Equivalence of Definitions of Metrizable Topology/Lemma 3
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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 topological spaces $\struct{S, \tau}$ and $\struct{A, \tau_d}$.
Let $d_\phi : S \times S \to \R_{\ge 0}$ be the metric defined by:
- $\forall s,t \in S: \map {d_\phi} {s,t} = \map d {\map \phi s, \map \phi t}$
Then:
- $\forall s \in S, \epsilon \in \R_{\ge 0} : \phi \sqbrk {\map {B_\epsilon} s} = \map {B_\epsilon} {\map \phi s}$
where
- $(1) \quad \map {B_\epsilon} s$ is the open ball in $\struct{S, d_\phi}$ with center $s$ and radius $\epsilon$
- $(2) \quad \map {B_\epsilon} {\map \phi s}$ is the open ball in $\struct{A, d}$ with center $\map \phi s$ and radius $\epsilon$
Proof
Let $s \in S$.
Let $\epsilon \in \R_{\ge 0}$.
We have:
\(\ds x \in \map {B_\epsilon} s\) | \(\leadstoandfrom\) | \(\ds \map {d_\phi} {s, x} < \epsilon\) | Definition of Open Ball in $\struct{S, d_\phi}$ | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \map d {\map \phi s, \map \phi x} < \epsilon\) | Definition of $d_\phi$ | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \map \phi x \in \map {B_\epsilon} {\map \phi s}\) | Definition of Open Ball in $\struct{A, d}$ | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds x \in \phi^{-1} \sqbrk {\map {B_\epsilon} {\map \phi s} }\) | Definition of Preimage of Mapping |
By set equality:
- $\map {B_\epsilon} s = \phi^{-1} \sqbrk {\map {B_\epsilon} {\map \phi s}}$
Hence:
- $\phi \sqbrk {\map {B_\epsilon} s} = \phi \sqbrk {\phi^{-1} \sqbrk {\map {B_\epsilon} {\map \phi s}}}$
By definition of homeomorphism:
- $\phi$ is a surjection
From Image of Preimage of Subset under Surjection equals Subset:
- $\phi \sqbrk {\phi^{-1} \sqbrk {\map {B_\epsilon} {\map \phi s}}} = \map {B_\epsilon} {\map \phi s}$
The result follows.
$\blacksquare$