Supremum Metric on Bounded Continuous Mappings is Metric
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Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $A$ be the set of all continuous mappings $f: M_1 \to M_2$ which are also bounded.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
Proof
The set $A$ is a subset of the set $A'$ of all bounded mappings $f: M_1 \to M_2$.
Let $d': A' \times A' \to \R$ be the supremum metric on $A'$.
From Supremum Metric is Metric, $\struct {A', d'}$ is a metric space.
By definition, $A$ is a metric subspace of $A'$.
Hence the result.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$