From Nesting to a Common Point
A sequence of sets can keep shrinking while every set remains nonempty. In a compact space, if the sets are also closed and nested, they cannot shrink past every point: there is at least one point belonging to all of them. The key is to turn nesting into a condition about every finite collection of sets, then use compactness.
Throughout, \((X,\mathcal{T})\) is a topological space. Recall that \(X\) is compact if every open cover of \(X\) has a finite subcover. A family of sets is nested, or linearly ordered by inclusion, if for any two members one is contained in the other. In particular, a decreasing sequence satisfies \(C_{n+1}\subseteq C_n\) for each \(n\in\mathbb{N}\).
For a nested family of nonempty sets, the finite intersection property follows from nesting. Indeed, among any finite selection of members there is one contained in all the others. Their intersection is exactly that smallest selected member, so it is nonempty. Compactness will be used to pass from these finite intersections to the intersection of the entire family.
The Closed-Set Form of Compactness
Proof. Suppose first that \(X\) is compact, and let \(\{C_i:i\in J\}\) be a nonempty family of closed sets with the finite intersection property. Suppose, for a contradiction, that \(\bigcap_{i\in J}C_i=\varnothing\). Then every point of \(X\) is outside at least one \(C_i\), so the open sets \(X\setminus C_i\) cover \(X\). Compactness gives a finite subcover \(X\setminus C_{i_1},\ldots,X\setminus C_{i_k}\). Taking complements, \(C_{i_1}\cap\cdots\cap C_{i_k}=\varnothing\), contrary to the finite intersection property. Hence the full intersection is nonempty.
Conversely, suppose every nonempty family of closed sets with the finite intersection property has nonempty intersection. Let \(\{U_i:i\in J\}\) be an open cover of \(X\). If it had no finite subcover, then the closed sets \(C_i=X\setminus U_i\) would have the finite intersection property: an empty finite intersection would mean that the corresponding finite collection of \(U_i\)'s covers \(X\). The assumed property would give a point in every \(C_i\), hence outside every \(U_i\), contradicting that the \(U_i\)'s cover \(X\). Thus the open cover has a finite subcover, and \(X\) is compact. \(\square\)
The Nested Closed Sets Theorem
Proof. Choose any finite nonempty subfamily \(C_{i_1},\ldots,C_{i_k}\). Since the family is nested, these finitely many sets are linearly ordered by inclusion. One of them, say \(C_{i_r}\), is contained in every other one. Therefore \(C_{i_1}\cap\cdots\cap C_{i_k}=C_{i_r}\), which is nonempty by hypothesis. This proves that the family has the finite intersection property. Theorem 1 now applies because all the \(C_i\) are closed and \(X\) is compact. It gives \(\bigcap_{i\in J}C_i\ne\varnothing\). \(\square\)
The proof applies to any nested family, not only a sequence. For a decreasing sequence \(C_1\supseteq C_2\supseteq\cdots\), every finite intersection is the set with the largest index among the finitely many selected indices. That set is nonempty, so the theorem gives a point in every \(C_n\) at once. The conclusion guarantees a common point; it does not say that the intersection has exactly one point.
Worked Example: A Decreasing Sequence in the Unit Interval
Give \(X=[0,1]\) the subspace topology inherited from the standard topology on \(\mathbb{R}\), and let \(C_n=[0,1/n]\) for \(n\in\mathbb{N}\). Each \(C_n\) is nonempty. It is closed in \(X\), since its complement in \(X\), \((1/n,1]\), is \(X\cap(1/n,2)\), which is open in the subspace topology. Also \(C_{n+1}\subseteq C_n\), because \(1/(n+1)\leq 1/n\).
To compute the intersection, \(0\) belongs to every \(C_n\). If \(x\in[0,1]\) and \(x>0\), choose a positive integer \(n>1/x\). Then \(1/n<x\), so \(x\notin[0,1/n]=C_n\). Thus no positive \(x\) belongs to every set, and \(\bigcap_{n\in\mathbb{N}}C_n=\{0\}\).
Check: the computed intersection is nonempty and is contained in every \(C_n\). The point \(0\) satisfies \(0\leq 1/n\) for each positive integer \(n\), while every \(x>0\) was excluded by choosing \(n>1/x\).
