Compact Sets

Compactness Has Consequences Beyond Finite Subcovers

In “Finite Subcovers,” compactness was used to pass from an open cover to a finite selection. That definition also has consequences for the shape of a set and for families of closed subsets. In this tutorial, we use compactness to prove that compact subsets of the real line are closed, and then to establish a useful intersection principle.

Recall that \(K\subseteq\mathbb{R}\) is compact if every open cover of \(K\) has a finite subcover. The empty set is compact, and every finite subset of \(\mathbb{R}\) is compact. Also, by the result in “What Is Compactness?,” every compact subset of \(\mathbb{R}\) is bounded. The results here add a different conclusion: compactness also ensures that points outside \(K\) have open neighborhoods that miss \(K\).

Every Compact Set Is Closed

To prove that a set is closed, it is enough to show that its complement is open. So fix a point \(x\) outside a compact set \(K\), and aim to find an open interval around \(x\) that does not meet \(K\). For each \(y\in K\), the points \(x\) and \(y\) are distinct. We can therefore choose disjoint open intervals around them. Compactness lets us retain only finitely many of the intervals around points of \(K\); the intersection of the corresponding intervals around \(x\) will still be a neighborhood of \(x\) and will miss all of \(K\).

Proof. If \(K=\varnothing\), then \(K\) is closed. Suppose \(K\neq\varnothing\), and fix any \(x\in\mathbb{R}\setminus K\). For each \(y\in K\), let \(d_y=|x-y|\), which is positive because \(x\notin K\). Define

Both intervals are open, and \(y\in V_y\). They are disjoint: if some \(z\) belonged to both, then

contradicting \(d_y=|x-y|\). The family \(\{V_y:y\in K\}\) is an open cover of \(K\). By compactness, finitely many of these intervals cover \(K\), say \(V_{y_1},\ldots,V_{y_m}\). Consider the finite intersection

The point \(x\) belongs to every \(W_{y_i}\), so \(x\in W\). By the theorem on finite intersections of open sets, \(W\) is open. Moreover, \(W\cap V_{y_i}=\varnothing\) for each \(i\): a point in that intersection would belong to \(W_{y_i}\cap V_{y_i}\), which is empty. Since \(K\subseteq\bigcup_{i=1}^{m}V_{y_i}\), it follows that \(W\cap K=\varnothing\). Thus every \(x\notin K\) has an open neighborhood contained in \(\mathbb{R}\setminus K\), so \(\mathbb{R}\setminus K\) is open and \(K\) is closed. \(\square\)

The finite selection is essential. The intervals \(W_y\) depend on \(y\), and an arbitrary intersection of open sets need not be open. Compactness reduces the argument to a finite intersection, where openness is guaranteed.

Closed Subsets of a Compact Set and Intersections

Compactness also puts a limit on how a family of closed subsets can behave. If finitely many closed sets always overlap inside a compact set, then the entire family must have a common point. Otherwise, their open complements would cover the compact set, and a finite subcover would produce a finite collection with no common point.

A family of sets has the finite intersection property if the intersection of every finite subfamily is nonempty. When discussing a family indexed by an empty set, its intersection is taken to be the whole underlying set. Thus, for a nonempty underlying set, the empty subfamily also satisfies the nonempty-intersection requirement.

Proof. Suppose, for contradiction, that \(\bigcap_{\alpha\in A}F_\alpha=\varnothing\). For each \(\alpha\in A\), the set \(\mathbb{R}\setminus F_\alpha\) is open because \(F_\alpha\) is closed. These open sets cover \(K\): given \(x\in K\), the assumed empty total intersection means that \(x\) fails to belong to at least one \(F_\alpha\), so \(x\in\mathbb{R}\setminus F_\alpha\) for that index.

Compactness gives a finite subcover, with indices \(\alpha_1,\ldots,\alpha_m\). Since \(K\) is nonempty, this selection cannot be empty. The fact that these complements cover \(K\) says that no point of \(K\) belongs to all of \(F_{\alpha_1},\ldots,F_{\alpha_m}\). But each \(F_{\alpha_i}\) is contained in \(K\), so their intersection is contained in \(K\). Consequently,

This contradicts the finite intersection property. The total intersection must therefore be nonempty. \(\square\)

The theorem’s hypotheses matter. The sets \(F_\alpha\) must be closed in \(\mathbb{R}\), so their complements are open sets to which compactness applies. They must also be contained in \(K\). And \(K\) must be nonempty: if \(K=\varnothing\), the empty finite intersection convention would not give the required nonempty intersection property.

Closedness Is Not the Same as Compactness

Compactness implies closedness, but closedness alone does not imply compactness. For example, \(\mathbb{R}\) is closed: its complement is empty, which is open. Consider the open cover \(U_n=(-n,n)\), indexed by integers \(n\geq1\). It covers \(\mathbb{R}\), because for each real \(x\), the Archimedean property gives \(n>|x|\), and then \(-n<x<n\).

No finite selection covers \(\mathbb{R}\). The empty selection fails because \(\mathbb{R}\) is nonempty. Any nonempty finite selection has a largest index \(N\); each selected interval is contained in \((-N,N)\). The point \(N+1\) lies in \(\mathbb{R}\) but not in \((-N,N)\), so it is not covered by the selection. Thus \(\mathbb{R}\) is closed but not compact. This example is consistent with the earlier result that compact subsets of \(\mathbb{R}\) are bounded: \(\mathbb{R}\) is unbounded.

The finite intersection theorem offers a second way to use compactness. Rather than searching directly for a common point in infinitely many closed sets, verify that every finite selection has a common point. Compactness then guarantees a point common to the entire family. The conclusion depends on checking every finite subfamily, including the empty one under the stated convention; checking only the first few intersections is not enough.