# How to prove the triangle inequality for this distance?

I'm studying a proof in 'An Introduction to Metric Spaces and Fixed Point Theory' (M. Khamsi, W. Kirk) that shows the equivalence of injectiveness and hyperconvexity for metric spaces. I stumbled over the following part of the proof.

Let $(X, d)$ be an injective metric space and suppose that $\{B^*(x_i, r_i)\mid i\in I\}$ is a set of closed balls each two of which intersect. Define $Y=\{x_i\mid i\in I\}$ and $Z=Y\cup \{z\}$ with $z\notin Y$. Define the distance on $Z$ by $$d'(x_i, x_j) = d(x_i, x_j) \quad\text{and}\quad d'(x_i, z) = \inf{\{r\mid\exists j\in I\colon B^*(x_j, r_j) \subseteq B^*(x_i, r)\}}.$$ I now have to show that $d'$ has the triangle inequality. I've already shown that $$\forall k, l\in I\colon d'(x_k, x_l)\leq d'(x_k, z)+d'(x_l, z),$$ but I'm stuck with the proof for the second non-trivial case: $$\forall k, l\in I\colon d'(x_k, z)\leq d'(x_k, x_l)+d'(x_l, z).$$

I tried to distinguish two cases. If $d'(x_k, x_l)>r_k$, then it follows immediately that $$d'(x_k, z)\leq r_k < d'(x_k, x_l)<d'(x_k, x_l)+d'(x_l, z).$$ What is the proof for the other case $d'(x_k, x_l)\leq r_k$? Is it even necessary to make this distinguishment?

• I might have an issue with understanding your definition of $d'(x_i,z)$. For instance, if the initial family contained all the balls centered in $x_i$ with radius $r\leq1$, then $d'(x_i,z)$ would be defined to be zero
– user228113
Commented Apr 19, 2015 at 19:34

Let us define

$$\rho(i,j) := \inf \{ r > 0 : B^\ast(x_j,r_j) \subseteq B^\ast(x_i,r)\}.$$

The triangle inequality for $d$ yields

$$\rho(k,j) \leqslant d(x_k,x_l) + \rho(l,j),$$

so the desired inequality follows when we have shown

$$d'(x_i,z) = \inf \{ \rho(i,j) : j\in I\}.\tag{1}$$

By definition of $\rho(i,j)$, for all $r > \rho(i,j)$ we have $B^\ast(x_j,r_j) \subseteq B^\ast(x_i,r)$, hence $d'(x_i,z) \leqslant r$, and therefore $d'(x_i,z) \leqslant \rho(i,j)$. Since that holds for all $j\in I$, we have

$$d'(x_i,z) \leqslant \inf \{\rho(i,j) : j\in I\}.$$

On the other hand, if $r > d'(x_i,z)$, then there is a $k\in I$ with $B^\ast(x_k,r_k) \subseteq B^\ast(x_i,r)$, whence $\rho(i,k) \leqslant r$ and a fortiori $\inf \{\rho(i,j) : j \in I\} \leqslant r$. This holds for all $r > d'(x_i,z)$, hence

$$\inf \{\rho(i,j) : j\in I\} \leqslant d'(x_i,z),$$

and $(1)$ is shown.

• Thank you for your answer. How did you prove $$\rho(k,j) \leqslant d(x_k,x_l) + \rho(l,j)?$$ Commented Apr 23, 2015 at 7:43
• Let $r > \rho(l,j)$. Then look at the balls $B^\ast(x_l,r)$ and $B^\ast(x_k, d(x_k,x_l)+r)$. The former is contained in the latter by the triangle inequality for $d$. Hence $d(x_k,x_l)+r \geqslant \rho(k,j)$. Commented Apr 23, 2015 at 9:52