Euclidean Geometry Adds Structure to Connectedness
Connectedness is defined by the absence of a separation; it does not, by itself, describe a route between points. The Topologist’s Sine Curve Is Not Path Connected Theorem from the previous tutorial showed that even a connected subset of the plane need not be path connected. In Euclidean space, however, openness supplies a useful extra condition: every point of an open set has a small ball around it that stays inside the set. Since balls are convex, short straight segments can be used to explore the set locally.
This local geometry leads to a strong conclusion. A connected open subset of Euclidean space is not merely path connected: any two of its points can be joined by a path made from finitely many straight segments. We will prove this by collecting all points that can be reached from one fixed point by such paths, then using connectedness to show there can be no unreachable points.
A polygonal path can be parametrized as a continuous path on \([0,1]\): divide the parameter interval into finitely many subintervals and travel affinely along one segment on each. Thus polygonal connectedness implies path connectedness, which in turn implies connectedness by the Path Connected Implies Connected Theorem. The result below establishes the important converse for open subsets of Euclidean space.
Convexity Gives Connectedness
A subset \(E\subseteq\mathbb{R}^n\) is convex if, whenever \(p,q\in E\) and \(0\leq t\leq1\), the point \((1-t)p+tq\) also belongs to \(E\). In other words, the entire straight line segment joining any two points of \(E\) lies in \(E\). Convexity immediately provides paths, and therefore connectedness.
Proof. The empty set is connected by convention. If \(E\) has one point, it has no separation into two nonempty subsets and is connected. Now suppose \(E\) is convex and contains at least two points. For any \(p,q\in E\), define \(\gamma:[0,1]\to E\) by \(\gamma(t)=(1-t)p+tq\). Convexity ensures \(\gamma(t)\in E\) for every \(t\in[0,1]\), and the coordinate functions of \(\gamma\) are continuous. Hence \(E\) is path connected. The Path Connected Implies Connected Theorem now gives that \(E\) is connected. \(\square\)
Worked Example: Euclidean Balls Are Connected
Let \(a\in\mathbb{R}^n\) and \(r>0\). The open ball \(B_r(a)\) is convex. Indeed, take \(x,y\in B_r(a)\) and \(t\in[0,1]\). The Euclidean norm satisfies the triangle inequality and homogeneity, so \[ \|(1-t)x+ty-a\| =\|(1-t)(x-a)+t(y-a)\| \leq (1-t)\|x-a\|+t\|y-a\|. \] Since \(\|x-a\|<r\) and \(\|y-a\|<r\), the right-hand side is strictly less than \((1-t)r+tr=r\). Therefore \((1-t)x+ty\in B_r(a)\). The ball is convex, so the Convex Subsets of Euclidean Space Are Connected Theorem shows it is connected.
The same argument works for a closed ball, since replacing the strict inequalities by \(\leq\) shows that every segment between points in the closed ball stays in the closed ball. More generally, a half-space such as \(\{x\in\mathbb{R}^n:x_1>0\}\) is convex and hence connected.
Connected Open Sets Are Polygonally Connected
The key argument uses only two features of Euclidean space: small balls are convex, and a connected set cannot be split into two nonempty relatively open pieces. Fix one point \(p\) of an open set \(E\), and consider all points that can be reached from \(p\) by polygonal paths lying in \(E\). A small ball around any reachable point consists entirely of reachable points. The same local observation shows that a small ball around an unreachable point cannot contain a reachable point. Thus the reachable and unreachable points would form a separation if both kinds existed.
Proof. If \(E\) is empty, the statement about every pair of points holds vacuously. Otherwise, choose \(p\in E\), and let \(A\) be the set of points \(x\in E\) for which there is a polygonal path in \(E\) from \(p\) to \(x\). The point \(p\) belongs to \(A\), using the constant path, so \(A\) is nonempty.
We first show that \(A\) is open relative to \(E\). Let \(x\in A\). Because \(E\) is open in \(\mathbb{R}^n\), there is \(r>0\) such that \(B_r(x)\subseteq E\). If \(y\in B_r(x)\), convexity of the ball gives \([x,y]\subseteq B_r(x)\). Append this segment to a polygonal path from \(p\) to \(x\). The result is a polygonal path in \(E\) from \(p\) to \(y\), so \(y\in A\). Therefore \(B_r(x)\subseteq A\), proving that \(A\) is relatively open in \(E\).
Next, \(E\setminus A\) is also open relative to \(E\). Let \(x\in E\setminus A\). Choose \(r>0\) with \(B_r(x)\subseteq E\). Suppose some \(y\in B_r(x)\) belonged to \(A\). The segment \([y,x]\) would lie in \(B_r(x)\), by convexity of the ball. Appending that segment to a polygonal path from \(p\) to \(y\) would give a polygonal path from \(p\) to \(x\), contradicting \(x\notin A\). Thus \(B_r(x)\cap A=\varnothing\), and \(B_r(x)\subseteq E\setminus A\). This proves relative openness.
