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 $\struct{U, \tau_U}$

$\blacksquare$