From Open Covers to Closed Intersections
Compactness is defined using open covers: every open cover must contain a finite subcover. There is a complementary way to express the same condition, using closed sets instead. It is useful when a problem gives a family of closed sets and asks whether they have a common point. The key translation is that a finite collection of open sets covers the space exactly when the corresponding finite collection of complements has empty intersection.
Throughout, \((X,\mathcal{T})\) is a topological space. A family of subsets of \(X\) may be indexed, so different indices are allowed to specify the same set. The total intersection of a family \(\{C_i:i\in J\}\) is \(\bigcap_{i\in J}C_i\).
The empty-subfamily convention makes the definition consistent even when \(X=\varnothing\). A family has the FIP only if every finite selection, including the empty selection, has nonempty intersection. Compactness means that every open cover of \(X\) has a finite subcover; in particular, the empty space is compact.
Proof by Taking Complements
Proof. Suppose first that \(X\) is compact, and let \(\{C_i:i\in J\}\) be a family of closed subsets of \(X\) with the FIP. We must show that \(\bigcap_{i\in J}C_i\ne\varnothing\). Assume instead that \(\bigcap_{i\in J}C_i=\varnothing\). For each \(i\in J\), the set \(U_i=X\setminus C_i\) is open. The empty total intersection means that every point of \(X\) is outside at least one \(C_i\), so it belongs to at least one \(U_i\). Therefore \(\{U_i:i\in J\}\) is an open cover of \(X\). By compactness, finitely many of these sets cover \(X\): for some finite \(J_0\subseteq J\), \(X=\bigcup_{i\in J_0}U_i\). Taking complements gives \(\bigcap_{i\in J_0}C_i=\varnothing\), contradicting the FIP. Hence the total intersection is nonempty.
Conversely, suppose every family of closed subsets of \(X\) with the FIP has nonempty total intersection. Let \(\{U_i:i\in J\}\) be an open cover of \(X\). We must show it has a finite subcover. Assume it has none. For each \(i\in J\), define \(C_i=X\setminus U_i\), which is closed. For any finite \(J_0\subseteq J\), the sets \(\{U_i:i\in J_0\}\) do not cover \(X\), since the original cover has no finite subcover. Thus there is a point of \(X\) outside their union, and \(\bigcap_{i\in J_0}C_i=X\setminus\bigcup_{i\in J_0}U_i\ne\varnothing\). This also holds for the empty selection: if \(X\) were empty, the empty subfamily of the open cover would already be a finite subcover. So the closed family has the FIP. But \(\bigcap_{i\in J}C_i=X\setminus\bigcup_{i\in J}U_i=\varnothing\), because the \(U_i\) cover \(X\). This contradicts the assumed closed-set condition. Therefore every open cover has a finite subcover, and \(X\) is compact. \(\square\)
The proof does not require \(X\) to be Hausdorff, or any of the closed sets to be nonempty individually. It uses only the definitions of closed sets, open covers, compactness, and the finite intersection property.
Using the Criterion
Worked Example: Closed Intervals in a Compact Space
Let \(X=[0,1]\) with the subspace topology inherited from the standard topology on \(\mathbb{R}\). For each \(n\in\mathbb{N}\), let \(C_n=[0,1/n]\), considered as a subset of \(X\). These sets are closed in \(X\): each is the intersection of \(X\) with the closed interval \([0,1/n]\) in \(\mathbb{R}\).
For any finite selection of indices \(n_1,\ldots,n_k\), let \(N\) be the largest selected index. Since \(n_j\le N\), we have \(1/N\le 1/n_j\) for every \(j\), and hence \(\bigcap_{j=1}^k C_{n_j}=[0,1/N]\ne\varnothing\). Thus the family has the FIP. For a point \(x\in X\), membership in the total intersection requires \(0\le x\le 1/n\) for every positive integer \(n\). The point \(0\) satisfies all these inequalities. If \(x>0\), choose \(n>1/x\); then \(1/n<x\), so \(x\notin C_n\). Therefore \(\bigcap_{n\in\mathbb{N}}C_n=\{0\}\), which is nonempty, as the criterion predicts. The compactness of \([0,1]\) follows from the open-cover result for closed intervals.
Check: for \(n=4\), \(C_4=[0,1/4]\); for \(n=7\), \(C_7=[0,1/7]\). Their intersection is \([0,1/7]\), since \(1/7<1/4\). This agrees with the finite-intersection calculation using the larger index \(N=7\).
Worked Example: An Infinite Discrete Space Fails the Criterion
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with the discrete topology, in which every subset is open and closed. For \(n\in\mathbb{N}\), set \(C_n=\{n,n+1,n+2,\ldots\}\). Every \(C_n\) is closed. For a finite nonempty selection of indices, let \(N\) be the largest selected index. The intersection of the selected sets is \(C_N\), which contains \(N\), so it is nonempty. The empty selection has intersection \(X\), also nonempty. Hence this family has the FIP.
