Can the interior of the intersection of a closed and an open set be empty if the intersection is non-empty?

Question

Both the sets are convex polytopes in $\mathbb{R}^n$, one open, the other closed. My question is (as the title says) - can the interior of their intersection be empty if the intersection is non-empty? Basically, can their intersection have measure zero if it is non-empty? I checked this out. But in the example given in the answer, while the intersection is not open, it contains an open set (i.e. has non-empty interior).

Answer 1

Let $X$ and $Y$ be convex $n$-dimensional polytopes in $\newcommand{\RR}{\mathbb{R}}\RR^n$. If $X$ is closed and $Y$ is open, then either the intersection is empty or has nonempty interior.

Proof: Assume $X$ and $Y$ intersect. Let $x\in X\cap Y$. if $x\in X^\circ$, then $X^\circ \cap Y$ is a nonempty open subset of $X\cap Y$, so let's assume $x\not\in X^\circ$, i.e., $x\in\partial X$. Now because $X$ is a polytope of dimension $n$, $\partial X = \partial (X^\circ)$, so if $B$ is a ball around $x$ contained in $Y$, $B\cap X^\circ \ne \varnothing$. But then there exists $x'\in B\cap X^\circ \subset Y\cap X^\circ$, so once again, $X^\circ \cap Y$ is a nonempty open subset of $X\cap Y$.

As for why $\partial X = \partial(X^\circ)$, well all I really need is that $X$ is convex and has nonempty interior. Let $c\in X^\circ$. Then for any $x\in \partial X$. Assume $x=0$ by a change of coordinates. Then the points of the segment $[c,x]$ are $\lambda c$ for $0\le \lambda\le 1$. Additionally, if $U$ is an open subset of $X$ containing $c$, then $\lambda U$ is an open subset of $X$ (by convexity again) containing $\lambda c$ for $0 < \lambda \le 1$, so $[c,x)\subseteq X^\circ$. Hence $x\in \overline{X^\circ}$. Since $x\in \overline{X^C}\subset \overline{X^{\circ C}}$, $x\in\partial (X^\circ)$. Thus $\partial X \subseteq \partial(X^\circ)$. However, rewriting $$\partial X = \overline{X} \cap \overline{X^C}=X^{\circ C}\cap X^{C\circ C}, $$ and $$\partial (X^\circ)=X^{\circ C}\cap X^{\circ C\circ C}, $$ so since $X^{\circ C\circ C} \subseteq X^{C \circ C}$, we get $\partial(X^\circ)\subseteq \partial X$ in general. Hence since in our case both containments hold, the desired equality holds, and $\partial X = \partial(X^\circ)$.

Summary: We've proved the stronger result that if $X$ is convex with nonempty interior, and $Y$ is open, then either $X\cap Y=\varnothing$, or $X\cap Y$ has nonempty interior.