Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 10E / Solution

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous, then every real value between $f(a)$ and $f(b)$ is attained at some $c\in [a,b]$. Here is a proof using completeness of the real numbers. Endpoint values are immediate, so suppose $f(a)<y<f(b)$; reversing the sign of $f$ handles the other ordering. LetThe set is nonempty and bounded above. Continuity at $a$ puts points to the right of $a$ in $S$, while continuity at $b$ excludes an entire interval to the left of $b$, so $a<c<b$. If $f(c)<y$, continuity puts some point larger than $c$ in $S$, contradicting its supremum. If $f(c)>y$, continuity excludes $S$ in a neighborhood of $c$, contradicting the existence of points of $S$ arbitrarily close to its supremum from below. Therefore $f(c)=y$, proving the theorem.

For the fixed-point assertion, put $g(x)=f(x)-x$. This is continuous, with $g(0)=f(0)\ge 0$ and $g(1)=f(1)-1\le 0$. If either endpoint value is zero it already gives a fixed point; otherwise the intermediate value theorem gives an interior zero. HenceThe proof needs both continuity and the closed-interval endpoint information; the following counterexamples separate those hypotheses.