Why Nestedness Preserves Connectedness
A union of connected sets need not be connected: separate intervals, for example, can leave a gap between them. But when the sets are ordered by inclusion, each new member contains or is contained in every other member. This restriction prevents a separation of the union from dividing different members onto opposite sides. The result applies to increasing sequences, but its natural form allows any chain of subspaces, including families with no first or last member.
Let \(X\) be a topological space, and give each subset \(C\subseteq X\) the subspace topology. A separation of a space \(S\) is a pair of disjoint, nonempty, relatively open subsets whose union is \(S\). A space is connected if it has no separation. We use the usual separation definition, under which the empty space is connected.
The chain condition compares every pair of members, not just consecutive ones. An increasing sequence \(C_1\subseteq C_2\subseteq\cdots\) is a chain, but a chain may be indexed in other ways and need not have a first member. The theorem below uses only pairwise comparability by inclusion.
The Increasing-Union Theorem
Proof. Suppose, for a contradiction, that \(S\) has a separation \(S=P\cup Q\), where \(P\) and \(Q\) are disjoint, nonempty, relatively open subsets of \(S\). Since the \(C_\alpha\) cover \(S\), choose \(p\in P\) and \(q\in Q\), and choose indices \(\alpha,\beta\in J\) such that \(p\in C_\alpha\) and \(q\in C_\beta\). The chain condition gives either \(C_\alpha\subseteq C_\beta\) or \(C_\beta\subseteq C_\alpha\). In the first case, \(C_\beta\) contains \(p\), because \(p\in C_\alpha\), and also contains \(q\). In the second case, \(C_\alpha\) contains both \(p\) and \(q\). Thus some member \(C_\gamma\) meets both \(P\) and \(Q\). But \(P\cap C_\gamma\) and \(Q\cap C_\gamma\) are disjoint relatively open subsets of \(C_\gamma\) and cover \(C_\gamma\). Since each is nonempty, they separate \(C_\gamma\), contrary to its connectedness. Therefore \(S\) has no separation and is connected. \(\square\)
The relative openness used in the proof follows directly from the subspace topology: if \(P=S\cap O\) for an open set \(O\subseteq X\), then \(P\cap C_\gamma=C_\gamma\cap O\), which is open in \(C_\gamma\). The same holds for \(Q\). So a separation of the union would induce a separation on any member that meets both sides.
Proof. For any \(m,n\in\mathbb{N}\), either \(m\le n\), in which case the inclusions give \(C_m\subseteq C_n\), or \(n\le m\), in which case \(C_n\subseteq C_m\). Thus the family is a chain. The theorem applies and gives the claimed connectedness. If all the sets are empty, the union is empty and is connected by our convention; otherwise the theorem’s separation argument applies as stated. \(\square\)
Worked Applications
Worked Example: An Increasing Sequence of Intervals
In \(\mathbb{R}\) with its standard topology, let \(C_n=[0,1-\frac{1}{n+1}]\) for \(n\in\mathbb{N}\). Find the union and determine whether it is connected.
Each \(C_n\) is an interval, so it is connected by the result that every interval in \(\mathbb{R}\) is connected. Its right endpoint is \(1-\frac{1}{n+1}\). The endpoint of \(C_{n+1}\) is \(1-\frac{1}{n+2}\), and \(1-\frac{1}{n+1}<1-\frac{1}{n+2}\), since \(\frac{1}{n+1}>\frac{1}{n+2}\). Hence \(C_n\subseteq C_{n+1}\).
No member contains \(1\), because \(1-\frac{1}{n+1}<1\) for every \(n\in\mathbb{N}\). Conversely, let \(x\in[0,1)\). Since \(1-x>0\), choose \(n\in\mathbb{N}\) with \(n+1\ge\frac{1}{1-x}\). Then \(\frac{1}{n+1}\le 1-x\), so \(x\le 1-\frac{1}{n+1}\), and \(x\in C_n\). Thus \(\bigcup_{n\in\mathbb{N}}C_n=[0,1)\). The corollary shows this union is connected.
Check: The right endpoint for \(n=1\) is \(1-\frac12=\frac12\), and for \(n=2\) it is \(1-\frac13=\frac23\), confirming the endpoints increase. The point \(1\) is not in the union, while every \(x<1\) in \([0,1)\) belongs to some \(C_n\) by the choice above.
Worked Example: A Chain of Rays with No Common Point
In \(\mathbb{R}\) with the standard topology, consider \(C_t=(t,\infty)\) for every \(t\in\mathbb{R}\). Show that this is a chain, that its members have no common point, and that its union is connected.
