Compactness Through Sequences
Compactness is defined using open covers: every open cover must have a finite subcover. The previous tutorial gave a different way to recognize compact sets in \(\mathbb{R}\), through accumulation points of infinite subsets. Here we use sequences instead. The key question is whether every sequence in a set has a subsequence that converges to a point still belonging to that set.
The location of the limit is essential. A sequence can have a convergent subsequence in \(\mathbb{R}\) whose limit lies outside the set. Such a subsequence does not establish the property we need. We will show that, in the real line, the requirement that subsequence limits remain in the set is equivalent to compactness.
The convergence in this definition is ordinary convergence in \(\mathbb{R}\), and the limit must be in \(K\). If \(K\) is empty, it is sequentially compact because there are no sequences with all terms in \(K\), so the defining condition holds vacuously.
A Useful Subsequence Technique
A sequence may repeat values many times, or it may keep taking new values. The following elementary fact separates those possibilities. It is useful both for understanding finite sets and for deciding what kinds of subsequences to look for.
Proof. Suppose first that some value \(c\) occurs infinitely often among the terms of the sequence. Choose increasing indices \(n_1<n_2<\cdots\) such that \(x_{n_j}=c\) for every \(j\). This gives a constant subsequence.
Now suppose no value occurs infinitely often. We choose indices recursively so that the values selected are all different. Choose any index \(n_1\). After choosing \(n_1,\ldots,n_j\), consider the tail of the sequence after \(n_j\). That tail must contain a value different from each of \(x_{n_1},\ldots,x_{n_j}\). Otherwise, every term in the tail would be among those finitely many values. Since each of these values occurs only finitely often in the whole sequence, they could account for only finitely many terms, whereas the tail has infinitely many terms. This is a contradiction. Choose \(n_{j+1}>n_j\) with \(x_{n_{j+1}}\) different from the previously selected values. Continuing in this way produces a subsequence with pairwise distinct terms. The two cases exhaust all possibilities. \(\square\)
Worked Example: A Sequence with Two Convergent Subsequences
Consider the sequence \(x_n=(-1)^n+1/n\). If \(n=2k\) is even, then \[ x_{2k}=(-1)^{2k}+\frac{1}{2k}=1+\frac{1}{2k}, \] which tends to \(1\) as \(k\to\infty\). If \(n=2k-1\) is odd, then \[ x_{2k-1}=(-1)^{2k-1}+\frac{1}{2k-1}=-1+\frac{1}{2k-1}, \] which tends to \(-1\) as \(k\to\infty\).
Every term belongs to \([-2,2]\). For even indices, \(1<1+1/(2k)\leq3/2\), and for odd indices, \(-1<-1+1/(2k-1)\leq0\). Thus the sequence lies in the compact set \([-2,2]\), and the two subsequences just calculated converge to points of that set. This example illustrates that the subsequence supplied by sequential compactness need not be unique.
The Sequential Characterization of Compactness
Proof. If \(K\) is compact, then it is sequentially compact by the theorem Compact Sets Are Sequentially Compact, proved earlier in this course.
Conversely, suppose \(K\) is sequentially compact. If \(K=\varnothing\), then \(K\) is compact: every open cover of the empty set has the empty finite subcover. Now suppose \(K\neq\varnothing\). By the earlier theorem Sequentially Compact Sets Are Bounded and Closed, \(K\) is bounded and closed in \(\mathbb{R}\). The Heine–Borel Theorem says that a subset of \(\mathbb{R}\) is compact if and only if it is closed and bounded. Therefore \(K\) is compact. This proves both directions. \(\square\)
The proof uses two results established earlier rather than re-proving them: compactness gives sequential compactness, and sequential compactness forces closedness and boundedness. Heine–Borel then connects those properties back to compactness. This is a characterization specific to the real line, where closedness and boundedness characterize compact sets.
