Sequences Connect Convergence to Closure
The Density of the Irrationals Theorem showed that points of a set can be found arbitrarily close to every real number. Closure records this kind of approximation: a point belongs to \(\overline{E}\) when every neighborhood of it meets \(E\). The Sequential Characterization of Closure from “Closure of a Set” expresses the same idea using sequences. Here we use that established result to study two related questions: when does a set contain limits of its convergent sequences, and how can subsequences describe the closure of a sequence’s range?
For a sequence \((a_n)\), its range is the set \(A=\{a_n:n\geq 1\}\). A subsequence is a sequence of the form \((a_{n_k})\), where \(n_1<n_2<\cdots\) are positive integers. A real number \(x\) is a subsequential limit of \((a_n)\) if some subsequence converges to \(x\). A subsequential limit is a limit of selected terms, not necessarily a term of the original sequence.
This definition asks whether a set retains the limits of sequences drawn entirely from it. It does not require the sequence to be monotone, or its terms to be distinct. Repeated terms are allowed, and the sequence may converge to a point that is not one of its terms.
Closed Sets and Limits of Sequences
In \(\mathbb{R}\), sequential closedness is exactly ordinary closedness. This gives a useful way to recognize closed sets through convergence, without needing to describe their complements directly.
Proof. First suppose \(F\) is closed, and let \((a_n)\) be a sequence with \(a_n\in F\) for every \(n\) and \(a_n\to x\). If \(x\notin F\), then \(x\in\mathbb{R}\setminus F\). Since \(F\) is closed, its complement is open. There is therefore an \(r>0\) such that \(B_r(x)\subseteq\mathbb{R}\setminus F\). Convergence gives an \(N\) such that \(n\geq N\) implies \(|a_n-x|<r\), so \(a_n\in B_r(x)\) and hence \(a_n\notin F\). This contradicts \(a_n\in F\). Thus \(x\in F\).
Conversely, suppose every convergent sequence of points of \(F\) has its limit in \(F\). If \(F\) were not closed, the Closure Is the Smallest Closed Superset Theorem from “Closure as the Smallest Closed Superset” would imply that \(\overline{F}\) is not contained in \(F\). Choose \(x\in\overline{F}\setminus F\). By the Sequential Characterization of Closure from “Closure of a Set,” there is a sequence \((a_n)\) with \(a_n\in F\) for every \(n\) and \(a_n\to x\). The assumed property would give \(x\in F\), a contradiction. Therefore \(F\) is closed. \(\square\)
The theorem also explains a common way to show that a set is not closed: find a sequence of points in the set that converges to a point outside it. One sequence is enough to disprove the required property. In the other direction, proving that a set is closed by this test requires considering every convergent sequence of its points.
Worked Example: A Sequence in an Interval with a Limit Outside
Let \(F=[0,1)\), and define \(a_n=1-1/n\) for \(n\geq 1\). For every \(n\geq 1\), \(0\leq 1-1/n<1\), so \(a_n\in F\). Also,
Thus \(a_n\to 1\), but \(1\notin F\). The Sequential Criterion for Closedness shows that \([0,1)\) is not closed. The sequence makes visible the missing endpoint: points of the interval approach \(1\), although the endpoint itself is excluded.
Worked Example: A Closed Interval Retains Sequence Limits
Consider \(F=[-1,2]\), and let \((a_n)\) be any sequence with \(a_n\in F\) for every \(n\) and \(a_n\to x\). If \(x<-1\), choose \(r=(-1-x)/2>0\). Then \(x+r=(-1+x)/2<-1\), so every point of \(B_r(x)\) is less than \(-1\). Convergence would put all sufficiently late \(a_n\) in that ball, contradicting \(a_n\geq-1\). If \(x>2\), choose \(r=(x-2)/2>0\); every point of \(B_r(x)\) is greater than \(2\), again contradicting that all terms lie in \(F\). Therefore \(-1\leq x\leq 2\), so \(x\in F\). This direct check agrees with the theorem, since \([-1,2]\) is closed.
The Closure of a Sequence’s Range
The sequential criterion concerns a fixed set and all sequences drawn from it. A related result describes the closure of the range of one particular sequence. In addition to the terms already in the range, the only points needed to form its closure are the limits of its subsequences.
Proof. First, every point of \(A\) belongs to \(\overline{A}\), since \(A\subseteq\overline{A}\). If \(x\) is a subsequential limit, there is a subsequence \((a_{n_k})\) converging to \(x\), with every \(a_{n_k}\in A\). The Sequential Characterization of Closure implies that \(x\in\overline{A}\). This proves that the right-hand side is contained in \(\overline{A}\).
For the reverse inclusion, take \(x\in\overline{A}\). If \(x\in A\), it is already in the right-hand side. Suppose instead that \(x\notin A\). We construct a subsequence converging to \(x\). Choose any \(n_1\) such that \(|a_{n_1}-x|<1\); such an index exists because \(x\in\overline{A}\), so \(B_1(x)\cap A\neq\varnothing\).
