Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 62 / 1 / a / Solution

Let $\mathcal{H}$ be the family of closed half-spaces containing $C$. Certainly $C\subseteq {\bigcap }_{H\in \mathcal{H}}H$. To prove the reverse inclusion, exclude an arbitrary $z\notin C$.

Suppose first that $C$ is nonempty. Its Euclidean projection onto a convex set $p$ exists: minimizing the continuous squared distance may be restricted to a sufficiently large closed ball, whose intersection with the closed set is compact. Put $d=z-p\ne 0$. For every $x\in C$, convexity puts $p+t(x-p)$ in $C$ for $0\le t\le 1$. Minimality of $p$ givesThus the closed half-spacecontains $C$, while $⟨d,z⟩=⟨d,p⟩+\Vert d{\Vert }^2$ excludes $z$. Since every point outside $C$ is excluded by some member of $\mathcal{H}$,This is the half-space representation of a closed convex set. If $C=\varnothing$, every point can again be excluded by a containing half-space, so the intersection is empty. If $C={\mathbb{R}}^n$, no proper half-space contains it and the empty intersection is, by convention, ${\mathbb{R}}^n$.