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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Sat, 7 Dec 2024 16:21:32 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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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.