# 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:

 $\ds \size {\map f x - \map g x}$ $=$ $\ds \size {\map f x + \paren {-\map g x} }$ $\ds$ $\le$ $\ds \size {\map f x} + \size {\paren {-\map g x} }$ Triangle Inequality for Real Numbers $\ds$ $=$ $\ds \size {\map f x} + \size {\map g x}$ Definition of Absolute Value $\ds$ $\le$ $\ds K + L$

and so the right hand side exists.

### Proof of Metric Space Axiom $\text M 1$

 $\ds \map d {f, f}$ $=$ $\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map f x}$ Definition of $d$ $\ds$ $=$ $\ds \sup_{x \mathop \in S} \size 0$ as $d'$ fulfils Metric Space Axiom $\text M 1$ $\ds$ $=$ $\ds 0$

So Metric Space Axiom $\text M 1$ holds for $d$.

$\Box$

### Proof of Metric Space Axiom $\text M 2$

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

Let $x, y \in S$.

 $\ds c$ $\in$ $\ds S$ $\ds \leadsto \ \$ $\ds \map {d'} {\map f c, \map h c}$ $\le$ $\ds \map {d'} {\map f c, \map g c} + \map {d'} {\map g c, \map h c}$ as $d'$ fulfils Metric Space Axiom $\text M 2$ $\ds$ $\le$ $\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x} + \sup_{x \mathop \in S} \map {d'} {\map g x, \map h x}$ Definition of Supremum of Real-Valued Function $\ds$ $=$ $\ds \map d {f, g} + \map d {g, h}$ Definition of $d$

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 Metric Space Axiom $\text M 2$ holds for $d$.

$\Box$

### Proof of Metric Space Axiom $\text M 3$

 $\ds \map d {f, g}$ $=$ $\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x}$ Definition of $d$ $\ds$ $=$ $\ds \sup_{x \mathop \in S} \map {d'} {\map g x, \map f x}$ as $d'$ fulfils Metric Space Axiom $\text M 3$ $\ds$ $=$ $\ds \map d {g, f}$ Definition of $d$

So Metric Space Axiom $\text M 3$ holds for $d$.

$\Box$

### Proof of Metric Space Axiom $\text M 4$

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:

 $\ds \map d {f, g}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x}$ $=$ $\ds 0$ Definition of $d$ $\ds \leadsto \ \$ $\ds \forall x \in S: \,$ $\ds \map {d'} {\map f x, \map g x}$ $=$ $\ds 0$ as $d'$ fulfils Metric Space Axiom $\text M 1$ and Metric Space Axiom $\text M 4$ $\ds \leadsto \ \$ $\ds f$ $=$ $\ds g$ as $d'$ fulfils Metric Space Axiom $\text M 4$

So Metric Space Axiom $\text M 4$ holds for $d$.

$\blacksquare$