Isometry of Metric Spaces is Homeomorphism

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Theorem

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

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


Then $f$ is a homeomorphism from $M_1$ to $M_2$.


Proof

By the definition of an isometry, $f$ is a bijection $f: A_1 \to A_2$ such that:

$\forall a, b \in A_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$


By Isometry between Metric Spaces is Continuous, $f$ is a continuous mapping from $M_1$ to $M_2$.

By the corollary to Isometry between Metric Spaces is Continuous, $f^{-1}$ is a continuous mapping from $M_2$ to $M_1$.


Thus $f$ is a homeomorphism from $M_1$ to $M_2$.

Hence the result.

$\blacksquare$