Three Points in Ultrametric Space have Two Equal Distances

Theorem
Let $\struct {X,d}$ be a non-Archimedean metric space.

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

Then:


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

Proof
Without loss of generality, suppose:
 * $d \paren {x,z} > d \paren {y,z} $

Then:

On the other hand:

Putting the two inequalities together it follows:
 * $d \paren {x,y} = d \paren {x,z} = \max \set {d \paren {x,z}, d \paren {y,z} }$