Proof That Compact Sets Are Sequentially Compact

From Open Covers to Subsequences

Sequential compactness describes a set by the behavior of sequences whose terms lie in it. Compactness, by contrast, is defined using open covers and finite subcovers. The Compactness Gives a Convergent Subsequence theorem was stated earlier in this course. Here we examine the finite-cover mechanism behind that conclusion: compactness lets us repeatedly keep infinitely many terms inside smaller and smaller neighborhoods. The selected terms form a Cauchy subsequence.

The argument has two important parts. First, finite subcovers let us find one small neighborhood that contains infinitely many terms from the indices retained so far. Repeating this produces increasingly tight control over the selected terms. Second, the completeness of \(\mathbb{R}\) turns the resulting Cauchy subsequence into a convergent one, and the closedness of a compact set ensures its limit belongs to the set.

The key point is that a sufficiently small ball has small diameter: if two points lie in the same open ball of radius \(r\), their distance is less than \(2r\). We will apply this observation at successively smaller radii. The centers of the balls can change from one stage to the next; what matters is that the subsequence’s later terms are all in the same ball at each fixed stage.

The Finite-Cover Extraction Lemma

Proof. If \(K=\varnothing\), there is no sequence with every term in \(K\), so the assertion is vacuous. Suppose \(K\neq\varnothing\). We construct nested infinite sets of indices \(I_0\supseteq I_1\supseteq I_2\supseteq\cdots\), beginning with \(I_0=\mathbb{N}\).

Suppose an infinite set \(I_{j-1}\) has been chosen, where \(j\geq1\). The family of open balls \[ \{B_{2^{-j}}(x):x\in K\} \] covers \(K\), because every point \(x\in K\) belongs to the ball \(B_{2^{-j}}(x)\). Compactness supplies a finite subcover, say \(B_{2^{-j}}(c_1),\ldots,B_{2^{-j}}(c_m)\), with each center \(c_i\in K\).

For every \(n\in I_{j-1}\), the point \(x_n\) lies in at least one of these finitely many balls. Assign \(n\) to the first ball in the listed subcover that contains \(x_n\). If each ball were assigned only finitely many indices, their finite union would account for only finitely many indices in \(I_{j-1}\). That contradicts the fact that \(I_{j-1}\) is infinite. Thus at least one ball contains \(x_n\) for infinitely many indices \(n\in I_{j-1}\). Let \(I_j\) be the set of those indices for one such ball. Then \(I_j\) is infinite and \(I_j\subseteq I_{j-1}\).

For each \(j\), choose \(n_j\in I_j\) so that \(n_j>n_{j-1}\), with the first index \(n_1\) chosen from \(I_1\). This is possible at every stage because an infinite subset of \(\mathbb{N}\) is unbounded. The indices satisfy \(n_1<n_2<\cdots\), so they define a subsequence. Fix \(j\). Whenever \(k,\ell\geq j\), the nesting gives \(n_k,n_\ell\in I_j\). By the construction of \(I_j\), both \(x_{n_k}\) and \(x_{n_\ell}\) lie in the same ball of radius \(2^{-j}\). If its center is \(c\), then \[ |x_{n_k}-x_{n_\ell}| \leq |x_{n_k}-c|+|x_{n_\ell}-c| <2\cdot 2^{-j}=2^{1-j}. \] Given any \(\varepsilon>0\), choose \(j\) large enough that \(2^{1-j}<\varepsilon\). For all \(k,\ell\geq j\), the displayed estimate then gives \(|x_{n_k}-x_{n_\ell}|<\varepsilon\). Therefore \((x_{n_k})\) is Cauchy. \(\square\)

The finite-subcover step is where compactness enters. Covering \(K\) by infinitely many small balls is straightforward, but that alone does not guarantee that any one ball captures infinitely many terms. A finite subcover makes the pigeonhole argument possible. Repeating the step matters too: one stage of selection only gives control at one scale, whereas the nested sets make the terms close at every sufficiently fine scale.

Compact Sets Are Sequentially Compact

Proof. The empty set is sequentially compact by definition. Suppose \(K\neq\varnothing\), and let \((x_n)\) be any sequence in \(K\). By the Compactness Produces a Cauchy Subsequence lemma, there is a subsequence \((x_{n_k})\) that is Cauchy. The real numbers are complete, so there is some \(x\in\mathbb{R}\) such that \(x_{n_k}\to x\). Every term \(x_{n_k}\) belongs to \(K\). By the theorem that every compact subset of \(\mathbb{R}\) is closed, and the Sequential Criterion for Closedness, the limit \(x\) belongs to \(K\). Thus every sequence in \(K\) has a subsequence converging to a point of \(K\), as required. \(\square\)

Notice that the final step uses more than the existence of a limit in \(\mathbb{R}\). A convergent subsequence might in principle approach a point outside the set. Compactness prevents this here because compact sets are closed. The proof therefore combines the finite-cover extraction argument with completeness and closedness; each has a distinct role.

Worked Applications of the Proof

What the Argument Does—and Does Not—Say

When applying the proof, begin with an arbitrary sequence in the compact set and cover the whole set by small open balls. Do not assume that a single ball will contain infinitely many terms at every chosen radius. Instead, at each stage use a finite subcover and retain an infinite collection of indices associated with one ball. The nested-index construction is what ensures that later selected terms satisfy all the earlier distance estimates.

There are also two different conclusions to keep separate. The extraction lemma gives a Cauchy subsequence, not yet a subsequence converging to a point of \(K\). Completeness of \(\mathbb{R}\) supplies a real limit; closedness of \(K\) puts that limit back in \(K\). If either step is omitted, the argument does not establish sequential compactness as defined in the previous tutorial.

This proof connects two ways of describing compactness. Open covers provide finite control at each scale, while sequences record the points selected from the set. Compactness converts the finite control into a Cauchy subsequence, and the familiar completeness and closedness properties finish the argument. In later results, this sequential viewpoint will let us investigate compactness through accumulation points and bounded infinite sets.