Filters and the Finite Intersection Property

Finite Intersections and Proper Filters

Compactness is defined using open covers, but it can also be expressed by asking whether certain families of closed sets have a common point. Filters package the same finite-intersection information in a form useful for studying products and compactness. The central point is to keep finite intersections distinct from the intersection of an entire family: the former may all be nonempty even when the latter is empty.

Let \(X\) be a set, usually equipped with a topology \(\mathcal{T}\) when we discuss closed sets or compactness. For a family \(\mathcal{A}\subseteq\mathcal{P}(X)\), say that \(\mathcal{A}\) has the finite intersection property (FIP) if the intersection of every finite subfamily is nonempty. The intersection of the empty subfamily is, by convention, \(X\). Thus, when \(X\ne\varnothing\), the empty family has the FIP and its total intersection is \(X\). When \(X=\varnothing\), the empty family does not have the FIP under this convention.

Definition: A proper filter on \(X\) is a collection \(\mathcal{F}\subseteq\mathcal{P}(X)\) such that (1) \(X\in\mathcal{F}\) and \(\varnothing\notin\mathcal{F}\); (2) if \(A,B\in\mathcal{F}\), then \(A\cap B\in\mathcal{F}\); and (3) if \(A\in\mathcal{F}\) and \(A\subseteq B\subseteq X\), then \(B\in\mathcal{F}\). The filter is proper because it excludes the empty set.

The axioms imply closure under every finite intersection: apply the intersection axiom repeatedly, with \(X\) serving as the intersection of no sets. Every proper filter has the FIP, since each finite intersection belongs to the filter and therefore cannot be empty. Conversely, a family with the FIP generates a proper filter.

Generating a Filter from a Family

Definition: For \(\mathcal{A}\subseteq\mathcal{P}(X)\), the filter generated by \(\mathcal{A}\), when it is proper, is \(\langle\mathcal{A}\rangle=\{B\subseteq X:\text{ for some finite }\mathcal{A}_0\subseteq\mathcal{A},\ \bigcap_{A\in\mathcal{A}_0}A\subseteq B\}\). The empty finite subfamily contributes the intersection \(X\).

Proposition: The collection \(\langle\mathcal{A}\rangle\) is a proper filter if and only if \(\mathcal{A}\) has the FIP.

Proof. Suppose first that \(\mathcal{A}\) has the FIP. Taking the empty finite subfamily shows \(X\in\langle\mathcal{A}\rangle\). No finite intersection of members of \(\mathcal{A}\) is empty, so none can be a subset of \(\varnothing\); hence \(\varnothing\notin\langle\mathcal{A}\rangle\). Upward closure follows directly: if a finite intersection is contained in \(B\) and \(B\subseteq D\), that same finite intersection is contained in \(D\). If \(B,D\in\langle\mathcal{A}\rangle\), choose finite subfamilies \(\mathcal{A}_1,\mathcal{A}_2\) whose intersections are contained in \(B,D\), respectively. The intersection over \(\mathcal{A}_1\cup\mathcal{A}_2\) is contained in \(B\cap D\), so \(B\cap D\in\langle\mathcal{A}\rangle\). Thus it is a proper filter. Conversely, if \(\mathcal{A}\) fails the FIP, some finite subfamily has empty intersection. Since \(\varnothing\) is a subset of every \(B\subseteq X\), in particular \(\varnothing\in\langle\mathcal{A}\rangle\). The generated collection is not proper. \(\square\)

This proposition also covers the empty family. If \(X\ne\varnothing\), its generated collection is \(\{X\}\), a proper filter. If \(X=\varnothing\), the generated collection is \(\{\varnothing\}\), which is not proper. Accordingly, there is no proper filter on the empty set.

Worked Example: A Filter on a Finite Set

Give \(X=\{a,b,c\}\) the discrete topology, in which every subset is open (and therefore every subset is also closed). Let \(\mathcal{A}=\{\{a,b\},\{b,c\}\}\).

The intersection of the first set alone is \(\{a,b\}\), and the second alone is \(\{b,c\}\). The intersection of both is \(\{b\}\), which is nonempty. The empty subfamily has intersection \(X\), also nonempty. These are all the finite subfamilies, so \(\mathcal{A}\) has the FIP.

