Chow's theorem states that an analytic subspace X of complex projective space $\mathbb{P}_r(\mathbb{C})$ that is closed in the topology defined by an analytic subspace is an algebraic subvariety.
My understanding of an algebraic subvariety is that defined in Serre's FAC page 4 section 34: http://achinger.impan.pl/fac/fac.pdf.
Part of the definition requires that there be a finite open cover of X = {$U_i$} such that each $U_i$ is biregularly isomorphic with $V_i \subset \mathbb{C}^n$ and $V_i$ being an intersection of a Zariski closed and Zariski open set.
If X is proven Zariski-closed in $\mathbb{P}_r(\mathbb{C})$ why is it an algebraic variety? Maybe this is obvious and we can let the finite cover be {X}, but I still don't understand why it should be isomorphic to a subset of $\mathbb{C}^n$.
Please clear up the confusion, thank you.