Inclusion Mapping on Metric Space is Continuous

Theorem

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

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

Then the inclusion mapping $i_H: H \to A$ is continuous.

Let $a \in H$.

Let $\epsilon \in \R_{>0}$.

Let $\delta = \epsilon$.

Then:

$\ds \map {d_H} {a, y}$

$<$

$\ds \delta$

for some $y \in H$

$\ds \leadsto \ \ $

$\ds \map d {\map {i_H} a, \map {i_H} y}$

$=$

$\ds \map d {a, y}$

$\ds $

$=$

$\ds \map {d_H} {a, y}$

$\ds $

$<$

$\ds \delta$

$\ds $

$=$

$\ds \epsilon$

So by definition $d_H$ is continuous at $a$.

As $a \in H$ is arbitrary, it follows that $d_H$ is continuous on $H$.

$\blacksquare$

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Theorem $7.2$