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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.logic
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Wed, 27 Nov 2024 12:04:35 +0100
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On 26.11.2024 20:44, Jim Burns wrote:
> On 11/26/2024 2:15 PM, WM wrote:
>> On 26.11.2024 19:49, Jim Burns wrote:
> 
>>> There are no last end.segments of ℕᶠⁱⁿ
>>> There are no finitely.sized end segments of ℕᶠⁱⁿ
>>> There are no finite cardinals common to
>>>  each end.segment of ℕᶠⁱⁿ
>>
>> That is a contradiction.
> 
> It contradicts ℕᶠⁱⁿ being finite, nothing else.

It contradicts inclusion monotony.
> 
>> If there are no common numbers,
>> then all numbers must have been lost.
>> But then no numbers are remaining.
> 
> Yes.

Then also no numbers are remaining in the endsegments.
> 
> Each finite.cardinal k is countable.past to
>   k+1 which indexes
>    Eᶠⁱⁿ(k+1) which doesn't hold
>     k which is not common to
>      all end segments.
> 
> Each finite.cardinal k is not.in
>   the intersection of all end segments,
>   the set of elements common to all end.segments,
>    which is empty.
> 
> No numbers are remaining.

That is true. But you claimed that every endsegment is infinite. In an 
infinite endsegment numbers are remaining. In many infinite endsegments 
infinitely many numbers are the same.
> 
>> Then there are finite endsegments because
>> ∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.
> 
> For each cardinal.which.can.change.by.1 j

That does not contradict the fact that infinite endsegments have 
infinitely many numbers in common and hence an infinite intersection.

Regards, WM