Convex Hull of a non finite set

Question

I know that if $S=\{s_1,...,s_n\}$ is a finite set, then the convex Hull is given by $$covHul(S)=\{\sum_{s_i\in S}\alpha _is_i\mid \alpha _1+...+\alpha _n=1\}.$$

But how do we define it if $S$ is not finite (or even uncountable) ? In the case of $$\mathcal C=\{(x,y)\mid x^2+y^2\leq 1, (x,y)\notin [0,1)\times [0,1)\}$$ (a sort of packman), then $$CovHul(\mathcal C)=\{\alpha x+(1-\alpha )y\mid \alpha \in [0,1]\}$$ looks to work. Actually, if a set is connected, the definition above looks to work... but maybe is not correct. Any suggestion ?

Answer 1

If $V$ is a vector space and $C \subset V$ then the convex hull of $C$ is defined as the set of all vectors of the form $ \sum\limits_{k=1}^{n} a_k c_k$ where $n$ is a positive integer, each $c_i \in C$, each $a_i >0$ and $ \sum\limits_{k=1}^{n} a_k=1$.

See my comment below for an example where we cannot use just two terms in the sum.

Answer 2

Generally, the convex hull of a set $S$ is the smallest convex superset of $S$.

Your ansatz for an explicit formula already goes into the right direction, but is not sufficient (apart from the fact that you nowhere stated what $x$ and $y$ are supposed to be). For example, consider the convex hull of three separate small non-intersecting circles located at the corners of an equilateral triangle (the triangle itself not being part of that set). Then your construction leaves a triangular hole.

For the connected case, just consider the three-dimensional set of a circle in the $xy$ plane with three "antennas" in the form of straight line segments in $z$ direction, starting at three different points of the circle. This set is clearly connected, even path connected (to get from a point on one antenna to a point on another, first go along the antenna to the circle, then move to the base point of the second antenna, and then move to the destination point along the antenna). But your construction leaves an excavation between the ends of the antennas.

However it is easy to fix: Instead of taking the convex-combinations of just two elements of you set, take all convex combinations of finitely many elements. That is, consider the set $$\left\{\sum_k=1^n p_k x_k\,\middle|\, n\in\mathbb N\setminus\{0\}, x_k\in S \text{ for all $k$}, \sum_{k=1}^n p_k=1, p_k\ge 0 \text{ for all $k$}\right\}.$$

Note that according to Carathéodory’s Theorem, in $\mathbb R^n$ you only need to consider convex combination of at most $n+1$ points.