Gus Gassmann schrieb am Freitag, 11. November 2022 um 13:32:57 UTC+1:
> On Friday, 11 November 2022 at 08:01:27 UTC-4, WM wrote:
> > zelos...@gmail.com schrieb am Freitag, 11. November 2022 um 07:17:38 UTC+1:
> > > onsdag 9 november 2022 kl. 16:27:45 UTC+1 skrev WM:
> >
> > > > > >"The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)] Why not take it as Cantor calls it?
> > > > > Notice he says nothing about "checking" like you imagine it!
> > > > But I require only to find "each of them only once at a determined place".
> > >
> > > Which does not help your case. It just means it needs to have a specific spot in the sequence, which all do
> > Not those which have come in the limit only.
> > All the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in trouble such that no individual leaves in a definable way.
> there is no last step!

In all steps the O's remain. Almost all fractions have not been indexed. But then comes the limit. After the limit has been attained, nothing else happens.

Regards, WM

zelos...@gmail.com schrieb am Freitag, 11. November 2022 um 14:26:57 UTC+1:
> fredag 11 november 2022 kl. 13:01:27 UTC+1 skrev WM:

> > All the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in trouble such that no individual leaves in a definable way. This implies that "in the limit" infinitely many fractions have to become indexed such that none of them can be checked - contrary to Cantor's opinion.
> Your O's do not fucking matter here!

Not indexed fractions do not matter when all fractions shall be indexed? Strange minority opinion.

Regards, WM

FromTheRafters schrieb am Freitag, 11. November 2022 um 15:28:17 UTC+1:
> WM used his keyboard to write :
> >
> > But the limit destroys the bijection.
> Huh?

We come to the conclusion that all the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in trouble such that no individual leaves in a definable way. This implies that "in the limit" infinitely many fractions have to become indexed such that none of them can be checked - contrary to Cantor's opinion.

Regards, WM

Jim Burns schrieb am Freitag, 11. November 2022 um 19:54:45 UTC+1:
> On 11/11/2022 7:05 AM, WM wrote:
> > Jim Burns schrieb am Freitag, a
> > 11. November 2022 um 00:43:15 UTC+1:
> >> I think that it is useful to keep in mind that,
> >> in this context, we get completeness _for free_
> >> We know _for free_ that our claims are true of
> >> individuals in the _complete_ set of natural numbers.
> >
> > For ℕ_def.
> For Cantor's ℕ

No.

But the dark numbers in Cantor's ℕ immediately become clear here: All the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in trouble such that no individual leaves in a definable way. This implies that "in the limit" infinitely many fractions have to become indexed such that none of them can be checked - contrary to Cantor's opinion.

Regards, WM

Chris M. Thomasson schrieb am Freitag, 11. November 2022 um 21:56:59 UTC+1:
> On 11/10/2022 2:20 AM, FromTheRafters wrote:
> > Chris M. Thomasson brought next idea :

> >>> There are an infinite number of FISONS, they miss nothing, right?
> >>
> >> They as the infinite number of FISONS, that represent N.
> >
> > That's better. Infinitely many FISONs is not the same as infinite FISONs.
> I see. Thank you. :^)

No, you don't see. An actual infinity of FISONs is impossible because of the pigeonhole principle. Only a potentially infinite collection is possible because the FISONs provide a potentially infinite collection of numbers in FISONs. In order to distinguish an actual infinity of FISONs an actually infinite FISON would be required.

Regards, WM

WM formulated on Saturday :
> zelos...@gmail.com schrieb am Freitag, 11. November 2022 um 14:26:57 UTC+1:
>> fredag 11 november 2022 kl. 13:01:27 UTC+1 skrev WM:
>
>>> All the O remain present in all definable matrices until "in the limit"
>>> these infinitely many O have to leave in trouble such that no individual
>>> leaves in a definable way. This implies that "in the limit" infinitely many
>>> fractions have to become indexed such that none of them can be checked -
>>> contrary to Cantor's opinion.
>> Your O's do not fucking matter here!
>
> Not indexed fractions do not matter when all fractions shall be indexed?
> Strange minority opinion.

Not strange at all, when all fractions are indexed. Why should
something which doesn't exist matter?

WM formulated the question :
> Chris M. Thomasson schrieb am Freitag, 11. November 2022 um 21:56:59 UTC+1:
>> On 11/10/2022 2:20 AM, FromTheRafters wrote:
>>> Chris M. Thomasson brought next idea :
>
>>>>> There are an infinite number of FISONS, they miss nothing, right?
>>>>
>>>> They as the infinite number of FISONS, that represent N.
>>>
>>> That's better. Infinitely many FISONs is not the same as infinite FISONs.
>> I see. Thank you. :^)
>
> No, you don't see. An actual infinity of FISONs is impossible because of the
> pigeonhole principle.

