A development sharing the initial Cauchy surface with a larger development is causally convex in it

(Setup and the maximal portion of $\nu$ in $U$.) Let $\nu :[a,b]\to M$ be a future causal curve with $\nu (a)\in \mathrm{\Sigma }$ and $\nu (b)\in U\cap I^{+}(\mathrm{\Sigma })$, and suppose for contradiction that $\nu ([a,b])\subseteq ̸U$. Since $\nu$ is continuous [L3] and $U$ is open, ${\nu }^{-1}(U)$ is relatively open in $[a,b]$ and contains $a$ and $b$; let $J$ be the connected component of ${\nu }^{-1}(U)$ containing $b$. Then $J=(c,b]$ for some $c\in [a,b)$, and $\nu ((c,b])\subseteq U$ while $\nu (c)\mathrm{\notin }U$; moreover $c>a$ because $\nu (a)\in \mathrm{\Sigma }\subseteq U$.

(A causal curve in $U$ that meets $\mathrm{\Sigma }$ exactly once.) The restriction $\beta :=\nu {\mathrm{\mid }}_J$ is a future causal curve in $U$ (Common subdevelopments, maximal common developments and corresponding boundary points clause (a)); as $t\to c$ from above, $\beta (t)\to \nu (c)\mathrm{\notin }U$, so $\beta$ does not converge to a point of $U$ at its past end, while at its future end it converges to $\nu (b)\in U$. Choose a future-directed timelike $w\in T_{\nu (b)}U$ (it exists because $N{\mathrm{\mid }}_U$ is a time orientation); by [L2] there is a future-directed causal geodesic $\rho :I_{\rho }\to U$ maximal in $U$ with $\rho (0)=\nu (b)$ and ${\rho }^{\mathrm{\prime }}(0)=w$, and $\rho$ converges to no point of $U$ at its future end. Define $\xi$ on the interval obtained by gluing $(c,b]$ and $[0,\sup I_{\rho })$ at $b\sim 0$ by $\xi (t):=\nu (t)$ for $t\le b$ and $\xi (t):=\rho (t-b)$ for $t\ge b$. By [L3] $\xi$ is a future causal curve in $U$; at its past end it does not converge in $U$ (limit $\nu (c)\mathrm{\notin }U$) and at its future end it does not converge in $U$ ([L2]), so $\xi$ is inextendible in $U$; since $\mathrm{\Sigma }$ is a Cauchy set of $U$ ([L1]), $\xi$ meets $\mathrm{\Sigma }$ in exactly one parameter value.

(Clause (b) in the fixed time orientation.) Let $\nu :[a,b]\to M$ be future causal with $\nu (a)\in U\cap I^{-}(\mathrm{\Sigma })$ and $\nu (b)\in \mathrm{\Sigma }$, and suppose it leaves $U$. The component of ${\nu }^{-1}(U)$ containing $a$ is $[a,c)$ for some $c<b$, with $\nu (c)\mathrm{\notin }U$. Choose a future timelike maximal geodesic $\rho$ in $U$ through $\nu (a)$, and concatenate its past half ending at $\nu (a)$ with $\nu {\mathrm{\mid }}_{[a,c)}$. The resulting future causal curve is inextendible in $U$: its past geodesic end has no limit there by [L2], and its future end converges in $M$ to $\nu (c)\mathrm{\notin }U$. Every point on the geodesic half is in $I^{-}(\mathrm{\Sigma })$ by push-up from $\nu (a)\in I^{-}(\mathrm{\Sigma })$; every point of $\nu ([a,c))$ is in $J^{-}(\mathrm{\Sigma })\setminus \mathrm{\Sigma }=I^{-}(\mathrm{\Sigma })$, since the remaining segment reaches $\nu (b)\in \mathrm{\Sigma }$ and [L1] gives the Cauchy-set trichotomy and acausality. Thus this inextendible causal curve in $U$ misses $\mathrm{\Sigma }$, contradicting the Cauchy property in [L1]. This proves (b) without changing $N$ or the matter data.

