Closure of an open set

Question

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

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Context of the question as given in the OP's answer:

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

Answer 1

No : take $A={]{-1},0[}$ and $B={]{-\infty},{0}[}$.

Answer 2

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 $X$.