The complement of an algebraic set in $\mathbb{P}^n$ is path-connected

Question

I'm studying the marvellous book "Algebraic Curves and Riemann Surfaces" of Rick Miranda and I've found this problem (III.5.Q pag.103). Actually it's not related to Riemann surfaces, but I think it's interesting anyway.

A subset $Z\subseteq \mathbb{P}^n$ is an algebraic set if there is a set of homogeneous polynomials $\{F_{\alpha} \}$ such that $Z=\{p\in \mathbb{P}^n | F_{\alpha}(p) = 0$ for every $\alpha\}$. Here $\mathbb{P}^n$ is the complex projective n-space, i.e. the set of 1-dimensional subspaces of $\mathbb{C}^{n+1}$.
Show that the complement of an algebraic set in $\mathbb{P}^n$ is path-connected.

Remarks.

1. Since an algebraic set is an intersection of hypersurfaces, if the proposition above holds for hypersurfaces, a fortiori holds for generic algebraic sets. So, it suffices to prove that the complement of the set of zeroes in $\mathbb{P}^n$ of a homogeneous polynomial $F$ is path-connected.
   Proof. (asked by ronno)
   Assume that the proposition holds for hypersurfaces. Let $Z_1$ and $Z_2$ be two hypersurfaces defined by the zeroes sets in $\mathbb{P^n}$ respectively of $F_1$ and $F_2$ homogeneous polynomials. Since $F_1F_2$ is a homogeneous polynomial, its zeroes in $\mathbb{P^n}$ define a hypersurface $Z_3 = Z_1 \cup Z_2$, so the finite union of hypersurfaces is a hypersurface. Since $\mathbb{P^n} -Z_3 = (\mathbb{P^n}-Z_1) \cap (\mathbb{P^n}-Z_2)$, the intersection of the complements of two hypersurfaces is path-connected. Therefore $(\mathbb{P^n}-Z_1) \cup (\mathbb{P^n}-Z_2)$ is path-connected. Hence the complement of an algebraic set, being union of sets two-by-two path-connected, is path-connected.
2. The problem is equivalent to say that the open sets in the Zariski topology on $\mathbb{P}^n$ are path-connected.

Thanks for any hint or answer.

Answer 1

Let $x,y\in\Bbb P^n$ be two distinct points. Let $L\cong\Bbb P^1$ be the unique line containing these points. If we can find a path in $L$ which joins $x$ and $y$ while missing $L\cap Z$, then we win - the image of this path in $\Bbb P^n$ will be a path between $x$ and $y$ missing $Z$. But $L\cap Z$ is a proper closed subset of $L$ (since it does not contain $x$ nor $y$) so $L\cap Z\subset L$ is a finite set of points. Since $\Bbb P^1$ in the Euclidean topology is homeomorphic to $S^2$ and $S^2$ minus a finite set of points is path connected, we win.