On 15.11.2023 13:28, FromTheRafters wrote:

>> 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,

But here:

"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)]

Celebrate 150 years of supertask, coming soon.

Regards, WM

On 15.11.2023 13:18, William wrote:
> 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,

Exactly like every element of A and the index of every algebraic numbers.

You billiards game only specified elements
> of A(n), You cannot and do not specifiy elements of your putative matrix A.

What element of the submatrix B of A is not specified?

Regards, WM

On Wednesday, November 15, 2023 at 1:40:15 PM UTC-4, WM wrote:
> On 15.11.2023 13:18, William wrote:
> > 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,
> Exactly like every element of A
Nope, Your billiard game only specifies elements of A(n) (every element of B is in some A(n), but there is no one A(n) that "works" for all elements of B). You cannot and do not specify the elements of your putative matrix A.

On Wednesday, November 15, 2023 at 6:34:23 PM UTC+1, WM wrote:

> Celebrate 150 years of supertask, coming soon.

Du redest wirres Zeug (saudummen Scheißdreck) daher, Mückenheim.

On Wednesday, November 15, 2023 at 6:40:15 PM UTC+1, WM wrote:

> What element of the submatrix B of A is not specified?

There is no matrix A in your "proof", you silly crank.

WM pretended :
> On 15.11.2023 13:28, FromTheRafters wrote:
>
> >> 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,
>
> But here:
>
> "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)]
>
> Celebrate 150 years of supertask, coming soon.

It still doesn't.

On 11/15/2023 5:50 AM, WM wrote:
> 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.

Yes.
Unless we are Chuck Norris.
I am not Chuck Norris. And you?

However,
we can describe an enumeration of infinitely-many,
and we can augment the description finitely with
only not-first-false claims, which,
because
a finite sequence of only not-first-false claims
is a finite sequence of only not-false claims,
we know are true.

On 15.11.2023 18:53, William wrote:
> On Wednesday, November 15, 2023 at 1:40:15 PM UTC-4, WM wrote:
>> On 15.11.2023 13:18, William wrote:
>>> 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,
>> Exactly like every element of A
> Nope, Your billiard game only specifies elements of A(n) (every element of B is in some A(n), but there is no one A(n) that "works" for all elements of B).

That is impossible. If every element of B is there, then all elements of
B are there, because the sequence of elements of B is inclusion-monotonic.

> You cannot and do not specify the elements of your putative matrix A.

I can and do like Dedekind does:

"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)]

And I do the same:

Push the natnumbers of the first column 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. That is best done by creating a pattern like

1, 2, 4, ...
3, 5, 8, ...
6, 9, 13, ...
.... ,

According to this simple rule it is impossible, in eternity, to remove a
fraction from the matrix or to attach a natnumber to a fraction.

Compare: "So the set of roots of all polynomials is countable. It is the
set A of all algebraic numbers."

So all natural numbers have been applied completing the matrix A.

Any difference?

Regards, WM

On 15.11.2023 18:58, Fritz Feldhase wrote:
> On Wednesday, November 15, 2023 at 6:40:15 PM UTC+1, WM wrote:
>
>> What element of the submatrix B of A is not specified?
>
> There is no matrix A in your "proof",

If there is a matrix B, then it exists because every element is
specified by k = (m + n - 1)(m + n - 2)/2 + m. Same is true for my
matrix A: Every element is specified:

Push the natnumbers of the first column 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. That is best done by creating a pattern like B.

Regards, WM

On 15.11.2023 19:08, FromTheRafters wrote:
> WM pretended :

>> Celebrate 150 years of supertask, coming soon.
>
> It still doesn't.

No enumeration of algebraics completed yet?

Regards, WM

On 15.11.2023 20:54, Jim Burns wrote:
> On 11/15/2023 5:50 AM, WM wrote:
>> 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.
>
> Yes.

But Cantor has claimed to have done it and all matheologians follow him.

"If we think the numbers p/q in such an order [...] then every number
p/q comes at an absolutely fixed position of a simple infinite sequence"
[E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 126]

"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

Note that he obtained completeness (indexing all fractions) from his
prescription to index every fraction.

