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

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

On 14.12.2024 05:54, Jim Burns wrote:

[...]

>
Explain your vision of the problem:
If
ℕ is a set, i.e.
 if it is complete such that
 all numbers can be used  for indexing sequences or in other mappings,
then
it can also be exhausted such that
no element remains.
Then
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.
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