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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, 16 Dec 2024 11:55:23 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Mon, 16 Dec 2024 09:30:18 +0100 schrieb WM:
> On 15.12.2024 21:21, joes wrote:
>> Am Sun, 15 Dec 2024 16:25:55 +0100 schrieb WM:
>>> On 15.12.2024 12:15, joes wrote:
>>>> Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:
>>>
>>>>>>>>> That pairs the elements of D with the elements of ℕ. Alas, it
>>>>>>>>> can be proved that for every interval [1, n] the deficit of hats
>>>>>>>>> amounts to at least 90 %. And beyond all n, there are no further
>>>>>>>>> hats.
>>>>>>>> But we aren't dealing with intervals of [1, n] but of the full
>>>>>>>> set.
>>>>>>> Those who try to forbid the detailed analysis are dishonest
>>>>>>> swindlers and tricksters and not worth to participate in
>>>>>>> scientific discussion.
>>>>>> No, we are not forbiding "detailed" analysis
>>>>> Then deal with all infinitely many intervals [1, n].
>>>> ??? The bijection is not finite.
>>> Therefore we use all [1, n].
>> Those are all finite.
> All n are finite.
Contrary to the bijection.

>>>>>>>> The problem is that you can't GET to "beyond all n" in the
>>>>>>>> pairing,
>>>>>>>> as there are always more n to get to.
>>>>>>> If this is impossible, then also Cantor cannot use all n.
>>>>>> Why can't he? The problem is in the space of the full set, not the
>>>>>> finite sub sets.
>>>>> The intervals [1, n] cover the full set.
>>>> Only in the limit.
>>> With and without limit.
>> Wonrg. There is no natural n that „covers N”.
> All intervals do it because there is no n outside of all intervals [1,
> n]. My proof applies all intervals.
It does not. It applies to every single finite „interval” (whyever those
matter), but not to the whole N.

-- 
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.