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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Tue, 26 Nov 2024 10:10:36 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Tue, 26 Nov 2024 10:11:22 +0100 schrieb WM:
> On 25.11.2024 22:05, joes wrote:
>> Am Mon, 25 Nov 2024 13:18:28 +0100 schrieb WM:
> 
>>>> But there is no finite set with ALL natural numbers.
>>>> Like usual, you mess up with your qualifiers.
>>> ℕ is fixed, that means |ℕ| is fixed.
>> What does that have to do with it?
> It is impossible to add or to delete an element.
> It is impossible to change |ℕ| by 1 or more.
It is possible to change N to N\0.
How does that relate to all infinitely many naturals being finite?

>>>>>> Limit theory only works if the limit actually exists
>>>>> If limits exist at all, then the limit of the sequence 1/10, 1/10,
>>>>> 1/10, ... does exist.
>>>> But the concept of 1/10th of an infinte set does not exist..
>>> It does.
>> It has the same cardinality.
> Yes, it is much.
Countably infinite.

>>>>>> You can get things that APPEAR to reach a limit, but actually
>>>>>> don't.
>>>>> But if infinite sets do exist, then the set ℕ does exist, and all
>>>>> its elements are members of finite intervals (0, n].
>>>> No, any given element is a member of a finite set, but you can't then
>>>> say that ALL are in such a set.
>>> All are in the union of all finite sets.
>> Why not just directly take N, made up of finite numbers?
> Why not? Do it. Consider the black hats at every 10 n and white hats at
> all other numbers n. It is possible to shift the black hats such that
> every interval (0, n] is completely covered by black hats. There is no
> first n discernible that cannot be covered by  black hat.
Cantor proved nothing more.

> But the origin
> of each used black hat larger than n is now covered by a white hat.
Not if you really coloured ALL n.

> Without deleting all white hats it is not possible to cover all n by
> black hats. But deleting white hats is prohibited by logic. Exchanging
> can never delete one of the exchanged elements.
An infinite exchange can.

> Therefore we have here,
> like in all Cantor-pairings, the same impediment and further disussion
> is futile.
Thanks for shutting up.

-- 
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.