Connectedness: Can a Set Be Split Apart?
Compactness describes a set through finite subcovers, convergent subsequences, and boundedness and closedness in the real line. Connectedness asks a different question: can a set be divided into two genuinely separate pieces? On the real line, the answer is closely tied to a familiar geometric feature. A set is connected exactly when it contains every point between any two of its points.
To make “divided into separate pieces” precise, we use relative openness. A subset \(A\) of \(E\subseteq\mathbb{R}\) is relatively open in \(E\) if there is an open set \(U\subseteq\mathbb{R}\) such that \(A=E\cap U\). Thus, a set can be open within \(E\) even when it is not open in all of \(\mathbb{R}\). For example, a singleton may be relatively open in a finite set.
The empty set is connected under this definition: it cannot be partitioned into two nonempty sets. A singleton is connected as well. The definition focuses not on the number of points in a set, but on whether the set can be partitioned into two nonempty relatively open pieces.
A Useful Equivalent Test
A subset of \(E\) is relatively closed in \(E\) when its complement in \(E\) is relatively open. A set that is both relatively open and relatively closed is sometimes called clopen. The next criterion reformulates connectedness in terms of such subsets.
Proof. Suppose first that \(E\) is disconnected. By definition, there is a separation \(E=A\cup B\). The sets \(A\) and \(B\) are nonempty and relatively open in \(E\), and \(B=E\setminus A\). Since \(B\) is relatively open, \(A\) is relatively closed. Thus \(A\) is a nonempty proper subset of \(E\) that is both relatively open and relatively closed.
Conversely, suppose \(C\) is a nonempty proper subset of \(E\) that is both relatively open and relatively closed in \(E\). Since \(C\) is relatively closed, its complement \(E\setminus C\) is relatively open. The sets \(C\) and \(E\setminus C\) are disjoint, nonempty, relatively open in \(E\), and their union is \(E\). They form a separation, so \(E\) is disconnected. This proves both directions. \(\square\)
The criterion is useful when a natural subset of \(E\) and its complement are easy to describe. It also highlights why ordinary openness in \(\mathbb{R}\) is not the right requirement: the pieces need only be open relative to the set being tested.
Intervals Are Connected
An interval \(J\subseteq\mathbb{R}\) has the interval property: whenever \(x,z\in J\) and \(x<y<z\), the point \(y\) also belongs to \(J\). This property prevents a separation. If two points of \(J\) were put into different pieces, the points between them would have to lie in one piece or the other, and relative openness would make a boundary between those pieces impossible.
Proof. Suppose, to obtain a contradiction, that \(J=A\cup B\) is a separation. Choose \(a\in A\) and \(b\in B\). If necessary, interchange the names of \(A\) and \(B\), so that \(a<b\). Because \(J\) is an interval, \([a,b]\subseteq J\). Consider \[ S=A\cap[a,b]. \] The set \(S\) is nonempty because \(a\in S\), and it is bounded above by \(b\). Let \(c=\sup S\). Since \(a\leq c\leq b\) and \([a,b]\subseteq J\), we have \(c\in J\). Since \(A\) and \(B\) partition \(J\), either \(c\in A\) or \(c\in B\).
First, \(c<b\). The set \(B\) is relatively open in \(J\), and \(b\in B\), so there is an \(\varepsilon>0\) such that every point of \(J\) within \(\varepsilon\) of \(b\) belongs to \(B\). Choose \(\eta>0\) with \(\eta<\varepsilon\) and \(\eta<b-a\). Every point of \(J\cap(b-\eta,b]\) then belongs to \(B\), so no point of \(S\) lies in \((b-\eta,b]\). Hence \(c\leq b-\eta<b\).
If \(c\in A\), relative openness of \(A\) gives a neighborhood of \(c\) whose points in \(J\) all belong to \(A\). Since \(c<b\), that neighborhood contains a point \(d\) with \(c<d<b\). The interval property gives \(d\in J\), so \(d\in A\cap[a,b]=S\). This contradicts that \(c\) is an upper bound for \(S\).
If \(c\in B\), relative openness of \(B\) gives a neighborhood of \(c\) whose points in \(J\) all belong to \(B\). But \(c=\sup S\), and \(c\notin S\) because \(S\subseteq A\) and \(A\cap B=\varnothing\). Therefore, for every \(\delta>0\), there is an \(s\in S\) with \(c-\delta<s\leq c\); otherwise \(c-\delta\) would be an upper bound for \(S\), smaller than \(c\). Taking \(\delta\) smaller than the neighborhood radius produces an \(s\in S\subseteq A\) that lies in the neighborhood of \(c\) contained in \(B\), a contradiction. Both cases are impossible, so \(J\) has no separation. Therefore \(J\) is connected. \(\square\)
Worked Example: A Closed Interval
Consider \(J=[-2,5]\). If \(x,z\in J\) and \(x<y<z\), then \(-2\leq x<y<z\leq5\), so \(-2<y<5\) and \(y\in J\). Thus \(J\) is an interval, and the theorem shows it is connected. In particular, there cannot be a partition of \([-2,5]\) into two disjoint, nonempty relatively open sets.
The theorem applies without requiring the interval to be closed or bounded. For example, \((1,4)\) and \([3,\infty)\) are intervals too, so each is connected. The endpoints included in an interval do not determine connectedness; the interval property does.
Connected Subsets of the Real Line Are Intervals
The converse also holds in \(\mathbb{R}\). If a set omits a point lying between two of its points, that missing point provides a cut: the set can be divided into the points to its left and the points to its right. These two pieces are relatively open because they are obtained by intersecting the set with open rays.
