Cauchy sections do not end at a one-sided chronological horizon

Under the AC assumption in [F1], [F8] supplies Countable Choice and Dependent Choice for all the causal suppliers below. By [F2], $P$ and $Q$ are open, hence $E$ is an open time-oriented Lorentzian submanifold. By [F4] and the inclusion of timelike reachability in causal reachability from [F7], $I^{-}(P)\subseteq P$ and $I^{+}(Q)\subseteq Q$: for example $x\ll y\ll u$ with $y\in P$ and $u\in U$ gives $x\ll u$. Thus $P$ is a past set and $Q$ a future set. Every causal segment from $a\in E$ to $b\in E$ stays in $E$: choose $u,v\in U$ with $u\ll a$ and $b\ll v$; every point $x$ of the segment satisfies $a\le x\le b$, so [F4] gives $u\ll x\ll v$. Finally apply [F5] in $E$ to the given Cauchy hypersurface $C$, obtaining acausality relative to $E$. This invokes the Cauchy-set theorem directly, rather than applying one-sided extension to a finite segment.

Suppose $p\in \bar{C}\cap Q\cap \mathrm{\partial }P$. Use [F6] at $p$ with initial vector $N_p$ to obtain a future timelike geodesic $\beta$ with $\beta (0)=p$. Since $Q$ is open, choose $\epsilon >0$ so that $\beta ([-\epsilon ,0])\subset Q$. For $-\epsilon \le t<0$, its smooth timelike segment from $\beta (t)$ to $p$ is an admissible timelike curve by [F7], so $\beta (t){\ll }_{M}p$. Since $p\in \mathrm{\partial }P\subseteq \bar{P}$ and $P$ is a past set, [F2] puts this entire preceding tail in $P$, hence in $E$. Reparametrize it by $t=-\epsilon \mathrm{/}(1+s)$, $s\in [0,\mathrm{\infty })$; this is an increasing smooth diffeomorphism onto $[-\epsilon ,0)$ with positive derivative. By [F7] the resulting curve $\gamma$ is future timelike. It has ambient limit $p$ as $s\to \mathrm{\infty }$, and $p\mathrm{\notin }P$ since $P$ is open. It therefore has no future endpoint in $E$, so is future-inextendible relative to $E$.

Apply [F3] in $E$ to $\gamma :[0,\mathrm{\infty })\to E$, obtaining an inextendible future causal curve ${\gamma }^{\ast }:I^{\ast }\to E$ that agrees with $\gamma$ for $s\ge 0$. The Cauchy condition gives $s_c\in I^{\ast }$ with $c={\gamma }^{\ast }(s_c)\in C$. Choose $s>\max \{s_c,0\}$. The segment of ${\gamma }^{\ast }$ gives $c{\le }_M\gamma (s)$, while the remaining timelike geodesic segment to $p$ gives $\gamma (s){\ll }_{M}p$. Hence [F4] gives $c{\ll }_{M}p$. By [F2], $I_M^{+}(c)$ is an open neighborhood of $p$; since $p\mathrm{\ne }c$, shrink it to a neighborhood $V$ of $p$ avoiding $c$. From $p\in \bar{C}$ choose $c^{\mathrm{\prime }}\in C\cap V$. Then $c^{\mathrm{\prime }}\mathrm{\ne }c$ and $c{\ll }_M{c}^{\mathrm{\prime }}$. Step 1.1 puts a joining timelike segment entirely in $E$, contradicting acausality of $C$ there. Reversing the time orientation interchanges $P$ and $Q$ and gives the same contradiction for $p\in \bar{C}\cap P\cap \mathrm{\partial }Q$.

Since $C\subset P\cap Q$, its closure lies in $\bar{P}\cap \bar{Q}$. At any point of that closure outside $E$, at least one of the two open-set memberships fails and hence the corresponding boundary occurs. If the point lies in exactly one of $P,Q$, step 3.1 excludes it. If it lies in neither, it belongs to $\mathrm{\partial }P\cap \mathrm{\partial }Q$. Thus $\bar{C}\setminus E\subset \mathrm{\partial }P\cap \mathrm{\partial }Q$, as claimed.