Cluster Points of Sequences

Limits That a Sequence Returns Near

A sequence can fail to converge while still returning arbitrarily close to particular real numbers. For instance, the terms of an alternating sequence may keep returning near two different values. To describe this recurring behavior, we focus not on a single limit for the whole sequence, but on limits of selected subsequences.

The previous tutorial, Proof Using Subsequences, used selected indices to retain properties that occur infinitely often. Here we name the real numbers that can be approached along such a selection. These numbers record the sequence’s finite limiting behavior, even when the sequence itself does not converge.

Definition: A real number $L$ is a cluster point of a real sequence $(a_n)_{n=0}^{\infty}$ if there is a subsequence $(a_{n_k})_{k=0}^{\infty}$ that converges to $L$. The set of all cluster points of $(a_n)$ is called its cluster-point set.

Thus, in this course, a cluster point is exactly a subsequential limit. The earlier theorem Infinite Nearness Criterion for Subsequential Limits gives an equivalent test: $L$ is a cluster point if and only if, for every $\varepsilon>0$, infinitely many indices $n$ satisfy $|a_n-L|<\varepsilon$. We will use that criterion throughout. The phrase “infinitely many indices” matters: it concerns how often terms occur, not just whether some term has a value close to $L$.

Key idea: A cluster point is a real number that the sequence visits arbitrarily close to infinitely often. It need not be a value that the sequence ever equals.

Finding Cluster Points in Examples

Worked Example: A Sequence Approaching Zero

Let $a_n=1/(n+1)$ for $n\in\mathbb{N}_0$. We show that its only cluster point is $0$. Given any $\varepsilon>0$, the sequence $1/(n+1)$ converges to $0$, so there is an $N$ such that $1/(n+1)<\varepsilon$ whenever $n\geq N$. There are infinitely many such indices, so infinitely many terms satisfy $|a_n-0|<\varepsilon$. By the infinite-nearness criterion, $0$ is a cluster point.

Now fix $c\neq0$, and choose $\varepsilon=|c|/2$. Since $a_n\to0$, for all sufficiently large $n$ we have $|a_n|<|c|/3$. The reverse triangle inequality then gives $$ |a_n-c|\geq |c|-|a_n|>\frac{2|c|}{3}>\frac{|c|}{2}=\varepsilon. $$ Only finitely many indices can therefore satisfy $|a_n-c|<\varepsilon$. The criterion shows that $c$ is not a cluster point. Hence the cluster-point set is exactly $\{0\}$. Notice that $0$ is not itself a term of the sequence.

Worked Example: Two Repeated Values

Define $a_n=5$ when $n$ is a square of a nonnegative integer, and $a_n=0$ otherwise. There are infinitely many square indices, so $a_n=5$ for infinitely many $n$. There are also infinitely many nonsquare indices, so $a_n=0$ for infinitely many $n$. Every neighborhood of $0$ and every neighborhood of $5$ therefore contains terms at infinitely many indices. Both $0$ and $5$ are cluster points.

For any $c\notin\{0,5\}$, set $r=\frac12\min\{|c|,|c-5|\}$, which is positive. Every term is either $0$ or $5$, and each of those values is at distance at least $2r$ from $c$. Thus no term satisfies $|a_n-c|<r$, so $c$ is not a cluster point. The cluster-point set is exactly $\{0,5\}$.

Worked Example: Alternating Values with Small Errors

Consider $$ a_n=2+(-1)^n+\frac{1}{n+1}. $$ At even indices $n=2k$, substitution gives $$ a_{2k}=2+1+\frac{1}{2k+1}=3+\frac{1}{2k+1}. $$ As $k$ increases, these terms converge to $3$, so $3$ is a cluster point. At odd indices $n=2k+1$, $$ a_{2k+1}=2-1+\frac{1}{2k+2}=1+\frac{1}{2k+2}, $$ and these terms converge to $1$. Thus $1$ is also a cluster point.

To see that there are no others, fix $c\neq1,3$, and let $d=\min\{|c-1|,|c-3|\}>0$. For all sufficiently large $n$, the term $a_n$ is within $d/3$ of the endpoint associated with its parity: within $d/3$ of $3$ at even indices and within $d/3$ of $1$ at odd indices. Each endpoint is at least $d$ from $c$. The triangle inequality therefore gives $|a_n-c|>2d/3$ for all sufficiently large $n$. Only finitely many indices can lie within $d/3$ of $c$. Hence $c$ is not a cluster point, and the cluster-point set is exactly $\{1,3\}$.

The Cluster-Point Set Is Closed

A cluster-point set cannot have a finite limiting point missing from it. The reason is that a point close to a cluster point inherits infinitely many nearby terms, with a slightly larger neighborhood. The following theorem makes this precise.

Theorem (The Cluster-Point Set Is Closed): For every real sequence, the complement of its cluster-point set is open. Equivalently, its cluster-point set is a closed subset of $\mathbb{R}$.

Proof. Let $C$ be the cluster-point set, and take any $x\notin C$. By the infinite-nearness criterion, there is an $r>0$ such that only finitely many indices satisfy $|a_n-x|<r$. We show that every $y$ with $|y-x|<r/2$ is also outside $C$.

