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 $\struct {S, \tau}$ 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
We note that by definition $d: M \to \R$ is a metric on $M$.
Hence $d$ satisfies all the metric space axioms.
$d_\phi$ satisfies Metric Space 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 $d_\phi$ | ||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | as $d$ satisfies Metric Space Axiom $(\text M 1)$ |
Hence $d_\phi$ satisfies Metric Space Axiom $(\text M 1)$.
$\Box$
$d_\phi$ satisfies Metric Space Axiom $(\text M 2)$: Triangle Inequality
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 $d_\phi$ | ||||||||||
\(\ds \) | \(\le\) | \(\ds \map d {\map \phi s, \map \phi u} + \map d {\map \phi u, \map \phi t}\) | as $d$ satisfies Metric Space Axiom $(\text M 2)$: Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {d_\phi} {s, u} + \map {d_\phi} {u, t}\) | Definition of $d_\phi$ |
Hence $d_\phi$ satisfies Metric Space Axiom $(\text M 2)$: Triangle Inequality.
$\Box$
$d_\phi$ satisfies Metric Space 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 $d_\phi$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \map d {\map \phi t, \map \phi s}\) | as $d$ satisfies Metric Space Axiom $(\text M 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map d {\map \phi t, \map \phi s}\) | Definition of $d_\phi$ |
Hence $d_\phi$ satisfies Metric Space Axiom $(\text M 1)$.
$\Box$
$d_\phi$ satisfies Metric Space 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 $d_\phi$ | ||||||||||
\(\ds \) | \(>\) | \(\ds 0\) | as $d$ satisfies Metric Space Axiom $(\text M 4)$ |
Hence $d_\phi$ satisfies Metric Space Axiom $(\text M 4)$.
$\Box$
It follows that $d_\phi$ is a metric on $S$ by definition.
$\blacksquare$