Local Basis Generated from Neighborhood Basis

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $x$ be an element of $S$.

Let $\BB$ be a neighborhood basis of $x$.

For any subset $A \subseteq S$, let $A^\circ$ denote the interior of $A$.

Then the set:
 * $\BB' = \set {H^\circ: H \in B}$

is a local basis of $x$.

Proof
First it must be shown that $\BB'$ is a set of open neighborhoods of $x$.

From the definition of the interior of a subset, $\BB'$ is a set of open sets.

Let $H \in \BB$.

By assumption, $H$ is a neighborhood of $x$.

From the definition of a neighborhood:
 * $\exists U \in \tau : x \in U \subseteq H$

From the definition of the interior of a subset:
 * $U \subseteq H^\circ$

Hence $x \in H^\circ$.

It follows that $\BB'$ is a set of open neighborhoods of $x$.

It remains to show that $\BB'$ is a neighborhood basis.

Let $N$ be a neighborhood of $x$.

Since $\BB$ is a neighborhood basis of $x$ then:
 * $\exists H \in \BB : H \subseteq N$

By definition of the interior, $H^\circ \subseteq H$.

From Subset Relation is Transitive $H^\circ \subseteq N$.

Hence $\BB'$ is a neighborhood basis.

The result follows.