Equivalence of Definitions of Metrizable Topology/Lemma 1

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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.

$\blacksquare$

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