Sequential Closure

Collecting Limits of Sequences from a Set

The previous tutorial connected closure with limits of sequences and showed that a set in $\mathbb{R}$ is closed precisely when it contains the limits of its convergent sequences. We now make the collection of those limits into a set of its own. This construction is called the sequential closure. It gives a sequence-based way to add to a set every point that its own sequences can approach.

Definition: For $E\subseteq\mathbb{R}$, the sequential closure of $E$, denoted $\operatorname{scl}(E)$, is the set of all limits of convergent sequences whose terms lie in $E$: $$ \operatorname{scl}(E)=\{x\in\mathbb{R}:\text{ there is a sequence }(x_n)\text{ with }x_n\in E\text{ for every }n\text{ and }x_n\to x\}. $$

The terms of such a sequence need not be distinct. In particular, for any $x\in E$, the constant sequence $x_n=x$ converges to $x$. Thus $E\subseteq\operatorname{scl}(E)$. If $E$ is empty, there is no sequence whose terms all lie in $E$, so $\operatorname{scl}(E)=\varnothing$.

The Sequential Characterization of Closure from “Closure of a Set” immediately identifies this construction with the ordinary closure: $x\in\overline{E}$ exactly when some sequence in $E$ converges to $x$. Hence $\operatorname{scl}(E)=\overline{E}$ in $\mathbb{R}$. We will use that established characterization rather than prove it again. The focus here is what the sequential definition lets us do, especially when limits are taken more than once.

Examples of Sequential Closure

Worked Example: The Sequential Closure of an Open Interval

Let $E=(2,5)$. We show that $\operatorname{scl}(E)=[2,5]$. First, each $x\in(2,5)$ belongs to $\operatorname{scl}(E)$ by the constant sequence. For the left endpoint, the sequence $x_n=2+1/n$ lies in $(2,5)$ for every $n\geq1$, since $2<2+1/n\leq3<5$, and $x_n\to2$. For the right endpoint, $y_n=5-1/n$ lies in $(2,5)$, since $4\leq5-1/n<5$, and $y_n\to5$.

Now suppose $z_n\in(2,5)$ for every $n$ and $z_n\to z$. We verify that $2\leq z\leq5$. If $z<2$, put $\varepsilon=(2-z)/2>0$. Convergence would imply that eventually $z_n<z+\varepsilon=(z+2)/2<2$, contradicting $z_n>2$. If $z>5$, put $\varepsilon=(z-5)/2>0$. Eventually $z_n>z-\varepsilon=(z+5)/2>5$, again a contradiction. Thus every such limit lies in $[2,5]$, proving the claimed equality.

Worked Example: The Integers Are Their Own Sequential Closure

Let $E=\mathbb{Z}$, and suppose a sequence of integers $m_n$ converges to $x\in\mathbb{R}$. By convergence, there is an $N$ such that $n\geq N$ implies $|m_n-x|<1/3$. For any $n,k\geq N$, the triangle inequality gives $$ |m_n-m_k|\leq |m_n-x|+|x-m_k|<\frac{2}{3}<1. $$ Since $m_n-m_k$ is an integer, the only way its absolute value can be less than $1$ is if $m_n-m_k=0$. Therefore all terms from index $N$ onward equal the same integer, say $m$. The sequence converges to $m$, and uniqueness of limits gives $x=m\in\mathbb{Z}$. Constant sequences show the reverse inclusion, so $\operatorname{scl}(\mathbb{Z})=\mathbb{Z}$.

Worked Example: A Countable Union Can Add a New Limit

For each positive integer $j$, let $E_j=\{3+1/j\}$. The sequential closure of a singleton is itself: every sequence with terms in $E_j$ is constant at $3+1/j$. Therefore $$ \bigcup_{j=1}^{\infty}\operatorname{scl}(E_j) =\{3+1/j:j\geq1\}. $$ But the sequence $x_j=3+1/j$, with $x_j\in\bigcup_{j=1}^{\infty}E_j$, converges to $3$. Hence $3\in\operatorname{scl}(\bigcup_{j=1}^{\infty}E_j)$, even though $3$ does not belong to $\bigcup_{j=1}^{\infty}\operatorname{scl}(E_j)$. The limit comes from moving through different sets in the union; it need not be a limit obtained within any one fixed member.

Sequential Closure Is Stable Under Further Limits

A central property of sequential closure is that taking limits of sequences from $\operatorname{scl}(E)$ does not produce points outside $\operatorname{scl}(E)$. The proof uses a diagonal choice: for each term of a sequence approaching $x$, choose a point of $E$ sufficiently close to that term. The selected points then form one sequence in $E$ approaching $x$.

Theorem (Sequential Closure Is Sequentially Closed): Let $E\subseteq\mathbb{R}$. If $x_n\in\operatorname{scl}(E)$ for every $n$ and $x_n\to x$, then $x\in\operatorname{scl}(E)$.

