# Composition of Distance-Preserving Mappings is Distance-Preserving

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## Theorem

Let:

- $\struct {X_1, d_1}$
- $\struct {X_2, d_2}$
- $\struct {X_3, d_3}$

be metric spaces.

Let:

- $\phi: \struct {X_1, d_1} \to \struct {X_2, d_2}$
- $\psi: \struct {X_2, d_2} \to \struct {X_3, d_3}$

be distance-preserving mappings.

Then the composite of $\phi$ and $\psi$ is also a distance-preserving mapping.

## Proof

Let $x,y \in X_1$.

Then:

\(\ds \map {d_1} {x,y}\) | \(=\) | \(\ds \map {d_2} {\map \phi x, \map \phi y}\) | $\phi$ is a distance-preserving mapping | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {d_3} {\map \psi {\map \phi x}, \map \psi {\map \phi y} }\) | $\psi$ is a distance-preserving mapping | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {d_3} {\map {\psi \circ \phi} x, \map {\psi \circ \phi} y}\) | Definition of Composition of Mappings |

By definition of a distance-preserving mapping, $\psi \circ \phi$ is distance-preserving.

$\blacksquare$