WM submitted this idea :
> FromTheRafters schrieb am Montag, 14. November 2022 um 12:45:23 UTC+1:
>> WM explained :
>
>>> A counterexample is an index not issued in a finite number of steps or a
>>> step where the number of O's has decreased.
>> There are no steps, verifying step-by-step is not needed to show the
>> bijection.
>>
> But it is possible, if there is a bijection, i.e., if "every number p/q comes
> at an absolutely fixed place of a simple infinite sequence"

No, it would be possible if an infinite step-by-step process could be
completed. It can't, so it isn't. For each p/q there is a unique next
p/q and we can label (or associate) each p/q with a next natural
number.

WM brought next idea :
> zelos...@gmail.com schrieb am Montag, 14. November 2022 um 14:28:48 UTC+1:
>> You cannot find one natural number excluded!
>
> But I have proved that many fractions are not indexed and cannot be found
> either.
>
> Here is the Outline of the proof:
>
> (1) We assume that all natural numbers are existing and are indexing all
> integer fractions in a matrix of all positive fractions.

Wrong approach. First, envision the naturals as being your 'matrix'
like this;

1, 2, 6, 7, 15
3, 5, 8, 14
4, 9,13
10,12
11

....

See the serpentine pattern? It could as easily be non-alternating
striped pattern, it doesn't matter.

Arrange your p/q elements of Cantor's sequence using the same pattern
too.

If you cannot see the one-to-one pairing of these elements then there
is no hope you will ever understand why what you are doing is just
insane.

WM has brought this to us :
>
> Jim Burns schrieb am Montag, 14. November 2022 um 17:20:58 UTC+1:
>> On 11/14/2022 5:01 AM, WM wrote:
>>> Jim Burns schrieb am Sonntag,
>
>>>> What you think Cantor says is not right.
>>>> The only things a sequence arrives at are
>>>> the items in the sequence.
>>> [E. Zermelo:
>>> "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)]
>>> [E. Zermelo:
>>
>>> [E. Zermelo:
>>> The clearness of these expressions is noteworthy:
>>> all and every,
>>> at an absolutely fixed place,
>>> at a definite position,
>>> not a single one has been forgotten.
>> If Cantor and Zermelo mean what you think they
>> mean, then Cantor and Zermelo are not right.
>
> They mean what they said. The clarity of these expressions is noteworthy.

They hadn't met you.

It's clear to others that it shows "none are missed" because there is a
place for everything and everything is in its place -- there cannot be
ones missed or left over at some fantasy ending point.

On Tuesday, November 15, 2022 at 4:30:25 AM UTC-8, FromTheRafters wrote:
> WM has brought this to us :
> >
> > Jim Burns schrieb am Montag, 14. November 2022 um 17:20:58 UTC+1:
> >> On 11/14/2022 5:01 AM, WM wrote:
> >>> Jim Burns schrieb am Sonntag,
> >
> >>>> What you think Cantor says is not right.
> >>>> The only things a sequence arrives at are
> >>>> the items in the sequence.
> >>> [E. Zermelo:
> >>> "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)]
> >>> [E. Zermelo:
> >>
> >>> [E. Zermelo:
> >>> The clearness of these expressions is noteworthy:
> >>> all and every,
> >>> at an absolutely fixed place,
> >>> at a definite position,
> >>> not a single one has been forgotten.
> >> If Cantor and Zermelo mean what you think they
> >> mean, then Cantor and Zermelo are not right.
> >
> > They mean what they said. The clarity of these expressions is noteworthy.
> They hadn't met you.
>
> It's clear to others that it shows "none are missed" because there is a
> place for everything and everything is in its place -- there cannot be
> ones missed or left over at some fantasy ending point.

Nobody has. Every time I see "WM" it reads "White Male".
Not all my heroes are Dead White Males.
Many are, though.

They earned it.

On 11/15/2022 2:49 AM, WM wrote:
> FromTheRafters schrieb am Montag, 14. November 2022 um 12:45:23 UTC+1:
>> WM explained :
>
>>> A counterexample is an index not issued in a finite number of steps or a step
>>> where the number of O's has decreased.
>> There are no steps, verifying step-by-step is not needed to show the
>> bijection.
>>
> But it is possible, if there is a bijection, i.e., if "every number p/q comes at an absolutely fixed place of a simple infinite sequence"
>
> Regards, WM

so you are saying it is possible to completly verify using "step by step" on an infinite sequence ?

note that:
"every number p/q comes at an absolutely fixed place of a simple infinite sequence"

is a bijection, as each "absolutely fixed place" has a unique natural number for it.

