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Question

Show that every open subset of a metric space can be expressed as a union of open balls.

So far I have the following:

"Let $U \subseteq X$. For each $a \in U$, choose $r_a > 0$ such that $B(a, r_a) \subseteq U$."

I'm just not sure what the next step to show that $\bigcup_{a \in U}B(a, r_a) = U$.

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Well, you've got two sets that you want to show are equal, so try showing that each is contained in the other. –  Tara B Apr 23 '12 at 11:04

2 Answers 2

up vote 2 down vote accepted

$\bigcup_{a \in U}B(a, r_a) \subseteq U$, because it is, by definition, an union of subsets $B(a, r_a)$ of $U$.

$U \subseteq \bigcup_{a \in U}B(a, r_a)$, because each $a \in U$ is also contained in a right-hand side.

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Balls are, by definition, a basis for the neighborhood system. This implies the result.

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When you write an answer it must be close to the level of OP –  no identity Apr 23 '12 at 11:35
    
Well, I guess the OP should know what a neighborhood basis is. In metric space theory, this is THE definition of an open subset. Actually, I think that the question is rather tautological: a set in a metric space is open when every point has a ball containing it and contained in the set. In some sense, the OP was asking what a union of sets is. –  Siminore Apr 23 '12 at 11:41
    
It depends on you math course. In some institutes neighborhoods bases and other topological notions are not studied at all –  no identity Apr 23 '12 at 11:52
    
Well, it's very hard to study the topology of metric spaces without a definition of neighborhoods! I still believe that the question is essentially a question about union of subsets. –  Siminore Apr 23 '12 at 12:47
    
@Siminore: I think you are basically right, but this doesn't change the fact that your answer is unlikely to be much help to the OP. Of course we might all be wrong about that, and so I certainly wouldn't go as far as downvoting your answer. –  Tara B Apr 23 '12 at 13:11

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