Filter base and the filter it generates

Let $X$ be a set. A family $\mathcal{B} \subseteq \mathcal{P}(X)$ is a filter base on $X$ when it satisfies:

• (B1) nonemptiness: $\mathcal{B} \neq \emptyset$;
• (B2) properness: $\emptyset \notin \mathcal{B}$;
• (B3) downward directedness: for all $B_1, B_2 \in \mathcal{B}$ there is $B_3 \in \mathcal{B}$ with $B_3 \subseteq B_1 \cap B_2$.

The upward closure of $\mathcal{B}$ in $X$ is

$\langle \mathcal{B} \rangle := \{\, A \subseteq X \;:\; B \subseteq A \text{ for some } B \in \mathcal{B} \,\}.$

This is a filter on $X$ (Filter on a set), indeed the smallest filter on $X$ containing $\mathcal{B}$, by The upward closure of a filter base is the smallest filter containing it ↗; it is called the filter generated by $\mathcal{B}$, and $\mathcal{B}$ is called a base of it. The definite article is licensed by that lemma and by nothing else: the notation $\langle \mathcal{B} \rangle$ is used only for a family already known to be a filter base.

• (B3) does not ask that $B_1 \cap B_2$ itself belong to $\mathcal{B}$, only that some member of $\mathcal{B}$ sit inside it. That is what makes a base usable: a family described by a shape, for instance the open discs of the plane that contain the origin, is generally not closed under intersection, since the intersection of two such discs is not a disc, yet it always contains a smaller such disc. Families that are closed under pairwise intersection, such as the tails $\{n \in \mathbb{N} : N \leq n\}$ of $\mathbb{N}$, are the special case $B_3 = B_1 \cap B_2$ of (B3).
• Every filter is a filter base, and it generates itself. A filter satisfies (B1) because it contains $X$, satisfies (B2) by (F2), and satisfies (B3) with $B_3 = B_1 \cap B_2$ by (F3). Both halves are proved in The upward closure of a filter base is the smallest filter containing it ↗, so "base" is a way of presenting a filter, never a different kind of object.
• When a subfamily is a base for the filter it sits in. If $\mathcal{B} \subseteq \mathcal{F}$ and every member of $\mathcal{F}$ contains a member of $\mathcal{B}$, then $\mathcal{B}$ is a filter base and $\langle \mathcal{B} \rangle = \mathcal{F}$: (B1) holds because $X \in \mathcal{F}$ contains some member of $\mathcal{B}$, (B2) because $\mathcal{B} \subseteq \mathcal{F}$, and (B3) because $B_1 \cap B_2 \in \mathcal{F}$ contains some member of $\mathcal{B}$. So a base is a presentation, not an invariant. What is determined is the generated filter, which is why every statement below is about $\langle \mathcal{B} \rangle$ and never about $\mathcal{B}$ itself. The criterion is sufficient, and it is not a claim that a filter has many bases: any base of $\mathcal{F}$ is a subfamily of it, so the filter $\{X\}$ on $X$ has exactly one base, namely $\{X\}$.
• (B2) is exactly the properness of the generated filter, and it is not automatic: dropping it would allow $\emptyset \in \mathcal{B}$ and then $\langle \mathcal{B} \rangle = \mathcal{P}(X)$, the improper filter excluded by the convention recorded in Filter on a set.