Suppose \(n_{k-1}\) has been chosen. Since \(x\notin A\), each of the finitely many distances \(|x-a_j|\), for \(1\leq j\leq n_{k-1}\), is positive. Their minimum \(d_k\) is therefore positive. Set \(\delta_k=\min\{d_k,1/k\}/2>0\). Because \(x\in\overline{A}\), there is a term \(a_{n_k}\in B_{\delta_k}(x)\). Its index must satisfy \(n_k>n_{k-1}\): if \(n_k\leq n_{k-1}\), then \(|a_{n_k}-x|\geq d_k>\delta_k\), contradicting its membership in that ball. Moreover, $$ |a_{n_k}-x|<\delta_k\leq\frac{1}{2k}<\frac{1}{k}. $$ These inequalities show that \(a_{n_k}\to x\), so \(x\) is a subsequential limit. Thus every point of \(\overline{A}\) belongs to the right-hand side, completing the proof. \(\square\)
The condition \(x\notin A\) matters in the construction. It ensures that \(x\) has positive distance from each of the finitely many terms whose indices have already been passed. That lets us choose a nearby term with a strictly larger index, as required for a subsequence. If \(x\in A\), no subsequence construction is needed to put \(x\) in the formula: it is already in \(A\).
Worked Example: The Closure of the Range of One Over n
Let \(a_n=1/n\) for \(n\geq1\), so \(A=\{1,1/2,1/3,\ldots\}\). The full sequence converges to \(0\), so \(0\) is a subsequential limit. Every subsequence \((a_{n_k})\) also converges to \(0\): because the indices increase, \(n_k\geq k\), and hence
No other point is a subsequential limit. The Closure of the Range of a Sequence Theorem therefore gives $$ \overline{A}=A\cup\{0\}. $$ In particular, \(0\) is in the closure even though \(0\) is not a term of the sequence.
Worked Example: A Sequence with Two Subsequential Limits
Define \(a_n=(-1)^n\). Its range is \(A=\{-1,1\}\). The even-indexed terms form the constant subsequence \(a_{2k}=1\), and the odd-indexed terms form the constant subsequence \(a_{2k-1}=-1\). Thus both \(1\) and \(-1\) are subsequential limits.
There are no others. If a subsequence converges and contains infinitely many terms equal to \(1\) and infinitely many equal to \(-1\), then it has a constant subsequence with limit \(1\) and another with limit \(-1\). This contradicts uniqueness of limits, since every subsequence of a convergent sequence has the same limit. Therefore any convergent subsequence must eventually have only one of these values, and its limit is that value. The theorem now gives \(\overline{A}=A=\{-1,1\}\).
Worked Example: Rational Terms Converging to an Irrational Limit
By Rational Approximation from “Density of the Rationals,” for each positive integer \(n\) we can choose \(q_n\in\mathbb{Q}\) with \(|q_n-\sqrt{2}|<1/n\). Since \(1/n\to0\), these inequalities give \(q_n\to\sqrt{2}\). The number \(\sqrt{2}\) is irrational, as established in “Density of the Irrationals,” so it is not in the range \(A=\{q_n:n\geq1\}\). Yet it is a subsequential limit: the full sequence is itself a subsequence. Therefore \(\sqrt{2}\in\overline{A}\), although \(\sqrt{2}\notin A\).
The same construction also shows that \(\mathbb{Q}\) is not closed. Each \(q_n\) belongs to \(\mathbb{Q}\), while their limit \(\sqrt{2}\) does not. The Sequential Criterion for Closedness applies to the sequence \((q_n)\) and rules out closedness.
What These Sequence Tests Do—and Do Not—Say
The two theorems address different sets of sequences. To test whether a set \(F\) is closed, the Sequential Criterion for Closedness asks about every convergent sequence whose terms lie in \(F\). To find the closure of the range of one sequence, the Closure of the Range of a Sequence Theorem asks about the subsequential limits of that sequence, together with its terms. Keeping these questions separate avoids a common confusion between a set and the range of a particular sequence.
Neither theorem says that a dense set must equal \(\mathbb{R}\). The irrationals are dense, but a rational number is not irrational. Density means that every neighborhood meets the set; closedness means that limits of sequences in the set remain in it. The rationals illustrate the distinction: they are dense, and rational sequences can converge to irrational numbers, so they are not closed.
In later work, it will be useful to collect the limits of sequences drawn from a set. The sequential criterion already gives a practical test: a set is closed precisely when it contains every such limit. The range theorem gives a complementary perspective for one sequence at a time, identifying which additional points must be added to its range to obtain its closure.
Check Your Understanding
Use the sequence and closure results to answer the following questions.
- What sequence property characterizes closed subsets of \(\mathbb{R}\)?
- Why does a sequence of points in \([0,1)\) converging to \(1\) show that \([0,1)\) is not closed?
- For a sequence with range \(A\), which points, besides points already in \(A\), are needed to describe \(\overline{A}\)?
- In the proof of the Closure of the Range of a Sequence Theorem, why must the chosen indices increase strictly?
- Why does the existence of rational numbers \(q_n\to\sqrt{2}\) show that \(\mathbb{Q}\) is not closed but does not contradict the density of \(\mathbb{Q}\)?