On 11/15/2022 3:26 AM, WM wrote:
> Gus Gassmann schrieb am Montag, 14. November 2022 um 18:58:15 UTC+1:
>> On Monday, 14 November 2022 at 05:57:23 UTC-4, WM wrote:
>> [...]
>>> You are seeing the light in the darkness. When I say "all", I mean, and should say, every definable.
>> Aside from the fact that there is no difference between "all natural numbers" and "all definable natural numbers",
>
> Wrong. See my proof. Here is the Outline of the proof
>
> (1) We assume that all natural numbers are existing and are indexing all integer fractions in a matrix of all positive fractions.

so you agree with Cantor's Enumeration of the rationals.

> (2) Then we distribute according to Cantor's prescription these indices over the whole matrix. We observe that in every step prescribed by Cantor the set of indices does not increase and the set of not indexed fractions does not decrease.

this was already done by Cantor, in his one to one mapping. see above (1)

> (3) Therefore it is impossible to index all fractions in a definable way. Indexing many fractions "in the limit" would be undefined and can be excluded according to the second section below.

you use the word "define" or "definable" however you mean "finite" which is wrong.
in (1) above an infinite set is mapped one to one to another infinite set, it is already done.

there is no limit either.

so (3) is 100% wrong

> (4) After having issued all indices only indexed fractions can be seen in the matrix.

Wrong. both sets are infinite and mapped one to one in (1) above. Every fraction is indexed by a natural number

> (5) We conclude from the existing but invisible not indexed fractions that they are sitting at invisible positions inside of the matrix.

Wrong. to have not indexed fractions, you must have stopped in your step by step. Failure.

Your invisibles is your imaginations.

> (6) Hence also the first column of the matrix and therefore also  has invisible, co-called dark elements.

Wrong. No darkies there. for clearity use another matrix or set instead of the first column.

> (7) Hence also the initial mapping of natural numbers and integer fractions cannot have been complete.

because you stopped in an infinite set, and that is obvious and silly.

> Bijections, i.e., complete mappings, of infinite sets and are impossible.

Wrong. your ability to comprehend "mappings" is impossible.

>
>> if you do not say what you mean
>
> I said: Every definable *FINITE* number obeys
> ∀n ∈ ℕ_def *FINITE*: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> All numbers however yield
> |ℕ \ {1, 2, 3, ...}| = 0 . *INFINITE*
>
> Regards, WM

1 out of 7 correct! 14/100 total Failure.

correct you paper, and submit it again, very poor work.

On 11/15/2022 4:37 AM, WM wrote:
> Jim Burns schrieb am Montag,
> 14. November 2022 um 17:20:58 UTC+1:

>> A number is _attainable_ iff
>> an ordered set exists
>> such that
>> for each split of that set,
>> a successor-step across that split exists,
>> and that set begins at 0
>> and that set ends at that number.

>> A set is _attaining_ iff
>> each attainable number is in that set.
>
> Infinite sets of this kind do not exist,
> because always ℵo successors remain unattainable.

No.

Here is the distinction you (WM) gave
between a collection and a set:
sets do not change.

Whether
something is an attainable number
or not
is determined by whether
a set of the given description exists
which it ends
or not.

Sets of the given description
do not change.
Not from not existing to existing.
Not from existing to not existing.
Not from having one end to having another.

Whether something is an attainable number
does not change.

Define a collection.
You call it what you like.
I'll call it { attainable n }

_Define_ { attainable n } so that
whether
something is in { attainable n }
is determined by
whether
it is an attainable number
-- a fact about it which doesn't change.

{ attainable n } exists,
by the axioms of collections (AKA pluralities)

{ attainable n } doesn't change.

{ attainable n } is a set,
by the condition you set.

The _set_ { attainable n }
contains each attainable number.

{ attainable n } is an attaining set.
==
{ attainable n } is an _attainer_

>> A set is _attaining_ iff
>> each attainable number is in that set.
>
> Infinite sets of this kind do not exist,
> because always ℵo successors remain unattainable.

No.
{ attainable n } is a set of this kind.

>> A number is _omnipresent_ iff
>> each attaining set contains that number.
>
> The set {1} does not contain 2
> but 2 is an attainable number

So NOT each attainable number is in {1}
and {1} is NOT an attaining set.

> and {1} is an attaining set.

No. 2 ∉ {1}

----
I like the sound of "attainer" more than
"attaining set", so you'll see me use both.
They are completely interchangeable.

