How Connected Sets Behave Under Basic Operations
The earlier characterization of connected subsets of \(\mathbb{R}\) gives a useful way to recognize them: a subset of the real line is connected exactly when it is an interval. In particular, a connected set cannot have a missing point strictly between two of its points. But connected sets also arise as pieces of larger sets, and it is useful to know when combining or enlarging those pieces preserves connectedness.
The key issue for unions is whether the sets meet. Two intervals that overlap, even at a single endpoint, combine without leaving a gap. By contrast, the union of two separated intervals can have a missing point between its elements. We will prove a general union result directly from the definition of a separation, then establish that taking the closure of a connected set preserves connectedness.
Unions of Pairwise-Intersecting Connected Sets
A common point shared by every set is one sufficient condition for a connected union. In fact, the sets need not all pass through the same point. It is enough that every pair of sets in the family intersects. The reason is that a separation of the union would force each connected set to lie entirely on one side, and intersections prevent different sets from lying on opposite sides.
Proof. Let \(S=\bigcup_{\lambda\in\Lambda}C_\lambda\). Suppose, for a contradiction, that \(S=P\cup Q\) is a separation. Thus \(P\) and \(Q\) are nonempty, disjoint, relatively open in \(S\), and their union is \(S\). By the earlier Theorem Connected Sets Lie on One Side of a Separation, each connected set \(C_\lambda\subseteq P\cup Q\) must be contained entirely in \(P\) or entirely in \(Q\).
Choose an index \(\lambda_0\in\Lambda\). Since \(C_{\lambda_0}\) is nonempty and lies on one side, exchange the names \(P,Q\) if necessary so that \(C_{\lambda_0}\subseteq P\). Now take any \(\lambda\in\Lambda\). By hypothesis, choose \(z\in C_\lambda\cap C_{\lambda_0}\). Since \(z\in C_{\lambda_0}\subseteq P\), the set \(C_\lambda\) meets \(P\). It cannot be contained in \(Q\), because \(P\cap Q=\varnothing\). As \(C_\lambda\) lies entirely in one side, it follows that \(C_\lambda\subseteq P\). This holds for every \(\lambda\), so \(S\subseteq P\). That contradicts the requirement that \(Q\) be nonempty. Therefore \(S\) has no separation and is connected. \(\square\)
Two connected sets with nonempty intersection are a particularly useful case. The theorem also applies to a family with a common point, since any two sets in the family then intersect. The proof shows why these hypotheses matter: intersections link the sets together and prevent a separation from assigning one set to each side.
Worked Example: Two Intervals Meeting at an Endpoint
Let \(C=[-3,1]\) and \(D=[1,5]\). Each is an interval, so each is connected by the earlier theorem that intervals are connected. They meet at \(1\), and $$ C\cup D=[-3,5]. $$ The union theorem therefore implies that \(C\cup D\) is connected. The endpoint \(1\) belongs to both intervals, so there is no omitted point between a point of \(C\) and a point of \(D\). In this example the union is also visibly an interval, which provides a direct check of the conclusion.
Worked Example: A Family of Intervals Whose Union Is the Real Line
For each positive integer \(n\), let \(I_n=(-n,n+1)\). Every \(I_n\) is an interval and therefore connected. Also \(0\in I_n\) for every \(n\), since \(-n<0<n+1\). Thus the intervals are pairwise-intersecting. Their union is all of \(\mathbb{R}\): given \(x\in\mathbb{R}\), choose a positive integer \(n>|x|+1\). Then \(-n<x<n+1\), so \(x\in I_n\). The union theorem yields that \(\bigcup_{n=1}^{\infty}I_n=\mathbb{R}\) is connected.
Why the Intersection Hypothesis Cannot Be Dropped
It is not enough for each set in a union to be connected. If the sets do not intersect, their union may have a gap. The union theorem gives a sufficient condition for connectedness, not a claim that every union of connected sets is connected.
