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

On 1/8/2025 4:22 AM, WM wrote:

On 08.01.2025 06:45, Jim Burns wrote:

On 1/7/2025 4:13 AM, WM wrote:

On 06.01.2025 23:43, Jim Burns wrote:

k ∈ ℕ  ⇒  k+1 ∈ ℕ
is true for both the darkᵂᴹ and the visibleᵂᴹ.

>
One exception exists: ω-1.

>
No.
ω-1 does not exist, darkᵂᴹ or visibleᵂᴹ,
because
⦃k: k < ω ≤ k+1⦄ = ⦃⦄

>
If ω exists, then ω-1 exists.
Then your claim is wrong.

If ω exists, then
⦃k: k < ω ≤ k+1⦄ = ⦃⦄
and ω-1 does not exist.
See below
⎛ Define  k < ω  ⇔  #⟦0,k⦆ < #⟦0,k+1⦆
Without that (or similar),
how could all this make sense?

⎛ Assume otherwise.
⎜ Assume ω-1 exists.
⎜ ω-1 < ω
⎜ ¬∃ᵒʳᵈψ: ω-1 < ψ < ω
⎜
⎜ However,
⎜ ω-1 ∉ ⦃⦄ = ⦃k: k < ω ≤ k+1⦄

>
⦃⦄ = ⦃k: k < ω ≤ k+1⦄
is a wrong presupposition.
>

⎜ ¬(w-1 < w ≤ (w-1)+1)
⎜ w-1 < (w-1)+1 < w
⎜ ∃ᵒʳᵈψ: ω-1 < ψ < ω
⎝ Contradiction.

⎛ Define  k < ω  ⇔  #⟦0,k⦆ < #⟦0,k+1⦆
⎜
⎜ Lemma:  #A < #B  ⇒  #(A∪{a}) < #(B∪{b})
⎜
⎜ A = ⟦0,k⦆
⎜ B = ⟦0,k+1⦆
⎜ A∪{a} = ⟦0,k+1⦆ = ⟦0,k⦆∪⦃k⦄
⎜ B∪{b} = ⟦0,k+2⦆ = ⟦0,k+1⦆∪⦃k+1⦄
⎜
⎜ #⟦0,k⦆ < #⟦0,k+1⦆  ⇒  #⟦0,k+1⦆ < #⟦0,k+2⦆
⎜
⎜ By lemma and definition
⎜ k < ω  ⇒  k+1 < ω
⎜
⎜ ¬(k < ω ≤ k+1)
⎜
⎝ ⦃k: k < ω ≤ k+1⦄ = ⦃⦄

⦃k: k < ω ≤ k+1⦄ has one element.
It is dark like ω.

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