Re: Incompleteness of Cantor's enumeration of the rational numbers

On 11/30/2024 6:57 AM, WM wrote:

On 30.11.2024 02:44, Jim Burns wrote:

The sets do not have any of
the cardinalities which would change.
The sets have a different cardinality,
 one which does not change when the set changes.

>
Then the intersection which is infinite
too remain infinite.

About end segments of the finite cardinals:
The new end.segment with one element removed
has the same infinite cardinality ℵ₀
That removed element
  is not in common with each end.segment
  is not.in the intersection of all.
Again.
A third end.segment, with another removed,
has the same infinite cardinality ℵ₀
Both removed elements
  are not in common with each end.segment
  are not.in the intersection of all.
After all the finitely.ranked removals,
each finitely.ranked finite.cardinal
  is not.in the intersection of all.
All finite.cardinals are finitely.ranked.
All are not.in.
The intersection of all the end segments of
  the finite.cardinals
contains only
finite.cardinals which aren't finitely.ranked.
There aren't any.
The intersection of all is empty.

After all the swaps
  (of which no swap is a change in cardinality)
what remains is a proper subset
  (which is not a change in cardinality).
Because infinite.

>
Like the intersection.

The intersection of a collection of end.segments
is particular to that collection.
The collection doesn't change.
Each end.segment doesn't change.
Each finite.cardinal doesn't change.
However,
we can consider other collections,
other end.segments,
other finite.cardinals.

Like the intersection.

Apparently, what you (WM) call "the intersection"
is each of infinitely.many intersections,
some infinite, some empty,
"changing" from one to another,
in a manner you accept or you do not accept.
It's a mess, but it's your mess.
Our sets do not change.