Difference of two regularly open sets, if non-empty, has non-empty interior Let $T$ be a metric space. A subset of $T$ is regularly open it is equal to the interior of its closure. Given a proper inclusion $A\subset B$ of two regularly open sets in $X$, must the difference $B\setminus A$ have non-empty interior?

The answer is yes. This holds in fact for any topological space (not just metric spaces).

First, note that we must have $\bar{A}\subsetneq \bar{B}$, for if $\bar{A}= \bar{B}$, then $A=\mathrm{int}(\bar{A})=\mathrm{int}(\bar{B})=B$, which contradicts $A\subsetneq B$.

Now $U:=\bar{A}^c$ (complement of $\bar{A}$) is a non-empty open set, which has non-empty intersection with $\bar{B}$. It follows that also $U\cap B\
eq \emptyset$ since $U$ is open. Now $U\cap B\subseteq B\setminus A$ is open, hence $B\setminus A$ has non-empty interior.