Recognizing Connected Sets in the Real Line
The Characterization of Connected Subsets of \(\mathbb{R}\), established earlier in this course, turns the question “Is this set connected?” into a question about its shape: a subset of the real line is connected exactly when it is an interval. This gives many examples immediately. Open and closed intervals are connected, as are half-open intervals and rays. A singleton is connected as well.
There are also useful ways to build or recognize connected sets beyond listing intervals. The continuous image of a connected set is connected, and intersections of connected sets can be analyzed by checking whether the points between any two members remain in every set. We will use these ideas to develop examples and two further consequences of the interval characterization.
Intervals, Rays, and Singletons
The endpoint brackets do not determine whether an interval is connected. What matters is that the set contains every point between any two of its points. For example, \((1,4]\) contains every point strictly between any two of its members, even though it includes only one endpoint. The same reasoning applies to \([1,4)\), \((1,4)\), and \([1,4]\). By the interval characterization, all four sets are connected.
Worked Example: A Half-Open Interval and Two Rays
Consider \(J=(-2,5]\), \(L=(-\infty,3)\), and \(R=[7,\infty)\). Each is an interval: if two points belong to any one of these sets, every real number between them belongs to that set as well. For example, if \(x,z\in(-2,5]\) and \(x<y<z\), then \(x>-2\) implies \(y>-2\), and \(z\leq5\) implies \(y<5\). Thus \(y\in(-2,5]\). The same between-points check applies to each ray. All three sets are therefore connected. The endpoint \(5\) is included in \(J\), whereas the endpoint \(3\) is excluded from \(L\); neither choice changes the conclusion.
A singleton \(\{c\}\) is an interval because it has no two distinct points with a point strictly between them. Thus it is connected. This is a useful reminder that connected sets need not contain a whole range of different values: a one-point set is connected, while a set with several points may or may not be.
Worked Example: Checking a Singleton Directly
Let \(E=\{6\}\). There cannot be a separation \(E=A\cup B\) into two nonempty disjoint subsets: any nonempty subset of \(E\) must contain \(6\), so two such subsets would not be disjoint. Hence \(E\) is connected. Equivalently, it is an interval, since there are no \(x,z\in E\) with \(x<z\) for which the between-points condition could fail.
Intersections of Connected Sets
Unions and intersections behave differently. Earlier, the Union of Pairwise-Intersecting Connected Sets Theorem showed how intersections among the sets can keep a union connected. For intersections, the interval characterization gives a direct criterion: if a point lies between two points in every set of a family, it lies in their intersection too. Nonemptiness of the intersection is not needed for the conclusion: an empty intersection is connected under our convention, but there are no points to analyze in that case.
Proof. If \(\bigcap_{\lambda\in\Lambda}C_\lambda=\varnothing\), the conclusion follows because the empty set is connected. Otherwise, let \(x,z\in\bigcap_{\lambda\in\Lambda}C_\lambda\) with \(x<z\), and choose any \(y\in\mathbb{R}\) such that \(x<y<z\). For every \(\lambda\in\Lambda\), both \(x\) and \(z\) belong to \(C_\lambda\). Since \(C_\lambda\) is connected, the Characterization of Connected Subsets of \(\mathbb{R}\) implies that \(C_\lambda\) is an interval. Therefore \(y\in C_\lambda\) for every \(\lambda\in\Lambda\), so \(y\in\bigcap_{\lambda\in\Lambda}C_\lambda\). We have shown that the intersection contains every point between any two of its points. It is an interval and hence connected. \(\square\)
The theorem concerns the intersection of sets that are each connected in \(\mathbb{R}\), not arbitrary subsets of a connected set. It also allows infinitely many sets: the proof checks membership in each member of the family and does not depend on there being only finitely many.
Worked Example: Intersecting Intervals with Different Endpoints
Take \(C_1=[-2,4)\), \(C_2=(0,6]\), and \(C_3=[1,3]\). Each is an interval and therefore connected. The first two intersect in \((0,4)\), and intersecting with \(C_3\) gives $$ C_1\cap C_2\cap C_3=[1,3]. $$ To verify the endpoints, \(1\) and \(3\) belong to all three sets, while every number strictly between them does too. Numbers outside \([1,3]\) fail to belong to \(C_3\). The result is an interval, so it is connected, in agreement with the theorem.
