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

On 12/8/2024 5:50 AM, WM wrote:

On 08.12.2024 00:38, Jim Burns wrote:

Each finite.cardinal is not in common with
more.than.finitely.many end.segments.

>
Of course not.

Of course not, because...
Each finite.cardinal k is only held in
end.segments ⟦j,ℵ₀⦆ with their minimum j ∈ ⟦0,k⟧
k is finite.
⟦0,k⟧ is finite.
The set of holding.k end.segments is finite.
A more.than.finitely.many set of end.segments
holds end.segments not in
the finite set of holding.k end.segments,
and holds end.segments not.holding.k.
Each finite.cardinal k is not in common with
each of a more.than.finite set of end.segments.

All non-empty endsegments belong to
a finite set with an upper bound.

Consider finite.cardinals and their end.segments:
Each non.empty end.segment belongs to
the set.of.all non.empty end.segments.
Each non.empty end.segment holds
its minimum and the successor of its minimum.
For each
non.empty end.segment in set.of.all,
there is
⎛ a 1.element.emptier non.empty end.segment
⎝  which is also in set.of.all
It is that.end.segment\{its.minimum}
Each non.empty end.segment for which there is
⎛ a 1.element.emptier non.empty end.segment
⎝  which is also in set.of.all
⎛ is not minimal in set.of.all.
⎜ is not the (upper? lower?) bound of set.of.all.
⎝ is not the intersection of set.of.all.
Each non.empty end.segment in set.of.all
is not the intersection of set.of.all.

Each end.segment has, for each finite.cardinal,
a subset larger than that cardinal.

>
That is not true for the last dark endsegments.
It changes at the dark finite cardinal ω/2.

All those claims are true for
each finite dark.finite.cardinal.
You (WM) are considering
infinite dark.finite.cardinals,
which do not exist.