A Practical Classification Strategy
For subsets of the real line, connectedness has a particularly concrete test: a connected set cannot have a gap between two of its points. This turns many abstract separation questions into order questions. Other useful approaches are to exhibit a subset that is both open and closed relative to the set, or to recognize the set as the image of a connected interval under a continuous map.
Throughout, \(\mathbb{R}\) has its standard topology, and every subset \(A\subseteq\mathbb{R}\) has the subspace topology unless another topology is specified. A set \(U\subseteq A\) is relatively open if \(U=A\cap O\) for some open \(O\subseteq\mathbb{R}\); relative closedness is defined in the same way using a closed subset of \(\mathbb{R}\). A separation of \(A\) is a pair of disjoint, nonempty, relatively open subsets of \(A\) whose union is \(A\). A space is connected if it has no separation.
The interval condition does not require that endpoints belong to the set. Thus open, closed, half-open, unbounded intervals, singletons, and the empty set all satisfy it. The central classification result is that, for subsets of \(\mathbb{R}\), being connected is exactly the same as being an interval.
Proof. Suppose first that \(A\) is connected. If \(A\) were not an interval, there would be \(x,y\in A\) and \(z\notin A\) with \(x<z<y\). Define \(U=A\cap(-\infty,z)\) and \(V=A\cap(z,\infty)\). The rays used here are open in \(\mathbb{R}\), so \(U\) and \(V\) are relatively open in \(A\). They are disjoint, and because \(z\notin A\), their union is \(A\). Also \(x\in U\) and \(y\in V\), so both are nonempty. They form a separation, contradicting connectedness. Therefore \(A\) is an interval.
Conversely, suppose \(A\) is an interval. If \(A\) is empty or has only one point, it cannot be separated into two nonempty subsets. In the remaining case, suppose for contradiction that \(A=U\cup V\) is a separation. Choose one point from each part; after interchanging the names of the parts if necessary, write \(u\in U\), \(v\in V\), with \(u<v\). Since \(A\) is an interval, \([u,v]\subseteq A\). The set \(U\cap[u,v]\) is nonempty, since it contains \(u\), and is bounded above by \(v\). Let \(s=\sup(U\cap[u,v])\). Because \(U\) is relatively open in \(A\), there is \(\delta>0\) such that \((s-\delta,s+\delta)\cap A\subseteq U\) if \(s\in U\). Because \(V\) is relatively open, the corresponding statement holds with \(V\) in place of \(U\) if \(s\in V\). As \(U\cup V=A\), \(s\) belongs to one of these two sets.
If \(s\in U\), then \(s\ne v\), since \(v\in V\) and the parts are disjoint. Thus \(s<v\). The interval property gives points of \(A\) strictly between \(s\) and \(v\), arbitrarily close to \(s\). Relative openness of \(U\) puts such points in \(U\), contradicting that \(s\) is an upper bound of \(U\cap[u,v]\). If \(s\in V\), then \(s\ne u\), since \(u\in U\); hence \(s>u\). By the definition of supremum, there is a point of \(U\cap[u,v]\) greater than \(s-\delta\), for every \(\delta>0\). Choose \(\delta\) small enough that \(s-\delta>u\). Such a point lies in \((s-\delta,s)\), but relative openness of \(V\) says that \((s-\delta,s+\delta)\cap A\subseteq V\). This contradicts \(U\cap V=\varnothing\). Both cases are impossible, so \(A\) is connected. \(\square\)
For a classification problem, the theorem gives a quick first test: look for two points of the set with a missing point between them. If you find them, the missing point gives an explicit separation. If there is no such gap, the set is an interval and hence connected.
Gaps and Relative Clopen Sets
Worked Example: A Missing Point Splits a Set
Let \(A=(-2,1)\cup(1,4]\subseteq\mathbb{R}\). Decide whether \(A\) is connected.
The points \(0\) and \(2\) belong to \(A\), because \(-2<0<1\) and \(1<2\leq4\). The point \(1\) lies strictly between them but is not in \(A\). Thus \(A\) is not an interval, so the classification theorem says it is disconnected.
Here is the corresponding separation. Set \(U=A\cap(-\infty,1)=(-2,1)\) and \(V=A\cap(1,\infty)=(1,4]\). The rays are open in \(\mathbb{R}\), so \(U,V\) are relatively open in \(A\). They are nonempty, disjoint, and their union is \(A\), since \(1\notin A\). In addition, \(A\setminus U=V\) and \(A\setminus V=U\); therefore each is also relatively closed. This verifies the separation directly.
A subset that is both relatively open and relatively closed is called clopen. The following test is useful when a convenient separation is already visible.
Proof. If \(X\) has a separation \(U,V\), then \(U\) is open and its complement \(V\) is open, so \(U\) is also closed. It is nonempty and proper. Conversely, if \(C\subseteq X\) is nonempty, proper, and clopen, then \(C\) and \(X\setminus C\) are disjoint, nonempty, open, and have union \(X\). They form a separation. Thus a space is disconnected exactly when it has a nonempty proper clopen subset. \(\square\)
In a subspace \(A\), “clopen” means open and closed relative to \(A\), not necessarily open and closed as a subset of \(\mathbb{R}\). This distinction matters: a set can fail to be open in the ambient real line and still be open in a subspace.
Worked Example: A Relative Separation of the Rationals
Let \(E=\mathbb{Q}\cap[0,1]\), with the subspace topology inherited from \(\mathbb{R}\). We show that \(E\) is disconnected by cutting at \(\alpha=\sqrt{2}/2\).