Any finite intersection from \(\mathcal{A}\) is one of \(X,\{a,b\},\{b,c\},\{b\}\). A set contains at least one of these intersections exactly when it contains \(b\). Therefore \(\langle\mathcal{A}\rangle=\{\{b\},\{a,b\},\{b,c\},X\}\). Every listed set contains \(b\), so none is empty; intersections of listed sets still contain \(b\); and every superset in \(X\) of a listed set is listed. These checks verify the filter axioms. The total intersection of the original family is \(\{b\}\), in agreement with the finite-intersection calculation.

Closed Families and Compactness

Theorem (Closed-set criterion for compactness): A topological space \(X\) is compact if and only if every nonempty family of closed subsets of \(X\) with the FIP has nonempty total intersection.

The theorem is stated for nonempty families to avoid a convention issue: when \(X\ne\varnothing\), the empty family has the FIP, and its total intersection is \(X\). When \(X=\varnothing\), the empty family does not have the FIP under the convention used here. The empty space is compact, and there is no nonempty family of its closed sets with the FIP, so the stated criterion applies to it as well.

Proof. Suppose \(X\) is compact, and let \(\mathcal{C}\) be a nonempty family of closed subsets with the FIP. If \(\bigcap_{C\in\mathcal{C}}C=\varnothing\), then the open sets \(X\setminus C\), for \(C\in\mathcal{C}\), cover \(X\): each point must be absent from at least one member of \(\mathcal{C}\). Compactness gives a finite subcover \(X\setminus C_1,\ldots,X\setminus C_n\). Taking complements shows \(C_1\cap\cdots\cap C_n=\varnothing\), contradicting the FIP. Thus the total intersection is nonempty.

Conversely, suppose every nonempty closed family with the FIP has nonempty total intersection. Let \(\mathcal{U}\) be an open cover of \(X\). If it had no finite subcover, the family \(\{X\setminus U:U\in\mathcal{U}\}\) would consist of closed sets and have the FIP: an empty finite intersection of complements would mean the corresponding finite collection of open sets covers \(X\). Its total intersection is empty because \(\mathcal{U}\) covers \(X\). This contradicts the assumed closed-set condition. Hence every open cover has a finite subcover, which is compactness. \(\square\)

For a family \(\mathcal{C}\) of closed sets with the FIP, the generated filter \(\langle\mathcal{C}\rangle\) is proper. Conversely, every proper filter is itself a family with the FIP. The generated filter need not consist only of closed sets, but it records exactly the finite intersections and supersets forced by the original family.

Worked Example: Closed Tails in the Real Line

Use the standard topology on \(\mathbb{R}\), and set \(C_n=[n,\infty)\) for \(n\in\mathbb{N}\). Each \(C_n\) is closed because its complement \((-\infty,n)\) is open. For a nonempty finite selection of indices, let \(m\) be the largest selected index. The intersection of the selected sets is \([m,\infty)\), which is nonempty. The empty selection has intersection \(\mathbb{R}\), also nonempty, so the family has the FIP.

Nevertheless, \(\bigcap_{n\in\mathbb{N}}C_n=\varnothing\). Indeed, for any \(x\in\mathbb{R}\), choose a natural number \(n>x\); then \(x\notin[n,\infty)\). The generated filter consists of all subsets of \(\mathbb{R}\) that contain \([N,\infty)\) for some \(N\in\mathbb{N}\). To check this description, any nonempty finite intersection of tails is \([m,\infty)\) for its largest index \(m\), and the empty intersection \(\mathbb{R}\) imposes no additional sets beyond the upward-closed collection. Thus this is a proper filter even though the original closed sets have empty total intersection. It demonstrates why an FIP alone does not guarantee a common point in a noncompact space.

Worked Example: Nested Closed Intervals

In \(\mathbb{R}\) with its standard topology, let \(D_n=[0,1/n]\), where \(n\in\mathbb{N}\). Each interval is closed. For any nonempty finite selection, let \(m\) be the largest selected index. Since \(1/k\ge 1/m\) whenever \(k\le m\), the intersection of those selected intervals is \([0,1/m]\), which is nonempty. Thus the family has the FIP.

