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

Am Thu, 28 Nov 2024 18:46:21 +0100 schrieb WM:

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.

Trademark ambiguous phrasing.
The intersection of a segment with its successor produces the successor.
Both are infinite.
The intersection of all infinitely many segments is empty.

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.

The infinite intersection is empty.

Every infinite endsegments has an infinite intersection with all its
predecessors.

Itself, yes.

If all endsegments are infinite, then this holds for all
endsegments. They simply had not the chance to lose these numbers.

It does. But it does not for the union of all of them - N itself,
or E(0). It doesn't even have a predecessor.

--
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.