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

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Theorem

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

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

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