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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?

2 Answers 2

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