# Equivalence of Definitions of Filter Basis

## Theorem

The following definitions of the concept of Filter Basis are equivalent:

Let $S$ be a set.

Let $\FF$ be a filter on $S$.

### Definition 1

Let $\BB \subset \powerset S$ such that $\O \notin \BB$ and $\BB \ne \O$.

Then $\FF := \set {V \subseteq S: \exists U \in \BB: U \subseteq V}$ is a filter on $S$ if and only if:

$\forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$

Such a $\BB$ is called a filter basis of $\FF$.

### Definition 2

Let $\BB$ be a subset of a filter $\FF$ on $S$ such that $\BB \ne \O$.

Then $\BB$ is a filter basis of $\FF$ if and only if:

$\forall U \in \FF: \exists V \in \BB: V \subseteq U$

## Proof

### $(1)$ implies $(2)$

Let $\BB$ be a filter basis of $\FF$ by definition $1$.

Then by definition:

$\BB \subset \powerset S$ such that $\O \notin \BB$ and $\BB \ne \O$

and $\FF := \set {V \subseteq S: \exists U \in \BB: U \subseteq V}$ is a filter on $S$ if and only if:

$\forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$

Let $U \in \FF$.

Then by definition of $\FF$:

$\exists V \in \BB: V \subseteq U$

Thus $\BB$ is a filter basis of $\FF$ by definition $2$.

$\Box$

### $(2)$ implies $(1)$

Let $\BB$ be a filter basis of $\FF$ by definition $2$.

By definition, $\BB$ is a filter basis of $\FF$ if and only if:

$\forall U \in \FF: \exists V \in \BB: V \subseteq U$

Let $V_1, V_2 \in \BB$.

Then:

$V_1, V_2 \in \FF$

By definition of filter:

$V_1 \cap V_2 \in \FF$

and so:

$\exists U \in \BB: U \subseteq V_1 \cap V_2$

Thus $\BB$ is a filter basis of $\FF$ by definition $1$.

$\blacksquare$