Examples of Disconnected Sets

How to Recognize Disconnected Sets

The Characterization of Connected Subsets of $\mathbb{R}$ says that a set is connected exactly when it is an interval. Thus a set that omits a point between two of its members is disconnected. The Gap Criterion for Separation, established earlier in this course, gives a direct way to use such a missing point. In examples, however, it is useful to exhibit the separation itself: two disjoint, nonempty pieces that are relatively open in the set and together cover it.

This tutorial develops several ways to do that. An isolated point gives an especially simple separation. A gap between intervals gives another. We will also see that a set can be disconnected even when its two pieces are not separated by a positive distance, as happens with the rational numbers.

Recall: A separation of $E\subseteq\mathbb{R}$ is a representation $E=A\cup B$, where $A$ and $B$ are disjoint, nonempty, and relatively open in $E$. A set is disconnected if it has a separation. By the Clopen Criterion for Connectedness, a set is connected exactly when it has no nonempty proper subset that is both relatively open and relatively closed.

An Isolated Point Forces Disconnectedness

A point $p\in E$ is isolated in $E$ if some open interval around $p$ contains no other points of $E$. In that case, the singleton $\{p\}$ is relatively open in $E$. It is also relatively closed, because $\{p\}$ is closed in $\mathbb{R}$. If $E$ contains another point, $\{p\}$ is a nonempty proper subset of $E$, so the Clopen Criterion applies.

Theorem (An Isolated Point Gives a Separation): Let $E\subseteq\mathbb{R}$ contain an isolated point $p$ and at least one other point. Then $E$ is disconnected.

Proof. Since $p$ is isolated in $E$, there exists $\varepsilon>0$ such that $$ E\cap(p-\varepsilon,p+\varepsilon)=\{p\}. $$ In particular, $\{p\}=E\cap(p-\varepsilon,p+\varepsilon)$, so $\{p\}$ is relatively open in $E$. It is relatively closed as well: $\{p\}$ is closed in $\mathbb{R}$, and its intersection with $E$ is itself. Since $E$ contains a point other than $p$, the subset $\{p\}$ is nonempty and proper. The Clopen Criterion for Connectedness therefore implies that $E$ is disconnected. More explicitly, $A=\{p\}$ and $B=E\setminus\{p\}$ are disjoint, nonempty, relatively open sets whose union is $E$. Thus they form a separation. $\square$

Worked Example: A Sequence Set with an Isolated Point

Let $E=\{-3\}\cup\{4+1/n:n\geq1\}$. The point $-3$ is isolated: the interval $(-4,-2)$ contains $-3$, and every point $4+1/n$ is greater than $4$, so none of those points lies in $(-4,-2)$. The set also contains $5=4+1/1$, which is different from $-3$. The Isolated Point Gives a Separation Theorem now implies that $E$ is disconnected. A separation is $$ A=\{-3\},\qquad B=\{4+1/n:n\geq1\}. $$ Indeed, $A=E\cap(-4,-2)$ and $B=E\cap(3,\infty)$, so both pieces are relatively open in $E$. They are nonempty, disjoint, and cover $E$.

The isolated-point criterion is sufficient, but it is not necessary. Some disconnected sets have no isolated points at all. The rational numbers provide an important example, and their separation also illustrates why the pieces of a separation need not be a positive distance apart.

Disconnected Sets with Explicit Separations

For a simple finite example, choose a real number that lies strictly between the points of the set. The two sides of that number give relatively open pieces. More generally, the Gap Criterion for Separation says that a set is disconnected exactly when there are two of its points with a missing point strictly between them. The examples below make the pieces of the separation visible rather than relying only on that criterion.

Worked Example: A Two-Point Set

Consider $F=\{-2,5\}$. Define $A=F\cap(-\infty,0)$ and $B=F\cap(0,\infty)$. Since $-2<0<5$, these pieces are $A=\{-2\}$ and $B=\{5\}$. They are disjoint and nonempty, and their union is $F$. Each is relatively open in $F$, because it is the intersection of $F$ with an open set in $\mathbb{R}$. Thus $A$ and $B$ form a separation, and $F$ is disconnected.

