Connectedness as the Absence of a Separation
Heine–Borel characterized compact subsets of \(\mathbb{R}^n\) through closedness and boundedness. Connectedness describes a different feature: whether a set can be divided into two separated pieces. The definition uses the topology that the set inherits from \(\mathbb{R}^n\), so the pieces must be open relative to the set, not necessarily open in the whole space.
A set \(A\subseteq E\) is open in the subspace \(E\) precisely when \(A=E\cap G\) for some open set \(G\subseteq\mathbb{R}^n\). The Clopen Characterization of Connectedness, established earlier in “Connectedness in Topological Spaces,” gives an equivalent test: \(E\) is connected exactly when its only subsets that are both open and closed in \(E\) are \(\varnothing\) and \(E\). In particular, the sets in a separation are both open and closed relative to \(E\), since each is the complement in \(E\) of the other.
Connectedness should not be confused with path connectedness. A path in \(E\) from \(a\) to \(b\) is a continuous function \(\gamma:[0,1]\to E\) with \(\gamma(0)=a\) and \(\gamma(1)=b\). A set is path connected if every pair of its points can be joined by such a path. Path connectedness implies connectedness, but connectedness is defined by the absence of a separation, not by the existence of paths. For open subsets of \(\mathbb{R}^n\), however, a useful stronger relationship holds: connectedness is equivalent to the existence of polygonal paths.
Convex Sets Are Connected
Convexity provides an immediate way to construct paths. A set \(C\subseteq\mathbb{R}^n\) is convex if, whenever \(x,y\in C\) and \(t\in[0,1]\), the point \((1-t)x+ty\) also belongs to \(C\). Thus a convex set contains the entire line segment joining each pair of its points.
Proof. The empty set is connected by definition, so suppose \(C\) is a nonempty convex set. For any \(a,b\in C\), define \[ \gamma(t)=(1-t)a+tb,\qquad 0\leq t\leq1. \] Convexity ensures that \(\gamma(t)\in C\) for every \(t\in[0,1]\). Each coordinate of \(\gamma(t)\) is a linear function of \(t\), so \(\gamma\) is continuous. Also, \(\gamma(0)=a\) and \(\gamma(1)=b\). Hence \(C\) is path connected.
The interval \([0,1]\) is connected, and continuous images of connected spaces are connected, by the results in “Connectedness in Topological Spaces.” To apply those results, fix \(a\in C\). For every \(b\in C\), the path just constructed has image \(\gamma([0,1])\) that is connected and contains \(a\) and \(b\). More directly, path connectedness implies connectedness: if \(C\) had a separation \(A,B\), choose \(a\in A\) and \(b\in B\). A path from \(a\) to \(b\) would give a continuous map from \([0,1]\) into \(C\); the preimages of \(A\) and \(B\) would be disjoint, nonempty, relatively open sets covering \([0,1]\). That would separate the connected interval \([0,1]\), a contradiction. Therefore \(C\) is connected. \(\square\)
Worked Example: An Open Ball Is Connected
Let \(a\in\mathbb{R}^n\) and \(r>0\), and consider the open Euclidean ball \[ B_r^{(2)}(a)=\{x\in\mathbb{R}^n:\|x-a\|_2<r\}. \] We check that it is convex. Take \(x,y\in B_r^{(2)}(a)\) and \(t\in[0,1]\). The triangle inequality and homogeneity of the Euclidean norm give \[ \|(1-t)x+ty-a\|_2 =\|(1-t)(x-a)+t(y-a)\|_2 \leq (1-t)\|x-a\|_2+t\|y-a\|_2. \] Both norms on the right are strictly less than \(r\), so \[ (1-t)\|x-a\|_2+t\|y-a\|_2 <(1-t)r+tr=r. \] Thus \((1-t)x+ty\in B_r^{(2)}(a)\), and the ball is convex. The theorem now shows that it is connected. The argument applies to every dimension \(n\geq1\).
The proof illustrates why convexity is sufficient, not necessary. To establish connectedness, it is enough to find paths; a set can still be connected even when some straight segments between its points leave the set. The next result gives a path criterion for open sets without requiring convexity.
Open Connected Sets and Polygonal Paths
A polygonal path is made from finitely many line segments joined end to end. A subset \(U\subseteq\mathbb{R}^n\) is polygonally path connected if any two of its points can be joined by a polygonal path lying entirely in \(U\). Every such path can be parametrized as a continuous path by traversing its segments successively.
Proof. First suppose that \(U\) is polygonally path connected. A polygonal path, parametrized on \([0,1]\), is continuous: on each of finitely many consecutive subintervals it is a linear parametrization of one segment, and the formulas agree at the shared endpoints. Thus \(U\) is path connected. As shown in the proof above, a path connected space is connected. Therefore \(U\) is connected.
Conversely, suppose \(U\) is nonempty, open, and connected. Fix \(p\in U\), and let \(A\) be the set of points \(x\in U\) that can be joined to \(p\) by a polygonal path in \(U\). The constant path shows that \(p\in A\), so \(A\neq\varnothing\).
We first show that \(A\) is open relative to \(U\). Take \(x\in A\). Since \(U\) is open, there is \(r>0\) such that \(B_r^{(2)}(x)\subseteq U\). If \(y\in B_r^{(2)}(x)\), the segment from \(x\) to \(y\) lies in this ball. Indeed, for \(t\in[0,1]\), \[ \|(1-t)x+ty-x\|_2=t\|y-x\|_2<r. \] There is a polygonal path in \(U\) from \(p\) to \(x\), and appending this segment gives a polygonal path in \(U\) from \(p\) to \(y\). Hence \(y\in A\), so \(B_r^{(2)}(x)\subseteq A\). This proves that \(A\) is open relative to \(U\).
