Two-endpoint limit curve theorem: bounded lengths give joining limits, unbounded lengths give two inextendible tails with the closure relation

(Common-domain reparametrization in the bounded positive case.) Suppose $0<b<\mathrm{\infty }$. Define ${\tilde{\alpha }}_k(t):={\alpha }_k\text{\hspace{0.17em}⁣}\left(\frac{b_k}{b}t\right),0\le t\le b\mathrm{.}$ Since $b_k\mathrm{/}b\to 1$, after discarding finitely many terms these curves are 2-Lipschitz for $d_h$: the arclength parametrization gives $d_h({\tilde{\alpha }}_k(s),{\tilde{\alpha }}_k(t))\le {\mathrm{\ell }}_h({\tilde{\alpha }}_k{\mathrm{\mid }}_{[s,t]})=\frac{b_k}{b}\mathrm{\mid }t-s\mathrm{\mid }\le 2\mathrm{\mid }t-s\mathrm{\mid }\mathrm{.}$ Their initial points converge to $x$, and their images lie in one closed $h$-ball about $x$ of radius $2b+1$, which is compact by [L4]. Applying [L6] gives a subsequence converging uniformly on $[0,b]$ to a 2-Lipschitz map $\alpha$. The sharper inequality $d_h({\tilde{\alpha }}_k(s),{\tilde{\alpha }}_k(t))\le \frac{b_k}{b}\mathrm{\mid }t-s\mathrm{\mid }$ and $b_k\mathrm{/}b\to 1$ imply $d_h(\alpha (s),\alpha (t))\le \mathrm{\mid }t-s\mathrm{\mid }$. Uniform convergence at the endpoints gives $\alpha (0)=x$ and $\alpha (b)=z$.

(The limit is future causal.) Let $[s,t]$ be a parameter interval such that $\alpha ([s,t])$ lies in a strictly convex normal set $C$. Since this image is compact and $C$ is open, uniform convergence implies ${\tilde{\alpha }}_k([s,t])\subset C$ for all sufficiently large $k$. By Equivalence of the absolutely-continuous and continuous causal conventions, and one-sided inextendible extensions clause (a), each approximating causal subarc has a future-causal position vector $w_k={\exp }_{{\tilde{\alpha }}_k(s)}^{-1}({\tilde{\alpha }}_k(t))\mathrm{.}$ Smoothness of the two-point inverse on $C\times C$ gives $w_k\to w={\exp }_{\alpha (s)}^{-1}(\alpha (t))$; closedness of the future cone gives that $w$ is future causal or zero. Thus $\alpha$ is geodesically connectible. It is locally Lipschitz in charts by its $d_h$-Lipschitz bound and [L7], hence locally absolutely continuous. At almost every differentiability parameter $t<b$, choose a relatively compact adapted chart $U$ around $\alpha (t)$ and $h_0>0$ so $\alpha ([t,t+h])\subset U$ for $0<h\le h_0$. For all sufficiently large $k$ the approximating arcs also lie in $U$ and have $h$-length $(b_k\mathrm{/}b)h$. By [L7], $\frac{b_k}{b}h\le C(x^0({\tilde{\alpha }}_k(t+h))-x^0({\tilde{\alpha }}_k(t)))$ for a chart constant $C$. Passing to the limit and using $b_k\mathrm{/}b\to 1$ gives $x^0(\alpha (t+h))-x^0(\alpha (t))\ge h\mathrm{/}C\mathrm{.}$ At differentiability points this yields $(x^0\circ \alpha {)}^{\mathrm{\prime }}(t)\ge 1\mathrm{/}C>0$. For small $h>0$ the position vector ${\exp }_{\alpha (t)}^{-1}(\alpha (t+h))$ is future causal or zero by the preceding connectibility argument; divided by $h$ it tends in coordinates to ${\alpha }^{\mathrm{\prime }}(t)$, because the differential of the second-variable inverse at the diagonal is the identity (The joint exponential map: smoothness, invertible differential at the diagonal, and the small-velocity two-point inverse clauses (b),(c)). The future cone is closed and ${\alpha }^{\mathrm{\prime }}$ has positive time component, so ${\alpha }^{\mathrm{\prime }}(t)$ is nonzero future causal almost everywhere. This proves $\alpha$ is a future causal curve and is not constant on any nondegenerate interval.

(The future tail in the unbounded case.) Suppose $b=\mathrm{\infty }$. For each fixed $T>0$, $[0,T]$ is contained in $[0,b_k]$ for all large $k$. Applying clause (b) of [L1] by diagonal extraction over integer $T$ gives a subsequence converging uniformly on every compact subinterval of $[0,\mathrm{\infty })$ to a future-inextendible future causal curve ${\alpha }^{+}:[0,\mathrm{\infty })\to M$ with ${\alpha }^{+}(0)=x$.

(The past tail in the unbounded case.) Define ${\delta }_k(s):={\alpha }_k(b_k+s)$ on $[-b_k,0]$. For each fixed $T>0$, $[-T,0]$ lies in $[-b_k,0]$ for all large $k$. Apply the time-reversed half-line result of [L1] and diagonalize to obtain a past-inextendible future causal curve ${\alpha }^{-}:(-\mathrm{\infty },0]\to M$ with ${\alpha }^{-}(0)=z$, such that ${\delta }_k$ converges uniformly to ${\alpha }^{-}$ on compact subintervals. A nested subsequence retains both tail limits.

(The zero-length case.) If $b=0$, then $d_h(x,z)\le d_h(x,{\alpha }_k(0))+d_h({\alpha }_k(0),{\alpha }_k(b_k))+d_h({\alpha }_k(b_k),z)\le o(1)+b_k+o(1),$ so $x=z$. For every $t\in [0,b_k]$, $d_h({\alpha }_k(t),x)\le d_h({\alpha }_k(0),x)+t\le o(1)+b_k\mathrm{.}$ Hence the curves converge uniformly on their varying domains to the constant curve at $x$.

(Imprisonment.) If the spacetime is strongly causal and all images lie in a compact $K$, [L5] gives $b_k\le L(K)$ for every $k$. Taking limits yields $b\le L(K)<\mathrm{\infty }$, excluding the unbounded case.

(The closure relation.) Fix $t_{+}\ge 0$ and $t_{-}\le 0$. For all sufficiently large $k$, $t_{+}\le b_k+t_{-}$, so by [L2], ${\alpha }_k(t_{+})\le {\alpha }_k(b_k+t_{-})={\delta }_k(t_{-})\mathrm{.}$ The uniform-on-compact limits in steps 1.3-1.4 give endpoints converging respectively to ${\alpha }^{+}(t_{+})$ and ${\alpha }^{-}(t_{-})$. This is the stated sequential closure relation.

(Coincident endpoints.) If $x=z$ and $0<b<\mathrm{\infty }$, step 1.2 gives a future causal curve $\alpha$ from $x$ to $x$ which is nonconstant on every nondegenerate interval. If $x=z$ and $b=\mathrm{\infty }$, concatenate ${\alpha }^{-}$ and ${\alpha }^{+}$ at their common endpoint $x$. The concatenation is future causal by [L3], and is past- and future-inextendible because its respective tails are; hence it is inextendible.

(Conclusion.) Clause (i) for $b>0$ is steps 1.1-1.2 and for $b=0$ is step 1.5; clause (ii) is steps 1.3, 1.4 and 2.1; clause (iii) is step 2.2; and clause (iv) is step 1.6.