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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: Wed, 27 Nov 2024 20:59:43 +0100
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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.

Don't blather nonsense. If all endsegments are infinite then infinitely 
many natbumbers remain in all endsegments.

Infinite endsegments with an empty intersection are excluded by 
inclusion monotony. Because that would mean infinitely many different 
numbers in infinite endsegments.
> 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.

> 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.
> 
>> 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.

Regards, WM