On 14.12.2023 15:23, William wrote:
> On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
>> On 13.12.2023 20:41, William wrote:
>> need dark numbers.
>>> Only producing the putative A shows that "dark" numbers are needed.
>> No. Producing B
>
> does not involve a limit

It involves the realization of all natural numbers
k = (m + n - 1)(m + n - 2)/2 + m
at matrix places
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, ...

This requires a limit or a last step. Your belief alone is not sufficient.

Regards, WM

On 2023-12-14 15:22:50 +0000, WM said:

> On 14.12.2023 15:23, William wrote:
>> On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
>>> On 13.12.2023 20:41, William wrote:
>>> need dark numbers.
>>>> Only producing the putative A shows that "dark" numbers are needed.
>>> No. Producing B
>>
>> does not involve a limit
>
> It involves the realization of all natural numbers
> k = (m + n - 1)(m + n - 2)/2 + m
> at matrix places
> 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, ...
>
> This requires a limit or a last step.

You say but don't show. Your belief alone is not sufficient.

Mikko

On 15.12.2023 10:58, Mikko wrote:
> On 2023-12-14 15:22:50 +0000, WM said:

>> This requires a limit or a last step.
>
> You say but don't show.
What proves completeness in your opinion?

Regards, WM

On 12/15/2023 7:12 AM, WM wrote:
> On 15.12.2023 10:58, Mikko wrote:
>> On 2023-12-14 15:22:50 +0000, WM said:

>>> This requires a limit or a last step.
>>
>> You say but don't show.
>
> What proves completeness in your opinion?

In my opinion,
the answer is as difficult as it is
because it's too obvious.
https://en.wikipedia.org/wiki/The_Purloined_Letter
| however, Dupin conjectured that it would instead
| be hidden in plain sight.

Write something which is true of
each one of multiple objects.
For example,
| | x is the point between the fore- and hindpart
| of a non-empty split F,H of the rationals ℚ

What's hidden in plain sight is:
when those are the points we're discussing,
| | x is the point between the fore- and hindpart
| of a non-empty split F,H of the rationals ℚ
| is true for each of what we're discussing.
Complete.

In instances involving infinity,
we do not _accomplish_ completeness.
We do not step
from incomplete to complete.
(Unless we are Chuck Norris.)
(Chuck Norris counted to infinity.)
(Twice.)

For lesser beings, our justified claim of
completeness involving infinity
has a too.obvious initial completeness
and only makes completeness.preserving
further claims.

Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1:
> On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
> >
> > What proves completeness [...]?
>
> A DEFINITION of /completeness/ and a PROOF (of a statement referring to completeness).

Completeness in counting the natural numbers cannot be proven because the contrary can be proven. And that is simple: For every eps > 0 there are only finitely many unit fractions in (eps, 1] counted, but infinitely many in (0, eps] uncounted and uncountable because eps cannot be made small enough to leave only finitely many uncounted. Therefore every definition of completeness in this respect is nonsense.

Regards, WM

On 15.12.2023 16:57, Jim Burns wrote:
> On 12/15/2023 7:12 AM, WM wrote:
>> On 15.12.2023 10:58, Mikko wrote:
>>> On 2023-12-14 15:22:50 +0000, WM said:
>
>>>> This requires a limit or a last step.
>>>
>>> You say but don't show.
>>
>> What proves completeness in your opinion?
>
> Write something which is true of
> each one of multiple objects.

That does not make it true. There are differences.
Example: Every fraction is the quotient of two natnumbers. That is true.
Every fraction can be isolated from all smaller fractions by an eps
between them. That is wrong.

> For example,
> |
> | x is the point between the fore- and hindpart
> | of a non-empty split F,H of the rationals ℚ
>
> What's hidden in plain sight is:
> when those are the points we're discussing,
> |
> | x is the point between the fore- and hindpart
> | of a non-empty split F,H of the rationals ℚ
> |
> is true for each of what we're discussing.
> Complete.

As far as I can understand, you say the same as I.
>
> In instances involving infinity,
> we do not _accomplish_ completeness.

That's it!

