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I want to show that $G$ is open in $X$ iff $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for every $A \subset X$.

This direction $\rightarrow$ is easy.

I'm stuck in showing that $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for every $A \subset X$ implies $G$ is open in $X$.

So I guess the game is to find suitable A and apply the hypothesis, I tried taking A as the interior of G and also as the interior of the complement of G but didn't work (at least I didn't see it).

Any ideas?

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@Paulo As was already requested by Pete Clark in another thread, could you please accept one of the answers if you have solved the problem? Otherwise people will keep coming back to this question, thinking that no satisfactory answer was provided. – Alex B. Dec 23 '10 at 1:19

Here are some hints:

  • $G$ is open if and only if the complement $G^C$ is closed.
  • A set is closed if and only if it contains all its limit points.

Now, see whether the assumption that $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for all $A$ and that $G^C$ is not closed (i.e. that there exists a limit point of $G^C$ that is not in $G^C$) leads you to a contradiction.

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@Paulo: Thank you. – Paulo Dec 22 '10 at 6:24

Trying $A=X\setminus G$ should do the trick.

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