# Composition of Distance-Preserving Mappings is Distance-Preserving

## 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:

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

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

\(\displaystyle \) | \(=\) | \(\displaystyle d_3 \paren {\map {\psi \circ \phi} x, \map {\psi \circ \phi} y }\) | $\quad$ Definition of composite mappings | $\quad$ |

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

$\blacksquare$