Zariski dense implies classically dense?

I was surprised that I wasn't able to find this question already posted; if it has been posted and I just didn't find the right search terms, let me know.

Let $X$ be any complex variety. A priori, any set which is dense in the classical topology on $X$ is automatically dense in the Zariski topology on $X$, just because the Zariski topology has fewer open/closed sets.

For the converse, this question, though distinct, does shed some light: in $\mathbb A_{\mathbb C}^1$, any infinite subset is Zariski dense, but certainly not necessarily classically dense.

But in my experience, it seems that any Zariski open subset of $X$ that is Zariski dense is also classically dense. Is this true? How do you prove it?

Solution 1:

Let $T$ be a locally constructible subset of a finite type $\mathbb C$-scheme $X$. (You can take $T$ to be a (Zariski) open of a complex algebraic variety, for instance.)

Then $T$ is dense in $X$ if and only if $T(\mathbb C)$ is dense in $X(\mathbb C)$. A reference for this is Expose XII, Cor. 2.3 p. 243 of SGA 1.

http://arxiv.org/pdf/math/0206203v2.pdf

Solution 2:

Let's start with the affine case:

Claim. If $A\subsetneq\mathbb A^n_{\mathbb C}$ is algebraic, then its complement is dense.

Proof. Let $A=\{\,x\in\mathbb A^n\mid \forall f\in S\colon f(x)=0\,\}$ where the $S\subseteq \mathbb C[X_1,\ldots, X_n]$. For $a\in A$ we have to exhibit points close to $a$ that are $\notin A$. Pick $b\in\mathbb A^n\setminus A$ and $f\in S$ with $f(b)\ne 0$. Then the polynomial $g(T)=f(a+(b-a)T)\in\mathbb C[T]$ is not the zero polynomial. Hence its root at $0$ is isolated and so $g(h)\ne 0$ for all sufficiently small nonzero $h$. Then $a+h(b-a)\notin A$ for such $h$, showing the claim. $_\square$

The extension to the projective case and then to the locally quasiprojective case should be clear