# Supremum Metric on Bounded Continuous Mappings is Metric

## Theorem

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $A$ be the set of all continuous mappings $f: M_1 \to M_2$ which are also bounded.

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

Then $d$ is a metric.

## Proof

The set $A$ is a subset of the set $A'$ of all bounded mappings $f: M_1 \to M_2$.

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

From Supremum Metric is Metric, $\left({A', d'}\right)$ is a metric space.

By definition, $A$ is a metric subspace of $A'$.

Hence the result.

$\blacksquare$