Supremum Metric on Continuous Real Functions is Subspace of Bounded

From ProofWiki
Jump to: navigation, search

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


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


Then $\left({\mathscr C \left[{a \,.\,.\, b}\right], d_{\mathscr C} }\right)$ is a subspace of $\left({\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right), d}\right)$.


Proof

Let $f \in \mathscr C \left[{a \,.\,.\, b}\right]$.

Then by Image of Closed Real Interval is Bounded, $f$ is bounded on $\left[{a \,.\,.\, b}\right]$.

Thus $f \in \mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ and the result follows.

$\blacksquare$


Sources