IMC 2007 · Day 2 · P9

Suppose $f(x) \ne x$ for all $x \in C$. Let $[a, b]$ be the smallest closed interval that contains $C$. Since $C$ is closed, $a, b \in C$. By our hypothesis $f(a) > a$ and $f(b) < b$. Let $p = \sup \{ x \in C : f(x) > x \}$. Since $C$ is closed and $f$ is continuous, $f(p) \ge p$, so $f(p) > p$. For all $x > p$, $x \in C$ we have $f(x) < x$. Therefore $f\bigl( f(p) \bigr) < f(p)$ contrary to the fact that $f$ is non-decreasing.