 + m ?
It describes every step.
>
>> Proof: There is no
>> obstacle or halt in the sequence of steps.
>
> So what? At no step is a matrix produdced such that B is a submatrix.

At no step all algebraic numbers have been indexed. Does this mean that
the algebraic numbers are not in bijection with ℕ, in your opinion too?

Regards, WM

On Sunday, November 12, 2023 at 6:41:15 AM UTC-4, WM wrote:
> On 12.11.2023 10:03, William wrote:
> > On Sunday, November 12, 2023 at 4:33:41 AM UTC-4, WM wrote:
> >> There is no
> >> obstacle or halt in the sequence of steps.
> >
> > So what? At no step is a matrix produced such that B is a submatrix.
> At no step

Correct. Your game of billiards does not produce B.

On 12.11.2023 21:10, William wrote:
> On Sunday, November 12, 2023 at 6:41:15 AM UTC-4, WM wrote:
>> On 12.11.2023 10:03, William wrote:
>>> On Sunday, November 12, 2023 at 4:33:41 AM UTC-4, WM wrote:
>
>>>> There is no
>>>> obstacle or halt in the sequence of steps.
>>>
>>> So what? At no step is a matrix produced such that B is a submatrix.
>> At no step
>
> Correct. Your game of billiards does not produce B.

That is certainly a possible and sensible standpoint. But it is
inconsistent then to believe that the rational numbers or the algebraic
numbers are countable.

I am afraid I have to suspect you of believing in this inconsistent
conception.

Regards, WM

On Sunday, November 12, 2023 at 5:28:31 PM UTC-4, WM wrote:
> On 12.11.2023 21:10, William wrote:
> > On Sunday, November 12, 2023 at 6:41:15 AM UTC-4, WM wrote:
> >> On 12.11.2023 10:03, William wrote:
> >>> On Sunday, November 12, 2023 at 4:33:41 AM UTC-4, WM wrote:
> >
> >>>> There is no
> >>>> obstacle or halt in the sequence of steps.
> >>>
> >>> So what? At no step is a matrix produced such that B is a submatrix.
> >> At no step
> >
> > Correct. Your game of billiards does not produce B.
> That is certainly a possible and sensible standpoint

Correct and, as you note, B exists. So the fact that your game of billiards does not produce X does not mean X does not exist.

