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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: Fri, 10 Jan 2025 09:15:02 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 09 Jan 2025 22:55:13 +0100 schrieb WM:
> On 09.01.2025 21:17, joes wrote:
>> Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:
> 
>>> Losing all numbers but keeping infinitely many is impossible in
>>> inclusion-monotonic sequences.
>> This case doesn't occur.
> Loss of all numbers is proven by the empty intersection.
> Keeping infinitely many is poved by Fritsche.
....for different cases. There is no empty segment, each is infinite.

>>>If all endsegments remain infinite, we have a
>>> contradiction.
>> No, they are subsets of the same cardinality. There is no
>> contradiction.
> They remain infinite. But infinitely many endsegments require all
> natnumbers as indices. What makes up their infinite content?
You may have noticed that every segment is different.

>>>> In the sequence of end.segments of ℕ there is no number which empties
>>>> an infinite set to a finite set.
>>> Then there cannot exist a sequence of endsegments obeying ∀k ∈ ℕ:
>>> E(k+1) = E(k) \ {k+1} for all k ∈ ℕ and getting empty.
>> No term of the sequence is empty, if you mean that.
> Then not all natnumbers are outside of content and inside of the set of
> indices.
Untrue. The sequence is, unfathomably, infinite.

>>>> and there is no number which is in common with all its endsegments.
>>> Therefore all numbers get lost from the content and become indices.
>> WDYM "become"? There is no point at which all naturals would be counted
>> - N being infinite.
> The endsegment E(n) loses its element n+1 ad becomes E(n+1).
>
>>>> ℕ has only infinite endsegments.
>>> Then it has only finitely many, because not all numbers get lost from
>>> the content.
>> Huh? No. Then not all numbers would be "indices".
> Then there are only finitely many indices.
Contradiction. There are inf. many.

>>>> The intersection of all (infinite) end.segments of ℕ is empty.
>>> What is the content if all elements of ℕ have become indices?
>> There is no such endsegment.
> What element of ℕ does not become an index?
omega is not an element of N.

-- 
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.