When Nested Sets Must Share a Point
A sequence of sets can become smaller at each step without any single point being selected in advance. Compactness can nevertheless ensure that the sets never lose all their common points. This is a useful companion to the Compactness and Subsequences result: rather than starting with a sequence of points and finding a convergent subsequence, we will choose one point from each set and use a convergent subsequence to find a point that belongs to every set.
The key hypotheses are that each set is nonempty and compact, and that the sets are nested in decreasing order. The conclusion is that their intersection is nonempty. If, in addition, the sets become arbitrarily small in diameter, the common point is unique, and any choice of one point from each set converges to it.
The direction of containment matters: each later set must be contained in the preceding one. In particular, if \(x\in K_n\), then \(x\in K_m\) for every \(m\leq n\). This simple observation will let us place the tail of a sequence of chosen points inside any fixed set \(K_m\).
The Nested Compact Sets Theorem
Compactness has already been shown to give convergent subsequences: every sequence in a compact set has a subsequence converging to a point of that set. We apply this result to points chosen successively from the nested sets. Closedness then ensures that the subsequential limit remains in each fixed set.
Proof. For each positive integer \(n\), choose \(x_n\in K_n\). Since \(K_n\subseteq K_1\), every \(x_n\) belongs to \(K_1\). The Compact Sets Are Sequentially Compact Theorem therefore gives a subsequence \((x_{n_j})\) that converges to some \(x\in K_1\).
Fix any positive integer \(m\). Since the subsequence indices increase, \(n_j\geq m\) for all sufficiently large \(j\). For each such \(j\), nesting gives \(K_{n_j}\subseteq K_m\), and \(x_{n_j}\in K_{n_j}\), so \(x_{n_j}\in K_m\). Thus a tail of the convergent subsequence lies in \(K_m\). The set \(K_m\) is compact, hence closed. By the Sequential Criterion for Closedness, its limit \(x\) belongs to \(K_m\).
The integer \(m\) was arbitrary, so \(x\) belongs to every \(K_m\). Therefore \(x\in\bigcap_{n=1}^{\infty}K_n\), proving that the intersection is nonempty. \(\square\)
The limit point is found by one subsequence, but the proof checks membership separately in each \(K_m\). It is essential that the subsequence indices eventually exceed \(m\): only then does nesting guarantee that the chosen points lie in \(K_m\). The argument does not require the original choices \(x_n\) to converge.
Worked Examples
Worked Example: A Nested Family with Two Common Points
For each positive integer \(n\), let
Each \(K_n\) is nonempty and compact: it is a bounded closed interval together with a single point. The intervals shrink as \(n\) increases, while the point \(2\) remains in every set. Hence \(K_{n+1}\subseteq K_n\) for every \(n\).
We can identify the intersection directly. Both \(0\) and \(2\) belong to every \(K_n\). If \(x\) belongs to every \(K_n\) and \(x\neq 2\), then \(x\in[-1/n,1/n]\) for every \(n\), so \(|x|\leq 1/n\) for every positive integer \(n\). This forces \(x=0\): if \(|x|>0\), the Archimedean property gives an \(n\) with \(1/n<|x|\), a contradiction. Thus
Nested compact sets need not have just one common point. The theorem guarantees nonemptiness, not uniqueness.
Worked Example: Reciprocal Tails with One Common Point
Let
Each \(K_n\) is a subset of the compact set \(\{0\}\cup\{1/k:k\geq1\}\), which is closed and bounded. It is also closed: the only limit point of the reciprocal tail is \(0\), and \(0\) is included. Therefore \(K_n\) is compact. Increasing \(n\) only removes reciprocal terms, so \(K_{n+1}\subseteq K_n\).
The point \(0\) belongs to every \(K_n\). On the other hand, for a fixed positive integer \(k\), the point \(1/k\) does not belong to \(K_{k+1}\), since \(k\) is not at least \(k+1\). So no reciprocal \(1/k\) belongs to every set. It follows that
Here the sets retain a common point even though their nonzero elements keep disappearing. The Nested Compact Sets Theorem guarantees that some point survives; the direct check identifies it.
Worked Example: Why Compactness Cannot Be Dropped
Consider the nested sets
Every \(E_n\) is nonempty and closed, and \(E_{n+1}\subseteq E_n\). But their intersection is empty. Indeed, if a real number \(x\) belonged to every \(E_n\), then \(x\geq n\) for every positive integer \(n\), which is impossible for a real number.
