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.
>
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.

KP


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