Worked Example: Two Connected Sets with a Disconnected Union
Let \(C=\{-2\}\) and \(D=\{3\}\). Each singleton is an interval, hence connected. They are disjoint, and their union \(S=\{-2,3\}\) omits \(0\), which lies strictly between its two points. By the Gap Criterion for Separation, \(S\) is disconnected. Explicitly, its pieces $$ A=S\cap(-\infty,0)=\{-2\}, \qquad B=S\cap(0,\infty)=\{3\} $$ are nonempty, disjoint, relatively open in \(S\), and cover \(S\). This example shows that connectedness of the individual sets alone does not guarantee connectedness of their union.
There is a useful contrast here. In the previous example, every interval contained \(0\), so the sets were linked by a common point. In this example, there is no intersection to stop the two pieces from lying on opposite sides of a separation. When applying the union theorem, check the intersection condition rather than relying only on the connectedness of each set.
The Closure of a Connected Set
Closing a set adds any points that can be approached arbitrarily closely from within it. On the real line, this operation does not destroy connectedness. The proof uses the interval characterization: if two points belong to the closure, every point between them must also belong to the closure.
Proof. If \(E=\varnothing\), then \(\overline{E}=\varnothing\), which is connected because it cannot be written as a union of two nonempty sets. Now suppose \(E\ne\varnothing\). By the Characterization of Connected Subsets of \(\mathbb{R}\), \(E\) is an interval.
We show that \(\overline{E}\) is also an interval. Take \(x,z\in\overline{E}\) with \(x<z\), and let \(y\) be any real number with \(x<y<z\). Because \(x\in\overline{E}\), every open interval about \(x\) meets \(E\). In particular, choose \(u\in E\) such that \(|u-x|<y-x\). This inequality implies \(u<y\). Likewise, because \(z\in\overline{E}\), choose \(v\in E\) such that \(|v-z|<z-y\); this implies \(y<v\). Therefore \(u<y<v\). Since \(E\) is an interval and \(u,v\in E\), it follows that \(y\in E\), and hence \(y\in\overline{E}\). We have proved that every point between any two points of \(\overline{E}\) belongs to \(\overline{E}\), so \(\overline{E}\) is an interval. The same characterization now implies that \(\overline{E}\) is connected. \(\square\)
Worked Example: Closing an Open Interval
Take \(E=(0,2)\). This is an interval, so it is connected. Its closure is \(\overline{E}=[0,2]\): every point strictly between \(0\) and \(2\) is already in \(E\), and every neighborhood of either endpoint contains points of \(E\); points outside \([0,2]\) have a neighborhood disjoint from \(E\). The Closure of a Connected Set Theorem guarantees that \([0,2]\) is connected. In this case, the result can also be checked directly because \([0,2]\) is an interval.
What the Closure Result Does—and Does Not—Say
The closure theorem gives a one-way implication: if \(E\) is connected, then \(\overline{E}\) is connected. It does not say that a set is connected whenever its closure is connected. For instance, let \(F=[0,1]\setminus\{1/2\}\). Its closure is \([0,1]\), which is connected, but \(F\) is disconnected: it has points on both sides of the missing intermediate point \(1/2\). Thus adding a missing limit point can join pieces in the closure even though the original set had a separation.
This is a general lesson about implications in analysis. A theorem that says an operation preserves a property need not say that the property can be recovered from the result of that operation. When using the closure theorem, start with a connected set and conclude that its closure is connected; do not reverse that reasoning without a separate argument.
The two results in this tutorial address different ways of building sets. A union of connected sets remains connected when their intersections link the family together. Taking the closure of one connected set also preserves connectedness, because the closure retains the between-points property of an interval. In each case, the interval characterization provides a useful guide, while the separation argument explains how connected pieces behave when they sit inside a larger set.
Check Your Understanding
Use the union and closure results to answer the following questions.
- Why does a nonempty common intersection imply the pairwise-intersection hypothesis for a family of connected sets?
- In the union theorem’s proof, why can no connected set intersect one side of a separation and be contained entirely in the other side?
- Give an example of two connected subsets of \(\mathbb{R}\) whose union is disconnected, and identify a missing point between points of the union.
- In the closure theorem’s proof, how do the choices of \(u\) and \(v\) ensure that \(u<y<v\)?
- Why does the fact that \(\overline{F}\) is connected not imply that \(F\) is connected for \(F=[0,1]\setminus\{1/2\}\)?