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

On 28.12.2024 18:31, Jim Burns wrote:

On 12/28/2024 9:03 AM, WM wrote:

[0, ω-1] = [0,ω⦆ = ℕ =/= [0, ω]

 Yes,
⟦0,ω⦆  =  ℕ  ≠  ⟦0,ω⟧
 However,
⎛ Assume k < ω
⎜ #⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
⎜ #⟦0,k+1⦆ ≠ #(⟦0,k+1⦆∪⦃k+1⦄)
⎝ k+1 < ω

That holds for almost all natural numbers k.
It cannot hold for an actually infinite system without disappearing Bob.
For example, all endsegments obey the sequence defined by
∀k ∈ ℕ : E(k+1) = E(k) \ {k+1}.
If all natnumbers exist, then all will leave the terms of the sequence, but cannot other than in single steps including the dark endsegments
E(ω-3) = {ω-2, ω-1}, E(ω-2) = {ω-1}, E(ω-1) = { }.
∀k ∈ ℕ : E(k+1) = E(k) \ {k+1} does not allow another alternative.
Regards, WM