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Suppose $(X,d)$ is metric space. I want to show that $A \subseteq X$ is open iff $A$ is union of open balls.

My attempt. suppose $A$ is open, then for every $x \in A$, there exists $r>0$ such that $B(x,r) \subset A$ by definition. We claim that $A = \bigcup_{x\in A} B(x,r) $. To see this, pick $x \in A$, then can find $r>0$ such that $x \in B(x,r) \subseteq \bigcup B(x,r)$. Conversely, suppose $y \in \bigcup B(x,r) \implies y \in B(x,r) $ for some $x$. But $B(x,r) \subseteq A$ for some $x$, hence $y \in A$. So, our claim is proved.

For the other direction, suppose $A = \bigcup_{\alpha} O_{\alpha} $ where $O_{\alpha}$ is open ball. Take $x \in A$, then $x \in O_{\alpha} $ for some $\alpha$. But $O_{\alpha} \subseteq \bigcup O_{\alpha} = A $. So, we have found an open ball inside $A$, and since $x$ was chosen arbitrary, then $A$ must be an open set by definition.

Is this correct? Any feedback would be greatly appreciated. thanks

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  • $\begingroup$ Looks good.$~~~~~~~$ $\endgroup$ – Sammy Black Feb 3 '14 at 8:10
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    $\begingroup$ Don't you need to show that if $x\in A$ then there is an $r>0$ with $B(x,r)\subset A$? All you have is that $x\in O_\alpha$, but $O_\alpha$ does not need to have center at $x$. $\endgroup$ – Andrés E. Caicedo Feb 3 '14 at 8:31
  • $\begingroup$ Can you explain more carefully? write it in the asnwer box, so I can give you points. thank you. $\endgroup$ – user124140 Feb 3 '14 at 9:06
  • $\begingroup$ You write: for every $x\in A$ there exists some $r>0$ such that $B\left(x,r\right)\subset A$. That is true, but it is useful to change it into: for every $x\in A$ there exists some $r_{x}>0$ such that $B\left(x,r_{x}\right)\subset A$. Then go on by showing that $A=\cup_{x\in A}B\left(x,r_{x}\right)$. If $x\in A$ then $x\in B\left(x,r_{x}\right)\subset A$ and conversely the fact $B\left(x,r_{x}\right)\subset A$ for every $x\in A$ implies directly that $\cup_{x\in A}B\left(x,r_{x}\right)\subset A$. $\endgroup$ – drhab Feb 3 '14 at 9:52
  • $\begingroup$ If $A=\cup_{\alpha}O_{\alpha}$ where every $O_{\alpha}$ is an open ball then $A$ is the union of a family of open sets. Consequently $A$ is open. So a more direct conclusion is possible and playing with open balls is not needed here. $\endgroup$ – drhab Feb 3 '14 at 9:56
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You are right, but it's even simpler. Firstly, any union of open sets is open by definition of topology (a set of subsets, which is closed under arbitrary unions).

Take your set $A'=\bigcup_{x\in A} B(x,r(x))$ (having chosen a $r(x)$ for each $x$).

Of course $A' \subseteq A$, because all of the $B(x,r(x))$ are (any point of $A'$ is a point of at least one $B(x,r(x))$).

On the other side, any point $x_0\in A$ belongs to its ball $B(x_0, r(x_0))$ which is a subset of $A'$. And that's all.

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  • $\begingroup$ Right, it was implicit in the sentence "having chosen a $r$ for each $x$, but it could have been more explicit. $\endgroup$ – rewritten Feb 3 '14 at 10:24
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    $\begingroup$ I made your repair complete. $\endgroup$ – drhab Feb 3 '14 at 10:28
  • $\begingroup$ Thanks @drhab. I appreciate it. $\endgroup$ – rewritten Feb 3 '14 at 11:34

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