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 $1$.
$\blacksquare$