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Here is a question from a past comprehensive exam:

Let $X$ be an arbitrary topological space. Prove that $G$ is open if and only if the closure of $G \cap \overline{A}$ and the closure of $G \cap A$ are equal for all $A \subset X$.

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  • $\begingroup$ Sorry, meant to say for every subset $A$ of $X$. $\endgroup$ – syxiao May 31 '11 at 21:29
  • $\begingroup$ The way I'm reading this, the condition is that $\mathrm{cl}(G \cap \mathrm{cl}(A)) = G \cap A$ for all $A \subset X$. In particular, this would imply that $G \cap G = G$ is closed, which is clearly false for most topological spaces. So I think the question could use clarification. $\endgroup$ – Charles Staats May 31 '11 at 21:38
  • $\begingroup$ @girdav - yes that was the intended question; sorry if the original phrasing was unclear. $\endgroup$ – syxiao May 31 '11 at 21:52
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If $cl (G\cap cl(A))=cl(G\cap A)$ for any subset $A\subset X$, pick $A=X\setminus G$. It would follow that $cl (G \cap cl(X\setminus G))=\emptyset$, which means that $cl(X\setminus G)\subset X\setminus G$ and $X\setminus G$ is closed, which means $G$ is open.

For the converse, pick $G$ open. One inclusion is obvious $cl(G\cap A)\subset cl(G\cap cl(A))$.

Pick $x \in G\cap cl(A)$. Then $x \in G$ and $x \in cl(A)$. Pick $V$ an open neighborhood of $x$. Then $V_1=G \cap V$ is still an open neighbourhood of $x$. Then $x \in V_1$, and $V_1 \cap A \neq \emptyset$. Therefore $V\cap (G\cap A)=V_1\cap A\neq \emptyset$, i.e. $x \in cl(G\cap A)$. We proved that $ G \cap cl(A) \subset cl(G\cap A)$. Take the closure and get the other inclusion.

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