Supremum Metric is Metric

Theorem
Let $S$ be a set.

Let $M = \left({A', d'}\right)$ be a metric space.

Let $A$ be the set of all bounded mappings $f: S \to M$.

Let $d: A \times A \to \R$ be the supremum metric on $A$.

Then $d$ is a metric.

Proof
We have that the supremum metric on $A \times A$ is defined as:


 * $\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right)$

where $f$ and $g$ are bounded mappings.

First note that we have:

and so the RHS exists.

Proof of $M1$
So axiom $M1$ holds for $d$.

Proof of $M2$
Let $f, g, h \in A$.

Let $x, y \in S$.

Thus $d \left({f, g}\right) + d \left({g, h}\right)$ is an upper bound for:
 * $X := \left\{ {d' \left({f \left({c}\right), g \left({c}\right)}\right): c \in X}\right\}$

So:
 * $d \left({f, g}\right) + d \left({g, h}\right) \ge \sup X = d \left({f, h}\right)$

So axiom $M2$ holds for $d$.

Proof of $M3$
So axiom $M3$ holds for $d$.

Proof of $M4$
As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$:
 * $\forall f, g \in A: d \left({f, g}\right) \ge 0$

Suppose $f, g \in A: d \left({f, g}\right) = 0$.

Then:

So axiom $M4$ holds for $d$.