# Open set is countable union of disjoint open balls

by dreamtheater
Tags: balls, countable, disjoint, union
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 P: 10 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.
 Sci Advisor HW Helper PF Gold P: 4,771 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 $\{I_x:x\in U\}$ 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 $\{C_x:x\in U\}$ is countable and its elements are disjoint open subsets of X whose union is U.
 Sci Advisor HW Helper PF Gold P: 4,771 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

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