Re: Separable Linear Order

K. P. Hart wrote:
> 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.
> >
> More directly:
> let D be a countable dense set in L that includes the end points, if any.
> Then L = bigcup{[a,b]: a,b in D; a<b} writes L as a union of countably
> many closed intervals, each of which is covered by countably many O's.

I don't get it. How do you show that a closed interval is countably
covered? Why is that easier than the original problem? What am I
missing?

.

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