First, \(\alpha\) is strictly between \(0\) and \(1\): it is positive, and \(\alpha^2=1/2<1\). It is irrational, since if \(\sqrt{2}/2\) were rational, then \(\sqrt{2}\) would be rational. To recall why that is impossible, a fraction in lowest terms \(m/n=\sqrt{2}\) would give \(m^2=2n^2\). Hence \(m\) is even; writing \(m=2k\) gives \(n^2=2k^2\), so \(n\) is even as well, contrary to the fraction being in lowest terms.
Define \(E_-=E\cap(-\infty,\alpha)\) and \(E_+=E\cap(\alpha,\infty)\). These are relatively open in \(E\), since both rays are open in \(\mathbb{R}\). Neither is empty: \(0\in E_-\) because \(0<\alpha\), and \(1\in E_+\) because \(\alpha<1\). They are disjoint, and their union is \(E\), because no rational point of \(E\) equals the irrational number \(\alpha\). Thus they form a separation. In particular \(E_-\) is clopen relative to \(E\), with relative complement \(E_+\).
This example also illustrates the interval test: \(0,1\in E\), but the irrational number \(\alpha\) between them is not in \(E\). The explicit separation records that gap in the topology of the subspace.
Using Continuous Images
A different strategy is useful when a subset of \(\mathbb{R}\) is presented as the range of a continuous function. We use the standard result that a continuous image of a connected space is connected; its short proof explains exactly how to apply it.
Proof. Suppose \(f(X)\) had a separation \(U,V\). Continuity means that \(f^{-1}(U)\) and \(f^{-1}(V)\) are open in \(X\), because \(U,V\) are open in the subspace \(f(X)\). They are disjoint and their union is \(X\). They are nonempty because every point of \(f(X)\) has a preimage in \(X\), and both \(U\) and \(V\) are nonempty. Thus they would separate \(X\), contradicting its connectedness. Therefore \(f(X)\) is connected. \(\square\)
Worked Example: Finding the Exact Range of a Continuous Function
Let \(f:[0,2]\to\mathbb{R}\) be \(f(t)=(t-1)^2-1\). Find \(f([0,2])\) and establish that the range is connected.
The domain \([0,2]\) is an interval, so it is connected by the classification theorem. The function is a polynomial expression in \(t\), hence continuous. The continuous-image theorem therefore shows that \(f([0,2])\) is connected.
To identify the range, take any \(t\in[0,2]\). Then \(-1\leq t-1\leq1\), so \(0\leq(t-1)^2\leq1\). Subtracting \(1\) gives \(-1\leq f(t)\leq0\); hence \(f([0,2])\subseteq[-1,0]\). The endpoint and midpoint calculations are \(f(0)=(0-1)^2-1=1-1=0\), \(f(1)=(1-1)^2-1=0-1=-1\), and \(f(2)=(2-1)^2-1=1-1=0\). Thus both \(-1\) and \(0\) belong to the range.
Because the range is a connected subset of \(\mathbb{R}\), it is an interval. An interval containing \(-1\) and \(0\) contains every point between them. Combining this with the bounds gives \(f([0,2])=[-1,0]\).
Check: the formula gives \(f(1)=-1\), the smallest permitted value, and \(f(0)=f(2)=0\), the largest permitted value. The bounds show that no value outside \([-1,0]\) can occur, while connectedness and the attained endpoint values ensure that none inside is skipped.
Choosing the Efficient Argument
Find \(x,y\in A\) and \(z\notin A\) with \(x<z<y\). If they exist, use the two sides of \(z\) to construct a separation.
If the set is described by a cut or a natural partition, verify relative openness and relative closedness inside the set itself.
If \(A=f(I)\) for an interval \(I\) and a continuous \(f\), conclude that \(A\) is connected. To identify the range exactly, also establish its bounds and show its endpoint values are attained.
The first two methods prove disconnectedness by producing a separation; the continuous-image method proves connectedness by transferring it from a known connected domain. Keep the hypotheses visible: a continuous map from a disconnected domain need not have a connected image, and merely being a subset of a connected space does not make a set connected.
Common pitfall. For \(A=(0,1)\cup(1,2)\), a mistaken relative-complement calculation is \(A\setminus\bigl(A\cap(-\infty,1)\bigr)=[1,2)\). That calculation incorrectly includes \(1\), even though \(1\notin A\). The correct calculation is \(A\setminus\bigl(A\cap(-\infty,1)\bigr)=(1,2)\). Consequently, \(A\cap(-\infty,1)=(0,1)\) and \((1,2)\) are disjoint nonempty relatively open sets covering \(A\); they form a separation. Equivalently, \(0.5,1.5\in A\) and the intervening point \(1\notin A\), so \(A\) is not an interval. Relative set operations must be computed using membership in \(A\), including whether the cut point itself belongs to \(A\).
Check Your Understanding
For each question, state which test you use and give the key justification.
- Is \([0,1)\cup[2,3]\) connected? If not, exhibit a separation.
- Let \(A=\mathbb{Z}\subseteq\mathbb{R}\). Use the interval criterion to show that \(A\) is disconnected.
- Explain why a nonempty proper clopen subset of a space gives a separation.
- Suppose \(g:[-1,1]\to\mathbb{R}\) is continuous and \(g(-1)=-3\), \(g(1)=2\). What interval must be contained in \(g([-1,1])\), and which results justify your answer?
- Why does the fact that \(A\subseteq\mathbb{R}\) fail, by itself, to imply that \(A\) is connected?