Choosing a Useful Connectedness Test
When proving that a metric space is connected, it is usually more effective to start by assuming a separation exists and then derive a contradiction than to inspect every possible pair of subsets directly. Earlier in this course, connectedness was characterized using separated pairs: a space is disconnected exactly when it is the union of two nonempty separated subsets. This workshop develops two practical ways to use that idea. One converts a separation into a nontrivial clopen set; the other shows how a separation of a larger set would force a separation of a connected subset.
A set is clopen in a space if it is both open and closed in that space. Openness and closedness here are relative to the space under discussion. That qualification is essential: a subset may be open in a subspace without being open in the ambient metric space.
Proof. Suppose first that \(X\) is disconnected. By the Characterization of Separated Sets and the Theorem on Connectedness and Separated Pairs, there are nonempty disjoint separated subsets \(A,B\subseteq X\) with \(X=A\cup B\). By the definition of separated sets, \(A\cap\overline{B}=\varnothing\) and \(B\cap\overline{A}=\varnothing\). Since \(X\setminus A=B\), the first equality gives \(A\cap\overline{X\setminus A}=\varnothing\), which means \(A\) is open in \(X\). The second equality gives \(B\cap\overline{X\setminus B}=\varnothing\), so \(B\) is open as well. Because each is the complement of the other, each is also closed. Thus \(A\) is a nonempty proper clopen subset of \(X\).
Conversely, suppose \(U\subseteq X\) is clopen and satisfies \(\varnothing\ne U\ne X\). Then \(U\) and \(X\setminus U\) are disjoint, nonempty, and have union \(X\). Both are open in \(X\), since \(U\) is open and \(X\setminus U\) is open because \(U\) is closed. They form a separation of \(X\), so \(X\) is disconnected. The two implications prove the characterization. \(\square\)
Worked Applications of the Clopen Test
Worked Example: Detecting a Separation in a Subspace
Let \(X=[-3,-1]\cup[2,5]\), equipped with the metric inherited from \(\mathbb{R}\), and let \(U=[-3,-1]\). We show that \(U\) is clopen in \(X\), even though it is not clopen in all of \(\mathbb{R}\).
For every \(x\in U\), the ball in \(X\) centered at \(x\) with radius \(1\) meets no point of \([2,5]\): the distance from \(x\leq -1\) to any \(y\geq 2\) is at least \(3\). Thus \(B_1(x)\cap X\subseteq U\), so \(U\) is open in \(X\). Its complement in \(X\) is \([2,5]\). For every \(y\in[2,5]\), the ball \(B_1(y)\cap X\) contains no point of \([-3,-1]\), because the distance between the two intervals is \(3\). Hence \(X\setminus U\) is open in \(X\), and \(U\) is closed in \(X\) as well.
The set \(U\) is nonempty and proper, so the Clopen Characterization of Connectedness shows that \(X\) is disconnected. The proof depends on the gap between the pieces, not on whether \(U\) is open in the ambient real line.
Worked Example: Why Relative Topology Cannot Be Ignored
Consider \(X=[0,2]\) and \(U=[0,1]\). The set \(U\) is closed in \(X\), but it is not open in \(X\): every relative ball centered at \(1\) contains points of \((1,2]\). Thus \(U\) is not a nontrivial clopen certificate for disconnectedness.
In fact, \(X\) is an interval in \(\mathbb{R}\), and intervals are connected. The failure of \(U\) to be open in \(X\) is exactly what prevents \(U\) and \(X\setminus U\) from forming a separation. Merely writing a set and its complement as a partition is not enough; both pieces must be open relative to the space.
Locally Constant Maps as Separation Detectors
A function is locally constant if its value stays fixed on some neighborhood of each point. Such a function cannot change value on a connected space. If it did, the points assigned one fixed value and the points assigned other values would yield a nontrivial clopen subset.
Proof. Fix \(x\in X\) and set \(U=\{z\in X:f(z)=f(x)\}\). Local constancy means that for every \(z\in X\), there is an open neighborhood \(V_z\) of \(z\) in \(X\) on which \(f\) is constant. If \(z\in U\), then \(V_z\subseteq U\), so \(U\) is open in \(X\). If \(z\notin U\), then the value of \(f\) on \(V_z\) is \(f(z)\ne f(x)\), so \(V_z\subseteq X\setminus U\). Therefore \(X\setminus U\) is open, and \(U\) is clopen.
The set \(U\) contains \(x\), so it is nonempty. By the Clopen Characterization of Connectedness, \(U=X\). Thus \(f(z)=f(x)\) for every \(z\in X\), and \(f\) is constant. \(\square\)
Worked Example: A Locally Constant Map Reveals Disconnectedness
Let \(X=\{-2\}\cup[1,2]\), with the metric inherited from \(\mathbb{R}\), and define \(f:X\to\{0,1\}\) by $$ f(x)= \begin{cases} 0,&x=-2,\\ 1,&x\in[1,2]. \end{cases} $$ We verify that \(f\) is locally constant. At the point \(-2\), the relative ball \(B_1(-2)\cap X\) is just \(\{-2\}\), so \(f\) is constant on that neighborhood. For any \(y\in[1,2]\), take \(r=(y+2)/2\), which is positive and strictly less than \(y+2=|y-(-2)|\). Then \(B_r(y)\cap X\) excludes \(-2\), so it lies in \([1,2]\), where \(f\) is constantly \(1\).
