Compactness and Subsequences

Compactness Organizes Subsequence Limits

The previous tutorials established that every sequence in a compact subset of \(\mathbb{R}\) has a convergent subsequence. That result guarantees at least one limit, but it does not describe all possible limits obtained by choosing different subsequences. A sequence can have one subsequential limit or many, and the full collection of these limits reveals whether the sequence settles toward a single value or continues to oscillate.

In this tutorial, we use compactness to study that collection. We will prove that a sequence lying in a compact set has a nonempty compact set of subsequential limits. We will then show that such a sequence converges exactly when this set contains just one point. The proofs use subsequences as a way to isolate persistent behavior: a subsequence can retain terms that stay away from a proposed limit, or can be chosen to approach a limit point.

The limit of a subsequence is always taken in \(\mathbb{R}\), even when the original sequence is required to lie in a particular set. When that set is compact, however, any such limit must remain in the set: compact subsets of \(\mathbb{R}\) are closed. This is one way compactness controls the limits we can obtain.

The Set of Subsequential Limits

We recall the Compact Sets Are Sequentially Compact Theorem: every sequence in a compact set has a subsequence converging to a point of that set. We also use the fact, established earlier, that compact subsets of \(\mathbb{R}\) are closed and that a closed subset of a compact set is compact. These results give nonemptiness and containment of the set \(C\). The less immediate part is proving that \(C\) is closed.

Proof. Since \((x_n)\) is a sequence in \(K\), the Compact Sets Are Sequentially Compact Theorem gives a subsequence that converges to some \(y\in K\). Thus \(y\in C\), so \(C\) is nonempty.

If \(z\in C\), then \(z\) is the limit of a subsequence of points in \(K\). The set \(K\) is closed, so the Sequential Criterion for Closedness implies that \(z\in K\). Hence \(C\subseteq K\).

It remains to prove that \(C\) is closed. Let \((y_j)\) be a sequence in \(C\) converging to a real number \(y\). For each \(j\), \(y_j\) is a subsequential limit of \((x_n)\). Consequently, arbitrarily far along the original sequence there are terms as close as we wish to \(y_j\). We can therefore choose indices \(n_1<n_2<\cdots\) such that

for every \(j\). More explicitly, after choosing \(n_{j-1}\), take a subsequence converging to \(y_j\); sufficiently far along that subsequence its index exceeds \(n_{j-1}\) and its term is within \(1/j\) of \(y_j\). This supplies the next index \(n_j\). For \(j=1\), choose any suitable index \(n_1\).

The triangle inequality gives

Both terms on the right tend to zero as \(j\) increases. Therefore \(x_{n_j}\to y\), so \(y\in C\). We have shown that every convergent sequence in \(C\) has its limit in \(C\); by the Sequential Criterion for Closedness, \(C\) is closed. Since \(C\) is a closed subset of the compact set \(K\), it is compact. \(\square\)

The diagonal choice of indices is the key technique in the closedness argument. Each \(y_j\) may come from a different subsequence, so the approximating terms must be chosen in increasing order to form one subsequence of the original sequence. Making the \(j\)-th approximation within \(1/j\) ensures that its error tends to zero. The triangle inequality then transfers the convergence \(y_j\to y\) to the selected terms \(x_{n_j}\to y\).

Worked Examples

Convergence and Uniqueness of the Subsequential Limit

A sequence can fail to converge because it repeatedly makes excursions away from a candidate limit. If its terms lie in a compact set, those excursions contain a subsequence that converges somewhere. The next theorem makes this reasoning precise: a sequence in a compact set converges if and only if its set of subsequential limits is a singleton.

Proof. Suppose first that \(x_n\to L\). Every subsequence of a convergent sequence also converges to \(L\), so every element of \(C\) equals \(L\). The Compact Sets Are Sequentially Compact Theorem guarantees that at least one subsequence converges, so \(C\) is nonempty. Hence \(C=\{L\}\).

Conversely, suppose \(C=\{L\}\). We prove that \(x_n\to L\). Assume instead that \((x_n)\) does not converge to \(L\). By the definition of convergence, there is an \(\varepsilon>0\) such that for every positive integer \(N\), some \(n\geq N\) satisfies

We can choose increasing indices \(n_1<n_2<\cdots\) with \(|x_{n_j}-L|\geq\varepsilon\) for every \(j\): after choosing \(n_j\), apply the failure of convergence with \(N=n_j+1\) to choose \(n_{j+1}\). The sequence \((x_{n_j})\) lies in \(K\). By compactness and sequential compactness, it has a convergent subsequence \((x_{n_{j_\ell}})\) with limit \(y\in K\). This is also a subsequence of the original sequence, so \(y\in C\). The assumption \(C=\{L\}\) gives \(y=L\).

But every selected term satisfies \(|x_{n_{j_\ell}}-L|\geq\varepsilon\). Since \(x_{n_{j_\ell}}\to y=L\), continuity of the absolute value, or the definition of convergence, gives \(|x_{n_{j_\ell}}-L|\to0\). This contradicts the lower bound \(\varepsilon>0\). Therefore \(x_n\to L\). \(\square\)

The theorem also gives a useful description of nonconvergence. A sequence in a compact set always has at least one subsequential limit. If it does not converge, it cannot have exactly one: its set of subsequential limits must contain at least two distinct points. In practice, it is often easier to find two subsequences with different limits than to analyze every term of the original sequence directly.

Why the Compact-Set Hypothesis Matters

The proof of the convergence criterion depends on compactness at a specific point: after selecting terms that stay at least \(\varepsilon\) away from \(L\), compactness provides a convergent subsequence. Its limit is still in \(K\), and the distance from \(L\) cannot suddenly become smaller than \(\varepsilon\). Without compactness, a sequence can keep making such excursions while having no convergent subsequence in the set.

A common pitfall is to check only one subsequence. Finding a subsequence that converges does not prove that the original sequence converges; the alternating sequence has two convergent subsequences with different limits. Conversely, failure of one particular subsequence to converge does not settle the behavior of all the others. The convergence theorem uses the entire set \(C\): it is the uniqueness of the subsequential limit, together with compactness, that forces convergence of the full sequence.

The compactness result also gives more structure than a single convergence test. The set \(C\) is itself compact, so it is closed and bounded. For example, all possible subsequential limits of a sequence in \(K\) lie within \(K\) and form a set with the same compactness safeguards. This is useful when a sequence has several limiting behaviors: rather than tracking every choice of subsequence, one can study the compact set of all limits together.