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

On 1/13/2025 12:29 PM, WM wrote:

On 13.01.2025 18:06, Jim Burns wrote:

On 1/13/2025 7:48 AM, WM wrote:

On 13.01.2025 05:51, Moebius wrote:

Es gilt dann also |ℕ|/n = |ℕ|.

>
Wrong. |ℕ| is a fixed number.

>
By 'fixed', you mean that #ℕ > #(ℕ\{0})
>
However,
ℕ is the set of all 'fixed' (finite) ordinals.

>
Which is ivariable.

ℕ is only invariable in the sense which we use.
Nothing changes status from in.ℕ to not.in.ℕ,
or the other direction.
However,
you (WM) are convinced that
a set (such as ℕ) larger than
any set with sets.different.in.size.by.one
changes (has elements inserted or deleted)
in order to not.change.in.size.by.one.
ℕ does not change.
ℕ changing is not the reason that
there aren't sets.different.in.size.by.one
ℕ is larger than any of those sets.
Yes,
it would be very weird for a finite set
to be larger than any of those sets.
However, ℕ is not a finite set.

⎛ Assume ℕ is 'fixed'.
⎜
⎜ A 'fixed' ordinal ⟦0,𝔑⦆ the size of ℕ exists.

>
|ℕ| = ω-1.

No.
j < ω  ⇒  j < j+1 < ω
j < k < ω  ⇒  j ≠ ω-1
j < ω  ⇒  j ≠ ω-1

⎜ #⟦0,𝔑⦆ = #ℕ
⎜
⎜ ⟦0,𝔑+1⦆ is 'fixed', too.

>
ω

Is ω = ⟦0,𝔑+1⦆ your 'fixed' (our 'finite')?
If ⟦0,𝔑⦆ is finite, then
⟦0,𝔑+1⦆, ⟦0,𝔑+2⦆, ⟦0,𝔑+3⦆ are finite.
What about ω+1 and ω+2?

⎜ ⟦0,𝔑+1⦆ ⊆ ℕ

>
No.

Yes.
𝔑 ∈ ℕ  ⇔  finite ⟦0,𝔑⦆
finite ⟦0,𝔑⦆  ⇔  finite ⟦0,𝔑+1⦆
finite ⟦0,𝔑+1⦆  ⇔  𝔑+1 ∈ ℕ
∀𝔑 ∈ ℕ: #⟦0,𝔑⦆ < #⟦0,𝔑+1⦆ ≤ #ℕ