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

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

On 09.01.2025 22:03, joes wrote:

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

 

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

No number is an element of all segments.

But what numbers are inside if all natural numbers are outside?

Inside what? Inside each segment are infinitely many naturals. Mind
your quantifiers: It is not required that the intersection be nonempty.

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.

This however does not imply and empty endsegment,

What remains?

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

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

Infinitely many numbers leave. All elements of ℕ leave.

Yes, in the limit.

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?

The law that the intersection gets empty but only by one element per
term.

Well, in the limit (sigh) infinitely many numbers have been "lost".

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.

It follows from the law that only one element per term can leave.

Care to write out that deduction?

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.

There is no limit involved when counting the fractions.

It is an infinite "process".

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.

An infinite set of infinite endsegments happens according to matheology.
It requires two infinite sequences.

No, it doesn't. There is no "infinitieth" segment.

The claim of infinitely many infinite endsegments is false.

There are infinitely many naturals though.

But all must be in the set of indices, if there are infinitely many
endsegments. But infinitely many must be in the set of contents, if all
endsegments are infinite.

Those are the same set N.

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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.