While pondering a problem on separating objects in $R^n$ by hyperplanes, I came across the following puzzle:Suppose we are given a convex subset of $S^n$ which we call $M$.  For each point $p \in S^n$, let $C(p)$ define the great circle obtained by considering the point $p$ as the north pole and $C(p)$ as the equator. (If we imagine $p$ to correspond to a unit vector $v(p)$ in the obvious manner, then $C(p)$ corresponds to the set of all unit vectors orthogonal to $v(p)$).Now, define the set $X = \{x \in S^n$ s.t $C(x) \cap M \neq \phi \}$  Is $X$ necessarily convex? The above definition is equivalent to saying that $X$ is the set of all unit vectors orthogonal to at least one unit vector in $M$. Edit: Sasho has clarified in his answer that $X$ need not be necessarily convex. I am wondering if we may say something weaker about the topology of $X$. I will conjecture that $X$ would be decomposable into $2$ separate convex components?
Determining the convexity of a set on $S^n$
cg.comp geom
Yes, at least if you define your terms correctly.Following standard usage in Riemannian geometry, a subset $M \subseteq S^n$ is convex if and only if, for every pair of points $x$ and $y$ in $M$, some shortest path from $x$ to $y$ lies entirely within $M$.  With this standard definition, any convex subset $M  \subseteq S^n$ is both the convex closure of the points in $M$ and the intersection of all closed hemispheres that contain $M$.  In particular, the only convex subset of $S^n$ that does not lie entirely in a closed hemisphere is the entire sphere $S^n$.  (And yes, $S^n$ is the intersection of the empty set of closed hemispheres containing $S^n$!) Your definition of convexity allows some non-standard convex sets, like the complement of a single point.Let's assume that the sphere $S^n$ is centered at the origin of $\mathbb{R}^{n+1}$.  Following the standard definitions, a subset $M\subseteq S^n$ is convex if and only if $M$ is the intersection of $S^n$ with a convex cone in $\mathbb{R}^{n+1}$.   So let $C$ be a convex cone in $\mathbb{R}^{n+1}$, and let $M = C \cap S^n$.  Let $C^*$ denote the dual cone of $C$; this is also a convex cone, so the intersection $S^n\cap C^*$ is convex.  Your set $X$ is $S^n \cap (C^* \cup -C^*)$ = $(S^n\cap C^*) \cup -(S^n\cap C^*)$.  Thus, $X$ is the union of two  antipodal convex subsets of $S^n$.