Closure of an open set Given an open set $A$ as a proper subset of an open set $B$, is the closure of $A$ necessarily contained in $B$.  I think it is, but would like a proof either way
 A: No : take $A={]{-1},0[}$ and $B={]{-\infty},{0}[}$.
A: Thanks.  The reason I asked the question in the first place was that I was trying to prove what I call the Russian Doll theorem to my satisfaction.   Given X a normal topological space A, a closed subset of X, U the open set containing A which normality guarantees.  Prove that U contains another open set (call it W) whose closure is contained in U. 
I started with X - U as another closed set and was able to prove to my satis faction that normality guaranteed the existence of V another open set containing X - U disjoint from another open set W also containing A and such that W was a proper open subset of U.  I needed to prove that the closure of W was contained in U and couldn't do it
A: I will prove that the only topology that satisfies the given property is the discrete topology.
Assume that $(X,T)$ is a topological space such that if $A$ and $B$ are open sets with $A$ a proper subset of $B$, then $\bar{A}\subseteq B$. Let $E$ be any subset of $X$, and consider the boundary of $E$, 
$$\partial E=\bar{E}\setminus E^{\mathrm{o}}$$ 
If $x\in E^{\mathrm{o}}$, then by assumption $x\in\bar{E}$, so $\partial E=\emptyset$. Note that $E$ is closed if and only if $\partial E \subseteq E$. Therefore $E$ is closed. Since $E$ was an arbitrary set, it follows that $T$ is the discrete topology on $X$.   