Regards, WM

On 12/15/2023 12:09 PM, WM wrote:
> On 15.12.2023 16:57, Jim Burns wrote:
>> On 12/15/2023 7:12 AM, WM wrote:
>>> On 15.12.2023 10:58, Mikko wrote:
>>>> On 2023-12-14 15:22:50 +0000, WM said:

>>>>> This requires a limit or a last step.
>>>>
>>>> You say but don't show.
>>>
>>> What proves completeness in your opinion?
>>
>> Write something which is true of
>> each one of multiple objects.
>
> That does not make it true.

No.
Something which is true of
each one of multiple objects
is
something which is true of
each one of multiple objects.

Perhaps this is too obvious,
hidden in plain sight.

If,
on the other hand,
you have picked something else to say,
something which _isn't_ true of
each one of the multiple objects
which you intend to discuss,
then
pick a different thing to say.

>> For example,
>> |
>> | x is the point between the fore- and hindpart
>> | of a non-empty split F,H of the rationals ℚ

I picked this thing to say because
a function of _all_ those points,
including points.between (real numbers),
which is continuous at each point
doesn't jump.

I'm familiar with contexts in which
it is useful to say of a function that
it doesn't jump.

This is not a fantasy.example,
not Scrooge McDuck's vaults.
How we say _what we intend to say_
which is that a function doesn't jump,
is to say something once which is
true in infinitely.many ways.

>> In instances involving infinity,
>> we do not _accomplish_ completeness.
>
> That's it!

Our methods take that into account.
We begin at completeness and
maintain completeness as we learn
by not-first-false augmenting.

We don't accomplish completeness and
we don't need to.

On 12/15/23 11:51 AM, WM wrote:
> Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1:
>> On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
>>>
>>> What proves completeness [...]?
>>
>> A DEFINITION of /completeness/ and a PROOF (of a statement referring to completeness).
>
> Completeness in counting the natural numbers cannot be proven because the contrary can be proven. And that is simple: For every eps > 0 there are only finitely many unit fractions in (eps, 1] counted, but infinitely many in (0, eps] uncounted and uncountable because eps cannot be made small enough to leave only finitely many uncounted. Therefore every definition of completeness in this respect is nonsense.
>
> Regards, WM
So NOTHING in your logic system can completely talk about the whole of
Natural Numbers, so it just can't handle them.

Your logic is BOUNDED, but the Numbers are UNBOUNDED.

Thus, your logic fails you.

On 2023-12-15 12:12:36 +0000, WM said:

> On 15.12.2023 10:58, Mikko wrote:
>> On 2023-12-14 15:22:50 +0000, WM said:
>
>>> This requires a limit or a last step.
>>
>> You say but don't show.
> What proves completeness in your opinion?

A proof of completeness. Details depend on what kind of completeness
you mean.

Mikko

Le 16/12/2023 à 09:07, Mikko a écrit :
> On 2023-12-15 12:12:36 +0000, WM said:
>
>> On 15.12.2023 10:58, Mikko wrote:
>>> On 2023-12-14 15:22:50 +0000, WM said:
>>
>>>> This requires a limit or a last step.
>>>
>>> You say but don't show.
>> What proves completeness in your opinion?
>
> A proof of completeness. Details depend on what kind of completeness
> you mean.

You "prove" that Cantor enumerates all fractions. That is wrong. The
matrix
XOOO...
XOOO...
XOOO...
XOOO...
...will never be covered by indeXes. The indeX transferred from matrix
position (k, j) to matrix position (m, n) implies that, in exchange, O is
transferred from matrix position (m, n) to matrix position (k, j). An O is
indicating a not indexed position. No O will ever get off of the matrix.
Hence not all positions will be indeXed. Therefore your "proofs" are
worthless.

Regards, WM

Le 16/12/2023 à 02:05, Richard Damon a écrit :
> On 12/15/23 11:51 AM, WM wrote:
>> Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1:
>>> On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
>>>>
>>>> What proves completeness [...]?
>>>
>>> A DEFINITION of /completeness/ and a PROOF (of a statement referring to
>>> completeness).
>>
>> Completeness in counting the natural numbers cannot be proven because the
>> contrary can be proven. And that is simple: For every eps > 0 there are only
>> finitely many unit fractions in (eps, 1] counted, but infinitely many in (0, eps]
>> uncounted and uncountable because eps cannot be made small enough to leave only
>> finitely many uncounted. Therefore every definition of completeness in this respect
>> is nonsense.
>>
> So NOTHING in your logic system can completely talk about the whole of
> Natural Numbers, so it just can't handle them.

