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

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.

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. No natural
is part of every segment (in particular not that of its successor,
which all naturals have infinitely many of). There is no natural
after which every number is "lost". All segments are infinite.

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.
But that is not itself a segment, therefore doesn't need a "content".

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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.