Three Points in Ultrametric Space have Two Equal Distances/Corollary

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

Let $x, y, z \in X$.

Then:


 * at least two of the distances $d \paren {x,y}$, $d \paren {x,z}$ and $d \paren {y,z}$ are equal.

Proof
Either:
 * $d \paren {x,z} = d \paren {y,z}$ or $d \paren {x,z} \ne d \paren {y,z}$.

By All "triangles" in Non-Archimedean Metric Space are Isoceles then:
 * $d \paren {x,z} = d \paren {y,z}$ or $d \paren {x,y} = \max \set {d \paren {x,z}, d \paren {y,z} }$

In either case two of the distances are equal.