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

On 11/30/2024 4:56 PM, WM wrote:

On 30.11.2024 22:45, Jim Burns wrote:

On 11/30/2024 2:07 PM, WM wrote:

But there is a sequence of endsegments
E(1), E(2), E(3), ...

>
With an empty set of common finite.cardinals.

>
Inclusion monotony prevents
an empty set of common finite cardinals without an empty endsegment.

No,
inclusion.monotony does not prevent
an empty set of common finite.ordinals
without an empty end segment.
For collection ENDS holding infinitely.many end.segments,
each finite ordinal ξ
⎛ is not.in only finitely.many fore.segments ⟦0,β⟧ ⊆ ⟦0,ξ-1⟧
⎜ is in only finitely.many end.segments E(β+1) = ℕ\⟦0,β⟧
⎜ is in fewer.than.infinitely.many end.segments in ENDS
⎜ is not in common with each end segment in ENDS
⎝ is not in ⋂ENDS
For a collection ENDS holding infinitely.many end.segments,
⎛ each finite ordinal ξ is not in ⋂ENDS
⎝ ⋂ENDS = {}

Inclusion monotony prevents
an empty set of common finite cardinals without an empty endsegment.

No.
For a collection ENDS holding infinitely.many end.segments,
ENDS′ = ENDS\{{}} is
⎛ a collection without an empty end.segment
⎜ holding infinitely.many end.segments,
⎜ having an empty intersection.
⎝ ⋂ENDS′ = {}

Yes, they are identical. And identically empty.

>
You said that the endsegments are never empty.

I intended to say that those intersections are empty.