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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?

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 $X^2$?

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    $\begingroup$ Even if $X$ is the unit circle with path-metric then there is no continuous map $S$, therefore, preimages of open sets will not be in general open. $\endgroup$
    – Misha
    Commented Oct 20, 2013 at 18:18
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    $\begingroup$ The question now makes less sense and should be edited: You have to explain which topology you put on the set of all subsets of $G(X)$ since your new map takes values not in $G(X)$ but in $2^{G(X)}$. If your topology is at all reasonable, the circle will provide a counter-example to the continuity question. $\endgroup$
    – Misha
    Commented Oct 20, 2013 at 18:34
  • $\begingroup$ I'm not interested in a topology on $2^{G(X)}$, I'm taking $S$ open in $G(X)$. $\endgroup$
    – User11111
    Commented Oct 20, 2013 at 19:56
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    $\begingroup$ Your map is $S: X^2\to 2^{G(X)}$, as you assign to $x,y$ the set of all geodesics connecting $x$ and $y$, just read what is written in your post. How are you planning to define continuity if you have no topology on the target space? $\endgroup$
    – Misha
    Commented Oct 21, 2013 at 3:44
  • $\begingroup$ @ User11111. If you have $S:X^2\rightarrow 2^{G(X)}$, then what do you mean by "preimage through $S$ of an open subset of $G(X)$" ? $\endgroup$
    – TaQ
    Commented Oct 21, 2013 at 3:58

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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.)

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  • $\begingroup$ I'm sorry, but what do you mean exactly by "Euclidean" geodesic? $\endgroup$
    – User11111
    Commented Oct 20, 2013 at 13:44
  • $\begingroup$ I mean a constant speed geodesic in the Euclidean distance $\sqrt{|x_1-y_1|^2+|x_2-y_2|^2}$ parametrized by the unit interval $[0.1]$. $\endgroup$
    – Tapio Rajala
    Commented Oct 20, 2013 at 14:16

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