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

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

On 08.01.2025 16:23, Alan Mackenzie wrote:

WM <wolfgang.mueckenheim@tha.de> wrote:

On 07.01.2025 12:36, Alan Mackenzie wrote:

then they would make a difference to some mathematical result.

For the inclusion-monotonic sequence of endsegments of natural numbers
E(k) = {k+1, k+2, k+3, ...} the intersection of all terms is empty.
But if every number k has infinitely many successors, as ZF claims,
then the intersection is not empty.

That is false.  The intersection of even just two infinite sets can be
empty.

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.

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
endsegmnet to endsegment only one number is lost, never more!

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

the general law of mathematics ∀k ∈ ℕ : E(k+1) = E(k) \ {k+1}
or ∀k ∈ ℕ : ∩{E(0), E(1), E(2), ..., E(k+1)} = ∩{E(0), E(1), E(2), ...,
E(k)} \ {k+1}
proves that the empty intersection requires finite intersections
preceding it.

On the contrary: for every (finite!) k, E(k) is not empty.

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

It does not hold for the infinite intersection.

Therefore set theory, claiming both, is false.

Set theory doesn't "claim" both.  Set theory doesn't "claim" at all.
It has axioms and theorems derived from those axioms.  If one accepts
the axioms, and nearly all mathematicians do, then one is logically
forced to accept the theorems, too.

But the theorems contradict the general law of mathematics.

No such thing.

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

False.
 Such sequences have an empty
intersection if there is no element which is a member of each set in
the sequence.  This is trivially true for the sequence of endsegments
of the natural numbers.

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

Which noone contradicted.

Therefore the empty intersection of all requires the existence of
finite terms which must be dark.

That isn't mathematics.  Jim proved some while ago that there are no
dark numbers, in as far as he could get a definition of them out of
you.

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.

Further there are not infinitely many infinite endsegments possible
because the indices of an actually infinite set of endsegements
without gaps must be all natural numbers.

That's meaningless gobbledegook.

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?

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???

--
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.