externally convex M-space A metric space (X, d) is called externally convex  if for all distinct points
x, y such that $d(x, y) = \lambda$, and $r >\lambda $ there exists a unique z of X such that
$$d(x, y) + d(y, z) = d(x, z) = r.$$
A convex metric space (X, d) is called an M-space  if for every two points
x, y ∈ X with $d(x, y) = \lambda$, and for every $r ∈ [0, \lambda ]$, there exists a unique $z_r \in X$
such that $B[x, r] \cap B[y, \lambda − r] = {z_r}$, where $B[x, r] = \{y ∈ X : d(x, y) \leq r\}$ 
Is sequence space $C_0$ a externally convex M-space?
 A: Here's a hint (i.e. it doesn't answer your question, but perhaps you may be able to use it to find/deduce the answer).

Claim: The condition that the metric space $(X,d)$ is an M-space is equivalent to the following: for every two distinct points $x,y \in X$ such that $d(x,y)=\lambda$, and for every $r \in [0, \lambda]$, there exists a unique $z_r \in X$ such that $d(x,z_r) =r$ and $d(x,z_r)+d(z_r,y)=d(x,y)$.

Proof of claim: Assume that $(X,d)$ is an M-space. Then for every two distinct points $x,y \in X$ with $d(x,y)=\lambda$, $\lambda > 0$ because $x,y$ are distinct and $X$ is a metric space. Moreover, for every $r \in [0, \lambda]$, there exists a unique $z_r \in X$ such that $B[x,r] \cap B[y, \lambda -r] = z_r$. This means that $d(x,z_r) \le r$ and $d(y,z_r) \le \lambda - r$. By the triangle inequality, $\lambda = d(x,y) \le d(x,z_r)+d(z_r,y) \le r+ (\lambda-r) = \lambda$. Therefore, we have that $d(x,y)=d(x,z_r)+d(z_r,y)=\lambda$.
Moreover, by the triangle inequality, $d(x,z_r) \ge d(x,y)-d(z_r,y)$. Because $z_r \in B[y, \lambda -r]$, we have that $d(z_r, y) \le \lambda - r \iff -d(z_r,y) \ge r - \lambda$. So $d(x, z_r) \ge d(x,y) + r - \lambda = \lambda + r - \lambda = r \implies d(x, z_r) \ge r$. Since $z_r \in B[x,r]$, we also have that $d(x,z_r) \le r$, thus $r \le d(x,z_r) \le r \iff d(x,z_r) =r$.  Thus $X$ satisfies the condition stated in the claim (since remember that $X$ being an M-space means that $z_r$ is unique).
Conversely, assume that $(X,d)$ satisfies the condition in the claim. For any two points $x,y \in X$, we have two cases. Case 1: $x=y$. Then $d(x,y)=\lambda = 0$ (because $X$ is a metric space) and setting $z_r = x = y$ satisfies the condition of being an M-space trivially. Case 2: $x$ and $y$ are distinct. Then $d(x,y)=\lambda > 0$. By assumption, for every $r \in [0, \lambda]$, there exists a unique $z_r$ such that $d(x,z_r) = r$ and $d(x,z_r) + d(z_r,y) = d(x,y)$. It follows immediately that $d(z_r,y)= \lambda -r$. Therefore, $z_r \in B[x,r]$ and $z_r \in B[y,\lambda -r]$, so $z_r \in B[x,r] \cap B[y,\lambda -r]$.
Assume that there were another $\xi \in X$ such that $\xi \in B[x,r] \cap B[y,\lambda -r]$. Then $d(x,\xi)\le r$ and $d(\xi,y) \le \lambda - r$. Therefore $\lambda = d(x,y) \le d(x,\xi) + d(\xi,y) = r + \lambda -r = \lambda$. So $d(x,\xi)+d(\xi,y-r)= \lambda$. Since $d(\xi, y) \le \lambda -r, -d(\xi,y) \ge r-\lambda$. By the triangle inequality, $d(x,\xi) \ge d(x,y) - d(\xi,y) \ge \lambda + r -\lambda =r$, i.e. $d(x,\xi) \ge r$. Since $\xi \in B[x,r]$, $d(x,\xi) \le r$. So $r \le d(x,\xi) \le r \iff d(x,\xi)=r$. But then by the uniqueness in our assumption (i.e. since $d(x,\xi) =r $ and $d(x,\xi) + d(\xi,y) = d(x,y)$), we have that $\xi = z_r$. Since $\xi \in B[x,r] \cap B[y, \lambda -r]$ was arbitrary, it follows that any element of $B[x,r] \cap B[y, \lambda -r]$ is equal to $z_r$. Therefore we have shown the necessary uniqueness of $z_r$ to be able to conclude that $(X,d)$ is in fact an M-space. $\square$

Thus a metric space $(X,d)$ is an externally convex M-space if and only if, for every two distinct points $x,y \in X$:

*

*for $r \in [0, d(x,y)]$, there exists a unique $z_r$ such that $d(x,z_r)=r$ and $d(x,z_r)+d(z_r,y)=d(x,y)$.

*for $r \in (d(x,y),\infty)$, there exists a unique $z_r$ such that $d(x,z_r)=r$ and $d(x,z_r)=d(x,y)+d(y,z_r)$.


