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

On 23.11.2024 16:33, Jim Burns wrote:

On 11/23/2024 3:54 AM, WM wrote:

∀k ∈ ℕ_def:
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀

 

For all endsegments:
∀k ∈ ℕ:
|E(k+1)| = |E(k)| - 1

Perhaps what you intend to say  is

precisely what I wrote.

⎛ For all post.definable end.segments
⎜ ∀k ∈ (ℕ\ℕ_def):
⎝ |E(k+1)| = |E(k)| - 1

|E(k+1)| = |E(k)| - 1 is true for _all_ endsegments.
Cardinality ℵo of infinite endsegements cannot describe this because
ℵo - 1 = ℵo.

 Perhaps what you intend to say  is
⎛ There are post.definable end.segments
⎝ which have finite cardinalities.

Otherwise the intersection is infinite.

 I nearly agreed with that,
because our ℕ = ℕ_def

The intersection of all definable endsegements is infinite.
The intersection of all endsegments is empty.

However,
your mention of 'E(k+1)' implies
⎛ ∀k ∈ (ℕ\ℕ_def):
⎝ (ℕ\ℕ_def) ∋ k+1
 For _your_ end.segments,
k ↦ k+1 : one.to.one
E(k) → E(k+1) : one.to.one
|E(k)| ≤ |E(k+1)|

No. |E(k)| ≥ |E(k+1)| = |E(k)| - 1.

 Also,

Only!

E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|

True.

 |E(k)| = |E(k+1)|

Wrong. True only when cardinality is used because
ℵo - 1 = ℵo.

Remember:
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.

>
No one should "remember" that.
It is incorrect.

>
∀k ∈ ℕ_def:
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
>
∩{E(1), E(2), ...} = { }.

⎜ ℕ_def ∋ 0

No.
∀k ∈ ℕ_def: ℕ_def ∋ k+1
Yes.

⎜ ℕ_def ⊆ S  ⇐  S ∋ 0 ∧ ∀k ∈ S: S ∋ k+1

There is no S.

⎝ Each end.segment of ℕ_def is countable.to.

Yes.

 ⋂{E(k):k∈ℕ_def} = {}

No.

 because
∀j ∈ ℕ_def:
j ∉ E(j+1) ∈ {E(k):k∈ℕ_def}
j ∉ ⋂{E(k):k∈ℕ_def}
 The end.segment.intersection is empty because
 

No, the intersection contains all dark numbers ℕ\ℕ_def.
 > each end.segment of ℕ_def is countable.past.
What is countable.past?
Regards, WM