Equivalence of Definitions of Metrizable Topology/Lemma 3

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:

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.