# Equivalence of Definitions of Local Basis

## Contents

## Theorem

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

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

The following definitions of the concept of **Local Basis** are equivalent:

### Local Basis for Open Sets

A **local basis** at $x$ is a set $\BB$ of open neighborhoods of $x$ such that:

- $\forall U \in \tau: x \in U \implies \exists H \in \BB: H \subseteq U$

That is, such that every open neighborhood of $x$ also contains some set in $\BB$.

### Neighborhood Basis of Open Sets

A **local basis** at $x$ is a set $\BB$ of open neighborhoods of $x$ such that every neighborhood of $x$ contains a set in $\BB$.

That is, a **local basis** at $x$ is a neighborhood basis of $x$ consisting of open sets.

## Proof

### Local Basis for Open Sets Implies Neighborhood Basis of Open Sets

Let $\mathcal B$ be a set of open neighborhoods of $x$ such that:

- $\forall U \in \tau: x \in U \implies \exists H \in \mathcal B: H \subseteq U$

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

Then there exists $U \in \tau$ such that $x \in U$ and $U \subseteq N$ by definition.

By assumption, there exists $H \in \mathcal B$ such that $H \subseteq U$.

From Subset Relation is Transitive, $H \subseteq N$.

The result follows.

$\Box$

### Neighborhood Basis of Open Sets Implies Local Basis for Open Sets

Let $\mathcal B$ be a set of open neighborhoods of $x$ such that:

- every neighborhood of $x$ contains a set in $\mathcal B$.

Let $U \in \tau$ such that $x \in U$.

From Set is Open iff Neighborhood of all its Points then $U$ is a neighborhood of $x$.

By assumption, there exists $H \in \mathcal B$ such that $H \subseteq U$.

The result follows.

$\blacksquare$