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

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

On 27.12.2024 22:00, Jim Burns wrote:

On 12/27/2024 5:14 AM, WM wrote:

They are invariable numbers like ω and ω+1.

>
ω is
the set of (well.ordered) ordinals k such that
#⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
(such that k is finite)

>
That is one interpretation.

Under that interpretation,
#⟦0,ω⦆ = #(⟦0,ω⦆∪⦃ω⦄)

My interpretation is Cantor's original one:
ω is the limit of the sequence 1, 2, 3, ... .

Your ω
ʸᵒᵘʳ( #⟦0,ω⦆ ≠ #(⟦0,ω⦆∪⦃ω⦄)
is infinitely.smaller.than
our ω
ᵒᵘʳ( k < ω  ⇔  #⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)

A separate fact is that
⟦0,ω⦆ ≠ ⟦0,ω⦆∪⦃ω⦄

>
[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
k < ω  ⇒  k < k+1 < ω
i < j < ω  ⇒  i ≠ ω-1
k < k+1 < ω  ⇒  k ≠ ω-1
¬∃⟦0,ω-1⟧