Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 71 / 1 / b / Solution

Take zero cosmological constant and an open geometry, which means $k<0$ in the metric convention. Evaluating the Friedmann equations today gives $H_0^2=H_0^2{\Omega }_0-k/a_0^2$, soOn a constant-time slice the radial line element is $d\ell =a_{0}dr/\sqrt{1-k{r}^2}$. ThereforeThis is a cosmological proper distance on today's slice, not the distance traveled during the photon flight. The printed condition $k<1$ is insufficient to specify an open model; the derivation uses its intended $k<0$.