On 11/12/2023 12:33 AM, WM wrote:
> On 11.11.2023 21:21, William wrote:
>> On Saturday, November 11, 2023 at 3:14:25 PM UTC-4, WM wrote:
>>
>>
>>>> The rules of your billiard game do not give a way of producing a
>>>> single matrix that has B as a submatrix. However we can specify a
>>>> bijection between |N and |Nx|N. This specifies B.
>>> This specification specifies that k enumerates [(m,n)]
>>
>> Nope.  By "k enumerates (m.n)" you mean there is a stepwise process
>> that produces B.   There is no stepwise process (like your game of
>> billiards) that produces B.
>
> Other ways of producing B are not existing.
>
>
>> This does not mean you cannot specify B (nor does it mean that there
>> cannot be a stepwise process involved in specifying B, it just means
>> you cannot use *only* a stepwise process. e.g for every pair (m,n)
>> there is an r(m,n) such that
>> B[m,n]=A(r(m,n))[m,n], as well as the stepwise process for producing
>> the A(k) you need to specify the function r(m,n)  )
>
> The function is only a shortcut describing the steps. Proof: There is no
> obstacle or halt in the sequence of steps.
> The function is described in the OP. It is given only in another
> language. Should set theory depend on the used language?
> The function for indexing the algebraic numbers for instance has been
> given by Dedekind and repeated by Cantor:
>     Gehen wir auf die Gleichung (1), welcher eine algebraischen Zahl 
> genügt und welche nach den gedachten Festsetzungen eine völlig bestimmte
> ist, zurück, so möge die Summe der absoluten Beträge ihrer
> Koeffizienten, vermehrt um die Zahl n - 1, wo n den Grad von  angibt,
> die Höhe der Zahl  genannt und mit N bezeichnet werden; es ist also,
> unter Anwendung der üblich gewordenen Bezeichnungsweise:
> N = n - 1 + |a0| + |a1| + ... + |an|
>     Die Höhe N ist danach für jede reelle algebraische Zahl  eine
> bestimmte positive ganze Zahl; umgekehrt gibt es zu jedem positiven
> ganzzahligen Werte von N nur eine endliche Anzahl algebraischer reeller
> Zahlen mit der Höhe N; die Anzahl derselben sei (N); es ist
> beispielsweise (1) = 1; (2) = 2; (3) = 4.  Es lassen sich alsdann die
> Zahlen des Inbegriffes (), d. h. sämtliche algebraischen reellen Zahlen
> folgendermaßen anordnen: man nehme als erste Zahl 1 die eine Zahl mit
> der Höhe N = 1; lasse auf sie, der Größe nach steigend, die (2) = 2
> algebraischen reellen Zahlen der Höhe N = 2 folgen, bezeichne sie mit
> 2, 3; an diese mögen sich die (3) = 4 Zahlen mit der Höhe N = 3,
> ihrer Größe nach aufsteigend, anschließen; allgemein mögen, nachdem in
> dieser Weise sämtliche Zahlen aus () bis zu einer gewissen Höhe N = N1
> abgezählt und an einen bestimmten Platz gewiesen sind, die reellen
> algebraischen Zahlen mit der Höhe N = N1 + 1 auf sie folgen, und zwar
> der Größe nach aufsteigend; so erhält man den Inbegriff () aller
> reellen algebraischen Zahlen in der Form
>
> ω1, ω2, ... ωn,...
>
> und kann mit Rücksicht auf diese Anordnung von der ten algebraischen
> Zahl reden, wobei keine einzige aus dem Inbegriffe () vergessen ist
> Use Google for translation. Only the last sentence is important:
>
> "thus we get the epitome (ω) of all real algebraic numbers [...] and
> with respect to this order we can talk about the th algebraic number
> where not a single one of this epitome () has been forgotten." [E.
> Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
> philosophischen Inhalts", Springer, Berlin (1932) p. 116]
>
> What is the sufficient condition here?

Cantor Pairing is not the only infinite pairing out there... Think of an
n-ary tree. I already gave you some examples of it.

wrt 2-ary, a natural number in the tree has two unique children. I can
derive these children from any natural number. I can get at a child's
parent just from its mapped natural. It's 100% full circle.

0
/ \
/ \
1 2
/ \ / \
3 4 5 6
............

The children of 1 are:

c[0] = 1 * 2 + 1 = 3
c[1] = c[0] + 1 = 4

Nice! Now, to map back

The parent of 3 is:

p = ceil(3 / 2) - 1 = 1

The parent of 4 is:

p = 4 / 2 - 1 = 1

The parent of 5 is:

p = ceil(5 / 2) - 1 = 2

The parent of 6 is:

p = 6 / 2 - 1 = 2

Notice I do not have to use ceil in the case of 2-ary when the natural
number in question is even? Premature optimization? ;^)

It works even with using ceil all the time:

Take the parent of 3 and 4:

p = ceil(3 / 2) - 1 = 1
p = ceil(4 / 2) - 1 = 1

Lets try a parent at zero with its 2-ary children of 1 and 2:

p = ceil(1 / 2) - 1 = 1
p = ceil(2 / 2) - 1 = 1

;^D

I need to adapt it for negative numbers. Think of the following 2-ary tree:

-1 -2
\ /
0
/ \
+1 +2

So, lets try it out... The children on the negative side of zero. Flip
things wrt +1 becomes -1:

c[0] = 0 * 2 - 1 = -1
c[1] = c[0] - 1 = -2

Well, that works! Let's get the parent of -2, and flip the sign on the
-1 to +1, should be zero: Also, lets flip ceil to floor:

p = floor(-2 / 2) + 1 = 0

Nice, lets try -1:

p = floor(-1 / 2) + 1 = 0

It works... Interesting to me.

Lets try -3, its parent should be -1:

p = floor(-3 / 2) + 1 = -1

Also, -4's parent should be -1:

p = floor(-4 / 2) + 1 = -1

Nice!

-5 and -6 should both have a parent of -2:

p = floor(-5 / 2) + 1 = -2
p = floor(-6 / 2) + 1 = -2

perfect.

