Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 1 / 10D / i / Solution

The fixed-point property fails on an open interval. The continuous functionmaps $(a,b)$ into itself, because $a<t<f(t)<b$. Its only possible fixed point satisfies $t=(t+b)/2$, hence $t=b$, which is outside the domain. It is therefore a counterexample even with the self-map requirement retained.