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:

\(\displaystyle d_1 \paren {x,y}\)

\(=\)

\(\displaystyle d_2 \paren {\map \phi x, \map \phi y }\)

$\phi$ is a distance-preserving mapping

\(\displaystyle \)

\(=\)

\(\displaystyle d_3 \paren {\map \psi {\map \phi x}, \map \psi {\map \phi y} }\)

$\psi$ is a distance-preserving mapping

\(\displaystyle \)

\(=\)

\(\displaystyle d_3 \paren {\map {\psi \circ \phi} x, \map {\psi \circ \phi} y }\)

Definition of composite mappings

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

$\blacksquare$