If $|a_n-y|<r/2$, then the triangle inequality gives $$ |a_n-x|\leq |a_n-y|+|y-x|<\frac{r}{2}+\frac{r}{2}=r. $$ Thus every index for which $a_n$ lies within $r/2$ of $y$ is among the finitely many indices for which $a_n$ lies within $r$ of $x$. Only finitely many terms lie within $r/2$ of $y$, so the infinite-nearness criterion implies $y\notin C$. We have shown that the whole neighborhood $(x-r/2,x+r/2)$ lies outside $C$. Since this holds for every $x\notin C$, the complement of $C$ is open, as required. $\square$

This result does not say that the cluster-point set must be an interval or contain more than one point. The examples above have finite cluster-point sets, which are closed. It says that if cluster points themselves approach a real number, that real number must also be a cluster point.

Finite Changes Do Not Affect Cluster Points

Cluster points describe behavior that persists across infinitely many indices. Changing a finite number of terms cannot remove infinitely many visits to any neighborhood, or create infinitely many visits where there were only finitely many. This gives a useful invariance principle.

Theorem (Finite Changes Preserve Cluster Points): Suppose two real sequences $(a_n)$ and $(b_n)$ agree at all but finitely many indices. Then they have exactly the same cluster-point set.

Proof. Fix a real number $L$ and $\varepsilon>0$. Because the sequences differ at only finitely many indices, the sets $$ \{n:|a_n-L|<\varepsilon\} \quad\text{and}\quad \{n:|b_n-L|<\varepsilon\} $$ can differ only at those finitely many indices. One of these sets is infinite if and only if the other is infinite: adding or removing finitely many indices cannot change whether a set is infinite. This holds for every $\varepsilon>0$. By the infinite-nearness criterion, $L$ is a cluster point of $(a_n)$ if and only if it is a cluster point of $(b_n)$. Since this is true for every real $L$, the cluster-point sets are equal. $\square$

This is a statement about cluster points, not about the terms themselves. A finite change can affect individual values and can affect whether a particular value appears in the sequence. It cannot change which real numbers are approached at infinitely many indices.

Worked Example: A Unique Cluster Point Does Not Guarantee Convergence

Define $a_{2k}=0$ and $a_{2k+1}=k+1$ for $k\in\mathbb{N}_0$. The even-indexed terms are all zero, so $0$ is a cluster point. We show that no other real number is one. Fix $c\neq0$ and use the neighborhood radius $\varepsilon=|c|/3$. At even indices, $|a_{2k}-c|=|c|>\varepsilon$. At odd indices, $a_{2k+1}=k+1$, which eventually exceeds $|c|+\varepsilon$. For all such $k$, $$ |a_{2k+1}-c|\geq a_{2k+1}-|c|>\varepsilon. $$ Consequently, only finitely many terms can lie within $\varepsilon$ of $c$. The cluster-point set is exactly $\{0\}$.

Nevertheless, the sequence does not converge. Its odd-indexed terms $k+1$ are unbounded, whereas every convergent sequence is bounded by Convergent Sequences Are Bounded. If the full sequence converged, that theorem would make it bounded too, a contradiction. So having exactly one cluster point does not, by itself, ensure convergence: terms can make increasingly large excursions that do not approach any finite real number.

Cluster Points of a Convergent Sequence

For a convergent sequence, there is exactly one cluster point: its limit. This follows directly from the infinite-nearness criterion and helps distinguish the cluster-point description from convergence itself.

Theorem (Cluster Points of a Convergent Sequence): If $a_n\to L$, then the cluster-point set of $(a_n)$ is $\{L\}$.

Proof. Given $\varepsilon>0$, convergence gives an $N$ such that $|a_n-L|<\varepsilon$ for every $n\geq N$. There are infinitely many such indices, so $L$ is a cluster point by the infinite-nearness criterion.

Now let $c\neq L$, and put $d=|c-L|>0$. For all sufficiently large $n$, convergence gives $|a_n-L|<d/3$. For those indices, $$ |a_n-c|\geq |c-L|-|a_n-L|>\frac{2d}{3}>\frac{d}{3}. $$ Only finitely many indices can therefore satisfy $|a_n-c|<d/3$, so $c$ is not a cluster point. Thus $L$ is the only cluster point. $\square$

The implication in this theorem goes from convergence to a unique cluster point; the example with increasingly large odd-indexed terms shows why the converse fails in general. Cluster points describe finite values reached as subsequential limits. They do not record every possible form of unbounded behavior.

Check Your Understanding

Use the definition, the infinite-nearness criterion, and the proved results to answer the following questions.

What does it mean for $L$ to be a cluster point of $(a_n)$ in terms of a subsequence?
Why is $0$ a cluster point of $a_n=1/(n+1)$, even though no term equals $0$?
What does the closedness theorem say about a sequence of cluster points that converges to a real number?
Why can changing finitely many terms not change a sequence’s cluster-point set?
Give a reason that a sequence with exactly one cluster point need not converge.

What You'll Learn