Neither can yours.
XOOO...
XOOO...
XOOO...
XOOO...
...
The indeX transferred from matrix position (k, j) to matrix position (m,
n) implies that, in exchange, O is transferred from matrix position (m, n)
to matrix position (k, j). An O is indicating a not indexed position. No O
will ever get off of the matrix. Hence not all positions will be indeXed.

But you believe that. So you are provably wrong.

Regards, WM

On 12/16/23 7:26 AM, WM wrote:
> Le 16/12/2023 à 02:05, Richard Damon a écrit :
>> On 12/15/23 11:51 AM, WM wrote:
>>> Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1:
>>>> On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
>>>>>
>>>>> What proves completeness [...]?
>>>>
>>>> A DEFINITION of /completeness/ and a PROOF (of a statement referring
>>>> to completeness).
>>>
>>> Completeness in counting the natural numbers cannot be proven because
>>> the contrary can be proven. And that is simple: For every eps > 0
>>> there are only finitely many unit fractions in (eps, 1] counted, but
>>> infinitely many in (0, eps] uncounted and uncountable because eps
>>> cannot be made small enough to leave only finitely many uncounted.
>>> Therefore every definition of completeness in this respect is nonsense.
>>>
>> So NOTHING in your logic system can completely talk about the whole of
>> Natural Numbers, so it just can't handle them.
>
> Neither can yours. XOOO...
> XOOO...
> XOOO...
> XOOO...
> ..
> The indeX transferred from matrix position (k, j) to matrix position (m,
> n) implies that, in exchange, O is transferred from matrix position (m,
> n) to matrix position (k, j). An O is indicating a not indexed position.
> No O will ever get off of the matrix. Hence not all positions will be
> indeXed.
>
> But you believe that. So you are provably wrong.
>
> Regards, WM
>

But that arguement is irrelevent. There are many ways that fail to index
the set, but the key is that they are shown equal if there exist An
indexing that matches them.

So, restricting yourself to only an indexing that doesn;t work is
meaningless.

Le 16/12/2023 à 13:48, Richard Damon a écrit :
> On 12/16/23 7:26 AM, WM wrote:
>> Le 16/12/2023 à 02:05, Richard Damon a écrit :
>>> On 12/15/23 11:51 AM, WM wrote:
>>>> Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1:
>>>>> On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
>>>>>>
>>>>>> What proves completeness [...]?
>>>>>
>>>>> A DEFINITION of /completeness/ and a PROOF (of a statement referring
>>>>> to completeness).
>>>>
>>>> Completeness in counting the natural numbers cannot be proven because
>>>> the contrary can be proven. And that is simple: For every eps > 0
>>>> there are only finitely many unit fractions in (eps, 1] counted, but
>>>> infinitely many in (0, eps] uncounted and uncountable because eps
>>>> cannot be made small enough to leave only finitely many uncounted.
>>>> Therefore every definition of completeness in this respect is nonsense.
>>>>
>>> So NOTHING in your logic system can completely talk about the whole of
>>> Natural Numbers, so it just can't handle them.
>>
>> Neither can yours. XOOO...
>> XOOO...
>> XOOO...
>> XOOO...
>> ..
>> The indeX transferred from matrix position (k, j) to matrix position (m,
>> n) implies that, in exchange, O is transferred from matrix position (m,
>> n) to matrix position (k, j). An O is indicating a not indexed position.
>> No O will ever get off of the matrix. Hence not all positions will be
>> indeXed.
>>
>> But you believe that. So you are provably wrong.
>>
> But that arguement is irrelevent.

No,it is exactly Cantor's way. Only one step is inserted: The first column
is put in bojection with the natural numbers: 1/n <--> n. But if that is
not posible, then all is impossible.

