Neighborhood Basis of Open Subspace iff Neighborhood Basis

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 $\struct{U, \tau_U}$ $\NN$ 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 User:Leigh.Samphier/Topology/Neighborhood in Open Subspace:
 * $\NN \subseteq \map \NN s$ $\NN \subseteq \map \MM s$

Necessary Condition
Let $\NN$ be a neighborhood basis of $s$ in $\struct{U, \tau_U}$.

By definition of neighborhood basis:
 * $\forall M \in \map \MM s : \exists N_M \in \NN : x \in N_M \subseteq M$

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 assumption that $\NN$ is a neighborhood basis of $s$ in $\struct{U, \tau_U}$:
 * $\exists N_V \in \NN : x \in N_V \subseteq V$

From Subset Relation is Transitive:
 * $N_V \subseteq K$

We have:
 * $\exists N_V \in \NN : x \in N_V \subseteq K$

It follows that:
 * $\NN$ is a neighborhood basis of $s$ in $\struct{S, \tau}$

Sufficient Condition
Let $\NN$ be a neighborhood basis of $s$ in $\struct{S, \tau}$.

By definition of neighborhood basis:
 * $\forall M \in \map \NN s : \exists N_M \in \NN : x \in N_M \subseteq M$

Let $K \in \map \MM s$.

From User:Leigh.Samphier/Topology/Neighborhood in Open Subspace:
 * $K \in \map \NN s$.

Hence:
 * $\exists N_K \in \NN : x \in N_K \subseteq K$

It follows that:
 * $\NN$ is a neighborhood basis of $s$ in $\struct{U, \tau_U}$