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

On 1/12/2025 2:39 PM, WM wrote:

On 12.01.2025 20:33, Jim Burns wrote:

On 1/12/2025 10:54 AM, WM wrote:

No, it depends on completeness.

>
It is completely true
that each natural number is a natural number   and
that only natural numbers are natural numbers.

>
and that nothing fits between them and ω.

Yes.
ω is first infinite ordinal.
if any infinite ordinal exists,
then, by well.order, ω exists.
ω is first infiniteᵒʳᵈ.
No infiniteᵒʳᵈ is before ω
No finiteᵒʳᵈ is after ω
⎛ Assume otherwise.
⎜ Assume ω < k and k finiteᵒʳᵈ.
⎜
⎜ ⟦0,k⦆ = ⟦0,ω⦆∪⟦ω,k⦆
⎜ ω not finiteᵒʳᵈ
⎜ ¬(#⟦0,ω⦆ > #⟦1,ω⦆)
⎜ ¬(#⟦0,ω⦆ < #⟦1,ω⦆)
⎜ #⟦0,ω⦆ = #⟦1,ω⦆
⎜ #(⟦0,ω⦆∪⟦ω,k⦆) = #(⟦1,ω⦆∪⟦ω,k⦆)
⎜ #⟦0,k⦆ = #⟦1,k⦆
⎜
⎜ However,
⎜ k finiteᵒʳᵈ.
⎜ #⟦0,k⦆ > #⟦1,k⦆
⎝ Contradiction.
Therefore,
no finiteᵒʳᵈ is after ω

ℕ is the set of finite ordinals.

>
such than none can be added.

No more are needed for arithmetic of finitesᵒʳᵈ.
⎛ Assume k < ω ≤ k+1
⎜
⎜ #⟦0,k⦆ > #⟦1,k⦆
⎜ #⟦0,k+1⦆ > #⟦1,k+1⦆
⎜ k+1 < ω
⎝ Contradiction.
¬(k < ω ≤ k+1)
⎛ Assume j,k < ω ≤ j+k
⎜
⎜ By well.order, first k₀ < ω exists
⎜ such that j+(k₀-1) < ω ≤ j+k₀
⎜
⎜ However,
⎜ from up.post,
⎜ j+(k₀-1) < (j+(k₀-1))+1 < ω
⎜ By definition of '+',
⎜ (j+(k₀-1))+1 = j+((k₀-1)+1) = j+k₀
⎜ j+k₀ < ω
⎝ Contradiction.
¬(j,k < ω ≤ j+k)
⎛ Assume j,k < ω ≤ j×k
⎜
⎜ By well.order, first k₀ < ω exists
⎜ such that j×(k₀-1) < ω ≤ j×k₀
⎜
⎜ However,
⎜ from up.post,
⎜ j×(k₀-1) < (j×(k₀-1))+j < ω
⎜ By definition of '×',
⎜ (j×(k₀-1))+j = j×((k₀-1)+1) = j×k₀
⎜ j×k₀ < ω
⎝ Contradiction.
¬(j,k < ω ≤ j×k)
⎛ Assume j,k < ω ≤ j^k
⎜
⎜ By well.order, first k₀ < ω exists
⎜ such that j^(k₀-1) < ω ≤ j^k₀
⎜
⎜ However,
⎜ from up.post,
⎜ j^(k₀-1) < (j^(k₀-1))×j < ω
⎜ By definition of '^',
⎜ (j^(k₀-1))×j = j^((k₀-1)+1) = j^k₀
⎜ j^k₀ < ω
⎝ Contradiction.
¬(j,k < ω ≤ j^k)

Regular distances in (0, ω) multiplied by 2
remain regular distances in (0, 2ω).

¬(j,2 < ω ≤ j×2)
Regular distances in ⦅0,ω⦆ multiplied by 2
remain regular distances in ⦅0,ω⦆, not.in ⟦ω,2ω⦆