Sequential Characterization of Closed Sets

Testing Closedness with Sequences

A set is closed when it contains all of its boundary points. That description can be difficult to apply directly: it may not be clear which points are boundary points, or how to check them all. Sequences provide a practical alternative. Instead of locating every possible boundary point, we examine sequences of points already in the set and ask where their limits lie.

The Sequential Characterization of Closed Sets, established in the tutorial “Completeness and Cauchy Sequences,” gives the key test. A subset of $\mathbb{R}$ is closed if and only if every convergent sequence whose terms belong to the set has its limit in the set. We will use that result here, not re-prove it. The emphasis is on applying the test and deriving useful facts about how closed sets combine.

Definition: A subset $E\subseteq\mathbb{R}$ is closed if its complement $\mathbb{R}\setminus E$ is open.

The sequential test can be used in either direction. To prove that a set is closed, begin with an arbitrary convergent sequence in the set and show that its limit also belongs to the set. To show that a set is not closed, it is enough to find one sequence in the set that converges to a point outside it. In both cases, the sequence must have all its terms in the set; the limit itself need not be in the set unless closedness has been established.

A Sequence That Detects a Missing Limit

Worked Example: A Set with a Missing Endpoint

Let $$ E=\left\{2+\frac{1}{n+1}: n\in\mathbb{N}_0\right\}. $$ For every $n\in\mathbb{N}_0$, $1/(n+1)>0$, so every element of $E$ is greater than $2$. In particular, $2\notin E$. Now set $x_n=2+1/(n+1)$. By its definition, $x_n\in E$ for every $n$, and $$ |x_n-2|=\frac{1}{n+1}\longrightarrow 0. $$ Thus $x_n\to2$, while $2\notin E$. The Sequential Characterization of Closed Sets therefore shows that $E$ is not closed.

The sequence gives a specific witness to the failure of closedness: its terms approach a missing point. It is not necessary to describe the complement of $E$ or identify every boundary point to reach the conclusion.

This method also clarifies why the choice of sequence matters. A sequence that merely has some terms in $E$ does not test the criterion: every term must lie in $E$. And a sequence in $E$ that fails to converge does not provide a missing limit to test. The useful combination is membership at every index and convergence to a point outside the set.

Closedness of Intersections

An intersection requires a point to satisfy every membership condition at once. If every set in a family is closed, then a limit of points satisfying all those conditions still satisfies each one. This works even when the family has infinitely many sets.

Theorem (Arbitrary Intersections of Closed Sets): Let $\{E_\alpha:\alpha\in A\}$ be any family of closed subsets of $\mathbb{R}$. Then $\bigcap_{\alpha\in A}E_\alpha$ is closed.

Proof. If the index set $A$ is empty, the intersection is understood to be $\mathbb{R}$, which is closed. Now suppose $A$ is nonempty, and let $(x_n)$ be any convergent sequence whose terms belong to $\bigcap_{\alpha\in A}E_\alpha$. Write $x_n\to x$. For each $\alpha\in A$, membership in the intersection implies $x_n\in E_\alpha$ for every $n$. Since $E_\alpha$ is closed, the Sequential Characterization of Closed Sets gives $x\in E_\alpha$. This holds for every $\alpha\in A$, so $x\in\bigcap_{\alpha\in A}E_\alpha$. The same characterization now shows that the intersection is closed. $\square$

The proof handles an infinite family without choosing one index that works uniformly across all the sets. For each fixed $\alpha$, the sequence lies in $E_\alpha$, so its limit lies in $E_\alpha$. Since this conclusion is valid for every $\alpha$, the limit belongs to the intersection.

Worked Example: Intersecting Closed Intervals

Consider the intervals $E_1=[-2,4]$ and $E_2=[1,6]$. Each is closed, so their intersection is closed by the theorem. Directly, a point belongs to both intervals precisely when it is at least $1$ and at most $4$, giving $$ [-2,4]\cap[1,6]=[1,4]. $$ The sequential argument verifies closedness without needing this particular endpoint calculation. If $x_n\in E_1\cap E_2$ and $x_n\to x$, closedness of $E_1$ gives $x\in E_1$, while closedness of $E_2$ gives $x\in E_2$. Hence $x\in E_1\cap E_2$.

Closedness of Finite Unions

For a union, a point need only belong to one of the sets. In a sequence of points from a finite union, the set containing the term can vary with the index. The important observation is that if there are only finitely many sets, at least one of them must contain terms at infinitely many indices. Those terms form a subsequence, and a subsequence of a convergent sequence has the same limit. Closedness of that one set then places the limit in the union.

