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.

$\blacksquare$