Is the closure of $ X \cap Y$ equal to $\bar{X} \cap \bar{Y}$? 
$U$ is a topological space. $X$ is an open subset of $U$, and $Y$ is a closed subset. Let $Z = X \cap Y$. Does $\bar{Z} = \bar{X} \cap \bar{Y}$.

Here, $\bar{X}$ denote the closure of $X$, and $\bar{Y}$, $\bar{Z}$ respectively. (So, $\bar{Y}=Y$.)
It is clear that $\bar{Z} \subseteq \bar{X} \cap \bar{Y}$, but is it true in the reversed direction?
 A: A word on intuition. At least for metric spaces (and more generally for first-countable spaces), you can think of $\bar{X}$ as the collection of all points which are limits of sequences of points in $X$ (and in general you can replace "sequence" with "net" or "filter"). Then $\overline{X \cap Y}$ is obviously contained in both $\overline X$ and $\overline Y$ (as sequences of points in $X \cap Y$ are sequences of points in $X$ and also sequences of points in $Y$), but on the other hand a point in $\bar{X} \cap \bar{Y}$ is


*

*a limit of a sequence of points in $X$ and

*a limit of a different sequence of points in $Y$


and there's no obvious way to use either of these sequences to cook up a sequence of points in $X \cap Y$; the two sequences above may be disjoint, and in fact $X \cap Y$ may be empty while $\bar{X} \cap \bar{Y}$ is non-empty. With that in mind it isn't hard to find a counterexample. 
A: Let $X$ be open and $Y = U \smallsetminus X$. Then $X \cap Y = \emptyset$. However, $\overline{X} \cap Y = \partial X$ won't be empty in general. Take $X$ to be an open ball in $\mathbb{R}^n$, for example.
