Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 311 / 3 / ii / Solution

Let $t^a$ exhibit the stable causality of $g$, so $g-t_a{t}_b$ has no closed timelike curves. SetThe vector $r^a$ remains timelike for $g^{\prime }$. Widening the cones of $g^{\prime }$ once more using $r$ giveswhich is causal by assumption. Hence