On continuously uniquely geodesic space II This question was inspired by this answer of @wspin.
Definition : A continuously uniquely geodesic space is a uniquely geodesic space whose geodesics vary continuously with endpoints.  

Question : Is a complete uniquely geodesic space, continuously uniquely geodesic ?

 A: Not a complete answer, but it might be useful:
First of all one needs to clarify what "geodesics vary continuous" means. I assume this means that for converging sequences $x_n \to x$ and $y_n \to y$ the corresponding sequence of geodesics $c_n$ from $x_n$ to $y_n$ converges to the geodesic $c$ connecting $x$ and $y$, that is if we parametrize $c_n$ and $c$ proportional to arc length on the interval $[0,1]$ we find that $c_n \to c$ uniformly.
The following seems correct: Let $(X,d)$ be a uniquely geodesic space. Then $X$ is continuous if $X$ is compact (which implies that $X$ is complete):
The proof is an easy consequence of the the Arzela Ascoli theorem for compact metric spaces: For a compact metric space $X$ any sequence of curves $c_n \subset X$ with uniformly bounded length contains a uniformly converging subsequence.
Now assume $X$ is not continuous. Choose a sequence of geodesics $c_n : [0,1] \to X$ such that the endpoints converge, $c_n(0) = x_n \to x$ and $c_n(1) = y_n \to y$, but $c_n$ does not converge to $c$, the geodesic from $x$ to $y$. We claim that there exists an $\varepsilon > 0$ and a subsequence $c_{n_k}$ of $c_n$ as well as a sequence $t_k \in [0,1]$ such that $$d(c_{n_k}(t_k),c(t_k)) > \varepsilon$$
for all $k$. It is easy to proof this claim from our assumption that $c_n$ does not converge to $c$. Maybe after taking another subsequence of $c_{n_k}$ we may assume that $c_{n_k} \to \tilde c$ for $k \to \infty$ by the Arzela-Ascoli theorem. Since geodesics converge to geodesics we obtain a contradiction as $\tilde c$ is a different geodesic connecting $x$ and $y$.
Some remarks:


*

*I have no idea whether this remains true for only complete spaces. But conversely to my previous guess i'd say no :)

*Observing this proof it is clear that it is enough to assume that for some $x \in X$ and all $r > 0$ the closed ball $\{y | d(x,y) \leq r\}$ is compact, which is weaker than compactness of $X$.

*A complete length space (a metric space where points are connected by geodesics) is not locally compact, i.e. small balls at a point need not be compact. See for example Complete implies locally compact in length metric space?

*A good reference for the Arzela-Ascoli thm used here is "Burago et alii: A course in metric geometry". Uniquely geodesic spaces are a special case of length spaces which are discussed there in great detail.
