Metric on Shift of Finite Type is Metric

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

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

Then the metric $d_\theta$ is indeed a metric on $X_\mathbf A$.

Well-defined
Since $x,y\in X_\mathbf A$ are sequences, we have:
 * $x\ne y$ $\exists i\in\Z : x_i\ne y_i$

Thus, for $x\neq y$:
 * $\max \set {n \in \N : x_i = y_i \text { for all } i \in \openint {-n} n}$

exists.

Therefore the mapping
 * $d _\theta : X_\mathbf A \times X_\mathbf A \to \R$

is well-defined.

M1, M3, M4
These follow directly from the definition.

M2
Let $x,y,z\in X _\mathbf A$.

If $x=y$ or $y=z$, then $\map {d _\theta} {x,y} = 0$ or $\map {d _\theta} {y,z} = 0$, respectively.

Hence,
 * $\map {d _\theta} {x,z} = \map {d _\theta} {x,y} + \map {d _\theta} {y,z}$

and (M2) is satisfied.

Now, let $x\ne y$ and $y\ne z$.

Let:
 * $N_1 := \max \set {n \in \N : x_i = y_i \text { for all } i \in \openint {-n} n}$

and:
 * $N_2 := \max \set {n \in \N : y_i = z_i \text { for all } i \in \openint {-n} n}$

so that:
 * $\map {d _\theta} {x,y} = \theta ^{N_1}$

and:
 * $\map {d _\theta} {y,z} = \theta ^{N_2}$

Without loss of generality, say $N_1 \leq N_2$.

Then for all $i \in \openint {-N_1} {N_1}$:
 * $x_i = y_i = z_i$

Hence:
 * $\max \set {n \in \N : x_i = z_i \text { for all } i \in \openint {-n} n} \geq N_1$

so that:
 * $\map {d _\theta} {x,z} \leq \theta ^{N_1}$

Therefore:
 * $\map {d _\theta} {x,z} \leq \theta ^{N_1} \leq \theta ^{N_1} + \theta ^{N_2} = \map {d _\theta} {x,y} + \map {d _\theta} {y,z}$