Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

Am Thu, 09 Jan 2025 10:34:42 +0100 schrieb WM:

On 09.01.2025 00:44, joes wrote:

Am Wed, 08 Jan 2025 15:31:13 +0100 schrieb WM:

 

(Losing all numbers but keeping infinitely many can only be possible
if new numbers are acquired.)

No, this isn't even the case. The infinite(!) intersection is empty.

Empty intersection means that all numbers of the contents have become
indices k of the endsegments E(k). What remains in the always infinite
E(k)?

You have a faulty mental image.There is an infinite sequence of
infinite segments. How do you reconcile that with an empty
infinite intersection?
WDYM "remains"? Every segment is "missing" only finitely many numbers
from N. There is also no "becoming". The intersection of inf. many
segments is not a segment.

--
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.