Connectedness Without an Order
On the real line, the Characterization of Connected Subsets of \(\mathbb{R}\) gives a useful picture: connected sets are intervals. That description depends on the order of the real numbers, so it is not the right definition for an arbitrary metric space. In a metric space, the basic question remains whether a set can be split into two separated pieces. The metric supplies open balls and distances, but the definition of connectedness itself uses the open sets generated by those balls.
Let \((X,d)\) be a metric space. For \(x\in X\) and \(r>0\), the open ball with center \(x\) and radius \(r\) is \(B(x,r)=\{y\in X:d(x,y)<r\}\). A subset \(E\subseteq X\) is considered with the subspace metric: distances between its points are still measured by \(d\), while its open sets are relative to \(E\). In particular, \(U\subseteq E\) is relatively open if, for every \(x\in U\), there is an \(r>0\) such that \(B(x,r)\cap E\subseteq U\).
Relative openness matters: a piece of \(E\) need not be open in all of \(X\). It must be open when viewed from within \(E\). Thus connectedness is a property of a set with its inherited metric topology, rather than a requirement that its pieces be open in the surrounding space.
The Clopen Test
A subset is called relatively closed in \(E\) if its complement in \(E\) is relatively open. A set that is both relatively open and relatively closed is often called relatively clopen. The following criterion is useful because it lets us test connectedness by looking at a single subset and its complement.
Proof. Suppose first that \(E=A\cup B\) is a separation. Since \(A\) and \(B\) are relatively open and \(E\setminus A=B\), the set \(A\) is also relatively closed. It is nonempty and proper because both pieces of the separation are nonempty. Thus a separation gives a nonempty proper relatively clopen subset.
Conversely, suppose \(C\subseteq E\) is nonempty, proper, and relatively clopen. The sets \(C\) and \(E\setminus C\) are disjoint, nonempty, and cover \(E\). The first is relatively open by assumption; the second is relatively open because \(C\) is relatively closed. They therefore form a separation. This proves both directions. \(\square\)
Worked Example: A Discrete Metric Has No Connected Sets With Two Points
Let \(X=\{a,b,c\}\) and give it the discrete metric: \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\). For any \(x\in X\), the ball \(B(x,1/2)\) is the singleton \(\{x\}\). Consequently every subset of \(X\) is open in \(X\), and every subset is open in each of its subspaces.
If \(E\subseteq X\) contains at least two points, choose \(p\in E\). Then \(\{p\}\) is a nonempty proper relatively open subset of \(E\), and its complement \(E\setminus\{p\}\) is relatively open as well. By the Clopen Criterion, \(E\) is disconnected. If \(E\) is empty or a singleton, it is connected. So, in this metric space, the only connected subsets are the empty set and the singletons.
The example illustrates that connectedness depends on the topology supplied by the metric. In the discrete metric, individual points are open, so a set with more than one point can be split immediately. A metric space need not resemble the real line, and connected subsets need not have an interval-like description.
Connectedness Passes to the Closure
The closure \(\overline{E}\) of a set \(E\) in \(X\) consists of \(E\) together with its limit points in \(X\). If \(E\) is dense in its closure, it is natural to ask whether adding those limit points can introduce a separation. It cannot. This extends the Closure of a Connected Set theorem from the real line to arbitrary metric spaces.
Proof. If \(E\) is empty, then \(\overline{E}\) is empty and is connected. Now suppose \(E\) is nonempty. Assume, for contradiction, that \(\overline{E}=A\cup B\) is a separation. Thus \(A\) and \(B\) are disjoint, nonempty, and relatively open in \(\overline{E}\).
Every nonempty relatively open subset of \(\overline{E}\) meets \(E\). Indeed, if \(a\in A\), relative openness supplies an open set \(U\subseteq X\) with \(a\in U\) and \(U\cap\overline{E}\subseteq A\). Since \(a\in\overline{E}\), every open neighborhood of \(a\) meets \(E\); hence \(U\cap E\ne\varnothing\), and \(U\cap E\subseteq A\cap E\). The same argument, starting with a point of \(B\), shows \(B\cap E\ne\varnothing\).
The sets \(A\cap E\) and \(B\cap E\) are disjoint and cover \(E\). They are relatively open in \(E\), since restricting a relatively open subset of \(\overline{E}\) to \(E\subseteq\overline{E}\) preserves relative openness. Both are nonempty by the preceding paragraph. They would form a separation of \(E\), contrary to its connectedness. Therefore \(\overline{E}\) is connected. \(\square\)
Worked Example: Closing a Parabolic Arc
Consider \(E=\{(t,t^2):0<t<1\}\) in the Euclidean plane. The map \(f:(0,1)\to\mathbb{R}^2\) defined by \(f(t)=(t,t^2)\) is continuous, since both coordinate functions are continuous. The interval \((0,1)\) is connected, and the Continuous Images of Connected Sets theorem applies to maps between metric spaces. Thus \(E=f((0,1))\) is connected.
Its closure is \(\overline{E}=\{(t,t^2):0\leq t\leq1\}\). To check this description, any sequence of points \((t_n,t_n^2)\in E\) converging to \((x,y)\) has \(t_n\to x\) and \(t_n^2\to y\). Since \(0<t_n<1\), we have \(0\leq x\leq1\), and continuity of squaring gives \(y=x^2\). Conversely, points with \(0<x<1\) already belong to \(E\); the two remaining points \((0,0)\) and \((1,1)\) are limits of points in \(E\), obtained by taking \(t\to0\) and \(t\to1\), respectively. The Closure Theorem therefore shows that this closed parabolic arc is connected.
