Distance-Preserving Image Isometric to Domain for Metric Spaces

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Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $\phi: M_1 \to M_2$ be a distance-preserving mapping.


Then:

$\phi: M_1 \to \Img \phi$

is an isometry.


Proof

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $\phi$ be a distance-preserving mapping from $M_1$ to $M_2$.

Let $A = \Img \phi$ be the image of $\phi$.

By Subspace of Metric Space is Metric Space, $\struct {A, d_2}$ is a metric space.

As $\phi$ is a distance-preserving mapping, by Distance-Preserving Mapping is Injection of Metric Spaces it is injective.

From Restriction of Injection is Injection, $\phi: M_1 \to \Img \phi$ is an injection.

From Restriction of Mapping to Image is Surjection, $\phi: M_1 \to \Img \phi$ is a surjection.

Thus $\phi \to \Img \phi$ is by definition a bijection.

Thus $\phi: M_1 \to \Img \phi$ is a bijective distance-preserving mapping.

Hence, by definition, $\phi: M_1 \to \Img \phi$ is an isometry.

$\blacksquare$