Proving that being CAT(0) implies being geodesic, my professor used the following fact:
If for every pair of points $x$ and $y$ in a complete metric space $X$ there exists a point $m \in X$ such that $d(x,m) = d(y,m) = d(x,y)/2$, then $X$ is a geodesic space.
I don't get how to use that m in creating the geodesic
Moreover, if such $m$ is unique, does it imply that the space is uniquely geodesic? I.e., does being CAT(0) implies being uniquely geodesic?
Thanks for your help!
If you are studying CAT(0) spaces, or geometry of metric spaces generally, the book A Course in Metric Geometry by Burago, Burago, and Ivanov should (ideally) always be in your vicinity. The statement is Theorem 2.4.16 there. Here is the proof.
Given $a,b$ at distance $L=d(a,b)$, we want an isometry $\gamma:[0,L]\to X$ such that $\gamma(0)=a$ and $\gamma(L)=b$. Note that we already have defined $\gamma$ on the subset $\{0,L\}$. Expand the definition iteratively: $\gamma(L/2)$ is a midpoint between $\gamma(0)$ and $\gamma(L)$, and so forth. Observe that $\gamma$ is an isometry from its domain into $X$. (Use the triangle inequality). So far we have $\gamma$ defined at all numbers of the form $kL/2^m$ with $m=0,1,2,\dots$ and $0\le k\le 2^m$. Extend to all of $[0,L]$ by Lipschitz continuity. The isometric property is preserved. So, we have a geodesic.
Also, uniqueness of midpoints implies the uniqueness of geodesics. Indeed, if midpoints are unique, then any other geodesic $\tilde \gamma$ would have to agree with our $\gamma$ at the points $kL/2^m$, and since they are dense, we have $\tilde \gamma\equiv \gamma$.