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

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

On 13.01.2025 20:31, Jim Burns wrote:

On 1/13/2025 12:17 PM, WM wrote:

[...]

>
A step is never from finite to infinite.
Therefore, a step never crosses ω
Therefore, a sum never crosses ω
Therefore, a product never crosses ω
Therefore, a power never crosses ω

>
All that is true in potential infinity,
however it is wrong in completed infinity.

All that is true in ⟦0,ω⦆
Steps, sums, products and powers
never cross out of ⟦0,ω⦆
∀ᵒʳᵈk:  k ∈ ⦅0,ω⦆  ⇔  #⟦0,k⦆ > #⟦1,k⦆
#⟦0,k⦆ > #⟦1,k⦆  ⇒
#⟦0,k+1⦆ > #⟦1,k+1⦆  ⇔  k+1 ∈ ⦅0,ω⦆
#⟦0,j⦆ > #⟦1,j⦆  ∧  #⟦0,k⦆ > #⟦1,k⦆  ⇒
#⟦0,j+k⦆ > #⟦1,j+k⦆  ⇔  j+k ∈ ⦅0,ω⦆
#⟦0,j⦆ > #⟦1,j⦆  ∧  #⟦0,k⦆ > #⟦1,k⦆  ⇒
#⟦0,j×k⦆ > #⟦1,j×k⦆  ⇔  j×k ∈ ⦅0,ω⦆
#⟦0,j⦆ > #⟦1,j⦆  ∧  #⟦0,k⦆ > #⟦1,k⦆  ⇒
#⟦0,j^k⦆ > #⟦1,j^k⦆  ⇔  j^k ∈ ⦅0,ω⦆
----

Doubling of all n
deletes the odd numbers
but cannot change the number of numbers,

>
ℕ is the set of finite ordinals.
>
There is no finite set larger than ℕ
thus ℕ is infinite.
There is no infinite set smaller than ℕ

𝔼 is the set of even finite ordinals.
There is no finite set larger than 𝔼
thus 𝔼 is infinite
𝔼 ⊆ ℕ
#𝔼 ≤ #ℕ
There is no infinite set smaller than ℕ
#𝔼 ≥ #ℕ

 That is obviously wrong.

For each finite set A, there is
a finite ordinal ⟦0,k[A]⦆ larger than A
For each infinite set Y, there isn't
a finite ordinal ⟦0,k[Y]⦆ larger than Y
#ℕ is an upper.bound of finite #⟦0,k⦆
∀ᵒʳᵈk: finite ⟦0,k⦆  ⇒  #⟦0,k⦆ ≤ #ℕ
#ℕ = #(⋃⦃finiteᵒʳᵈ⦄) is
the least.upper.bound of finite #⟦0,k⦆
∀ᵒʳᵈk: finite ⟦0,k⦆  ⇒  #⟦0,k⦆ ≤ #Y
⇒  #ℕ ≤ #Y
If Y is an infinite set,
then #Y is an upper.bound of finite #⟦0,k⦆
such that #Y is not.smaller.than #ℕ
Otherwise,
#ℕ wouldn't be the least.upper.bound.
There is no infinite set smaller than ℕ
----
𝔼 is the set of even finite (<ω) ordinals.
There is no finite set larger than 𝔼
thus 𝔼 is infinite
𝔼 ⊆ ℕ
#𝔼 ≤ #ℕ
There is no infinite set smaller than ℕ
#𝔼 ≥ #ℕ
#𝔼 = #ℕ