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

On 09.01.2025 01:00, joes wrote:

Am Wed, 08 Jan 2025 22:45:26 +0100 schrieb WM:

On 08.01.2025 16:23, Alan Mackenzie wrote:

Of course. But do you know what inclusion-monotony means? E(n+1) is a
proper subset of E(n): {n+2, n+3, n+4, ...} c {n+1, n+2, n+3, n+4, ...}.
Here the intersection cannot be empty unless there is an empty
endsegment.

Only valid for finite sets.

Why? What changes the basics? The intersection is only empty, when no natural number remains in in all endsegments. If none is empty, then other numbers must be inside. Contradiction.

 

As for the intersection of all endsegments of natural numbers, this is
obviously empty.

But for all definable endsegments the intersection is infinite, and from
endsegmenet to endsegment only one number is lost, never more!

And nobody said otherwise, since there are infinitely many segments.

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

Unless you claim that the general law does not hold for ∀k ∈ ℕ.

It does not hold for the infinite intersection.

Why not?

Inclusion monotonic sequences can only have an empty intersection if
they have an empty term.

False.

Simple logic. For an empty intersection, there must be infinitely many endsegments. That means no natural number can remain in the content. If a number n remained, then it would impose an upper limit on the sequence of indices.

It is trivially true that only one element can vanish with each
endsegment.

Which noone contradicted.

Then the empty intersection is preceded by finite intersections.

Jim "proved" that when exchanging two elements O and X, one of them can
disappear. His "proofs" violate logic which says that lossless exchange
will never suffer losses.

Wrong. The limit of the harmonic series is zero, even though none of the
terms are.

There is no exchange involved.

That's a simple fact. The sequence of natural numbers 1, 2, 3, ..., n,
n+1, ...
cannot be cut into two actually infinite sequences, namely indices and
contents.

Why should it?

Because an infinite sequence of indices followed by an infinite sequence of contentent would require two infinite sequences.

 

When all contents is appearing as an infinite sequence of indices then
no number can remain in the contents.

Yes, there are no more numbers after the naturals???

So it is. The claim of infinitely many infinite endsegments is false.
Regards, WM