Homeomorphic Metric Spaces are not necessarily Isometric

Theorem

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

Let $M_1$ and $M_2$ be homeomorphic.

Then it is not necessarily the case that $M_1$ and $M_2$ are isometric.

Consider the spaces $\struct {\openint 0 4, d}$ and $\struct {\openint 0 1, d}$, where $d$ is the Euclidean Metric.

Let $\phi: \openint 0 4 \to \openint 0 1$ be a mapping.

Then:

$\ds \map d {\map \phi 1, \map \phi 3}$

$\le$

$\ds \map d {0, 1}$

$\ds $

$=$

$\ds 1$

$\ds $

$<$

$\ds 2$

$\ds $

$=$

$\ds \map d {1, 3}$

showing that $\phi$ cannot be an isometry.

Therefore the two spaces above are not isometric.

$\blacksquare$

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces