Convex independence

Remarks

Example: the points $(0,0)$, $(1,0)$, $(0,1)$, $(1,1)$ of $\R^2$ are in convex position. Each of them is a vertex of the unit square, and no vertex is a convex combination of the other three. For instance, the other three points satisfy $x + y \le 1$, so all their convex combinations do, while $(1,1)$ does not. The other vertices are handled by symmetry.

Non-example: add the point $(\frac12, \frac12)$. It equals $\frac12 (1,0) + \frac12 (0,1)$, so it lies in the convex hull of the other points.

A convexly independent family is injective, because $x_i = x_j$ with $i \neq j$ puts $x_i$ in the hull of $\set{x_j}$. A finite set $S$ is in convex position if and only if every point of $S$ is an extreme point of $\operatorname{conv}(S)$, that is, a vertex of the polytope $\operatorname{conv}(S)$. Affinely independent families are convexly independent, but not conversely: the four corners of a square are affinely dependent.