Isometry between Metric Spaces is Continuous
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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 $\phi: M_1 \to M_2$ be an isometry.
Then $\phi: M_1 \to M_2$ is a continuous mapping.
Corollary
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.
Proof
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 $H$.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Lemma $7.5$