Nonsense. Each FISON represents a natural number, and the size of the
set of natural numbers is aleph_zero which is a countable infinity.

Absolutely no pigeons needed.

WM laid this down on his screen :
> FromTheRafters schrieb am Freitag, 11. November 2022 um 15:28:17 UTC+1:
>> WM used his keyboard to write :
>>>
>>> But the limit destroys the bijection.
>> Huh?
>
> We come to the conclusion that all the O remain present

Oh, that crap again.

FromTheRafters schrieb am Samstag, 12. November 2022 um 12:35:07 UTC+1:
> WM formulated on Saturday :

> > Not indexed fractions do not matter when all fractions shall be indexed?
> > Strange minority opinion.
> Not strange at all, when all fractions are indexed.

But they are not as long as one O is indicating a not indexed fraction.

Regards, WM

FromTheRafters schrieb am Samstag, 12. November 2022 um 12:41:24 UTC+1:
> WM formulated the question :
> > Chris M. Thomasson schrieb am Freitag, 11. November 2022 um 21:56:59 UTC+1:
> >> On 11/10/2022 2:20 AM, FromTheRafters wrote:
> >>> Chris M. Thomasson brought next idea :
> >
> >>>>> There are an infinite number of FISONS, they miss nothing, right?
> >>>>
> >>>> They as the infinite number of FISONS, that represent N.
> >>>
> >>> That's better. Infinitely many FISONs is not the same as infinite FISONs.
> >> I see. Thank you. :^)
> >
> > No, you don't see. An actual infinity of FISONs is impossible because of the
> > pigeonhole principle.
> Each FISON represents a natural number,

but not vice versa

> and the size of the
> set of natural numbers is aleph_zero which is a countable infinity.

that is more than any FISON can supply.
>
> Absolutely no pigeons needed.

No, but holes.

Regards, WM

FromTheRafters schrieb am Samstag, 12. November 2022 um 12:42:18 UTC+1:
> WM laid this down on his screen :
> > FromTheRafters schrieb am Freitag, 11. November 2022 um 15:28:17 UTC+1:
> >> WM used his keyboard to write :
> >>>
> >>> But the limit destroys the bijection.
> >> Huh?
> >
> > We come to the conclusion that all the O remain present
> Oh, that crap again.

Not desired but existing as long as defined indexes are existing.

Regards, WM

On 11/12/2022 4:16 AM, WM wrote:
> Gus Gassmann schrieb am Freitag, 11. November 2022 um 13:32:57 UTC+1:
>> On Friday, 11 November 2022 at 08:01:27 UTC-4, WM wrote:
>>> zelos...@gmail.com schrieb am Freitag, 11. November 2022 um 07:17:38 UTC+1:
>>>> onsdag 9 november 2022 kl. 16:27:45 UTC+1 skrev WM:
>>>
>>>>>>> "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)] Why not take it as Cantor calls it?
>>>>>> Notice he says nothing about "checking" like you imagine it!
>>>>> But I require only to find "each of them only once at a determined place".
>>>>
>>>> Which does not help your case. It just means it needs to have a specific spot in the sequence, which all do
>>> Not those which have come in the limit only.
>>> All the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in trouble such that no individual leaves in a definable way.
>> there is no last step!
>
> In all steps the O's remain. Almost all fractions have not been indexed. But then comes the limit. After the limit has been attained, nothing else happens.
>
> Regards, WM

4 wrongs;

1. there are no Os
2. All fractions have been indexed
3. there is no limit
4. Life after 'WMs Math' continues

On 11/12/2022 8:08 AM, WM wrote:
> FromTheRafters schrieb am Samstag, 12. November 2022 um 12:35:07 UTC+1:
>> WM formulated on Saturday :
>
>>> Not indexed fractions do not matter when all fractions shall be indexed?
>>> Strange minority opinion.
>> Not strange at all, when all fractions are indexed.
>
> But they are not as long as one O is indicating a not indexed fraction.

if you are having that problem, you did it wrong.

>
> Regards, WM

On 11/12/2022 6:10 AM, WM wrote:
> FromTheRafters schrieb am Samstag, 12. November 2022 um 12:41:24 UTC+1:
>> WM formulated the question :
>>> Chris M. Thomasson schrieb am Freitag, 11. November 2022 um 21:56:59 UTC+1:
>>>> On 11/10/2022 2:20 AM, FromTheRafters wrote:
>>>>> Chris M. Thomasson brought next idea :
>>>
>>>>>>> There are an infinite number of FISONS, they miss nothing, right?
>>>>>>
>>>>>> They as the infinite number of FISONS, that represent N.
>>>>>
>>>>> That's better. Infinitely many FISONs is not the same as infinite FISONs.
>>>> I see. Thank you. :^)
>>>
>>> No, you don't see. An actual infinity of FISONs is impossible because of the
>>> pigeonhole principle.
>> Each FISON represents a natural number,
>
> but not vice versa

A natural number can represent a FISON. Say:

i[0] = { 0 }
i[1] = { 0, 1 }
i[2] = { 0, 1, 2 }
i[3] = { 0, 1, 2, 3 }
....

