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

On 08.01.2025 16:23, Alan Mackenzie wrote:

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

On 07.01.2025 12:36, Alan Mackenzie wrote:

 

If there were such
things as "potential" and "actual" infinity in maths,

 

Your comments about my quotes show that you have lost all contact with
mathematics.

 

then they would make a difference to some mathematical result.

 

Of course. Here is a simple example, accessible to every student who is
not yet stultified by matheology.

 

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.

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

Of course.  ∩{E(k) : k ∈ ℕ} = { }.
But for all definable endsegments the intersection is infinite, and from endsegmnet to endsegment only one number is lost, never more! Therefore 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. Unless you claim that the general law does not hold for ∀k ∈ ℕ.

 

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.

 

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

 False.  Where do you get such an idea from?

See above.

 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.

 

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.

 

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