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

Am Mon, 02 Dec 2024 17:46:04 +0100 schrieb WM:

On 02.12.2024 15:52, joes wrote:

Am Mon, 02 Dec 2024 15:28:30 +0100 schrieb WM:

On 02.12.2024 12:53, FromTheRafters wrote:
>

Endsegment E(n) = {n, n+1, n+2, ...}

This is his definition of endsegment, which as almost anyone can see,
has no last element, so yes it is infinite. He says 'infinite
endsegment' as if there were a choice, only to add confusion.

Infinite endsegments contain an infinite set each

They ARE infinite sets.

Nevertheless they also contain an infinite set

No, they contain numbers.

infinitely many elements of which are in the intersection.

The intersection of all infinite segments?

So it is.

Ah, no. The intersection is infinite only for a finite number of segments.

An empty intersection cannot come before an empty endsegment has been
produced by losing one element at every step.

Which happens only in the limit.
 

E(1), E(2), E(3), ...
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n)
= E(n).

What do you reckon the limit is?

Whatever the limit is, it is the same for the sequence of endsegments
and the sequence of intersections.

It is the empty set.

--
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.