Convex hull of finite set in $\mathbb{R^2}$ is a polygon

Question

How can I prove that a convex hull of a finite set of points in the plane is always a convex polygon? This result looks very intuitive, but how can you prove it formally? I thought about taking any convex polygon that contains in it all of the points, and then make it smaller and smaller until one of it's sides meets one of the points, and then do the same to the other sides, but how do you make it formal?

Answer 1

Hint

Denote by $S_n = \{p_1, \dots, p_n\}$ the finite set of points and $\overline{S_n} = \mathrm{Conv}(S_n)$.

Let $\mathcal M = \{M \subseteq S_n \mid \mathrm{Conv}(M) = \mathrm{Conv}(S_n)\}$ and $P$ an element of $\mathcal M$ with the minimum cardinal.

Prove that $P$ is a polygon when ordering its points in an appropriate way, i.e rotation around a point in $\mathrm{Conv}(P)$.

Answer 2

Let $S\subset\Bbb{R}^2$ finite and $C$ the convex hull of $S$. Because $C$ is convex and contains $S$, it also contains every line segment connecting pairs of points in $S$. It follows that $C$ contains every polygon with its vertices in $S$.

Let $P$ be the union of all polygons with vertices in $S$, so that $S\subset P\subset C$. Note that $P$ is itself a polygon because $S$ is finite. Then $P$ is itself convex because it contains every polygon on its vertices. It follows that $C\subset P$ and hence $C=P$, so the convex hull is a polygon.