TITLE: Zariski dense contains Zariski open?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.)