(The image of $\xi$ lies in $I^{+}(\mathrm{\Sigma })$.) For $t\in (c,b]$: $\nu (t)$ is the endpoint of the future causal segment $\nu {\mathrm{\mid }}_{[a,t]}$ started at $\nu (a)\in \mathrm{\Sigma }$, so $\nu (t)\in J^{+}(\mathrm{\Sigma })$; if $\nu (t)\in \mathrm{\Sigma }$ then the distinct points $\nu (a),\nu (t)\in \mathrm{\Sigma }$ would be causally related, contradicting [L5], so $\nu (t)\mathrm{\notin }\mathrm{\Sigma }$ and hence $\nu (t)\in J^{+}(\mathrm{\Sigma })\setminus \mathrm{\Sigma }=I^{+}(\mathrm{\Sigma })$ by [L1]. For $s\in [0,\sup I_{\rho })$: $\rho (s)\in J^{+}(\nu (b))$, and $\nu (b)\in I^{+}(\mathrm{\Sigma })$; by push-up [L4] $J^{+}(\nu (b))\subseteq I^{+}(\mathrm{\Sigma })$, so $\rho (s)\in I^{+}(\mathrm{\Sigma })$. Thus $\xi ((c,b+\sup I_{\rho }))\subseteq I^{+}(\mathrm{\Sigma })$, which is disjoint from $\mathrm{\Sigma }$.

(Contradiction and conclusion of (a).) By step 2.1 the causal curve $\xi$ of step 1.2 misses $\mathrm{\Sigma }$, contradicting its established property of meeting $\mathrm{\Sigma }$ exactly once. Hence $\nu ([a,b])\subseteq U$, which is clause (a).

(Clause (c).) Let $\nu :[a,b]\to M$ have both endpoints in $U$. [step 3.1, step 1.3, L1, L3, L4, L5, algebra] If $a=b$, there is nothing to prove. Otherwise use the disjoint trichotomy $M=I^{-}(\mathrm{\Sigma })\bigsqcup \mathrm{\Sigma }\bigsqcup I^{+}(\mathrm{\Sigma })$ of [L1]. If both endpoints are in $I^{+}(\mathrm{\Sigma })$, choose a future timelike curve from a point of $\mathrm{\Sigma }$ to $\nu (a)$ and concatenate it with $\nu$; clause (a) applies to this entire curve ending at $\nu (b)\in U\cap I^{+}(\mathrm{\Sigma })$. If both endpoints are in $I^{-}(\mathrm{\Sigma })$, choose a future timelike curve from $\nu (b)$ to a point of $\mathrm{\Sigma }$ and append it to $\nu$. Clause (b) applies to that concatenation, starting at $\nu (a)\in U\cap I^{-}(\mathrm{\Sigma })$. In either case its restriction $\nu$ lies in $U$. A future causal curve cannot go from $I^{+}(\mathrm{\Sigma })$ to $\mathrm{\Sigma }\cup I^{-}(\mathrm{\Sigma })$, or from $\mathrm{\Sigma }$ to $I^{-}(\mathrm{\Sigma })$: push-up would give a chronological relation between points of $\mathrm{\Sigma }$, contradicting [L1]. Both endpoints on $\mathrm{\Sigma }$ are excluded for a nonconstant segment by [L5]. If the first endpoint lies on $\mathrm{\Sigma }$ and the second in $I^{+}(\mathrm{\Sigma })$, apply (a); if the first lies in $I^{-}(\mathrm{\Sigma })$ and the second on $\mathrm{\Sigma }$, apply (b). The remaining case has endpoints in $I^{-}(\mathrm{\Sigma })$ and $I^{+}(\mathrm{\Sigma })$. Continuity and connectedness force an intermediate value $\nu (c)\in \mathrm{\Sigma }$, for otherwise the image would lie in the disjoint open union of those two sides. Apply (b) to $\nu {\mathrm{\mid }}_{[a,c]}$ and (a) to $\nu {\mathrm{\mid }}_{[c,b]}$. Thus every allowable case gives $\nu ([a,b])\subseteq U$, proving (c).