Re: Separable Linear Order

fred.galvin@xxxxxxxxx wrote:
> 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.

PS. Note that, instead of assuming that L is separable, you can get by
with the weaker assumption that L satisfies the "countable chain
condition", i.e., every collection of disjoint open sets is countable.

.

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