Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 311 / 3 / e / Solution

Suppose instead that every future-directed normal timelike geodesic were complete. Choose a point more than $3/c_0$ proper time to the future of $S$. Global hyperbolicity supplies a longest timelike curve from $S$ to that point; it is a geodesic orthogonal to $S$ and has no focal point before its endpoint. But part d makes the convergence of every such normal congruence diverge within proper time $3/c_0$. Nearby geodesics then intersect, producing a focal point after which the geodesic cannot maximize proper time. This contradiction proves that at least one timelike geodesic has finite maximal proper time, so $(M,g)$ has timelike geodesic incompleteness.