When a Set Controls Its Sequences
The previous tutorial showed that every sequence in a compact subset of \(\mathbb{R}\) has a subsequence converging to a point in that set. This motivates a property stated directly in terms of sequences. Rather than begin with open covers, we can ask whether the sequences drawn from a set always admit a convergent subsequence whose limit remains in the set.
The distinction between a convergent sequence and a convergent subsequence is important. A sequence may oscillate indefinitely and still have several convergent subsequences. Sequential compactness requires only one such subsequence for each sequence in the set, but it also requires the subsequence limit to belong to the set.
The definition has two distinct requirements: a subsequence must converge, and its limit must be in \(E\). The first requirement rules out sequences whose terms escape in every subsequence; the second rules out a sequence that approaches a missing boundary point. In \(\mathbb{R}\), the Bolzano–Weierstrass Theorem and closedness will help explain these two requirements.
Testing the Definition on Sequences
Worked Example: A Finite Set
Let \(E=\{-3,4,9\}\). Take any sequence \((x_n)\) whose terms lie in \(E\). If each of the three possible values occurred only finitely many times, then the sequence would have only finitely many terms: it would be a union of three finite collections of indices. That is impossible, since the sequence has one term for every positive integer. Therefore at least one value, say \(c\in E\), occurs infinitely many times.
Choose the indices \(n_1<n_2<\cdots\) at which \(x_{n_k}=c\). Then \(x_{n_k}=c\) for every \(k\), so this subsequence converges to \(c\), and \(c\in E\). Since the original sequence was arbitrary, \(E\) is sequentially compact. The same argument works for every finite subset of \(\mathbb{R}\).
Worked Example: A Closed Interval
Let \(E=[2,5]\), and take any sequence \((x_n)\) in \(E\). The sequence is bounded, since \(2\leq x_n\leq5\) for every \(n\). We explain how the Bolzano–Weierstrass Theorem gives a convergent subsequence.
If the range \(\{x_n:n\geq1\}\) is finite, some value occurs infinitely often, and those occurrences form a constant subsequence. If the range is infinite, the Bolzano–Weierstrass Theorem gives an accumulation point \(p\) of the range. Every neighborhood of \(p\) contains infinitely many distinct values of the range, and hence terms occurring at arbitrarily large indices. We can therefore choose indices successively so that \[ n_1<n_2<\cdots \qquad\text{and}\qquad |x_{n_k}-p|<\frac{1}{k}. \] The inequality implies \(x_{n_k}\to p\). Since \(E\) is closed, the Sequential Criterion for Closedness shows that \(p\in E\). Thus every sequence in \([2,5]\) has a subsequence converging to a point of \([2,5]\).
Worked Example: A Sequence Approaching a Missing Endpoint
Let \(E=(2,5)\) and define \(x_n=2+1/(n+1)\). For every positive integer \(n\), we have \(0<1/(n+1)\leq1/2\), so \(2<x_n\leq5/2<5\), and therefore \(x_n\in E\). The sequence converges to \(2\): given \(\varepsilon>0\), choose \(N\) so that \(1/(N+1)<\varepsilon\). For \(n\geq N\), \[ |x_n-2|=\frac{1}{n+1}\leq\frac{1}{N+1}<\varepsilon. \]
Every subsequence also converges to \(2\), since its indices increase and the same estimate applies to its sufficiently late terms. But \(2\notin E\). No subsequence can converge to a point of \(E\), so \((2,5)\) is not sequentially compact. This example shows why a limit outside the set does not meet the definition.
Worked Example: A Set Containing Its Accumulation Point
Consider \(A=\{2\}\cup\{2+1/m:m\geq1\}\). Take any sequence in \(A\). If one value occurs infinitely many times, its occurrences give a constant subsequence converging to a point of \(A\). Suppose instead that each value occurs only finitely many times. Then the sequence has infinitely many distinct terms of the form \(2+1/m\).
We can choose successive indices \(n_k\) so that \(x_{n_k}=2+1/m_k\) with \(m_k>k\). To see that the indices can be chosen successively, note that there are only finitely many possible values with \(m\leq k\), and each occurs only finitely many times. Thus only finitely many terms of the sequence have those values; a later term has a denominator larger than \(k\). It follows that \[ 0<|x_{n_k}-2|=\frac{1}{m_k}<\frac{1}{k}. \] Hence \(x_{n_k}\to2\), and \(2\in A\). In either case, a subsequence converges to a point of \(A\), so \(A\) is sequentially compact.
