Supremum Metric on Bounded Real Functions on Closed Interval is Metric/Proof 1

Theorem

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

Let $A$ be the set of all bounded real functions $f: \closedint a b \to \R$.

Let $d: A \times A \to \R$ be the supremum metric on $A$.

Then $d$ is a metric.

Proof

The interval is an instance of a set.

Hence Supremum Metric on Bounded Real-Valued Functions is Metric can be directly applied.

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$