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

On 1/15/2025 1:13 PM, WM wrote:

On 14.01.2025 19:41, Jim Burns wrote:

On 1/14/2025 4:07 AM, WM wrote:

On 13.01.2025 20:31, Jim Burns wrote:

A step is never from finite to infinite.

>
The dark realm is appears infinite.

Nowhere,
among what appears and
among what doesn't appear,
is there a step from finite to infinite.
⎛ There is a route of steps.0.to any finiteᵒʳᵈ.
⎜ There isn't a route of steps.0.to any infiniteᵒʳᵈ.
⎜ If you disagree, re.consider what infinitesᵒʳᵈ are.
⎜
⎜ There is no pairᵒʳᵈ ω-1,(ω-1)+1 such that
⎜⎛ there is a route 0.to ω-1 and
⎜⎝ there isn't a route 0.to (ω-1)+1
⎜ because
⎜ continuing an assumed route 0.to ω-1 with
⎜ a step ω-1 to (ω-1)+1 produces
⎝ a contradicting route 0.to (ω-1)+1
Nowhere,
among what appears and
among what doesn't appear,
is there finite ω-1 and infinite (ω-1)+1

There is no infinite set smaller than ℕ
#𝔼 ≥ #ℕ

>
That is obviously wrong.

Yes,
𝔼 is an infiniteˢᵉᵗ emptier than ℕ
However,
𝔼 is not an infiniteˢᵉᵗ smaller than ℕ
⎛ Define f(n) = 2×n
⎜ f one.to.one: ℕ ⇉ 𝔼
⎝ #ℕ ≤ #𝔼
Also, more generally,
there is no infiniteˢᵉᵗ smaller than ℕ
⎛ Assume that 𝕌 is an infiniteˢᵉᵗ.
⎜ 𝕌 is larger.than.any.finiteᵒʳᵈ
⎜( Otherwise, 𝕌 is finiteˢᵉᵗ.
⎜
⎜ For each finiteᵒʳᵈ k,
⎜ 𝕌 is larger.than.⟦0,k⦆
⎜ ∃fₖ one.to.one: ⟦0,k⦆ ⇉ 𝕌
⎜
⎜ By using all fₖ  there can be defined g
⎜ g one.to.one: ℕ ⇉ 𝕌
⎜ [1]
⎜
⎝ Therefore, #ℕ ≤ #𝕌
If 𝕌 is an infinite set,
then #ℕ ≤ #𝕌
[1]
⎛ For each finiteᵒʳᵈ k
⎜ ∃fₖ one.to.one: ⟦0,k⦆ ⇉ 𝕌
⎜
⎜ fₖ⟦0,k⦆ =
⎜ {fₖ(j) ∈ 𝕌: j ∈ ⟦0,k⦆} =
⎜ ⟨u₀,u₁,u₂,...,uₖ₋₁⟩ ⊆ 𝕌
⎜
⎜ Fₖ is the multi.concatenation ('&')
⎜ f₀⟦0,0⦆ & f₁⟦0,1⦆ & f₂⟦0,2⦆ & ... & fₖ⟦0,k⦆
⎜
⎜ |fₖ⟦0,k⦆| = #fₖ⟦0,k⦆ = k
⎜ There are at least k unique elements of 𝕌 in Fₖ
⎜
⎜ Define g(k) to be
⎜ the kᵗʰ initial occurrence in Fₖ, which is to say,
⎜ kᵗʰ when skipping any second or later re.occurrences
⎜
⎜ The kᵗʰ initial occurrence in Fₖ is also
⎜ the kᵗʰ initial occurrence in Fₘ   m > k
⎜ g(k) is one.to.one
⎜
⎝ g one.to.one: ℕ ⇉ 𝕌

 The rule of subset proves that
every proper subset has
fewer elements than its superset.

Lagniappe.
For each finiteˢᵉᵗ which has
emptier.by.one subsets which are smaller,
there is a finiteᵒʳᵈ of that size.
ℕ is the set of finitesᵒʳᵈ which have
emptier.by.one subsets which are smaller.
None of those finitesᵒʳᵈ is the size of ℕ
ℕ is not a finiteˢᵉᵗ which has
emptier.by.one subsets which are smaller.