Why a Shared Point Controls a Union
Connectedness is defined by the absence of a separation: a connected space cannot be divided into two nonempty, disjoint open pieces. This definition can make a union of many sets difficult to analyze directly. A shared point gives a way to control any proposed separation: every connected set in the family must lie on the same side as that point.
Throughout, subsets of a topological space \((X,\mathcal{T})\) carry the subspace topology. A separation of a space \(Y\) is a pair of disjoint, nonempty sets \(U,V\) that are open in \(Y\) and satisfy \(Y=U\cup V\). A space is connected if it has no separation. A subset is called connected when it is connected with its subspace topology.
Proof. Write \(Y=\bigcup_{\alpha\in J}C_\alpha\). The point \(p\) belongs to \(Y\), so \(Y\) is nonempty. Suppose, for a contradiction, that \(Y\) has a separation \(Y=U\cup V\). Since \(p\in U\cup V\), we may label the two sets so that \(p\in U\). Fix any \(\alpha\in J\). The sets \(C_\alpha\cap U\) and \(C_\alpha\cap V\) are disjoint, cover \(C_\alpha\), and are open in \(C_\alpha\), because \(U\) and \(V\) are open in \(Y\). Moreover, \(C_\alpha\cap U\) is nonempty because it contains \(p\). If \(C_\alpha\cap V\) were also nonempty, these two intersections would be a separation of \(C_\alpha\), contrary to its connectedness. Therefore \(C_\alpha\cap V=\varnothing\), so \(C_\alpha\subseteq U\). This holds for every \(\alpha\), and hence \(Y\subseteq U\). It contradicts that \(V\) is nonempty and \(Y=U\cup V\). Thus \(Y\) has no separation and is connected. \(\square\)
The essential point in the proof is not that the sets overlap pairwise. It is that every member contains the same point \(p\). That point forces each connected member to lie on the same side of any proposed separation of the union.
Finite Unions and Intervals
The finite version follows immediately. If \(C_1,\ldots,C_m\) are connected and all contain \(p\), their union is connected. There is no need to combine the sets two at a time: the theorem already applies to the whole finite family. The same proof also applies when the family is infinite.
Worked Example: A Finite Union of Intervals
In \(\mathbb{R}\) with its standard topology, let \(C_1=[-2,1]\), \(C_2=[1,4]\), and \(C_3=[1,3]\). Determine their union and establish that it is connected.
Each \(C_i\) is an interval, hence is connected by the connectedness of intervals in \(\mathbb{R}\). All three contain \(1\), so the common-point union theorem applies. To compute the union, \(C_1\cup C_2=[-2,4]\): every \(x\in[-2,1]\) belongs to \(C_1\), and every \(x\in[1,4]\) belongs to \(C_2\); neither interval contains a point outside \([-2,4]\). Also \(C_3=[1,3]\subseteq[1,4]\). Therefore
and this union is connected. Check: the left endpoint \(-2\) belongs to \(C_1\), the right endpoint \(4\) belongs to \(C_2\), and every point between them belongs to at least one of those two intervals. Neither contributes points beyond those endpoints, and \(C_3\) adds no new points.
Infinite Families: The Shared Point Still Suffices
For an infinite family, the conclusion does not require the union to be an interval, closed, or equal to any one member. The theorem only needs each member to be connected and to contain the same point. Computing the union as a set can nevertheless require a separate argument.
Worked Example: An Increasing Family of Intervals
In standard \(\mathbb{R}\), for each integer \(n\ge2\), let \(I_n=[0,1-\frac1n]\). Find \(\bigcup_{n\ge2}I_n\) and prove it is connected.
Each \(I_n\) is an interval and therefore connected, and \(0\in I_n\) for every \(n\ge2\). The common-point union theorem gives connectedness. To identify the union, first note that \(1-\frac1n<1\) for every \(n\ge2\), so no \(I_n\) contains \(1\) or any number greater than \(1\). All the intervals contain only nonnegative numbers. Thus the union is contained in \([0,1)\).
Conversely, let \(x\in[0,1)\). Then \(1-x>0\). Choose an integer \(n\ge2\) with \(n>\frac1{1-x}\). This inequality gives \(\frac1n<1-x\), hence \(x<1-\frac1n\), so \(x\in I_n\). Therefore every point of \([0,1)\) is in the union, and
Check: for \(x=0\), membership holds in every \(I_n\). For any \(x\) with \(0\le x<1\), the choice of \(n\) above gives strict inequality \(x<1-\frac1n\), which is sufficient for membership in the closed interval \(I_n\). The point \(1\) is excluded because \(1-\frac1n<1\) for every permitted \(n\).
Worked Example: Segments in Infinitely Many Directions
Give a connected subset of the Euclidean plane obtained as a union of infinitely many different connected sets, and describe the union explicitly.
For \(n\in\mathbb{N}\), define \(C_n=\{(t,t/n):0\le t\le1\}\subseteq\mathbb{R}^2\), where \(\mathbb{R}^2\) has its Euclidean topology. The map \(f_n:[0,1]\to\mathbb{R}^2\) given by \(f_n(t)=(t,t/n)\) is continuous: for \(s,t\in[0,1]\),
For fixed \(n\), the factor \(\sqrt{1+\frac{1}{n^2}}\) is constant, so if \(|s-t|\) is sufficiently small then \(\|f_n(s)-f_n(t)\|\) is as small as prescribed. The interval \([0,1]\) is connected, and continuous images of connected spaces are connected; hence \(C_n=f_n([0,1])\) is connected. (The continuous-image result is the standard connectedness theorem for continuous maps.) Also \(f_n(0)=(0,0)\), so all the \(C_n\) contain the origin. Their union is connected by the common-point union theorem.
