From Open Covers to Closed Sets
Compactness is defined using open covers: every open cover must have a finite subcover. There is an equivalent way to express the same condition using closed sets. Instead of asking whether some finite collection of open sets covers the space, we ask whether finite collections of closed sets can have a common point. This translation is particularly useful for products, where closed conditions on several coordinates can be gathered into one family.
Throughout, \((X,\mathcal{T})\) is a topological space. A family \(\{F_\alpha:\alpha\in A\}\) of subsets of \(X\) has the finite intersection property, abbreviated FIP, if the intersection of every finite subfamily is nonempty. We require \(A\ne\varnothing\) when using the closed-set characterization below. For a nonempty finite set of indices \(J\subseteq A\), the condition is \(\bigcap_{\alpha\in J}F_\alpha\ne\varnothing\).
Proof. Suppose first that \(X\) is compact, and let \(\{F_\alpha:\alpha\in A\}\) be a nonempty family of closed sets with the FIP. Assume, to obtain a contradiction, that \(\bigcap_{\alpha\in A}F_\alpha=\varnothing\). Taking complements gives \(\bigcup_{\alpha\in A}(X\setminus F_\alpha)=X\), so the open sets \(X\setminus F_\alpha\) form an open cover of \(X\). Compactness gives a finite subcover, say \(X\setminus F_{\alpha_1},\ldots,X\setminus F_{\alpha_k}\). Taking complements of that finite union gives \(F_{\alpha_1}\cap\cdots\cap F_{\alpha_k}=\varnothing\), contrary to the FIP. Thus the total intersection is nonempty.
Conversely, suppose every nonempty family of closed sets with the FIP has nonempty total intersection. Let \(\{U_\alpha:\alpha\in A\}\) be an open cover of \(X\). Suppose it has no finite subcover. Then \(A\ne\varnothing\), and \(F_\alpha=X\setminus U_\alpha\) is closed for every \(\alpha\). For any finite nonempty set \(J\subseteq A\), the sets \(U_\alpha\), \(\alpha\in J\), do not cover \(X\); otherwise they would give a finite subcover. Therefore \(\bigcap_{\alpha\in J}F_\alpha=X\setminus\bigcup_{\alpha\in J}U_\alpha\ne\varnothing\). The family \(\{F_\alpha:\alpha\in A\}\) has the FIP. By the assumed property, its total intersection is nonempty. But that intersection equals \(X\setminus\bigcup_{\alpha\in A}U_\alpha=\varnothing\), since the \(U_\alpha\) cover \(X\). This contradiction proves that every open cover has a finite subcover, so \(X\) is compact. \(\square\)
The two directions use complements, but their logic runs in opposite ways. A cover with no finite subcover produces closed sets whose finite intersections are all nonempty. A family of closed sets with the FIP but empty total intersection produces an open cover with no finite subcover.
Tychonoff’s Theorem in Closed-Set Form
For a family of spaces \(\{X_i:i\in I\}\), the product \(\prod_{i\in I}X_i\) is understood to have the product topology: a basic open set restricts coordinates in only finitely many positions and leaves the other coordinates unrestricted. Tychonoff’s theorem says that if every \(X_i\) is compact, then the product is compact in this topology. Its open-cover form is the standard result proved by the ultrafilter method in the earlier material.
Proof. By the open-cover form of Tychonoff’s theorem, the product is compact. Apply the closed-set characterization of compactness to this product: every nonempty family of its closed subsets with the FIP has nonempty total intersection. \(\square\)
This formulation does not say that an arbitrary family of closed sets has a common point. The finite intersection property is essential: every finite selection must have a common point. Nor does the conclusion say that a particular finite intersection is enough; it concerns the intersection of the entire family.
There are harmless degenerate cases to keep in view. If some factor is empty, the product is empty. With our convention that the family in the closed-set statement is nonempty, there is no FIP family of closed subsets of the empty product: every subset is empty, and a one-set subfamily already has empty intersection. Thus the closed-set statement holds vacuously, just as the empty space is compact by the open-cover definition. If the index set \(I\) is empty, the product is a singleton, which is compact.
Worked Example: Nested Closed Sets in an Interval
In \(X=[0,1]\) with its subspace topology from \(\mathbb{R}\), consider the family \(F_n=[0,1/n]\), for \(n\in\mathbb{N}\). Each \(F_n\) is closed in \(X\): its complement in \(X\) is \((1/n,1]\), which is open in the subspace topology because it is \(X\cap(1/n,2)\).
Take any finite nonempty selection of indices \(J\subseteq\mathbb{N}\), and let \(N\) be its largest index. If \(n\in J\), then \(n\le N\), so \(1/N\le 1/n\). Consequently, every point of \([0,1/N]\) belongs to each \(F_n\) selected, and the selected intersection is exactly \(\bigcap_{n\in J}F_n=[0,1/N]\). It is nonempty, since it contains \(0\). Thus the family has the FIP.
