Cauchy sets are closed, split the manifold into past, set and future, and are locally Lipschitz graphs

(No closed causal curve.) Let $\delta :[0,1]\to M$ be a future causal curve with $\delta (0)=\delta (1)$, nonconstant. Define the periodic extension $\beta :\mathbb{R}\to M$ by $\beta (t):=\delta (t-\lfloor t\rfloor )$. On each interval $(n,n+1)$ the curve $\beta$ is the restriction of $\delta$ up to an increasing affine reparametrization, so it is locally absolutely continuous with future causal velocity almost everywhere, and by [L5] this persists across the finitely many junctions of every compact interval: $\beta$ is a future causal curve. It is future-inextendible: if $\beta (t)\to z$ as $t\to +\mathrm{\infty }$, then for every $s\in (0,1)$ the sequence $\beta (s+n)=\beta (s)$ converges to $z$, so $\beta (s)=z$ for all $s\in (0,1)$, contradicting the nonconstancy of $\delta$ on $[0,1]$; the same argument at $-\mathrm{\infty }$ (using $\beta (s-n)=\beta (s)$) gives past-inextendibility. Hence $\beta$ is an inextendible causal curve, and by the Cauchy property ${\beta }^{-1}[S]$ has exactly one element $t_0$; but then $t_0+1\in {\beta }^{-1}[S]$ as well, since $\beta (t_0+1)=\beta (t_0)\in S$, and $t_0+1\mathrm{\ne }{t}_0$. This contradiction shows that no such $\delta$ exists.

(A vertical line meets $S$ at most once.) Fix $p\in S$ and the chart $(U,x)$, constants $K\ge 1$ and $c=16$, and minorant ${\eta }_c$ of [L4]. Write $x(U)=B_r(0)$. Choose $0<a<r\mathrm{/}2$, $0<\rho <a\mathrm{/}(4\sqrt{c})$ and $\eta :=a\mathrm{/}2$. Then $\sqrt{a^2+{\rho }^2}<r$ and $\sqrt{c}\rho +\eta <a$, so $W:=x^{-1}((-a,a)\times B_{\rho }(0))\subseteq U$. These choices can be made arbitrarily small: for any prescribed neighborhood $V$ of $p$, first reduce $r$ so that $x^{-1}(B_r(0))\subseteq V$, retaining the same adapted coordinate functions, cone constants and $g_{00}=-1$, and then choose $a,\rho ,\eta$ as above. For $z\in B_{\rho }(0)$ the vertical line $L_z$ is the image of a future timelike curve [L4]; two distinct points $x^{-1}(s,z)$ and $x^{-1}(s^{\mathrm{\prime }},z)$ of $S$ with $s<s^{\mathrm{\prime }}$ would be chronologically related, contradicting [L1]. Hence $S\cap L_z$ has at most one point.

(Disjointness and the decomposition.) Let $x\in M$. Choose a future timelike vector $w\in T_{x}M$ and let $\gamma :={\gamma }_{x,w}:I\to M$ be the maximal geodesic through $x$ with ${\gamma }^{\mathrm{\prime }}(0)=w$; by [L2] it is an inextendible future timelike curve, and $\gamma (0)=x$. By the Cauchy property there is exactly one $t_0\in I$ with $\gamma (t_0)\in S$. If $x=\gamma (t_0)$, then $x\in S$. If $x\mathrm{\ne }\gamma (t_0)$ and $0>t_0$, then $\gamma {\mathrm{\mid }}_{[t_0,0]}$ is a future timelike curve from a point of $S$ to $x$, so $x\in I^{+}(S)$; if $0<t_0$, then $\gamma {\mathrm{\mid }}_{[0,t_0]}$ is a future timelike curve from $x$ to a point of $S$, so $x\in I^{-}(S)$. Hence $M=I^{-}(S)\cup S\cup I^{+}(S)$. The sets are pairwise disjoint: if $s\in S\cap I^{+}(S)$, there is $s^{\mathrm{\prime }}\in S$ with $s^{\mathrm{\prime }}\ll s$, and $s^{\mathrm{\prime }}\mathrm{\ne }s$ because $s^{\mathrm{\prime }}\ll s$ forces $s^{\mathrm{\prime }}\mathrm{\ne }s$; this contradicts acausality [L1]. If $x\in I^{+}(S)\cap I^{-}(S)$, there are $s_1,s_2\in S$ with $s_1\ll x\ll s_2$; if $s_1\mathrm{\ne }{s}_2$ this contradicts acausality [L1], while if $s_1=s_2$ then transitivity of $\ll$ gives $s_1\ll s_1$, that is, a closed future timelike curve through $s_1$, which is excluded by step 1.1 applied to that closed curve. This proves (a), including the characterization of membership in $I^{\pm }(S)$.

(Closedness.) Let $s_k\in S$ with $s_k\to q\in M$, and suppose $q\mathrm{\notin }S$. Choose a maximal geodesic $\gamma :={\gamma }_{q,w}$ with future timelike $w$ as in [L2]; it is an inextendible future timelike curve and meets $S$ in exactly one point $r:=\gamma (t_0)$, with $r\mathrm{\ne }q$ because $q\mathrm{\notin }S$. Let $t_q:=0$ be the parameter of $q$. If $t_0<0$ then $\gamma {\mathrm{\mid }}_{[t_0,0]}$ is a future timelike curve from $r$ to $q$, so $r\ll q$; since $I^{+}(r)$ is open [L3] and $s_k\to q$, for all large $k$ we have $s_k\in I^{+}(r)$, that is, $r\ll s_k$ with $r,s_k\in S$; since $s_k\mathrm{\ne }r$ for large $k$ (as $s_k\to q\mathrm{\ne }r$), this contradicts the acausality of $S$ [L1]. If $t_0>0$ then $\gamma {\mathrm{\mid }}_{[0,t_0]}$ is a future timelike curve from $q$ to $r$, so $q\ll r$; since $I^{-}(r)$ is open [L3], for large $k$ we have $s_k\in I^{-}(r)$, that is, $s_k\ll r$ with $s_k,r\in S$ and $s_k\mathrm{\ne }r$, again contradicting acausality. Hence $q\in S$: $S$ is closed. This proves (b).

