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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: Thu, 9 Jan 2025 11:54:27 +0100
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On 09.01.2025 01:00, joes wrote:
> Am Wed, 08 Jan 2025 22:45:26 +0100 schrieb WM:
>> On 08.01.2025 16:23, Alan Mackenzie wrote:

>> Of course. But do you know what inclusion-monotony means? E(n+1) is a
>> proper subset of E(n): {n+2, n+3, n+4, ...} c {n+1, n+2, n+3, n+4, ...}.
>> Here the intersection cannot be empty unless there is an empty
>> endsegment.
> Only valid for finite sets.

Why? What changes the basics? The intersection is only empty, when no 
natural number remains in in all endsegments. If none is empty, then 
other numbers must be inside. Contradiction.
> 
>>> As for the intersection of all endsegments of natural numbers, this is
>>> obviously empty.
>> But for all definable endsegments the intersection is infinite, and from
>> endsegmenet to endsegment only one number is lost, never more!
> And nobody said otherwise, since there are infinitely many segments.

Infinitely many endsegments need infinitely many indices. Therefore no 
natural number must remain as content in the sequence of endsegments.

>> Unless you claim that the general law does not hold for ∀k ∈ ℕ.
> It does not hold for the infinite intersection.

Why not?

>>>> Inclusion monotonic sequences can only have an empty intersection if
>>>> they have an empty term.
>>> False.

Simple logic. For an empty intersection, there must be infinitely many 
endsegments. That means no natural number can remain in the content. If 
a number n remained, then it would impose an upper limit on the sequence 
of indices.

>> It is trivially true that only one element can vanish with each
>> endsegment.
> Which noone contradicted.

Then the empty intersection is preceded by finite intersections.

>> Jim "proved" that when exchanging two elements O and X, one of them can
>> disappear. His "proofs" violate logic which says that lossless exchange
>> will never suffer losses.
> Wrong. The limit of the harmonic series is zero, even though none of the
> terms are.

There is no exchange involved.

>> That's a simple fact. The sequence of natural numbers 1, 2, 3, ..., n,
>> n+1, ...
>> cannot be cut into two actually infinite sequences, namely indices and
>> contents.
> Why should it?

Because an infinite sequence of indices followed by an infinite sequence 
of contentent would require two infinite sequences.
> 
>> When all contents is appearing as an infinite sequence of indices then
>> no number can remain in the contents.
> Yes, there are no more numbers after the naturals???

So it is. The claim of infinitely many infinite endsegments is false.

Regards, WM
>