On 12.11.2023 22:46, William wrote:
> On Sunday, November 12, 2023 at 5:28:31 PM UTC-4, WM wrote:
>> On 12.11.2023 21:10, William wrote:

>>> Correct. Your game of billiards does not produce B.
>> That is certainly a possible and sensible standpoint
>
> Correct and, as you note, B exists.

I note: B exists if and only if A exists.

> So the fact that your game of billiards does not produce X does not mean X does not exist.

Cantor's indexing of the rationals is same as my billiards.
But may be you want to explain how the algebraic numbers get indexed in
a new thread that I will open today.

Regards, WM

On Monday, November 13, 2023 at 4:32:22 AM UTC-4, WM wrote:
> On 12.11.2023 22:46, William wrote:
> > On Sunday, November 12, 2023 at 5:28:31 PM UTC-4, WM wrote:
> >> On 12.11.2023 21:10, William wrote:
>
> >>> Correct. Your game of billiards does not produce B.
> >> That is certainly a possible and sensible standpoint
> >
> > Correct and, as you note, B exists.
> I note: B exists if and only if A exists.
Nope, B exists, But your billiards game does not produce it. The putative "matrix" A does not exist. Your billiards game does not produce A.

On 13.11.2023 12:05, William wrote:
> On Monday, November 13, 2023 at 4:32:22 AM UTC-4, WM wrote:
>> On 12.11.2023 22:46, William wrote:
>>> On Sunday, November 12, 2023 at 5:28:31 PM UTC-4, WM wrote:
>>>> On 12.11.2023 21:10, William wrote:
>>
>>>>> Correct. Your game of billiards does not produce B.
>>>> That is certainly a possible and sensible standpoint
>>>
>>> Correct and, as you note, B exists.
>> I note: B exists if and only if A exists.
> Nope, B exists,

May be, but it does not exist as a result of enumerating the fractions.

> But your billiards game does not produce it. The putative "matrix" A does not exist. Your billiards game does not produce A.

Either both of us produce a complete application of natural indices, or
none of us.
If describing the supertask is sufficient to achieve completeness, then
A and B are reached. If *performing* the supertask is necessary, then
neither A nor B are reached.
Please find a significant difference and note it here:
https://groups.google.com/g/sci.math/c/2CPLkDDWsSs

Regards, WM

On 13.11.2023 12:05, William wrote:
> On Monday, November 13, 2023 at 4:32:22 AM UTC-4, WM wrote:
>> On 12.11.2023 22:46, William wrote:
>>> On Sunday, November 12, 2023 at 5:28:31 PM UTC-4, WM wrote:
>>>> On 12.11.2023 21:10, William wrote:
>>
>>>>> Correct. Your game of billiards does not produce B.
>>>> That is certainly a possible and sensible standpoint
>>>
>>> Correct and, as you note, B exists.
>> I note: B exists if and only if A exists.
> Nope, B exists,

May be, but it does not exist as a result of enumerating the fractions.

> But your billiards game does not produce it. The putative "matrix" A does not exist. Your billiards game does not produce A.

Either both of us produce a complete application of natural indices, or
none of us.
If describing the supertask is sufficient to achieve completeness, then
A and B are reached. If *performing* the supertask is necessary, then
neither A nor B are reached.
Please find a significant difference and note it here:
https://groups.google.com/g/sci.math/c/2CPLkDDWsSs

Regards, WM

On 11/13/2023 1:23 PM, WM wrote:
> On 13.11.2023 12:05, William wrote:

>> Nope,   B exists,
>
> May be,
> but it does not exist as a result of
> enumerating the fractions.

Some sets are finite.
Some sets are not finite.

The Finite continues and continues,
but it only gets to more of the Finite.
The fact that the Finite can get there is why
where it can get is only more of the Finite.

We describe one of the Finite so:
"We can get there"
We augment the description with
only not-first-false claims.
The augmenting claims are true of
wherever we can get,
true even if where we can get is in the Darkᵂᴹ
true even if it is ridiculously far away.

We have not gone out and checked in the Darkᵂᴹ
That isn't how we know the claims.
We check that
the claims are only not-first-false.
The claims are not in the Darkᵂᴹ
They're on the page or screen in front of us.

