Let $X$ be a complete and separable metric space, let $G(X) \subset C([0,1],X)$ be the space of continuous curves from $[0,1]$ to $X$ with constant speed, i.e.
$$ d(f(t),f(s)) = |t-s| d(f(0), f(1)). $$
Let $S : X^2 \rightarrow 2^{G(X)}$ be such that at every couple of points $(x,y)$ it associates a curve starting at $x$, i.e. $f(0) = x$, and ending at $y$, i.e. $f(1) = y$. We define the preimage of a set $T \subseteq G(X)$ through $S$ as
$$ S^{-1}(T) := \lbrace (x,y) \in X^2 : F(x,y) \cap T \neq \emptyset \rbrace. $$


It can be proven that $G(X)$ endowed with the distance
$$\delta(f,g) = \sup\_{x \in [0,1]} d(f(x), g(x))$$
is still a complete and separable metric space.


Now take $A \subset G(X)$ an open set. Is $S^{-1} (\lbrace A \rbrace) $ an open set? Is it at least a Borel set?




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Take now, in the same situation as before, $S : X^2 \rightarrow 2^{G(X)}$ s.t. at every couple of points $(x,y)$ it associates **all the curves** with constant speed starting at $x$, i.e. $f(0) = x$, and ending at $y$, i.e. $f(1) = y$.


In this situation is the preimage through $S$ of an open subset of $G(X)$ open or Borel in $X^2$?



Open set of geodesics implies the set of starting points is open
It need not be even Borel. Take $X = \mathbb{R}^2$ with $d((x\_1,x\_2),(y\_1,y\_2)) = \max(|x\_1-y\_1|,|x\_2-y\_2|)$. Then it is easy to construct a selection $S$ with $S^{-1}(A)$ not Borel for some open set $A$. (For example select for a suitable non-Borel set of doubles of points a non-Euclidean geodesic and for the rest a Euclidean one.)


