Zariski dense contains Zariski open? Let $U\subseteq\mathbb A_K^n$ be a Zariski dense subset (here, $K$ is an algebraically closed field). I was told that $U$ then contains a Zariski open and dense subset. However, I can't seem to come up with an easy proof. 
Edit: The above is wrong, and not equivalent to what we really used in the course: In fact, we need to prove that there exists a non-constant polynomial $f\in\Bbbk[X_1,\ldots,X_n]\setminus\{0\}$ with $D(f)\cap U$ dense.
 A: No, an arbitrary Zariski dense subset need not contain any non-empty Zariski open subset.  Take $S = \mathbb{Z} = \{ \dots, -2, -1, 0, 1, 2, \dots \} \subset \mathbb{A}^1_{\mathbb{C}}$.  Since the non-trivial Zariski closed subsets of $\mathbb{A}^1_{\mathbb{C}}$ are the finite sets, it is easy to check that the Zariski closure of $S$ is $\mathbb{A}^1_{\mathbb{C}}$ and that the only open set $U \subset S$ is $U = \emptyset$.
Edit.  You probably want to prove the following stronger statement:  If $U$ is Zariski dense in $\mathbb{A}^n_K$ and $f$ is any non-constant polynomial, then $D(f) \cap U$ is Zariski dense in $\mathbb{A}^n_K$.  Indeed, suppose for contradiction that it were not.  Then $D(f) \cap U \subseteq Z(g)$ for some polynomial $g$ (where $Z(g)$ is the zero locus of $g$).  But then $$U = (D(f) \cap U) \cup (U \setminus D(f)) \subseteq Z(g) \cup Z(f) = Z(fg),$$
contradicting the assumed density of $U$.
A: If $f$ is any non-constant polynomial, then its zero locus is non-empty, and so $D(f)$ is a proper subset of $\mathbb A^n_K$.  So if $U$ is equal to all of $\mathbb A^n_K$ then it is not contained in any such $D(f)$.
If $n > 1$, and if we take $U = \mathbb A^n_K \setminus \text{ the origin }$ (or more generally, the complement of any closed subset of codimension at least two),
then again $U$ is Zariski dense in $\mathbb A^n_K$, but is not contained in $D(f)$ for any non-constant $f$.  
So the statement you ask about is not true.  (My guess is that you've misunderstood something.)
