Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

Am Thu, 05 Dec 2024 22:31:01 +0100 schrieb WM:

On 05.12.2024 21:11, joes wrote:

Am Thu, 05 Dec 2024 20:30:22 +0100 schrieb WM:

On 05.12.2024 18:12, Jim Burns wrote:

On 12/5/2024 4:00 AM, WM wrote:

On 04.12.2024 21:36, Jim Burns wrote:

On 12/4/2024 12:29 PM, WM wrote:

>

No intersection of more.than.finitely.many end.segments of the
finite.cardinals holds a finite.cardinal,  or is non.empty.

Small wonder.
More than finitely many endsegments require infinitely many indices,
i.e., all indices. No natnumbers are remaining in the contents.

⎛ That's the intersection.

And it is the empty endsegment.

There is no empty segment.

If all natnumbers have been lost, then nothing remains. If there are
infinitely many endsegments, then all contents has become indices.

Yes, and then we don't need any more "contents".

The contents cannot disappear "in the limit". It has to be lost one by
one if ∀k ∈ ℕ : E(k+1) = E(k) \ {k} is really true for all natnumbers.

Same thing. Every finite number is "lost" in some segment.
All segments are infinite.

Two identical sequences have the same limit.
As long as all endsegments are infinite so is their intersection.

As long as you intersect only finitely many segments.

More than finitely many endsegments require infinitely many indices,
i.e., all indices. No natnumbers are remaining in the contents.

I really don't understand this connection. First, this also makes every
segment infinite. The set of all indices is the infinite N.

Yes. It is E(1) having all natnumbers as its content.

And there is no empty segment.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.