Proof. If \(E\) is an interval, it is connected by the Intervals Are Connected Theorem.
For the converse, suppose \(E\) is connected. If \(E\) is empty or contains at most one point, it is an interval. Otherwise, take any \(a,b\in E\) with \(a<b\), and let \(c\) satisfy \(a<c<b\). We show that \(c\in E\). If \(c\notin E\), define \[ A=E\cap(-\infty,c) \quad\text{and}\quad B=E\cap(c,\infty). \] Both sets are nonempty: \(a\in A\) and \(b\in B\). They are disjoint, and because \(c\notin E\), their union is \(E\). The rays \((-\infty,c)\) and \((c,\infty)\) are open in \(\mathbb{R}\), so \(A\) and \(B\) are relatively open in \(E\). They form a separation of \(E\), contradicting connectedness. Thus every point strictly between any two points of \(E\) also belongs to \(E\), which is the interval property. \(\square\)
Worked Example: A Set with a Gap
Let \(E=\{0\}\cup[2,4]\). The points \(0\) and \(2\) belong to \(E\), but \(1\) lies between them and does not belong to \(E\). The interval characterization already shows that \(E\) is not connected. We can also exhibit a separation directly: \[ A=E\cap(-\infty,1)=\{0\}, \qquad B=E\cap(1,\infty)=[2,4]. \] The open rays used in these intersections show that \(A\) and \(B\) are relatively open in \(E\). They are disjoint and nonempty, and \(A\cup B=E\). Hence \(E\) is disconnected.
Worked Example: The Rational Numbers
The set \(\mathbb{Q}\) is not an interval: \(1,2\in\mathbb{Q}\), but \(\sqrt{2}\) lies between them and is irrational, so \(\sqrt{2}\notin\mathbb{Q}\). Therefore the characterization shows that \(\mathbb{Q}\) is disconnected.
For a direct separation, take \[ A=\mathbb{Q}\cap(-\infty,\sqrt{2}) \quad\text{and}\quad B=\mathbb{Q}\cap(\sqrt{2},\infty). \] The Irrationality of the Square Root of Two, established earlier in this course, ensures that \(\sqrt{2}\notin\mathbb{Q}\), so \(A\cup B=\mathbb{Q}\). Both sets are nonempty, since \(1\in A\) and \(2\in B\). They are disjoint and relatively open in \(\mathbb{Q}\), as each is the intersection of \(\mathbb{Q}\) with an open ray. Thus they form a separation.
Why Connectedness Is Useful
The characterization makes connectedness particularly concrete in the real line: to show a set is connected, verify that it is an interval; to show it is disconnected, it is enough to find a missing point between two of its points. This is also a useful warning. A set can be large, dense, or unbounded and still be disconnected. The rational numbers, for instance, are dense in \(\mathbb{R}\), but they have gaps in the interval-property sense because they omit irrational numbers between rational numbers.
Connectedness also behaves well under continuous maps. A continuous function cannot turn a connected set into two pieces that can be separated by relatively open sets. The following theorem makes that statement precise.
Proof. Suppose instead that \(f(E)\) is disconnected, with a separation \(C,D\). Then \(f(E)=C\cup D\), and \(C,D\) are disjoint, nonempty, relatively open in \(f(E)\). Define \[ A=f^{-1}(C)=\{x\in E:f(x)\in C\}, \qquad B=f^{-1}(D)=\{x\in E:f(x)\in D\}. \] Since \(C\) and \(D\) are nonempty subsets of \(f(E)\), each contains an output \(f(x)\) for some \(x\in E\). Thus \(A\) and \(B\) are nonempty. They are disjoint, and every \(x\in E\) has \(f(x)\in C\cup D\), so \(A\cup B=E\).
It remains to check relative openness. Since \(C\) is relatively open in \(f(E)\), there is an open set \(U\subseteq\mathbb{R}\) such that \(C=f(E)\cap U\). Therefore \[ A=f^{-1}(C)=\{x\in E:f(x)\in U\}=f^{-1}(U). \] Continuity of \(f\) means that \(f^{-1}(U)\) is relatively open in \(E\). The same argument shows that \(B\) is relatively open in \(E\). Hence \(A,B\) form a separation of \(E\), contradicting its connectedness. Thus \(f(E)\) is connected. \(\square\)
Worked Example: A Continuous Image of an Interval
Let \(f:[-1,2]\to\mathbb{R}\) be given by \(f(x)=x^2\). The function is continuous, and \([-1,2]\) is an interval, hence connected. The continuous-image theorem shows that \(f([-1,2])\) is connected. Directly, its values range from \(0\) to \(4\): every square is nonnegative, \(f(0)=0\), \(f(2)=4\), and for every \(y\in[0,4]\), the point \(x=\sqrt{y}\) belongs to \([0,2]\subseteq[-1,2]\) and satisfies \(f(x)=y\). Thus \[ f([-1,2])=[0,4], \] which is indeed an interval.
A common pitfall is to confuse connectedness with being an interval of a particular endpoint type. Open, closed, half-open, unbounded, and singleton intervals are all connected. Another is to check only that a set is nonempty or dense; neither property rules out a separation. In the real line, the decisive test is whether every point between two points of the set also belongs to it.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What conditions must two sets satisfy to form a separation of \(E\)?
- Why does a nonempty proper subset of \(E\) that is both relatively open and relatively closed imply that \(E\) is disconnected?
- In the proof that intervals are connected, why is the supremum \(c\) guaranteed to lie in \(J\)?
- How does a point missing between two points of \(E\) produce a separation?
- Where does continuity enter the proof that a continuous image of a connected set is connected?