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

On 27.11.2024 22:07, joes wrote:

Am Wed, 27 Nov 2024 20:59:43 +0100 schrieb WM:

 

Infinite endsegments with an empty intersection are excluded by
inclusion monotony.

Not by intersecting infinitely many segments.

In every case! But infinitely many infinite endsegments cannot exist because if there are ℵo indices then nothing can remain as contents because after ℵo no natural numbers exist. But endsegments contain only natural numbers.

Each end.segment is infinite.

That means it has infinitely many numbers in common with every other
infinite endsegment. If not, then there is an infinite endsegment with
infinitely many numbers but not with infinitely many numbers in common
with other infinite endsegments. Contradiction by inclusion monotony.

That is true.
 

Their intersection of all is empty. These claims do not conflict.

It conflicts with the fact, that the endsegments can lose elements but
never gain elements.

They are infinite, they don't need to gain elements.

As long as they are infinite they have an infinite intersection with their predecessors. But they are only finitely many.

 

In an infinite endsegment numbers are remaining.
In many infinite endsegments infinitely many numbers are the same.

And the intersection of all, which isn't any end.segment, is empty.

Wrong. Up to every endsegment the intersection is this endsegment. Up to
every infinite endsegment the intersection is infinite. This cannot
change as long as infinite endsegments exist.

I.e. never.

So it is. Infinite endsegments have an infinite intersection with all predecessors and with all their infinite successors.
Regards, WM