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

On 09.01.2025 21:17, joes wrote:

Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:

Losing all numbers but keeping infinitely many is impossible in
inclusion-monotonic sequences.

This case doesn't occur.

Loss of all numbers is proven by the empty intersection.
Keeping infinitely many is poved by Fritsche.

 

If all endsegments remain infinite, we have a
contradiction.

No, they are subsets of the same cardinality. There is no contradiction.

They remain infinite. But infinitely many endsegments require all natnumbers as indices. What makes up their infinite content?

 

In the sequence of end.segments of ℕ there is no number which empties
an infinite set to a finite set.

Then there cannot exist a sequence of endsegments obeying
∀k ∈ ℕ: E(k+1) = E(k) \ {k+1} for all k ∈ ℕ and getting empty.

No term of the sequence is empty, if you mean that.

Then not all natnumbers are outside of content and inside of the set of indices.

 

and there is no number which is in common with all its end.segments.

Therefore all numbers get lost from the content and become indices.

WDYM "become"? There is no point at which all naturals would be
counted - N being infinite.

The endsegment E(n) loses its element n+1 ad becomes E(n+1).

 

ℕ has only infinite end.segments.

Then it has only finitely many, because not all numbers get lost from
the content.

Huh? No. Then not all numbers would be "indices".

Then there are only finitely many indices.

 

The intersection of all (infinite) end.segments of ℕ is empty.

What is the content if all elements of ℕ have become indices?

There is no such endsegment.

What element of ℕ does not become an index?
Regards, WM