Lemma $14.2(5)$ of O'Neill's book Semi-Riemannian geometry with applications to relativity states the following:
If $\mathcal{C}$ is a convex open set in $M$, then, a causal curve $\alpha$ contained in a compact subset $K$ of $\mathcal C$ is (continuously) extendible,
meaning by that (Def. $1.55$) that if $\alpha:[0,B) \rightarrow M$, $B\leq \infty$, then, $\alpha$ has a continuous extension $\tilde \alpha:[0,B]\rightarrow M$, with $q=\tilde\alpha(B)$ called an endpoint.
The proof is constructed by contradiction and it looks clear to me until the last point. In short, it goes like this: let us consider two different sequences $\{s_i\}\rightarrow B$ and $\{t_i\}\rightarrow B$, with $\{\alpha(s_i)\}\rightarrow p$ and $\{\alpha(t_i)\}\rightarrow q$. Because $K$ is compact $p,q \in K$. Now, let us assume that $p\neq q$. Then, because both sequences tend to $B$, we can choose sub-sequences $\{s_k\}\rightarrow B$ and $\{t_k\}\rightarrow B$ so that $s_k < t_k < s_k+1$. Thus, it turns out that $p\in J^+(q,\mathcal C)$ and analogously $q\in J^+(p,\mathcal C)$ so, it is concluded, $p = q$.
My objection is that $p\in J^+(q,\mathcal C)$ and $q\in J^+(p,\mathcal C)$ implies $p=q$ provided that there aren't closed causal curves in M, isn't it? Otherwise it could well happen that $J^+(q) = J^+(p)$ with $p \neq q$, as the following (counter) example shows:
Let $M$ be $ \mathbb R\times S^1$ described by coordinates $(t,x)$ with the points $(t,x)$ and $(t,x+1)$ identified, and with the following Lorentzian metric $$ ds^2 = -2 dtdx , $$ and the causal (time-like) curve $\alpha: [0,\infty)\rightarrow M$, given by $$ \alpha(s) = (\tanh s, s) . $$ In that case, let us consider $K = [0,1]\times S^1$ and the following sequences (e.g.) $$ s_n = \frac{1}{2} + n \rightarrow \infty \ \ , \ \ t_n = \frac{1}{3} + n \rightarrow \infty , $$ for which $$ \alpha(s_n) \rightarrow (1, \frac{1}{2}) \equiv p \ \ , \ \ \alpha(t_n) \rightarrow (1, \frac{1}{3}) \equiv q , $$ and, $p\neq q$. What is wrong with my reasoning? Thanks in advance
The main point is to understand what convexity means in this context. We can follow O'Neill's definitions for that. He defines a convex open set as being one that is a normal neighborhood of each of its points (Def. 5.5). A normal neighborhood $U$ of some $p\in M$ must such that $\mathrm{exp}_p\colon V\to U$ is a diffeomorphism for some starshaped neighborhood $V$ of $0$ in $T_pM$.
Essentially, this is already enough to see that the compact set $ K = [0,1]\times S^1$ in the given example can never be contained in a convex open set, because of the topology of $K$. But we can make this more concrete:
(cf. O'Neil Prop. 3.31) If $q$ is in a normal neighborhood $U$ of some $p\in U\subset M$, then there is a unique geodesic segment $\sigma\colon [0,1]\to U$, such that $\sigma(0) = p$ and $\sigma(1) = q$.
Now in the given example consider $\gamma_1\colon [0,1]\to K$ and $\gamma_{-1}\colon [0,1]\to K$ given by $\gamma_1(t) = (0,t)$ and $\gamma_2(t)=(0,-t)$. It is easy to check that these are two (lightlike) geodesic segments contained in $K$. Due to the circular nature of the $x$-coordinate, we have $\gamma_1(1) = \gamma_{-1}(1) = (0,0)$, so $(0,0)$ is joined to itself by at least two different geodesic segments. So $K$ cannot be contained in a convex open set as it contains points joined by non-unique geodesic segments.
All this burns down to the fact that O'Neil Lemma 14.2 is about "very simple" subsets of Lorentzian manifolds (convex ones), which exhibit the same simple causality structure as Minkowski space. More complex situations (as in the given example) are not part of this Lemma, as they are ruled out by convexity.