Compact Metric Space Proof Mastery

A Proof Strategy for Compactness Arguments

Compactness proofs often begin with a family of conditions that must hold simultaneously. Each condition may define a closed subset of the space, and the goal is to show that some point satisfies all of them. A useful change of viewpoint is to ask whether every finite collection of conditions can be satisfied. In a compact space, that finite consistency is enough to guarantee a point satisfying the whole family.

This principle is a practical companion to the open-cover definition of compactness. The definition asks whether every open cover has a finite subcover; the finite intersection property turns the same idea around by taking complements. We will prove that equivalence, then use it to establish a nested-set theorem and a uniqueness criterion. These results provide a reusable structure for proofs involving infinitely many closed constraints.

Definition: A family \(\mathcal{F}\) of subsets of a nonempty set \(X\) has the finite intersection property if the intersection of every finite subfamily is nonempty. The empty finite subfamily has intersection \(X\), so it causes no difficulty when \(X\) is nonempty.

The key word is “finite.” A family can have nonempty intersections for every finite selection even though its full intersection is empty in a space that is not compact. Compactness is precisely what prevents that failure for families of closed subsets.

The Finite Intersection Property

Theorem (Finite Intersection Property): Let \(X\) be a nonempty compact metric space, and let \(\mathcal{F}\) be a family of closed subsets of \(X\). Then \(\bigcap_{F\in\mathcal{F}}F\ne\varnothing\) if and only if \(\mathcal{F}\) has the finite intersection property.

Proof. Suppose first that the full intersection is nonempty. Choose \(x\in\bigcap_{F\in\mathcal{F}}F\). For any finite selection \(F_1,\ldots,F_m\) from the family, \(x\in F_j\) for every \(j\), so \(x\in F_1\cap\cdots\cap F_m\). Thus every finite intersection is nonempty.

Conversely, suppose that \(\mathcal{F}\) has the finite intersection property. If its full intersection were empty, then every \(x\in X\) would fail to belong to at least one member of \(\mathcal{F}\). Equivalently, the open sets \(X\setminus F\), for \(F\in\mathcal{F}\), would cover \(X\). They are open because each \(F\) is closed. Compactness gives a finite subcover, say \(X\setminus F_1,\ldots,X\setminus F_m\). Taking complements of this covering statement gives \(F_1\cap\cdots\cap F_m=\varnothing\), contradicting the finite intersection property. Therefore the full intersection is nonempty. \(\square\)

This proof is a useful pattern: assume the desired common point does not exist, take complements, and obtain an open cover. Compactness then reduces the contradiction to finitely many of the original conditions. Notice that the closedness hypothesis matters because it makes the complements open. The theorem does not say that an arbitrary family of subsets with the finite intersection property has nonempty intersection.

Worked Example: Intersecting Infinitely Many Closed Constraints

For each positive integer \(n\), let \(F_n=[0,1/n]\), regarded as a closed subset of the compact metric space \(X=[0,1]\). The sets are nested: \(F_{n+1}\subseteq F_n\), since \(1/(n+1)\leq1/n\). Any finite selection has a smallest interval among its members, namely the one with the largest index, so every finite intersection is nonempty. The finite intersection property theorem therefore guarantees a point in \(\bigcap_{n=1}^{\infty}F_n\).

We can identify the intersection exactly. The point \(0\) belongs to every \(F_n\). If \(x>0\), choose a positive integer \(n>1/x\). Then \(1/n<x\), so \(x\notin[0,1/n]\). Hence no positive \(x\) belongs to every set, and

\bigcap_{n=1}^{\infty}[0,1/n]=\{0\}.

The theorem supplies existence from finite consistency; the final calculation identifies the point in this particular example.

Nested Closed Sets and Common Points

A particularly common way to obtain the finite intersection property is through nesting. A sequence \((K_n)\) is nested decreasing if \(K_{n+1}\subseteq K_n\) for every \(n\). In any finite selection \(K_{n_1},\ldots,K_{n_m}\), the set with the largest index is contained in all the others. If each set is nonempty, that finite intersection is therefore nonempty.

Theorem (Nested Closed Sets in a Compact Space): Let \(X\) be a compact metric space, and let \((K_n)\) be a sequence of nonempty closed subsets of \(X\) such that \(K_{n+1}\subseteq K_n\) for every \(n\). Then \(\bigcap_{n=1}^{\infty}K_n\ne\varnothing\).

Proof. Consider the family \(\{K_n:n\geq1\}\). Take any finite subfamily, with indices \(n_1,\ldots,n_m\), and let \(N=\max\{n_1,\ldots,n_m\}\). Repeated nesting gives \(K_N\subseteq K_{n_j}\) for every \(j\). Since \(K_N\ne\varnothing\), the finite intersection \(K_{n_1}\cap\cdots\cap K_{n_m}\) is nonempty. The family thus has the finite intersection property. Each \(K_n\) is closed in the compact space \(X\), so the Finite Intersection Property Theorem gives \(\bigcap_{n=1}^{\infty}K_n\ne\varnothing\). \(\square\)

The hypotheses are easy to check but should not be blurred. The sets must be nonempty, closed in the compact space, and nested in the stated direction. Nonemptiness of each individual set does not by itself imply a common point; nesting is what ensures finite consistency. Closedness and compactness then turn finite consistency into full consistency.

Worked Example: Nested Intervals Inside the Unit Interval

For each positive integer \(n\), define

K_n=\left[\frac13,\frac13+\frac1{n+1}\right]\subseteq[0,1].

These are nonempty closed subsets of \([0,1]\). The upper endpoint is at most \(1/3+1/2=5/6<1\), so each interval really is contained in the unit interval. Also \(1/(n+2)<1/(n+1)\), which gives \(K_{n+1}\subseteq K_n\). The nested closed sets theorem guarantees a common point.

