Finite Intersection Property

From Nested Intervals to Finite Intersections

The Nested Interval Theorem gives a common point when a particular sequence of closed intervals is nested. Compactness provides a broader principle: a family of closed sets in a compact set has a common point whenever every finite selection from the family has a common point. The sets need not be intervals, and they need not be nested. What matters is compatibility across every finite subfamily.

This principle is useful in both directions. It can establish the existence of a point satisfying an entire collection of closed constraints, and it can show that a compactness question has a finite witness: if all the constraints cannot be satisfied together, then some finite selection already cannot be satisfied together.

Definition (Finite Intersection Property): A family $\{F_\alpha:\alpha\in A\}$ of subsets of a set $K$ has the finite intersection property if the intersection of every finite subfamily is nonempty. We interpret the intersection of the empty subfamily as $K$. In particular, if $K$ is nonempty, the empty family has the finite intersection property.

The phrase “every finite subfamily” includes the intersection of one member and of any larger finite selection. It does not mean merely that each pair of sets intersects. For a family indexed by $A$, the condition says that for every finite set of indices $\{\alpha_1,\ldots,\alpha_m\}$, the intersection $F_{\alpha_1}\cap\cdots\cap F_{\alpha_m}$ is nonempty.

The Finite Intersection Theorem

We first state the result for closed subsets of a fixed compact set. “Closed in $K$” means closed in the relative topology on $K$, so the sets in the family need not be closed in all of $\mathbb{R}$. We assume $K$ is nonempty; this ensures that the intersection of an empty family is nonempty under the convention in the definition.

Theorem (Finite Intersection Theorem): Let $K\subseteq\mathbb{R}$ be nonempty and compact, and let $\{F_\alpha:\alpha\in A\}$ be a family of subsets closed in $K$. If the family has the finite intersection property, then $$ \bigcap_{\alpha\in A}F_\alpha\neq\varnothing. $$

Proof. If $A$ is empty, then the intersection is $K$, which is nonempty. Suppose $A$ is nonempty. Assume, for a contradiction, that $\bigcap_{\alpha\in A}F_\alpha=\varnothing$. For each $\alpha$, the complement $K\setminus F_\alpha$ is open in $K$. These relative open sets cover $K$: for any $x\in K$, the empty total intersection means that $x\notin F_\alpha$ for at least one index $\alpha$, so $x\in K\setminus F_\alpha$.

Compactness gives a finite subcover. To apply compactness in $\mathbb{R}$, write each relative open set as $K\setminus F_\alpha=K\cap V_\alpha$ for some open $V_\alpha\subseteq\mathbb{R}$. The corresponding finite collection of $V_\alpha$'s covers $K$, and therefore some finite collection of the sets $K\setminus F_\alpha$ covers $K$. For those finitely many indices $\alpha_1,\ldots,\alpha_m$, this says

K\setminus F_{\alpha_1},\ldots,K\setminus F_{\alpha_m} \text{ cover }K,

or equivalently $F_{\alpha_1}\cap\cdots\cap F_{\alpha_m}=\varnothing$. This contradicts the finite intersection property. Hence the total intersection is nonempty. $\square$

The proof is a complement argument: an empty intersection becomes a cover by complements. Compactness reduces that cover to finitely many members, and taking complements back gives a finite intersection that is empty. The finite intersection property rules out precisely that possibility.

Compactness Characterized by Closed Families

The same argument also works in reverse. If closed families with the finite intersection property always have a common point, then every open cover must have a finite subcover. Thus the finite intersection property is not just a consequence of compactness; for nonempty subsets of $\mathbb{R}$, it gives an equivalent formulation.

Theorem (Finite Intersection Characterization of Compactness): A nonempty set $K\subseteq\mathbb{R}$ is compact if and only if every family of subsets closed in $K$ that has the finite intersection property has nonempty total intersection.

Proof. The forward direction is the Finite Intersection Theorem. For the reverse direction, suppose every family of subsets closed in $K$ with the finite intersection property has nonempty total intersection. Let $\{U_\alpha:\alpha\in A\}$ be an open cover of $K$, with each $U_\alpha$ open in $\mathbb{R}$. Define

F_\alpha=K\setminus U_\alpha.

Each $F_\alpha$ is closed in $K$. If this family had the finite intersection property, the assumed property would give a point in $\bigcap_{\alpha\in A}F_\alpha$. Such a point belongs to $K$ and to none of the $U_\alpha$, contradicting that the $U_\alpha$'s cover $K$. Therefore the $F_\alpha$'s do not have the finite intersection property. Some finite selection $F_{\alpha_1},\ldots,F_{\alpha_m}$ has empty intersection. Taking complements relative to $K$, we find that $U_{\alpha_1},\ldots,U_{\alpha_m}$ cover $K$. Thus every open cover of $K$ has a finite subcover, so $K$ is compact. $\square$

The nonempty hypothesis is relevant to this formulation. For an empty set $K$, the empty subfamily has intersection $K=\varnothing$, so no family has the finite intersection property under our convention. The condition that every family with the finite intersection property has nonempty total intersection is therefore vacuously true for $K=\varnothing$, and compactness itself also allows the empty set. Stating the characterization for nonempty $K$ avoids this vacuous case without changing the useful content of the theorem.