On and on:

https://youtu.be/oRrlRbGT-LU

;^)

>
>> and the size of the
>> set of natural numbers is aleph_zero which is a countable infinity.
>
> that is more than any FISON can supply.
>>
>> Absolutely no pigeons needed.
>
> No, but holes.
>
> Regards, WM

On 11/12/2022 5:25 AM, WM wrote:
> Jim Burns schrieb am Freitag,
> 11. November 2022 um 19:54:45 UTC+1:
>> On 11/11/2022 7:05 AM, WM wrote:
>>> Jim Burns schrieb am Freitag, a
>>> 11. November 2022 um 00:43:15 UTC+1:

>>>> I think that it is useful to keep in mind that,
>>>> in this context, we get completeness _for free_
>>>> We know _for free_ that our claims are true of
>>>> individuals in the _complete_ set of
>>>> natural numbers.
>>>
>>> For ℕ_def.
>>
>> For Cantor's ℕ
>
> No.

Cantor:
> Wir wollen nun eine vollständige Induktion ausführen,

| We now want to carry out a complete induction

What Cantor carried out is _reasoning_ about
the _complete_ set of natural numbers.

Consider _attainable_ and _omnipresent_ numbers.

An _attainable_ number k ends some FISON 𝐹

FISON 𝐹 is
an ordered collection 𝐹 such that
0 begins 𝐹 and 𝐹 ends somewhere and
each split of 𝐹 has
a successor-step across that split.

An _omnipresent_ number exists in
each collection in which there is
at least each attainable number.

Lemma.
Each attainable number is omnipresent.

I think that what you are calling dark numbers
are those which are omnipresent-but-unattainable.

Omnipresent-but-unattainable numbers do not exist.

Lemma.
Each omnipresent number is attainable.

| Assume otherwise.
| Assume n is omnipresent-but-unattainable.
| | Let 𝑆 contain each attainable number and n
| n is not attainable.
| 𝑆\{n} also contains each attainable number,
| but 𝑆\{n} doesn't contain n
| | n is not omnipresent.
| Contradiction.

Therefore,
omnipresent-but-unattainable numbers do not exist.

Each omnipresent number is attainable.

Lemma.
Attainable numbers and omnipresent numbers
are the same numbers.

> But the dark numbers in Cantor's ℕ
> immediately become clear here:
> All the O remain present in all definable matrices
> until "in the limit"

A limit is not something a sequence arrives at.
The existence of a limit, the value of a limit,
if it exists, are facts about a sequence

The only things a sequence arrives at are
the items in the sequence.

If the value of the limit is the value of
one of the items, then the sequence arrives at
its limit. If not, then not.

The matrix-sequence of X's and O's which starts
XOOO⋯
XOOO⋯
XOOO⋯
⋮ ⋮ ⋮ ⋱

has a limit
XXXX⋯
XXXX⋯
XXXX⋯
⋮ ⋮ ⋮ ⋱

The sequence does not arrive at its limit.

> these infinitely many O have to leave in trouble
> such that no individual leaves in a definable way.
> This implies that
> "in the limit" infinitely many fractions
> have to become indexed such that
> none of them can be checked
> - contrary to Cantor's opinion.

Consider Cantor's sequence as fraction-index pairs.
𝑪 = ⟨ ⟨1/1 1⟩ ⟨1/2 2⟩ ⟨2/1 3⟩ ⟨1/3 4⟩ ⋯ ⟩

For each split of 𝑪
there is a step ⟨p/q k⟩ ⟨p'/q' k'⟩ across that split
such that
if q = 1 then p'/q' = 1/p⁺⁺
otherwise p'/q' = p⁺⁺/q⁻⁻
k' = k⁺⁺

For each omnipresent-and-attainable fraction p/q
𝑪 has a unique fraction-index pair ⟨p/q k⟩
s = p+q
k = (s-2)(s-2)/2+p

For each omnipresent-and-attainable index k
𝑪 has a unique fraction-index pair ⟨p/q k⟩
s = ⌈(((8k+1)¹ᐟ²+1)/2)⌉
p = k-(s-1)(s-2)/2
q = s-p

Since 𝑪 is a sequence,
if 𝑪 arrived at any omnipresent-but-
-unattainable fraction p/q or index k
there would need to be a _first_ omnipresent-but-
-unattainable fraction p₁/q₁ or index k₁

However,
since p₁/q₁ or k₁ would be first omnipresent-but-
unattainable, that implies
q₁⁻⁻/1 is omnipresent-and-attainable,
so 1/q₁⁻⁻⁺⁺ = 1/q₁ is attainable
or
p₁⁻⁻/q₁⁺⁺ is omnipresent-and-attainable,
so p₁⁻⁻⁺⁺/q₁⁺⁺⁻⁻ = p1/q1 is omnipresent-and-attainable
or
k₁⁻⁻ is omnipresent-and-attainable,
so k₁⁻⁻⁺⁺ = k₁ is omnipresent-and-attainable.
Contradiction.