We next show that \(U\setminus A\) is open relative to \(U\). Take \(x\in U\setminus A\). Choose \(r>0\) such that \(B_r^{(2)}(x)\subseteq U\). Suppose, for contradiction, that some \(y\in B_r^{(2)}(x)\) belongs to \(A\). There is a polygonal path in \(U\) from \(p\) to \(y\). The segment from \(y\) to \(x\) lies in \(B_r^{(2)}(x)\), by the same estimate used above. Appending that segment gives a polygonal path in \(U\) from \(p\) to \(x\), contradicting \(x\notin A\). Therefore \(B_r^{(2)}(x)\subseteq U\setminus A\), and \(U\setminus A\) is open relative to \(U\).
If \(U\setminus A\) were nonempty, then \(A\) and \(U\setminus A\) would be disjoint, nonempty, relatively open sets whose union is \(U\). They would form a separation of \(U\), contradicting connectedness. Thus \(U\setminus A=\varnothing\), so \(A=U\). Every point of \(U\) can therefore be joined to \(p\) by a polygonal path in \(U\). Given any \(x,y\in U\), reverse a polygonal path from \(p\) to \(x\) and concatenate it with a polygonal path from \(p\) to \(y\); this yields a polygonal path from \(x\) to \(y\). Hence \(U\) is polygonally path connected. \(\square\)
Worked Example: A Connected Open Set That Is Not Convex
In \(\mathbb{R}^2\), let \[ U=\{(x,y):x>0\}\cup\{(x,y):y>0\}. \] This set is open because it is a union of two open half-planes. It is not convex: \(u=(1,-2)\) belongs to \(U\), and \(v=(-2,1)\) belongs to \(U\), but their midpoint is \[ \frac{u+v}{2}=\left(-\frac12,-\frac12\right), \] which belongs to neither half-plane and therefore is not in \(U\).
Nevertheless, \(U\) is connected. To verify this, take \(z\in U\). If its first coordinate is positive, the segment from \(z\) to \(q=(1,1)\) stays in the half-plane \(x>0\): every point on the segment has first coordinate \((1-t)z_1+t>0\). If its first coordinate is not positive, membership in \(U\) forces its second coordinate to be positive; then the segment from \(z\) to \(q\) stays in the half-plane \(y>0\), since its second coordinate is \((1-t)z_2+t>0\). Thus every point of \(U\) can be joined to \(q\) by a segment in \(U\). Reversing one such segment and appending another gives a polygonal path between any two points of \(U\). The Polygonal Path Characterization shows that \(U\) is connected.
How a Separation Can Be Verified
To prove that a set is disconnected, it is not enough to notice a visible gap in a drawing. One must exhibit a separation and check its conditions: both pieces are nonempty, disjoint, cover the set, and are open in the relative topology. When the set lies on opposite sides of a coordinate hyperplane, intersections with suitable open half-spaces often provide the required pieces.
Worked Example: Two Separated Balls Form a Disconnected Set
In \(\mathbb{R}^n\), let \(e_1=(1,0,\ldots,0)\) and define \[ E=\overline{B_1^{(2)}(-3e_1)}\cup\overline{B_1^{(2)}(3e_1)}. \] Write \(E_-= \overline{B_1^{(2)}(-3e_1)}\) and \(E_+=\overline{B_1^{(2)}(3e_1)}\). If \(x\in E_-\), then \[ |x_1+3|\leq \|x+3e_1\|_2\leq1, \] so \(x_1\leq-2\). If \(x\in E_+\), then \[ |x_1-3|\leq \|x-3e_1\|_2\leq1, \] so \(x_1\geq2\). Thus \(E_-\) and \(E_+\) are both nonempty, disjoint, and their union is \(E\).
To check relative openness, observe that \[ E\cap\{x:x_1<0\}=E_- \quad\text{and}\quad E\cap\{x:x_1>0\}=E_+. \] The half-spaces \(\{x:x_1<0\}\) and \(\{x:x_1>0\}\) are open in \(\mathbb{R}^n\), so these equalities show that \(E_-\) and \(E_+\) are open relative to \(E\). They form a separation, and therefore \(E\) is disconnected. This verifies the separation directly, rather than relying on the appearance of the two balls.
The open-set hypothesis in the Polygonal Path Characterization matters. Its proof uses small balls around points of \(U\) to extend a polygonal path and to show that the points not reachable by such paths also form an open set. For arbitrary subsets, connectedness still means that no separation exists, but the theorem does not say that every connected subset of \(\mathbb{R}^n\) is polygonally path connected. Use convexity or openness when applying the path results above; otherwise, check connectedness from its definition or from an appropriate theorem.
Check Your Understanding
Use the definitions and results in this tutorial to answer each question.
- What conditions must two sets satisfy to form a separation of \(E\subseteq\mathbb{R}^n\)?
- Why does convexity imply path connectedness? Which earlier result then gives connectedness?
- Where does openness enter the proof that a connected open set is polygonally path connected?
- Give two points in the open set \(\{(x,y):x>0\}\cup\{(x,y):y>0\}\) whose midpoint is outside the set.
- In the two-ball example, why are the pieces open relative to their union?