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

On 1/8/2025 4:16 AM, WM wrote:

On 08.01.2025 00:50, Jim Burns wrote:

The cardinal:ordinal distinction
-- which does not matter in the finite domain
matters in the infinite domain.

>
The reason is that
the infinite cardinal ℵ₀ is based on
the mapping of the potentially infinite collection of
natural numbers n,
all of which have
infinitely many successors.
The cardinal ℵ₀ is not based on
the mapping of
the actually infinite set ℕ where
ℕ \ {1, 2, 3, ...} = { }.

For each set smaller.than a fuller.by.one set,
the cardinal:ordinal distinction doesn't matter.
Cardinals and ordinals always go together.
For each set smaller.than a fuller.by.one set
there is an ordinal of its size in
the set ℕ of all finite ordinals.
Each set for which
there is NOT an ordinal of its size in
the set ℕ of all finite ordinals
is NOT a set smaller.than a fuller.by.one set.
For a set NOT smaller.than a fuller.by.one set,
cardinals and ordinals don't always go together.
For those (infinite) sets,
the cardinal:ordinal distinction matters.
For example,
there is NOT an ordinal of size #ℕ in
the set ℕ of all finite ordinals.
⎛ Assume otherwise.
⎜ Assume 𝔊 ∈ ℕ: #⟦0,𝔊⦆ = #ℕ
⎜
⎜ 𝔊 ∈ ℕ
⎜ ⟦0,𝔊⦆ is finite
⎜ #⟦0,𝔊⦆ < #⟦0,𝔊+1⦆ < #⟦0,𝔊+2⦆
⎜ ⟦0,𝔊+1⦆ is finite.
⎜ ⟦0,𝔊+1⦆ ⊆ ℕ
⎜ #⟦0,𝔊+1⦆ ≤ #ℕ
⎜ #ℕ = #⟦0,𝔊⦆ < #⟦0,𝔊+1⦆ ≤ #ℕ
⎜ #ℕ < #ℕ
⎝ Contradiction.
Therefore,
there is NOT an ordinal of size #ℕ in
the set ℕ of all finite ordinals.
ℕ is NOT smaller.than a fuller.by.one set.
The cardinal:ordinal distinction matters for ℕ