Accumulation Points and Isolated Points: Definition, Properties and Examples

Accumulation point

Let \( A\subseteq\mathbb R \). A point \( x_0\in\mathbb R \) is called an accumulation point of \( A \) if every neighborhood of \( x_0 \) contains at least one element of \( A \) different from \( x_0 \).

Equivalently,

\[ \forall r>0, \qquad \Bigl((x_0-r,x_0+r)\setminus\{x_0\}\Bigr)\cap A \neq\varnothing. \]

In other words, no matter how small we make the neighborhood centered at \( x_0 \), we can always find points of \( A \) arbitrarily close to \( x_0 \).

It is important to note that \( x_0 \) need not belong to the set \( A \). The definition only requires that elements of \( A \) different from \( x_0 \) be present arbitrarily close to it.

Moreover, every neighborhood of an accumulation point necessarily contains infinitely many points of the set different from \( x_0 \). Indeed, if some neighborhood contained only finitely many of them, we could construct a smaller neighborhood excluding them all, contradicting the definition.

In particular, no finite set can have accumulation points.

Geometrically, an accumulation point is a point around which the set clusters. If we imagine zooming in repeatedly on the real line near \( x_0 \), we would always keep seeing elements of the set arbitrarily close to that point.

An accumulation point may or may not belong to the set. For instance, if \( A=(0,1) \), the point \( \displaystyle \frac12 \) belongs to \( A \) and is an accumulation point. The points \( 0 \) and \( 1 \) are accumulation points as well, even though they do not belong to the set, since every neighborhood of either one contains elements of \( A \).

Isolated point

Let \( A\subseteq\mathbb R \). A point \( x_0\in A \) is called an isolated point of \( A \) if there exists a neighborhood of \( x_0 \) containing no other element of the set.

In symbols,

\[ \exists r>0 \quad\text{such that}\quad (x_0-r,x_0+r)\cap A=\{x_0\}. \]

This means that \( x_0 \) is separated from the other elements of \( A \) by a positive distance. In a sufficiently small neighborhood of \( x_0 \), the only point of the set present is \( x_0 \) itself.

Geometrically, an isolated point can be thought of as a "lonely" element of the set, surrounded by a region of the real line free of any other points belonging to \( A \).

The notions of isolated point and accumulation point are closely related. If \( x_0\in A \), then exactly one of the following holds:

• \( x_0 \) is an isolated point;
• \( x_0 \) is an accumulation point.

The two properties are mutually exclusive. Indeed, if there exists a neighborhood containing only \( x_0 \), then \( x_0 \) cannot be an accumulation point. Conversely, if \( x_0 \) is an accumulation point, every neighborhood of it contains infinitely many points of the set different from \( x_0 \), and therefore \( x_0 \) cannot be isolated.

Comparison between isolated points and accumulation points

A single set may contain both isolated points and accumulation points. For example, in the set

\[ \left\{0\right\} \cup \left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}, \]

the point \(0\) is an accumulation point, whereas all the points of the form \( \displaystyle \frac1n \) are isolated.

Fundamental examples

Let us look at some fundamental examples, useful for telling accumulation points and isolated points precisely apart.

Intervals

Let \( A=(0,1) \). Every point of \( (0,1) \) is an accumulation point of \( A \), since every neighborhood of it contains infinitely many points of the interval. The endpoints \(0\) and \(1\) are accumulation points too, even though they do not belong to \(A\). Indeed, every neighborhood of \(0\) contains positive points less than \(1\), while every neighborhood of \(1\) contains points of the interval less than \(1\).

Hence the set of accumulation points of \( (0,1) \) is

\[ [0,1]. \]

Finite sets and discrete sets

If \( A=\{1,3,7\} \), then all the elements of \(A\) are isolated points. For instance, around the point \(3\) we can choose a sufficiently small interval that contains neither \(1\) nor \(7\). More generally, every finite set of real numbers consists solely of isolated points and has no accumulation points.

The set of integers \( \mathbb Z \) likewise consists only of isolated points. Indeed, for every \( n\in\mathbb Z \), the neighborhood

\[ \left(n-\frac12,n+\frac12\right) \]

contains the point \(n\) as its only integer. Hence every integer is isolated, and \( \mathbb Z \) has no accumulation points in \( \mathbb R \).

A set with isolated points and an accumulation point

Consider the set

\[ A=\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}. \]

Every point of the form \(\displaystyle \frac1n \) is isolated. Indeed, once a value of \(n\) is fixed, we can choose a neighborhood of \( \displaystyle \frac1n \) small enough to contain no other element of the sequence \(1,\displaystyle\frac12,\displaystyle\frac13,\ldots\).