On 11/14/2022 11:20 AM, Jim Burns wrote:
> On 11/14/2022 5:01 AM, WM wrote:

>> "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)]

I am describing Cantor's sequence 𝐶
which has the non-standard order '⩹'
𝐶 = ⟨ 1/1 1/2 2/1 1/3 ⋯ cₖ ⋯ ⟩

Most of this description works equally well for any
sequence. I make it explicit because
the existence of a sequence can lead to
consequences which are interesting. Unbelievable,
for some people.

An item cₖ is _attainable_ iff
an ordered-by-⩹ set 𝐹ₖ exists
such that
for each split 𝐵ᵢ 𝐹ₖ\𝐵ᵢ there is a
a ⩹-successor-step cᵢ cᵢ₊₊ across it
and 1/1 begins 𝐹ₖ
and cₖ ends 𝐹ₖ

A set 𝑆 is an _attainer_ iff
each attainable item cₖ is in 𝑆

An item cₖ is _omnipresent_ iff
each attainer 𝑆 contains cₖ

Lemma.
The attainable items and the omnipresent items
are the same items.
No omnipresent-but-not-attainable items exist.

Definition.
Cantor's sequence 𝐶 is the ordered-by-⩹ set
of attainable-and-omnipresent items.

Consider an item cₖ of 𝐶 as a fraction-index-pair
cₖ = ⟨p/q k⟩

Definition.
cₖ₊₊ is the ⩹-successor of cₖ where
cₖ = ⟨p/q k⟩
cₖ₊₊ = ⟨p'/q' k'⟩

p'/q' = 1/p⁺⁺ if q = 1

p'/q' = p⁺⁺/q⁻⁻ otherwise

k' = k⁺⁺

For each item cₖ = ⟨p/q k⟩
p q k are attainable-and-omnipresent items
of the natural-number sequence ℕ⁺
ℕ⁺ = ⟨ 1 2 3 ⋯ n ⋯ ⟩

"Attainable", "attainer", "omnipresent",
"sequence" are as above.

There are several descriptions of the
natural-number successor i⁺⁺ of i
The one for the von Neumann ordinals
might be easiest to describe.
i⁺⁺ ≝ i ∪ {i}

The essential properties of '⁺⁺' are:
each successor has a successor,
each successor is not 0,
each successor is different.

0 is not a successor.
0 has a successor.

FromTheRafters schrieb am Dienstag, 15. November 2022 um 13:10:53 UTC+1:
> WM submitted this idea :
> > FromTheRafters schrieb am Montag, 14. November 2022 um 12:45:23 UTC+1:
> >> WM explained :
> >
> >>> A counterexample is an index not issued in a finite number of steps or a
> >>> step where the number of O's has decreased.
> >> There are no steps, verifying step-by-step is not needed to show the
> >> bijection.
> >>
> > But it is possible, if there is a bijection, i.e., if "every number p/q comes
> > at an absolutely fixed place of a simple infinite sequence"
> No, it would be possible if an infinite step-by-step process could be
> completed.

That is: "every number p/q comes at an absolutely fixed place of a simple infinite sequence"

> It can't, so it isn't.

That's what I prove.

> For each p/q there is a unique next
> p/q and we can label (or associate) each p/q with a next natural
> number.

That is not sufficient. For a bijection nothing must remain. All pairs have to sit at fixed finite places. In particular Cantor's diagonal number requires this completeness.

Regards, WM

FromTheRafters schrieb am Dienstag, 15. November 2022 um 13:23:57 UTC+1:
> WM brought next idea :
> > zelos...@gmail.com schrieb am Montag, 14. November 2022 um 14:28:48 UTC+1:
> >> You cannot find one natural number excluded!
> >
> > But I have proved that many fractions are not indexed and cannot be found
> > either.
> >
> > Here is the Outline of the proof:
> >
> > (1) We assume that all natural numbers are existing and are indexing all
> > integer fractions in a matrix of all positive fractions.
> Wrong approach. First, envision the naturals as being your 'matrix'
> like this;
>
> 1, 2, 6, 7, 15
> 3, 5, 8, 14
> 4, 9,13
> 10,12
> 11
>
> ...
>
> See the serpentine pattern? It could as easily be non-alternating
> striped pattern, it doesn't matter.

It will never fill the whole matrix, never cross the diagonal.
But in order to see the defect, it is better to proceed as shown below:
> Arrange your p/q elements of Cantor's sequence using the same pattern
> too.

Never fill the whole matrix.
>
> If you cannot see the one-to-one pairing of these elements then there
> is no hope you will ever understand why what you are doing is just
> insane.

I can see it for the tiny left upper corner. But that does not prove anything. My proof covers all the matrix.

If all positive fractions m/n are existing, then they all are contained in the matrix

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
...