A point \(x\) belongs to the total intersection precisely when \(0\le x\le 1/n\) for every \(n\in\mathbb{N}\). The point \(0\) satisfies all these inequalities. If \(x>0\), choose \(n\in\mathbb{N}\) with \(n>1/x\); then \(1/n<x\), so \(x\notin F_n\). Hence \(\bigcap_{n\in\mathbb{N}}F_n=\{0\}\), which is nonempty. This agrees with the closed-set characterization, since \([0,1]\) is compact by the Heine–Borel theorem.
Worked Example: Closed Coordinate Conditions in a Product
Let \(X=\prod_{n\in\mathbb{N}}[0,1]\) with the product topology, and write a point of \(X\) as \(x=(x_1,x_2,\ldots)\). Each factor is compact by the Heine–Borel theorem, so Tychonoff’s theorem applies. For each \(n\), define \(C_n=\{x\in X:x_n\in[0,1/n]\}\).
Each \(C_n\) is closed. Indeed, the coordinate projection \(\pi_n:X\to[0,1]\) is continuous: the preimage of an open set \(V\subseteq[0,1]\) is the basic open set restricting coordinate \(n\) to \(V\). Since \([0,1/n]\) is closed in \([0,1]\), its preimage \(C_n=\pi_n^{-1}([0,1/n])\) is closed.
For any finite nonempty set \(J\subseteq\mathbb{N}\), the point \(x=(0,0,\ldots)\) belongs to \(C_n\) for every \(n\in J\), because its \(n\)-th coordinate is \(0\) and \(0\in[0,1/n]\). Thus every finite intersection is nonempty. The same point belongs to every \(C_n\), so the total intersection is nonempty as well. More precisely, it consists of all points whose \(n\)-th coordinate lies in \([0,1/n]\) for every \(n\). The example illustrates how closed conditions can be imposed coordinate by coordinate while using the product theorem to control their common intersection.
Check: for the finite selection \(J=\{2,5\}\), the point \((0,0,\ldots)\) has second coordinate \(0\in[0,1/2]\) and fifth coordinate \(0\in[0,1/5]\), so it belongs to \(C_2\cap C_5\). It also belongs to every other \(C_n\), verifying membership in the total intersection directly.
What Fails Without Compactness
Worked Example: A Closed Family on the Real Line
Give \(X=\mathbb{R}\) its standard topology, and let \(U_n=(-n,n)\) for \(n\in\mathbb{N}\). These sets form an open cover: for any \(x\in\mathbb{R}\), the Archimedean property gives an integer \(n>|x|\), so \(x\in(-n,n)\).
No finite selection covers \(\mathbb{R}\). For a finite nonempty set \(J\subseteq\mathbb{N}\), let \(N\) be its largest element. The selected intervals have union \((-N,N)\), which omits \(N\). Thus the cover has no finite subcover and \(\mathbb{R}\) is not compact.
Taking complements gives closed sets \(F_n=\mathbb{R}\setminus U_n=(-\infty,-n]\cup[n,\infty)\). For finite nonempty \(J\) with largest element \(N\), their intersection is \(\bigcap_{n\in J}F_n=(-\infty,-N]\cup[N,\infty)\), which is nonempty. The family has the FIP. Its total intersection, however, is empty: given any \(x\in\mathbb{R}\), choose \(n>|x|\); then \(-n<x<n\), so \(x\notin F_n\). This is exactly the closed-set failure that the characterization predicts for a noncompact space.
A Common Mistake: Pairwise Is Not Finite
The FIP requires nonempty intersections for all finite subfamilies, not merely for pairs. For a concrete check, in \(\mathbb{R}\) with the standard topology let \(F_1=\{0,1\}\), \(F_2=\{1,2\}\), and \(F_3=\{0,2\}\). These sets are closed. The pairwise intersections are \(F_1\cap F_2=\{1\}\), \(F_1\cap F_3=\{0\}\), and \(F_2\cap F_3=\{2\}\), so every pair intersects. But \(F_1\cap F_2\cap F_3=\varnothing\). Therefore this family does not have the FIP. The incorrect calculation is to check the three pairwise intersections and conclude that the family has the FIP; the correction is to check the three-set subfamily too.
The closed-set formulation matters because it makes the finite-versus-total distinction explicit. In a compact space, finite consistency of a nonempty family of closed conditions guarantees a point satisfying all of them. In a noncompact space, as the intervals \((-n,n)\) demonstrate, finite consistency can hold while the full family has no common point.
Check Your Understanding
Use the complement correspondence between open sets and closed sets to justify your answers.
- State the finite intersection property and explain why it must be checked for every finite subfamily.
- Let \(X\) be compact and let \(\{F_\alpha\}\) be a nonempty FIP family of closed subsets. What would a finite subcover of \(\{X\setminus F_\alpha\}\) imply?
- For \(F_n=[1/n,1]\subseteq[0,1]\), determine whether the family has the FIP and compute its total intersection.
- Why does the family \(\{(-\infty,-n]\cup[n,\infty):n\in\mathbb{N}\}\) not contradict the closed-set characterization of compactness?
- What convention about the family is used in the closed-set characterization, and why is it relevant when \(X=\varnothing\)?