> There are many ways that fail to index
> the set,

There are only ways that fail. Otherwise every injection would be a
bijection.

> but the key is that they are shown equal if there exist An
> indexing that matches them.

It is hard to understand how Cantor could convince the matheologians of
this silly idea. But in order to help you I did use Cantor's formula k =
(m + n - 1)(m + n - 2)/2 + m with the resulting sequence 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, ...

Regards, WM

Le 15/12/2023 à 19:31, Jim Burns a écrit :
> On 12/15/2023 12:09 PM, WM wrote:

> Something which is true of
> each one of multiple objects
> is
> something which is true of
> each one of multiple objects.

Never an O leaves the matrix.

> I'm familiar with contexts in which
> it is useful to say of a function that
> it doesn't jump.

Here for instance: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
>
> This is not a fantasy.example,
> not Scrooge McDuck's vaults.
> How we say _what we intend to say_
> which is that a function doesn't jump,
> is to say something once which is
> true in infinitely.many ways.
>
>>> In instances involving infinity,
>>> we do not _accomplish_ completeness.
>>
>> That's it!
>
> Our methods take that into account.
> We begin at completeness and
> maintain completeness as we learn
> by not-first-false augmenting.

The set of O remains completely in the matrix.
>
> We don't accomplish completeness and
> we don't need to.

A bijection woul require it.

Regards, WM

On 12/16/23 8:20 AM, WM wrote:
> Le 16/12/2023 à 13:48, Richard Damon a écrit :
>> On 12/16/23 7:26 AM, WM wrote:
>>> Le 16/12/2023 à 02:05, Richard Damon a écrit :
>>>> On 12/15/23 11:51 AM, WM wrote:
>>>>> Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48
>>>>> UTC+1:
>>>>>> On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
>>>>>>>
>>>>>>> What proves completeness [...]?
>>>>>>
>>>>>> A DEFINITION of /completeness/ and a PROOF (of a statement
>>>>>> referring to completeness).
>>>>>
>>>>> Completeness in counting the natural numbers cannot be proven
>>>>> because the contrary can be proven. And that is simple: For every
>>>>> eps > 0 there are only finitely many unit fractions in (eps, 1]
>>>>> counted, but infinitely many in (0, eps] uncounted and uncountable
>>>>> because eps cannot be made small enough to leave only finitely many
>>>>> uncounted. Therefore every definition of completeness in this
>>>>> respect is nonsense.
>>>>>
>>>> So NOTHING in your logic system can completely talk about the whole
>>>> of Natural Numbers, so it just can't handle them.
>>>
>>> Neither can yours. XOOO...
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> ..
>>> The indeX transferred from matrix position (k, j) to matrix position
>>> (m, n) implies that, in exchange, O is transferred from matrix
>>> position (m, n) to matrix position (k, j). An O is indicating a not
>>> indexed position. No O will ever get off of the matrix. Hence not all
>>> positions will be indeXed.
>>>
>>> But you believe that. So you are provably wrong.
>>>
>> But that arguement is irrelevent.
>
> No,it is exactly Cantor's way. Only one step is inserted: The first
> column is put in bojection with the natural numbers: 1/n <--> n. But if
> that is not posible, then all is impossible.

Yes, you can biject n <--> 1/n, you can also biject k <--> n/m with a
different mapping, This is NOT a "Contradiction"

The fact that the first mapping missed an infinite number of terms of
the second is irrelvent.

>
>> There are many ways that fail to index the set,
>
> There are only ways that fail. Otherwise every injection would be a
> bijection.

Nope, you are confusing your qualifiers. If there exists at least one
mapping that is ONE to ONE covering all, we have a bijection.

If we can establish a mapping of every element of the first set to some
unique element in the second (maybe with some left over) and another
mapping of every element of the second to a unique element of the first
(maybe with some left over) we can also establish they are of the same size.

>
>> but the key is that they are shown equal if there exist An indexing
>> that matches them.
>
> It is hard to understand how Cantor could convince the matheologians of
> this silly idea. But in order to help you I did use Cantor's formula k =
> (m + n - 1)(m + n - 2)/2 + m with the resulting sequence 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,  ...
>
> Regards, WM
>

So, there exists A mapping of N to N/N so they are the same size.

