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

Am Sat, 07 Dec 2024 22:37:04 +0100 schrieb WM:

On 07.12.2024 20:59, Jim Burns wrote:

On 12/7/2024 6:09 AM, WM wrote:

On 06.12.2024 19:17, Jim Burns wrote:

On 12/6/2024 3:19 AM, WM wrote:

On 05.12.2024 23:20, Jim Burns wrote:

 

⎜ With {} NOT as an end.segment,

all endsegments hold content.

But no common.to.all finite.cardinals.

Show two endsegments which do not hold common content.

I will, after you show me a more.than.finitely.many two.

There are no more than finitely many natural numbers which can be shown.
All which can be shown have common content.

Actually there are infinitely many naturals, and of every two segments
one is a superset, but not for the set of all segments.

⎜ there STILL are  more.than.finite.many end.segments,

Not actually infinitely many however.

More.than.finitely.many are enough to break the rules we devise for
finitely.many.

More than finitely many are finitely many,

You (WM) define it that way,
which turns your arguments into gibberish.

This is not gibberish but mathematics:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
Every counter argument has to violate this. That is inacceptable.

Incorrect. "More than finite" means infinite. Your formula only
talks about finite intersections.

Each finite.cardinality cannot be more.than.finitely.many

So it is. Each finite cardinal cannot turn a finite set into an infinite
set. But even for infinite sets we have ∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) =
E(n).

That is not an infinite intersection.

⎜ If n is a finite.cardinal, then ⎜ n is.less.than n+1, and n+1 is
finite.
⎝ n cannot be more.than.finitely.many.
 

More than finitely many are *finitely many,

*in

unless they are actually infinitely many. Therefore they are not
enough.

Call it 'potential'.

It is.

None of that changes that each finite.cardinal is followed by a
finite.cardinal,

and leaves the set finite. But even for infinitely many n:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
No empty intersection without an empty endsegment.

That happens, as you know, only in the limit.

--
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.