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Open subsets are a union of disjoint open intervals

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that any open subset of [itex]\Real[/itex] can be written as an at most countable union of disjoint open intervals.


    2. Relevant equations
    An at most countable set is either finite or infinitely countable.


    3. The attempt at a solution
    It seems very intuitive but I am at lost where to even start. We're doing compactness in metric spaces so I would assume it must apply. But I thought a set has to be closed in order to be compact and this deals with an open subset so it can't possibly be compact. Any help would be much appreciated!
     
  2. jcsd
  3. Nov 17, 2009 #2

    Office_Shredder

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    This has pretty much nothing to do with compactness. Try this:
    1) Show you can write it as a union of disjoint intervals.
    2) Think about rational numbers
     
  4. Nov 17, 2009 #3
    Hmmm, I think I've made some development but I'm still unsure how solid it is.

    Let [itex]E[/itex] be an open subset of [itex]\Real[/itex]. Define the equivalence relation [itex]x \sim y \Longleftrightarrow \exists (a,b)[/itex] such that [itex]{x,y} \in (a,b) \subseteq E[/itex]. The equivalence classes will be distinct. By the archimedean property, there exists a distinct rational number in each interval. Since the rationals are countable, we have a countable number of intervals. The countable union of these intervals will equal the set E. Therefore, every subset of R can be represented by a countable union of disjoint open intervals.
     
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