TITLE: Union of a countable collection of open balls
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: In every topological space, every union of open sets is open. This is one of the axioms of topology.
However countable unions are easier to handle, compared to to uncountable unions, and they still have more expressive power than finite unions (e.g. the unit "square" $D=\{(x_1,\ldots,x_n)\mid-1<x_i<1,i=1,\ldots n\}$ is not a finite union of balls, and vice versa).
So we are particularly interested in spaces that have the property that some nicely described collection is sufficient to generate all the topology very "quickly", i.e. every open set is expressible as a countable union of these sets. In our case open balls, and we can reduce this even more: open balls around points in $\mathbb Q^n$ whose radii is rational.