If \(s<t\), every \(x\in(t,\infty)\) also satisfies \(x>s\), so \(C_t\subseteq C_s\). Thus every pair is comparable by inclusion. Each \(C_t\) is an interval and is connected.
To find the union, let \(x\in\mathbb{R}\). Taking \(t=x-1\) gives \(x>x-1=t\), so \(x\in C_{x-1}\). Every real number is therefore in at least one member, and \(\bigcup_{t\in\mathbb{R}}C_t=\mathbb{R}\). On the other hand, no real number belongs to every member: for any \(x\in\mathbb{R}\), the set \(C_x=(x,\infty)\) excludes \(x\). The theorem nevertheless applies and confirms that the union \(\mathbb{R}\) is connected.
Check: For \(s=0\) and \(t=2\), \(C_2=(2,\infty)\subseteq(0,\infty)=C_0\), as predicted. The point \(3\) belongs to \(C_2\), whereas \(2\) does not; more generally, the strict inequality in \(C_t=(t,\infty)\) is why \(x\notin C_x\).
Worked Example: Nested Segments in the Plane
Give \(\mathbb{R}^2\) its standard Euclidean topology, and for \(n\in\mathbb{N}\) let \(C_n=[-n,n]\times\{0\}\). Find the union and establish its connectedness.
For each \(n\), define \(f_n:[-n,n]\to C_n\) by \(f_n(t)=(t,0)\). The domain is an interval in \(\mathbb{R}\), hence connected. The map is continuous: for \(s,t\in[-n,n]\), the Euclidean distance between \(f_n(s)\) and \(f_n(t)\) is \(\|f_n(s)-f_n(t)\|=\|(s-t,0)\|=|s-t|\). Thus inputs close together have images equally close together. Since continuous images of connected spaces are connected, \(C_n=f_n([-n,n])\) is connected.
If \(n\le m\), then \([-n,n]\subseteq[-m,m]\), so \(C_n\subseteq C_m\). A point in the union must have second coordinate \(0\). Conversely, for any \(x\in\mathbb{R}\), choose \(n\in\mathbb{N}\) with \(n\ge |x|\); then \((x,0)\in C_n\). Therefore \(\bigcup_{n\in\mathbb{N}}C_n=\mathbb{R}\times\{0\}\). By the corollary, this union is connected.
Check: For \(n=2\), \(C_2=[-2,2]\times\{0\}\), so \((-2,0)\) and \((2,0)\) are included, while \((0,1)\) is excluded because its second coordinate is not \(0\). Every point \((x,0)\) is included at a stage with \(n\ge|x|\).
What the Hypothesis Rules Out
The theorem is not a claim that every union of connected subspaces is connected. The chain condition is what ensures that any two members chosen from opposite sides of a proposed separation are contained one in the other, forcing one member to meet both sides. Without that condition, the argument has no such member to contradict connectedness.
Common pitfall. A tempting but incorrect argument is: “Each set is connected, so their union is connected.” In \(\mathbb{R}\) with the standard topology, take \(A=[0,1]\) and \(B=[2,3]\). Both are intervals and hence connected, but neither contains the other. Their union \(S=A\cup B\) is separated by \(P=S\cap(-1,\frac32)=[0,1]\) and \(Q=S\cap(\frac32,4)=[2,3]\). These sets are disjoint, nonempty, relatively open in \(S\), and cover \(S\), so \(S\) is disconnected. The correction is to check the theorem’s actual hypothesis: the family must be a chain under inclusion. A common point is another sufficient condition for a union of connected sets, but this theorem does not require one, as the family of rays demonstrates.
| Family | What to verify | Conclusion |
|---|---|---|
| An increasing sequence \(C_1\subseteq C_2\subseteq\cdots\) | Every member is connected | The union is connected |
| A chain indexed in any way | Every pair is comparable by inclusion, and every member is connected | The union is connected |
| A family of connected sets with no inclusion ordering | The chain hypothesis may fail | The theorem alone gives no conclusion |
Check Your Understanding
In each question, distinguish the chain condition from connectedness of the individual members.
- Let \(C_n=[-n,0]\subseteq\mathbb{R}\) for \(n\in\mathbb{N}\). Find the union and explain why it is connected.
- For \(D_t=(-\infty,t)\subseteq\mathbb{R}\), with \(t\in\mathbb{R}\), verify that the family is a chain and determine its union.
- Does the theorem apply to \(A=[0,1]\) and \(B=[1,2]\)? State which condition you check.
- Give a point that belongs to \(C_3=[-3,3]\times\{0\}\) but not to \(C_2=[-2,2]\times\{0\}\), and verify both membership claims.
- Explain why connectedness of every member, without nestedness or another sufficient hypothesis, does not guarantee that the union is connected.