Worked Example: A Finite Set Is Sequentially Compact
Let \(K=\{-2,4,7\}\), and take any sequence \((x_n)\) whose terms lie in \(K\). At least one of the three possible values must occur infinitely often. If each occurred only finitely many times, then the sequence would have only finitely many terms in total, which is impossible. Choose a value \(c\in K\) that occurs infinitely often and select the terms at those indices. The resulting subsequence is constant at \(c\), so it converges to \(c\in K\).
Thus \(K\) is sequentially compact. By the theorem, it is compact as well. The argument does not require the original sequence to have any particular pattern: the repeated value might be \(-2\), \(4\), or \(7\).
Worked Example: A Sequence in an Open Interval Has No Suitable Subsequence
Let \(K=(0,1)\), and define \(x_n=1-1/(n+1)\) for \(n\geq1\). Since \(n+1\geq2\), we have \(0<1/(n+1)\leq1/2\), and therefore \(1/2\leq x_n<1\). In particular, every \(x_n\) lies in \(K\).
Let \((x_{n_j})\) be any subsequence. Its indices increase, so \(n_j\to\infty\), and \[ \left|x_{n_j}-1\right|=\frac{1}{n_j+1}\longrightarrow0. \] Hence every subsequence converges to \(1\), which is not in \(K\). No subsequence can converge to a point of \(K\): if a subsequence also converged to \(x\in K\), then the triangle inequality would give \[ |x-1|\leq |x-x_{n_j}|+|x_{n_j}-1|\longrightarrow0, \] so \(x=1\), contradicting \(x\in(0,1)\). Therefore \(K\) is not sequentially compact, and the theorem confirms that it is not compact.
Worked Example: An Unbounded Set Fails the Sequential Test
Consider the set of positive integers \(K=\{1,2,3,\ldots\}\) and the sequence \(x_n=n\) in \(K\). For any subsequence, the indices satisfy \(n_j\geq j\), so \(x_{n_j}=n_j\geq j\). Thus the terms of every subsequence are unbounded.
A convergent real sequence is bounded: if \(a_j\to a\), then for all sufficiently large \(j\), \(|a_j-a|<1\), so \(|a_j|<|a|+1\); the finitely many earlier terms are bounded as well. Consequently, no subsequence of \((n)\) can converge in \(\mathbb{R}\). The positive integers are not sequentially compact and, by the theorem, are not compact.
What the Criterion Does—and Does Not—Say
The sequential criterion is often more convenient than the open-cover definition when a problem already concerns sequences. To prove that a set is compact, one can begin with an arbitrary sequence in the set and find a subsequence converging to a point in the set. To disprove compactness, it is enough to exhibit a single sequence for which no such subsequence exists, as in the open-interval and positive-integer examples.
A frequent error is to find a convergent subsequence but neglect to check where its limit lies. In \((0,1)\), the sequence in the example has subsequences converging in \(\mathbb{R}\), but all of them converge to the missing endpoint \(1\). Sequential compactness requires a limit belonging to the set itself. This requirement reflects closedness, while the need to prevent sequences from escaping without a convergent subsequence reflects boundedness in \(\mathbb{R}\).
The theorem therefore offers a practical test, not a change in the meaning of compactness. Compactness still means that every open cover has a finite subcover. The equivalence says that, for subsets of the real line, this open-cover condition can be recognized through subsequences. Its proof relies on the real-line results already established: the equivalence of compactness with closedness and boundedness, and the consequences of sequential compactness for those properties.
Check Your Understanding
Use the definition and the sequential characterization to answer the following questions.
- In the definition of sequential compactness, why must the subsequence limit belong to \(K\), rather than merely to \(\mathbb{R}\)?
- How does the proof of the constant-or-pairwise-distinct subsequence lemma handle the case in which no value appears infinitely often?
- Which earlier theorem gives closedness and boundedness when \(K\) is sequentially compact?
- For the sequence \(x_n=1-1/(n+1)\) in \((0,1)\), what is the limit of every subsequence, and why does that rule out sequential compactness?
- Why can no subsequence of the positive-integer sequence \(x_n=n\) converge in \(\mathbb{R}\)?