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:

\(\displaystyle d_1 \paren {x,y}\) \(=\) \(\displaystyle d_2 \paren {\map \phi x, \map \phi y }\) $\phi$ 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.

$\blacksquare$