Proof. For each $n\geq1$, the membership $x_n\in\operatorname{scl}(E)$ means there is a sequence $(y_{n,k})_{k\geq1}$ with $y_{n,k}\in E$ for every $k$ and $y_{n,k}\to x_n$ as $k\to\infty$. Since this sequence converges to $x_n$, there is an index $k(n)$ such that $$ |y_{n,k(n)}-x_n|<\frac{1}{n}. $$ Set $y_n=y_{n,k(n)}$. Then $y_n\in E$ for every $n$, and $$ |y_n-x|\leq |y_n-x_n|+|x_n-x|<\frac{1}{n}+|x_n-x|. $$ Both terms on the right tend to $0$. Thus $y_n\to x$, with every $y_n\in E$. By the definition of sequential closure, $x\in\operatorname{scl}(E)$. $\square$

The array $(y_{n,k})$ in this proof contains one approximating sequence for each $x_n$. The diagonal choice takes one sufficiently accurate term from each row. It is essential that the accuracy improves with $n$: the bound $1/n$ ensures that the error from replacing $x_n$ by $y_n$ vanishes as the outer sequence approaches $x$.

Corollary (Sequential Closure Is the Smallest Sequentially Closed Superset): For every $E\subseteq\mathbb{R}$, $\operatorname{scl}(E)$ is sequentially closed, contains $E$, and is contained in every sequentially closed set that contains $E$.

Proof. The theorem proves that $\operatorname{scl}(E)$ is sequentially closed. The constant-sequence observation gives $E\subseteq\operatorname{scl}(E)$. Now let $F$ be any sequentially closed set with $E\subseteq F$. If $x\in\operatorname{scl}(E)$, there is a sequence $(x_n)$ in $E$ converging to $x$. Since each $x_n\in F$, sequential closedness of $F$ gives $x\in F$. Thus $\operatorname{scl}(E)\subseteq F$, as required. $\square$

This smallest-superset property explains why the construction is useful: it is not merely a list of some limits. It is a set that already contains all limits of sequences from itself, and it is the least set with that property that contains $E$. In $\mathbb{R}$, the Sequential Characterization of Closure also tells us that this smallest sequentially closed superset is exactly $\overline{E}$.

Finite Unions and the Countable-Union Pitfall

Sequential closure behaves particularly simply for a union of two sets. A sequence in $A\cup B$ has infinitely many terms in at least one of $A$ or $B$. Those terms form a subsequence, and a subsequence of a convergent sequence has the same limit. This observation gives the following result directly from the sequential definition.

Theorem (Sequential Closure of a Finite Union): For any $A,B\subseteq\mathbb{R}$, $$ \operatorname{scl}(A\cup B)=\operatorname{scl}(A)\cup\operatorname{scl}(B). $$

Proof. If $x\in\operatorname{scl}(A)$, a sequence in $A$ converges to $x$, and that sequence is also in $A\cup B$. Hence $x\in\operatorname{scl}(A\cup B)$. The same argument applies to $x\in\operatorname{scl}(B)$, proving one inclusion.

For the other inclusion, let $x\in\operatorname{scl}(A\cup B)$, and choose a sequence $x_n\in A\cup B$ with $x_n\to x$. If infinitely many terms lie in $A$, select those terms in increasing index order. They form a subsequence in $A$ that converges to $x$, so $x\in\operatorname{scl}(A)$. If only finitely many terms lie in $A$, then all sufficiently late terms lie in $B$; the tail is a sequence in $B$ converging to $x$, so $x\in\operatorname{scl}(B)$. In either case $x\in\operatorname{scl}(A)\cup\operatorname{scl}(B)$, completing the proof. $\square$

The same reasoning applies to any finite number of sets: among finitely many sets, at least one contains infinitely many terms of a given sequence. It does not extend to an arbitrary countable union. In the worked example, the term $3+1/j$ comes from the $j$-th singleton, so no single singleton supplies a subsequence converging to $3$. This is why a limit from a countable union can fail to belong to the union of the individual sequential closures.

Why the Diagonal Argument Matters

The diagonal argument is a useful technique whenever each point in an approximating sequence comes with its own sequence of approximations. The inner sequences may vary from one outer index to the next, so one cannot simply treat them as a single sequence. Choosing the $n$-th approximation within $1/n$ controls this variation and allows the triangle inequality to transfer convergence from $x_n$ to the selected points of $E$.

A common pitfall is to assume that a limit of points in $\operatorname{scl}(E)$ automatically belongs to it just because each point is already a limit of points in $E$. That conclusion requires combining the separate approximations. The diagonal proof supplies the missing construction. A different pitfall is to assume the same finite-union identity holds for countably many sets. The singleton example shows exactly how the limit can move from one set to another at each stage.

For subsets of the real line, sequential closure and ordinary closure ultimately describe the same set, by the established Sequential Characterization of Closure. The sequential viewpoint nevertheless provides concrete tools: it identifies approximating sequences, proves stability under further limits, and reveals why finite and countable unions behave differently at the level of individual members.

Check Your Understanding

Use the definition and results about sequential closure to answer the following questions.

Why does every point of $E$ belong to $\operatorname{scl}(E)$?
In the diagonal proof, why is it useful to choose an approximation within $1/n$ of $x_n$?
What does the smallest-sequentially-closed-superset property say about any sequentially closed set containing $E$?
Why does a sequence in $A\cup B$ have a subsequence lying entirely in one of $A$ or $B$?
How does the family of singleton sets in the countable-union example produce a limit absent from each individual sequential closure?

What You'll Learn