Sequential Compactness

Compactness Tested by Sequences

Open-cover compactness asks whether an arbitrary family of open sets can be reduced to a finite cover. There is another way to express finite control in a metric space: examine sequences and ask whether they always contain a convergent subsequence. This property is called sequential compactness. It focuses on the behavior of points chosen one after another, rather than on families of open sets.

A convergent sequence already has a limit in the space, but sequential compactness asks for less of each individual sequence: the original sequence need not converge, provided that some subsequence does. The distinction matters. A sequence can move back and forth indefinitely while still having a convergent subsequence.

Definition: A metric space \((X,d)\) is sequentially compact if every sequence \((x_n)\) in \(X\) has a subsequence \((x_{n_k})\) that converges to a point \(x\in X\). Here the indices satisfy \(n_1<n_2<\cdots\), and convergence is with respect to the metric on \(X\).

The limit is required to belong to \(X\). A sequence might converge in a larger metric space to a point outside \(X\), but that does not provide a convergent subsequence in \(X\). The theorem Convergence in a Subspace established that convergence for a sequence in a subspace is measured using the restricted metric; the requirement that the limit lie in the subspace is therefore essential.

Three Tests Using Sequences

Worked Example: An Infinite Discrete Space Is Not Sequentially Compact

Let \(X\) be an infinite set equipped with the discrete metric \(\delta\), where \(\delta(x,y)=1\) for distinct points and \(\delta(x,x)=0\). Since \(X\) is infinite, we can choose a sequence \(x_1,x_2,\ldots\) of distinct points. For any two different indices \(j\) and \(k\), \(\delta(x_j,x_k)=1\).

No subsequence of this sequence can converge. Indeed, if a subsequence converged to some \(x\in X\), the theorem Eventual Exclusion of a Different Point would imply that its terms are eventually equal to \(x\): in the discrete metric, every point different from \(x\) is at distance \(1\) from \(x\). But the terms of the subsequence are distinct, so at most one of them can equal \(x\). This is a contradiction. Thus \(X\) is not sequentially compact.

Worked Example: A Space with Two Subsequence Limits

Consider the subset

K=\{-1,1\}\cup\{(-1)^n(1+1/n):n\geq1\}

of the real line with its usual metric. We show directly that \(K\) is sequentially compact. Take any sequence \((y_j)\) in \(K\). If some point of \(K\) occurs infinitely many times, those occurrences form a constant subsequence, which converges to that point in \(K\).

It remains to consider the case in which every point occurs only finitely many times. In particular, the terms \(1\) and \(-1\) occur only finitely many times. After discarding those terms, each remaining term has the form \((-1)^{m_j}(1+1/m_j)\) for an integer \(m_j\geq1\). Since each fixed point of \(K\) occurs only finitely often, the indices \(m_j\) in this remaining sequence cannot stay in a finite set: there are only finitely many points corresponding to any finite set of indices. We can therefore select a subsequence whose associated indices tend to infinity.

Among these selected indices, at least one parity occurs infinitely often. Pass to that further subsequence. If its indices are even, its terms have the form \(1+1/m_j\) and converge to \(1\). If its indices are odd, its terms have the form \(-1-1/m_j\) and converge to \(-1\). Both possible limits belong to \(K\). Every sequence in \(K\) therefore has a subsequence converging in \(K\), as required.

Worked Example: The Open Interval \((0,1)\) Is Not Sequentially Compact

Use the usual metric on \(X=(0,1)\), and consider \(x_n=1/(n+1)\). Every term lies in \(X\), and \(x_n\to0\) in the real line. Every subsequence \(x_{n_k}\) also tends to \(0\), since \(n_k\to\infty\).

If some subsequence converged to a point \(x\in(0,1)\), it would also converge to \(0\) in the ambient real line. The uniqueness of metric limits would then give \(x=0\), which is not in \(X\). Hence no subsequence converges to a point of \(X\), and \((0,1)\) is not sequentially compact.

These examples illustrate two different obstructions. In the discrete space, distinct points remain separated by a fixed positive distance, so no subsequence can settle near a limit. In \((0,1)\), the sequence approaches a point that has been left out of the space. When testing sequential compactness, it is not enough to find a subsequence that converges somewhere; its limit must be in the space.

Basic Consequences of Sequential Compactness

A sequentially compact metric space cannot spread out without bound. It also passes sequential compactness to closed subsets, and continuous maps preserve the property in their images. Each fact follows by starting with a sequence and extracting a suitable subsequence.

Theorem (Sequentially Compact Metric Spaces Are Bounded): Every nonempty sequentially compact metric space is bounded.

Proof. Let \((X,d)\) be nonempty and sequentially compact, and choose a point \(x_0\in X\). Suppose, for contradiction, that \(X\) is not bounded. Then for every positive integer \(n\), there is \(x_n\in X\) such that \(d(x_n,x_0)>n\). Sequential compactness gives a subsequence \((x_{n_k})\) converging to some \(x\in X\).

