Supremum Metric on Continuous Real Functions is Metric/Proof 2

Proof
Let $A := \mathscr D^r \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$ which are of differentiability class $r$.

Let $d_r: A \times A \to \R$ be the supremum metric on $A$, defined as:
 * $\ds \forall f, g \in A: \map d {f, g} := \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {g^{\paren i} } x}$

where:
 * $f$ and $g$ are continuous functions on $\closedint a b$ which are $r$ times differentiable
 * $r \in \N$ is a natural number.

From Supremum Metric on Differentiability Class is Metric, $d_r$ is a metric on $A$.

By definition, real functions which are continuous on $\closedint a b$ are of differentiability class $0$.

Thus setting $r = 0$ it follows that $\mathscr C \closedint a b$ is $\mathscr D^0 \closedint a b$.

The result follows.