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

On 1/4/2025 3:42 AM, WM wrote:

On 1/3/2025 3:56 PM, Jim Burns wrote:

All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.

>
But removing
every ordinal that you can define
(and all its predecessors) from ℕ leaves
almost all ordinals in ℕ.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

ℕ is the set of each and only finite.ordinals.
A finite.ordinal is larger.by.one than emptier.by.one
  unless it is 0 = ⦃⦄
⎛ finite k  ⇔
⎝ #⟦0,k⦆ > #(⟦0,k⦆\⦃0⦄)  ∨  k = ⦃⦄
ℕ = ⦃k: finite k⦄
By definition of ℕ
for each set A larger.by.one than emptier.by.one,
(for each finite set)
there is an ordinal k in ℕ, ⟦0,k⦆ the same size as A
⎛ #A > #(A\{a})  ⇒
⎝ ∃k ∈ ℕ: #⟦0,k⦆ = #A
There isn't an ordinal #ℕ in ℕ of the same size as ℕ
Set ℕ is not larger.by.one.than.emptier.by.one.
⎛ ¬∃k ∈ ℕ: #⟦0,k⦆ = #ℕ  ⇒
⎝ ¬(#ℕ > #(ℕ\{0}))
For each set Y the same size as ℕ
There isn't an ordinal #Y in ℕ of the same size as Y
Y is not larger.by.one.than.emptier.by.one
⎛ ¬∃k ∈ ℕ: #⟦0,k⦆ = #Y  ⇒
⎝ ¬(#Y > #(Y\{y}))
|ℕ| := ℵ₀ = |ℕ\{0}| = |ℕ\{0,1}| = ... =
|ℕ\{0,1,...,n}| = ...
The sequence of end.segments of ℕ
grows emptier.one.by.one but
it doesn't grow smaller.one.by.one.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

ℕ is the set of each and only finite.ordinals.
Each finite.ordinal is not weird.
Even an absurdly.large one like Avogadroᴬᵛᵒᵍᵃᵈʳᵒ
is not weird.
However,
ℕ is weird.
Declaring that ℕ is not weird, but
like an absurdly.large finite.ordinal,
does not make ℕ
like an absurdly.large finite.ordinal,
because
ℕ is the set of each and only finite.ordinals,
which is not any finite.ordinal,
not even an absurdly.large one.