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

On 11/30/2024 03:54 AM, FromTheRafters wrote:

on 11/30/2024, WM supposed :

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.

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

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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, ...}

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There it is!! Don't you see that the ellipsis means that endsegments are
defined as infinite?

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

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

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In what manner are they decreasing?

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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.

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But how is that related to decreasing? What has decreased?
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When you filter out the FISON, the rest, the tail, as a set, stays
the same size of aleph_zero.

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For all endsegments which are infinite

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Which they all are, see above.
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and therefore have an infinite intersection.

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

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It doesn't decrease in size at all.

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Then also the size of the intersection does not decrease.

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Of course not, since it stays at emptyset unless there is a last element
-- which there is not since endsegments are infinite.
>

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?

>
Finite sets versus infinite sets. Finite ordered sets have a last
element which can be in the intersection of all previously considered
finite sets. Infinite ordered sets have no such last element.

 What about the "infinite-middle" models?

 This is simply about a symmetric rather than a-symmetric
outset of integers, for example.
 As "sets", with their "ordering", and by that I mean sets
with an ordering, there's first and last, alpha and omega
as it were.