4
$\begingroup$

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?

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.