If all natural numbers k are existing, then they can be used as indices to index the integer fractions m/1 of the first column. Denoting indexed fractions by X and not indexed fractions by O, we obtain the following matrix:

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

Cantor claimed that all natural numbers k are existing and can be applied to index all positive fractions m/n. They are distributed according to

k = (m + n - 1)(m + n - 2)/2 + m.

The result is a sequence of fractions

1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .

This sequence is modelled here in the language of matrices. The indices are taken from their initial positions in the first column and are distributed in the given order.

Index 1 remains at the first term 1/1. The next term 1/2 is indexed with 2 which is taken from position 2/1.

XXOO...
OOOO...
XOOO...
XOOO...
XOOO...
...

Then index 3 it taken from 3/1 and attached to 2/1:

XXOO...
OOOO...
XOOO...
XOOO...
XOOO...
...

Then index 4 it taken from 4/1 and attached to 1/3:

XXXO...
XOOO...
OOOO...
OOOO...
XOOO...
...

Then index 5 it taken from 5/1 and attached to 2/2:

XXXO...
XXOO...
OOOO...
OOOO...
OOOO...
...

And so on. When finally all exchanges of X and O have been carried through and, according to Cantor, all fractions have been enumerated, we get the matrix

XXXX...
XXXX...
XXXX...
XXXX...
XXXX...
...

When, according to Cantor, all indices have been issued, it turns out that no fraction without index is visible any longer, but by the technique of lossless exchange of X and O no O can have left the matrix as long as finite natural numbers are issued as indexes. Therefore there are not less fractions without index than at the beginning.

Regards, WM

FromTheRafters schrieb am Dienstag, 15. November 2022 um 13:30:25 UTC+1:
> WM has brought this to us :
> >
> > Jim Burns schrieb am Montag, 14. November 2022 um 17:20:58 UTC+1:
> >> On 11/14/2022 5:01 AM, WM wrote:
> >>> Jim Burns schrieb am Sonntag,
> >
> >>>> What you think Cantor says is not right.
> >>>> The only things a sequence arrives at are
> >>>> the items in the sequence.
> >>> [E. Zermelo:
> >>> "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)]
> >>> [E. Zermelo:
> >>
> >>> [E. Zermelo:
> >>> The clearness of these expressions is noteworthy:
> >>> all and every,
> >>> at an absolutely fixed place,
> >>> at a definite position,
> >>> not a single one has been forgotten.
> >> If Cantor and Zermelo mean what you think they
> >> mean, then Cantor and Zermelo are not right.
> >
> > They mean what they said. The clarity of these expressions is noteworthy.
> They hadn't met you.
>
> It's clear to others that it shows "none are missed" because there is a
> place for everything and everything is in its place -- there cannot be
> ones missed or left over at some fantasy ending point.

"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."

Therefore every place can be checked step by step because it is a finite place.

Regards, WM

On 11/15/2022 1:37 AM, WM wrote:
>
> Jim Burns schrieb am Montag, 14. November 2022 um 17:20:58 UTC+1:
>> On 11/14/2022 5:01 AM, WM wrote:
>>> Jim Burns schrieb am Sonntag,
>
>>>> What you think Cantor says is not right.
>>>> The only things a sequence arrives at are
>>>> the items in the sequence.
>>> [E. Zermelo:
>>> "The infinite sequence thus defined has the
>>> peculiar property to contain all positive
>>> rational numbers and each of them only once
>>> at a determine> at a definite position,
>>> not a single one has been forgotten.
>> If Cantor and Zermelo mean what you think they
>> mean, then Cantor and Zermelo are not right.
>
> They mean what they said. The clarity of these expressions is noteworthy.
>
>> The only things a sequence arrives at are
>> the items in the sequence.
>> "All of the sequence" does not include anything
>> which is not in the sequence.
>
> It includes all natnumbers and all fractions, each one at e definite place. No limits.
>>
>> If I tell you I am talking about the widgets
>> produced at the widget factory you and I are
>> standing in, "all the widgets" does not include
>> widgets made at any other factory.
>
> The mapping includes all natnumbers and all fractions.
>>
>> A number is _attainable_ iff
>> an ordered set exists
>> such that
>> for each split of that set,
>> a successor-step across that split exists,
>> and that set begins at 0
>> and that set ends at that number.
>
> That is the last number of a FISON. It is attainable or definable and has
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo successors. Only when we drop the precondition of beeing attainable, we can get rid of the not attainable numbers, but only collectively
> |ℕ \ {1, 2, 3, ...}| = 0 .
>>
>> A set is _attaining_ iff
>> each attainable number is in that set.
>
> Infinite sets of this kind do not exist, because always ℵo successors remain unattainable.
>>d place."
>>> [G. Cantor, letter to R. Lipschitz
>>> (19 Nov 1883)]
>>> [E. Zermelo:
>>
>>> [E. Zermelo:
>>> The clearness of these expressions is noteworthy:
>>> all and every,
>>> at an absolutely fixed place,
>>
>> A number is _omnipresent_ iff
>> each attaining set contains that number.
>
> The set {1} does not contain 2 but 2 is an attainable number and {1} is an attaining set.

