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

Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:

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.

This case doesn't occur.

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.

Especially not of the same cardinality as n+1.

Much waffle deleted.

Honest thanks for the note.

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

 

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

No.

Don't be silly.

It is possible with infinite sets, which can't be reduced by a finite
number.

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.

No, they are subsets of the same cardinality. There is no 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.

No term of the sequence is empty, if you mean that.

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.

WDYM "become"? There is no point at which all naturals would be
counted - N being infinite.

ℕ has only infinite end.segments.

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

Huh? No. Then not all numbers would be "indices".

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

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

There is no such endsegment.

Sets do not change.

--
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.