Standard Bounded Metric is Metric

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $\bar d: A^2 \to \R$ be the standard bounded metric of $d$:
 * $\forall \tuple {x, y} \in A^2: \map {\bar d} {x, y} = \min \set {1, \map d {x, y} }$

Then $\bar d$ is a metric for $A$.

Proof
It is to be demonstrated that $\bar d$ satisfies all the metric space axioms.

Proof of
So holds for $\bar d$.

Proof of
Suppose that $\map d {x, y} < 1$ and $\map d {y, z} < 1$.

Then:

Now suppose that either $\map d {x, y} > 1$ or $\map d {x, y} > 1$.

, suppose $\map d {x, y} > 1$.

Then:

The same argument applies for $\map d {y, z} > 1$

So holds for $\bar d$.

Proof of
So holds for $\bar d$.

Proof of
So holds for $\bar d$.

Thus $\bar d$ satisfies all the metric space axioms and so is a metric.