Connected Spaces Admit Arbitrarily Fine Chains
A finite \(\varepsilon\)-chain from \(x\) to \(y\) is a finite list \(x=x_0,x_1,\ldots,x_n=y\) of points of \(E\) such that each successive distance is less than \(\varepsilon\). The points need not lie on a path, and a chain is not required to vary continuously: it is simply a finite sequence of small steps. Connectedness guarantees that such chains can be found at every positive scale.
Proof. Fix \(\varepsilon>0\) and \(x\in E\). Let \(C\) be the set of points of \(E\) that can be joined to \(x\) by a finite \(\varepsilon\)-chain. The chain of length zero shows that \(x\in C\), so \(C\) is nonempty.
We show that \(C\) is relatively open in \(E\). If \(z\in C\) and \(w\in E\) satisfies \(d(z,w)<\varepsilon\), append \(w\) to a chain from \(x\) to \(z\). The result is a finite \(\varepsilon\)-chain from \(x\) to \(w\), so \(w\in C\). Thus \(B(z,\varepsilon)\cap E\subseteq C\) for every \(z\in C\).
The complement \(E\setminus C\) is also relatively open. If \(z\in E\setminus C\) and \(w\in E\) satisfies \(d(z,w)<\varepsilon\), then \(w\) cannot belong to \(C\): otherwise a chain from \(x\) to \(w\), followed by the step from \(w\) to \(z\), would put \(z\) in \(C\). Therefore \(B(z,\varepsilon)\cap E\subseteq E\setminus C\) for each \(z\in E\setminus C\).
If \(E\setminus C\) were nonempty, \(C\) and \(E\setminus C\) would be disjoint, nonempty, relatively open sets covering \(E\). That would be a separation, contradicting connectedness. Hence \(C=E\). Since \(y\in E\) was arbitrary, every \(y\) can be joined to \(x\) by a finite \(\varepsilon\)-chain. As \(x\) was arbitrary too, the result follows. \(\square\)
Worked Example: Every Two Rational Numbers Have Fine Chains
Give \(\mathbb{Q}\) the metric inherited from \(\mathbb{R}\). Take \(p,q\in\mathbb{Q}\) and \(\varepsilon>0\). If \(p=q\), the chain consisting of the single point \(p\) suffices. If \(p\ne q\), choose a positive integer \(n\) large enough that \(|q-p|/n<\varepsilon\), and set \(x_k=p+k(q-p)/n\) for \(k=0,\ldots,n\). Each \(x_k\) is rational, since \(p,q\), and \(k/n\) are rational.
For each \(k=1,\ldots,n\), direct subtraction gives \(x_k-x_{k-1}=(q-p)/n\), so \(d(x_k,x_{k-1})=|q-p|/n<\varepsilon\). Also \(x_0=p\) and \(x_n=q\). Thus \(\mathbb{Q}\) has an \(\varepsilon\)-chain between every pair of its points, for every \(\varepsilon>0\). This property alone does not imply connectedness.
Worked Example: The Rationals Are Still Disconnected
Choose an irrational number \(c\), for example \(c=\sqrt{2}\), and define \(A=\{q\in\mathbb{Q}:q<c\}\) and \(B=\{q\in\mathbb{Q}:q>c\}\). Since no rational number equals \(c\), the sets are disjoint and cover \(\mathbb{Q}\). They are nonempty: \(1\in A\) and \(2\in B\), because \(1<\sqrt{2}<2\).
For any \(q\in A\), the radius \(r=(c-q)/2\) is positive. If \(z\in\mathbb{Q}\) and \(|z-q|<r\), then \(z<q+r=(q+c)/2<c\), so \(z\in A\). This proves that \(A\) is relatively open in \(\mathbb{Q}\). For \(q\in B\), take \(r=(q-c)/2>0\). If \(z\in\mathbb{Q}\) and \(|z-q|<r\), then \(z>q-r=(q+c)/2>c\), so \(z\in B\). Thus \(B\) is relatively open as well. They form a separation, and \(\mathbb{Q}\) is disconnected despite having arbitrarily fine chains between every pair of points.
What the Metric Tells Us
The fine-chain theorem gives a necessary consequence of connectedness: at every scale, a connected metric space can be traversed by finitely many steps smaller than that scale. The rational-number example warns against reversing the implication. Small-step chains can cross a cut even when there is no point of the set at the cut, so the existence of chains does not by itself rule out a separation.
The Clopen Criterion and the closure theorem are useful in spaces where no order-based description is available. The first turns connectedness into a question about subsets and their complements. The second allows limit points to be added without losing connectedness. Alongside these tools, remember that continuous images of connected sets are connected, as established in the previous tutorial: this often supplies connected examples in metric spaces even when their geometry is not interval-like.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why does a separation give a nonempty proper subset that is both relatively open and relatively closed?
- In the closure theorem, why must each nonempty relatively open piece of \(\overline{E}\) meet \(E\)?
- In the fine-chain proof, why are both the chain-reachable set and its complement relatively open?
- How can the rationals have arbitrarily fine chains between all pairs of points and still be disconnected?
- Which hypotheses and metric-space construction make the discrete-metric example work?