Worked Example: The Integers

Let $\mathbb{Z}$ denote the set of integers, and split it at $1/2$: $$ A=\mathbb{Z}\cap(-\infty,1/2),\qquad B=\mathbb{Z}\cap(1/2,\infty). $$ No integer equals $1/2$, so $A\cup B=\mathbb{Z}$. Both pieces are nonempty: $0\in A$ and $1\in B$. They are disjoint and relatively open in $\mathbb{Z}$, since each is the intersection of $\mathbb{Z}$ with an open interval or open ray. This is a separation, so $\mathbb{Z}$ is disconnected. Notice that the cut need not itself belong to the set; here $1/2$ is a convenient point in the gap between $0$ and $1$.

Worked Example: The Rational Numbers

The rational numbers can be separated at the irrational number $\sqrt{2}$: $$ A=\mathbb{Q}\cap(-\infty,\sqrt{2}),\qquad B=\mathbb{Q}\cap(\sqrt{2},\infty). $$ Because $\sqrt{2}\notin\mathbb{Q}$, every rational number belongs to exactly one of $A$ and $B$. The sets are nonempty: $1\in A$ and $2\in B$, since $1<\sqrt{2}<2$. Both are relatively open in $\mathbb{Q}$, being intersections with open rays. Therefore they form a separation, and $\mathbb{Q}$ is disconnected.

This separation does not come from a positive-width interval containing no rational numbers. The rational numbers are dense in $\mathbb{R}$, so rational numbers occur arbitrarily close to $\sqrt{2}$ on both sides. The two pieces can have distance zero and still form a separation: the definition requires relative openness and a disjoint covering, not positive distance.

A Gap Between Closed Intervals

A particularly transparent family of disconnected sets is formed by two nonempty closed intervals with a positive gap between them. Their union fails to contain points in the gap, and cutting at any point strictly inside that gap separates the two intervals. The following result records this construction.

Theorem (Two Closed Intervals with a Gap): If $a\leq b<c\leq d$, then $E=[a,b]\cup[c,d]$ is disconnected.

Proof. Choose $m$ with $b<m<c$; for example, $m=(b+c)/2$. Put $$ A=E\cap(-\infty,m),\qquad B=E\cap(m,\infty). $$ Every point of $[a,b]$ is less than $m$, so $[a,b]\subseteq A$; every point of $[c,d]$ is greater than $m$, so $[c,d]\subseteq B$. Since $E$ is the union of those two intervals, this gives $A=[a,b]$ and $B=[c,d]$. In particular, both sets are nonempty. The inequalities $b<m<c$ show that they are disjoint, and their union is $E$. Each is relatively open in $E$, because $A=E\cap(-\infty,m)$ and $B=E\cap(m,\infty)$, with both rays open in $\mathbb{R}$. Hence $A,B$ form a separation of $E$, proving that $E$ is disconnected. $\square$

Worked Example: Two Closed Intervals

Take $E=[-5,-2]\cup[1,4]$. The point $m=-1/2$ lies strictly between $-2$ and $1$, since $-2<-1/2<1$. Define $$ A=E\cap(-\infty,-1/2)=[-5,-2],\qquad B=E\cap(-1/2,\infty)=[1,4]. $$ The equalities follow because every point of the first interval is at most $-2<-1/2$, while every point of the second is at least $1>-1/2$. The pieces are nonempty, disjoint, relatively open in $E$, and cover $E$. Thus this union of intervals is disconnected.

What the Examples Show

In each construction, it is useful to check the same points explicitly: the pieces are nonempty, disjoint, cover the set, and are relatively open. A missing intermediate point often suggests where to place the cut. An isolated point suggests the singleton separation. These methods give a concrete separation, rather than merely the conclusion that a set is not an interval.

Do not confuse disconnectedness with having a positive distance between two pieces. The rational-number example rules out that shortcut: its separating pieces approach the cut arbitrarily closely. Conversely, an isolated point is a powerful sufficient condition, but its absence does not imply connectedness. Connectedness is determined by whether a separation exists, not by whether the set has isolated points or whether its separated pieces are a positive distance apart.

Check Your Understanding

For each question, identify the separation or the specific condition that guarantees one.

Why does a set with an isolated point $p$ and another point have a nonempty proper subset that is both relatively open and relatively closed?
Give the two pieces of a separation of $\{-2,5\}$, and explain why each is relatively open.
Why does cutting $\mathbb{Z}$ at $1/2$ produce two nonempty pieces that cover all the integers?
Why does the rational-number separation at $\sqrt{2}$ not require the two pieces to have positive distance?
For $E=[-5,-2]\cup[1,4]$, why does $-1/2$ provide a valid cut?

What You'll Learn