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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Fri, 22 Nov 2024 15:48:55 +0100
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On 22.11.2024 13:39, joes wrote:
> Am Fri, 22 Nov 2024 13:08:28 +0100 schrieb WM:

>> All endsegments have an empty intersection.
> No, all segments have an infinite intersection with each other,
> namely the one that comes "later", with a larger index.

The set of all endsegmente has an empty intersection.
> 
>> Since every endsegment can lose only one
>> number, there must be infinitely many endsegments involved in reducing
>> the intersection from infinite to empty.
> Exactly. That is all of them, there are infinitely segments.

Up to every endsegemnet with visible index, the intersection is 
infinite: ∩{E(1), E(2), ..., E(k)} = E(k) with |E(k)| = ℵo.
> 
>>>> What one cannot be counted to?
>> Just the indices involved in reducing the intersection from infinite to
>> empty. They are dark.
> They don't exist.

The decrease from ℵo to 0 can only be accomplished by ℵo endsegments 
each of which loses one number until all numbers are gone.

Regards, WM