Accumulation Point in a Metric Space¶

Definition¶

Let $(X,d)$ be a metric space and let $E\subseteq X$ $X$ . A point $x\in X$ is called an accumulation point (or cluster point, limit point) of $E$ if every metric ball around $x$ contains a point of $E$ different from $x$ , i.e.,

\forall \varepsilon>0,\quad \left(B_d(x,\varepsilon)\setminus\{x\}\right)\cap E\neq\varnothing.

Let $(X,d)$ be a metric space and let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $X$ . A point $x\in X$ is called an accumulation point (or cluster point) of $(x_n)$ if, for every $\varepsilon>0$ $0$ , there exist infinitely many $n\in\mathbb{N}$ such that $x_n\in B_d(x,\varepsilon)$ $B_{d} (x, ε)$ , i.e., for every $\varepsilon>0$ $0$ ,

\left|\{n\in\mathbb{N}\colon x_n\in B_d(x,\varepsilon)\}\right|=\infty.

Interpretation¶

An accumulation point of a set is a location you can approach arbitrarily closely using other points from the set. For a sequence, it is a point to which the sequence returns again and again: however small the ball around it, infinitely many terms lie in that ball.

Both definitions ask something of every $\varepsilon>0$ $0$ , so only small balls matter. For a set, the words "different from $x$ " rule out counting $x$ itself, and $x$ need not belong to $E$ : for $E=(0,1)$ in $\mathbb{R}$ , every point of $[0,1]$ is an accumulation point, including $0$ and $1$ . For a sequence, indices $n$ are counted, not distinct values. The constant sequence $x_n=c$ has $c$ as an accumulation point, although the set $\{c\}$ of its terms has none.

The accumulation points of a sequence in a metric space are exactly the limits of its convergent subsequences, which is the form in which they appear in the Bolzano-Weierstrass theorem. For a bounded real sequence, the limit superior and limit inferior are its largest and smallest accumulation points. The accumulation points of a set are what a closed set must contain: a set is closed if and only if it contains all its accumulation points, and its closure is the set together with them.

A common slip is to treat a single accumulation point as a limit. The sequence $0,1,0,2,0,3,\dots$ has $0$ as its only accumulation point, yet it does not converge, because it is unbounded.

Key Results¶

A limit point can be characterized as an adherent point that is not an isolated point (a point $x\in E$ such that some ball $B_d(x,\varepsilon)$ meets $E$ only in $x$ ).

Examples¶

For $E=\{1,\,1/2,\,1/3,\,\dots\}$ ${1, 1/2, 1/3, \dots}$ , the only accumulation point is $0$ . Each point $1/n$ is isolated.
The sequence $a_n=(-1)^n$ has accumulation points 1 and -1. It does not converge.
Consider the sequence $(a_n)_{n=1}^\infty$ that is defined by $a_n=\sin n$ $sin n$ . We can show that any number $A\in[-1,1]$ is an accumulation point of this sequence.

Among the first $50$ terms, four lie in the band $[0.4,0.6]$ (red); infinitely many terms fall in every band around $0.5$ , so $0.5$ is an accumulation point.

Prerequisites¶

Open Metric Ball

Sources¶

Math III: Analysis · Lecture 9

Aliases: cluster point, limit point, subsequential limit Symbols: element of, epsilon, for all, infinity, intersection, natural numbers, real numbers, subset

Last updated: 20 October 2026 03:02