Dense subset in product of varieties Given two (algebraic) varieties $X,Y$ (not necessarily irreducible) and $D,E$ dense subsets of $X$ and $Y$ respectively, I've read something saying that the product $D \times E$ is dense in $X \times Y$ (given the Zariski topology of course).
Is it true ? And how do you prove it ?
I've tried to look for it in different books but I couldn't find anything.
Thank you ! 
 A: Suppose $X,Y$ are varieties over an algebraically closed field $k$.
Claim
If $D\subset X$ and  $ E\subset Y$ are dense subsets, then $D\times E\subset X\times Y$ is dense too.
Proof
 Let $Z$ be the closure of $D\times E$ in $X\times Y$.
 Fix a poind $d_0\in D$ and consider the closed subvariety $\lbrace d_0\rbrace \times Y \subset X\times Y$.
The  subset $\lbrace d_0\rbrace \times E \subset \lbrace d_0\rbrace \times Y$ has closure $\lbrace d_0\rbrace \times Y \;$  [ because  $\lbrace d_0\rbrace \times Y$ is isomorphic to $Y$ and $E$ is dense in $Y$]
 so that $\lbrace d_0\rbrace \times Y \subset Z $ .    We have proved the  
Partial Result:
For all $d_0\in D $ we have  $\lbrace d_0\rbrace \times Y \subset Z$
End of proof:
Consider an arbitrary $y\in Y$ .
 For any $d\in D$ we know, thanks to the Partial Result,  that  $(d,y)\in Z$.      Hence $D\times \lbrace y \rbrace \subset Z$.
Since the closure of $D\times \lbrace y \rbrace $ is $X\times \lbrace y \rbrace $ [ because  $X \times \lbrace y \rbrace  $ is isomorphic to $X$ and $D$ is dense in $X$], we have proved   that  $X\times \lbrace y \rbrace \subset Z$.
Since $y\in Y$ was arbitrary,  this implies that $Z=X\times Y $ i.e. that $D\times E$ is dense in $X\times Y $ 
