I´m stuck with this problem:
Let be $A \subseteq X$.
Prove that $A$ is an open set in $X$ if and only if $C\operatorname{cl}(A \cap \operatorname{cl}(B))=\operatorname{cl}(A \cap B)$ for all $B \subseteq X$, where $\operatorname{cl}(A)$ means the topological closure.
I have no idea how to use that $A$ is an open set.
Help me please.
On
If $A$ is open, let $B \subseteq X$. Then $B \subseteq \operatorname{cl}(B)$, so $A \cap B \subseteq A \cap \operatorname{cl}(B)$ and thus, taking closures:
$$\operatorname{cl}(A \cap B) \subseteq \operatorname{cl}(A \cap \operatorname{cl}(B))$$
Note that we did not use anything on $A$ yet; this inclusion always holds.
Now, let $x \in \operatorname{cl}(A \cap \operatorname{cl}(B))$, we want to show that $X \in \operatorname{cl}(A \cap B)$, so let $O$ be any open neighbourhood of $x$. Then $O \cap (A \cap \operatorname{cl}(B)) \neq \emptyset$, so that there is some $x' \in O \cap A \cap \operatorname{cl}(B)$. Then $O \cap A$ is open (here we use $A$ is open!) and contains $x'$ and $x' \in \operatorname{cl}(B)$ so that $(O \cap A) \cap B \neq \emptyset$ (here and before I use that $p \in \operatorname{cl}(C)$ iff every open neighbourhood of $p$ intersects $C$). But the last non-empty intersection can also be written as $O \cap (A \cap B) \neq \emptyset$ and so $O$ intersects $A \cap B$, and as $O$ was an arbitary open neighbourhood of $x$ we have shown that $x \in \operatorname{cl}(A \cap B)$, showing $$\operatorname{cl}(A \cap \operatorname{cl}(B)) \subseteq \operatorname{cl}(A \cap B)$$
and we have equality of closures, when $A$ is open. This is one half of the statement.
Suppose we have that
$$\forall B\subseteq X: \operatorname{cl}(A \cap B) = \operatorname{cl}(A \cap \operatorname{cl}(B))$$
and we'll show that $A$ is open. Suppose $A$ were not open. Then there is a point $x \in A$ such that $x$ is not an interior point of $A$, which means that every open neighbourhood of $x$ intersects $X \setminus A$, so defining $B = X\setminus A$ we see that $x \in \operatorname{cl}(B)$.
So on the one hand, $x \in A \cap \operatorname{cl}(B)$ so $$\operatorname{cl}(A \cap \operatorname{cl}(B)) \neq \emptyset$$ while also
$$\emptyset = A \cap B \text{ so } \operatorname{cl}(A \cap B) = \emptyset$$
which is a contradiction with the assumed property for $A$: we have found a subset $B$ where the equality of closures fails. So the assumption that $A$ was not open was false, hence $A$ is open.
We could also use $B = X \setminus A$ immediately and conclude that
$$\operatorname{cl}(A \cap \operatorname{cl}(B)) = \operatorname{cl}(A \cap B) = \operatorname{cl}(\emptyset) = \emptyset$$ and so $$A \cap \operatorname{cl}(X \setminus A) = \emptyset$$ from which it follows that $$\operatorname{cl}(X \setminus A) \subseteq X \setminus A$$ so $X \setminus A$ is closed and $A$ is open. This avoids a proof by contradiction.
Clearly $cl(A \cap B) \subset cl(A\cap cl(B))$. Suppose $A$ is open and $x \in cl(A\cap cl(B))$. If $x \notin cl(A \cap B) $ then there is an open set $U$ conatining $x$ such that $U \cap A \cap B=\emptyset$. This gives $B \subset (U\cap A)^{c}$ Since the right side closed we get $cl(B) \subset (U\cap A)^{c}$ which means $U \cap A \cap cl(B)=\emptyset$. This contradicts the fact that $x \in cl(A\cap cl(B))$. Hence $cl(A \cap B) = cl(A\cap cl(B))$ if $A$ is open. for the converse take $B=A^{c}$ as suggested by Mirko.