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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, 28 Nov 2024 16:45:48 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:
> On 28.11.2024 09:34, Jim Burns wrote:
> 
>> Consider the sequence of claims.
>> ⎛⎛ [∀∃] for each end.segment ⎜⎜ there is an infinite set such that ⎜⎝
>> the infinite set subsets the end.segment
> and its predecessors!
Naturally.

> If each endsegment is infinite, then this is valid for each endsegment
> with no exception. because all are predecessors of an infinite
> endsegment. That means it is valid for all endsegments.
That is what "every" means.

> The trick here is that the infinite set has no specified natural number
> (because all fall out at some endsegment) but it is infinite without any
> other specification.
Yes. You can call it omega or N.

>> ⎜⎛ [∃∀] there is an infinite set such that ⎜⎜ for each end.segment ⎝⎝
>> the infinite set subsets the end.segment
>> We cannot SEE,
>> just by looking at the claims,
>> that, after [∀∃], [∃∀] is not.first.false.
> 
> I have proved above that [∃∀] is true for all infinite endsegments.
Uh, that is wrong.

> A simpler arguments is this: All endsegments are in a decreasing
> sequence.
There is no decrease, they are all infinite.

> Before the decrease has reached finite endsegments, all are
> infinite and share an infinite contents from E(1) = ℕ on. They have not
> yet had the chance to reduce their infinite subset below infinity.
All segments are infinite. Nothing can come "afterwards".

-- 
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.