Therefore,
there are no omnipresent-but-unattainable
fractions or indexes in Cantor's sequence 𝑪

On Saturday, November 12, 2022 at 2:52:20 PM UTC-8, Jim Burns wrote:
> On 11/12/2022 5:25 AM, WM wrote:
> > Jim Burns schrieb am Freitag,
> > 11. November 2022 um 19:54:45 UTC+1:
> >> On 11/11/2022 7:05 AM, WM wrote:
> >>> Jim Burns schrieb am Freitag, a
> >>> 11. November 2022 um 00:43:15 UTC+1:
>
> >>>> I think that it is useful to keep in mind that,
> >>>> in this context, we get completeness _for free_
> >>>> We know _for free_ that our claims are true of
> >>>> individuals in the _complete_ set of
> >>>> natural numbers.
> >>>
> >>> For ℕ_def.
> >>
> >> For Cantor's ℕ
> >
> > No.
> Cantor:
> > Wir wollen nun eine vollständige Induktion ausführen,
>
> | We now want to carry out a complete induction
> What Cantor carried out is _reasoning_ about
> the _complete_ set of natural numbers.
> Consider _attainable_ and _omnipresent_ numbers.
>
> An _attainable_ number k ends some FISON 𝐹
>
> FISON 𝐹 is
> an ordered collection 𝐹 such that
> 0 begins 𝐹 and 𝐹 ends somewhere and
> each split of 𝐹 has
> a successor-step across that split.
>
> An _omnipresent_ number exists in
> each collection in which there is
> at least each attainable number.
>
> Lemma.
> Each attainable number is omnipresent.
>
> I think that what you are calling dark numbers
> are those which are omnipresent-but-unattainable.
>
> Omnipresent-but-unattainable numbers do not exist.
>
> Lemma.
> Each omnipresent number is attainable.
>
> | Assume otherwise.
> | Assume n is omnipresent-but-unattainable.
> |
> | Let 𝑆 contain each attainable number and n
> | n is not attainable.
> | 𝑆\{n} also contains each attainable number,
> | but 𝑆\{n} doesn't contain n
> |
> | n is not omnipresent.
> | Contradiction.
>
> Therefore,
> omnipresent-but-unattainable numbers do not exist.
>
> Each omnipresent number is attainable.
>
> Lemma.
> Attainable numbers and omnipresent numbers
> are the same numbers.
> > But the dark numbers in Cantor's ℕ
> > immediately become clear here:
> > All the O remain present in all definable matrices
> > until "in the limit"
> A limit is not something a sequence arrives at.
> The existence of a limit, the value of a limit,
> if it exists, are facts about a sequence
>
> The only things a sequence arrives at are
> the items in the sequence.
>
> If the value of the limit is the value of
> one of the items, then the sequence arrives at
> its limit. If not, then not.
>
> The matrix-sequence of X's and O's which starts
> XOOO⋯
> XOOO⋯
> XOOO⋯
> ⋮ ⋮ ⋮ ⋱
>
> has a limit
> XXXX⋯
> XXXX⋯
> XXXX⋯
> ⋮ ⋮ ⋮ ⋱
>
> The sequence does not arrive at its limit.
> > these infinitely many O have to leave in trouble
> > such that no individual leaves in a definable way.
> > This implies that
> > "in the limit" infinitely many fractions
> > have to become indexed such that
> > none of them can be checked
> > - contrary to Cantor's opinion.
> Consider Cantor's sequence as fraction-index pairs.
> 𝑪 = ⟨ ⟨1/1 1⟩ ⟨1/2 2⟩ ⟨2/1 3⟩ ⟨1/3 4⟩ ⋯ ⟩
>
> For each split of 𝑪
> there is a step ⟨p/q k⟩ ⟨p'/q' k'⟩ across that split
> such that
> if q = 1 then p'/q' = 1/p⁺⁺
> otherwise p'/q' = p⁺⁺/q⁻⁻
> k' = k⁺⁺
>
> For each omnipresent-and-attainable fraction p/q
> 𝑪 has a unique fraction-index pair ⟨p/q k⟩
> s = p+q
> k = (s-2)(s-2)/2+p
>
> For each omnipresent-and-attainable index k
> 𝑪 has a unique fraction-index pair ⟨p/q k⟩
> s = ⌈(((8k+1)¹ᐟ²+1)/2)⌉
> p = k-(s-1)(s-2)/2
> q = s-p
>
> Since 𝑪 is a sequence,
> if 𝑪 arrived at any omnipresent-but-
> -unattainable fraction p/q or index k
> there would need to be a _first_ omnipresent-but-
> -unattainable fraction p₁/q₁ or index k₁
>
> However,
> since p₁/q₁ or k₁ would be first omnipresent-but-
> unattainable, that implies
> q₁⁻⁻/1 is omnipresent-and-attainable,
> so 1/q₁⁻⁻⁺⁺ = 1/q₁ is attainable
> or
> p₁⁻⁻/q₁⁺⁺ is omnipresent-and-attainable,
> so p₁⁻⁻⁺⁺/q₁⁺⁺⁻⁻ = p1/q1 is omnipresent-and-attainable
> or
> k₁⁻⁻ is omnipresent-and-attainable,
> so k₁⁻⁻⁺⁺ = k₁ is omnipresent-and-attainable.
> Contradiction.
>
> Therefore,
> there are no omnipresent-but-unattainable
> fractions or indexes in Cantor's sequence 𝑪

