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 $A \subset X$.

  • $\begingroup$ Sorry, meant to say for every subset $A$ of $X$. $\endgroup$
    – syxiao
    May 31, 2011 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. $\endgroup$ May 31, 2011 at 21:38
  • $\begingroup$ @girdav - yes that was the intended question; sorry if the original phrasing was unclear. $\endgroup$
    – syxiao
    May 31, 2011 at 21:52

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.