Re: Incompleteness of Cantor's enumeration of the rational numbers

On 22.11.2024 13:39, joes wrote:

Am Fri, 22 Nov 2024 13:08:28 +0100 schrieb WM:

 

All endsegments have an empty intersection.

No, all segments have an infinite intersection with each other,
namely the one that comes "later", with a larger index.

 The set of all endsegmente has an empty intersection.

>

Since every endsegment can lose only one
number, there must be infinitely many endsegments involved in reducing
the intersection from infinite to empty.

Exactly. That is all of them, there are infinitely segments.

 Up to every endsegemnet with visible index, the intersection is infinite: ∩{E(1), E(2), ..., E(k)} = E(k) with |E(k)| = ℵo.

>

What one cannot be counted to?

Just the indices involved in reducing the intersection from infinite to
empty. They are dark.

They don't exist.

 The decrease from ℵo to 0 can only be accomplished by ℵo endsegments each of which loses one number until all numbers are gone.