Nevertheless, \(0\) is an accumulation point of \(A\). Indeed, for every \(r>0\) there exists \(n\in\mathbb N\) large enough that

\[ 0<\frac1n<r. \]

Hence every neighborhood of \(0\) contains elements of \(A\) different from \(0\). Note that \(0\notin A\): this shows that an accumulation point need not belong to the set.

If instead we consider

\[ B=\{0\}\cup\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}, \]

the point \(0\) remains an accumulation point, but it now belongs to the set \(B\) as well. The points \( \displaystyle\frac1n \), on the other hand, are still isolated points.

The rational numbers

Consider the set of rational numbers \( \mathbb Q \). By the density of the rationals in \( \mathbb R \), between any two distinct real numbers there always lie infinitely many rationals. Consequently, every neighborhood of any point of the real line contains infinitely many elements of \( \mathbb Q \).

Therefore every real number is an accumulation point of \( \mathbb Q \); that is,

\[ \mathbb Q' = \mathbb R. \]

Moreover, \( \mathbb Q \) has no isolated point whatsoever. This example strikingly shows that a set can have accumulation points at every point of the real line, even though it is not an interval and is, in fact, totally disconnected.

Characterization via sequences

Accumulation points can be characterized by means of sequences. This result is especially important because it lets us translate a geometric property of sets into a property of convergence.

Theorem. Let \( A\subseteq\mathbb R \) and let \( x_0\in\mathbb R \). Then \( x_0 \) is an accumulation point of \( A \) if and only if there exists a sequence \( (x_n) \subseteq A\setminus\{x_0\} \) such that

\[ \lim_{n\to\infty}x_n=x_0. \]

Proof. Suppose that \( x_0 \) is an accumulation point of \( A \). For every \( n\in\mathbb N \), the neighborhood

\[ \left(x_0-\frac1n,x_0+\frac1n\right) \]

contains at least one point \( x_n\in A\setminus\{x_0\} \). It follows that

\[ 0<|x_n-x_0|<\frac1n. \]

Since \( \displaystyle \frac1n\to0 \), the squeeze theorem gives \( x_n\to x_0 \).

Conversely, suppose there exists a sequence \( (x_n)\subseteq A\setminus\{x_0\} \) such that \( x_n\to x_0 \). Let \( r>0 \). By the definition of limit there exists \( N\in\mathbb N \) such that

\[ n\ge N \quad\Longrightarrow\quad |x_n-x_0|<r. \]

In particular \( x_N\in(x_0-r,x_0+r) \), with \( x_N\in A \) and \( x_N\neq x_0 \). Therefore every neighborhood of \( x_0 \) contains an element of \( A \) different from \( x_0 \), and hence \( x_0 \) is an accumulation point of \( A \).

This characterization is one of the principal links between the theory of sets and the study of sequences.

The derived set and closed sets

The set of all accumulation points of a set \( A \subseteq \mathbb R \) is called the derived set of \( A \) and is denoted by \( A' \).

In symbols,

\[ A'=\{x\in\mathbb R : x \text{ is an accumulation point of } A\}. \]

The derived set describes the behavior of \( A \) in its surroundings and gathers all the points around which the set clusters.

Note that the points of \( A' \) need not belong to \( A \). For example, if \( A=(0,1) \), then \( 0 \) and \( 1 \) belong to \( A' \) even though they do not belong to the set.

Let us look at a few examples.

• If \( A=(0,1) \), then \[ A'=[0,1]. \]
• If \( A=\mathbb Z \), then \[ A'=\varnothing, \] since every integer is an isolated point.
• If \[ A=\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}, \] then \[ A'=\{0\}. \]

The notion of derived set makes it easy to characterize closed sets.

Theorem. A set \( A\subseteq\mathbb R \) is closed if and only if it contains all of its accumulation points.

Equivalently,

\[ A \text{ is closed} \quad\Longleftrightarrow\quad A'\subseteq A. \]

In other words, a set is closed precisely when it does not "lose" any point toward which its elements may accumulate.

For example, the closed interval \( [0,1] \) contains all of its accumulation points and is therefore a closed set. By contrast,

\[ (0,1) \]

is not closed, since the points \(0\) and \(1\) are accumulation points but do not belong to the set.

Accumulation points play a fundamental role in mathematical analysis. One of the most important theorems of real analysis, the Bolzano–Weierstrass theorem, states that every infinite, bounded subset of \( \mathbb R \) has at least one accumulation point.

This result underscores how the presence of accumulation points is an intrinsic property of bounded infinite sets, and it stands as one of the pillars of mathematical analysis.