Compactness and Sequences

From Open Covers to Sequences

The previous tutorial showed how a missing closure point can obstruct compactness by producing an open cover with no finite subcover. There is another way to understand what compactness controls: it restricts how sequences of points in a set can behave. A sequence may oscillate and fail to converge, but if all its terms lie in a compact subset of the real line, some subsequence must converge to a point that still belongs to the set.

This conclusion combines familiar ideas from earlier tutorials. Compact subsets of \(\mathbb{R}\) are bounded and closed by the Heine–Borel Theorem. Boundedness keeps a sequence from escaping arbitrarily far, while closedness ensures that a limit reached by points of the set is not missing. We will also see how the open-cover definition of compactness produces finite approximations at every chosen scale.

Compact Sets Have Finite Approximations

Fix a compact set \(K\) and a positive distance \(\varepsilon\). Consider the open balls of radius \(\varepsilon\) centered at points of \(K\). They cover \(K\), because each point is the center of one of these balls. Compactness says that a finite number of them already cover \(K\). Thus, although \(K\) may have infinitely many points, at the chosen scale finitely many centers suffice to get within distance \(\varepsilon\) of every point of \(K\).

Proof. If \(K=\varnothing\), the assertion holds with no centers, since there are no points that need to be covered. Suppose \(K\neq\varnothing\). For each \(c\in K\), let \(B_\varepsilon(c)=(c-\varepsilon,c+\varepsilon)\). The family \(\{B_\varepsilon(c):c\in K\}\) is an open cover of \(K\): if \(x\in K\), then \(x\in B_\varepsilon(x)\), since \(|x-x|=0<\varepsilon\). By compactness, finitely many of these balls cover \(K\). Write them as \(B_\varepsilon(c_1),\ldots,B_\varepsilon(c_m)\), with each \(c_j\in K\). Every \(x\in K\) belongs to one of these balls, which means \(|x-c_j|<\varepsilon\) for some \(j\). This proves the assertion. \(\square\)

A finite collection of centers with this property is called a finite \(\varepsilon\)-net for \(K\). The centers need not be unique, and the number of centers may depend on \(\varepsilon\). Smaller radii may require more centers. The important point is that for each fixed positive radius, compactness guarantees a finite collection.

Extracting a Convergent Subsequence

Finite approximations can be applied repeatedly at finer and finer scales. At the first scale, infinitely many terms of a sequence must lie near one of finitely many centers. Keep those terms. At the next, finer scale, infinitely many of the retained terms must lie near one of the new finite collection of centers. Continuing this process produces nested infinite collections of indices. Choosing one successively later index from each collection gives a subsequence whose terms become arbitrarily close to one another.

Proof. If \(K=\varnothing\), there is no sequence with all its terms in \(K\), so the statement is vacuously true. Now suppose \(K\neq\varnothing\), and let \((x_n)\) be a sequence in \(K\).

We construct nested infinite sets of indices. Let \(I_0=\mathbb{N}\). For each positive integer \(m\), use the Finite Approximation Property with \(\varepsilon=1/m\) to cover \(K\) by finitely many open balls of radius \(1/m\). Since \(I_{m-1}\) is infinite and the corresponding terms \(x_n\), for \(n\in I_{m-1}\), lie in the union of finitely many balls, at least one ball contains \(x_n\) for infinitely many indices \(n\in I_{m-1}\). Otherwise each ball would contain terms for only finitely many such indices, and the finite union of those finite index sets would be finite, contradicting that \(I_{m-1}\) is infinite. Let \(I_m\) be the infinite set of those indices in one such ball. Then \(I_m\subseteq I_{m-1}\), and there is a center \(c_m\in K\) such that \(|x_n-c_m|<1/m\) for every \(n\in I_m\).

Choose indices \(n_m\in I_m\) successively so that \(n_1<n_2<\cdots\). This is possible because each \(I_m\) is an infinite subset of the positive integers, hence contains an index larger than any previously chosen index. For any \(j,k\geq m\), nesting gives \(n_j,n_k\in I_m\). Therefore both \(x_{n_j}\) and \(x_{n_k}\) are within distance \(1/m\) of \(c_m\), and the triangle inequality gives \[ |x_{n_j}-x_{n_k}|\leq |x_{n_j}-c_m|+|c_m-x_{n_k}|<\frac{2}{m}. \] Given any \(\eta>0\), choose \(m\) so large that \(2/m<\eta\). Then \(|x_{n_j}-x_{n_k}|<\eta\) whenever \(j,k\geq m\). Thus \((x_{n_m})\) is a Cauchy sequence.

The real numbers are complete, so this Cauchy sequence converges to some \(x\in\mathbb{R}\). Each term \(x_{n_m}\) belongs to \(K\), and \(K\) is closed because every compact subset of \(\mathbb{R}\) is closed. The Sequential Criterion for Closedness therefore implies that \(x\in K\). Hence the original sequence has a subsequence converging to a point of \(K\). \(\square\)

The proof makes the roles of the ingredients visible. Finite covers provide a way to retain infinitely many terms at each finer scale. Completeness supplies a limit for the resulting Cauchy subsequence. Closedness keeps that limit inside the set. In \(\mathbb{R}\), compactness supplies all of these ingredients.

What the Sequence Viewpoint Does—and Does Not—Say

The sequence theorem gives a practical consequence of compactness: if points are chosen from a compact set again and again, one can always select infinitely many of those choices that settle toward a point in the set. This is useful when a problem involves a sequence of approximations, candidate solutions, or points selected under changing conditions. Instead of trying to prove that the entire sequence converges, one can first seek a convergent subsequence.

A common pitfall is to confuse “has a convergent subsequence” with “converges.” The sequence \((-1)^n\) in \([-1,1]\) demonstrates the distinction: it does not converge, but it has constant subsequences. Another pitfall is to ignore where the subsequence limit lies. A sequence in a nonclosed set can converge to a point outside it, as the sequence in \((0,1)\) above does. Compactness rules out that outcome for sequences whose terms all lie in the compact set.

The finite-net argument is also a technique worth keeping. At each scale, a compact set can be covered by finitely many small neighborhoods. If infinitely many sequence terms are present, one of those neighborhoods must contain infinitely many of them. Repeating this observation at finer scales creates a subsequence with terms arbitrarily close to one another. The next tutorial develops the sequence-based notion of compactness more systematically.