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

On 30.11.2024 11:57, FromTheRafters wrote:

WM explained :

On 29.11.2024 22:50, FromTheRafters wrote:

WM wrote on 11/29/2024 :

>

The size of the intersection remains infinite as long as all endsegments remain infinite (= as long as only infinite endsegments are considered).

>
Endsegments are defined as infinite,

>
Endsegments are defined as endsegments. They have been defined by myself many years ago.

>
As what is left after not considering a finite initial segment in your new set and considering only the tail of the sequence.

 Not quite but roughly. The precise definitions are:
Finite initial segment F(n) = {1, 2, 3, ..., n}.

Endsegment E(n) = {n, n+1, n+2, ...}

 

Almost all elements are considered in the new set, which means all endsegments are infinite.

 Every n that can be chosen has infinitely many successors. Every n that can be chosen therefore belongs to a collection that is finite but variable.
 

Try to understand inclusion monotony. The sequence of endsegments decreases.

>
In what manner are they decreasing?

 They are losing elements, one after the other:
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
But each endsegment has only one element less than its predecessor.
 

When you filter out the FISON, the rest, the tail, as a set, stays the same size of aleph_zero.

 For all endsegments which are infinite and therefore have an infinite intersection.

>

As long as it has not decreased below ℵo elements, the intersection has not decreased below ℵo elements.

>
It doesn't decrease in size at all.

 Then also the size of the intersection does not decrease.
Look: when endsegments can lose all elements without becoming empty, then also their intersection can lose all elements without becoming empty. What would make a difference?
 Regards, WM