From Open Covers to Closed Intersections
Compactness is defined using open covers: every open cover has a finite subcover. Sometimes, however, the problem describes closed sets and asks whether they have a common point. Taking complements converts one question into the other. The key is to retain the quantifier over every finite subfamily; pairwise intersections alone do not suffice.
Let \((X,\mathcal{T})\) be a topological space. A family of subsets of \(X\) has the finite-intersection property if the intersection of every finite subfamily is nonempty. We include the empty finite subfamily in this definition: its intersection is \(X\). Thus a family has the finite-intersection property only if \(X\ne\varnothing\). This convention also handles the empty space: its empty family does not have the finite-intersection property.
Proof. Suppose first that \(X\) is compact, and let \(\{F_\alpha\}_{\alpha\in J}\) be a family of closed subsets with the finite-intersection property. Assume, for contradiction, that \(\bigcap_{\alpha\in J}F_\alpha=\varnothing\). Taking complements gives \(\bigcup_{\alpha\in J}(X\setminus F_\alpha)=X\), so the open sets \(X\setminus F_\alpha\) form an open cover of \(X\). Compactness supplies a finite subcover, say \(X\setminus F_{\alpha_1},\ldots,X\setminus F_{\alpha_k}\). Their union being \(X\) means \(F_{\alpha_1}\cap\cdots\cap F_{\alpha_k}=\varnothing\), contrary to the finite-intersection property. Hence the total intersection is nonempty.
Conversely, suppose every family of closed subsets with the finite-intersection property has nonempty total intersection. Let \(\{U_\alpha\}_{\alpha\in J}\) be an open cover of \(X\). If it had no finite subcover, the closed sets \(F_\alpha=X\setminus U_\alpha\) would have the finite-intersection property: for any finite selection, an empty intersection of those complements would mean that the corresponding finite selection of the \(U_\alpha\) covers \(X\). But \(\bigcap_{\alpha\in J}F_\alpha=\varnothing\), since the \(U_\alpha\) cover \(X\). This contradicts the assumed closed-set condition. Therefore every open cover has a finite subcover, so \(X\) is compact. If \(X=\varnothing\), every open cover already has the empty finite subcover, so the same conclusion holds. \(\square\)
The proof is a translation, not a new compactness assumption: an open cover has no finite subcover exactly when its complementary closed sets have the finite-intersection property but empty total intersection. Use the theorem when the closed sets are easier to describe than the open cover.
Applying the Criterion to Nested Closed Sets
Proof. Consider any finite selection \(F_{n_1},\ldots,F_{n_k}\), where \(k\ge 1\), and let \(N\) be the largest of the selected indices. Since the sets decrease as the index increases, \(F_N\) is contained in every selected set. Therefore \(F_N\subseteq F_{n_1}\cap\cdots\cap F_{n_k}\). The intersection is nonempty because \(F_N\) is nonempty. The empty finite selection has intersection \(X\), which is nonempty because it contains \(F_1\). Thus the family has the finite-intersection property, and compactness plus the theorem gives a nonempty total intersection. \(\square\)
Worked Example: Nested Intervals in the Compact Interval
Use the criterion to find the intersection of \(F_n=[0,1/n]\), for \(n\in\mathbb{N}\), regarded as subsets of \([0,1]\) with its subspace topology from \(\mathbb{R}\).
Each \(F_n\) is closed in \([0,1]\): its relative complement is \((1/n,1]\), which is open in the subspace topology. Each is nonempty because it contains \(0\), and \(F_{n+1}\subseteq F_n\). The space \([0,1]\) is compact by the Heine–Borel theorem. Therefore the nested-set corollary shows that the total intersection is nonempty.
To determine it exactly, \(0\) belongs to every \(F_n\), so \(0\in\bigcap_{n\in\mathbb{N}}F_n\). If \(x\in[0,1]\) and \(x>0\), choose \(n\in\mathbb{N}\) with \(n>1/x\). Then \(1/n<x\), so \(x\notin[0,1/n]=F_n\). No positive \(x\) lies in every set. Consequently, \(\bigcap_{n\in\mathbb{N}}F_n=\{0\}\).
Check: The point \(0\) satisfies \(0\le 1/n\) for every \(n\in\mathbb{N}\). Every other point \(x\in[0,1]\) is positive and is excluded by an index \(n>1/x\). This verifies the computed intersection.
Using Closed Families to Show Compactness
The criterion can also prove compactness: instead of starting with an arbitrary open cover, start with an arbitrary family of closed sets that has the finite-intersection property. On a finite set this verification is short, because there are only finitely many points to check.
Worked Example: A Finite Discrete Space
Let \(X=\{a,b,c,d\}\) have the discrete topology, in which every subset is open. Show that \(X\) is compact using the closed-set criterion, and calculate the intersection of \(F_1=\{a,b,c\}\), \(F_2=\{b,c,d\}\), and \(F_3=\{b,c\}\).