Worked Example: An Infinite Intersection
For each positive integer \(n\), define \(I_n=[-1/n,1+1/n]\). Every \(I_n\) is a connected interval. We claim that $$ \bigcap_{n=1}^{\infty} I_n=[0,1]. $$ Every \(x\in[0,1]\) belongs to each \(I_n\), since \(-1/n\leq0\leq x\leq1\leq1+1/n\). If \(x<0\), choose a positive integer \(n>1/(-x)\). Then \(1/n<-x\), so \(-1/n>x\), and \(x\notin I_n\). If \(x>1\), choose \(n>1/(x-1)\). Then \(1/n<x-1\), so \(1+1/n<x\), and again \(x\notin I_n\). This proves the claimed equality. The intersection is the connected interval \([0,1]\).
Finite Connected Sets
The interval characterization has a strong consequence for finite sets. A finite set cannot contain two distinct points and also contain every point between them. In fact, any interval with two distinct points contains infinitely many points. This observation completely classifies finite connected subsets of the real line.
Proof. The empty set and every singleton are connected, as noted above. For the other direction, suppose \(E\subseteq\mathbb{R}\) is finite and connected. If \(E\) contained two distinct points, there would be \(x,z\in E\) with \(x<z\). Since \(E\) is connected, it is an interval by the Characterization of Connected Subsets of \(\mathbb{R}\). Thus every point strictly between \(x\) and \(z\) would belong to \(E\).
For each positive integer \(n\), define \(t_n=x+(z-x)/(n+1)\). Because \(z-x>0\) and \(0<1/(n+1)<1\), we have \(x<t_n<z\), so \(t_n\in E\). If \(m\ne n\), then \(1/(m+1)\ne1/(n+1)\); multiplying by \(z-x>0\) and adding \(x\) gives \(t_m\ne t_n\). Hence \(E\) contains infinitely many distinct points, contradicting that \(E\) is finite. Therefore \(E\) has at most one point and is either empty or a singleton. \(\square\)
Worked Example: A Two-Point Set Is Not Connected
Let \(F=\{2,7\}\). If \(F\) were connected, the interval characterization would require every point between \(2\) and \(7\) to lie in \(F\). But \(4\) satisfies \(2<4<7\), and \(4\notin F\). Thus \(F\) is not connected. More generally, the Finite Connected Subsets Theorem shows that no finite set with two or more points is connected; a finite connected set must be empty or a singleton.
Connected Sets as Continuous Images
Not every connected set is most naturally presented as an interval. Sometimes it appears as the set of values of a function. The Continuous Images of Connected Sets Theorem, proved earlier in this course, says that if \(E\) is connected and \(f:E\to\mathbb{R}\) is continuous, then \(f(E)\) is connected. Since connected subsets of the real line are intervals, this theorem can identify the shape of a function's range without writing every value down in advance.
Worked Example: The Range of a Continuous Function
Let \(f(x)=x^2\) on the interval \([-2,1]\). The function is continuous, so the Continuous Images of Connected Sets Theorem implies that \(f([-2,1])\) is connected and therefore an interval. For \(x\in[-2,1]\), we have \(-2\leq x\leq1\), so \(0\leq x^2\leq4\). The value \(0\) is attained at \(x=0\), and the value \(4\) is attained at \(x=-2\). Thus the range contains both endpoints \(0\) and \(4\), and it lies within \([0,4]\). Because it is an interval, it contains every number between \(0\) and \(4\). Consequently, $$ f([-2,1])=[0,4]. $$ This gives a connected set described as a function's range, rather than initially as an interval.
Using These Examples Carefully
The interval characterization is especially efficient when a set is already given by inequalities. Check whether every point between two members is also in the set; if so, the set is connected. For a function range, check instead whether the domain is connected and the function is continuous, then use the continuous-image theorem. For an intersection, verify that each set is connected and apply the intersection result.
Keep the hypotheses in view. The intersection result does not assert that an arbitrary intersection of sets is connected; it uses the fact that each set is connected, and hence an interval. Nor does connectedness of each set in a union by itself guarantee a connected union. The pairwise-intersection condition in the earlier union theorem addresses that different operation. Finally, the finite-set theorem does not say that every set with more than one point is disconnected: an interval such as \([2,7]\) has infinitely many points and is connected. It says that a finite connected set cannot have two distinct points.
Check Your Understanding
Use the interval characterization and the results proved here to answer the following questions.
- Why are both \((3,8]\) and \([3,8)\) connected, even though they include different endpoints?
- What feature of the proof shows that an intersection of infinitely many connected subsets of \(\mathbb{R}\) is still connected?
- Why must the empty-intersection case be considered separately in the proof of the intersection theorem?
- In the finite connected-set theorem, why are the points \(t_n=x+(z-x)/(n+1)\) all distinct and between \(x\) and \(z\)?
- For \(f(x)=x^2\) on \([-2,1]\), which points of the domain show that the range contains both \(0\) and \(4\)?