# 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?

• I think $B\setminus A$ needn't be closed. – Aweygan Aug 28 '16 at 19:47
• $B\setminus A$ need not be closed - consider $(0, 1)$ and $(0, 2)$ in the usual topology on $\mathbb{R}$. $[1, 2)$ is neither open nor closed. – Noah Schweber Aug 28 '16 at 19:49

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 $U\cap B\neq \emptyset$ since $U$ is open. Now $U\cap B\subseteq B\setminus A$ is open, hence $B\setminus A$ has non-empty interior.