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

On 08.12.2024 19:01, Jim Burns wrote:

On 12/8/2024 5:50 AM, WM wrote:

∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't you understand here?

 {E(i):i} is the set.of.all non.empty end.segments.
 ⋂{E(i):i} is the intersection.of.all

non.empty end.segments.

 ∀n ∈ ℕ:
{E(i):i}∪{E(n+1)} = {E(i):i}
Each is "already" in.

Not the empty endsegment.
∀n ∈ ℕ: E(n) is non-empty. But not every E(n+1).

∀n ∈ ℕ:
(⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}

No. E(n+1) = { } is possible.

∀n ∈ ℕ:
⎛ E(1)∩E(2)∩...∩E(n) = E(n)
⎜ E(n)∩E(n+1) ≠ E(n)

E(n)∩E(n+1) = E(n+1)

⎜ (⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}

No. ⋂{E(i):i} is the intersection of non-empty endsegments and as such non-empty.

⎝ E(n) ≠ ⋂{E(i):i}

Not for all n.
Regards, WM

 ⋂{E(i):i} ∉ {E(i):i}
The intersection.of.all non.empty end.segments
isn't any non.empty end.segment.