What "complete set of natural numbers"?

Or, "which"?

On 11/12/2022 3:50 PM, Ross A. Finlayson wrote:
> On Saturday, November 12, 2022 at 2:52:20 PM UTC-8, Jim Burns wrote:
>> On 11/12/2022 5:25 AM, WM wrote:
>>> Jim Burns schrieb am Freitag,
>>> 11. November 2022 um 19:54:45 UTC+1:
>>>> On 11/11/2022 7:05 AM, WM wrote:
>>>>> Jim Burns schrieb am Freitag, a
>>>>> 11. November 2022 um 00:43:15 UTC+1:
>>
>>>>>> I think that it is useful to keep in mind that,
>>>>>> in this context, we get completeness _for free_
>>>>>> We know _for free_ that our claims are true of
>>>>>> individuals in the _complete_ set of
>>>>>> natural numbers.
>>>>>
>>>>> For ℕ_def.
>>>>
>>>> For Cantor's ℕ
>>>
>>> No.
>> Cantor:
>>> Wir wollen nun eine vollständige Induktion ausführen,
>>
>> | We now want to carry out a complete induction
>> What Cantor carried out is _reasoning_ about
>> the _complete_ set of natural numbers.
>> Consider _attainable_ and _omnipresent_ numbers.
>>
>> An _attainable_ number k ends some FISON 𝐹
>>
>> FISON 𝐹 is
>> an ordered collection 𝐹 such that
>> 0 begins 𝐹 and 𝐹 ends somewhere and
>> each split of 𝐹 has
>> a successor-step across that split.
>>
>> An _omnipresent_ number exists in
>> each collection in which there is
>> at least each attainable number.
>>
>> Lemma.
>> Each attainable number is omnipresent.
>>
>> I think that what you are calling dark numbers
>> are those which are omnipresent-but-unattainable.
>>
>> Omnipresent-but-unattainable numbers do not exist.
>>
>> Lemma.
>> Each omnipresent number is attainable.
>>
>> | Assume otherwise.
>> | Assume n is omnipresent-but-unattainable.
>> |
>> | Let 𝑆 contain each attainable number and n
>> | n is not attainable.
>> | 𝑆\{n} also contains each attainable number,
>> | but 𝑆\{n} doesn't contain n
>> |
>> | n is not omnipresent.
>> | Contradiction.
>>
>> Therefore,
>> omnipresent-but-unattainable numbers do not exist.
>>
>> Each omnipresent number is attainable.
>>
>> Lemma.
>> Attainable numbers and omnipresent numbers
>> are the same numbers.
>>> But the dark numbers in Cantor's ℕ
>>> immediately become clear here:
>>> All the O remain present in all definable matrices
>>> until "in the limit"
>> A limit is not something a sequence arrives at.
>> The existence of a limit, the value of a limit,
>> if it exists, are facts about a sequence
>>
>> The only things a sequence arrives at are
>> the items in the sequence.
>>
>> If the value of the limit is the value of
>> one of the items, then the sequence arrives at
>> its limit. If not, then not.
>>
>> The matrix-sequence of X's and O's which starts
>> XOOO⋯
>> XOOO⋯
>> XOOO⋯
>> ⋮ ⋮ ⋮ ⋱
>>
>> has a limit
>> XXXX⋯
>> XXXX⋯
>> XXXX⋯
>> ⋮ ⋮ ⋮ ⋱
>>
>> The sequence does not arrive at its limit.
>>> these infinitely many O have to leave in trouble
>>> such that no individual leaves in a definable way.
>>> This implies that
>>> "in the limit" infinitely many fractions
>>> have to become indexed such that
>>> none of them can be checked
>>> - contrary to Cantor's opinion.
>> Consider Cantor's sequence as fraction-index pairs.
>> 𝑪 = ⟨ ⟨1/1 1⟩ ⟨1/2 2⟩ ⟨2/1 3⟩ ⟨1/3 4⟩ ⋯ ⟩
>>
>> For each split of 𝑪
>> there is a step ⟨p/q k⟩ ⟨p'/q' k'⟩ across that split
>> such that
>> if q = 1 then p'/q' = 1/p⁺⁺
>> otherwise p'/q' = p⁺⁺/q⁻⁻
>> k' = k⁺⁺
>>
>> For each omnipresent-and-attainable fraction p/q
>> 𝑪 has a unique fraction-index pair ⟨p/q k⟩
>> s = p+q
>> k = (s-2)(s-2)/2+p
>>
>> For each omnipresent-and-attainable index k
>> 𝑪 has a unique fraction-index pair ⟨p/q k⟩
>> s = ⌈(((8k+1)¹ᐟ²+1)/2)⌉
>> p = k-(s-1)(s-2)/2
>> q = s-p
>>
>> Since 𝑪 is a sequence,
>> if 𝑪 arrived at any omnipresent-but-
>> -unattainable fraction p/q or index k
>> there would need to be a _first_ omnipresent-but-
>> -unattainable fraction p₁/q₁ or index k₁
>>
>> However,
>> since p₁/q₁ or k₁ would be first omnipresent-but-
>> unattainable, that implies
>> q₁⁻⁻/1 is omnipresent-and-attainable,
>> so 1/q₁⁻⁻⁺⁺ = 1/q₁ is attainable
>> or
>> p₁⁻⁻/q₁⁺⁺ is omnipresent-and-attainable,
>> so p₁⁻⁻⁺⁺/q₁⁺⁺⁻⁻ = p1/q1 is omnipresent-and-attainable
>> or
>> k₁⁻⁻ is omnipresent-and-attainable,
>> so k₁⁻⁻⁺⁺ = k₁ is omnipresent-and-attainable.
>> Contradiction.
>>
>> Therefore,
>> there are no omnipresent-but-unattainable
>> fractions or indexes in Cantor's sequence 𝑪
>
> What "complete set of natural numbers"?
>
> Or, "which"?
>

