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

On 11/26/2024 3:45 AM, WM wrote:

On 26.11.2024 06:58, Jim Burns wrote:

On 11/25/2024 8:52 AM, WM wrote:

Finite cardinalities belong to dark endsegments.

>
Finite cardinals can change by 1

>
Yes.
The last endsegments have 3, 2, 1, 0 elements.
>

Each end.segment Eᶠⁱⁿ(k) of the finite.cardinalities ℕᶠⁱⁿ
holds a countable.to.from.0 least.element

>
No.

Yes.
⎛ If
⎜  anything is in ℕᶠⁱⁿ which
⎜   is NOT a finite cardinal,
⎜ then
⎜  ℕᶠⁱⁿ is NOT the set of finite cardinals.
⎜
⎜ If
⎜  anything is NOT in ℕᶠⁱⁿ which
⎜   is a finite cardinal,
⎜ then
⎜  ℕᶠⁱⁿ is NOT the set of finite cardinals.
⎜
⎜ However,
⎜ ℕᶠⁱⁿ IS the set of finite cardinals
⎜
⎜ Therefore, extensionality:
⎜( A thing is in/not.in ℕᶠⁱⁿ
⎜ is equivalent to
⎝( A thing is/is.not a finite cardinal.
⎛ Our finite cardinals do not change.
⎜ But one can consider other finite cardinals.
⎜
⎜ Our set of finite cardinals does not change.
⎜ But one can consider other sets, each of which
⎝ is not the set of finite cardinals.

All elements of finite endsegments
 (and almost all of infinite endsegments)
are dark.

Each element k of each end.segment Efin(j) of Nfin
is a finite cardinal
is countable.to.from.0
is not.dark
----

Each end.segment Eᶠⁱⁿ(k) of the finite.cardinalities ℕᶠⁱⁿ
holds a countable.to.from.0 least.element

>
No.

Yes.
Each non.empty.subset S of ℕᶠⁱⁿ
holds an element k which
⎛ is a finite.cardinal
⎜ is countable.to.from.0
⎝ ends finite sequence ⟦0,k⟧ ⊆ ℕᶠⁱⁿ
The non.empty intersection ⟦0,k⟧∩S
holds a least.element i ∈ ℕᶠⁱⁿ which
⎛ is the least.element of S
⎝ is countable.to.from.0
Each end.segment Eᶠⁱⁿ(k) of ℕᶠⁱⁿ
⎛ is a non.empty.subset S of ℕᶠⁱⁿ
⎝ holds a countable.to.from.0 least.element
Each end.segment Eᶠⁱⁿ(k) of ℕᶠⁱⁿ
⎛ has a countable.to.from.0 least.element.index
⎜ is countable.to.from.ℕᶠⁱⁿ
⎝ is not.dark

The endsegments
only can have an empty intersection
if there are endsegments with 3, 2, 1, 0 elements.

>
The end.segments
can only have a non.empty intersection
if there is an element which is in each end.segment.

>
That is the case for every non-empty endsegment
before all elements are lost.

That is what 'intersection' means.
Each finite cardinal in ℕᶠⁱⁿ
⎛ is countable.to.from.0
⎝ is countable.past
Each end.segment of ℕᶠⁱⁿ
⎛ is indexed by a finite cardinal
⎜ is countable.to.from.ℕᶠⁱⁿ
⎝ is countable.past
Each finite cardinal k in ℕᶠⁱⁿ
⎛ indexes end.segment Eᶠⁱⁿ(k) of ℕᶠⁱⁿ
⎜ is in Eᶠⁱⁿ(k)
⎜ is not.in successor Eᶠⁱⁿ(k+1)
⎜ is not in common with each end.segment
⎝ is not in their intersection
Their intersection is empty.

Otherwise
two endsegments with different elements
must exist.
That is impossible by inclusion monotony.

----

Finite cardinals can change by 1

>
Yes.
The last endsegments have 3, 2, 1, 0 elements.

For each end.segnent Eᶠⁱⁿ(k) of ℕᶠⁱⁿ
for each finite.cardinal j in ℕᶠⁱⁿ
Eᶠⁱⁿ(k) has a more.than.j.sized subset
|Eᶠⁱⁿ(k)\Eᶠⁱⁿ(k+j+1)| = j+1 > j
|Eᶠⁱⁿ(k)| ≠ j
For each end.segnent Eᶠⁱⁿ(k) of ℕᶠⁱⁿ
|Eᶠⁱⁿ(k)| isn't any finite cardinal.
|Eᶠⁱⁿ(k)| can't change by 1
There are no last end.segments of ℕᶠⁱⁿ
There are no finitely.sized end segments of ℕᶠⁱⁿ
There are no finite cardinals common to
  each end.segment of ℕᶠⁱⁿ