> However,
> we can describe an enumeration of infinitely-many,

So do I: Push the natnumbers of the first column 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. That is best done by creating a pattern like

1, 2, 4, ...
3, 5, 8, ...
6, 9, 13, ...
.... ,

According to this simple rule it is impossible, in eternity, to remove a
fraction from the matrix or to attach a natnumber to a fraction.

Do you understand this simple proof?

Regards, WM

WM pretended :
> On 15.11.2023 19:08, FromTheRafters wrote:
>> WM pretended :
>
>>> Celebrate 150 years of supertask, coming soon.
>>
>> It still doesn't.
>
> No enumeration of algebraics completed yet?

It still doesn't mean or say anything about steps.

On Thursday, November 16, 2023 at 4:42:39 AM UTC-4, WM wrote:
> On 15.11.2023 18:53, William wrote:
> > On Wednesday, November 15, 2023 at 1:40:15 PM UTC-4, WM wrote:
> >> On 15.11.2023 13:18, William wrote:
> >>> 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,
> >> Exactly like every element of A
> > Nope, Your billiard game only specifies elements of A(n) (every element of B is in some A(n), but there is no one A(n) that "works" for all elements of B).
> That is impossible. If every element of B is there, then all elements of
> B are there,

Correct. All elements of B "are there" but not in one matrix. Your putative matrix A does not exist.

On 16.11.2023 15:11, William wrote:
> On Thursday, November 16, 2023 at 4:42:39 AM UTC-4, WM wrote:

>> That is impossible. If every element of B is there, then all elements of
>> B are there,
>
> Correct. All elements of B "are there" but not in one matrix.

Precisely that is the case for Cantor's enumeration. But he claims that
from "every" it follows completeness:

"then every number p/q comes at an absolutely fixed position of a simple
infinite sequence"

"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely"

>Your putative matrix A does not exist.

Then matrix B does not exist either. In all cases we must be allowed to
conclude "all" from the given "every".

But independent of that is the enumeration of all algebraics:

All indices are applied to matrix positions.
All indices are applied to roots of polynomials.

What is the difference?

Regards, WM

On Thursday, November 16, 2023 at 10:33:22 AM UTC-4, WM wrote:
> On 16.11.2023 15:11, William wrote:
> > On Thursday, November 16, 2023 at 4:42:39 AM UTC-4, WM wrote:
>
> >> That is impossible. If every element of B is there, then all elements of
> >> B are there,
> >
> > Correct. All elements of B "are there" but not in one matrix.
> Precisely

Each element of B is specified, B exists. No element of the putative matrix A is specified. A does not exist.

On 16.11.2023 16:20, William wrote:
> On Thursday, November 16, 2023 at 10:33:22 AM UTC-4, WM wrote:
>> On 16.11.2023 15:11, William wrote:
>>> On Thursday, November 16, 2023 at 4:42:39 AM UTC-4, WM wrote:
>>
>>>> That is impossible. If every element of B is there, then all elements of
>>>> B are there,
>>>
>>> Correct. All elements of B "are there" but not in one matrix.
>> Precisely
>
> Each element of B is specified, B exists. No element of the putative matrix A is specified. A does not exist.
>
Each element of B is specified by matrices*) as well as by Cantor. B
exists in both cases. But in case of matrices we know that the elements
of B are not the only elements because all fractions are remaining.

*)Remember, you said: "every element of B is in some A(n)". That means
each element of B is specified by my matrices.

Regards, WM

On 16.11.2023 14:00, FromTheRafters wrote:
> WM pretended :
>> On 15.11.2023 19:08, FromTheRafters wrote:
>>> WM pretended :
>>
>>>> Celebrate 150 years of supertask, coming soon.
>>>
>>> It still doesn't.
>>
>> No enumeration of algebraics completed yet?
>
> It still doesn't mean or say anything about steps.

Can't you read?

ordering the polynomials by their height H
for every height H there is a finite number of polynomials.

n different roots, every root can be inserted

Do you know what it means to insert something?

into a sequence containing all of them.

Every root can be inserted, that means that all are contained.

Regards, WM

On Thursday, November 16, 2023 at 2:24:56 PM UTC-4, WM wrote:
> you said: "every element of B is in some A(n)". That means
> each element of B is specified by my matrices.

