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

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

On 08.12.2024 00:38, Jim Burns wrote:

Each end.segment is more.than.finite and
the intersection of the end.segments is empty.

>
∀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.
∀n ∈ ℕ:
{E(i):i}∪{E(n+1)} = {E(i):i}
Each is "already" in.
Our sets do not change.
∀n ∈ ℕ:
(⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}
Our intersections do not change.
∀n ∈ ℕ:
⎛ E(1)∩E(2)∩...∩E(n) = E(n)
⎜ E(n)∩E(n+1) ≠ E(n)
⎜ (⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}
⎝ E(n) ≠ ⋂{E(i):i}
⋂{E(i):i} ∉ {E(i):i}
The intersection.of.all non.empty end.segments
isn't any non.empty end.segment.