# Equivalence of Definitions of Metrizable Topology/Lemma 1

## Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $M = \struct {A, d}$ be a metric space.

Let $\tau_d$ be the topology induced by $d$ on $A$.

Let $\phi : \struct{S, \tau} \to \struct{A, \tau_d}$ be a homeomorphism between topological spaces $\struct{S, \tau}$ and $\struct{A, \tau_d}$.

Let $d_\phi : S \times S \to \R_{\ge 0}$ be the mapping defined by:

$\forall s,t \in S: \map {d_\phi} {s,t} = \map d {\map \phi s, \map \phi t}$

Then:

$d_\phi$ is a metric on $S$

## Proof

##### $d_\phi$ satisfies Metric Axiom $(\text M 1)$

We have:

 $\ds \forall s \in S: \,$ $\ds \map {d_\phi} {s,s}$ $=$ $\ds \map d {\map \phi s, \map \phi s}$ Definition of $d_\phi$ $\ds$ $=$ $\ds 0$ Metric axiom $(\text M 1)$

Hence $d_\phi$ satisfies metric axiom $(\text M 1)$.

$\Box$

##### $d_\phi$ satisfies Metric Axiom $(\text M 2)$

We have:

 $\ds \forall s, t, u \in S: \,$ $\ds \map {d_\phi} {s,t}$ $=$ $\ds \map d {\map \phi s, \map \phi t}$ Definition of $d_\phi$ $\ds$ $\le$ $\ds \map d {\map \phi s, \map \phi u} + \map d {\map \phi u, \map \phi t}$ metric axiom $(\text M 1)$ $\ds$ $=$ $\ds \map {d_\phi} {s, u} + \map {d_\phi} {u, t}$ Definition of $d_\phi$

Hence $d_\phi$ satisfies metric axiom $(\text M 2)$.

$\Box$

##### $d_\phi$ satisfies Metric Axiom $(\text M 3)$

We have:

 $\ds \forall s, t \in S: \,$ $\ds \map {d_\phi} {s,t}$ $=$ $\ds \map d {\map \phi s, \map \phi t}$ Definition of $d_\phi$ $\ds$ $=$ $\ds \map d {\map \phi t, \map \phi s}$ Metric axiom $(\text M 3)$ $\ds$ $=$ $\ds \map d {\map \phi t, \map \phi s}$ Definition of $d_\phi$

Hence $d_\phi$ satisfies metric axiom $(\text M 3)$.

$\Box$

##### $d_\phi$ satisfies Metric Axiom $(\text M 4)$

By definition of homeomorphism:

$\phi$ is an injection

By definition of injection:

$\forall s, t \in S : s \ne t \implies \map \phi s \ne \map \phi t$

We have:

 $\ds \forall s, t \in S : s \ne t: \,$ $\ds \map {d_\phi} {s,t}$ $=$ $\ds \map d {\map \phi s, \map \phi t}$ Definition of $d_\phi$ $\ds$ $>$ $\ds 0$ Metric axiom $(\text M 4)$

Hence $d_\phi$ satisfies metric axiom $(\text M 4)$.

$\Box$

It follows that $d_\phi$ is a metric on $S$ by definition.

$\blacksquare$