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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: Wed, 27 Nov 2024 21:07:10 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Wed, 27 Nov 2024 20:59:43 +0100 schrieb WM:
> On 27.11.2024 16:57, Jim Burns wrote:
>> On 11/27/2024 6:04 AM, WM wrote:
> 
>> However,
>> one makes a quantifier shift, unreliable, to go from that to ⛔⎛ there
>> is an end segment such that ⛔⎜ for each number (finite cardinal) ⛔⎝ 
the
>> number isn't in the end segment.
> 
> If all endsegments are infinite then infinitely
> many natbumbers remain in all endsegments.
Yes.

> Infinite endsegments with an empty intersection are excluded by
> inclusion monotony.
Not by intersecting infinitely many segments.

> Because that would mean infinitely many different
> numbers in infinite endsegments.
Weird way to put it.

>> Each end.segment is infinite.
> That means it has infinitely many numbers in common with every other
> infinite endsegment. If not, then there is an infinite endsegment with
> infinitely many numbers but not with infinitely many numbers in common
> with other infinite endsegments. Contradiction by inclusion monotony.
That is true.

>> Their intersection of all is empty. These claims do not conflict.
> It conflicts with the fact, that the endsegments can lose elements but
> never gain elements.
They are infinite, they don't need to gain elements.

>>> In an infinite endsegment numbers are remaining.
>>> In many infinite endsegments infinitely many numbers are the same.
>> And the intersection of all, which isn't any end.segment, is empty.
> Wrong. Up to every endsegment the intersection is this endsegment. Up to
> every infinite endsegment the intersection is infinite. This cannot
> change as long as infinite endsegments exist.
I.e. never.

-- 
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.