Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 2 / 13B (Analysis II)

Choose $a_0\in A$ and put $R=\Vert a_0-y\Vert$. The intersection $K=A\cap \overline{B}(y,R)$ is nonempty, closed and bounded in ${\mathbb{R}}^n$, hence compact. The continuous distance function $a\mapsto \Vert a-y\Vert$ attains a minimum at some $x\in K$. For $a\in A\setminus K$ its value is strictly greater than $R$, while the value at $x$ is at most that at $a_0$, namely $R$. Thus this minimum on $K$ is a minimum on all of $A$:The set $A$ itself need not be bounded; only the compact portion needed to find the nearest point is bounded.

Suppose two distinct minimizers $x_1,x_2$ have the same minimum distance $d$ from $y$. Convexity puts their midpoint $z=(x_1+x_2)/2$ in $A$. Expanding the Euclidean norm givescontradicting minimality. Therefore the nearest point is unique. Together with the preceding existence proof, this defines the metric projection onto a nonempty closed convex set.