$A$ 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

Solution 1:

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.