Theorem (Finite Unions of Closed Sets): If $E_1,\ldots,E_r$ are closed subsets of $\mathbb{R}$, where $r$ is a positive integer, then $\bigcup_{j=1}^{r}E_j$ is closed.

Proof. Let $(x_n)$ be a convergent sequence with $x_n\in\bigcup_{j=1}^{r}E_j$ for every $n$, and suppose $x_n\to x$. For each index $n$, choose one $j\in\{1,\ldots,r\}$ for which $x_n\in E_j$. This assigns every index to one of finitely many sets of indices. At least one set, say the indices assigned to $E_k$, must be infinite; otherwise each of the $r$ sets of indices would be finite, and their finite union could not contain every $n\in\mathbb{N}_0$.

List those infinitely many indices in increasing order as $n_0<n_1<n_2<\cdots$. Then $x_{n_m}\in E_k$ for every $m$, and $(x_{n_m})$ is a subsequence of $(x_n)$. Since $x_n\to x$, every subsequence also converges to $x$. Because $E_k$ is closed, the Sequential Characterization of Closed Sets gives $x\in E_k$. Therefore $x\in\bigcup_{j=1}^{r}E_j$. The sequential characterization proves that this union is closed. $\square$

Finiteness is essential to this argument. If there are infinitely many sets, the terms can keep moving to new sets so that no single set contains infinitely many terms. Indeed, the union of the closed singleton sets $\{2+1/(n+1)\}$, for $n\in\mathbb{N}_0$, is the set in the first worked example, which is not closed. Arbitrary intersections preserve closedness, but arbitrary unions need not.

Worked Example: A Union of Two Closed Rays

Let $F=(-\infty,0]\cup[3,\infty)$. Each ray is closed: if a sequence of numbers at most $0$ converges to $x$, order preservation for limits gives $x\leq0$; if a sequence of numbers at least $3$ converges to $x$, it gives $x\geq3$. Thus both rays are closed by the sequential characterization, and their finite union $F$ is closed by the theorem.

The union argument matters here because a sequence in $F$ need not stay on one side of the gap. For example, terms may alternate between the two rays. The proof does not assume the entire sequence lies in one ray: it uses infinitely many terms from at least one ray and the corresponding convergent subsequence.

Applying the Test to Inequalities

Many sets are described by inequalities rather than by intervals or unions. The sequential method can handle these directly: take a convergent sequence satisfying the inequality, pass the relevant expression to the limit, and verify that the limiting inequality still holds. The limit laws and order-preservation results from earlier tutorials provide the needed steps.

Worked Example: A Set Defined by a Square Inequality

Consider $$ G=\{x\in\mathbb{R}:x^2\leq 7\}. $$ Let $(x_n)$ be any sequence in $G$ that converges to $x$. For every $n$, $x_n^2\leq7$. By the limit law for a fixed positive integer power, $x_n^2\to x^2$. Order preservation for limits therefore yields $x^2\leq7$. Hence $x\in G$, and the Sequential Characterization of Closed Sets proves that $G$ is closed.

The key step is to pass the inequality to the limit, not to assume that the sequence itself is monotone. No monotonicity is needed: every term satisfies the same upper bound, and the squares converge to the square of the limit.

What the Sequential Test Does—and Does Not—Say

The test is useful because it converts a global question about a set into a statement about arbitrary convergent sequences. To establish closedness, the sequence must be arbitrary; checking one convenient sequence is not enough. To disprove closedness, however, a single counterexample sequence suffices. These two tasks are logically different and call for different amounts of evidence.

It is also important to keep the direction of the test straight. For a closed set, sequences that start inside the set cannot converge to a point outside it. A sequence approaching a point from outside the set does not, by itself, contradict closedness. In the first example, the terms were deliberately chosen from $E$, and their limit was shown to be missing.

The results about intersections and finite unions illustrate how the sequential viewpoint supports proofs about set operations. In an intersection, the same sequence lies in every set, so closedness can be applied separately to each one. In a finite union, the set containing a term may change, so an infinite-subsequence selection is needed. Recognizing which of these patterns applies is often the main step in a proof.

Check Your Understanding

Use the sequential characterization and the set-operation results to answer the following questions.

What properties must a sequence have to show that a set is not closed?
Why do arbitrary intersections of closed sets remain closed even when there are infinitely many sets?
In the proof for a finite union, why must some one set contain terms at infinitely many indices?
Why does the finite-union proof not establish that arbitrary unions of closed sets are closed?
How would you use a convergent sequence to test whether $\{x\in\mathbb{R}:x^2\leq 10\}$ is closed?