Solution
= Solution
\b[The fixed-point property fails on an open interval.] The <continuous function>
$$
\boxed{f(t)=\frac{t+b}{2}}
$$
maps $(a,b)$ 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.