Metric on Shift of Finite Type is Metric

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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 $X_\mathbf A$.

That is, $\struct {X _\mathbf A, d _\theta} $ is a metric space.

Proof

M1

Let $x\in X _\mathbf A$

Obviously, $x_i = x_i$ for all $i \in \openint {-\infty} \infty$.

Therefore:

$\map {d _\theta} {x,x} = \theta ^\infty = 0$.

$\Box$

M3

This is clear from the definition.

$\Box$

M4

Let $x,y\in X _\mathbf A$ such that $x\ne y$.

That is:

$\exists k\in\Z : x_k\ne y_k$.

Thus:

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

Therefore:

$\map {d _\theta} {x,x} = \theta ^N > 0$.

$\Box$

M2

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:

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

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:

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

so that:

$\map {d _\theta} {x,z} \leq \theta ^{N_1}$

Therefore:

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

$\blacksquare$