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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.logic
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Thu, 12 Dec 2024 15:33:20 +0100
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On 12.12.2024 15:23, joes wrote:
> Am Thu, 12 Dec 2024 10:12:26 +0100 schrieb WM:

>>>> The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
> The sequence is endless, has no end, is infinite.

If a bijection with ℕ is possible, the sequence can be exhausted so that 
no natural numbers remains in an endsegment.
> 
>>> None of which are an infinite sets, so trying to take a "limit" of
>>> combining them is just improper.
>>
>> Most endsegments are infinite. But if Cantor can apply all natural
>> numbers as indices for his sequences, then all must leave the sequence
>> of endsegments. Then the sequence (E(k)) must end up empty. And there
>> must be a continuous staircase from E(k) to the empty set.

> It makes no sense not being able to „apply” numbers. Clearly Cantor does.

He claims it. That means no numbers remain unpaired in endsegments.

> The sequence IS continuous. It’s just that you misconceive of the
> limit as reachable.

Cantor does. If the limit is not reachable, then complete bijections 
cannot be established.

"If we think the numbers p/q in such an order [...] then every number 
p/q comes at an absolutely fixed position of a simple infinite sequence" 
[E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und 
philosophischen Inhalts", Springer, Berlin (1932) p. 126]

"The infinite sequence thus defined has the peculiar property to contain 
the positive rational numbers completely, and each of them only once at 
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

If you accept these claims, then no number must remain in an endsegment.

Regards, WM
>