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

On 28.11.2024 18:36, joes wrote:

Am Thu, 28 Nov 2024 18:09:16 +0100 schrieb WM:

On 28.11.2024 17:45, joes wrote:

Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:

>

A simpler arguments is this: All endsegments are in a decreasing
sequence.

There is no decrease, they are all infinite.

Every endsegment has one number less than its predecessor.
That is called decrease.

It is called a subset. It is still infinite

Yes this decrease produces subsets. All infinite subsets produce infinite intersections.

 

Before the decrease has reached finite endsegments, all are infinite
and share an infinite contents from E(1) = ℕ on. They have not yet had
the chance to reduce their infinite subset below infinity.

All segments are infinite. Nothing can come "afterwards".

Then the intersection is never empty.

No finite intersection anyway.

No intersection of infinite endsegments is finite.
Every infinite endsegments has an infinite intersection with all its predecessors. If all endsegments are infinite, then this holds for all endsegments. They simply had not the chance to lose these numbers.
Regards, WM