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

On 12/19/2024 1:29 PM, Jim Burns wrote:

On 12/19/2024 9:50 AM, WM wrote:

The empty term
has lost all natnumbers one by one.

>
One.by.one and (darkᵂᴹ or visibleᵂᴹ) endlessly.
>
For each set smaller.than fuller.by.one sets,
fuller.by.ONE sets are smaller.than
fuller.by.TWO sets.

For b ∉ A ∌ c ≠ b
#A < #(A∪{b})  ⇒  #(A∪{b}) < #(A∪{b,c})
⎛ Assume otherwise.
⎜ Assume #A < #(A∪{b}) ≥ #(A∪{b,c})
⎜
⎜ #(A∪{b}) ≥ #(A∪{b,c})
⎜ one.to.one g: A∪{b,c} ⇉ A∪{b}
⎜ Define
⎜⎛ f(g⁻¹(b)) = g(c)
⎜⎜ otherwise f(x) = g(x)
⎜⎝ one.to.one f: A∪{b} ⇉ A
⎜ #A ≥ #(A∪{b})
⎜
⎜ However,
⎜ #A < #(A∪{b})
⎝ Contradiction.
Therefore, for b ∉ A ∌ c ≠ b
#A < #(A∪{b})  ⇒  #(A∪{b}) < #(A∪{b,c})
For each set smaller.than fuller.by.one sets,
fuller.by.ONE sets are smaller.than
fuller.by.TWO sets.

No set is largest among
the sets smaller.than fuller.by.one sets.