In coordinates, the union is exactly
Indeed, a point of \(C_n\) has \(x=t\in[0,1]\) and \(y=t/n=x/n\). Conversely, if \(0\le x\le1\) and \(y=x/n\) for some \(n\in\mathbb{N}\), then setting \(t=x\) places \((x,y)\) in \(C_n\). At \(x=0\), every segment contributes only \((0,0)\). For \(x>0\), the points on different segments have different second coordinates, since \(x/n\ne x/m\) when \(n\ne m\). This verifies that the family contains infinitely many distinct segments and that the coordinate description accounts for all of them.
A Connected Core Can Replace One Shared Point
Sometimes the sets in a family do not all share a single point, but each meets the same connected set. The connected core then takes the role of the common point.
Proof. If \(J\) is empty, the union in the conclusion is \(C\), which is connected by assumption. Suppose \(J\) is nonempty and set \(Y=C\cup\bigcup_{\alpha\in J}D_\alpha\). If \(Y\) had a separation \(Y=U\cup V\), the connected set \(C\) would have to lie entirely in one side: otherwise \(C\cap U\) and \(C\cap V\) would be a separation of \(C\). Label the sides so that \(C\subseteq U\). Each \(D_\alpha\) meets \(C\), and therefore meets \(U\). If it also met \(V\), then \(D_\alpha\cap U\) and \(D_\alpha\cap V\) would separate \(D_\alpha\). Thus \(D_\alpha\subseteq U\) for every \(\alpha\). This gives \(Y\subseteq U\), contradicting that \(V\) is nonempty. Hence \(Y\) is connected. \(\square\)
Worked Example: Vertical Segments Meeting a Horizontal Core
In \(\mathbb{R}^2\) with its Euclidean topology, let \(C=[0,1]\times\{0\}\), and for each \(n\in\mathbb{N}\), let \(D_n=\{1/n\}\times[0,1]\). Prove that \(C\cup\bigcup_{n\in\mathbb{N}}D_n\) is connected.
The set \(C\) is the continuous image of \([0,1]\) under \(t\mapsto(t,0)\), and each \(D_n\) is the continuous image of \([0,1]\) under \(t\mapsto(1/n,t)\). These maps are continuous: changing the parameter by an amount \(r\) changes the Euclidean distance between image points by exactly \(|r|\). Since \([0,1]\) is connected and continuous images preserve connectedness, \(C\) and every \(D_n\) are connected.
For each \(n\), the point \((1/n,0)\) lies in \(C\), because \(0\le1/n\le1\), and also lies in \(D_n\), because \(0\in[0,1]\). Thus \(D_n\cap C\ne\varnothing\) for every \(n\), and the connected-core corollary proves that the full union is connected. Check: the intersection point depends on \(n\); for example, \(D_1\) has first coordinate \(1\), while \(D_2\) has first coordinate \(1/2\), so these two vertical segments have no common point. The corollary applies because each meets \(C\), not because all the vertical segments share one point.
The Common-Point Hypothesis and a Common Pitfall
Connectedness of the individual sets alone is not enough. The common-point theorem cannot be applied unless there really is a point belonging to every member of the family. A quick membership check can prevent a false use of the result.
Common pitfall. Consider \(A=(-1,0)\) and \(B=(0,1)\) in standard \(\mathbb{R}\). Both are connected intervals, and their union is sometimes incorrectly treated as a common-point union on the grounds that the intervals “meet at \(0\).” But \(0\notin A\) and \(0\notin B\), since both intervals are open at that endpoint. In fact, \(A\cup B\) is disconnected: its subsets \(A\) and \(B\) are disjoint and nonempty, cover the union, and are open in the union because they are intersections of that union with the open sets \((-\infty,0)\) and \((0,\infty)\), respectively. The correction is to check membership in the actual sets before invoking the theorem. Here there is no shared point, so the theorem makes no claim about this union.
The common-point argument is useful because it reduces a global question about a large union to a local constraint on each connected member. The proof does not require a finite number of sets, nor does it require the members to be closed or open in the ambient space. It requires connectedness in the subspace topology and a shared point (or, in the connected-core version, an intersection with the core). Keeping these hypotheses explicit is what makes the method reliable.
Check Your Understanding
For each question, identify the connected sets and verify exactly which hypothesis permits the conclusion.
- Let \(A=[-3,0]\) and \(B=[0,5]\) in standard \(\mathbb{R}\). Find \(A\cup B\) and explain why it is connected.
- For \(n\ge2\), let \(J_n=[-1+\frac1n,1-\frac1n]\). Do all the \(J_n\) contain a common point? Use the common-point union theorem if its hypotheses hold.
- State the common-point union theorem using a family indexed by an arbitrary nonempty set \(J\). Which part of its proof would fail if the sets did not share a point?
- Let \(C\) be connected and let \(D_1,D_2\) be connected, with each \(D_i\) meeting \(C\). Explain why \(C\cup D_1\cup D_2\) is connected even if \(D_1\cap D_2=\varnothing\).
- In standard \(\mathbb{R}\), check whether \(0\) belongs to either \((-2,0)\) or \((0,2)\). What does this imply about applying the common-point union theorem to their union?