Worked Example: A Nested Family with No Last Member
Again take \(X=[0,1]\) with its subspace topology. For each real \(t\in[0,1)\), set \(C_t=[t,1]\). These sets are nonempty and closed in \(X\): their complements are \(X\cap(-1,t)\), open in \(X\). If \(s\leq t\), then \(C_t\subseteq C_s\), so the family is nested. There is no largest index in \([0,1)\), and hence no set \(C_t\) at index \(t=1\).
Take any finite nonempty selection of indices and let \(m\) be their maximum. The finite intersection of the corresponding sets is \([m,1]\), which is nonempty because \(m<1\). For the full intersection, \(1\) belongs to every \(C_t\). If \(x\in[0,1]\) and \(x<1\), choose \(t=(x+1)/2\). Then \(x<t<1\), so \(x\notin C_t\). Consequently \(\bigcap_{t\in[0,1)}C_t=\{1\}\), as required by the theorem.
Check: every finite intersection has the form \([m,1]\) for a selected maximum \(m<1\), and contains \(1\). The only point that survives all indices is \(1\), since each \(x<1\) is excluded by the index \(t=(x+1)/2\).
Why Compactness and the Full Finite Intersection Property Matter
Worked Example: Nested Closed Sets Can Miss in a Noncompact Space
Give \(X=\mathbb{R}\) its standard topology, and define \(C_n=[n,\infty)\) for \(n\in\mathbb{N}\). Each \(C_n\) is nonempty and closed, since its complement \((-\infty,n)\) is open. The sets are decreasing because \(n+1\geq n\) implies \([n+1,\infty)\subseteq[n,\infty)\).
No real number belongs to every \(C_n\): for any \(x\in\mathbb{R}\), choose \(n\in\mathbb{N}\) with \(n>x\), and then \(x\notin[n,\infty)\). Therefore \(\bigcap_{n\in\mathbb{N}}C_n=\varnothing\). This does not contradict Theorem 2, because \(\mathbb{R}\) is not compact. For example, the open sets \(U_n=(-n,n)\) cover \(\mathbb{R}\), but a finite selection has a largest index \(N\) and its union is \((-N,N)\), which does not contain \(N+1\).
Check: every finite intersection \(C_1\cap\cdots\cap C_k\) is \([k,\infty)\), so it is nonempty. Thus the family has the finite intersection property even though its full intersection is empty; the missing hypothesis is compactness.
Common pitfall: checking only pairs. The finite intersection property requires every finite intersection to be nonempty, not merely every pairwise intersection. For a concrete counterexample, let \(X=\{a,b,c\}\) have the discrete topology, in which every subset is open. This space is compact: from any open cover, choose one member containing each of the finitely many points, producing a finite subcover. The closed sets \(A=\{a,b\}\), \(B=\{b,c\}\), and \(C=\{a,c\}\) have pairwise intersections \(A\cap B=\{b\}\), \(A\cap C=\{a\}\), and \(B\cap C=\{c\}\), all nonempty. But \(A\cap B\cap C=\varnothing\). They cannot be a nested family, and they fail the finite intersection property. Pairwise checks alone therefore do not justify applying Theorem 1.
The useful chain of reasoning is precise: nesting makes every finite intersection nonempty; closedness lets the closed-set form of compactness apply; compactness then supplies a point in the intersection of the entire family. In applications, check each of these conditions separately. A family of nonempty sets need not have nonempty finite intersections, and even a nested family of nonempty closed sets can have empty intersection when the ambient space is not compact.
Check Your Understanding
For each question, identify which hypotheses of the nested closed sets theorem are relevant.
- In \(X=[0,1]\) with the subspace topology, compute the intersection of \(C_n=[1-1/n,1]\), \(n\in\mathbb{N}\).
- Explain why the finite intersection of any finite selection from a decreasing sequence \(C_1\supseteq C_2\supseteq\cdots\) equals the set with the largest selected index.
- In the discrete topology on \(\{a,b,c\}\), find three nonempty closed sets whose pairwise intersections are nonempty but whose triple intersection is empty.
- For \(C_n=(-\infty,-n]\) in standard \(\mathbb{R}\), verify nesting and determine the full intersection.
- Does the theorem guarantee that the intersection is a singleton? Give a nested family of nonempty closed sets in a compact space whose intersection has more than one point.