If $A \cap B =\emptyset $ then $A^- \cap B^\circ =\emptyset$?

Question

Let $X$ be a topological space and $A,B\subset X$ If $A \cap B =\emptyset $ then $A^- \cap B^\circ =\emptyset$ ? We have that $A \cap B^\circ \subset A\cap B = \emptyset$ and $A \cap B^\circ \subset A^- \cap B^\circ $ ...? How can I finish it. Please

Answer 1

The other way is to use the definition of closure and interior:

Suppose $A \cap B = \varnothing$. Could $A^- \cap B^o$ be nonempty? Suppose $x \in A^- \cap B^o$. Then in particular $x \in B^o$. Therefore, there is a neighborhood $U$ of $x$ with $U \subseteq B$. But since $A \cap B = \varnothing$, we have $U \cap A = \varnothing$. So: $x$ has a neighborhood disjoint from $A$, which means $x \notin A^-$. This contradicts $x \in A^- \cap B^o$.

Answer 2

Use that $A^-$ is the smallest closed set that contains $A$; also $X\setminus B^\circ$ is some closed set that contains $A$. Hence $A^-\subseteq X\setminus B^\circ$.