# Equivalence of Definitions of Metrizable Topology

## Theorem

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

The following definitions of the concept of Metrizable Topology are equivalent:

### Definition 1

$T$ is said to be metrizable if and only if there exists a metric $d$ on $S$ such that:

$\tau$ is the topology induced by $d$ on $S$.

### Definition 2

$T$ is said to be metrizable if and only if there exists a metric space $M = \struct{A, d}$ such that:

$T$ is homeomorphic to the topological space $\struct{A, \tau_d}$

where $\tau_d$ is the topology induced by $d$ on $A$.

## Proof

### Definition 1 implies Definition 2

Let $d$ be a metric on $S$ such that $\tau$ is the topology induced by $d$.

$T$ is homeomorphic to a topological space with a topology induced by a metric.

$\Box$

### Definition 2 implies Definition 1

Let $M = \struct{A, d}$ be a metric space such that $T$ is homeomorphic to $\struct{A,\tau_d}$ where $\tau_d$ is the topology induced by $d$.

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

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

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

#### Lemma 1

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

$\Box$

#### Lemma 2

$\forall U \subseteq S : U$ is open in $\struct{S, d_\phi}$ if and only if $\phi \sqbrk U$ is open in $\struct{A, d}$

$\Box$

It remains to show that $\tau$ is the topology induced by the metric $d_\phi$.

We have:

 $\ds U \text{ is open in } \struct{S, \tau}$ $\leadstoandfrom$ $\ds \phi \sqbrk U \text{ is open in } \struct{A, \tau_d}$ Definition of Homeomorphism $\ds$ $\leadstoandfrom$ $\ds \phi \sqbrk U \text{ is open in } \struct{A, d}$ Definition of Topology Induced by Metric $\ds$ $\leadstoandfrom$ $\ds U \text{ is open in } \struct{S, d_\phi}$ Lemma 2

Hence $\tau$ is a topology induced by a metric by definition.

$\blacksquare$