# Isometry of Metric Spaces is Equivalence Relation

## Theorem

Let $M_1$ and $M_2$ be metric spaces.

Let $M_1 \sim M_2$ denote that $M_1$ and $M_2$ are isometric.

The relation $\sim$ is an equivalence relation.

## Proof

Checking in turn each of the criteria for equivalence:

### Reflexivity

Let $M$ be a topological space.

From Identity Mapping on Metric Space is Isometry, the identity mapping $I_M: M \to M$ is an isometry.

So $M \sim M$, and $\sim$ has been shown to be reflexive.

$\Box$

### Symmetry

Let $M_1$ and $M_2$ be topological spaces such that $M_1 \sim M_2$.

By definition, there exists an isometry $f: M_1 \to M_2$.

From Inverse of Isometry of Metric Spaces is Isometry it follows that $f^{-1}: M_2 \to M_1$ is also an isometry.

So $M_2 \sim M_1$, and $\sim$ has been shown to be symmetric.

$\Box$

### Transitivity

Let $M_1, M_2, M_3$ be metric spaces such that $M_1 \sim M_2$ and $M_2 \sim M_3$.

By definition, there exist isometries $f: M_1 \to M_2$ and $g: M_2 \to M_3$.

From Composite of Isometries is Isometry it follows that $g \circ f: M_1 \to M_3$ is also an isometry.

So $M_1 \sim M_3$, and $\sim$ has been shown to be transitive.

$\Box$

$\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$

## Sources

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 2.7$: Subspaces and Equivalence of Metric Spaces