Prove there exists an open set containing a closed set disjoint with another closed set Let $F$ and $G$ be closed sets in a metric space $X$ and $F \cap G = \emptyset$. Show that there exists an open set $U$ such that $F \subseteq U$ and $\bar{U} \cap G = \emptyset$.
I tried proving this in general terms, using sequences, by contradiction... yet still failed. Could somebody please give me a hint in which direction I should go in this proof?
 A: It's tempting to do something like saying "find the distance $s$ between $F$ and $G$," because you're in a metric space, and then let $U$ be all points whose distance to $F$ is less than $s/2$. 
That doesn't work, because the distance between two closed sets can be zero: think of $F$ being the $x$-axis and $G$ being the graph of $y  = 1/x$ in the plane, for instance. 
By at a point like $x = 3$, the idea sort of works: the distance from $(3, 0)$ to the graph $G$ is strictly positive because the point $(3, 0)$ is a compact set, and the distance from a compact set to a closed set from which it is disjoint is always positive. 
This means that around the point $a = (3, 0)$, you can find an open ball $B_a$ that misses $G$. You could even, by using as the ball-radius half the distance to $G$, ensure that the closure of your ball was disjoint from $G$. 
Well, that's all very well for a single point, but what do you do if the set $X$ is larger, i.e., contains many points? Do this for each of them, and take a union. Because a union of open balls is an open set. :) 
