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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: Mon, 2 Dec 2024 19:08:44 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Mon, 02 Dec 2024 17:46:04 +0100 schrieb WM:
> On 02.12.2024 15:52, joes wrote:
>> Am Mon, 02 Dec 2024 15:28:30 +0100 schrieb WM:
>>> On 02.12.2024 12:53, FromTheRafters wrote:
>>>
>>>>>> Endsegment E(n) = {n, n+1, n+2, ...}
>>>> This is his definition of endsegment, which as almost anyone can see,
>>>> has no last element, so yes it is infinite. He says 'infinite
>>>> endsegment' as if there were a choice, only to add confusion.
>>> Infinite endsegments contain an infinite set each
>> They ARE infinite sets.
> Nevertheless they also contain an infinite set
No, they contain numbers.

>>> infinitely many elements of which are in the intersection.
>> The intersection of all infinite segments?
> So it is.
Ah, no. The intersection is infinite only for a finite number of segments.

>>> An empty intersection cannot come before an empty endsegment has been
>>> produced by losing one element at every step.
>> Which happens only in the limit.
>> 
>>> 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).
>> What do you reckon the limit is?
> Whatever the limit is, it is the same for the sequence of endsegments
> and the sequence of intersections.
It is the empty set.

-- 
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.