On Tuesday, November 14, 2023 at 12:18:44 AM UTC+1, Jim Burns wrote:

> We describe [...] the [In]finite so:
> "We can['t] get there"

Good one.

I'd like to add: in finitely many steps. ;-P

> The claims are not in the Darkᵂᴹ

Agree.

Hint: Nonsense(Darkᵂᴹ) is true.

On Monday, November 13, 2023 at 2:23:16 PM UTC-4, WM wrote:
> Either both of us produce a complete application of natural indices, or
> none of us.

Nope. B can be specified without using a stepwise process.. Your billiards game does not produce A

On 14.11.2023 09:32, William wrote:
> On Monday, November 13, 2023 at 2:23:16 PM UTC-4, WM wrote:
>> Either both of us produce a complete application of natural indices, or
>> none of us.
>
> Nope. B can be specified without using a stepwise process..

Yes, only so. But enumerating the fractions is a stepwise process like
enumerating the algebraic numbers.

> Your billiards game does not produce A

Can the algebraic numbers be enumerated completely? And why do you think so?

Regards, WM

On 14.11.2023 00:25, Fritz Feldhase wrote:
> On Tuesday, November 14, 2023 at 12:18:44 AM UTC+1, Jim Burns wrote:
>
>> We describe [...] the [In]finite so:
>> "We can['t] get there"
>
> I'd like to add: in finitely many steps.

Can we get there in infinitely many steps? Can we do infinitely many
steps? Can Cantor enumerate the rationals / the algebraics in infinitely
many steps? Can the game get there in infinitely many steps? Why can
Cantor reach B but the game not reach A?

Regards, WM

On Tuesday, November 14, 2023 at 7:05:08 AM UTC-4, WM wrote:
> On 14.11.2023 09:32, William wrote:
> > On Monday, November 13, 2023 at 2:23:16 PM UTC-4, WM wrote:
> >> Either both of us produce a complete application of natural indices, or
> >> none of us.
> >
> > Nope. B can be specified without using a stepwise process.
> Yes

So the fact that your billiards game (a stepwise process) does not produce an A has nothing to do with the fact that B exists.

After serious thinking WM wrote :
> On 14.11.2023 09:32, William wrote:
>> On Monday, November 13, 2023 at 2:23:16 PM UTC-4, WM wrote:
>>> Either both of us produce a complete application of natural indices, or
>>> none of us.
>>
>> Nope. B can be specified without using a stepwise process..
>
> Yes, only so. But enumerating the fractions is a stepwise process like
> enumerating the algebraic numbers.

No it isn't. The process by which it is shown to happen, uses steps.

On 14.11.2023 12:58, William wrote:

> So the fact that your billiards game (a stepwise process) does not produce an A has nothing to do with the fact that B exists.

Irrelevant. What is relevant is this:
If Cantor produces B by the definition k = (m + n - 1)(m + n - 2)/2 + m,
then I produce A by the OP.
I Cantor produces B by a supertask, then I produce A by a supertask.
If supertasks don't get complete, then an enumeration of the algebraic
numbers is impossible.
But note that Cantor and Dedekind were of opposite opinion.

Regards, WM

On 14.11.2023 15:22, FromTheRafters wrote:
> After serious thinking WM wrote :
>> On 14.11.2023 09:32, William wrote:
>>> On Monday, November 13, 2023 at 2:23:16 PM UTC-4, WM wrote:
>>>> Either both of us produce a complete application of natural indices, or
>>>> none of us.
>>>
>>> Nope.  B can be specified without using a stepwise process..
>>
>> Yes, only so. But enumerating the fractions is a stepwise process like
>> enumerating the algebraic numbers.
>
> No it isn't.

Then there is no enumeration of the algebraics?

>The process by which it is shown to happen, uses steps.

If Cantor produces B by the definition k = (m + n - 1)(m + n - 2)/2 + m,
then I produce A by the OP.
I Cantor produces B by a supertask, then I produce A by a supertask.
If supertasks don't get complete, then an enumeration of the algebraic
numbers is impossible.
But note that Cantor and Dedekind were of opposite opinion.