Convergence implies that, for all sufficiently large \(k\), \(d(x_{n_k},x)<1\). The triangle inequality then gives

d(x_{n_k},x_0)\leq d(x_{n_k},x)+d(x,x_0)<1+d(x,x_0).

The right-hand side is a fixed finite number, whereas the construction gives \(d(x_{n_k},x_0)>n_k\). Since \(n_k\to\infty\), these distances cannot eventually be bounded by \(1+d(x,x_0)\). This contradiction proves that \(X\) is bounded. \(\square\)

The nonempty hypothesis avoids having to choose a point \(x_0\). The conclusion is also specifically about metric spaces: the proof uses distances from a fixed point to detect a sequence escaping every bounded region.

Theorem (Closed Subsets Inherit Sequential Compactness): Let \(X\) be a sequentially compact metric space, and let \(A\subseteq X\) be closed. With the restricted metric, \(A\) is sequentially compact.

Proof. If \(A\) is empty, there are no sequences in \(A\), so the definition holds vacuously. Suppose \(A\) is nonempty and take any sequence \((a_n)\) in \(A\). It is also a sequence in \(X\). By sequential compactness of \(X\), it has a subsequence \((a_{n_k})\) converging in \(X\) to some \(x\in X\).

The subsequence consists of points of \(A\), and \(A\) is closed in \(X\). By the Sequential Characterization of Closed Sets, the limit \(x\) belongs to \(A\). The Convergence in a Subspace theorem shows that this same subsequence converges to \(x\) in \(A\) with its restricted metric. Thus every sequence in \(A\) has a subsequence converging to a point of \(A\), and \(A\) is sequentially compact. \(\square\)

Closedness is doing real work here: it ensures that limits found in the surrounding space do not escape the subset. The example \((0,1)\subseteq[0,1]\) shows why this condition cannot simply be dropped: \([0,1]\) is sequentially compact, but \((0,1)\) is not.

Theorem (Continuous Images Preserve Sequential Compactness): Let \(X\) be a sequentially compact metric space, let \(Y\) be a metric space, and let \(f:X\to Y\) be continuous. Then \(f(X)\), equipped with the restricted metric from \(Y\), is sequentially compact.

Proof. Take any sequence \((y_n)\) in \(f(X)\). For each \(n\), choose \(x_n\in X\) such that \(f(x_n)=y_n\). Sequential compactness of \(X\) gives a subsequence \((x_{n_k})\) converging to some \(x\in X\). By continuity of \(f\), the sequence \((f(x_{n_k}))\) converges in \(Y\) to \(f(x)\). Since each \(f(x_{n_k})=y_{n_k}\), this is a subsequence of \((y_n)\), and its limit \(f(x)\) belongs to \(f(X)\). The convergence also holds in the restricted metric on \(f(X)\). Therefore \(f(X)\) is sequentially compact. \(\square\)

This proof is a useful general method: to study a sequence in an image, choose preimages, extract a convergent subsequence in the domain, and then use continuity to transfer convergence to the image.

What the Definition Does—and Does Not—Say

Sequential compactness guarantees a convergent subsequence, not convergence of the original sequence. For example, in \(K\) above, a sequence can alternate between terms approaching \(1\) and terms approaching \(-1\); that sequence need not converge, even though one of its parity subsequences does. The definition also does not require a single point to serve as the limit of subsequences extracted from different sequences.

A practical way to disprove sequential compactness is to construct one sequence for which every subsequence fails. Two common strategies are to choose points whose mutual distances stay bounded away from zero, as in the infinite discrete space, or to choose a sequence whose subsequences can only converge to points missing from the space, as in \((0,1)\). To prove sequential compactness, by contrast, one must start with an arbitrary sequence and explain how a convergent subsequence can always be found.

The properties proved above are useful checkpoints, but boundedness alone is not a test for sequential compactness. The open interval \((0,1)\) is bounded and still fails the definition. The next step in this sequence of ideas is to connect sequence-based compactness with the open-cover formulation already introduced for metric spaces.

Check Your Understanding

Use the definition and the arguments above to answer the following questions.

What two requirements must a subsequence satisfy to demonstrate sequential compactness?
Why can no subsequence of distinct points in an infinite discrete metric space converge?
In the example \(K\), why does a subsequence with indices of fixed parity converge to one of the two points \(-1\) or \(1\)?
What prevents the sequence \(1/(n+1)\) from having a convergent subsequence in \((0,1)\)?
Where is closedness used in proving that a closed subset of a sequentially compact space is sequentially compact?
How does continuity help prove that the image of a sequentially compact space is sequentially compact?

What You'll Learn