# In a metric space, is every open set the countable union of closed sets?

In a metric space, it is true that every open set can be represented as the union of closed sets (e.g. the union of closed balls).

I am wondering if every open set can be represented as the countable union of closed sets?

Say $U$ is an open set. If $U$ is empty or if it is the whole space $X$ then $U$ is also closed, so we are done. Now assume that $U$ and its complement $U^c=X\setminus U$ are both non-empty.
Let $A_n = \bigcup_{x\in U^c} B(x,\dfrac1n)$. Notice that $A_n$ is open and its complement $A_n^c$ (which is closed) is contained in $U$. To finish off the problem show that $U$ is the union of the $A_n^c$.
A set which is the union of countably many closed sets is called an $F_\sigma$ set. It is known that every open set in a metric space is an $F_\sigma$ set.