Worked Examples: Applying the Property

Worked Example: A Decreasing Family of Closed Intervals

Let $K=[0,1]$, and for each positive integer $n$ set $F_n=[0,1/(n+1)]$. Each $F_n$ is closed in $K$. If $n_1,\ldots,n_m$ are any finite selection of indices and $N$ is the largest of them, then the intervals decrease as the index increases, so

F_{n_1}\cap\cdots\cap F_{n_m} = [0,1/(N+1)].

This intersection contains $0$, so every finite subfamily has nonempty intersection. The Finite Intersection Theorem therefore guarantees a point common to all the $F_n$. In fact, $0$ is the only such point: it belongs to every interval, and if $x>0$, choose $n$ large enough that $1/(n+1)<x$. Then $x\notin F_n$. Consequently,

\bigcap_{n=1}^{\infty}F_n=\{0\}.

Worked Example: Pairwise Intersections Do Not Suffice

Take $K=\{1,2,3\}$, which is compact because every finite subset of $\mathbb{R}$ is compact. Consider the subsets

F_1=\{1,2\},\qquad F_2=\{2,3\},\qquad F_3=\{1,3\}.

Every subset of $K$ is closed in $K$. Each pair intersects: $F_1\cap F_2=\{2\}$, $F_1\cap F_3=\{1\}$, and $F_2\cap F_3=\{3\}$. But the intersection of all three is empty. Indeed, $F_1\cap F_2=\{2\}$, and $2\notin F_3$, so $F_1\cap F_2\cap F_3=\varnothing$.

Thus pairwise intersection does not give the finite intersection property: the intersection of this three-member subfamily is empty. There is no contradiction with the theorem, because its hypothesis requires every finite selection, not only every pair, to have a common point.

Worked Example: Why Compactness Is Needed

Let $K=(0,1)$, which is not compact, and define $F_n=(0,1/(n+1)]$ for each positive integer $n$. These sets are subsets of $K$ and are closed relative to $K$: their complements in $K$ are $(1/(n+1),1)$, which are open relative to $K$. Every finite intersection is the set with the largest index among the chosen indices. In particular, for indices $n_1,\ldots,n_m$, with $N=\max\{n_1,\ldots,n_m\}$,

F_{n_1}\cap\cdots\cap F_{n_m}=(0,1/(N+1)]\neq\varnothing.

The family has the finite intersection property. Yet its total intersection is empty. If $x$ belonged to every $F_n$, then $x>0$ and $x\leq1/(n+1)$ for every $n$. Choose $n$ so large that $1/(n+1)<x$; this contradicts $x\leq1/(n+1)$. Thus the common-point conclusion can fail when the containing set is not compact.

A Finite Witness for Failure

One useful consequence reverses the viewpoint. In a compact set, if a family of relatively closed subsets has empty total intersection, then some finite subfamily already has empty intersection. This does not say how many sets are needed or identify them in advance. It says that an obstruction to satisfying all the closed constraints can always be detected using finitely many constraints.

Corollary (Finite Witness for an Empty Intersection): Let $K\subseteq\mathbb{R}$ be nonempty and compact, and let $\{F_\alpha:\alpha\in A\}$ be closed in $K$. If $\bigcap_{\alpha\in A}F_\alpha=\varnothing$, then there are finitely many indices $\alpha_1,\ldots,\alpha_m$ such that $F_{\alpha_1}\cap\cdots\cap F_{\alpha_m}=\varnothing$.

Proof. If every finite subfamily had nonempty intersection, the family would have the finite intersection property. The Finite Intersection Theorem would then imply that the total intersection is nonempty, contrary to the hypothesis. Therefore at least one finite subfamily has empty intersection. $\square$

This finite-witness form is often the convenient way to use the theorem. Rather than trying to find a point in an infinite intersection directly, one can show that every finite set of constraints is consistent. Compactness then supplies a point satisfying all of them at once. Conversely, if no point satisfies all the constraints, compactness guarantees a finite selection that is already inconsistent.

The key checks are to identify the compact containing set, verify that the family is closed relative to it, and verify the finite intersection property for every finite selection. Checking only pairs is not enough, and the example in $(0,1)$ shows that the compactness hypothesis cannot simply be omitted.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What does the finite intersection property require of a family, and how is the intersection of the empty subfamily interpreted?
In the proof of the Finite Intersection Theorem, why do the complements of the closed sets cover $K$ when the total intersection is empty?
Why does compactness turn that cover into a finite subfamily whose intersection is empty?
In the three-subset example in $K=\{1,2,3\}$, why do pairwise intersections fail to establish the finite intersection property?
What feature of $K=(0,1)$ permits the closed sets in the final worked example to have the finite intersection property but empty total intersection?

What You'll Learn