Penrose null geodesic incompleteness

Assume for contradiction that every future-directed null geodesic is future complete. By [F1], Countable Choice and Dependent Choice hold under the assumed Axiom of Choice, so the hypotheses of the volume-time, Cauchy-splitting and causal-boundary suppliers are satisfied.

Apply [F2] to obtain a volume time input, a temporal function $T$, a smooth Cauchy slice $S_0=T^{-1}(0)$, and a complete timelike product flow $\mathrm{\Psi }$. The splitting satisfies $T(\mathrm{\Psi }(s,z))=s$. Hence $T$ tends to $+\mathrm{\infty }$ and $-\mathrm{\infty }$ along the two ends of every orbit; continuity prevents an endpoint in $M$, so each orbit is an inextendible timelike curve.

Apply [F3] to identify the given hypersurface $\mathrm{\Sigma }$ diffeomorphically with $S_0$ by projection along these flow lines. Therefore $S_0$ is connected and noncompact.

The restriction $T{\mathrm{\mid }}_S$ is continuous and $S$ is nonempty compact, so [F8] gives a point $s_0\in S$ at which $T{\mathrm{\mid }}_S$ is minimal. If $s_0\in I^{+}(S)$, some $s_1\in S$ satisfies $s_1\ll s_0$; since $T$ is temporal, [F6] gives $T(s_1)<T(s_0)$, contradicting minimality. Thus $s_0\mathrm{\notin }{I}^{+}(S)$. As $s_0\in S\subseteq J^{+}(S)$ by reflexivity, $s_0\in E^{+}(S)$, so this boundary is nonempty.

Define $\pi :M\to S_0$ by $\pi (\mathrm{\Psi }(s,z))=z$. This is continuous because ${\mathrm{\Psi }}^{-1}$ is smooth. If $x,y\in E^{+}(S)$ have $\pi (x)=\pi (y)$ and $x\mathrm{\ne }y$, their flow coordinates differ, so the timelike orbit between them makes them chronologically related. This contradicts the achronality in [F5]. Hence $\pi {\mathrm{\mid }}_{E^{+}(S)}$ is injective.

The domain $E^{+}(S)$ is a topological three-manifold without boundary by [F4], and $S_0$ is a smooth three-manifold by [F2]. For each $x\in E^{+}(S)$, choose manifold charts about $x$ and $\pi (x)$ and shrink the source chart so its image lies in the target chart. In these coordinates $\pi$ is continuous and injective between open subsets of ${\mathbb{R}}^3$; [F7] makes its image open. Taking the union over $x$ shows $\pi (E^{+}(S))$ is open in $S_0$.

By [F4], $E^{+}(S)$ is compact, and $\pi$ is continuous, so [F8] makes $\pi (E^{+}(S))$ compact. The smooth manifold $S_0$ is Hausdorff; [F9] therefore makes this image closed in $S_0$.

The image $\pi (E^{+}(S))$ is nonempty by step 3.2 and both open and closed by steps 4.1 and 4.2. Since $S_0$ is connected by step 3.1, [F9] implies $\pi (E^{+}(S))=S_0$. But $\pi (E^{+}(S))$ is compact by step 4.2, which would make $S_0$ compact, contradicting step 3.1. Thus the assumed future null completeness is false. By [F10], at least one maximal future-directed null geodesic has finite upper affine endpoint, which is the asserted conclusion.