Correct. Your billiards game specifies every element of B. Your billiards game does not specify any element of the putative matrix A.

On 16.11.2023 20:01, William wrote:
> On Thursday, November 16, 2023 at 2:24:56 PM UTC-4, WM wrote:
>
>> you said: "every element of B is in some A(n)". That means
>> each element of B is specified by my matrices.
>
> Correct. Your billiards game specifies every element of B.

With every element of B, B is specified, according to Cantor.

> Your billiards game does not specify any element of the putative matrix A.

I specify B like Cantor does (in fact he specifies only every matrix B_n
and *claims* to have specified B), but I show that there is more in the
specified matrix. Matrix A need not be specified. One more element than
B is enough.

Regards, WM

On Thursday, November 16, 2023 at 3:21:30 PM UTC-4, WM wrote:
> I show that there is more in the specified matrix.

Nope. Every element specified is an element of |N. Your matrices A(n) each have elements that are "fractions" but
these are not used to specify elements of B.

On 16.11.2023 20:37, William wrote:
> On Thursday, November 16, 2023 at 3:21:30 PM UTC-4, WM wrote:
>> I show that there is more in the specified matrix.
>
> Nope. Every element specified is an element of |N.

Every element of B is a natural number.

> Your matrices A(n) each have elements that are "fractions" but
> these are not used to specify elements of B.

Nevertheless all natnumbers specifying all of B are there.

Cantor specifies B_1, B_2, B_3, ... and claims to have specified B.
I specifiy A_1, A_2, A_3, ... with B_n submatrix of A_n for every n.

But your misrepresentation of these facts will be proven when you try to
answer this question: Is there a difference in completeness of the
result between
All indices are applied to matrix positions.
and
All indices are applied to roots of polynomials.

Regards, WM

> I specify A_1, A_2, A_3, ...

Correct. Each is an |Nx|N matrix with the upper left corner composed of elements of |N and it is this part that is used to specify the elements of B . You cannot use this sequence to specify the elements of a putative larger matrix of which B is a proper submatix.

WM formulated on Thursday :
> On 16.11.2023 14:00, FromTheRafters wrote:
>> WM pretended :
>>> On 15.11.2023 19:08, FromTheRafters wrote:
>>>> WM pretended :
>>>
>>>>> Celebrate 150 years of supertask, coming soon.
>>>>
>>>> It still doesn't.
>>>
>>> No enumeration of algebraics completed yet?
>>
>> It still doesn't mean or say anything about steps.
>
> Can't you read?
>
> ordering the polynomials by their height H
> for every height H there is a finite number of polynomials.
>
> n different roots, every root can be inserted
>
> Do you know what it means to insert something?
>
> into a sequence containing all of them.
>
> Every root can be inserted, that means that all are contained.

I still see no steps. Are you ordering your polynomials one at a time?

On 16.11.2023 22:08, FromTheRafters wrote:
> WM formulated on Thursday :

>>
>> ordering the polynomials by their height H
>> for every height H there is a finite number of polynomials.
>>
>> n different roots, every root can be inserted
>>
>> Do you know what it means to insert something?
>>
>> into a sequence containing all of them.
>>
>> Every root can be inserted, that means that all are contained.
>
> I still see no steps.

Inserting one of n different roots into the sequence is done in the same
way as inserting a natural number n into its final position in the
matrix. Do you see a difference?

> Are you ordering your polynomials one at a time?

I am doing all that in the same way as Dedekind and Cantor did.

Regards, WM

On 16.11.2023 21:18, William wrote:
>
>
>> I specify A_1, A_2, A_3, ...
>
> Correct. Each is an |Nx|N matrix with the upper left corner composed of elements of |N and it is this part that is used to specify the elements of B . You cannot use this sequence to specify the elements of a putative larger matrix of which B is a proper submatix.

I do not specify the larger matrix. Only B is specified as completed in
the same manner as Cantor did so. But I know that B is surrounded by more.
Not specified but proven by applying logic, namely: no fractions leaves
the matrix but no fraction sits at an indexed position.
>
The question is about completing.
What is the difference in
All indices are applied to matrix positions
and
All indices are applied to roots of polynomials?

Regards, WM