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

On 14.12.2024 23:04, Jim Burns wrote:

On 12/14/2024 5:26 AM, WM wrote:

the set of what remains unused, i.e.,
 of intersections of endsegments
 (1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
Then,
by the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the content must become finite.

 

Explain your vision of the problem:

 A finite member ⟦0,ψ⦆ of the (well.ordered) ordinals
is smaller.than its successor ⟦0,ψ⦆∪{ψ}
 If ⟦0,ψ⦆ is smaller than its successor ⟦0,ψ⦆∪{ψ}
then ⟦0,ψ+1⦆ = ⟦0,ψ⦆∪{ψ} is smaller.than
its successor ⟦0,ψ+1⦆∪{ψ+1}
  which means
If ψ is finite, then ψ+1 is finite.
If ψ+1 is finite, then ψ+2 is finite.

Yes, that is the potentially infinite collection of definable numbers. But it explains nothing.

 ω is the first upper bound of finite ordinals.
If ψ < ω, then ψ < ψ+1 < ψ+2 ≤ ω
 If ω-1 exists
then
ω-1 is last.before.ω
α < β < ω  ⇒  α ≠ ω-1
 If ω-1 exists
then
ω-1 < (ω-1)+1 < (ω-1)+2 ≤ ω
ω-1 ≠ ω-1
 Therefore,
ω-1 doesn't exist
 

Not as a definable number. That is common knowledge. But you should not only say what not exists.
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ... loses all content.
By the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} = ∩{E(1),E(2),...,E(k)}\{k}
the sequence gets empty one by one.
Regards, WM