Definition 2

Given two filters $\mathfrak{F}, \mathfrak{F}'$ on the same set $X$, $\mathfrak{F}'$ is said to be finer than $\mathfrak{F}$, or $\mathfrak{F}$ is coarser than $\mathfrak{F}'$, if $\mathfrak{F} \subset \mathfrak{F}'$. If also $\mathfrak{F} \neq \mathfrak{F}'$, then $\mathfrak{F}'$ is said to be strictly finer than $\mathfrak{F}$, or $\mathfrak{F}$ strictly coarser than $\mathfrak{F}'$.

Two filters are said to be comparable if one is finer than the other. The set of all filters on $X$ is ordered by the relation " $\mathfrak{F}$ is coarser than $\mathfrak{F}'$"; this relation is induced by the inclusion relation in $\mathfrak{P}(\mathfrak{P}(X))$.

Let $(\mathfrak{F}_i)_{i \in I}$ be any non-empty family of filters on a set $X$ (which must therefore be non-empty); then the set

satisfies axioms $(F_I), (F_{II})$ and $(F_{III})$ and is therefore a filter; $\mathfrak{F}$ is called the intersection of the family of filters $(\mathfrak{F}_i)_{i \in I}$ and is obviously the greatest lower bound of the set of the $\mathfrak{F}_i$ in the ordered set of all filters on $X$.

The filter formed by the single set $X$ is the smallest element of the ordered set of all filters on $X$. We shall see in no. 4 that, if $X$ has more than one element, the set of all filters on $X$ has no greatest element.

Given a set $\mathfrak{G}$ of subsets of a set $X$, let us consider whether there are any filters on $X$ which contain $\mathfrak{G}$. If such a filter exists then by $(F_{II})$ it contains also the set $\mathfrak{G}'$ of finite intersections of sets of $\mathfrak{G}$ (including $X$, which is the intersection of the empty subset of $\mathfrak{G}$); hence a necessary condition for such a filter to exist is that the empty subset of $X$ is not in $\mathfrak{G}'$. This condition is also sufficient, for by $(F_I)$ any filter which contains $\mathfrak{G}'$ also contains the set $\mathfrak{G}''$ of subsets of $X$ which contain a set of $\mathfrak{G}'$. Now $\mathfrak{G}''$ clearly satisfies $(F_I)$; it satisfies $(F_{II})$ by reason of the definition of $\mathfrak{G}'$; and finally it satisfies $(F_{III})$ because the empty subset of $X$ does not belong to $\mathfrak{G}'$. Hence $\mathfrak{G}''$ is the coarsest filter which contains $\mathfrak{G}$, and we have proved: