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

On 11/29/2024 2:41 PM, WM wrote:

On 29.11.2024 19:53, Jim Burns wrote:

Yes,
if each end.segment is infinite
then each end.segment is infinite.

>
And their intersection is infinite.

Each finite.cardinal k is countable.past to
  k+1 which indexes
   Eᶠⁱⁿ(k+1) which doesn't hold
    k which is not common to
     all end segments.
Each finite.cardinal k is not.in
  the set of elements common to all end.segments,
  the intersection of all end segments,
   which is empty.
No numbers are remaining.
----
For each end.segment of finite.cardinals,
for each finite cardinal,
that end.segment has a subset larger than that cardinal.
Each end.segment of finite.cardinals
does not have any finite cardinality
is infinite.

because all are
predecessors of an infinite endsegment.

 

>

That means it is valid for all endsegments.
>
The trick here is that
the infinite set has no specified natural number
(because all fall out at some endsegment)

>
all fall out == empty

>
== empty endsegment.

No.
Empty intersection of all and only infinite end.segments.
Of finite cardinals.
Your "dark cardinals" are not here.

but it is infinite

>
infinite and empty

>
According to set theory.

You imagine that's what set theory says.
It doesn't say that.

==> Set theory is wrong.

Right or wrong,
set theory doesn't say that.
Maybe some day, you (WM) will learn
what set theory is, and,
from that point on,
you (WM) will be able to look for
issues to raise concerning set theory.
You (WM) have not yet reached that day, though.