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
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.