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

On 10.01.2025 10:03, joes wrote:

Am Thu, 09 Jan 2025 23:27:21 +0100 schrieb WM:

Mind
your quantifiers: It is not required that the intersection be nonempty.

Quantifiers cannot resolve the contradiction: The sequence is infinite only if all contents is transformed into indices.

 

Infinitely many endsegments need infinitely many indices. Therefore no
natural number must remain as content in the sequence of endsegments.

Only insofar as every number eventually "leaves" the endsegments.

But what remains?

In what? The intersection is empty.

When all content has gone inclsuion monotony requires that no content remains.

 

This however does not imply and empty endsegment,

What remains?

Nothing "remains". There is no end, only a limit.

There is no limit because we remain in the domain of natural numbers:
∀k ∈ ℕ : ∩{E(0), E(1), E(2), ..., E(k+1)} = ∩{E(0), E(1), E(2), ..., E(k)} \ {k+1}.
A limit is not in this domain.

 

since there inf. many of both naturals and therefore endsegments.

Infinitely many numbers leave. All elements of ℕ leave.

Yes, in the limit.

There is no limit in counting elements. There are all natural numbers.
"The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
Therefore there must all content be transformed into indices without limit.
Regards, WM