Supremum Metric on Continuous Real Functions is Metric/Proof 1

Theorem

Let $\closedint a b \subseteq \R$ be a closed real interval.

Let $\mathscr C \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$.

Let $d$ be the supremum metric on $\mathscr C \closedint a b$.

Then $d$ is a metric.

Proof

Let $\map {\mathscr B} {\closedint a b, \R}$ be the set of all bounded real functions $f: \closedint a b \to \R$.

From Supremum Metric on Continuous Real Functions is Subspace of Bounded, $\struct {\mathscr C \closedint a b, d_{\mathscr C} }$ is a (metric) subspace of $\struct {\map {\mathscr B} {\closedint a b, \R}, d}$.

The result follows from Subspace of Metric Space is Metric Space.

$\blacksquare$