Connected, but Not Path-Connected
Path-connectedness is a strong way for points to belong to the same connected piece: any two points can be joined by a continuous path. Connectedness asks for less. It rules out a decomposition into two disjoint, nonempty relatively open sets, but it does not promise paths between points. The topologist’s sine curve makes the difference explicit: it is connected, yet it has two path components.
Give \(\mathbb{R}^2\) its Euclidean topology, and give each of its subsets the subspace topology. Define the graph \(G\), the vertical segment \(V\), and their union \(T\) by
The set \(T\) is called the topologist’s sine curve. The graph oscillates infinitely often as \(x\) approaches \(0\), while \(V\) fills in the vertical segment at \(x=0\).
Every path-connected space is connected, but the converse does not hold. Here we will show why \(T\) is connected and why paths in it nevertheless cannot join a point of \(G\) to a point of \(V\).
Why the Whole Curve Is Connected
We use two facts about connectedness, with their justifications. First, a continuous image of a connected space is connected. Indeed, if the image had a separation into disjoint, nonempty relatively open sets, their preimages would be disjoint, nonempty, relatively open sets separating the domain. Second, the closure of a connected subset is connected:
Proof. Suppose, to the contrary, that \(\overline{A}=C\cup F\) is a separation: \(C\) and \(F\) are disjoint, nonempty, relatively open subsets of \(\overline{A}\). Each of \(C\) and \(F\) must meet \(A\). For example, if \(C\cap A=\varnothing\), then \(F\) is relatively closed as well as relatively open in \(\overline{A}\), so it contains the closure in \(\overline{A}\) of \(A\). That closure is all of \(\overline{A}\), contradicting that \(C\) is nonempty and disjoint from \(F\). The same reasoning shows \(F\cap A\ne\varnothing\). Now \(C\cap A\) and \(F\cap A\) are disjoint, nonempty, relatively open subsets of \(A\), and their union is \(A\). This separates \(A\), contradicting its connectedness. Therefore \(\overline{A}\) is connected. \(\square\)
The interval \((0,1]\) is connected, and the map \(x\mapsto(x,\sin(1/x))\) from \((0,1]\) to \(\mathbb{R}^2\) is continuous. Thus its image \(G\) is connected. We can now identify its closure exactly.
Worked Example: Computing the Closure of the Graph
We show that \(\overline{G}=T\), where the closure is taken in \(\mathbb{R}^2\).
First, every point of \(G\) has \(0<x\leq1\) and \(-1\leq\sin(1/x)\leq1\). Since the rectangle \([0,1]\times[-1,1]\) is closed, \(\overline{G}\) is contained in that rectangle. If \((x,y)\in\overline{G}\) has \(x>0\), then points of \(G\) approaching \((x,y)\) have first coordinate approaching \(x\). Continuity of \(u\mapsto\sin(1/u)\) at \(x>0\) forces \(y=\sin(1/x)\). Thus every such point is in \(G\). Any remaining point of \(\overline{G}\) has first coordinate \(0\), so it belongs to \(V\). This proves \(\overline{G}\subseteq G\cup V\).
For the reverse inclusion, \(G\subseteq\overline{G}\) by definition. Fix any \(y\in[-1,1]\), and let \(\arcsin y\in[-\pi/2,\pi/2]\) be the angle whose sine is \(y\). For each \(n\in\mathbb{N}\), set \(\theta_n=2\pi n+\arcsin y\) and \(x_n=1/\theta_n\). Since \(n\geq1\), \(\theta_n\geq2\pi-\pi/2=3\pi/2>1\), so \(0<x_n\leq1\). Also, \(\sin(1/x_n)=\sin(\theta_n)=y\), and \(x_n\to0\). Therefore \((x_n,\sin(1/x_n))=(x_n,y)\to(0,y)\). Every point of \(V\) lies in \(\overline{G}\), proving \(G\cup V\subseteq\overline{G}\).
Check: both inclusions give \(\overline{G}=G\cup V=T\). Since \(G\) is connected, the closure theorem proves that \(T\) is connected.
So the added vertical segment does not split the graph into separate connected pieces. It is precisely part of the graph’s closure, and taking that closure preserves connectedness.
Why No Path Joins the Two Parts
Within \(G\), any two points can be joined by a path: move the first coordinate linearly between their positive \(x\)-coordinates and use the graph formula for the second coordinate. Within \(V\), straight-line segments give paths. The obstruction is between the two sets.
Proof. Suppose there were a path \(\gamma(t)=(u(t),v(t))\), \(0\leq t\leq1\), with \(\gamma(0)\in V\) and \(\gamma(1)\in G\). Then \(u\) and \(v\) are continuous, \(u(0)=0\), and \(u(1)>0\). The set of zeros of \(u\) in \([0,1]\) is nonempty and closed, so it has a largest element \(s\). Because \(u(1)>0\), we have \(s<1\). For \(t>s\), \(u(t)\ne0\); since \(u\) is continuous and \(u(1)>0\), it follows that \(u(t)>0\) throughout \((s,1]\). Consequently \(\gamma(t)\in G\) there and \(v(t)=\sin(1/u(t))\).