Every subset of \(X\) is also closed, since its complement is a subset of \(X\) and is therefore open. To verify compactness by the criterion, take any family \(\mathcal{F}\) of closed subsets with the finite-intersection property. The empty finite selection has intersection \(X\), so the property applies as stated. Since the whole family has the finite-intersection property, its intersection is nonempty: if it were empty, then for each of the four points \(x\in X\), at least one member \(C_x\in\mathcal{F}\) would omit \(x\). The at most four selected sets \(C_a,C_b,C_c,C_d\) would then have empty intersection, contradicting the finite-intersection property. The closed-set characterization now proves \(X\) compact.
For the listed sets, \(F_1\cap F_2=\{b,c\}\), and \(\{b,c\}\cap F_3=\{b,c\}\). Thus \(F_1\cap F_2\cap F_3=\{b,c\}\). Each single set is nonempty, each pairwise intersection is \(\{b,c\}\), and the triple intersection is \(\{b,c\}\). The three sets therefore have the finite-intersection property and their common intersection is nonempty.
Check: The point \(b\) belongs to all three sets: it is in \(\{a,b,c\}\), in \(\{b,c,d\}\), and in \(\{b,c\}\). The point \(c\) does as well, while neither \(a\) nor \(d\) belongs to all three.
When the Property Fails: A Noncompact Example
To use the criterion in the reverse direction, find closed sets whose every finite intersection is nonempty but whose total intersection is empty. Such a family certifies that the space is not compact. Relative closedness matters: a set can be closed in a subspace even if it is not closed in the ambient real line.
Worked Example: Closed Sets in the Open Interval
For each integer \(n\ge2\), let \(F_n=(0,1/n]\) in \(X=(0,1)\), with the subspace topology from \(\mathbb{R}\). Show that the family has the finite-intersection property but empty total intersection.
First, \(F_n\) is closed relative to \(X\). Indeed, its complement in \(X\) is \((1/n,1)\), which is open in \(X\). Each \(F_n\) is nonempty because it contains \(1/n\), and \(1/n\in(0,1)\) for \(n\ge2\).
Take any nonempty finite selection of these sets, and let \(N\) be the largest selected index. The intervals decrease with the index, so the intersection of the selected sets is \(F_N=(0,1/N]\), which is nonempty. The empty finite selection has intersection \(X\), also nonempty. Hence the family has the finite-intersection property.
Now let \(x\in X\). Since \(x>0\), there is an integer \(n\ge2\) with \(n>1/x\). This gives \(1/n<x\), so \(x\notin(0,1/n]=F_n\). Thus no \(x\in X\) lies in every \(F_n\), and \(\bigcap_{n\ge2}F_n=\varnothing\). The closed-set characterization therefore shows that \((0,1)\) is not compact.
Check by complements: \(X\setminus F_n=(1/n,1)\). For each \(x\in(0,1)\), choosing \(n\ge2\) with \(n>1/x\) gives \(1/n<x<1\), so \(x\in(1/n,1)\). These open sets cover \(X\). No finite selection covers \(X\), because the complement of any finite selection is the nonempty intersection of its corresponding \(F_n\). This agrees with the closed-set calculation.
A Common Pitfall: Pairwise Is Not Enough
The finite-intersection property requires all finite intersections to be nonempty, not merely all pairwise intersections. For two sets, pairwise intersection is the whole condition; for three or more sets, it is only part of the condition.
Common pitfall. In the discrete space \(X=\{a,b,c\}\), consider \(A=\{a,b\}\), \(B=\{b,c\}\), and \(C=\{a,c\}\). A mistaken argument says that all three have a common point because every pair intersects: \(A\cap B=\{b\}\), \(A\cap C=\{a\}\), and \(B\cap C=\{c\}\). But the full intersection is \(A\cap B\cap C=\varnothing\): the only point in \(A\cap B\) is \(b\), and \(b\notin C\). The family does not have the finite-intersection property, despite its nonempty pairwise intersections. Since this finite discrete space is compact, the example does not contradict the theorem; it shows that the theorem’s hypothesis has not been met.
The practical test is therefore to examine an arbitrary finite selection, or to use a structural feature such as nestedness that determines every finite intersection. Once that check is complete, compactness can supply a point common to the entire family. Conversely, an empty total intersection from a family that passes every finite check is a direct compactness contradiction.
Check Your Understanding
For each question, specify the finite intersections being used and distinguish relative closedness from closedness in \(\mathbb{R}\) when needed.
- State the closed-set characterization of compactness, including what the empty finite intersection means.
- In \([0,1]\), let \(G_n=[1/n,1]\) for \(n\in\mathbb{N}\). Determine the intersection of the first \(N\) sets and the total intersection.
- Why does a family of nested nonempty closed sets in a compact space have the finite-intersection property?
- For \(X=(0,1)\), verify that \(H_n=(0,1/n]\), \(n\ge2\), is closed relative to \(X\), and explain why its total intersection is empty.
- Give three subsets of a finite discrete space that intersect pairwise but have empty triple intersection.