Let $(M,g)$ be a connected, smooth, strongly causal Lorentzian manifold, and consider an inextendible causal geodesic $\sigma : [0,b) \to M$ (a priori, $b$ may be $\infty$) with the following property:
For every sequence $(s_n)_{n=1}^\infty \subset [0,b)$ such that $s_n \to b$, there exists a subsequence $(s_{n_k})$ and an inextendible, incomplete geodesic $\gamma:[0,1) \to M$ such that there's a sequence $(t_k) \subset [0,1)$ with $t_k \to 1$ and $\sigma(s_{n_k}) = \gamma(t_k)$. In words: given any sequence whatsoever along the end of $\sigma$, there's an incomplete geodesic that intersects $\sigma$ infinitely many times along some subsequence.
Question: Must $\sigma$ be incomplete, i.e. must $b$ be finite?
Notes:
Of course, if $\sigma$ is incomplete, the property is satisfied trivially by taking $\gamma$ to be the appropriate reparameterization of $\sigma$.
Inextendible here means that $\displaystyle \lim_{s \to b} \sigma(s)$ does not exist (and similarly for $\gamma$), so that $\sigma$ cannot be future-extended as continuous curve (or, equivalently, as a smooth geodesic).
While I'm not sure if the causality and inextendibility conditions are necessary for the result, together they imply that $\sigma(s_n)$ can never converge, and hence that the image of $\sigma$ is closed, and further that $\sigma$ eventually exits and remains outside of any compact set.
My intuition is that the answer is positive, but I'm not sure how to go about a proof-- it seems one would want to somehow bound the parameter of $\sigma$ in terms of that of some $\gamma$.
If absolutely necessary, I'd also be interested in a positive answer under the additional hypothesis that the $\gamma$ are causal.
Edit: Since no answers have been given after a bit of time, I've now cross-posted this question on MathOverflow.