Truncated causal cones J^+(K) cap J^-(S_b) over compact sets and Cauchy levels are compact

(The sublevel and superlevel descriptions.) Let $q\in J^{-}(S_b)$: there is $s\in S_b$ with $q\le s$; if $q\mathrm{\ne }s$ then $t(q)<t(s)=b$ by strict monotonicity [L1], and if $q=s$ then $t(q)=b$. Conversely, if $t(q)\le b$: choose a maximal geodesic through $q$ with future timelike tangent (it is an inextendible future timelike curve, [L5]) and let $s$ be a point of it; by [L1] the values of $t$ along it run from $-\mathrm{\infty }$ to $+\mathrm{\infty }$, so there is a point $s$ on it with $t(s)=b$; since the geodesic is future-directed, $q\le s$. Hence $J^{-}(S_b)\subseteq \{t\le b\}\subseteq J^{-}(S_b)$, and the dual argument gives $J^{+}(S_a)=\{t\ge a\}$. In particular $J^{-}(S_b)$ is closed as a level sublevel set of the continuous function $t$.

(Each truncated one-point cone is compact.) Fix $x\in M$ with $t(x)<b$ and set $A:=J^{+}(x)\cap J^{-}(S_b)=J^{+}(x)\cap \{t\le b\}$; it is closed, being the intersection of the closed sets $J^{+}(x)$ and $\{t\le b\}$ [L2, step 1.1]. It is compact: otherwise, by [L4] applied in the connected component of $x$ (which contains $A$), there are $r_k\in A$ with no convergent subsequence. For each $k$ choose a future causal curve ${\gamma }_k:[0,1]\to M$ from $x$ to $r_k$ and reparametrize it by $h$-arclength: ${\alpha }_k:[0,L_k]\to M$, ${\alpha }_k(0)=x$, ${\alpha }_k(L_k)=r_k$. The images satisfy ${\alpha }_k([0,L_k])\subseteq J^{+}(x)\cap \{t\le b\}$: indeed every point of ${\gamma }_k$ lies causally between $x$ and $r_k$, so its $t$-value lies between $t(x)$ and $t(r_k)\le b$ by monotonicity [L1]. If the lengths $L_k$ were bounded, then $d_h(x,r_k)\le L_k$ would be bounded, so the $r_k$ would have a convergent subsequence by [L4]: contradiction. Hence $L_k\to \mathrm{\infty }$ along a subsequence, and the half-line limit curve lemma [L3] gives a future-inextendible future causal curve ${\alpha }^{+}:[0,\mathrm{\infty })\to M$ with ${\alpha }^{+}(0)=x$ that is a uniform limit on compact parameter intervals of the curves ${\alpha }_k$. Every point of ${\alpha }^{+}$ is a limit of points with $t$-values at most $b$, so $t({\alpha }^{+}(\tau ))\le b$ for all $\tau$ by continuity of $t$. But along the inextendible extension of ${\alpha }^{+}$ (which exists by [L5]) the volume time tends to $+\mathrm{\infty }$ at the future end [L1], hence $t({\alpha }^{+}(\tau ))\to +\mathrm{\infty }$ as $\tau \to \mathrm{\infty }$: contradiction. Therefore $A$ is compact.

(Finitely many past-based cones over $K$.) Consider all pairs $(x,z)$ with $x\in K$ and $z\ll x$; [L5] supplies at least one such $z$ for every $x$. The family of open sets $I^{+}(z)$ over these pairs covers $K$, so the indexed ambient-cover criterion [L6] selects finitely many pairs $(x_i,z_i)$ with $K\subseteq {\bigcup }_i{I}^{+}(z_i)$. No family of past points is chosen before taking this finite subcover. If $y\in J^{+}(K)$, choose $x\in K$ with $x\le y$ and $i$ with $x\in I^{+}(z_i)$; then $z_i\le x\le y$ by transitivity, so $y\in J^{+}(z_i)$. Hence $J^{+}(K)\subseteq {\bigcup }_i{J}^{+}(z_i)$ and $J^{+}(K)\cap J^{-}(S_b)\subseteq {\bigcup }_i(J^{+}(z_i)\cap \{t\le b\})$. Each set on the right is compact by step 2.1 if $t(z_i)<b$, empty if $t(z_i)>b$, and the singleton $\{z_i\}$ if $t(z_i)=b$, since strict causal monotonicity excludes any other point. The finite union is compact by [L6]. By [L2] and step 1.1 the set on the left is closed, hence compact by [L6]. If $K=\mathrm{\varnothing }$, the empty subcover and empty union give the same conclusion. Together with step 1.1 this proves the statement.