If \(E\setminus A\) were nonempty, then \(A\) and \(E\setminus A\) would be disjoint, nonempty, relatively open subsets of \(E\) whose union is \(E\). They would give a separation of \(E\), contrary to its connectedness. Hence \(E\setminus A=\varnothing\), so every point of \(E\) is reachable from \(p\). Since \(p\) was arbitrary, any two points of \(E\) can be joined by a polygonal path in \(E\). \(\square\)
Worked Example: An Open Annulus Is Connected
Consider the planar annulus \[ E=\{x\in\mathbb{R}^2:1<\|x\|<2\}. \] It is open because the norm is continuous and the inequalities are strict. It is not convex: for example, \((3/2,0)\) and \((-3/2,0)\) belong to \(E\), but their midpoint \((0,0)\) does not. So the convexity result does not apply directly.
We can nevertheless verify connectedness by constructing paths. Take \(x,y\in E\), and write \(u=x/\|x\|\) and \(v=y/\|y\|\), which are unit vectors. First move radially from \(x\) to \((3/2)u\), then follow a circular arc of radius \(3/2\) from \((3/2)u\) to \((3/2)v\), and finally move radially to \(y\). Each radial segment has norm between its endpoint norms, which are both strictly between \(1\) and \(2\); every point on the arc has norm \(3/2\). Thus the whole path stays in \(E\).
For the arc, choose an angle \(\theta\) with \(v=(\cos\theta)u+(\sin\theta)w\), where \(w\) is a unit vector perpendicular to \(u\). The map \(s\mapsto (3/2)((\cos(s\theta))u+(\sin(s\theta))w)\), for \(0\leq s\leq1\), is continuous, has the required endpoints, and stays at norm \(3/2\). Joining the three pieces gives a continuous path in \(E\). Therefore \(E\) is path connected and hence connected.
The annulus is also polygonally connected by the Connected Open Subsets of Euclidean Space Theorem. The explicit curved path proves connectedness, while the theorem guarantees that a route made of finitely many straight segments exists as well.
Connected Components of Open Sets
The polygonal-path theorem gives a useful description of components. A connected component is a maximal connected subset. In an open subset of Euclidean space, every component is itself open: around each of its points, a small ball is connected and must remain inside that component. The polygonal-path theorem then applies to each component.
Proof. Let \(C\) be a connected component of \(E\), and take \(x\in C\). Since \(E\) is open, choose \(r>0\) such that \(B_r(x)\subseteq E\). The ball \(B_r(x)\) is convex, hence connected by the Convex Subsets of Euclidean Space Are Connected Theorem. It meets \(C\) at \(x\). By the Union with a Common Point Theorem, \(C\cup B_r(x)\) is connected. Since \(C\) is a maximal connected subset of \(E\), and \(C\cup B_r(x)\subseteq E\), it follows that \(C\cup B_r(x)=C\). Thus \(B_r(x)\subseteq C\). Every point of \(C\) has a ball contained in \(C\), so \(C\) is open in \(\mathbb{R}^n\).
The component \(C\) is connected by the Construction of Components Theorem, and it is open by the preceding argument. The Connected Open Subsets of Euclidean Space Are Polygonally Connected Theorem therefore shows that \(C\) is polygonally connected. \(\square\)
Worked Example: Removing a Line Can Disconnect the Plane
Let \(E=\{(x,y)\in\mathbb{R}^2:x\ne0\}\), the plane with the vertical axis removed. This set is open, but it is not connected. Define \[ A=\{(x,y)\in E:x>0\},\qquad B=\{(x,y)\in E:x<0\}. \] Both sets are nonempty, disjoint, and have union \(E\). They are relatively open in \(E\): each is the intersection of \(E\) with an open half-plane. Their complements in \(E\) are the other relatively open half-plane, so each is also relatively closed in \(E\). Hence \(A\) and \(B\) form a separation of \(E\).
Each half-plane is convex, so each is connected. They are exactly the connected components of \(E\): a connected subset meeting both would, by the Connected Subset Lies on One Side Theorem, have to lie entirely on one side of the separation, which is impossible. This example shows that openness alone does not imply connectedness. It does show the conclusion of the components theorem in action: the two components are open and polygonally connected.
What the Euclidean Hypotheses Contribute
The conclusions in this tutorial depend on distinct hypotheses. Convexity gives a direct segment between any two points, with no connectedness assumption needed. Openness gives small balls contained in the set, but does not guarantee the set is connected. Connectedness rules out a separation, but without openness it need not provide paths, as the topologist’s sine curve demonstrates.
Together, openness and connectedness force polygonal connectedness because local straight-line moves can be extended throughout the set, and connectedness prevents the collection of reachable points from stopping partway. This is a particularly useful proof technique: define a reachability set, show it and its complement are relatively open, and then invoke connectedness. The same idea helps identify the components of open Euclidean sets, even when the overall set is disconnected.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why does convexity imply that the line segment between two points of a set stays in that set?
- In the polygonal-path theorem, why is the set of points reachable from a fixed point relatively open?
- Why would a reachable point in a small ball around an unreachable point lead to a contradiction?
- Which hypothesis fails for the open annulus when one tries to apply the convexity theorem directly?
- Why must every connected component of an open subset of Euclidean space be open?