Equivalence of Definitions of Metrizable Topology/Lemma 2

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:
 * $\forall U \subseteq S : U$ is open in $\struct{S, d_\phi}$ $\phi \sqbrk U$ is open in $\struct{A, d}$

Lemma 3
Let $U \subseteq S$.

Necessary Condition
Let $U$ be open in $\struct{S, d_\phi}$.

By definition of open set:
 * $\forall s \in U: \exists \epsilon_s : \map {B_{\epsilon_s}} s \subseteq U$

From Lemma 3:
 * $\forall s \in U: \exists \epsilon_s : \map {B_{\epsilon_s}} {\map \phi s} = \phi \sqbrk {\map {B_{\epsilon_s}} s} \subseteq \phi \sqbrk U$

By definition of open set:
 * $\phi \sqbrk U$ is open in $\struct{A, d}$.

Sufficient Condition
Let $\phi \sqbrk U$ be open in $\struct{A, d}$.

By definition of open set:
 * $\forall s \in U: \exists \epsilon_s : \map {B_{\epsilon_s}} {\map \phi s} \subseteq \phi \sqbrk U$

From Lemma 3:
 * $\forall s \in U: \exists \epsilon_s : \phi \sqbrk {\map {B_{\epsilon_s}} s} = \map {B_{\epsilon_s}} {\map \phi s} \subseteq \phi \sqbrk U$

From Preimage of Subset is Subset of Preimage:
 * $\forall s \in U: \exists \epsilon_s : \phi^{-1} \sqbrk {\phi \sqbrk {\map {B_{\epsilon_s}} s}} \subseteq \phi^{-1} \sqbrk {\phi \sqbrk {U}}$

By definition of homeomorphism:
 * $\phi$ is an injection

From Preimage of Image of Subset under Injection equals Subset:
 * $\forall s \in U: \exists \epsilon_s : \map {B_{\epsilon_s}} s = \phi^{-1} \sqbrk {\phi \sqbrk {\map {B_{\epsilon_s}} s}} \subseteq \phi^{-1} \sqbrk {\phi \sqbrk {U}} = U$

By definition of open set:
 * $U$ is open in $\struct{S, d_\phi}$.