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

On 09.01.2025 18:52, Jim Burns wrote:

On 1/8/2025 9:31 AM, WM wrote:

On 08.01.2025 14:31, Jim Burns wrote:

 

⦃k: k < ω ≤ k+1⦄ = ⦃⦄
ω-1 does not exist.

>
Let us accept this result.
>
Then
the sequence of endsegments
loses every natnumber but
not a last one.
Then
the empty intersection of
infinite but
inclusion monotonic endsegments
is violating basic logic.
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
Then the only possible way
to satisfy logic is
the non-existence of ω and
of endsegments as complete sets.

 

(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)

 No.

Losing all numbers but keeping infinitely many is impossible in inclusion-monotonic sequences.

Sets do not change.

But the terms (E(n)) differ from their successors by one number.

Not all sets are finite.
⎛ By 'finite', I mean
⎝ 'smaller.than fuller.by.one sets'

Spare your gobbledegook. Finite means like a natural number.
Much waffle deleted.

 

It is useless to prove your claim
as long as you cannot solve this problem.

More waffle deleted.

(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)

 No.

Don't be silly.

Sets emptier.by.one than ℕ are not smaller.

They are. But that is irrelevant here. The sequence of endsegments loses all numbers. If all endsegments remain infinite, we have a contradiction.

 In the sequence of end.segments of ℕ
there is no number which
empties an infinite set to a finite set.

Then there cannot exist a sequence of endsegments obeying
∀k ∈ ℕ : E(k+1) = E(k) \ {k+1}
for all k ∈ ℕ and getting empty.

and
there is no number which
is in common with all its end.segments.

Therefore all numbers get lost from the content and become indices.

ℕ has only infinite end.segments.

Then it has only finitely many, because not all numbers get lost from the content.

The intersection of
all (infinite) end.segments of ℕ
is empty.

What is the content if all elements of ℕ have become indices?

 Sets do not change.

The sequence of endsegments is defined by
E(0) = ℕ
and
∀k ∈ ℕ : E(k+1) = E(k) \ {k+1}.

Not all sets are finite

The sequence ℕ is infinite, but when cut at any n, then the first part is finite and the second part is actually infinite.
Infinitely many infinite endsegments are impossible in actual infinity.
Regards, WM