Equivalence of Definitions of Metrizable Topology/Lemma 1

From ProofWiki
Jump to navigation Jump to search


This article needs proofreading.
Please check it for mathematical errors.
If you believe there are none, please remove {{Proofread}} from the code.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Proofread}} from the code.


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 topological spaces $\struct{S, \tau}$ 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

$d_\phi$ satisfies Metric 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 $d_\phi$
\(\ds \) \(=\) \(\ds 0\) Metric axiom $(\text M 1)$

Hence $d_\phi$ satisfies metric axiom $(\text M 1)$.

$\Box$

$d_\phi$ satisfies Metric Axiom $(\text M 2)$

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 $d_\phi$
\(\ds \) \(\le\) \(\ds \map d {\map \phi s, \map \phi u} + \map d {\map \phi u, \map \phi t}\) metric axiom $(\text M 1)$
\(\ds \) \(=\) \(\ds \map {d_\phi} {s, u} + \map {d_\phi} {u, t}\) Definition of $d_\phi$

Hence $d_\phi$ satisfies metric axiom $(\text M 2)$.

$\Box$

$d_\phi$ satisfies Metric 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 $d_\phi$
\(\ds \) \(=\) \(\ds \map d {\map \phi t, \map \phi s}\) Metric axiom $(\text M 3)$
\(\ds \) \(=\) \(\ds \map d {\map \phi t, \map \phi s}\) Definition of $d_\phi$

Hence $d_\phi$ satisfies metric axiom $(\text M 3)$.

$\Box$

$d_\phi$ satisfies Metric 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 $d_\phi$
\(\ds \) \(>\) \(\ds 0\) Metric axiom $(\text M 4)$

Hence $d_\phi$ satisfies metric axiom $(\text M 4)$.

$\Box$


It follows that $d_\phi$ is a metric on $S$ by definition.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Metrizable_Topology/Lemma_1&oldid=616290"