6. Filters

1. Definition of a filter

$(\mathrm{F}_1)$ Every subset of $X$ which contains a set of $\mathfrak{F}$ belongs to $\mathfrak{F}$. $(\mathrm{F}_\mathrm{II})$ Every finite intersection of sets of $\mathfrak{F}$ belongs to $\mathfrak{F}$. $(\mathrm{F}_\mathrm{III})$ The empty set is not in $\mathfrak{F}$.

It follows from $(\mathrm{F}_\mathrm{II})$ and $(\mathrm{F}_\mathrm{III})$ that every finite intersection of sets of $\mathfrak{F}$ is non-empty.

A filter $\mathfrak{F}$ on $X$ defines a structure on $X$, the axioms of which are $(\mathrm{F}_1)$, $(\mathrm{F}_\mathrm{II})$ and $(\mathrm{F}_\mathrm{III})$; this structure is called a structure of a filtered set, and the set $X$ endowed with this structure is called a set filtered by $\mathfrak{F}$.

Axiom $(\mathrm{F}_\mathrm{II})$ is equivalent to the conjunction of the following two axioms: $(\mathrm{F}_{\mathrm{II}\,\mathrm{a}})$ The intersection of two sets of $\mathfrak{F}$ belongs to $\mathfrak{F}$. $(\mathrm{F}_{\mathrm{II}\,\mathrm{b}})$ $X$ belongs to $\mathfrak{F}$.

Axioms $(\mathrm{F}_{\mathrm{II}\,\mathrm{b}})$ and $(\mathrm{F}_\mathrm{III})$ show that there is no filter on the empty set.

In order for a set of subsets which satisfies $(\mathrm{F}_1)$ also to satisfy $(\mathrm{F}_{\mathrm{II}\,\mathrm{b}})$ it is necessary and sufficient that it is not empty. A set of subsets which satisfies $(\mathrm{F}_1)$ also satisfies $(\mathrm{F}_\mathrm{III})$ if and only if it is different from $\mathfrak{P}(X)$.

Examples of filters. 1) If $X \neq \varnothing$, the set of subsets consisting of $X$ alone is a filter on $X$. More generally, the set of all subsets of $X$ which contain a given non-empty subset $A$ of $X$ is a filter on $X$.

2) In a topological space $X$, the set of all neighbourhoods of an arbitrary non-empty subset $A$ of $X$ (and in particular the set of all neighbourhoods of a point of $X$) is a filter, called the neighbourhood filter of $A$.

3) If $X$ is an infinite set, the complements of the finite subsets of $X$ are the elements of a filter. The filter of complements of finite subsets of the set $\mathbf{N}$ of integers $\geqslant 0$ is called the Fréchet filter.

2. Comparison of filters

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:

3. Bases of a filter

If $\mathfrak{G}$ is a subbase of a filter $\mathfrak{F}$ on $X$ (no. 2), then $\mathfrak{F}$ is not in general the set of subsets of $X$ which contain a set of $\mathfrak{G}$; for $\mathfrak{G}$ to have this property it is necessary and sufficient that every finite intersection of sets of $\mathfrak{G}$ should contain a set of $\mathfrak{G}$. Hence the following proposition:

If $\mathfrak{G}$ is a subbase of a filter $\mathfrak{F}$, then the set $\mathfrak{G}'$ of finite intersections of sets of $\mathfrak{G}$ is a base of $\mathfrak{F}$ (no. 2).

4. Ultrafilters

Since the ordered set of all filters on $X$ is inductive (no. 2, Proposition 1, Corollary 3), Zorn's lemma (Set Theory, R, § 6, no. 10) shows that:

5. Induced filter

If a filter $\mathfrak{F}$ on $X$ induces a filter on $A \subset X$, then the trace on $A$ of a base of $\mathfrak{F}$ is a base of $\mathfrak{F}_A$, by reason of Proposition 3 of no. 3.

6. Direct image and inverse image of a filter base

Let $\mathfrak{B}$ be a filter base on a set $X$, and let $f$ be a mapping of $X$ into a set $X'$; then $f(\mathfrak{B})$ is a filter base on $X'$, for the relation $M \neq \emptyset$ implies $f(M) \neq \emptyset$, and we have $f(M \cap N) \subset f(M) \cap f(N)$. If $\mathfrak{B}_1$ is a base of a filter which is finer than the filter of base $\mathfrak{B}$, then $f(\mathfrak{B}_1)$ is a base of a filter finer than the filter of base $f(\mathfrak{B})$ (no. 3, Proposition 4).

7. Product of filters

Let $(X_i)_{i \in I}$ be a family of sets, and for each $i \in I$ let $\mathfrak{B}_i$ be a filter base on $X_i$. Let $\mathfrak{B}$ be the set of subsets of the product set $X = \prod_{i \in I} X_i$ which are of the form $\prod_{i \in I} M_i$, where $M_i = X_i$ except for a finite number of indices and where $M_i \in \mathfrak{B}_i$ for each $i$ such that $M_i \neq X_i$. The formula $\left( \prod_{i \in I} M_i \right) \cap \left( \prod_{i \in I} N_i \right) = \prod_{i \in I} (M_i \cap N_i)$ shows that $\mathfrak{B}$ is a filter base on $X$. Note that the filter of base $\mathfrak{B}$ is also generated by the

sets $\overline{\mathrm{pr}}_x^{-1}(M_x)$, where $M_x \in \mathfrak{B}_x$ and $x$ runs through $I$, since

The reader may easily verify that the product of the filters $\mathfrak{F}_i$ can also be defined as the coarsest filter $\mathfrak{G}$ on $X$ such that $\mathrm{pr}_i(\mathfrak{G}) = \mathfrak{F}_i$ for each $i \in I$.

