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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, 9 Dec 2024 19:54:55 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sun, 08 Dec 2024 11:50:08 +0100 schrieb WM:
> On 08.12.2024 00:38, Jim Burns wrote:
>> On 12/7/2024 4:37 PM, WM wrote:
>>> On 07.12.2024 20:59, Jim Burns wrote:
>>>> On 12/7/2024 6:09 AM, WM wrote:
>>>>> On 06.12.2024 19:17, Jim Burns wrote:
>>>>>> On 12/6/2024 3:19 AM, WM wrote:
>>>>>>> On 05.12.2024 23:20, Jim Burns wrote:
>> 
>>>>>>>> ⎜ With {} NOT as an end.segment,
>>>>>>> all endsegments hold content.
>>>>>> But no common.to.all finite.cardinals.
>>>>> Show two endsegments which do not hold common content.
>>>> I will, after you show me a more.than.finitely.many two.
>>> There are no more than finitely many natural numbers which can be
>>> shown.
>>> All which can be shown have common content.
> All endsegments which can be shown (by their indices) have common
> content.
Nope, not in common with all, only for finite intersections.

>> Each end.segment is more.than.finite and the intersection of the
>> end.segments is empty.
> ∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
> What can't I understand here?
That all those n only produce finite intersections.

>>> This is not gibberish but mathematics:
>>> ∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
>>> Every counter argument has to violate this.
>> False. For example, see above.
>
>> Each finite.cardinal is not in common with more.than.finitely.many
>> end.segments.
> Of course not. All non-empty endsegments belong to a finite set with an
> upper bound.
There are definitely more than finitely many naturals and segments
(none are empty).

>> Each end.segment has, for each finite.cardinal,
>> a subset larger than that cardinal.
> That is not true for the last dark endsegments. It changes at the dark
> finite cardinal ω/2.
What changes to what? Why exactly there?

-- 
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.