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

On 1/8/2025 9:31 AM, WM wrote:

On 08.01.2025 14:31, Jim Burns wrote:

⦃k: k < ω ≤ k+1⦄ = ⦃⦄
ω-1 does not exist.

>
Let us accept this result.
>
Then
the sequence of endsegments
loses every natnumber but
not a last one.
Then
the empty intersection of
infinite but
inclusion monotonic endsegments
is violating basic logic.
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
Then the only possible way
to satisfy logic is
the non-existence of ω and
of endsegments as complete sets.

(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)

No.
Sets do not change.
Not all sets are finite.
⎛ By 'finite', I mean
⎝ 'smaller.than fuller.by.one sets'
⎛ 'Finite' and 'infinite' are not
⎜ distance markers on the ordinal.road.
⎜ Finite is a quality.
⎜ Infinite is the negation of that quality.
⎜
⎜ You, poor finite being, <Jovian.laugh>
⎜ will not pass all finites.
⎜ That's what we mean by 'finite'.
⎜
⎜ However, you can describe each finite
⎜ truly.without.exception -- as finite.
⎜ You can follow that description with
⎜ true.or.not.first claims --
⎜ no claim of which can be false.
⎜
⎜ Each of those following.claims must be
⎜ true.without.exception of each finite.
⎜
⎜ Some of those following.claims are
⎜ more interesting than "It is finite".
⎜ Even the more.interesting ones are
⎝ true.without.exception of each finite.

It is useless to prove your claim
as long as you cannot solve this problem.

For each finite.set A, there is
an ordinal ⟦0,k⦆ finite and larger.than A
  ∀ˢᵉᵗA ≠ A∪{a}:
  #A < #(A∪{a})  ⇒
  ∃ᵒʳᵈk:  #⟦0,k⦆ < #⟦0,k+1⦆  ∧  #A < #⟦0,k⦆
That is validly.manipulable (preserving truth) to
  ∀ˢᵉᵗA ≠ A∪{a}:
  ∀ᵒʳᵈk:  #⟦0,k⦆ < #⟦0,k+1⦆  ∧  #⟦0,k⦆ ≤ #A
  ⇒  ¬(#A < #(A∪{a}))
If set A is at least as large as
each finite ordinal
then A is not finite.
⎛ By 'not finite', I mean
⎝ 'NOT smaller.than fuller.by.one sets'
ℕ is the set of all finite ordinals.
ℕ is at least as large as
each finite ordinal ⟦0,k⦆
ℕ is not finite.
ℕ is not smaller.than fuller.by.one sets.
Sets emptier.by.one than ℕ are not smaller.

(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)

No.
Sets emptier.by.one than ℕ are not smaller.
In the sequence of end.segments of ℕ
there is no number which
empties an infinite set to a finite set.
and
there is no number which
is in common with all its end.segments.
ℕ has only infinite end.segments.
The intersection of
all (infinite) end.segments of ℕ
is empty.
Sets do not change.
Not all sets are finite
⎛ By 'finite', I mean
⎝ 'smaller.than fuller.by.one sets'