# Are any (non-empty) Euclidean open sets dense in the Zariski topology?

It's well known and easy to show that every Zariski open set is dense in the Zariski topology. However I search the web and didn't find an answer to my question, which I believe is true.

My argument is not strict but only intuitive: given an Euclidean open set $U\subset \mathbb{A}^n$, a point $p\in \mathbb{A}^n \setminus U$, and any Zariski open set $\mathbb{A}^n\setminus V$ containing $p$. $V$ is a sub-variety, having no interior point( neither in the Euclidean topology nor the Zariski topology), so $V$ cannot cover an Euclidean open set like $U$, which implies $(\mathbb{A}^n\setminus V) \cap U\ne\emptyset$. Q.E.D.

It there any problem of the above argument? How can I prove that a sub-variety doesn't have interior points in both topology?

Let $V \subseteq \mathbb{A}^n$ be a subvariety which contains an interior point $p$ with respect to any of those topologies (Euclidean or Zariski). Let $L$ be a line through $p$. Then $V \cap L$ is a Zariski-closed subset of $L \cong \mathbb{A}^1$ which contains infinitely many points. But since $\mathbb{A}^1$ has the cofinite topology there is only one closed subset with infinitely many points, namely the whole space. So we conclude $V \cap L = L$, i.e. $L \subseteq V$.
Since this holds for all lines through $p$ we have $\mathbb{A}^n \subseteq V$.
• @XipanXiao: Your argument can also be shortened. If $\emptyset \neq U \subseteq \mathbb{A}^n$ is an Euclidean-open subset then $\overline{U}$ (Zariski-closure) is a subvariety with an interior point, so it must be the whole space. Hence $U$ is Zariski-dense. – Dune Sep 28 '14 at 23:30