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

Am Thu, 09 Jan 2025 11:54:27 +0100 schrieb WM:

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 all endsegments.

The empty (inf.) intersection is not a segment.

If none is empty, then other numbers must be inside. Contradiction.

No number is an element of all segments.

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.

Only insofar as every number eventually "leaves" the endsegments.
This however does not imply and empty endsegment, since there inf. many
of both naturals and therefore endsegments.

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

It does not hold for the infinite intersection.

Why not?

Sorry, I lost track. Which law?

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.

Does not follow.

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.

Same reason: the limit may have different properties than the terms.

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.

But this never happens. There can be no infinite starting segment short
of N itself, and nothing can follow that.

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.

There are infinitely many naturals though. I fail to picture your
imagination.

E(1) = {1, 2, 3, ...}
E(2) = {2, 3, 4, ...}
...
E(k-1) = {k-1, k, k+1, ...}
E(k) = {k, k+1, k+2, ...}
are all infinite and then you just give up?

--
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.