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Open set is countable union of disjoint open balls

  1. Dec 21, 2009 #1
    In R, every nonempty open set is the disjoint union of a countable collection of open intervals. (Royden/Fitzpatrick, 4th edition)

    What is the most general setting in which every open set is a disjoint union of countable collection of open balls (or bases)? In R^n? In metric spaces? In second countable topological spaces?

    The above theorem in R took me by surprise when I saw it for the first time.
     
  2. jcsd
  3. Dec 22, 2009 #2

    quasar987

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    What is the proof of this fact in R? Let U be some open set in R. For every x in U, let I_x be the largest open interval containing x and contained entirely in U. Then for x and y in U, either I_x and I_y are equal or disjoint. Moreover, if I_x and I_y are disjoints, then by density of Q in R, they both contain a rational number that the other does not contain. So the set [itex]\{I_x:x\in U\}[/itex] is countable and its elements are disjoint open intervals whose union is U. q.e.d.

    (By the way, notice that I_x is just the connected component of U containing x.)

    In light of this, it seems to me that the only ingredient necessarily is separability:

    Let X be a separable topological space. This means X admits a dense countable subset D. Let U be open in X. For each x in X, let C_x be the connected component of U containing x. Then for x and y in U, either C_x and C_y are equal or disjoint. Moreover, if C_x and C_y are disjoints, then by density of D] in X, they both contain an element of D that the other does not contain. So the set [itex]\{C_x:x\in U\}[/itex] is countable and its elements are disjoint open subsets of X whose union is U.
     
  4. Dec 22, 2009 #3

    quasar987

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    Ah, no. For this proof, we also need to add on X the hypothesis that the connected components of U will be open. This will be the case for instance if X is locally path connected.

    Indeed, consider the following counter-example. Let Q have the subspace topology inherited by R. Let r<s be any two irrational numbers. Then [itex]U:=]r,s[\cap \mathbb{Q}[/itex] is open in Q but its connected components are all the singletons {q} with q rational and contained in ]r,s[. By by density of the irrationals in R, none of the {q} are open in Q.
     
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