Topology open balls and open sets

Question

Let $A\subseteq \mathbb{R}^n$ be an open set

Prove that A can be written as a countable union of open balls.

We thought about it having to do with creating a ball with a radius of $r$ ($r>0$) around every point in $A$ that has rational coordinates. We think its the right direction but we're struggling with the formality.

Thanks

Answer 1

Consider the set $S$ of open balls of $\mathbb R^n$ whose centers are having rational coordinates and whose radii belong to the set $\{\frac{1}{n} : n \in \mathbb N, \ n\ge 1\}$. $S$ is countable.

Now take $x \in A$. As $A$ is open you can find an open ball $B(x,\frac{1}{n_x}) \subset A$, with $n_x \ge 1$ integer . As $\mathbb Q^n$ is dense in $\mathbb R^n$, it is possible to find $q_x \in \mathbb Q^n$ with $\Vert x - q_x \Vert < \frac{1}{2n_x}$. Then $x \in B(q_x,\frac{1}{2n_x}) \subset B(x,\frac{1}{n_x}) \subset A$

The set of open balls $\{B(q_x,\frac{1}{2n_x}) : x \in A\}$ is included in $S$, hence is countable and the union of those open balls is equal to $A$.