Every Cantor Pairing of a natural number results in a unique pair.

Chris M. Thomasson wrote :
> On 11/12/2022 6:10 AM, WM wrote:
>> FromTheRafters schrieb am Samstag, 12. November 2022 um 12:41:24 UTC+1:
>>> WM formulated the question :
>>>> Chris M. Thomasson schrieb am Freitag, 11. November 2022 um 21:56:59
>>>> UTC+1:
>>>>> On 11/10/2022 2:20 AM, FromTheRafters wrote:
>>>>>> Chris M. Thomasson brought next idea :
>>>>
>>>>>>>> There are an infinite number of FISONS, they miss nothing, right?
>>>>>>>
>>>>>>> They as the infinite number of FISONS, that represent N.
>>>>>>
>>>>>> That's better. Infinitely many FISONs is not the same as infinite
>>>>>> FISONs.
>>>>> I see. Thank you. :^)
>>>>
>>>> No, you don't see. An actual infinity of FISONs is impossible because of
>>>> the
>>>> pigeonhole principle.
>>> Each FISON represents a natural number,
>>
>> but not vice versa
>
> A natural number can represent a FISON. Say:
>
> i[0] = { 0 }
> i[1] = { 0, 1 }
> i[2] = { 0, 1, 2 }
> i[3] = { 0, 1, 2, 3 }
> ...
>
> On and on:

But off by one.

On 11/12/2022 6:50 PM, Ross A. Finlayson wrote:
> On Saturday, November 12, 2022
> at 2:52:20 PM UTC-8, Jim Burns wrote:

>> Cantor:
>>> Wir wollen nun eine vollständige Induktion ausführen,
>>
>> | We now want to carry out a complete induction
>>
>> What Cantor carried out is _reasoning_ about
>> the _complete_ set of natural numbers.
>> Consider _attainable_ and _omnipresent_ numbers.

> What "complete set of natural numbers"?
>
> Or, "which"?

Cantor's ℕ
_AKA_
our ℕ
_AKA_
the unique ℕ with 0, and all its successors,
and induction
_AKA_
0 ∈ ℕ ∧ ∀j ∈ ℕ : j⁺⁺ ∈ ℕ
0 ∈ 𝑆 ∧ ∀j ∈ 𝑆 : j⁺⁺ ∈ 𝑆 ⟹ ℕ ⊆ S
_AKA_
the _attainable_ numbers
in the union of FISONs
_AKA_
the _omnipresent_ numbers
in the unique intersection of inductive subsets
of any inductive set

I think that it is useful to keep in mind that,
in this context, we get completeness _for free_
We know _for free_ that our claims are true of
individuals in the _complete_ set of natural numbers.

