Homeomorphic Image of Local Basis is Local Basis

Theorem
Let Let $T_\alpha = \struct{S_\alpha, \tau_\alpha}$ and $T_\beta = \struct{S_\beta, \tau_\beta}$ be topological spaces.

Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.

Let $s \in S_\alpha$.

Let $\BB$ be a local basis of $s$ in $T_\alpha$.

Then:
 * $\BB' = \set{ \phi \sqbrk B : B \in \BB}$ is a local basis of $\map \phi s$ in $T_\beta$

Proof
By definition of homeomorphism:
 * $\forall U \in \tau_\alpha : \phi \sqbrk U \in \tau_\beta$

Hence:
 * $\BB'$ is a set of open sets in $T_\beta$ containing $\map \phi s$

Let $U \in \tau_\beta$ containing $\map \phi s$.

By definition of homeomorphism:
 * $\phi^{-1} \sqbrk U \in \tau_\alpha$ containing $s$

By definition of local basis:
 * $\exists V \in \BB: s \in V \subseteq \phi^{-1} \sqbrk U$

By definition of image of subset:
 * $\map \phi s \in \phi \sqbrk V$

From Subset Maps to Subset:
 * $\phi \sqbrk V \subseteq \phi \sqbrk {\phi^{-1} \sqbrk U}$

From Image of Preimage under Mapping:
 * $\phi \sqbrk {\phi^{-1} \sqbrk U} \subseteq U$

From Subset Relation is Transitive]:
 * $\phi \sqbrk V \subseteq U$

Hence:
 * $\exists \phi \sqbrk V \in \BB': \map \phi s \in \phi \sqbrk V \subseteq U$