# Open superset of a closed set

Let $(X,d)$ be a compact metric space and $A\subset X$ is closed, $A\neq X$. Does it necessary exist an open $B$ such that

1. $A\subset B$;

2. $\overline{B}\neq X$?

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Does $\subset$ mean proper containment here? – Jonas Meyer Jun 24 '11 at 12:14
@Jonas I hope not, else take two points and $A$ one of them. – t.b. Jun 24 '11 at 12:15
@Theo: That's precisely why I asked. Sorry about duplicating your example without acknowledgment. I've been having connection problems that acted up right when I was trying to post. – Jonas Meyer Jun 24 '11 at 12:49

If you want the containment $A\subset B$ to be proper, then no. E.g., $X=\{0,1\}$, $A=\{0\}$.
If you don't need the containment to be proper, then yes. Let $x$ be an element of $X\setminus A$. Let $r>0$ be such that the ball of radius $r$ centered at $x$ is disjoint from $A$, and let $B$ be the complement of the closed ball of radius $\frac{r}{2}$ centered at $x$. (Notice that compactness isn't needed.)
As a compact metric space $(X,d)$ is clearly regular. If $A \subset X$ is closed and $A \not = X$, then for every point $x \in X \setminus A$ we can find disjoint open neighborhoods $A \subset U$ and $x \in V$. Since $U$ and $V$ are disjoint, we see, that $\overline{U}$ does not contain $x$ and therefore $\overline{U}\not = X$.