These sets are unbounded, so they are not compact. They show why closedness and nesting alone do not ensure a common point: the sets can move arbitrarily far to the right instead of retaining a point in a bounded region.
Diameters and a Unique Common Point
To quantify how large a bounded set is, use its diameter. For a nonempty bounded set \(E\subseteq\mathbb{R}\), its diameter is the supremum of all distances between two of its points. If a nested family has diameters tending to zero, then no two distinct points can belong to every set.
Proof. By the Nested Compact Sets Theorem, there is at least one point \(p\) in the intersection. Suppose \(p\) and \(q\) both belong to the intersection. For every \(n\), both points belong to \(K_n\), so
The right side tends to zero. Since \(|p-q|\geq0\), this implies \(|p-q|=0\), and therefore \(p=q\). The intersection has exactly one point; call it \(p\).
Now choose any \(x_n\in K_n\). Since \(p\) belongs to every \(K_n\), both \(x_n\) and \(p\) belong to \(K_n\), giving
Given \(\varepsilon>0\), choose \(N\) such that \(\operatorname{diam}(K_n)<\varepsilon\) whenever \(n\geq N\). Then for every \(n\geq N\), \(|x_n-p|<\varepsilon\). This is precisely \(x_n\to p\). \(\square\)
The conclusion applies to every choice of \(x_n\in K_n\); the points do not need to be selected by a special rule. The diameter bound controls their distance from the common point directly.
Worked Example: Shrinking Intervals and Convergence of Choices
Let \(K_n=[-1/n,1/n]\). These sets are nonempty and compact, and \(K_{n+1}\subseteq K_n\). Their intersection is \(\{0\}\): a point in every interval satisfies \(|x|\leq1/n\) for all \(n\), which forces \(x=0\). Also,
Choose any \(x_n\in K_n\). Then \(|x_n|\leq1/n\). Given \(\varepsilon>0\), choose \(N\) such that \(1/N<\varepsilon\). For \(n\geq N\),
Thus every possible sequence of choices converges to the unique common point \(0\), as the Shrinking-Diameter Theorem predicts.
What Nesting Contributes
Compactness alone does not guarantee that an infinite collection of sets has a common point. For instance, consider the three compact sets \(\{0,1\}\), \(\{1,2\}\), and \(\{0,2\}\). Each pair intersects, but the intersection of all three is empty. The nested compact sets theorem uses more than pairwise overlap: containment ensures that every finite initial collection has the same intersection as its smallest set, and that later choices remain inside each earlier set.
There is also an important distinction between the two conclusions proved here. Nested nonempty compact sets have at least one common point, but that point need not be unique, as the family in the first worked example shows. To force uniqueness, it is enough to add the shrinking-diameter condition. Conversely, for nested nonempty compact sets, a singleton intersection does force the diameters to tend to zero: since the diameters decrease, if they did not tend to zero then there would be a \(d>0\) such that \(\operatorname{diam}(K_n)\geq d\) for every \(n\); choose \(p_n,q_n\in K_n\) with \(|p_n-q_n|\geq d/2\). Compactness of \(K_1\), applied successively to these two sequences, gives a common subsequence along which both converge; nesting and closedness put both limits in every \(K_m\), and their distance is at least \(d/2\), contradicting that the intersection is a singleton.
These results are useful whenever a problem constructs increasingly restrictive compact sets. The first theorem guarantees that the constraints are jointly consistent. The diameter refinement gives a way to locate the resulting point: choose one point at each stage, then use the shrinking size of the sets to prove convergence. This approach will be especially useful when nested sets are intervals.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What containment condition makes a sequence of sets nested decreasing?
- In the proof of the Nested Compact Sets Theorem, why do sufficiently late terms of the chosen subsequence belong to a fixed set \(K_m\)?
- Which hypothesis fails for the sets \(E_n=[n,\infty)\), whose intersection is empty?
- Does the Nested Compact Sets Theorem guarantee that the intersection contains exactly one point?
- Why does a diameter tending to zero imply that any two points in the intersection must be equal?
- Under the Shrinking-Diameter Theorem, what can be said about any choice of points \(x_n\in K_n\)?