Standard Bounded Metric is Metric

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

Let $d_2: A^2 \to \R$ be the mapping defined as:
 * $\forall \tuple {x, y} \in A^2: \map {d_2} {x, y} = \min \set {1, \map d {x, y} }$

Then $d_2$ is a metric for $A$.

Proof
It is to be demonstrated that $d_2$ satisfies all the metric space axioms.

Proof of
So holds for $d_2$.

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 $d_2$.

Proof of
So holds for $d_2$.

Proof of
So holds for $d_2$.

Thus $d_2$ satisfies all the metric space axioms and so is a metric.