Composition of Distance-Preserving Mappings is Distance-Preserving
Jump to navigation
Jump to search
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:
\(\ds d_1 \paren {x,y}\) | \(=\) | \(\ds d_2 \paren {\map \phi x, \map \phi y }\) | $\phi$ is a distance-preserving mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds d_3 \paren {\map \psi {\map \phi x}, \map \psi {\map \phi y} }\) | $\psi$ is a distance-preserving mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds d_3 \paren {\map {\psi \circ \phi} x, \map {\psi \circ \phi} y }\) | Definition of composite mappings |
By the definition of a distance-preserving mapping then $\psi \circ \phi$ is distance-preserving.
$\blacksquare$