Connected Components Can Contain Several Path Components
The previous tutorial established that every path-connected set is connected. Consequently, a path component lies inside a connected component. The converse question is more delicate: must all the points in a connected component be joinable by paths? In general, no. Connectedness rules out a separation, while path connectedness requires a continuous route between every pair of points. These conditions need not coincide.
We use \(C_X(x)\) for the connected component of \(x\) in a metric space \(X\), and \(P_X(x)\) for its path component. Earlier, the Components Partition Theorem and the Path Components Partition Theorem established that each kind of component partitions a nonempty space. The previous tutorial also established that \(P_X(x)\subseteq C_X(x)\). The first result below organizes these facts: a connected component is made up of the path components contained in it.
Proof. If \(y\in C_X(x)\), the Path Components Are Contained in Connected Components Corollary gives \(P_X(y)\subseteq C_X(y)\). Since \(y\in C_X(x)\), the Components Partition Theorem implies \(C_X(y)=C_X(x)\). Thus every set in the union on the right is contained in \(C_X(x)\), so the union is contained in \(C_X(x)\).
For the reverse inclusion, let \(z\in C_X(x)\). Every point belongs to its own path component, so \(z\in P_X(z)\). Since \(z\) is one of the indices in the union, \(z\) belongs to that union. Hence \(C_X(x)\) is contained in the union as well, proving equality. Finally, if a path component \(P_X(y)\) meets \(C_X(x)\), choose \(z\in P_X(y)\cap C_X(x)\). Points in the same path component have the same path component, so \(P_X(y)=P_X(z)\). The first part of the proof gives \(P_X(z)\subseteq C_X(x)\), as required. \(\square\)
This theorem gives a precise comparison. A connected component is path connected exactly when it consists of just one path component. In that case, if \(x\in X\), the component \(C_X(x)\) is path connected exactly when \(C_X(x)=P_X(x)\). This equality is not automatic: the theorem describes a connected component as a union, and that union may have more than one member.
Worked Example: Components in a Discrete Metric Space
Let \(X=\{a,b,c\}\) with the discrete metric \(d(u,v)=1\) whenever \(u\ne v\), and \(d(u,u)=0\). For each \(u\in X\), the ball \(B_{1/2}(u)\) is the singleton \(\{u\}\). Thus every singleton is open in \(X\). If a subset \(E\subseteq X\) contains at least two points, choose \(u\in E\). Then \(\{u\}\) and \(E\setminus\{u\}\) are disjoint, nonempty, relatively open subsets of \(E\) whose union is \(E\). Therefore \(E\) is disconnected.
It follows that the connected components are the singletons. A continuous path in \(X\) must also be constant: if a path \(\gamma:[0,1]\to X\) took distinct values, the inverse images of \(\{\gamma(0)\}\) and its complement would separate \([0,1]\). Hence the path components are also singletons. Here every connected component consists of exactly one path component.
A Connected Set Need Not Be Path Connected
To see why connected components can be larger than path components, consider the topologist’s sine curve: $$ S=\{(t,\sin(1/t)):0<t\leq 1\}\cup(\{0\}\times[-1,1])\subseteq\mathbb{R}^2. $$ The first part is the graph of a continuous function on an interval. As \(t\) tends to zero, the graph oscillates through all heights between \(-1\) and \(1\). The vertical segment records the limit points created by this oscillation.
Worked Example: The Graph Is Connected
Write \(G=\{(t,\sin(1/t)):0<t\leq1\}\). The map \(f:(0,1]\to\mathbb{R}^2\) given by \(f(t)=(t,\sin(1/t))\) is continuous, since both coordinate functions are continuous on \((0,1]\). The interval \((0,1]\) is connected, so the Continuous Images of Connected Spaces Are Connected Theorem shows that \(G=f[(0,1]]\) is connected.
We check that \(\overline{G}=S\). If a sequence \((t_n,\sin(1/t_n))\) in \(G\) converges to \((a,b)\), then \(0\leq a\leq1\) and \(-1\leq b\leq1\). If \(a>0\), continuity of \(t\mapsto\sin(1/t)\) at \(a\) gives \(b=\sin(1/a)\), so \((a,b)\in G\). If \(a=0\), then \((a,b)\in\{0\}\times[-1,1]\). This proves \(\overline G\subseteq S\), using the Sequential Characterization of Closure.
Conversely, every point of \(G\) is in \(\overline G\). For any \(b\in[-1,1]\), choose \(\theta\in[-\pi/2,\pi/2]\) with \(\sin\theta=b\). For all sufficiently large positive integers \(n\), set \(t_n=1/(2\pi n+\theta)\). Then \(0<t_n\leq1\), \(t_n\to0\), and $$ \sin(1/t_n)=\sin(2\pi n+\theta)=\sin\theta=b. $$ Thus \((t_n,\sin(1/t_n))\to(0,b)\), proving that every point of the vertical segment belongs to \(\overline G\). Therefore \(\overline G=S\). By the Closure of a Connected Set Theorem, \(S\) is connected.
