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

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.

 

⎜ 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.

 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).

⎜ 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,
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.
Regards, WM