Connected Pieces Do Not Guarantee a Connected Union
A useful theorem says that a union of connected sets is connected when all the sets share a common point. Without that condition, the conclusion can fail: connected pieces may sit on opposite sides of a separation of their union. But missing a common point does not force disconnection either. The key is not simply whether the pieces intersect; it is whether the union can be split into two nonempty relatively open parts.
Throughout, subsets of \(\mathbb{R}\) have the subspace topology inherited from the standard topology on \(\mathbb{R}\). A set is open in a subspace \(X\subseteq\mathbb{R}\) when it is the intersection of \(X\) with an open subset of \(\mathbb{R}\). Connectedness is always understood in the topology of the space being discussed.
In a separation, each part is also closed in \(X\): for example, \(X\setminus U=V\), and \(V\) is open. Thus a separation divides the space into two nonempty pieces that are both open and closed relative to that space. The relative qualification matters: a subset can fail to be open in the ambient real line but be open in the union under consideration.
Proof. The assumptions say that \(C\) and \(D\) are disjoint, nonempty, open subsets of \(X\), and that their union is \(X\). They therefore form a separation of \(X\) by definition. Hence \(X\) is disconnected. \(\square\)
This criterion is simple, but it identifies the work required in examples: we must check that the pieces cover the union, are disjoint, are nonempty, and are open in the union. Showing only that the pieces themselves are connected is not enough.
Separated Intervals
Worked Example: Two Closed Intervals with a Gap
Let \(A=[0,1]\), \(B=[2,3]\), and \(X=A\cup B\), all with the subspace topology from \(\mathbb{R}\). Each of \(A\) and \(B\) is connected, because every interval in \(\mathbb{R}\) is connected, as established in the classification of connected subsets of the real line.
We check that \(X\) is disconnected. The sets \(A\) and \(B\) are nonempty: \(0\in A\) and \(2\in B\). They are disjoint because every element of \(A\) is at most \(1\), while every element of \(B\) is at least \(2\). Their union is \(X\) by its definition.
To check relative openness, write \(A=X\cap(-1,3/2)\) and \(B=X\cap(3/2,4)\). Both intervals on the right are open in \(\mathbb{R}\), so \(A\) and \(B\) are open in \(X\). They form a separation of \(X\), and the proposition gives that \(X\) is disconnected.
Check: \(3/2\) lies strictly between \(1\) and \(2\), so \((-1,3/2)\) contains all of \([0,1]\) and none of \([2,3]\), while \((3/2,4)\) contains all of \([2,3]\) and none of \([0,1]\). The two relatively open sets therefore cover exactly \(X\), with no overlap.
The theorem about unions with a common point gives a useful sufficient condition: if every set in a family is connected and all share one point, their union is connected. The example above shows why some condition on how the pieces fit together is needed. It does not say that all unions without a common point are disconnected; a gap that produces a separation is the decisive feature here.
Worked Example: A Union of Singleton Sets
First, every singleton \(\{x\}\) is connected, regardless of the space containing it. In its subspace topology, the only subsets are \(\varnothing\) and \(\{x\}\). There cannot be two disjoint, nonempty subsets whose union is \(\{x\}\), so there is no separation.
Now give \(\mathbb{Q}\) the subspace topology inherited from \(\mathbb{R}\). It is the union of its singleton subsets: \(\mathbb{Q}=\bigcup_{q\in\mathbb{Q}}\{q\}\). Each set in this family is connected. Nevertheless, \(\mathbb{Q}\) is disconnected.
To verify this, let \(U=\{q\in\mathbb{Q}:q<\sqrt{2}\}\) and \(V=\{q\in\mathbb{Q}:q>\sqrt{2}\}\). The number \(\sqrt{2}\) is irrational, so every rational number is either less than or greater than \(\sqrt{2}\). Thus \(U\cup V=\mathbb{Q}\), and \(U\cap V=\varnothing\). Both sets are nonempty: \(1\in U\), since \(1<\sqrt{2}\), and \(2\in V\), since \(\sqrt{2}<2\).
Moreover, \(U=\mathbb{Q}\cap(-\infty,\sqrt{2})\) and \(V=\mathbb{Q}\cap(\sqrt{2},\infty)\). The two real intervals are open in \(\mathbb{R}\), so \(U\) and \(V\) are open in \(\mathbb{Q}\). They form a separation, proving that \(\mathbb{Q}\) is disconnected.
