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

On 12/15/24 7:00 AM, WM wrote:

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.

That is the collection of numbers known as the Natural Numbers, so I guess you are admitting that your "Definable Numbers" include *ALL* of the Natural Numbers, so NONE of your "Dark Numbers" are what are called Natural Numbers, but something else.

>
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.

It doesn't exist in the defined number space, just like sqrt(-1) doesn't exist in the reals.
Yes, there may be a super-trans-finite numbering system that defines it, but it isn't in the Number Set that you claim to be talking about.
There is not series of successions from 0 that will get you to ω-1.
There isn't even a series of successions from 0 that will get you to ω, as ω is the first ordinal of the new type.

 (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