Question. Let $V_1,\ldots,V_n$ be open, bounded and convex subsets of $\mathbb R^2$. Show that $F=\mathbb R^2\smallsetminus\bigcup_{i=1}^n V_i$ possesses only finitely many connected components.
I have managed to produced a rather long proof in the case when the $V_i$'s are rectangles, with sides parallel to the axes. As this looks intuitively almost obvious, I am wondering whether there is some straight-forward way to show this.
This follows from Alexander duality, and works with $\mathbb{R}^2$ replaced by $\mathbb{R}^d$ for any $d$. In more detail, let us compactify $\mathbb{R}^d$ to $S^d$ and consider $F'=S^d\setminus\bigcup V_i$ (this does not change the number of components*, since each $V_i$ is bounded). It is easy to see that $F'$ is locally contractible (in fact, it is locally star-convex). So by Alexander duality, there is an isomorphism $\tilde{H}_0(F')\cong H^{d-1}(\bigcup V_i)$. But by an easy induction on $n$ using Mayer-Vietoris sequences (or more simply, by using the Cech spectral sequence), the cohomology of $\bigcup V_i$ is finitely generated. So $\tilde{H}_0(F')$ is finitely generated, so $F'$ has only finitely many components.
*unless $d\leq 1$, in which case it changes the number of connected components by at most $1$.