Connectedness alone does not show whether two chosen points can be joined by a path. For this set, the oscillations near the vertical segment prevent any path from joining a point of that segment to a point of the graph. The details matter: a continuous path might move back and forth in its first coordinate, so simply asserting that its height oscillates is not enough. We isolate an interval on which the first coordinate is positive and use continuity at the endpoint of that interval.
Proof. Connectedness was proved above. Choose \(p\in\{0\}\times[-1,1]\) and \(q\in G\). Suppose there were a continuous path \(\gamma:[0,1]\to S\) with \(\gamma(0)=p\) and \(\gamma(1)=q\). Write \(\gamma(t)=(x(t),y(t))\). Both coordinate functions are continuous, \(x(0)=0\), and \(x(1)>0\).
The set \(U=\{t\in[0,1]:x(t)>0\}\) is relatively open in \([0,1]\). Consider the interval component of \(U\) containing \(1\). It has a left endpoint \(a<1\), with \(x(a)=0\), and \(x(t)>0\) for \(a<t\leq1\). Choose \(c\) with \(a<c<1\); then \(x(c)>0\). For every \(r\) with \(0<r<x(c)\), continuity and the Intermediate Value Theorem give at least one \(t\in[a,c]\) with \(x(t)=r\). Let \(\tau(r)\) be the least such \(t\); it exists because the set of such \(t\) is a nonempty closed subset of the compact interval \([a,c]\).
We claim that \(\tau(r)\to a\) as \(r\to0\) through positive values. Given \(\varepsilon>0\), choose \(0<\delta<\min(\varepsilon,c-a)\). Since \(x(a+\delta)>0\), whenever \(0<r<x(a+\delta)\), the Intermediate Value Theorem applied on \([a,a+\delta]\) gives a time \(t\leq a+\delta\) with \(x(t)=r\). By the definition of the least such time, \(a\leq\tau(r)\leq a+\delta<a+\varepsilon\). This proves the claim.
For every \(t\in(a,c]\), the point \(\gamma(t)\) has positive first coordinate and belongs to \(S\), so it lies on the graph \(G\). Therefore \(y(t)=\sin(1/x(t))\). For all sufficiently large \(n\), the positive numbers $$ r_n^+=\frac{1}{\pi/2+2\pi n} \qquad\text{and}\qquad r_n^-=\frac{1}{3\pi/2+2\pi n} $$ are both less than \(x(c)\), and both tend to zero. By construction, $$ y(\tau(r_n^+))=\sin(\pi/2+2\pi n)=1, \qquad y(\tau(r_n^-))=\sin(3\pi/2+2\pi n)=-1. $$ The times in both sequences tend to \(a\), but their \(y\)-coordinates do not converge to the same value. This contradicts continuity of \(y\) at \(a\). No such path exists. Since \(p\) and \(q\) cannot be joined by a path, \(S\) is not path connected. Finally, \(S\) is connected, so its connected component is all of \(S\), whereas its path component containing \(p\) does not contain \(q\). \(\square\)
Reading the Relationship Correctly
The examples show both possibilities. In the discrete metric example, connected and path components coincide. In the topologist’s sine curve, the whole set is one connected component but contains more than one path component. In general, the Connected Components Are Unions of Path Components Theorem says that each connected component is assembled from path components; it does not say that these pieces can be joined to one another by paths.
A common pitfall is to reverse the implication from the previous tutorial. Path connectedness implies connectedness, but a proof of connectedness does not automatically produce paths. Another is to assume that a path can be constructed by tracing the visible shape of a set. A path must be a continuous map on the entire interval, and in examples with infinitely rapid oscillation, continuity can rule out an apparently plausible route.
When comparing the two kinds of components, first identify which relation is being used: connected subsets determine connected components, while paths determine path components. Then use the established containment \(P_X(x)\subseteq C_X(x)\). Equality holds precisely when the connected component has no additional path components; without further hypotheses, the topologist’s sine curve shows that strict containment can occur.
Check Your Understanding
Use the component comparison and the topologist’s sine curve argument to answer these questions.
- Why does every path component that meets a connected component lie wholly inside that connected component?
- What condition makes a connected component equal to the path component of one of its points?
- Why is the graph \(G=\{(t,\sin(1/t)):0<t\leq1\}\) connected?
- In the non-path-connectedness proof, why do the selected times \(\tau(r)\) approach the left endpoint \(a\)?
- Which continuity requirement is contradicted by the alternating values \(1\) and \(-1\)?