Regards, WM

On Tuesday, November 14, 2023 at 12:29:12 PM UTC-4, WM wrote:
> On 14.11.2023 12:58, William wrote:
>
> > So the fact that your billiards game (a stepwise process) does not produce an A has nothing to do with the fact that B exists.
> Irrelevant.

Hardly. One can produce B (this is not a "supertask"). You cannot and do not produce A.

On 11/14/2023 6:47 AM, WM wrote:
> On 14.11.2023 00:25, Fritz Feldhase wrote:
>> On Tuesday, November 14, 2023
>> at 12:18:44 AM UTC+1, Jim Burns wrote:

>>> We describe [...] the [In]finite so:
>>> "We can['t] get there"
>>
>> I'd like to add: in finitely many steps.
>
> Can we get there in infinitely many steps?

We can't do infinitely-many steps.

> Can we do infinitely many steps?

No.
Well, Chuck Norris can.
Otherwise, no.

> Can Cantor enumerate
> the rationals / the algebraics in
> infinitely many steps?

No.
Who do you think Cantor is, Chuck Norris?

Cantor can describe, in
finitely-many finite-length claims,
a set which enumerates the rationals/algebraics.

The description describes infinitely-many steps.
The description does not have infinitely-many steps.

> Can the game get there in infinitely many steps?
> Why can Cantor reach B but the game not reach A?

Cantor reaches the description of B in
finitely-many linguistic actions.
That _task_ is sufficient to reason about B

Cantor does not reach B by
infinitely-many indexing actions.
That would be a _supertask_

Reasoning about a supertask can be a task.
For example,
finitely-many claims, each not-first-false,
are each not-false,
even if they are about infinitely-many.

On 14.11.2023 19:32, William wrote:
> On Tuesday, November 14, 2023 at 12:29:12 PM UTC-4, WM wrote:
>> On 14.11.2023 12:58, William wrote:
>>
>>> So the fact that your billiards game (a stepwise process) does not produce an A has nothing to do with the fact that B exists.
>> Irrelevant.
>
> Hardly. One can produce B (this is not a "supertask"). You cannot and do not produce A.

Producing A is a supertask: Every step defines a next step. What is
different in enumerating all algebraic numbers?

Regards, WM

On 14.11.2023 22:52, Jim Burns wrote:
> On 11/14/2023 6:47 AM, WM wrote:
>> On 14.11.2023 00:25, Fritz Feldhase wrote:
>>> On Tuesday, November 14, 2023
>>> at 12:18:44 AM UTC+1, Jim Burns wrote:
>
>>>> We describe [...] the [In]finite so:
>>>> "We can['t] get there"
>>>
>>> I'd like to add: in finitely many steps.
>>
>> Can we get there in infinitely many steps?
>
> We can't do infinitely-many steps.

Then we can't enumerate infinitely many elements. Every attaching of an
index n is a step. The formula k = (m + n - 1)(m + n - 2)/2 + m cannot
fool us. Every k has infinitely many successors.
>
>> Can we do infinitely many steps?
>
> No.
> Well, Chuck Norris can.
> Otherwise, no.

Cantor and Dedekind claimed that the could by enumerating the algebraic
numbers.

"But when ordering the polynomials by their height H, i.e., the sum of
their degree n and all absolute values of their coefficients a_nu
H = n + |a0| + |a1| + |a2| + ... + |an|
then for every height H there is a finite number of polynomials. Every
polynomial gets its place in the enumeration, and since every polynomial
of degree n has at most n different roots, every root can be inserted
into a sequence containing all of them. So the set of roots of all
polynomials is countable. It is the set A of all algebraic numbers." [R.
Dedekind, private note (29 Nov 1873). Cantor, p. 116)]

"Cantor observed that many infinite sets of numbers are countable: the
set of all integers, the set of all rational numbers, and also the set
of all algebraic numbers." [T. Jech: "Set theory", Stanford Encyclopedia
of Philosophy (2002)]
>
>> Can Cantor enumerate
>> the rationals / the algebraics in
>> infinitely many steps?
>
> No.
> Who do you think Cantor is, Chuck Norris?