The left endpoint \(1/3\) belongs to every interval. If \(x>1/3\), then \(x-1/3>0\), so there is a positive integer \(n\) large enough that \(1/(n+1)<x-1/3\). For this \(n\), the upper endpoint of \(K_n\) is less than \(x\), so \(x\notin K_n\). Points smaller than \(1/3\) lie in none of the intervals. Consequently,

\bigcap_{n=1}^{\infty}K_n=\left\{\frac13\right\}.

The inclusion check at the first index is part of the proof: since \(n\geq1\), the largest added length is \(1/2\), not \(1\).

When the Common Point Must Be Unique

Nesting guarantees at least one common point, but it does not usually guarantee only one. To force uniqueness, it is enough for the sets to become arbitrarily small in diameter. For a nonempty subset \(A\) of a metric space, its diameter is \(\operatorname{diam}(A)=\sup\{d(x,y):x,y\in A\}\), when this supremum is finite. Subsets of a compact metric space are bounded, so the diameters of the nonempty \(K_n\) in the theorem above are finite.

Corollary: Under the hypotheses of the Nested Closed Sets Theorem, if \(\operatorname{diam}(K_n)\to0\), then \(\bigcap_{n=1}^{\infty}K_n\) consists of exactly one point.

Proof. The nested closed sets theorem gives at least one point in the intersection. Let \(p\) and \(q\) be any two points in that intersection. Then \(p,q\in K_n\) for every \(n\), and so \(d(p,q)\leq\operatorname{diam}(K_n)\) for every \(n\). Since these diameters tend to zero, \(d(p,q)=0\). The defining property of a metric gives \(p=q\). Thus the intersection has exactly one point. \(\square\)

This corollary separates two tasks that are sometimes accidentally combined. Compactness and finite consistency prove existence. The diameter estimate proves uniqueness. A complete argument should state which step supplies each conclusion.

Worked Example: Shrinking Intervals Determine One Point

Use the intervals \(K_n=[1/3,\,1/3+1/(n+1)]\) from the preceding example. Their diameters are

\operatorname{diam}(K_n) =\left(\frac13+\frac1{n+1}\right)-\frac13 =\frac1{n+1}.

For any \(\varepsilon>0\), choose a positive integer \(N\) such that \(N+1>1/\varepsilon\). If \(n\geq N\), then \(1/(n+1)\leq1/(N+1)<\varepsilon\), so the diameters tend to zero. The corollary therefore proves that the intersection has exactly one point. The earlier endpoint calculation identifies it as \(1/3\).

The diameter calculation also shows why an interval family with a fixed positive length would not establish uniqueness: two distinct points could remain inside every set.

Why Compactness Cannot Be Omitted

The finite intersection property theorem depends on compactness, not merely on having closed sets or nested sets. The next example gives a decreasing sequence of nonempty relatively closed sets in a metric space whose full intersection is empty. Each finite stage remains consistent, but there is no compactness principle available to produce a point satisfying every constraint.

Worked Example: Nested Closed Sets with Empty Intersection

Take the noncompact space \(X=(0,1)\) with its usual metric, and for each integer \(n\geq2\) set \(F_n=[1/n,1)\), viewed as a subset of \(X\). Its complement in \(X\) is \((0,1/n)\), which is open in \(X\); thus \(F_n\) is closed relative to \(X\). Each \(F_n\) is nonempty, and \(F_{n+1}\subseteq F_n\).

Every finite intersection is nonempty: for indices \(n_1,\ldots,n_m\), the intersection is \(F_N\), where \(N\) is the largest index, and \(1/N\in F_N\). But the full intersection is empty. Indeed, given any \(x\in(0,1)\), choose an integer \(n\geq2\) with \(n>1/x\). Then \(1/n<x\), so \(x\notin F_n\). Therefore no \(x\) belongs to all the sets.

There is no contradiction with the nested closed sets theorem, because \(X=(0,1)\) is not compact. This example also illustrates why a proof must check that closedness is relative to the space being used: the sets are closed in \(X\), even though they are not closed as subsets of the real line.

A Compactness Proof Checklist

When a problem asks for a point satisfying infinitely many conditions, the following sequence of questions can make the proof more reliable:

Identify the ambient space.
State which space is compact, and check that the sets are closed in that space rather than only in a larger surrounding space.

Check finite consistency.
For a general family, verify the finite intersection property. For a nested sequence, show explicitly that a finite intersection reduces to the set with the largest index.

Invoke compactness for existence.
Use the finite intersection property theorem, or its nested-set consequence, to obtain a common point.

Prove uniqueness separately if needed.
Show that two common points have distance at most \(\operatorname{diam}(K_n)\) for every \(n\), and use a diameter limit when one is available.

A common pitfall is to treat “each set is nonempty” as if it implied “their intersection is nonempty.” It does not. Another is to use a conclusion about compact spaces when the sets are closed only in a different ambient space. Keeping the ambient space, finite intersections, and limiting argument visible makes these proofs both shorter and safer.

Check Your Understanding

Use the finite intersection property and nested-set arguments to answer the following questions.

Why are complements of closed sets the right objects to use when proving the finite intersection property theorem?
For a finite collection chosen from a decreasing sequence \(K_1\supseteq K_2\supseteq\cdots\), which set is contained in every member of that collection?
Which part of the nested closed sets argument proves existence, and what additional condition proves uniqueness?
In the example on \((0,1)\), why is each \(F_n=[1/n,1)\) closed relative to \(X\), and why is the full intersection empty?
What must be checked before applying a compactness theorem to closed subsets of a metric space?