This is just a FACT of the behavior of infinite sets.

On 12/16/2023 8:27 AM, WM wrote:
> Le 15/12/2023 à 19:31, Jim Burns a écrit :
>> On 12/15/2023 12:09 PM, WM wrote:

>>>> In instances involving infinity,
>>>> we do not _accomplish_ completeness.
>>>
>>> That's it!
>>
>> Our methods take that into account.
>> We begin at completeness and
>> maintain completeness as we learn
>> by not-first-false augmenting.
>
> The set of O remains completely
> in the matrix.
>
>> We don't accomplish completeness and
>> we don't need to.
>
> A bijection [would] require it.

⟨i,j⟩ ⟼ ⟨k,1⟩
k = i+(i+j-1)(i+j-2)/2

⟨k,1⟩ ⟼ ⟨i,j⟩
sₖ = max{h| (H-1)(h-2)/2 < k }
i+j = sₖ
i = k-(sₖ-1)(sₖ-2)/2
j = sₖ(sₖ-1)/2-k+1

That's complete.
We didn't accomplishᵂᴹ that completeness.
We didn't need to accomplishᵂᴹ it.
It's never incomplete.

We began with some statements true of
each one which we're discussing, such as
| k is in
| ordered ⟨1,…,k⟩ such that,
| for each non-empty split of ⟨1,…,k⟩
| i‖i+1 is last‖first in F‖H and
| 1‖k is first‖last in ⟨1,…,k⟩
| Initially.complete.

We augmented with only
claims not.first.false of
each one of what we're discussing.

If the descriptive claims are true,
as they are selected to be for
each one of what we're discussing,
then
the not.first.false augmenting claims are in
a finite sequence of only not.first.false

A finite sequence of only not.first.false
cannot include a false claim.

If the descriptive claims are true,
as they are selected to be true for
each one of what we're discussing,
then
the not.first.false augmenting claims are
also true of each one of what we're discussing.
Maintained.complete.

>> Something which is true of
>> each one of multiple objects
>> is
>> something which is true of
>> each one of multiple objects.
>
> Never an O leaves the matrix.

If O is at ⟨i,j⟩
then ⟨i,j⟩↔⟨kᵢⱼ,1⟩ is not swapped
kᵢⱼ = i+(i+j-1)(i+j-2)/2

If, for each ⟨i,j⟩
⟨i,j⟩↔⟨kᵢⱼ,1⟩ is swapped
then, for each ⟨i,j⟩
O is not at it.

After all ⟨i,j⟩↔⟨kᵢⱼ,1⟩ are swapped
O is not at each ⟨i,j⟩

The O's are overwritten by X's from ⟨k,1⟩
from proper.subset first.column,
which is same.sized as the whole matrix.

Herds of sheep and bags of pebbles
do not have same.sized proper.subsets

However,
these rows, columns, fractions, and swaps
are not.like.sheep in that respect.
They do have same.sized proper.subsets.

On 2023-12-16 12:18:21 +0000, WM said:

> Le 16/12/2023 à 09:07, Mikko a écrit :
>> On 2023-12-15 12:12:36 +0000, WM said:
>>
>>> On 15.12.2023 10:58, Mikko wrote:
>>>> On 2023-12-14 15:22:50 +0000, WM said:
>>>
>>>>> This requires a limit or a last step.
>>>>
>>>> You say but don't show.
>>> What proves completeness in your opinion?
>>
>> A proof of completeness. Details depend on what kind of completeness
>> you mean.
>
> You "prove" that Cantor enumerates all fractions.

No, I don't. I prefer to prove that one can associate a natural
number (or several natural numbers) to each positive rational number
so that every natural number is associated to at most one raltional
number. Or, if that is not sufficient, that there is a bijection
between natural numbers and rationals.

> That is wrong. The matrix
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ..will never be covered by indeXes.

Yes it is. Every position in the matrix has a row index and a column
indes. Positions not covered are not part of the matrix.

> The indeX transferred from matrix position (k, j) to matrix position (m, n)

What does "transferred" mean?

> implies that,

and how would a "transferred" index imply anything?

