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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Fri, 22 Nov 2024 12:39:17 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 22 Nov 2024 13:08:28 +0100 schrieb WM:
> On 22.11.2024 08:49, Moebius wrote:
>> Am 22.11.2024 um 03:58 schrieb Chris M. Thomasson:
>>> On 11/21/2024 1:45 PM, WM wrote:
>>>> On 21.11.2024 22:05, joes wrote:
> 
>>>>>> Counting concerns every single number.
>>>>> Every single natural can be counted to.
>>>> Nonsense.
>> Proof by induction:
> Induction proves that every initial segment of endsegments has an
> infinite intersection.
True. But not the intersection of all of them, since that is the (ugh)
infinite initial segment, and "infinite" itself is not a natural number
covered by induction - only all the naturals themselves.

>> 1 can be counted to (obviously). If n (where n is a natural number) can
>> be counted to, then n+1 can be counted to (obviously). Hence for each
>> and every natural numbers n: n can be counted too. qed
> But not all endsegments have an infinite intersection.
Yes they do (with what?).

> All endsegments have an empty intersection.
No, all segments have an infinite intersection with each other,
namely the one that comes "later", with a larger index.

> Since every endsegment can lose only one
> number, there must be infinitely many endsegments involved in reducing
> the intersection from infinite to empty.
Exactly. That is all of them, there are infinitely segments.

>>> What one cannot be counted to?
> Just the indices involved in reducing the intersection from infinite to
> empty. They are dark.
They don't exist. That is why they are disregarded.

-- 
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.