Countable means that the set can be indexed in finitely many or
infinitely many steps. Numbers are counting steps.
>
> Cantor can describe, in
> finitely-many finite-length claims,
> a set which enumerates the rationals/algebraics.
>
> The description describes infinitely-many steps.
> The description does not have infinitely-many steps.
>
>> Can the game get there in infinitely many steps?
>> Why can Cantor reach B but the game not reach A?
>
> Cantor reaches the description of B in
> finitely-many linguistic actions.
> That _task_ is sufficient to reason about B

That is similar to my game of billiards. The reasoning is very short,
cp. the OP.

Regards, WM

On Wednesday, November 15, 2023 at 6:35:04 AM UTC-4, WM wrote:
> On 14.11.2023 19:32, William wrote:
> > On Tuesday, November 14, 2023 at 12:29:12 PM UTC-4, WM wrote:
> >> On 14.11.2023 12:58, William wrote:
> >>
> >>> So the fact that your billiards game (a stepwise process) does not produce an A has nothing to do with the fact that B exists.
> >> Irrelevant.
> >
> > Hardly. One can produce B (this is not a "supertask"). You cannot and do not produce A.
> Producing A is a supertask:
Correct. However, specifying B is not a "supertask". Every element of B is specified, You billiards game only specified elements
of A(n), You cannot and do not specifiy elements of your putative matrix A..

WM was thinking very hard :
> On 14.11.2023 22:52, Jim Burns wrote:
>> On 11/14/2023 6:47 AM, WM wrote:
>>> On 14.11.2023 00:25, Fritz Feldhase wrote:
>>>> On Tuesday, November 14, 2023
>>>> at 12:18:44 AM UTC+1, Jim Burns wrote:
>>
>>>>> We describe [...] the [In]finite so:
>>>>> "We can['t] get there"
>>>>
>>>> I'd like to add: in finitely many steps.
>>>
>>> Can we get there in infinitely many steps?
>>
>> We can't do infinitely-many steps.
>
> Then we can't enumerate infinitely many elements. Every attaching of an index
> n is a step. The formula k = (m + n - 1)(m + n - 2)/2 + m cannot fool us.
> Every k has infinitely many successors.
>>
>>> Can we do infinitely many steps?
>>
>> No.
>> Well, Chuck Norris can.
>> Otherwise, no.
>
> Cantor and Dedekind claimed that the could by enumerating the algebraic
> numbers.
>
> "But when ordering the polynomials by their height H, i.e., the sum of their
> degree n and all absolute values of their coefficients a_nu
> H = n + |a0| + |a1| + |a2| + ... + |an|
> then for every height H there is a finite number of polynomials. Every
> polynomial gets its place in the enumeration, and since every polynomial of
> degree n has at most n different roots, every root can be inserted into a
> sequence containing all of them. So the set of roots of all polynomials is
> countable. It is the set A of all algebraic numbers." [R. Dedekind, private
> note (29 Nov 1873). Cantor, p. 116)]
>
>
> "Cantor observed that many infinite sets of numbers are countable: the set of
> all integers, the set of all rational numbers, and also the set of all
> algebraic numbers." [T. Jech: "Set theory", Stanford Encyclopedia of
> Philosophy (2002)]
>>
>>> Can Cantor enumerate
>>> the rationals / the algebraics in
>>> infinitely many steps?
>>
>> No.
>> Who do you think Cantor is, Chuck Norris?
>
> Countable means that the set can be indexed in finitely many or infinitely
> many steps. Numbers are counting steps.

Nothing about steps at all there, nor here:

https://en.wikipedia.org/wiki/Bijection

On 14.11.2023 15:22, FromTheRafters wrote:
> After serious thinking WM wrote :
>> On 14.11.2023 09:32, William wrote:
>>> On Monday, November 13, 2023 at 2:23:16 PM UTC-4, WM wrote:
>>>> Either both of us produce a complete application of natural indices, or
>>>> none of us.
>>>
>>> Nope.  B can be specified without using a stepwise process..
>>
>> Yes, only so. But enumerating the fractions is a stepwise process like
>> enumerating the algebraic numbers.
>
> No it isn't. The process by which it is shown to happen, uses steps.

The proof uses steps. Enumerating the algebraic numbers and constructing
the matrix A.

Regards, WM