For \(n\in\mathbb{N}\), define positive numbers \(\alpha_n=1/(\pi/2+2\pi n)\) and \(\beta_n=1/(3\pi/2+2\pi n)\). Both sequences tend to \(0\), and \(\sin(1/\alpha_n)=1\), while \(\sin(1/\beta_n)=-1\). For all sufficiently large \(n\), both numbers are less than \(u(1)\). By continuity of \(u\), each value \(\alpha_n\) and \(\beta_n\) is attained between \(s\) and \(1\). Choose corresponding times \(t_n,r_n\) in \((s,1]\). These times tend to \(s\): for any \(\delta>0\) with \(s+\delta\leq1\), \(u\) is positive on the compact interval \([s+\delta,1]\), so it has a positive minimum there. Once \(\alpha_n\) or \(\beta_n\) is below this minimum, its corresponding time must lie in \((s,s+\delta)\). It follows that \(t_n\to s\) and \(r_n\to s\). But \(v(t_n)=1\) and \(v(r_n)=-1\) for every chosen \(n\), contradicting continuity of \(v\) at \(s\), which requires both sequences of values to converge to \(v(s)\). Hence no such path exists. Reversing a path would also reverse its endpoints, so no path joins a point of \(G\) to a point of \(V\) in either direction. \(\square\)
Worked Example: Finding the Path Components
We determine the path components of \(T\). The projection \((x,y)\mapsto x\) shows that every point of \(T\) belongs to exactly one of \(G\) and \(V\). Points in either part can be joined within that part, as follows.
For \(p=(a,\sin(1/a))\) and \(q=(b,\sin(1/b))\) in \(G\), where \(0<a,b\leq1\), define \(\gamma(t)=((1-t)a+tb,\sin(1/((1-t)a+tb)))\). The first coordinate is positive for all \(t\in[0,1]\), so this is a continuous path in \(G\) from \(p\) to \(q\). For \(p=(0,c)\) and \(q=(0,d)\) in \(V\), the path \(\eta(t)=(0,(1-t)c+td)\) stays in \(V\), since \(-1\leq c,d\leq1\) implies \(-1\leq(1-t)c+td\leq1\).
The theorem just proved rules out paths between the two parts. Hence the path components of \(T\) are exactly \(G\) and \(V\). But \(T\) itself is connected, as shown above.
Check: the two sets are disjoint and cover \(T\); paths connect every pair within each set; and no path crosses between them. Those facts establish maximality as well as path-connectedness of each component.
What Connectedness Still Guarantees
The example separates two kinds of information. Path-connectedness organizes points according to whether paths can join them. Connectedness says that the entire space cannot be separated by relatively open sets. The existence of two path components does not by itself give a separation.
Worked Example: Why the Path Components Are Not a Separation
In \(T\), the set \(V\) is closed relative to \(T\): it is the intersection of \(T\) with the closed line \(\{0\}\times\mathbb{R}\). The set \(G\) is not closed relative to \(T\), because its closure in \(T\) is \(T\), not \(G\). In fact \(G\) is relatively open: for a point \((a,\sin(1/a))\in G\), choose an open disk centered there with radius less than \(a\). Every point of \(T\) in that disk has positive first coordinate, so lies in \(G\).
Thus \(G\) and \(V\) are disjoint, nonempty path components, but they are not both relatively open. In particular they do not form a separation. The calculation \(\overline{G}=T\) also shows directly why \(G\) cannot be a closed part separated from \(V\): points of \(V\) are limits of points of \(G\).
Check: a separation requires both pieces to be relatively open, whereas here \(V\) is not relatively open. Indeed, any relative neighborhood of \((0,y)\in V\) contains points of \(G\), by the explicit approximating sequence in the closure calculation.
More generally, connectedness rules out a nonconstant continuous map from the space to the discrete two-point space \(\{0,1\}\). If such a map existed, the preimages of \(\{0\}\) and \(\{1\}\) would be disjoint, open, nonempty sets covering the domain; they would form a separation. This illustrates what connectedness guarantees: continuous data with two separated discrete values cannot distinguish two nonempty open pieces of a connected space.
Common pitfall. A tempting but incorrect argument is: “There are two path components, \(G\) and \(V\), so \(T\) is disconnected.” The conclusion does not follow: path components need not form a separation. Here the mistaken argument overlooks the closure computation \(\overline{G}=T\), which shows that \(G\) is not closed in \(T\), and overlooks that \(V\) is not relatively open. The correct conclusion is that \(T\) is connected but not path-connected, with one connected component and two path components.
The example also warns against inferring paths from density. The graph \(G\) is dense in \(T\), and \(G\) is path-connected, but \(T\) is not path-connected. Density says that every open neighborhood of every point of \(T\) meets \(G\); it does not supply a continuous path from a point of \(G\) to a point added in its closure.
Check Your Understanding
Use the definitions and the arguments above to answer these questions.
- Why does the continuous image of \((0,1]\) give a connected graph \(G\)?
- For \(y\in[-1,1]\), verify that \(\theta_n=2\pi n+\arcsin y\) gives graph points approaching \((0,y)\).
- Which two sets are the path components of \(T\), and what argument rules out a path between them?
- Why do the two path components fail to be a separation of \(T\)?
- State one property connectedness guarantees that path-connectedness also guarantees, and explain how connectedness differs from a guarantee of paths.