Three Points in Ultrametric Space have Two Equal Distances

Theorem
Let $\struct {X, d}$ be an ultrametric space.

Let $x, y, z \in X$ with $\map d {x, z} \ne \map d {y, z}$.

Then:


 * $\map d {x, y} = \max \set {\map d {x, z}, \map d {y, z} }$

Proof
, let $\map d {x, z} > \map d {y, z}$.

Then:

On the other hand:

Putting the two inequalities together then:
 * $\map d {x, y} = \map d {x, z} = \max \set {\map d {x, z}, \map d {y, z} }$