Sequential Compactness Forces Boundedness and Closedness
The examples suggest two basic restrictions. A sequence selected from an unbounded set can be chosen to escape farther and farther from the origin. A sequence selected from a set that is not closed can converge to a missing point. Sequential compactness rules out both behaviors.
Proof. If \(E=\varnothing\), then it is both bounded and closed. Suppose \(E\neq\varnothing\).
First we prove boundedness. Suppose, to the contrary, that \(E\) is unbounded. For each positive integer \(n\), choose \(x_n\in E\) with \(|x_n|>n\). Sequential compactness gives a subsequence \((x_{n_k})\) converging to some \(x\in E\). Every convergent real sequence is bounded. But \(n_k\geq k\), so \[ |x_{n_k}|>n_k\geq k \] for every \(k\); this subsequence is unbounded, a contradiction. Therefore \(E\) is bounded.
To prove that \(E\) is closed, let \((y_n)\) be any convergent sequence with \(y_n\in E\) for every \(n\), and write \(y_n\to y\in\mathbb{R}\). By sequential compactness, some subsequence \((y_{n_k})\) converges to a point \(z\in E\). A subsequence of a convergent sequence has the same limit, so \(y_{n_k}\to y\) as well. Limits of real sequences are unique; hence \(y=z\in E\). The Sequential Criterion for Closedness now implies that \(E\) is closed. \(\square\)
The two parts use sequential compactness in different ways. In the boundedness argument, it prevents a sequence from having every subsequence escape to infinity. In the closedness argument, it supplies a subsequence limit inside \(E\); uniqueness of limits then identifies that point with the original sequence’s limit. Closedness and boundedness are necessary conditions here. The theorem does not itself assert that either condition alone is sufficient.
Two Useful Preservation Results
Sequential compactness also passes to closed subsets and finite unions. These facts allow the property to be checked or used in pieces, while keeping the requirement that limits stay in the relevant set.
Proof. If \(F=\varnothing\), the result follows from the definition. Otherwise, take any sequence \((x_n)\) in \(F\). It is also a sequence in \(E\), so there is a subsequence \((x_{n_k})\) converging to some \(x\in E\). Every term of this subsequence lies in \(F\), and \(F\) is closed. By the Sequential Criterion for Closedness, its limit \(x\) belongs to \(F\). Thus every sequence in \(F\) has a subsequence converging to a point of \(F\), as required. \(\square\)
Proof. Let \((x_n)\) be a sequence in \(\bigcup_{j=1}^{m}E_j\). Assign each index \(n\) to the least \(j\) for which \(x_n\in E_j\). This divides the positive integers into \(m\) collections of indices. At least one collection is infinite; if all were finite, their finite union would be finite and could not contain every positive integer.
Let \(j\) be an index whose collection is infinite, and list that collection in increasing order as \(n_1<n_2<\cdots\). Then \((x_{n_k})\) is a sequence in \(E_j\). Since \(E_j\) is sequentially compact, it has a subsequence converging to a point of \(E_j\), and therefore to a point of \(\bigcup_{j=1}^{m}E_j\). This proves the assertion. \(\square\)
How to Use the Definition
When testing whether a set is sequentially compact, begin with an arbitrary sequence in the set. A proof must explain how to choose infinitely many of its terms, in their original order, and show that the resulting subsequence converges to a point in the set. It is not enough to exhibit one convergent sequence, or even to show that the original sequence is bounded. Boundedness is necessary, but an unclosed set may still contain sequences converging only to points outside it.
Conversely, when trying to show that a set fails to be sequentially compact, it is enough to find one sequence in the set for which no subsequence converges to a point in the set. The sequence approaching \(2\) in \((2,5)\) works because every subsequence has the same limit outside the interval. For an unbounded set, a sequence whose terms have absolute values tending to infinity provides a different obstruction.
The previous tutorial established that compact subsets of \(\mathbb{R}\) have the subsequence property in this definition. The results here give further structure to sets with that property: they must be bounded and closed, and the property survives passage to closed subsets and finite unions. The next tutorial proves directly that compact sets are sequentially compact.
Check Your Understanding
Use the definition and the proved results to answer the following questions.
- What two conditions must a subsequence satisfy in order to establish that a set is sequentially compact?
- Why does an unbounded set contain a sequence that cannot have a convergent subsequence?
- In the proof that a sequentially compact set is closed, why must the limit of the extracted subsequence equal the limit of the original sequence?
- Why does a finite set of values force some value to occur infinitely many times in any sequence taking values in that set?
- How does the proof for finite unions ensure that all terms of the selected subsequence lie in one sequentially compact member?