Difference of closures and closure of difference

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.

Solution 1:

If $x\notin\operatorname{cl}(A\setminus B)$, then $x$ has an open nbhd $U$ such that $U\cap(A\setminus B)=\varnothing$. And $$U\cap(A\setminus B)=(U\cap A)\setminus(U\cap B)\;,$$ so $U\cap A\subseteq U\cap B$. If $x\notin\operatorname{cl}B$, then $x$ has an open nbhd $V$ such that $V\cap B=\varnothing$. Let $W=U\cap V$. Then

$$W\cap A=W\cap(U\cap A)\subseteq W\cap(U\cap B)=W\cap B=\varnothing\;,$$

so $W\cap A=\varnothing$, and $x\notin\operatorname{cl}A$. Thus, if $x\in(\operatorname{cl}A)\setminus\operatorname{cl}B$, then $x\in\operatorname{cl}(A\setminus B)$.

Solution 2:

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 :)