Homeomorphic Image of Neighborhood Basis is Neighborhood Basis
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 $\NN$ be a neighborhood basis of $s$ in $T_\alpha$.
Then:
- $\NN' = \set {\phi \sqbrk N : N \in \NN}$ is a neighborhood basis of $\map \phi s$ in $T_\beta$.
Proof
Let $N$ be a neighborhood of $s$ in $T_\alpha$.
By definition of neighborhood:
- $\exists V \in \tau_\alpha : s \in V \subseteq N$
By definition of image of subset:
- $\map \phi s \in \phi \sqbrk V$
From Subset Maps to Subset:
- $\phi \sqbrk V \subseteq \phi \sqbrk N$
By definition of homeomorphism:
- $\phi \sqbrk V \in \tau_\beta$
It follows that $\phi \sqbrk N$ is a neighborhood of $\map \phi s$ in $T_\beta$.
Hence:
- $\NN' = \set{ \phi \sqbrk N : N \in \NN}$ is a set of neighborhoods of $\map \phi s$ in $T_\beta$
Let $M$ be a neighborhood of $\map \phi s$ in $T_\beta$.
By definition of neighborhood:
- $\exists U \in \tau_\beta : \map \phi s \in U \subseteq M$
By definition of homeomorphism:
- $\phi^{-1} \sqbrk U \in \tau_\alpha$ containing $s$
By definition of neighborhood basis:
- $\exists N \in \NN: s \in N \subseteq \phi^{-1} \sqbrk U$
By definition of image of subset:
- $\map \phi s \in \phi \sqbrk N$
From Subset Maps to Subset:
- $\phi \sqbrk N \subseteq \phi \sqbrk {\phi^{-1} \sqbrk U}$
From Image of Preimage under Mapping:
- $\phi \sqbrk {\phi^{-1} \sqbrk U} \subseteq U$
From Subset Relation is Transitive:
- $\phi \sqbrk N \subseteq M$
Hence:
- $\exists \phi \sqbrk N \in \BB': \map \phi s \in \phi \sqbrk N \subseteq M$
The result follows.
$\blacksquare$