# $A$ is open iff it is union of open balls

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

• Looks good.$~~~~~~~$ – Sammy Black Feb 3 '14 at 8:10
• 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 $x$. – Andrés E. Caicedo Feb 3 '14 at 8:31
• Can you explain more carefully? write it in the asnwer box, so I can give you points. thank you. – user124140 Feb 3 '14 at 9:06
• 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 $\cup_{x\in A}B\left(x,r_{x}\right)\subset A$. – drhab Feb 3 '14 at 9:52
• 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. – drhab Feb 3 '14 at 9:56

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.
• Right, it was implicit in the sentence "having chosen a $r$ for each $x$, but it could have been more explicit. – rewritten Feb 3 '14 at 10:24