Isometry between Metric Spaces is Continuous

Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $\phi: M_1 \to M_2$ be an isometry.

Then $\phi: M_1 \to M_2$ is a continuous mapping.

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $\phi: M_1 \to M_2$ be an isometry.

Then its inverse $\phi^{-1}: M_2 \to M_1$ is a continuous mapping.

Let $a \in A_1$.

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

Let $\delta = \epsilon$.

Then:

$\ds \map {d_1} {a, y}$

$<$

$\ds \delta$

for some $y \in A_1$

$\ds \leadsto \ \ $

$\ds \map {d_2} {\map \phi a, \map \phi y}$

$<$

$\ds \delta$

as $\map {d_2} {\map \phi a, \map \phi y} = \map {d_1} {a, y}$

$\ds $

$=$

$\ds \epsilon$

So by definition $\phi$ 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: Lemma $7.5$