# Small codimension 1 ball on the boundary of metric ball in Busemann G-spaces

Let $(X,d)$ be a metric space. $X$ is said to be a Busemann $G$-space provided it satisfies the following axioms:

(1) Menger Convexity: Given distinct points $x,y\in X$, there is a point $z\in X-\{x,y\}$, so that $d(x,z)+d(z,y)=d(x,y)$.

(2) Finite Compactness: Every $d$-bounded infinite set has at least one accumulation point.

(3) Local Extendibility: For each point $p\in X$, there is a positive radius $\delta$, such that for any pair of distinct points $x,y\in B_p(\delta)$, there is a point $z\in \operatorname{Int}B_p(\delta)-\{x,y\}$ such that $d(x,y)+d(y,z)=d(x,z)$, where $B_p(\delta)$ denotes a closed metric ball centered at $p$ with radius $\delta$ and $\operatorname{Int}B_p(\delta)$ denotes its interior.

(4) Uniqueness of the Extension: Given distinct points $x,y\in X$, if there are points $z_1,z_2\in X$ for which both $$d(x,y)+d(y,z_i)=d(x,z_i) \text{ for }i=1,2$$ and $$d(y,z_1)=d(y,z_2)$$ hold, then $z_1=z_2$.

Question: Is sufficiently small (closed) codimension 1 metric ball on $\operatorname{Bd}B_p(\delta)$ with induced metric from $X$ also a Busemann $G$-space? If not, how about with intrinsic metric?

The ball with induced metric generally will not be a G-space. Take $R^2$ with the Euclidean metric, and consider the unit ball $S^1$ with the induced metric. Let $x=(0,1)$ and $y=(1,0)$; then $d(x,y)=\sqrt{2}$ in that metric, but there is no $z$ in $S^1$ with $d(x,z)+d(z,y)=d(x,y)$. So Menger convexity fails and that ball is not a G-space.