First, we know for free the claims are _complete_
because we know what it is we are talking about.

Second, third, and thereafter, we know for free
those claims are _complete_ because we keep the
completeness we had gotten for free.

We didn't need to work for it, and that's
just as well. We are finite, and they are infinite,
and any cost at all would be too much for us.

Compare to:
We know for free that a right triangle has
three corners and a right angle.
There is no imaginable method to "check"
each right triangle. But what would we check?
How we know is by knowing the meaning of
"right triangle", not by looking at even one
right triangle.

My point is exactly as trivial as it sounds.
Each right triangle is a right triangle.
Each natural number is a natural number.

But it is a trivial and _important_ point.
We are able to begin with _complete_ claims
by describing what we are talking about.
Because we know what we are talking about,
we are able to do this for free. Emphatically,
we are able to do it without infinite cost.

On 11/12/2022 4:20 PM, FromTheRafters wrote:
> Chris M. Thomasson wrote :
>> On 11/12/2022 6:10 AM, WM wrote:
>>> FromTheRafters schrieb am Samstag, 12. November 2022 um 12:41:24 UTC+1:
>>>> WM formulated the question :
>>>>> Chris M. Thomasson schrieb am Freitag, 11. November 2022 um
>>>>> 21:56:59 UTC+1:
>>>>>> On 11/10/2022 2:20 AM, FromTheRafters wrote:
>>>>>>> Chris M. Thomasson brought next idea :
>>>>>
>>>>>>>>> There are an infinite number of FISONS, they miss nothing, right?
>>>>>>>>
>>>>>>>> They as the infinite number of FISONS, that represent N.
>>>>>>>
>>>>>>> That's better. Infinitely many FISONs is not the same as infinite
>>>>>>> FISONs.
>>>>>> I see. Thank you. :^)
>>>>>
>>>>> No, you don't see. An actual infinity of FISONs is impossible
>>>>> because of the
>>>>> pigeonhole principle.
>>>> Each FISON represents a natural number,
>>>
>>> but not vice versa
>>
>> A natural number can represent a FISON. Say:
>>
>> i[0] = { 0 }
>> i[1] = { 0, 1 }
>> i[2] = { 0, 1, 2 }
>> i[3] = { 0, 1, 2, 3 }
>> ...
>>
>> On and on:
>
> But off by one.

I started my indexing at zero, well, zero is not a natural?

On Saturday, November 12, 2022 at 4:32:53 PM UTC-8, Chris M. Thomasson wrote:
> On 11/12/2022 4:20 PM, FromTheRafters wrote:
> > Chris M. Thomasson wrote :
> >> On 11/12/2022 6:10 AM, WM wrote:
> >>> FromTheRafters schrieb am Samstag, 12. November 2022 um 12:41:24 UTC+1:
> >>>> WM formulated the question :
> >>>>> Chris M. Thomasson schrieb am Freitag, 11. November 2022 um
> >>>>> 21:56:59 UTC+1:
> >>>>>> On 11/10/2022 2:20 AM, FromTheRafters wrote:
> >>>>>>> Chris M. Thomasson brought next idea :
> >>>>>
> >>>>>>>>> There are an infinite number of FISONS, they miss nothing, right?
> >>>>>>>>
> >>>>>>>> They as the infinite number of FISONS, that represent N.
> >>>>>>>
> >>>>>>> That's better. Infinitely many FISONs is not the same as infinite
> >>>>>>> FISONs.
> >>>>>> I see. Thank you. :^)
> >>>>>
> >>>>> No, you don't see. An actual infinity of FISONs is impossible
> >>>>> because of the
> >>>>> pigeonhole principle.
> >>>> Each FISON represents a natural number,
> >>>
> >>> but not vice versa
> >>
> >> A natural number can represent a FISON. Say:
> >>
> >> i[0] = { 0 }
> >> i[1] = { 0, 1 }
> >> i[2] = { 0, 1, 2 }
> >> i[3] = { 0, 1, 2, 3 }
> >> ...
> >>
> >> On and on:
> >
> > But off by one.
> I started my indexing at zero, well, zero is not a natural?

It's usually included. Don't let anyone tell you otherwise.

