Supremum Metric on Bounded Real-Valued Functions is Metric/Proof 1

Theorem

Let $X$ be a set.

Let $A$ be the set of all bounded real-valued functions $f: X \to \R$.

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

Then $d$ is a metric.

Proof

We have that the supremum metric on $A \times A$ is defined as:

$\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in X} \size {\map f x - \map g x}$

where $f$ and $g$ are bounded real-valued functions.

From Real Number Line is Metric Space, the real numbers $\R$ together with the absolute value function form a metric space.

The result follows by Supremum Metric is Metric.

$\blacksquare$