Show that a subset of $\mathbb{R^n}$ is open if and only if it is the
union of a countable collection of open balls.
Attempt:
I know a set G is open if there exists a $\epsilon >0 $ such that every point $x\in A$ satisfies $||x-y||<\epsilon$. Then I must show that $x$ is in some ball in $A$ which will make every point an interior point. What I don't understand in this question is the use of the words "countable collection" why does it have to be countable? Is it because an infinite collection of open sets is not open?
Best Answer
($\rightarrow$) Let $Y \subset \mathbb{R}^n$ be open, then for every $x \in Y$, there exists an open $U_x \subset Y$ such that $x \in U$. Clearly the set $\left\{ \displaystyle\bigcup U_x \colon x \in Y \right\} = Y$. However the collection $\mathscr{F} := \left\{ U_x \colon x \in Y \right\}$ may not be countable. Since $\mathbb{Q}^n$ is dense in $\mathbb{R}^n$, it is also dense in $Y$. Restrict $\mathscr{F}$ by considering then $\mathscr{G} := \left\{ U_x \in \mathscr{F} \colon x \in Y \cap \mathbb{Q}^n \right\}$, which is countable and covers $Y$ since $\mathbb{Q}^n \cap Y$ is dense in $Y$.
($\leftarrow$) Suppose that $Y \subset \mathbb{R}^n$ is the union of a countable collection $\mathscr{C}$ of open sets. By hypothesis, for every $x \in Y$, there exists some open set $U_x \in \mathscr{C}$ such that $x \in U_x$ and $U_x \subset Y$. Therefore by definition, $Y$ is open.