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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Thu, 5 Dec 2024 22:31:01 +0100
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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