I'm posting my answer to [this question](https://mathoverflow.net/questions/5146/algebraic-geometry-versus-complex-geometry) as its own question:


Let $V$ be an irreducible projective variety over $\mathbb{C}$. Let $U$ be a Zariski open set in $V$. I'll use $V(\mathbb{C})$ and $U(\mathbb{C})$ to mean $V$ and $U$ equipped with their Euclidean topologies, respectively.


What is the easiest proof that $U(\mathbb{C})$ is connected?


Here's the proof I know: Suppose that $U(\mathbb{C})$ can be written as a disjoint union of two open sets $A$ and $B$. Since the complement of $U$ in $V$ is a variety of smaller dimension than $V$, a theorem of Remmert and Stein implies that the closures $\overline{A}$ and $\overline{B}$ of $A$ and $B$ in $V(\mathbb{C})$ are projective analytic sets. By Chow's theorem that projective analytic sets are algebraic, $\overline{A}$ and $\overline{B}$ are subvarieties of $V$. Since they're proper, $V$ is not irreducible, and we have a contradiction.


I guess I'm really asking for the most elementary argument, as I think the above argument is nice intuitively. A reference would be fine.


(To avoid going through the same discussion in the comments that happened at the other question, let me point out that I am aware that irreducible varieties are connected and that $U$ is itself a variety in the sense that it is locally affine. It is just not obvious to me that it is irreducible (without appealing to the above argument).)



Why can't subvarieties separate?
[This has been completely rewritten at the request of Autumn Kent.]


Let $X$ be an irreducible topological space and $U$ a non-empty open subset of $X$. Then $U$ is also irreducible -- see e.g. Proposition 141 on page 88 [here](http://alpha.math.uga.edu/%7Epete/integral.pdf). (Surely it's also in Hartshorne and lots of other places, but one of the advantages of typing up your own notes is to be able to easily point to a reference because you know exactly where it is.)


Thus the question reduces to the fact that if $X\_{/\mathbb{C}}$ is an irreducible complex variety, then $X(\mathbb{C})$ with its "Euclidean topology" is connected. For this, see e.g. Section VII.2 of Shafarevich's *Basic Algebraic Geometry II*. (Again, there are other places, but I think his discussion is especially good.) He gives two different proofs, one of which is a simple induction on the dimension.


