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

On 14.12.2024 11:42, joes wrote:

Am Thu, 12 Dec 2024 15:25:59 +0100 schrieb WM:

But 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.

Yes, in the limit. There is no natural k such that E(k) is empty.

The sequence of intersections loses all content but only according to the rule
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}.

 

And there must be a continuous staircase from E(k) to the empty set.

This is where you are mistaken.

It is the definition of endegments. If endsegments exist, then they follow this rule.

How do you get from naturals to
finite dark numbers anyway?

That is the same as without dark numbers. The visible numbers are potentially infinite.
Regards, WM