The preceding remarks show that if $\mathfrak{B}_i$ is a base of $\mathfrak{F}_i$ for each $i \in I$, then $\mathfrak{B}$ is a base of the product filter $\prod_{i \in I} \mathfrak{F}_i$ (no. 3, Proposition 3).

On a product $X = \prod_{i \in I} X_i$ of topological spaces, the neighbourhood filter of any point $x = (x_i)$ is the product of the neighbourhood filters of the $x_i$ (§ 4, no. 1).

The construction of a product filter $\mathfrak{F} = \prod_{i \in I} \mathfrak{F}_i$ is simpler when the index set $I$ is finite: a base of $\mathfrak{F}$ is then formed by all products $\prod_{i \in I} M_i$, where $M_i \in \mathfrak{F}_i$ for each $i \in I$. If $I = \{ 1, 2, \ldots, n \}$ we write

8. Elementary filters

It comes to the same thing to say that the elementary filter associated with $(x_n)$ is the set of subsets $M$ of $X$ such that $x_n \in M$ except for a finite number of values of $n$. If $S_n$ denotes the set of all $x_p$ such that $p \geq n$, then the sets $S_n$ form a base of the elementary filter associated with the sequence $(x_n)$.

The elementary filter associated with an infinite subsequence of a sequence $(x_n)$ is finer than the elementary filter associated with $(x_n)$ (cf. Exercise 15).

By definition, every elementary filter has a countable base. Conversely:

9. Germs with respect to a filter

Let $\mathfrak{F}$ be a filter on a set $X$. On the set $\mathfrak{P}(X)$ of all subsets of $X$, the relation

"there exists $V \in \mathfrak{F}$ such that $M \cap V = N \cap V$"

between $M$ and $N$ is an equivalence relation $R$, for $R$ is obviously reflexive and symmetric, and if $M, N, P$ are three subsets of $X$ such that $M \cap V = N \cap V$ and $N \cap W = P \cap W$, where $V$ and $W$ belong to $\mathfrak{F}$, it follows that $M \cap (V \cap W) = N \cap (V \cap W) = P \cap (V \cap W)$ and $V \cap W \in \mathfrak{F}$, so that $R$ is transitive. The equivalence class mod $R$ of a subset $M$ of $X$ is called the germ of $M$ with respect to $\mathfrak{F}$; the quotient set $\mathfrak{P}(X)/R$ is called the set of germs of subsets of $X$ (with respect to $\mathfrak{F}$). The mappings $(M, N) \to M \cap N$ and $(M, N) \to M \cup N$ of

into $\mathfrak{P}(X)$ are compatible with the equivalence relations $R \times R$ and $R$

(Set Theory, R, , no. 8). For if $M \equiv M' \pmod{R}$ and $N \equiv N' \pmod{R}$ then there exist $V$ and $W$ in $\mathfrak{F}$ such that

Passing to the quotients, these mappings induce two mappings of $(\mathfrak{P}(X)/R) \times (\mathfrak{P}(X)/R)$ into $\mathfrak{P}(X)/R$, which we denote (by abuse of language) by $(\xi, \eta) \to \xi \cap \eta$ and $(\xi, \eta) \to \xi \cup \eta$ respectively. It is a straightforward exercise to verify that with respect to these laws of composition [defined throughout $\mathfrak{P}(X)/R$] every element is idempotent, and that each law is commutative and associative and distributive with respect to the other. Further, the relations $\xi = \xi \cap \eta$ and $\eta = \xi \cup \eta$ are equivalent; if we denote them (by abuse of language) by $\xi \subset \eta$, it is easily verified that this relation is an ordering on $\mathfrak{P}(X)/R$, with respect to which $\mathfrak{P}(X)/R$ is a lattice which has the germ of $\varnothing$ as least element and the germ of $X$ as greatest element. Note that the relation $\xi \subset \eta$ means that there exist $M \in \xi$, $N \in \eta$ and $V \in \mathfrak{F}$ such that $M \cap V \subset N \cap V$.

Now let $X'$ be another set, and let $\Phi$ be the set of all mappings of a set of $\mathfrak{F}$ into $X'$. The relation on $\Phi$

"there exists $V \in \mathfrak{F}$ such that $f$ and $g$ are defined and agree on $V$"

between $f$ and $g$ is an equivalence relation $S$; it is clear that $S$ is reflexive and symmetric, and if $f, g, h$ are three elements of $\Phi$ such that $f$ and $g$ are defined on $V \in \mathfrak{F}$, and $g$ and $h$ are defined and agree on $W \in \mathfrak{F}$, then $f$ and $h$ are defined and agree on $V \cap W \in \mathfrak{F}$, so that $S$ is transitive. The equivalence class mod $S$ of a mapping $f$ of a set $V \in \mathfrak{F}$ into $X'$ is called the germ of $f$ (with respect to $\mathfrak{F}$), and the quotient set $\tilde{\Phi} = \Phi/S$ is called the set of germs of mappings of $X$ into $X'$ (with respect to $\mathfrak{F}$).