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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 19:45:55 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
Message-ID: <13fd8b44b8179f3e839e233ece99dfeb9d67db52@i2pn2.org>
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Am Thu, 28 Nov 2024 18:46:21 +0100 schrieb WM:
> On 28.11.2024 18:36, joes wrote:
>> Am Thu, 28 Nov 2024 18:09:16 +0100 schrieb WM:
>>> On 28.11.2024 17:45, joes wrote:
>>>> Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:
>>>
>>>>> A simpler arguments is this: All endsegments are in a decreasing
>>>>> sequence.
>>>> There is no decrease, they are all infinite.
>>> Every endsegment has one number less than its predecessor.
>>> That is called decrease.
>> It is called a subset. It is still infinite
> Yes this decrease produces subsets. All infinite subsets produce
> infinite intersections.
Trademark ambiguous phrasing.
The intersection of a segment with its successor produces the successor.
Both are infinite.
The intersection of all infinitely many segments is empty.

>>>>> 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".
>>> Then the intersection is never empty.
>> No finite intersection anyway.
> No intersection of infinite endsegments is finite.
The infinite intersection is empty.

> Every infinite endsegments has an infinite intersection with all its
> predecessors.
Itself, yes.

> If all endsegments are infinite, then this holds for all
> endsegments. They simply had not the chance to lose these numbers.
It does. But it does not for the union of all of them - N itself,
or E(0). It doesn't even have a predecessor.

-- 
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.