In a globally hyperbolic manifold the causal relation is closed and truncated causal cones over compact sets are compact

(Compact enclosure of the joining curves.) If $p_k=q_k$ for all $k$ in an infinite set, then passing to that subsequence and using uniqueness of limits gives $p=q$, so $p\le q$ by reflexivity; the claim is then proved. Hence, passing to a subsequence, we may assume $p_k\mathrm{\ne }{q}_k$ for all $k$, and choose a future causal curve ${\gamma }_k:[0,1]\to M$ with ${\gamma }_k(0)=p_k$, ${\gamma }_k(1)=q_k$ for each $k$. Choose $\epsilon >0$ so small that $p^{-}:={\exp }_p(-\epsilon N_p)$ and $q^{+}:={\exp }_q(\epsilon N_q)$ are defined; by [L4], $p^{-}\ll p$ and $q\ll q^{+}$. Since $I^{+}(p^{-})$ and $I^{-}(q^{+})$ are open and contain $p$, respectively $q$ [L1], there is $k_0$ with $p_k\in I^{+}(p^{-})$ and $q_k\in I^{-}(q^{+})$ for $k\ge k_0$. Fix such a $k$. For every point $x$ of the image of ${\gamma }_k$ one has $p_k\le x\le q_k$, so $p^{-}\le x$ and $x\le q^{+}$ by transitivity [L1]; hence ${\gamma }_k([0,1])\subseteq D:=J^{+}(p^{-})\cap J^{-}(q^{+}),$ and $D$ is compact by [L1].

(Bounded reparametrizations.) Let ${\alpha }_k:[0,L_k]\to M$ be ${\gamma }_k$ reparametrized by $h$-arclength, so $L_k={\mathrm{\ell }}_h({\gamma }_k)$, ${\alpha }_k(0)=p_k$ and ${\alpha }_k(L_k)=q_k$. By step 1.1 and [L2], $L_k\le L(D)<\mathrm{\infty }$ for all $k\ge k_0$; in particular the sequence $L_k$ is bounded and has a convergent subsequence, with limit $b\le L(D)$.

(Closedness.) Passing to a subsequence, $L_k\to b<\mathrm{\infty }$. If $b>0$, clause (i) of [L3] gives a future causal curve $\alpha :[0,b]\to M$ with $\alpha (0)=p$ and $\alpha (b)=q$, so $p\le q$. If $b=0$, clause (i) gives $p=q$, so again $p\le q$. This proves the closedness of $\le$.

(Closedness of the cones.) Let $y_j\to y$ with $y_j\in J^{+}(K)$ for a compact $K$: choose $x_j\in K$ with $x_j\le y_j$. Passing to a subsequence, $x_j\to x\in K$ because $K$ is compact, so step 3.1 applied to $x_j\le y_j$ gives $x\le y$, that is, $y\in J^{+}(K)$; hence $J^{+}(K)$ is closed. The same argument with the order reversed shows that $J^{-}(K)$ is closed, and with $K=\{x\}$ this gives closedness of $J^{\pm }(x)$.

(Compact truncated cones.) Let $K_1,K_2$ be compact. If either compact set is empty, the intersection is empty and compact. Otherwise form the family of all $I^{+}({\exp }_x(-\epsilon N_x))$ with $x\in K_1$ and every positive $\epsilon$ for which the past timelike segment of [L4] is defined. This family covers $K_1$ by [L4], with no simultaneous selection of parameters. The ambient-cover criterion [L5] gives a finite subcover; finite choice names its representing pairs $(x_i,{\epsilon }_i)$ and puts $x_i^{-}:={\exp }_{x_i}(-{\epsilon }_i{N}_{x_i})$. If $y\in J^{+}(K_1)$, choose $x\in K_1$ with $x\le y$ and $i$ with $x\in I^{+}(x_i^{-})$; then $x_i^{-}\le x\le y$, so $y\in J^{+}(x_i^{-})$. Hence $J^{+}(K_1)\subseteq {\bigcup }_{i=1}^m{J}^{+}(x_i^{-})$. Likewise form the entire family of admissible future segments for $K_2$, and use [L5] to take a finite subcover by sets $I^{-}(y_j^{+})$ with $y_j\in K_2$ and $J^{-}(K_2)\subseteq {\bigcup }_{j=1}^n{J}^{-}(y_j^{+})$. Therefore $J^{+}(K_1)\cap J^{-}(K_2)\subseteq {\bigcup }_{i,j}(J^{+}(x_i^{-})\cap J^{-}(y_j^{+})),$ a finite union of compact causal diamonds, hence compact by [L5]; and $J^{+}(K_1)\cap J^{-}(K_2)$ is closed by step 4.1. A closed subset of a compact set is compact [L5], so $J^{+}(K_1)\cap J^{-}(K_2)$ is compact.