Did you get dropped on your head as a baby?

Sergi o schrieb am Dienstag, 15. November 2022 um 15:23:38 UTC+1:
> On 11/15/2022 2:49 AM, WM wrote:
> > FromTheRafters schrieb am Montag, 14. November 2022 um 12:45:23 UTC+1:
> >> WM explained :
> >
> >>> A counterexample is an index not issued in a finite number of steps or a step
> >>> where the number of O's has decreased.
> >> There are no steps, verifying step-by-step is not needed to show the
> >> bijection.
> >>
> > But it is possible, if there is a bijection, i.e., if "every number p/q comes at an absolutely fixed place of a simple infinite sequence"
> >
> so you are saying it is possible to completly verify using "step by step" on an infinite sequence ?

Cantor is saying so. He uses finished infinity, "das vollendete Unendliche".. das vollendete Unendliche, oder auch das Transfinite, Überendliche) ausspreche. Das Irrthümliche in jener Gauss'schen Stelle besteht darin, dass er sagt, das Vollendetunendliche könne nicht Gegenstand mathematischer Betrachtungen werden; dieser Irrthum hängt mit dem andern Irrthum zusammen, dass er [...] das Vollendetunendliche mit dem Absoluten, Göttlichen identificirt, [...] Das Vollendetunendliche findet sich allerdings in gewissem Sinne in den Zahlen ,  + 1, ..., , ...; sie sind Zeichen für gewisse Modi des Vollendetunendlichen und weil das Vollendetunendliche in verschiedenen, von einander mit der äussersten Schärfe durch den sogenannten 'endlichen, menschlichen Verstand' unterscheidbaren Modificationen auftreten kann, so sieht man hieraus deutlich wie weit man vom Absoluten entfernt ist, obgleich man das Vollendetunendliche sehr wohl fassen und sogar mathematisch auffassen kann." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

"da nun jeder Typus auch im letzteren Falle etwas in sich Bestimmtes, vollendetes ist, so gilt ein gleiches von der zu ihm gehörigen Zahl. [...] 'Eigentlichunendlichem = Transfinitum = Vollendetunendlichem = Unendlichseiendem = kategorematice infinitum' [...] dieser Unterschied ändert aber nichts daran, daß  als ebenso bestimmt und vollendet anzusehen ist, wie 2," [G. Cantor, letter to K. Laßwitz (15 Feb 1884). Cantor, p. 395]

> note that:
> "every number p/q comes at an absolutely fixed place of a simple infinite sequence"
> is a bijection, as each "absolutely fixed place" has a unique natural number for it.

which can be reached and checked in a finite number if steps. There are no infinite numbers of steps. There are no limits.

Regards, WM

On 11/15/2022 9:15 AM, WM wrote:
> FromTheRafters schrieb am Dienstag, 15. November 2022 um 13:10:53 UTC+1:
>> WM submitted this idea :
>>> FromTheRafters schrieb am Montag, 14. November 2022 um 12:45:23 UTC+1:
>>>> WM explained :
>>>
>>>>> A counterexample is an index not issued in a finite number of steps or a
>>>>> step where the number of O's has decreased.
>>>> There are no steps, verifying step-by-step is not needed to show the
>>>> bijection.
>>>>
>>> But it is possible, if there is a bijection, i.e., if "every number p/q comes
>>> at an absolutely fixed place of a simple infinite sequence"
>> No, it would be possible if an infinite step-by-step process could be
>> completed.
>
> That is: "every number p/q comes at an absolutely fixed place of a simple infinite sequence"
>
>> It can't, so it isn't.
>
> That's what I prove.
>
>> For each p/q there is a unique next
>> p/q and we can label (or associate) each p/q with a next natural
>> number.
>
> That is not sufficient. For a bijection nothing must remain. All pairs have to sit at fixed finite places. In particular Cantor's diagonal number requires this completeness.

You truly are a full blown moron. Sorry, but it sure seems to be
99.9...% true.

