Neighborhood Basis of Open Subspace iff Neighborhood Basis
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Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $U \subseteq S$ be an open subset
Let $\tau_U$ denote the subspace topology on $U$.
Let $s \in U$.
Let $\NN \subseteq \powerset U$.
Then:
- $\NN$ is a neighborhood basis of $s$ in $\struct{U, \tau_U}$ if and only if $\NN$ is a neighborhood basis of $s$ in $\struct{S, \tau}$
Proof
Let $\map \NN s$ denote the set of neighborhoods of $s$ in $\struct{S, \tau}$
Let $\map \MM s$ denote the set of neighborhoods of $s$ in $\struct{U, \tau_U}$
From Neighborhood in Open Subspace:
- $\NN \subseteq \map \NN s$ if and only if $\NN \subseteq \map \MM s$
Necessary Condition
Let $\NN$ be a neighborhood basis of $s$ in $\struct{U, \tau_U}$.
Let $K \in \map \NN s$.
By definition of neighborhood:
- $\exists W \in \tau : s \in W \subseteq K$
By definition of subspace topology:
- $V = W \cap U \in \tau_U$
From Set is Open iff Neighborhood of all its Points:
- $V \in \map \MM s$
By definition of neighborhood basis:
- $\exists N \in \NN : x \in N \subseteq V$
From Subset Relation is Transitive:
- $N \subseteq K$
We have:
- $\exists N \in \NN : x \in N \subseteq K$
It follows that:
- $\NN$ is a neighborhood basis of $s$ in $\struct{S, \tau}$
$\Box$
Sufficient Condition
Let $\NN$ be a neighborhood basis of $s$ in $\struct{S, \tau}$.
Let $K \in \map \MM s$.
From Neighborhood in Open Subspace:
- $K \in \map \NN s$.
Hence:
- $\exists N \in \NN : x \in N \subseteq K$
It follows that:
- $\NN$ is a neighborhood basis of $s$ in $\struct{U, \tau_U}$
$\blacksquare$