On Saturday, November 12, 2022 at 4:30:12 PM UTC-8, Jim Burns wrote:
> On 11/12/2022 6:50 PM, Ross A. Finlayson wrote:
> > On Saturday, November 12, 2022
> > at 2:52:20 PM UTC-8, Jim Burns wrote:
>
> >> Cantor:
> >>> Wir wollen nun eine vollständige Induktion ausführen,
> >>
> >> | We now want to carry out a complete induction
> >>
> >> What Cantor carried out is _reasoning_ about
> >> the _complete_ set of natural numbers.
> >> Consider _attainable_ and _omnipresent_ numbers.
> > What "complete set of natural numbers"?
> >
> > Or, "which"?
> Cantor's ℕ
> _AKA_
> our ℕ
> _AKA_
> the unique ℕ with 0, and all its successors,
> and induction
> _AKA_
> 0 ∈ ℕ ∧ ∀j ∈ ℕ : j⁺⁺ ∈ ℕ
> 0 ∈ 𝑆 ∧ ∀j ∈ 𝑆 : j⁺⁺ ∈ 𝑆 ⟹ ℕ ⊆ S
> _AKA_
> the _attainable_ numbers
> in the union of FISONs
> _AKA_
> the _omnipresent_ numbers
> in the unique intersection of inductive subsets
> of any inductive set
> I think that it is useful to keep in mind that,
> in this context, we get completeness _for free_
> We know _for free_ that our claims are true of
> individuals in the _complete_ set of natural numbers.
> First, we know for free the claims are _complete_
> because we know what it is we are talking about.
> Second, third, and thereafter, we know for free
> those claims are _complete_ because we keep the
> completeness we had gotten for free.
> We didn't need to work for it, and that's
> just as well. We are finite, and they are infinite,
> and any cost at all would be too much for us.
>
> Compare to:
> We know for free that a right triangle has
> three corners and a right angle.
> There is no imaginable method to "check"
> each right triangle. But what would we check?
> How we know is by knowing the meaning of
> "right triangle", not by looking at even one
> right triangle.
>
> My point is exactly as trivial as it sounds.
> Each right triangle is a right triangle.
> Each natural number is a natural number.
>
> But it is a trivial and _important_ point.
> We are able to begin with _complete_ claims
> by describing what we are talking about.
> Because we know what we are talking about,
> we are able to do this for free. Emphatically,
> we are able to do it without infinite cost.

Oh, you mean Russell's Frege's not-Frege's?

Notice if you just Russell-ize the finites it includes itself.

Russell did - ..., that hypocrite.

On Saturday, November 12, 2022 at 4:32:53 PM UTC-8, Chris M. Thomasson wrote:
> On 11/12/2022 4:20 PM, FromTheRafters wrote:
> > Chris M. Thomasson wrote :
> >> On 11/12/2022 6:10 AM, WM wrote:
> >>> FromTheRafters schrieb am Samstag, 12. November 2022 um 12:41:24 UTC+1:
> >>>> WM formulated the question :
> >>>>> Chris M. Thomasson schrieb am Freitag, 11. November 2022 um
> >>>>> 21:56:59 UTC+1:
> >>>>>> On 11/10/2022 2:20 AM, FromTheRafters wrote:
> >>>>>>> Chris M. Thomasson brought next idea :
> >>>>>
> >>>>>>>>> There are an infinite number of FISONS, they miss nothing, right?
> >>>>>>>>
> >>>>>>>> They as the infinite number of FISONS, that represent N.
> >>>>>>>
> >>>>>>> That's better. Infinitely many FISONs is not the same as infinite
> >>>>>>> FISONs.
> >>>>>> I see. Thank you. :^)
> >>>>>
> >>>>> No, you don't see. An actual infinity of FISONs is impossible
> >>>>> because of the
> >>>>> pigeonhole principle.
> >>>> Each FISON represents a natural number,
> >>>
> >>> but not vice versa
> >>
> >> A natural number can represent a FISON. Say:
> >>
> >> i[0] = { 0 }
> >> i[1] = { 0, 1 }
> >> i[2] = { 0, 1, 2 }
> >> i[3] = { 0, 1, 2, 3 }
> >> ...
> >>
> >> On and on:
> >
> > But off by one.
> I started my indexing at zero, well, zero is not a natural?

(Finite positive integers are usually called whole numbers,
which is more brief than 'non-zero naturals'.)

On Saturday, November 12, 2022 at 10:17:27 PM UTC+1, Chris M. Thomasson wrote:

> A natural number can represent a FISON.

If we define "the natural numbers" due to von Neumann, the natural Numbers ARE FISONs.

0 = {}
1 = {0}
2 = {0, 1}
3 = {0, 1, 2}
:

Which means: IN = Set_of_FISONs.

On Saturday, November 12, 2022 at 12:41:24 PM UTC+1, FromTheRafters wrote:

> Each FISON represents a natural number, and [....]

If we define "the natural numbers" due to von Neumann, the natural numbers ARE FISONs.

0 = {}
1 = {0}
2 = {0, 1}
3 = {0, 1, 2}
:

Which means: IN = Set_of_FISONs.