Check: The point \(\sqrt{2}\) is not a rational number, so it is not a missing point of \(\mathbb{Q}\); the two sets really do cover all of \(\mathbb{Q}\). Their nonemptiness and relative openness have also been checked, so this is a separation, not merely a partition into two sets.
This example emphasizes that arbitrary unions require particular care. Even if every member of a family is as simple as a singleton, the union can have a separation. In the rational-number example, the separation is induced by the irrational cut at \(\sqrt{2}\). The connectedness of the individual pieces provides no barrier to such a split.
Disjointness Alone Does Not Decide the Outcome
Worked Example: Disjoint Intervals with a Connected Union
Let \(C=[0,1)\), \(D=[1,2]\), and \(X=C\cup D\), with the subspace topology from \(\mathbb{R}\). The sets \(C\) and \(D\) are disjoint: every point of \(C\) is strictly less than \(1\), and every point of \(D\) is at least \(1\). Each is an interval, hence connected by the classification of connected subsets of \(\mathbb{R}\).
In this case the union is \([0,2]\). Indeed, \(C\) contains all points \(x\) with \(0\leq x<1\), and \(D\) contains all points \(x\) with \(1\leq x\leq2\); together these ranges contain exactly the points satisfying \(0\leq x\leq2\). Thus \(X\) is an interval and is connected.
Check: The endpoint \(1\) belongs to \(D\), since \(1\leq1\), and not to \(C\), since \(C\) uses the strict inequality \(x<1\). Hence no point is omitted at the join, and \(C\cup D=[0,2]\). This union is connected even though its two connected pieces are disjoint.
For a family of exactly two disjoint subsets that cover a union \(X\), both pieces are open in \(X\) precisely when each is also closed in \(X\): each piece is the complement of the other. The half-open intervals above do not give a separation. In particular, \(C\) is not closed in \(X=[0,2]\), since its complement \(D=[1,2]\) is not open in \(X\): every open neighborhood in \(X\) of \(1\) contains points less than \(1\). So disjoint pieces need not be relatively open in their union.
A Common Pitfall
Common pitfall. One incorrect argument says, “\(A\) and \(B\) are connected, so \(A\cup B\) is connected.” Take \(A=[0,1]\) and \(B=[2,3]\). Both are connected intervals, but the calculation \(A\cup B=[0,1]\cup[2,3]\) leaves the gap \((1,2)\). The relatively open sets \(A=X\cap(-1,3/2)\) and \(B=X\cap(3/2,4)\) separate \(X=A\cup B\), so the union is disconnected. The correction is to check a condition that actually links the pieces, such as the common-point condition, or to test directly for a separation.
The reverse overstatement is also wrong: “\(A\) and \(B\) are disjoint, so their union is disconnected.” The disjoint intervals \([0,1)\) and \([1,2]\) provide a counterexample; their union is the connected interval \([0,2]\). Disjointness concerns whether the pieces share points, while connectedness concerns whether the resulting space has a separation. Those are different questions.
When assessing a proposed union, use the following checks:
- Are the individual sets connected in their own subspace topologies?
- Do they cover the claimed union, including any endpoints?
- Can the union be divided into disjoint, nonempty sets open in the union?
- Does an available sufficient condition, such as a common point, apply?
The common-point theorem is sufficient, not necessary: the disjoint intervals in the last example have a connected union. Conversely, connectedness of every piece alone is not sufficient: the separated intervals and the singleton family show that their union can be disconnected.
Check Your Understanding
For each question, distinguish properties of the pieces from properties of their union.
- Give a separation of \([0,1]\cup[2,3]\) in its subspace topology from \(\mathbb{R}\), and verify relative openness.
- Why is each singleton connected? State which part of the definition of separation rules one out.
- In the rational-number example, why is it necessary to check that \(\sqrt{2}\notin\mathbb{Q}\) when verifying the union of the two sets?
- Explain why \([0,1)\) and \([1,2]\) are disjoint but their union is connected.
- Is the common-point condition necessary for a union of connected sets to be connected? Use an example to justify your answer.