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.
$\blacksquare$