Gus Gassmann schrieb am Dienstag, 15. November 2022 um 15:38:24 UTC+1:
> On Tuesday, 15 November 2022 at 05:26:45 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Montag, 14. November 2022 um 18:58:15 UTC+1:
> > > On Monday, 14 November 2022 at 05:57:23 UTC-4, WM wrote:
> > > [...]
> > > > You are seeing the light in the darkness. When I say "all", I mean, and should say, every definable.
> > > Aside from the fact that there is no difference between "all natural numbers" and "all definable natural numbers",
> > Wrong. See my proof. Here is the Outline of the proof
> > (1) We assume that all natural numbers are existing and are indexing all integer fractions in a matrix of all positive fractions.
> > (2) Then we distribute according to Cantor's prescription these indices over the whole matrix. We observe that in every step prescribed by Cantor the set of indices does not increase and the set of not indexed fractions does not decrease.
> > (3) Therefore it is impossible to index all fractions in a definable way. Indexing many fractions "in the limit" [...]
>
> Nobody is doing that,

You tried it until I showed that you are wrong.

> For every fraction there is a natural number (that is, a finite! index) which indexes it.

That excludes limits. But it allows to issue every index by exchanging X and O.

> For any 'O' in your matrix there is a finite index (a natural number --- even definable in your sense, as it turns out!) when this 'O' is turned into an 'X'.

Pure wrong belief. That can be excluded for all finite exchanges.

Regards, WM

Sergi o schrieb am Dienstag, 15. November 2022 um 15:40:36 UTC+1:
> On 11/15/2022 3:26 AM, WM wrote:
> > Gus Gassmann schrieb am Montag, 14. November 2022 um 18:58:15 UTC+1:
> >> On Monday, 14 November 2022 at 05:57:23 UTC-4, WM wrote:
> >> [...]
> >>> You are seeing the light in the darkness. When I say "all", I mean, and should say, every definable.
> >> Aside from the fact that there is no difference between "all natural numbers" and "all definable natural numbers",
> >
> > Wrong. See my proof. Here is the Outline of the proof
> >
> > (1) We assume that all natural numbers are existing and are indexing all integer fractions in a matrix of all positive fractions.
> so you agree with Cantor's Enumeration of the rationals.

No. Only the integer fractions. And even that is only assumed and later on disproved.

Regards, WM

Jim Burns schrieb am Dienstag, 15. November 2022 um 17:02:20 UTC+1:
> On 11/15/2022 4:37 AM, WM wrote:

> Whether something is an attainable number
> does not change.

Wrong. Attainable means potential infinity.

> > The set {1} does not contain 2
> > but 2 is an attainable number
> So NOT each attainable number is in {1}
> and {1} is NOT an attaining set.

It attains 1.

> > and {1} is an attaining set.
> No. 2 ∉ {1}

Then there is no infinite attainer.

Regards, WM

WM explained :
> FromTheRafters schrieb am Dienstag, 15. November 2022 um 13:10:53 UTC+1:
>> WM submitted this idea :
>>> FromTheRafters schrieb am Montag, 14. November 2022 um 12:45:23 UTC+1:
>>>> WM explained :
>>>
>>>>> A counterexample is an index not issued in a finite number of steps or a
>>>>> step where the number of O's has decreased.
>>>> There are no steps, verifying step-by-step is not needed to show the
>>>> bijection.
>>>>
>>> But it is possible, if there is a bijection, i.e., if "every number p/q
>>> comes at an absolutely fixed place of a simple infinite sequence"
>> No, it would be possible if an infinite step-by-step process could be
>> completed.
>
> That is: "every number p/q comes at an absolutely fixed place of a simple
> infinite sequence"
>
>> It can't, so it isn't.
>
> That's what I prove.
>
>> For each p/q there is a unique next
>> p/q and we can label (or associate) each p/q with a next natural
>> number.
>
> That is not sufficient. For a bijection nothing must remain.

Wrong.

On 11/15/2022 11:15 AM, WM wrote:
> FromTheRafters schrieb am Dienstag, 15. November 2022 um 13:10:53 UTC+1:
>> WM submitted this idea :
>>> FromTheRafters schrieb am Montag, 14. November 2022 um 12:45:23 UTC+1:
>>>> WM explained :
>>>
>>>>> A counterexample is an index not issued in a finite number of steps or a
>>>>> step where the number of O's has decreased.
>>>> There are no steps, verifying step-by-step is not needed to show the
>>>> bijection.
>>>>
>>> But it is possible, if there is a bijection, i.e., if "every number p/q comes
>>> at an absolutely fixed place of a simple infinite sequence"
>> No, it would be possible if an infinite step-by-step process could be
>> completed.
>
> That is: "every number p/q comes at an absolutely fixed place of a simple infinite sequence"

correct, Each place is fixed and numbered by a natural number, like the seats at a football game.

