G-spaces and manifolds In his book "The geometry of geodesics" H. Busemann defines the notion of a G-space to be a space which satisfies the following axioms:


*

*The space is metric

*The space is finitely compact, i.e., a bounded infinite set has at least one accumulation point

*[metric convexity] For every $x\neq z$ there exists a third point $y$ different from $x$ and $z$ such that $d(x,y)+d(y,z)=d(x,z)$

*[local prolongation] To every point $p$ there corresponds $\rho_p>0$ such that for every two point $x,y\in S(p,\rho_p)$ there exists a point $z$ such that $d(x,y)+d(y,z)=d(x,z)$

*[uniqueness of prolongation] If $d(x,y)+d(y,z_1)=d(x,z_1)$ and $d(x,y)+d(y,z_2)=d(x,z_2)$ and $d(y,z_1)=d(y,z_2)$ then $z_1=z_2$.


Busemann conjectured that every $G$-space is a topological manifold. My question is does every topological/smooth/Riemannian manifold is also a $G$-space?
As for connected complete Riemannian manifold, I figured out that 1 holds since by the metric. 3 holds since every two points can be joined by a minimal geodesic, and then we can pick $y$ to be a point on it. 4 holds since it is a manifold and locally it is homeomorphic to some Euclidean space. Unfortunately, even in this case, I couldn't figure out 5 and 2.
 A: If I am not wrong, I believe that every topological manifold admits a complete metric. With this metric, it is possible to give a Path Metric Space structure (in the sense of Gromov, see the book Metric Structures for Riemannian and non-Riemannian spaces). I guess that this structure allows to make the same arguments as for Riemannian manifolds for topological manifolds. 
Anyway, I recomend this survey about Busemann conjecture which also discusses a stronger conjecture (the Bing-Borsuk conjecture). 
A: On a *complete*smooth Riemannian manifold,


*

*Any bounded (with respect to the distance function induced by the Riemannian metric) closed set in a manifold is compact.

*This is telling you that there is a minimal geodesic joining $x$ to $y$ that, when extended, is also a minimal geodesic joining $x$ to $z_1$. And there is another minimal geodesic joining $x$ to $y$ that when extended is a minimal geodesic joining $x$ to $z_2$. But if there are two distinct geodesics joining $x$ to $y$, neither is minimal beyond $y$. So the two geodesics have to be the same and therefore $z_1 = z_2$.
CORRECTION: "complete" added to assumption above.
For a smooth manifold, you need to construct a distance function to get a G-space. One way to do this is to construct a complete Riemannian metric. I'm not certain that this can be done, but offhand if you take a locally finite covering by open sets diffeomorphic to the Euclidean ball, use the standard Euclidean metric on each ball (where each ball has radius $1$), and use a partition of unity subordinate to this cover to glue together these metrics, it seems to me that the resulting metric is complete.
For a topological manifold, I don't know. 
