# Equivalence of Definitions of Metrizable Topology

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## 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 $d$.

From Identity Mapping is Homeomorphism:

- $T$ 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.

$\blacksquare$