>
>> It can't, so it isn't.
>
> That's what I prove.

no, it was no proof. you gave an idea, and illustration that failed.

>
>> For each p/q there is a unique next
>> p/q and we can label (or associate) each p/q with a next natural
>> number.
>
> That is not sufficient.

wrong, that is all that is necessary.

> For a bijection nothing must remain.

all natural numbers are paired with specific fractions, all of them.

> All pairs have to sit at fixed finite places.

and that place is numbered with a specific natural number.

> In particular Cantor's diagonal number requires this completeness.

moot.

> Regards, WM

On Tuesday, 15 November 2022 at 13:28:09 UTC-4, WM wrote:
> FromTheRafters schrieb am Dienstag, 15. November 2022 um 13:30:25 UTC+1:
> > WM has brought this to us :
> > >
> > > Jim Burns schrieb am Montag, 14. November 2022 um 17:20:58 UTC+1:
> > >> On 11/14/2022 5:01 AM, WM wrote:
> > >>> Jim Burns schrieb am Sonntag,
> > >
> > >>>> What you think Cantor says is not right.
> > >>>> The only things a sequence arrives at are
> > >>>> the items in the sequence.
> > >>> [E. Zermelo:
> > >>> "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)]
> > >>> [E. Zermelo:
> > >>
> > >>> [E. Zermelo:
> > >>> The clearness of these expressions is noteworthy:
> > >>> all and every,
> > >>> at an absolutely fixed place,
> > >>> at a definite position,
> > >>> not a single one has been forgotten.
> > >> If Cantor and Zermelo mean what you think they
> > >> mean, then Cantor and Zermelo are not right.
> > >
> > > They mean what they said. The clarity of these expressions is noteworthy.
> > They hadn't met you.
> >
> > It's clear to others that it shows "none are missed" because there is a
> > place for everything and everything is in its place -- there cannot be
> > ones missed or left over at some fantasy ending point.
>
> "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."
> Therefore every place can be checked step by step because it is a finite place.

Of course. And whenever you have checked that finite place, you have infinitely many more finite places left to check. This seems to be an insurmountable problem to you alone. Why are you so dense?

On Tuesday, November 15, 2022 at 10:36:08 AM UTC-8, Gus Gassmann wrote:
> On Tuesday, 15 November 2022 at 13:28:09 UTC-4, WM wrote:
> > FromTheRafters schrieb am Dienstag, 15. November 2022 um 13:30:25 UTC+1:
> > > WM has brought this to us :
> > > >
> > > > Jim Burns schrieb am Montag, 14. November 2022 um 17:20:58 UTC+1:
> > > >> On 11/14/2022 5:01 AM, WM wrote:
> > > >>> Jim Burns schrieb am Sonntag,
> > > >
> > > >>>> What you think Cantor says is not right.
> > > >>>> The only things a sequence arrives at are
> > > >>>> the items in the sequence.
> > > >>> [E. Zermelo:
> > > >>> "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)]
> > > >>> [E. Zermelo:
> > > >>
> > > >>> [E. Zermelo:
> > > >>> The clearness of these expressions is noteworthy:
> > > >>> all and every,
> > > >>> at an absolutely fixed place,
> > > >>> at a definite position,
> > > >>> not a single one has been forgotten.
> > > >> If Cantor and Zermelo mean what you think they
> > > >> mean, then Cantor and Zermelo are not right.
> > > >
> > > > They mean what they said. The clarity of these expressions is noteworthy.
> > > They hadn't met you.
> > >
> > > It's clear to others that it shows "none are missed" because there is a
> > > place for everything and everything is in its place -- there cannot be
> > > ones missed or left over at some fantasy ending point.
> >
> > "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."
> > Therefore every place can be checked step by step because it is a finite place.
> Of course. And whenever you have checked that finite place, you have infinitely many more finite places left to check. This seems to be an insurmountable problem to you alone. Why are you so dense?

Well, this thread's about reached an end.

1000 posts, coming up.

No news here.

On 11/15/2022 11:36 AM, WM wrote:
> Sergi o schrieb am Dienstag, 15. November 2022 um 15:23:38 UTC+1:
>> On 11/15/2022 2:49 AM, WM wrote:
>>> FromTheRafters schrieb am Montag, 14. November 2022 um 12:45:23 UTC+1:
>>>> WM explained :
>>>
>>>>> A counterexample is an index not issued in a finite number of steps or a step
>>>>> where the number of O's has decreased.
>>>> There are no steps, verifying step-by-step is not needed to show the
>>>> bijection.
>>>>
>>> But it is possible, if there is a bijection, i.e., if "every number p/q comes at an absolutely fixed place of a simple infinite sequence"
>>>
>> so you are saying it is possible to completly verify using "step by step" on an infinite sequence ?
>
> Cantor is saying so.

totally Wrong.