Mikko

Le 17/12/2023 à 09:53, Mikko a écrit :
> On 2023-12-16 12:18:21 +0000, WM said:

>> That is wrong. The matrix
>> XOOO...
>> XOOO...
>> XOOO...
>> XOOO...
>> ..will never be covered by indeXes.
>
> Yes it is. Every position in the matrix has a row index and a column
> indes. Positions not covered are not part of the matrix.

There are not enough indices.
>
>> The indeX transferred from matrix position (k, j) to matrix position (m, n)
>
> What does "transferred" mean?

If (m, n) is to be indexed according to Cantors formula, then the indeX is
taken from its position (k, j).

Regards, WM

Le 16/12/2023 à 20:27, Jim Burns a écrit :
> On 12/16/2023 8:27 AM, WM wrote:

>> Never an O leaves the matrix.

> The O's are overwritten by X's from ⟨k,1⟩

But they remain in the matrix. The set of positione covered by O minus the
set of positions covered by X is infinite and will never decrease.

Regards, WM

Le 16/12/2023 à 15:25, Richard Damon a écrit :

> Yes, you can biject n <--> 1/n, you can also biject k <--> n/m with a
> different mapping, This is NOT a "Contradiction"
>
> The fact that the first mapping missed an infinite number of terms of
> the second is irrelvent.

Not in mathematics!

But the fact that the second mapping misses an infinite number of terms,
proved by the eternal presence of O, proves that Cantor's way fails.

Regards, WM

On 12/17/23 2:44 PM, WM wrote:
> Le 16/12/2023 à 15:25, Richard Damon a écrit :
>
>> Yes, you can biject n <--> 1/n, you can also biject k <--> n/m with a
>> different mapping, This is NOT a "Contradiction"
>>
>> The fact that the first mapping missed an infinite number of terms of
>> the second is irrelvent.
>
> Not in mathematics!
>
> But the fact that the second mapping misses an infinite number of terms,
> proved by the eternal presence of O, proves that Cantor's way fails.
>
> Regards, WM

Yes, in mathematics of infinite sets, the fact that one mapping misses
even an infinite number of members means nothing, only the existance of
some mapping that meets the requirements.

You just don't understand the infinite.

Le 17/12/2023 à 21:23, Richard Damon a écrit :

> Yes, in mathematics of infinite sets, the fact that one mapping misses
> even an infinite number of members means nothing,

But here Cantor's mapping misses an infinite number of fractions.

Regards, WM

On 12/17/23 4:18 PM, WM wrote:
> Le 17/12/2023 à 21:23, Richard Damon a écrit :
>
>> Yes, in mathematics of infinite sets, the fact that one mapping misses
>> even an infinite number of members means nothing,
>
> But here Cantor's mapping misses an infinite number of fractions.
>
> Regards, WM
>
>
>
>

You aren't using the right mapping, showing your ignorance.

On 2023-12-17 19:36:59 +0000, WM said:

> Le 17/12/2023 à 09:53, Mikko a écrit :
>> On 2023-12-16 12:18:21 +0000, WM said:
>
>>> That is wrong. The matrix
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> ..will never be covered by indeXes.
>>
>> Yes it is. Every position in the matrix has a row index and a column
>> indes. Positions not covered are not part of the matrix.
>
> There are not enough indices.

Then the matrix is too small, but every entry in the matrix has
tu indices.

>>> The indeX transferred from matrix position (k, j) to matrix position (m, n)
>>
>> What does "transferred" mean?
>
> If (m, n) is to be indexed according to Cantors formula, then the indeX
> is taken from its position (k, j).

How does that imply anything?

Mikko

Le 18/12/2023 à 01:09, Richard Damon a écrit :
> On 12/17/23 4:18 PM, WM wrote:
>> Le 17/12/2023 à 21:23, Richard Damon a écrit :
>>
>>> Yes, in mathematics of infinite sets, the fact that one mapping misses
>>> even an infinite number of members means nothing,
>>
>> But here Cantor's mapping misses an infinite number of fractions.
>
> You aren't using the right mapping, showing your ignorance.

Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
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, ...
for the first time?

Regards, WM