The function takes two different values, so it is nonconstant. The Locally Constant Maps on Connected Spaces Theorem therefore implies that \(X\) is not connected. Equivalently, its two pieces \(\{-2\}\) and \([1,2]\) are separated by a gap. This example illustrates a useful proof strategy: instead of constructing a separation directly, one can sometimes define a nonconstant locally constant function and then invoke the theorem.
Connectedness Between a Set and Its Closure
The Closure of a Connected Set Theorem says that the closure of a connected set is connected. A useful extension is that connectedness also holds for every set lying between the connected set and its closure. This can prove connectedness even when the intermediate set is neither closed nor equal to the original set.
Proof. If \(C=\varnothing\), then \(\overline{C}=\varnothing\), so the inclusions force \(E=\varnothing\), which is connected by convention. Suppose now that \(C\ne\varnothing\). Assume for contradiction that \(E\) is disconnected. By the Theorem on Connectedness and Separated Pairs, there are nonempty separated subsets \(A,B\subseteq E\) such that \(E=A\cup B\). In particular, \(A\) and \(B\) are disjoint and relatively open in \(E\).
The sets \(C\cap A\) and \(C\cap B\) are disjoint, relatively open in \(C\), and have union \(C\). Since \(C\) is connected, one of them must be empty. As \(C\ne\varnothing\), this means either \(C\subseteq A\) or \(C\subseteq B\). Interchanging the names of \(A\) and \(B\) if necessary, suppose \(C\subseteq A\). Choose \(b\in B\), which is possible because \(B\ne\varnothing\). Since \(B\) is open relative to \(E\), there is an open set \(O\subseteq X\) with \(b\in O\) and \(O\cap E\subseteq B\). But \(b\in E\subseteq\overline{C}\), so \(O\cap C\ne\varnothing\). Any point in \(O\cap C\) belongs to \(A\), because \(C\subseteq A\), and belongs to \(B\), because \(C\subseteq E\) and \(O\cap E\subseteq B\). This contradicts \(A\cap B=\varnothing\). Hence \(E\) is connected. \(\square\)
Worked Example: Adding a Limit Point Without Losing Connectedness
Take \(C=(1,4)\subseteq\mathbb{R}\). It is an interval, so it is connected, and its closure in \(\mathbb{R}\) is \([1,4]\). The set \(E=[1,4)\) satisfies $$ (1,4)\subseteq[1,4)\subseteq[1,4]. $$ The Intermediate Sets Between a Connected Set and Its Closure Theorem therefore shows that \([1,4)\) is connected. The theorem applies equally to \(E=(1,4]\) and \(E=[1,4]\). In each case, the set may include either, both, or neither of the two boundary points, but it must remain between \(C\) and \(\overline C\).
The inclusions are doing real work. The theorem does not say that every set containing \(C\) is connected. For example, adjoining a point outside \([1,4]\) could create a disconnected set. Nor does it say an arbitrary subset of \(\overline C\) is connected; the intermediate set must contain all of \(C\).
Proof Choices and Common Pitfalls
The three techniques in this workshop address different proof situations. Use the clopen characterization when a candidate subset and its complement are easy to show relatively open. Use the locally constant map theorem when a problem naturally assigns labels or discrete values to points. Use the intermediate-set theorem when a connected set is dense in the larger set under consideration, meaning its closure contains that larger set.
In a separation proof, keep track of the ambient space for every openness claim. If \(E\subseteq X\), a set open in \(E\) has the form \(E\cap O\) for some open \(O\subseteq X\); it need not itself be open in \(X\). In the intermediate-set proof, that distinction is exactly what allows the point \(b\in B\) to supply an ambient neighborhood \(O\) whose intersection with \(E\) lies in \(B\). Since \(b\) lies in the closure of \(C\), this neighborhood must also meet \(C\), producing the contradiction.
A second common mistake is to assume that a union of connected sets is connected without checking how the sets overlap. The product connectedness proof used the Union with a Common Point Theorem precisely because the slices shared points. In the intermediate-set argument, the closure condition plays a different role: it prevents one side of a proposed separation from being isolated from the connected core.
Check Your Understanding
Use the clopen, locally constant, and intermediate-set tests to answer the following questions.
- Why does a nonempty proper clopen subset of a metric space give a separation?
- In the proof for a locally constant map, why is the complement of a fiber open?
- Where is the inclusion \(E\subseteq\overline{C}\) used in the intermediate-set theorem?
- Why must openness in a separation of \(E\) be interpreted relative to \(E\)?
- If \(C\) is connected and dense in \(E\), which inclusions allow the intermediate-set theorem to be applied?