Supremum Metric is Metric

Theorem
Let $S$ be a set.

Let $M = \struct {A', d'}$ 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:


 * $\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x}$

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

First note that we have:

and so the exists.

Proof of
So holds for $d$.

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

Let $x, y \in S$.

Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:
 * $X := \set {\map {d'} {\map f c, \map g c}: c \in X}$

So:
 * $\map d {f, g} + \map d {g, h} \ge \sup X = \map d {f, h}$

So holds for $d$.

Proof of
So holds for $d$.

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

Suppose $f, g \in A: \map d {f, g} = 0$.

Then:

So holds for $d$.