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 $\bigcup_{a \in U}B(a, r_a) = U$.

  • 1
    $\begingroup$ Well, you've got two sets that you want to show are equal, so try showing that each is contained in the other. $\endgroup$ – Tara B Apr 23 '12 at 11:04

$\bigcup_{a \in U}B(a, r_a) \subseteq U$, 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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