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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.

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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 $x\in\operatorname{cl}(A\setminus B)$.

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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 :)

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  • $\begingroup$ You seem to have an error, you say that $U\cap A\not=\varnothing$ and $U\cap(A-B)=\varnothing$ implies that $U\subset B$, however I seem to be able to come up with a counter example. Let $A = (0,2)$, $B = (1,3)$, $x = 1.5$. Now we must choose an open set $U$ that contains $x$ such that it does not intersect $A-B$, let us choose the set $(1.2,1.7)\cup(4,5)$. Now we have an open set that contains $x$, does not intersect $A-B$, does intersect $A$, and is not a subset of $B$. $\endgroup$ – Benji Altman Jun 18 '17 at 15:19
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I think I got it.

All open sets containing $X$ also contains an element of $A$. There exists an open set $U$ containing $X$ that contains no element of $B$.

Now take any open set $G$ containing $X$. Then $G\cap U$ is also an open set containing $X$. It contains no element from $B$. It must contain an element from $A$. So it contains an element from $A-B$.

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