Homeomorphic Image of Local Basis is Local Basis
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Theorem
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:
- $\exists \phi \sqbrk V \in \BB': \map \phi s \in \phi \sqbrk V \subseteq U$
$\blacksquare$