(A vertical line meets $S$ at least once.) With $p,W,a,\rho ,\eta$ as in step 1.2, fix $z\in B_{\rho }(0)$ and put $q^{-}:=x^{-1}(-\sqrt{c}\mathrm{\mid }z\mathrm{\mid }-\eta ,z)$ and $q^{+}:=x^{-1}(\sqrt{c}\mathrm{\mid }z\mathrm{\mid }+\eta ,z)$. Both are in $W$ since $\sqrt{c}\mathrm{\mid }z\mathrm{\mid }+\eta <\sqrt{c}\rho +\eta <a$. In coordinate differences from $p$ one has ${\eta }_c(q^{+}-p,q^{+}-p)=-(\sqrt{c}\mathrm{\mid }z\mathrm{\mid }+\eta {)}^2+c\mathrm{\mid }z{\mathrm{\mid }}^2=-2\sqrt{c}\mathrm{\mid }z\mathrm{\mid }\eta -{\eta }^2<0$, and the same value for ${\eta }_c(p-q^{-},p-q^{-})$. Thus [L4] gives $q^{-}\ll p\ll q^{+}$, so $q^{-}\in I^{-}(S)$ and $q^{+}\in I^{+}(S)$. The vertical segment $J_z$ joining them is connected. If it avoided $S$, step 2.1 would partition it into the disjoint relatively open sets $J_z\cap I^{-}(S)$ and $J_z\cap I^{+}(S)$, open by [L3] and both nonempty by their endpoints. This contradicts connectedness. Hence $J_z\cap S\mathrm{\ne }\mathrm{\varnothing }$, and step 1.2 gives exactly one intersection.

(The graphing function is Lipschitz.) For $z\in B_{\rho }(0)$ let $\phi (z)\in (-a,a)$ be the $x^0$-coordinate of the unique point of $S\cap L_z$, so that $S\cap W=\{x^{-1}(\phi (z),z):z\in B_{\rho }(0)\}$, and $\mathrm{\mid }\phi (z)\mathrm{\mid }\le \sqrt{c}\mathrm{\mid }z\mathrm{\mid }+\eta <a$ by step 3.2. Since $p=x^{-1}(0,0)\in S$, uniqueness on $L_0$ gives $\phi (0)=0$. If $z,z^{\mathrm{\prime }}\in B_{\rho }(0)$ with $\mathrm{\mid }\phi (z)-\phi (z^{\mathrm{\prime }})\mathrm{\mid }>\sqrt{c}\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }$, set $\delta :=(\phi (z)-\phi (z^{\mathrm{\prime }}),z-z^{\mathrm{\prime }})$ and suppose $\phi (z)>\phi (z^{\mathrm{\prime }})$ (the other case is symmetric); then ${\eta }_c(\delta ,\delta )=-(\phi (z)-\phi (z^{\mathrm{\prime }}){)}^2+c\mathrm{\mid }z-z^{\mathrm{\prime }}{\mathrm{\mid }}^2<0$, and both $x^{-1}(\phi (z),z)$ and $x^{-1}(\phi (z^{\mathrm{\prime }}),z^{\mathrm{\prime }})$ lie in $S\cap W\subseteq U$, so [L4] gives $x^{-1}(\phi (z^{\mathrm{\prime }}),z^{\mathrm{\prime }})\ll x^{-1}(\phi (z),z)$: two distinct points of $S$ chronologically related, contradicting acausality [L1]. Hence $\mathrm{\mid }\phi (z)-\phi (z^{\mathrm{\prime }})\mathrm{\mid }\le \sqrt{c}\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }$. Taking $z^{\mathrm{\prime }}=0$ and using $\phi (0)=0$ gives $\mathrm{\mid }\phi (z)\mathrm{\mid }\le \sqrt{c}\mathrm{\mid }z\mathrm{\mid }<a$.

(Topological hypersurface.) For $p\in S$ and its chart as above, the map $\psi :B_{\rho }(0)\to S\cap W$, $\psi (z)=x^{-1}(\phi (z),z)$ is continuous (a Lipschitz function into Euclidean space followed by the continuous chart inverse), injective and surjective onto $S\cap W$ by step 4.1, and the restriction of the continuous coordinate projection $\pi :x^{-1}(s,z)\mapsto z$ to $S\cap W$ is its inverse, because a point of $S\cap W$ lies on exactly one vertical line and has the form $x^{-1}(\phi (\pi (\cdot )),\pi (\cdot ))$ by step 1.2; hence $\psi$ is a homeomorphism onto the relatively open set $S\cap W$. For two such graph charts $\psi ,{\psi }^{\mathrm{\prime }}$ the transition ${\psi }^{\mathrm{\prime }-1}\circ \psi ={\pi }^{\mathrm{\prime }}\circ \psi$ is continuous on its domain, so these charts form a topological atlas on $S$; consequently $S$ is a topological $3$-manifold in the subspace topology and its inclusion into $M$ is a topological embedding. This proves (c) and completes the proof.