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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: Wed, 4 Dec 2024 18:29:39 +0100
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On 04.12.2024 18:16, Jim Burns wrote:
> On 12/4/2024 8:31 AM, WM wrote:

> Below: two is finite.

No, they may be finite or infinite.
> 
>> In two sets A and B which
>> are non-empty both
>> but have an empty intersection,
>> there must be at least
>> two elements a and b which are
>> in one endsegment but not in the other:
>> a ∈ A but a ∉ B and b ∉ A but b ∈ B.
>>
>> Same with a set of endsegments.
>> It can be divided into two sets
>> for both of which the same is required.
> 
> No.
> Not all sets of end.segments
> can be subdivided into two FINITE sets.

The set of all endsegments can be subdivided into two sets, one of which 
is finite and the other is infinite.

> GREATERS is inclusion.monotonic and {}.free.

Then

> ⋂GREATERS = {}

is wrong.

E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because
E(1)∩E(2)∩...∩E(n) = E(n).

For very naive readers I recommend the bathtub. All its states have a 
non-empty intersection unless one of the states is the empty state.

Regards, WM



Regards, WM