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).

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Endsegments are defined as infinite,

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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