TITLE: $X$ topological space. $A$ open $A \cap Y = \emptyset \ \ \Longrightarrow A \cap \bar{Y} = \emptyset$?
QUESTION [5 upvotes]: I know this is an easy question, but I cannot demonstrate it properly.
Suppose by contradiction that $A \cap \bar{Y} \neq \emptyset$. Then $\exists \ x \in A \cap \bar{Y}$. 
I need help formalizing this reasoning  (or correcting if it is wrong)
By the definition of $A$ open, $\exists \ U \subset A$ open neighborhood of $x$. Then my idea is to prove that $U \cap Y \neq \emptyset$ because $U$ in somewhat sense "touches" $Y$, because it's a neighborhood of $x$. But i can't write down a formal demonstration.
Any help is appreciated.
Thanks^^

REPLY [3 votes]: Another type of proof of it:

If $A \cap \overline {Y} \neq \emptyset$, then there exists $x \in A \cap \overline {Y}$. Since $x \in \overline {Y}$, which implies that for any open set $U$ containing $x$, $U \cap Y \neq \emptyset$, and hence $A \cap Y \neq \emptyset$. A contradiction!