Metric on Shift of Finite Type is Non-Archimedean

Theorem

Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.

Let $\theta \in \openint 0 1$.

Then the metric $d_\theta$ on $X_\mathbf A$ is non-archimedean metric.

That is, $\struct {X_\mathbf A, d _\theta} $ is an ultrametric space.

Proof

Let $x, y, z \in X_\mathbf A$.

Let:

$N_1 := \sup \set {n \in \N \cup \set \infty : x_i = y_i \text { for all } i \in \openint {-n} n}$

and:

$N_2 := \sup \set {n \in \N \cup \set \infty : y_i = z_i \text { for all } i \in \openint {-n} n}$

so that:

$\map {d_\theta} {x, y} = \theta^{N_1}$

and:

$\map {d_\theta} {y, z} = \theta^{N_2}$

Let $M := \min \set {N_1, N_2}$.

Then for all $i \in \openint {-M} M$:

$x_i = y_i = z_i$

Hence:

$\sup \set {n \in \N \cup \set \infty : x_i = z_i \text { for all } i \in \openint {-n} n} \ge M$

so that:

$\map {d_\theta} {x, z} \le \theta ^M$

On the other hand:

$\theta^M = \theta^{\min \set {N_1, N_2} } = \max \set {\theta^{N_1}, \theta^{N_2} }$

Therefore:

$\map {d_\theta} {x, z} \le \max \set {\theta^{N_1}, \theta^{N_2} } = \max \set {\map {d_\theta} {x, y}, \map {d_\theta} {y, z} }$

$\blacksquare$