Given any two sets $X, Y \subset \mathbb{A}^n$, do we have $\overline{X} \times \overline {Y} = \overline {X \times Y}$? 
Given any two sets $X, Y \subset \mathbb{A}^n$, do we have $\overline{X} \times \overline {Y} = \overline {X \times Y}$?

We use Zarski topology on $\mathbb{A}^n$ and $\mathbb A^{2n}$. Since product topology on $\mathbb A^{2n}$ is contained in the Zarski topology, we know $\overline{X} \times \overline {Y} \supset \overline {X \times Y}$ because $\overline{X} \times \overline {Y}$ is a closed in product topology thus it is a closed set in Zarski topology containing $X\times Y$. 
I am having trouble proving the other side, and it is also hard to find counter examples, because this equality holds for closed (algebraic) sets or quasi varieties $X,Y$, and every subset of $\mathbb{A}^1$.
 A: Fix any $y\in Y$, we first show that $\overline{X}\times \{y\}\subset \overline{X\times\{y\}}$, which is the same as proving that any polynomial $f$ that vanishes on $\overline{X\times\{y\}}$ also vanish on ${\overline{X}\times\{y\}}$.
Notice first that $I(A)=I(\overline{A})$ for any set $A$ in some affine space, so it suffices to show that if $f$ vanishes on ${X\times\{y\}}$ then it also vanishes on ${\overline{X}\times\{y\}}$. To see this, we think of $f$ as an element in $k[x_1,x_2,\cdots,x_n,y_1\cdots,y_m]$. Therefore, $f(x_1,\cdots,x_n,y)\in k[x_1,\cdots,x_n]$ which vanishes on $X$. By the observation at the beginning of this paragraph again, we know $f(x_1,x_2,\cdots,x_n,y)$ vanishes on $\overline{X}$. This proves that $f$ vanishes on ${\overline{X}\times \{y\}}$.
Since $y$ is arbitrary, this means that ${\overline{X}\times Y}\subset\overline {X\times Y}$.
Repeat this argument, one can show that ${\overline{X}\times \overline{Y}}\subset \overline{{\overline{X}\times Y}}\subset \overline{X\times Y}$.