Its total intersection is empty. Indeed, for any \(m\in\mathbb{N}\), the set \(C_{m+1}\) does not contain \(m\). Thus no \(m\) belongs to every \(C_n\). The criterion shows that \(X\) is not compact. Directly, the open cover \(\{\{n\}:n\in\mathbb{N}\}\) has no finite subcover: a finite union of singletons contains only finitely many natural numbers, whereas \(X\) is infinite.
Check: \(C_2=\{2,3,4,\ldots\}\) and \(C_5=\{5,6,7,\ldots\}\), so \(C_2\cap C_5=C_5\), which is nonempty. Yet \(1\notin C_2\), \(2\notin C_3\), and, in general, \(m\notin C_{m+1}\); the total intersection has no point.
Worked Example: Checking the Criterion in the Cofinite Topology
Let \(X\) be an infinite set with the cofinite topology: the open sets are \(\varnothing\) and the sets whose complements in \(X\) are finite. The closed sets are \(X\) and the finite subsets of \(X\). We verify directly that every family of closed sets with the FIP has nonempty total intersection.
Let \(\mathcal{C}\) be such a family. If every member of \(\mathcal{C}\) is \(X\), then its total intersection is \(X\), which is nonempty; this also covers the empty family. Otherwise, choose a finite member \(F_0\in\mathcal{C}\). The FIP implies \(F_0\ne\varnothing\). For each \(C\in\mathcal{C}\), define \(G_C=F_0\cap C\). Every finite intersection of these sets is nonempty: for \(C_1,\ldots,C_k\in\mathcal{C}\), \(G_{C_1}\cap\cdots\cap G_{C_k}=F_0\cap C_1\cap\cdots\cap C_k\ne\varnothing\), by the FIP for \(\mathcal{C}\).
We now show \(\bigcap_{C\in\mathcal{C}}G_C\ne\varnothing\). If it were empty, then for each \(x\in F_0\) there would be some \(C_x\in\mathcal{C}\) with \(x\notin G_{C_x}\). There are only finitely many points in \(F_0\), so the selected sets \(G_{C_x}\), one for each \(x\in F_0\), form a finite subfamily. Their intersection is empty: it is contained in \(F_0\), and it omits every \(x\in F_0\). This contradicts the finite-intersection property just proved for the \(G_C\). Therefore the total intersection is nonempty. Since \(\bigcap_{C\in\mathcal{C}}G_C=F_0\cap\bigcap_{C\in\mathcal{C}}C\), the original family also has nonempty total intersection.
The criterion now implies that this infinite cofinite space is compact. The argument also shows why finite closed sets are especially useful here: once one such member \(F_0\) is chosen, the question of a common point reduces to checking subsets of the finite set \(F_0\).
Finite Intersections Are Not Pairwise Intersections
A frequent error is to check only that every two members of a family intersect. The FIP requires nonempty intersection for every finite subfamily, including selections of three or more sets. Here is a finite example where pairwise intersections all exist but the FIP fails.
Worked Example: Pairwise Intersections Do Not Suffice
Give \(X=\{1,2,3\}\) the discrete topology, so all its subsets are closed. Let \(A=\{1,2\}\), \(B=\{2,3\}\), and \(C=\{1,3\}\). The pairwise intersections are \(A\cap B=\{2\}\), \(A\cap C=\{1\}\), and \(B\cap C=\{3\}\). Each is nonempty. But \(A\cap B\cap C=\varnothing\): the only point in \(A\cap B\) is \(2\), and \(2\notin C\). Thus this family does not have the FIP. There is no contradiction with the criterion: \(X\) is compact, but the criterion applies only to families satisfying the full finite intersection property.
Correction to the mistaken argument: pairwise checks establish only the condition for subfamilies of size two. They do not establish the condition for the three-set selection \(\{A,B,C\}\). The computation of the triple intersection is necessary, and it is empty.
In applications, the complement method works in either direction. To prove that a space is compact, start with a closed family having the FIP and seek a common point. To prove that a space is not compact, it is enough to give a closed family with the FIP and empty total intersection; equivalently, its complements form an open cover with no finite subcover. The criterion concerns arbitrary families, not only sequences of sets, and it does not assert that the closed sets must be nested.
Check Your Understanding
For each question, identify the finite subfamilies that must be checked; pairwise intersections alone are not enough.
- In the discrete space \(X=\{a,b,c\}\), can the family \(\{\{a,b\},\{b,c\},\{b\}\}\) have the FIP? Find its total intersection.
- For \(C_n=[-1/n,1/n]\subseteq\mathbb{R}\), determine every finite intersection and the total intersection. Does this family have the FIP?
- Explain why the complements of a closed family with empty total intersection form an open cover.
- In the infinite discrete space \(\mathbb{N}\), use the sets \(D_n=\{1,\ldots,n\}\) to test whether the family \(\{D_n:n\in\mathbb{N}\}\) has the FIP and whether its total intersection is empty.
- Why does a family of three sets require checking its triple intersection in addition to its three pairwise intersections?