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

On 12/27/24 4:48 AM, WM wrote:

On 26.12.2024 20:59, joes wrote:

Am Tue, 24 Dec 2024 11:42:56 +0100 schrieb WM:

 

The sets E(n) decrease. If the sequence (E(n)) could not get empty one
by one then Cantor could not set up an infinite sequence using all
indices n of that sequence.

The „sequence” doesn’t „get” empty. No element is empty.

 The sequence does not get empty in the visible domain. That proves that not all indices can be applied and therefore there is no bijection with ℕ.
 Regards, WM
 

No, that does NOT prove what you claim.
The problem is that we can not "see" all the numbers at once, not because they are not all visible, but because we have finite vision, and there are an infinite number of the finite numbers n.