Re: Separable Linear Order

William Elliot wrote:
> Given a linear order L with the usual interval topology,
> how does one show when L is separable, L is Lindelof?

Let C be any open cover of L; we may assume that the members of C are
open intervals (a, b). Define an equivalence relation on L so that,
when x < y, the points x and y are equivalent just in case the closed
interval [x, y] is covered by countably many members of C. Each
equivalence class is open, since it is the union of some subcollection
of C. So the equivalence classes form a collection of disjoint nonempty
open subsets of L. Since L is separable, it follows that there are only
countably many equivalence classes. To finish the proof, we have to
show that each equivalence class is covered by countably many members
of C.

>>From the separability of L, it follows that L cannot have a subset of
order type omega_1. Hence each equivalence class must have a countable
cofinal subset, and (dually) a countable coinitial subset. Hence each
equivalence class is the union of countably many closed intervals. By
the definition of the equivalence relation, each closed interval
(contained in an equivalence class) is covered by countably many
members of C. That does it, I think.

.

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