Convex hull in projective space Let $S \subseteq \mathbb R\mathbb P^n$ be a closed connected set that does not intersect every hyperplane. If I choose any affine chart containing $S$, I can consider its convex hull, and it seems reasonable that this operation should not depend on the particular chart. Is there some good way to see this?
Thanks
 A: Consider two hyperplanes $[X],[Y] \subset \mathbb{R}P^n$, defined by two linear hyperplanes $X,Y \subset \mathbb{R}^{n+1}$, such that $S$ is contained in each of the two affine charts $\mathbb{R}P^n - [X]$, $\mathbb{R}P^n - [Y]$. It follows that $S \subset \mathbb{R}P^n - ([X] \cup [Y])$. 
The subset $\mathbb{R}^{n+1} - (X \cup Y)$ has four components, and these determine in a two-to-one fashion the two components of $\mathbb{R}P^n - ([X] \cup [Y])$: each component $W$ of $\mathbb{R}^{n+1} - (X \cup Y)$ determines a component of  $\mathbb{R}P^n - ([X] \cup [Y])$ that I'll denote $[W]$, consisting of all points $[A] \in \mathbb{R}P^n$ represented by an open ray $ A \subset W$ based at the origin.
Of the two components of $\mathbb{R}P^n - ([X] \cup [Y])$, one of them, say $[W]$, contains your set $S$. Given two points $[A],[B] \in [W]$ represented by rays $A,B \subset W$ as above, the set of positive linear combinations $sa+tb$, $a \in A$, $b \in B$, $s,t>0$ also lies in $W$, and so each of the affine charts $\mathbb{R}P^n - [X]$, $\mathbb{R}P^n - [Y]$ defines the same segment $\overline{[A][B]} \subset \mathbb{R}P^n - ([X] \cup [Y])$, consisting of all points in $\mathbb{R}P^n$ of the form $[sa+tb]$. 
This shows that the convex hull of $S$ as defined in each of the affine charts $\mathbb{R}P^n - [X]$, $\mathbb{R}P^n - [Y]$ is the same.
