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Let $A$ and $B$ be subsets of $\mathbb{R}^n$. If $A$ is open, and $B$ is arbitrary, does one always have the following inclusion $A \cap \operatorname{cl}(B) \subseteq \operatorname{cl}(A\cap B)$.

Is there a counter-example? I can't seem to find a way to prove it.

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up vote 2 down vote accepted

The statement is true. The proof just uses the relevant definitions.

Suppose $x \in A \cap \operatorname{cl} (B)$, i.e., $x \in A$ and $x \in \operatorname{cl}(B)$. The latter condition means that if $U$ is an open set containing $x$, then $U$ intersects $B$ (nontrivially).

Now, pick any open set $V$ containing $x$. Then $V \cap A$ is an open set containing $x$. Then by the preceding paragraph, $(V \cap A) \cap B \neq \emptyset$, which is equivalent to $V \cap (A \cap B) \neq \emptyset$. This means that $x \in \operatorname{cl}(A \cap B)$.

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I tried to use that "the interior of an intersection is the intersection of the interiors", but that lead nowhere (for me). Thanks for the nice explaination! – Maxim Oct 24 '11 at 22:00
    
@Maxim: This is because there is no guarantee that $B$ even has an interior to begin with. – Asaf Karagila Oct 24 '11 at 22:31

the closure is closed, and the intersection of a closed set and an open set is not always a closed set

a conter exemple may be: A=]0,2[ B=]1,3[

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Um, so does that actually answer the question? – Henning Makholm Oct 24 '11 at 21:51
    
$A=(0,2)$, $B=(1,3)$ is not a counterexample. We get $A\cap\mathrm{cl}(B)=[1,2)$ and $\mathrm{cl}(A\cap B)=[1,2]$. – Henning Makholm Oct 24 '11 at 21:57
    
i just noticed that its $A \cap \operatorname{cl}(B) \subseteq \operatorname{cl}(A\cap B)$ and not $A \cap \operatorname{cl}(B) = \operatorname{cl}(A\cap B)$ – Hassan Oct 24 '11 at 22:00

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