Intersection of closed and open set

Question

In arbitrary topological space, let $A$ be open and $B$ be closed. If $\text{int}(B) \neq \emptyset$ and $A \cap B \neq \emptyset$, is it guaranteed that $A \cap \text{int}(B) \neq \emptyset$? If not, is there any reasonably weak additional condition that can assure it?

Answer 1

There are already two answers showing the answer is no, here is one more example.
In the plane let $B$ be the closed unit disk together with the positive x-axis. Let $A$ be the half-plane $x>2$. Their intersection (is non-empty but) has empty interior.

Re "reasonably weak additional condition", you may impose the condition that $B$ is regularly closed, that is $B=\text{cl}(\text{int}(B))$.
Pick any $p\in A \cap B$ and assume that $A \cap \text{int}(B) = \emptyset$.
But then, since $A$ is open, we have that $A \cap \text{cl}(\text{int}(B))=\emptyset$, a contradiction.

Answer 2

No. It doesn't even hold in $\mathbb{R}$, even under a stronger cardinality assumption of $A\cap B$.

For example, $A=(-1,2)$, $B=K\cup [2,3]$, where $K$ is the Cantor set. Then $\operatorname{int}B=(2,3)$, $A\cap B=K$ has cardinality $c$, but $A\cap\operatorname{int}B=\varnothing$.

Answer 3

The described property (let's call it the Zapala property), will not hold in many "typical" spaces. The following proposition shows that Zapala is far from what one might call a weak property:

Proposition. If $X$ is Hausdorff and Zapala, then $X$ is discrete.

Proof. Assume $X$ is a counterexample to the claim. As $X$ is not discrete, let $a\in X$ such that $\{a\}$ is not open. Let $b\in X\setminus\{a\}$ (this is possible because the one-point space is discrete). By the Hausdorff property, there exist open sets $U,V$ with $a\in U$, $b\in V$, $U\cap V=\emptyset$. Now let $A=U$ and $B=(X\setminus U)\cap\{a\}$. Then $A$ is open, $B$ is closed with $b\in V\subseteq\operatorname{int}(B)$, and $A\cap B=\{a\}\ne\emptyset$. However, $a\notin\operatorname{int}(B)$ as that would make $\{a\}=U\cap\operatorname{int}(B)$ open. $\square$