Subspace of Metric Space is Metric Space

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $H \subseteq A$.

Let $d_H: H \times H \to \R$ be the restriction $d \restriction_{H \times H}$ of $d$ to $H$.

Let $\struct {H, d_H}$ be a metric subspace of $\struct {A, d}$.

Then $d_H$ is a metric on $H$.

By definition of restriction:

$\forall x, y \in H: \map {d_H} {x, y} = \map d {x, y}$

As $d$ is a metric, the metric space axioms are all fulfilled by all $x, y \in A$ under $d$.

As $H \subseteq A$, by definition of subset, all $x, y \in H$ are also elements of $A$.

Therefore the metric space axioms are all fulfilled by all $x, y \in H$ under $d_H$.

$\blacksquare$

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces