Turning Closedness into a Sequence Test
The previous tutorial showed how compactness can produce a convergent subsequence. Here we use sequences for a different purpose: to recognize closed sets. A closed set contains the limits of convergent sequences whose terms lie in the set. The more revealing direction is the converse: if a set is not closed, we can construct a sequence in it that converges to a point outside it.
The key construction comes from the definition of closure. A point is in the closure of a set when every open interval around it meets the set. Choosing a point from the set in each progressively smaller interval gives a sequence converging to the original point. This converts a statement about all neighborhoods into a concrete sequence.
In particular, every point of \(A\) belongs to \(\overline{A}\), since every interval centered at such a point contains the point itself. The closure can also include points not in \(A\), when points of \(A\) occur arbitrarily close to them. If \(A\) is empty, no interval meets \(A\), so \(\overline{A}\) is empty.
A Sequence Characterizes Closure
Proof. Suppose first that \(x\in\overline{A}\). Then \(A\) is nonempty, because every interval centered at \(x\) must meet \(A\). For each positive integer \(n\), the interval of radius \(1/n\) centered at \(x\) meets \(A\). Choose \(a_n\in A\) such that \(|a_n-x|<1/n\). To verify convergence, let \(\varepsilon>0\). Choose a positive integer \(N\) such that \(1/N<\varepsilon\). For every \(n\geq N\), \[ |a_n-x|<1/n\leq1/N<\varepsilon. \] Thus \(a_n\to x\).
Conversely, suppose there is a sequence \((a_n)\) in \(A\) with \(a_n\to x\). Let \(\varepsilon>0\). By convergence, there is a positive integer \(N\) such that \(|a_n-x|<\varepsilon\) for every \(n\geq N\). In particular, \(a_N\in A\) lies in the interval of radius \(\varepsilon\) centered at \(x\). Since this holds for every \(\varepsilon>0\), every such interval meets \(A\), so \(x\in\overline{A}\). \(\square\)
The proof in the first direction uses a specific choice at each stage: take a point within distance \(1/n\). The intervals shrink to \(x\), and the chosen points therefore converge to \(x\). In the reverse direction, convergence guarantees that sufficiently late terms lie in any prescribed interval around the limit. The terms belong to \(A\), so they witness that the interval meets \(A\).
Worked Example: A Point in the Closure That Is Not in the Set
Let \(A=\{2+1/n:n\in\mathbb N\}\). The point \(2\) is not in \(A\), because \(2+1/n>2\) for every positive integer \(n\). But the sequence \(a_n=2+1/n\) lies in \(A\) and converges to \(2\), since \[ |a_n-2|=|(2+1/n)-2|=1/n\longrightarrow0. \] The Sequential Characterization of Closure gives \(2\in\overline{A}\).
The same conclusion follows directly from neighborhoods. Given \(\varepsilon>0\), choose \(n\) with \(1/n<\varepsilon\). Then \(a_n=2+1/n\in A\) and \(|a_n-2|<\varepsilon\), so the interval around \(2\) of radius \(\varepsilon\) meets \(A\). This example illustrates why a set can fail to contain all of its closure points.
The Closed-Set Characterization
Proof. If \(F\) is closed and \((x_n)\) is a sequence in \(F\) converging to \(x\), then \(x\in F\) by the previously established Limits of Sequences in Closed Sets lemma.
For the converse, suppose every convergent sequence in \(F\) has its limit in \(F\). We show that \(F\) is closed. Suppose, to the contrary, that \(F\) is not closed. Then its complement \(\mathbb R\setminus F\) is not open. By the definition of an open set, there is some \(x\in\mathbb R\setminus F\) such that no open interval centered at \(x\) is contained in \(\mathbb R\setminus F\). Consequently, for every \(\varepsilon>0\), the interval of radius \(\varepsilon\) centered at \(x\) meets \(F\).
For each positive integer \(n\), choose \(x_n\in F\) with \(|x_n-x|<1/n\). Given \(\varepsilon>0\), choose \(N\) with \(1/N<\varepsilon\). Then, for every \(n\geq N\), \[ |x_n-x|<1/n\leq1/N<\varepsilon. \] Thus \(x_n\to x\). By the assumed sequence property, \(x\in F\). This contradicts the choice \(x\in\mathbb R\setminus F\). Therefore \(F\) is closed. \(\square\)
The contradiction produces exactly the sequence needed to challenge the proposed property: if the set is not closed, a point outside the set has every neighborhood meeting it. Choosing one point of the set from each shrinking neighborhood gives a sequence converging to that outside point. The sequential condition rules out this possibility.
Worked Example: The Integers Are Closed
We verify the sequence condition for \(\mathbb Z\). Suppose \((m_n)\) is a sequence of integers and \(m_n\to x\in\mathbb R\). Convergence gives a positive integer \(N\) such that \[ |m_n-x|<1/3 \] whenever \(n\geq N\). If \(n,k\geq N\), the triangle inequality gives \[ |m_n-m_k|\leq|m_n-x|+|x-m_k|<1/3+1/3=2/3. \] But \(m_n-m_k\) is an integer. The only integer whose absolute value is less than \(2/3\) is \(0\), so \(m_n=m_k\). Thus all terms from index \(N\) onward equal one integer, say \(m\).
Since the sequence is eventually equal to \(m\), it converges to \(m\). It also converges to \(x\) by assumption. Uniqueness of limits gives \(x=m\in\mathbb Z\). Every convergent sequence in \(\mathbb Z\) therefore has its limit in \(\mathbb Z\), and the Sequential Characterization of Closed Sets shows that \(\mathbb Z\) is closed.
Worked Example: A Sequence Detects a Missing Endpoint
Let \(F=(-\infty,4)\), and define \(x_n=4-1/n\) for each positive integer \(n\). Since \(1/n>0\), every term satisfies \(x_n<4\), so \(x_n\in F\). At the same time, \[ |x_n-4|=|(4-1/n)-4|=1/n\longrightarrow0. \] Thus a sequence in \(F\) converges to \(4\), which is not in \(F\). The Sequential Characterization of Closed Sets implies that \(F\) is not closed.
This reasoning is often the quickest way to show that a set is not closed: it is enough to exhibit one sequence in the set and compute its limit outside the set. To prove that a set is closed, by contrast, the sequence condition must be checked for every convergent sequence in the set.
Worked Example: A Closed Ray
Let \(F=[-2,\infty)\), and suppose \(x_n\in F\) for every \(n\), with \(x_n\to x\). Then \(-2\leq x_n\) for all \(n\). The Order Is Preserved Under Limits theorem, established earlier in the course, gives \(-2\leq x\). Hence \(x\in[-2,\infty)=F\). Every convergent sequence in \(F\) has its limit in \(F\), so the Sequential Characterization of Closed Sets shows that \(F\) is closed.
The order theorem makes this verification short: the defining inequality passes to the limit. The characterization then turns that sequence argument into a conclusion about closedness.
How to Use the Characterization
The two directions have different proof patterns. To prove a set is closed using sequences, start with an arbitrary convergent sequence in the set and show that its limit still satisfies the set’s defining condition. To show a set is not closed, look for one sequence in the set whose limit is missing. In both cases, the quantifiers matter: the closed-set condition concerns every convergent sequence in the set, while a single counterexample sequence disproves it.
The Sequential Characterization of Closure and the Sequential Characterization of Closed Sets fit together. A point \(x\) belongs to \(\overline{A}\) exactly when some sequence in \(A\) converges to \(x\). Therefore \(F\) is closed exactly when every point in \(\overline{F}\) already belongs to \(F\). Since every point of \(F\) belongs to its closure, this is equivalent to \(\overline{F}=F\).
A common pitfall is to confuse “a sequence in the set converges” with “the limit is in the set.” Convergence alone does not ensure that the limit belongs to the set. The set must have the additional property of being closed, or the sequence must be checked directly. The examples above show both possibilities: the integers and the closed ray retain sequence limits, while the open ray omits one.
If every interval around \(x\) meets \(A\), choose \(a_n\in A\) within distance \(1/n\) of \(x\).
Assume a sequence lies in \(F\) and converges; prove its limit satisfies the condition defining \(F\).
Find \(x\notin F\) with points of \(F\) arbitrarily close, and select one such point at each scale \(1/n\).
Check Your Understanding
Use the sequence and closure characterizations to answer the following questions.
- How does membership of \(x\) in \(\overline{A}\) allow you to choose a point of \(A\) within distance \(1/n\) of \(x\) for each \(n\)?
- Why does a sequence in \(A\) converging to \(x\) imply that every interval centered at \(x\) meets \(A\)?
- In the proof of the closed-set characterization, why does failure of openness of \(\mathbb R\setminus F\) provide points of \(F\) arbitrarily close to some \(x\notin F\)?
- What sequence shows that \((-\infty,4)\) is not closed?
- What must be proved about an arbitrary convergent sequence in \(F\) to establish that \(F\) is closed?