Proposition: Intrinsic distance and causality

The intrinsic distance of $N$ is

It is realised by a spatial geodesic lying entirely in $N$. Consequently,

For every compact $K\subset M$, ${J_M}^{\pm }(K)$ is closed in $M$.

Proof. First consider the completed two-dimensional cone. It is a proper length space: bounded closed sets are compact in polar coordinates, with all angular coordinates identified at $r=0$. A length-minimising curve therefore exists, by the usual compactness argument for equi-Lipschitz curves.

The developed straight segment with angular separation $\psi$ avoids the origin because $|\psi |\le \pi \alpha <\pi$. It has length

A minimiser cannot pass through the apex, since every such curve has length at least $r+r^{\prime }$. Away from the apex a minimiser develops to a straight segment. Its continuous angular variation has absolute value less than $\pi$ and is congruent to $\psi$ modulo $2\pi \alpha$. The smallest possible absolute variation is $|\psi |$. Since the chord length is nondecreasing in the absolute angle on $[0,\pi ]$, the displayed segment is minimising.

Taking the product with ${\mathbb{R}}_z$ gives (eq:R0): for every spatial curve, the integral triangle inequality bounds its length below by the Euclidean norm of its cone-length contribution and its total $z$-displacement. Equality is obtained by traversing the cone minimiser and the $z$-segment with constant relative speeds.

A future causal curve in the ultrastatic metric satisfies $|\dot{x}{|}_q\le \dot{t}$, hence $t-t^{\prime }\ge d(x,x^{\prime })$. Conversely, a distance-realising spatial curve traversed at speed at most one gives a causal curve whenever this inequality holds. This proves (eq:causal).

Finally, if $p_j\in J_M^{+}(K)$ and $p_j\to p\in M$, choose $k_j\in K$ witnessing the relation. A subsequence has $k_j\to k\in K$; continuity of $d$ preserves the inequality. The past case is identical.