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

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.

 

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.

And in all succeeding endsegments.

All segments are infinite.

Try to find a way to think straight. Two identical sequences have the same limit.
E(1)∩E(2)∩...∩E(n) = E(n).
As long as all endsegments are infinite so is their intersection.

 

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.
Regards, WM