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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: Sat, 14 Dec 2024 10:26:16 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 13 Dec 2024 11:55:16 +0100 schrieb WM:
> On 13.12.2024 03:23, Richard Damon wrote:
>> On 12/12/24 9:25 AM, WM wrote:

>>> if Cantor can apply all natural numbers as indices for his bijections,
>>> 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.
>> But a segment that is infinite in length is, by definiton, missing at
>> least on end.
> That means that the premise "if Cantor can apply all natural numbers as
> indices for his bijections" is false.
Nah. Just imagine all of the inf.many steps as a whole - y’know, ACTUAL
infinity.

>> So, which bijection from Cantor are you talking about? Of are you
>> working on a straw man that Cantor never talked about?
> 
> There are many. The mapping from natumbers to the rationals, for
> instance, needs all natural numbers. That means all must leave the
> endsegments. Another example is Cantor's list "proving" uncountable
> sets. If not every natural number has left the endsegment and is applied
> as an index of a line of the list, the list is useless.
Then the list were finite. It isn’t, though.

> But if every natural number has left the endsegments, then the
> intersection of all endsegments is empty.
Yes.
> Then the infinite sequence of
> endegments has a last term (and many finite predecessors, because of ∀k
> ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}).
No. It is literally „without an end”, and yet can be „completed”, if
only you were able to conceive of infinity.

-- 
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.