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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: Thu, 9 Jan 2025 20:17:39 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:
> On 09.01.2025 18:52, Jim Burns wrote:
>> On 1/8/2025 9:31 AM, WM wrote:
>>> On 08.01.2025 14:31, Jim Burns wrote:
>> 
>>>> ⦃k: k < ω ≤ k+1⦄ = ⦃⦄
>>>> ω-1 does not exist.
>>>
>>> Let us accept this result.
>>> Then the sequence of endsegments loses every natnumber but not a last
>>> one.
>>> Then the empty intersection of infinite but inclusion monotonic
>>> endsegments is violating basic logic.
>>> (Losing all numbers but keeping infinitely many can only be possible
>>> if new numbers are acquired.)
>>> Then the only possible way to satisfy logic is the non-existence of ω
>>> and of endsegments as complete sets.
>> 
>>> (Losing all numbers but keeping infinitely many can only be possible
>>> if new numbers are acquired.)
>> No.
> Losing all numbers but keeping infinitely many is impossible in
> inclusion-monotonic sequences.
This case doesn't occur.

>> Sets do not change.
> But the terms (E(n)) differ from their successors by one number.
>
>> Not all sets are finite.
>> ⎛ By 'finite', I mean ⎝ 'smaller.than fuller.by.one sets'
> Spare your gobbledegook. Finite means like a natural number.
Especially not of the same cardinality as n+1.

> Much waffle deleted.
Honest thanks for the note.

>>> It is useless to prove your claim as long as you cannot solve this
>>> problem.
> 
>>> (Losing all numbers but keeping infinitely many can only be possible
>>> if new numbers are acquired.)
>> No.
> Don't be silly.
It is possible with infinite sets, which can't be reduced by a finite
number.

>> Sets emptier.by.one than ℕ are not smaller.
> They are. But that is irrelevant here. The sequence of endsegments loses
> all numbers. If all endsegments remain infinite, we have a
> contradiction.
No, they are subsets of the same cardinality. There is no contradiction.

>> In the sequence of end.segments of ℕ there is no number which empties
>> an infinite set to a finite set.
> Then there cannot exist a sequence of endsegments obeying
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k+1} for all k ∈ ℕ and getting empty.
No term of the sequence is empty, if you mean that.

>> and there is no number which is in common with all its end.segments.
> Therefore all numbers get lost from the content and become indices.
WDYM "become"? There is no point at which all naturals would be
counted - N being infinite.

>> ℕ has only infinite end.segments.
> Then it has only finitely many, because not all numbers get lost from
> the content.
Huh? No. Then not all numbers would be "indices".

>> The intersection of all (infinite) end.segments of ℕ is empty.
> What is the content if all elements of ℕ have become indices?
There is no such endsegment.

>> Sets do not change.
-- 
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.