Supremum Metric on Continuous Real Functions is Metric/Proof 1

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Theorem

Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a closed real interval.

Let $\mathscr C \left[{a \,.\,.\, b}\right]$ be the set of all continuous functions $f: \left[{a \,.\,.\, b}\right] \to \R$.


Let $d$ be the supremum metric on $\mathscr C \left[{a \,.\,.\, b}\right]$.


Then $d$ is a metric.


Proof

Let $\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ be the set of all bounded real functions $f: \left[{a \,.\,.\, b}\right] \to \R$.

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

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

$\blacksquare$