Supremum Metric on Continuous Real Functions is Subspace of Bounded
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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 $\map {\mathscr B} {\closedint a b, \R}$ be the set of all bounded real functions $f: \closedint a b \to \R$.
Let $d$ be the supremum metric on $\map {\mathscr B} {\closedint a b, \R}$.
Then $\struct {\mathscr C \closedint a b, d_{\mathscr C} }$ is a subspace of $\struct {\map {\mathscr B} {\closedint a b, \R}, d}$.
Proof
Let $f \in \mathscr C \closedint a b$.
Then by Image of Closed Real Interval is Bounded, $f$ is bounded on $\closedint a b$.
Thus $f \in \map {\mathscr B} {\closedint a b, \R}$ and the result follows.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.8$