"every number p/q comes at an absolutely fixed place of a simple infinite sequence"
is a bijection, as each "absolutely fixed place" has a unique natural number for it.

steps do not apply. You are doing it wrong.

Cantor gave us the formula to map one to one the rationals; all, each and every fractions are directly related to a specific natural number.

applying step by step does not apply, nor does a limit.

since you do not know algebra, you cannot grasp the concept that an equation holds.

>
> Regards, WM

On 11/15/2022 11:45 AM, WM wrote:
> Sergi o schrieb am Dienstag, 15. November 2022 um 15:40:36 UTC+1:
>> On 11/15/2022 3:26 AM, WM wrote:
>>> Gus Gassmann schrieb am Montag, 14. November 2022 um 18:58:15 UTC+1:
>>>> On Monday, 14 November 2022 at 05:57:23 UTC-4, WM wrote:
>>>> [...]
>>>>> You are seeing the light in the darkness. When I say "all", I mean, and should say, every definable.
>>>> Aside from the fact that there is no difference between "all natural numbers" and "all definable natural numbers",
>>>
>>> Wrong. See my proof. Here is the Outline of the proof
>>>
>>> (1) We assume that all natural numbers are existing and are indexing all integer fractions in a matrix of all positive fractions.
>> so you agree with Cantor's Enumeration of the rationals.
>
> No. Only the integer fractions. And even that is only assumed and later on disproved.
>
> Regards, WM

is 1/2 an integer fraction ?

On 11/15/2022 12:50 PM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 15. November 2022 um 17:02:20 UTC+1:

>> Whether something is an attainable number
>> does not change.
>
> Wrong.
> Attainable means potential infinity.

I have told you what I mean by "attainable".
If you or Cantor or Zermelo say otherwise,
then you are not correct.

A number is _attainable_ iff
an ordered set exists
such that
for each split of that set,
a successor-step across that split exists,
and that set begins at 0
and that set ends at that number.

Whether a set of the given description exists
does not change.

Whether a number is attainable
does not change.

Whether something is in
the collection {attainable n} of attainable numbers
does not change.

The collection {attainable n} is a set which
contains each attainable number.

There is no omnipresent-by-not-attainable
number in {attainable n}

>>> The set {1} does not contain 2
>>> but 2 is an attainable number
>>
>> So NOT each attainable number is in {1}
>> and {1} is NOT an attaining set.
>
> It attains 1.

{1} doesn't attain 2, so {1} isn't an attainer.

A set is an _attainer_ iff
each attainable number is in that set.

>>> and {1} is an attaining set.
>>
>> No. 2 ∉ {1}
>
> Then there is no infinite attainer.

No.
{attainable n} is an attainer.

The collection (ie, plurality) {attainable n} exists
because, for collections,
∃𝐴 ∀n : n ∈ 𝐴 ⟺ Attainable(n)

and
{attainable n} is such a collection.

The collection {attainable n} is a set
because whether or not something is attainable
does not change.

{attainable n} contains each attainable number.

{attainable n} is an attainer.

There is no omnipresent-by-not-attainable
number in {attainable n}

On 11/15/2022 10:41 AM, Sergi o wrote:
> On 11/15/2022 11:45 AM, WM wrote:
>> Sergi o schrieb am Dienstag, 15. November 2022 um 15:40:36 UTC+1:
>>> On 11/15/2022 3:26 AM, WM wrote:
>>>> Gus Gassmann schrieb am Montag, 14. November 2022 um 18:58:15 UTC+1:
>>>>> On Monday, 14 November 2022 at 05:57:23 UTC-4, WM wrote:
>>>>> [...]
>>>>>> You are seeing the light in the darkness. When I say "all", I
>>>>>> mean, and should say, every definable.
>>>>> Aside from the fact that there is no difference between "all
>>>>> natural numbers" and "all definable natural numbers",
>>>>
>>>> Wrong. See my proof. Here is the Outline of the proof
>>>>
>>>> (1) We assume that all natural numbers are existing and are indexing
>>>> all integer fractions in a matrix of all positive fractions.
>>> so you agree with Cantor's Enumeration of the rationals.
>>
>> No. Only the integer fractions. And even that is only assumed and
>> later on disproved.
>>
>> Regards, WM
>
> is 1/2 an integer fraction ?
>

The numerator and denominator are integers. Since Cantor Pairing maps
the naturals to all of the positive fractions, it contains all of the
fractions that can be used to represent positive irrational numbers via
convergents of continued fractions... Fair enough?