The total intersection is \(\{0\}\). The point \(0\) belongs to every \(D_n\). A negative real number belongs to none of them, and if \(x>0\), choose \(n>1/x\); then \(1/n<x\), so \(x\notin D_n\). This also computes the generated filter: it consists of all subsets of \(\mathbb{R}\) containing some interval \([0,1/n]\). The point \(0\) belongs to every such interval, so it belongs to every member of the generated filter. The example contrasts with the tails: both families have the FIP, but their total intersections differ.

Filters and Common Points of Closures

For a proper filter \(\mathcal{F}\) on a topological space \(X\), call \(x\in X\) an adherent point of \(\mathcal{F}\) if \(x\in\overline{A}\) for every \(A\in\mathcal{F}\). Equivalently, every open neighborhood of \(x\) meets every member of \(\mathcal{F}\): the neighborhood test for closure says exactly that an open neighborhood of \(x\) meets \(A\) if and only if \(x\in\overline{A}\).

Proposition: A space \(X\) is compact if and only if every proper filter on \(X\) has an adherent point.

Proof. Suppose \(X\) is compact and \(\mathcal{F}\) is a proper filter. The family \(\{\overline{A}:A\in\mathcal{F}\}\) consists of closed sets. For any finite list \(A_1,\ldots,A_n\in\mathcal{F}\), the filter axioms give \(D=A_1\cap\cdots\cap A_n\in\mathcal{F}\), so \(D\ne\varnothing\). Since \(D\subseteq\overline{A_i}\) for every \(i\), the finite intersection of these closures is nonempty. The empty finite intersection is \(X\), which is nonempty because a proper filter exists. The closed-set criterion for compactness gives a point in all the closures, an adherent point of \(\mathcal{F}\).

Conversely, suppose every proper filter has an adherent point, and let \(\mathcal{C}\) be a nonempty closed family with the FIP. By the generating proposition, \(\langle\mathcal{C}\rangle\) is a proper filter. Let \(x\) be an adherent point of it. Every \(C\in\mathcal{C}\) belongs to the generated filter, so \(x\in\overline{C}=C\), since \(C\) is closed. Hence \(x\in\bigcap_{C\in\mathcal{C}}C\). The closed-set criterion now gives compactness. \(\square\)

This criterion asks for a common point of the closures of filter members, not necessarily a point in their intersection. The tail filter in the previous example has no common point in all its members: a real number \(x\) fails to belong to \([N,\infty)\) for any natural number \(N>x\). This distinction is important when using filters to express compactness.

A Common Pitfall: Pairwise Is Not Finite

Worked Example: Pairwise Intersections Can Mislead

In the discrete space \(X=\{1,2,3\}\), consider \(A=\{1,2\}\), \(B=\{2,3\}\), and \(C=\{1,3\}\). Each pair intersects: \(A\cap B=\{2\}\), \(A\cap C=\{1\}\), and \(B\cap C=\{3\}\). But \(A\cap B\cap C=\varnothing\), since no element belongs to all three sets. The family is pairwise intersecting but does not have the FIP. The incorrect calculation is to check only pairs and conclude that every finite intersection is nonempty; the correction is to check every finite subfamily, including the triple here.

A second tempting but false conclusion is that an FIP family must have nonempty total intersection. The closed tails \([n,\infty)\) provide the correction: every finite intersection is a nonempty tail, while the full intersection is empty. Compactness is precisely what makes the closed-family conclusion valid.

Takeaway: A family has the FIP when every finite intersection, not merely every pairwise intersection, is nonempty. Such a family generates a proper filter. In a compact space, every nonempty closed family with the FIP has a common point; without compactness, the real-line tails show that the conclusion can fail.

Check Your Understanding

For each question, distinguish the finite-intersection condition from a claim about the total intersection.

On \(X=\{a,b,c\}\), does the family \(\{\{a,b\},\{b,c\},\{a,c\}\}\) have the FIP? Identify the finite subfamily that decides the answer.
Describe the filter generated on \(\mathbb{R}\) by the family \(\{[n,\infty):n\in\mathbb{N}\}\).
Why does a proper filter necessarily have the FIP? Which filter axiom rules out an empty finite intersection?
State the closed-set criterion for compactness and explain how taking complements turns an open cover with no finite subcover into a closed family with the FIP.
For \(D_n=[0,1/n]\), find the total intersection and verify that every finite intersection is nonempty.