TITLE: Difference of closures and closure of difference
QUESTION [10 upvotes]: Let $A,B$ be subsets of a topological space $X$. Is it true that $\overline{A}-\overline{B}\subseteq\overline{A-B}$?

Suppose $x\in\overline{A}-\overline{B}$. So all open sets containing $X$ also contains an element of $A$. And there exists an open set $U$ containing $X$ that contains no element of $B$. So $U$ contains an element of $A-B$. But this is not enough to conclude that $x\in\overline{A-B}$. So I'm thinking the answer might be negative, but cannot find a counterexample.

REPLY [2 votes]: It is true actually, here is a proof: let $x \in \overline{A} - \overline{B}$ and assume for the sake of contradiction that $x \notin \overline{A-B}$. This last condition is equivalent to "there exists an open set $U$ such that $x \in U$ and $U \cap (A - B) = \emptyset $". Since $x \in \overline{A} - \overline{B} \subset \overline{A}$ then $U \cap A \neq \emptyset$. This implies that $U \subset B$ but then we have a contradiction since $x \in U \subset B$ and $x \in \overline{A} - \overline{B}$. 
What is false is the other inclusion: take $A = (0,2)$ and $B = (0,1)$. Then $\overline{A - B} = [1, 2]$ which is not a subset of $\overline{A} - \overline{B} = (1,2]$.
I hope it helps :)