On Monday, May 23, 2022 at 1:25:22 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:00:16 UTC+2:
> > On Sunday, May 22, 2022 at 3:29:12 PM UTC+2, WM wrote:
> > >
> > > Do you know an example of a finite [bunch of 'would be' elements] which is not a set?
> > >
> > Sure, consider the two proper classes ORD (the class of all ordinal numbers) and R (the Russell class).
> >
> They are a set of two elements.

No, the aren't. Actually,

> > these two (proper) classes do not form a set. [Got that you idiot?!]
> >
> > i.e. there's no set X such that Ax(x e X <-> x = ORD v x = R).*)
> >
> An element of a set need not consist of elements.

Then it is empty. Neither ORD nor R are empty.

So these two (proper) classes deliver the example you were asking for.

__________________________________________________________

*) At least not in the common set theories NBG or MK which allow for "proper classes" besides sets.
See: https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory
and: https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory

WM wrote on 5/23/2022 :
> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:00:16 UTC+2:
>> On Sunday, May 22, 2022 at 3:29:12 PM UTC+2, WM wrote:
>
>>> Do you know an example of a finite [bunch of 'would be' elements] which is
>>> not a set?
>>
>> Sure, consider the two proper classes ORD (the class of all ordinal numbers)
>> and R (the Russell class).
>
> They are a set of two elements.

Don't elements have to be sets?

>> These two (proper) classes do not form a set. I.e. There's no set X such
>> that Ax(x e X <-> x = ORD v x = R).*)
>
> An element of a set need not consist of elements.

Correct, but do you know why?

On Monday, May 23, 2022 at 2:23:15 PM UTC+2, FromTheRafters wrote:
> WM wrote on 5/23/2022 :
> > Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:00:16 UTC+2:
> >> On Sunday, May 22, 2022 at 3:29:12 PM UTC+2, WM wrote:
> > > >
> > > > Do you know an example of a finite [bunch of 'would be' elements] which is
> > > > not a set?
> > > >
> > > Sure, consider the two proper classes ORD (the class of all ordinal numbers)
> > > and R (the Russell class).
> > >
> > They are a set of two elements.
> >
> Don't elements have to be sets?

Sure, at least in the common set theories NBG or MK which allow for "proper classes" besides sets.

See: https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory
and: https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory

Especially NBG is closely related to our "standard" set theory ZFC.

> >> These two (proper) classes do not form a set. I.e. There's no set X such
> >> that Ax(x e X <-> x = ORD v x = R).*)
> >
> > An element of a set need not consist of elements.
> >
> Correct, but do you know why?

Hell!

He just claimed that the empty set is not actually a set, quoting Zermelo:

"It is not a genuine set and was introduced by me only for formal reasons." [E. Zermelo, letter to A. Fraenkel (1 Mar 1921)] "I increasingly doubt the justifiability of the 'null set'. Perhaps one can dispense with it by restricting the axiom of separation in a suitable way. Indeed, it serves only the purpose of formal simplification." [E. Zermelo, letter to A. Fraenkel (9 May 1921)]

Alas, more than 100 years later, the empty set is still with us. (Actually, Zermelos statements are quite nonsensical.)

On Monday, May 23, 2022 at 8:31:05 AM UTC-3, WM wrote:
> William schrieb am Sonntag, 22. Mai 2022 um 17:07:48 UTC+2:
> > On Sunday, May 22, 2022 at 10:24:53 AM UTC-3, WM wrote:
> > > All O's will vanish. This is obviously impossible
> > Piffle. The intersection of a sequence of sets, each of cardinality alpeh_0, can be empty.
> Here we are not at all interested in any intersection

Piffle. We are looking at the sets indexes if the 0's. The intersection of the of the sequence of sets of indexes is empty. This is not a contradiction, it is one of the results that you do not like.

--
William Hughes

On 5/23/2022 6:37 AM, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 19:04:52 UTC+2:
>> On Sunday, May 22, 2022 at 5:09:11 PM UTC+2, Fritz Feldhase wrote:
>>> On Sunday, May 22, 2022 at 3:24:53 PM UTC+2, WM wrote:
>>>
>>>> If the fractions m/n are enumerated by the natural numbers k according to Cantor's function
>>>> k = (m + n - 1)(m + n - 2)/2 + m
>>>> then all the fractions of the sequence
>>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
>>>> are enumerated.

>>
> That appears so to you because you don't know dark numbers. But they are proved by the fact that the share of enumerated fractions does never change. Your claim would require that the series -1 + 1 - 1 + 1 - 1 + 1 -+ = oo.
>
> Regards, WM

"dark numbers" is where you, WM, put your unknowns. your claims are bogus.

William schrieb am Montag, 23. Mai 2022 um 15:18:21 UTC+2:
> On Monday, May 23, 2022 at 8:31:05 AM UTC-3, WM wrote:
> > William schrieb am Sonntag, 22. Mai 2022 um 17:07:48 UTC+2:
> > > On Sunday, May 22, 2022 at 10:24:53 AM UTC-3, WM wrote:
> > > > All O's will vanish. This is obviously impossible
> > > Piffle. The intersection of a sequence of sets, each of cardinality alpeh_0, can be empty.
> > Here we are not at all interested in any intersection
> Piffle. We are looking at the sets indexes if the 0's.

We are looking at the final .result
If the natural numbers first are in bijection with the integer fractions of the first column of 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, ...
....

then they must be distributed over the matrix such that no fraction remains without index. That means, there is a permutation such that the X of the first column

XOOOO...
XOOOO...
XOOOO...
XOOOO...
XOOOO...
....

by being exchanged with the O's cover all matrix positions. All O's will vanish. This is obviously impossible because exchanging cannot reduce them. The number of not indexed fractions, represented by O's, will remain constant forever, in infinity.

Regards, WM

On Tuesday, May 24, 2022 at 7:55:26 AM UTC-3, WM wrote:
> William schrieb am Montag, 23. Mai 2022 um 15:18:21 UTC+2:
>> We are looking at the sets indexes if the 0's.
> We are looking at the final .result

Indeed, and the final result is that the final set of indexes of the O's (an intersection of a sequence of sets each of which has cardinality aleph_0) is empty.

--
William Hughes