Harnack's curve theorem for curves in $\textit{complex}$ projective plane? The wikipedia page gives the statement for algebraic curves in real projective plane. 
Is the statement also true in $\textit{complex}$ projective plane? If not, is there a similar statement about number of connected components of an algebraic curve in $\mathbb{P}^2$? Is it true that number of connected components are finitely many? 
 A: Let $k$ be an algebraically closed field. By Bézout's theorem, if $C$ and $D$ are two projective curves in the projective plane $\mathbb{P}_k^2$ such that $C$ and $D$ do not have a common irreducible component, then $C$ and $D$ intersect at exactly $\deg(C) \cdot \deg(D)$ points, counting multiplicity. In particular, any two projective plane curves have nonempty intersection over an algebraically closed field. This implies that all projective plane curves over an algebraically closed field are connected (in the Zariski topology, but as Hoot's comment notes, this implies connectedness in the complex topology when $k = \mathbb{C}$).
More generally, dimension plays nicely with (set-theoretic) intersection over algebraically closed fields: Suppose $V$ and $W$ are irreducible closed subvarieties of $\mathbb{P}_k^n$. Then every irreducible component of $V \cap W$ has dimension at least $\dim(V) + \dim(W) - n$, and if $\dim(V) + \dim(W) - n \geq 0$, then $V \cap W$ is nonempty. Hence, any closed subvariety of $\mathbb{P}_k^n$ with all irreducible components of dimension at least $\frac{n}{2}$ is connected.
