Two Ways to Describe a Set That Cannot Be Split
For subsets of the real line, connectedness and path connectedness give the same classification: both describe intervals. That agreement comes from the order structure of \(\mathbb{R}\) and from the fact that straight-line interpolation stays inside an interval. In the plane, the two ideas separate. A set can be connected as a whole even though some of its points cannot be joined by a continuous path that stays in the set.
We use the usual Euclidean topology on \(\mathbb{R}^2\). A separation of a set \(E\subseteq\mathbb{R}^2\) is a representation \(E=A\cup B\), where \(A\) and \(B\) are disjoint, nonempty, and relatively open in \(E\). As before, \(E\) is connected if it has no separation. A path in \(E\) from \(p\) to \(q\) is a continuous function \(\gamma:[0,1]\to E\) with \(\gamma(0)=p\) and \(\gamma(1)=q\); \(E\) is path connected if every pair of its points can be joined by such a path.
Proof. Suppose \(E\) is path connected, and assume for contradiction that \(E=A\cup B\) is a separation. Choose \(p\in A\) and \(q\in B\). There is a path \(\gamma:[0,1]\to E\) from \(p\) to \(q\). The preimages \(\gamma^{-1}(A)\) and \(\gamma^{-1}(B)\) are relatively open in \([0,1]\), because \(A\) and \(B\) are relatively open in \(E\) and \(\gamma\) is continuous. They are disjoint and cover \([0,1]\). Both are nonempty, since \(0\in\gamma^{-1}(A)\) and \(1\in\gamma^{-1}(B)\). They would therefore give a separation of \([0,1]\), contradicting the fact that intervals are connected. Thus \(E\) is connected. \(\square\)
This implication does not require a path to be one-to-one, or to move in any particular direction. The essential fact is that a path is continuous: a separation of its image would pull back to a separation of the parameter interval. The reverse implication, however, does not hold in general.
Examples Where the Two Notions Agree
On the real line, the Characterization of Connected Subsets of \(\mathbb{R}\) and the Path Connectedness Characterizes Intervals in \(\mathbb{R}\) theorem already establish that connectedness and path connectedness are equivalent. For example, \([4,9)\) is an interval, so it is connected and path connected. Given \(x,y\in[4,9)\), the path \(\gamma(t)=(1-t)x+ty\) stays between \(x\) and \(y\), and hence remains in \([4,9)\).
Worked Example: Paths in an Open Disk
Let \(D=\{(x,y)\in\mathbb{R}^2:x^2+y^2<9\}\), the open disk of radius \(3\) centered at the origin. Take any \(p,q\in D\) and set \(\gamma(t)=(1-t)p+tq\) for \(0\leq t\leq1\). This function is continuous, with \(\gamma(0)=p\) and \(\gamma(1)=q\).
To check that it stays inside the disk, use the triangle inequality for the Euclidean norm: \(\|\gamma(t)\|\leq(1-t)\|p\|+t\|q\|\). Since \(\|p\|<3\), \(\|q\|<3\), and \(0\leq t\leq1\), the right-hand side is strictly less than \(3\). Thus \(\gamma(t)\in D\) for every \(t\), so \(D\) is path connected. The theorem above then gives that \(D\) is connected as well.
The disk example works because the line segment between any two of its points stays in the disk. This is a useful sufficient condition: if a set in the plane contains the line segment between every pair of its points, the straight-line formula gives paths between all pairs. But connected sets need not have this property, and they need not admit other paths either.
A Connected Set With No Path Across It
A classic example is the topologist’s sine curve. Start with the graph of \(y=\sin(1/x)\) for \(0<x\leq1\), then include all the limiting points on the vertical segment at \(x=0\). Write
As \(x\) approaches \(0\) from above, \(\sin(1/x)\) oscillates between \(-1\) and \(1\) without settling down. The vertical segment includes all possible limiting values of that oscillation. We will show that \(T\) is connected but that no path in \(T\) joins \((0,0)\) to \((1,\sin 1)\).
Proof of connectedness. First, \(T\) is the closure of \(G\) in \(\mathbb{R}^2\). To see why, a limit of points of \(G\) cannot have a negative first coordinate or a first coordinate greater than \(1\). If its first coordinate is \(x>0\), continuity of \(x\mapsto\sin(1/x)\) forces its second coordinate to be \(\sin(1/x)\). If its first coordinate is \(0\), its second coordinate must lie in \([-1,1]\), since every graph point has second coordinate in that interval.
Conversely, for any \(y\in[-1,1]\), choose \(\theta\in[-\pi/2,\pi/2]\) with \(\sin\theta=y\). For all sufficiently large positive integers \(n\), \(x_n=1/(\theta+2\pi n)\) lies in \((0,1]\), and \(\sin(1/x_n)=\sin(\theta+2\pi n)=y\). Hence \((x_n,y)\in G\) and \((x_n,y)\to(0,y)\). This verifies that every point on the vertical segment is in the closure of \(G\), so \(\overline{G}=T\).
The function \(f:(0,1]\to\mathbb{R}^2\) defined by \(f(x)=(x,\sin(1/x))\) is continuous, and its image is \(G\). The interval \((0,1]\) is connected. A continuous image of a connected set is connected: if the image had a separation, its two pieces would have disjoint, nonempty, relatively open preimages that separate the domain. Thus \(G\) is connected.
The closure of a connected set is connected, in \(\mathbb{R}^2\) as well as in \(\mathbb{R}\). Here is the argument. If \(\overline{G}\) had a separation \(A\cup B\), then each nonempty relatively open piece would meet \(G\), because \(G\) is dense in \(\overline{G}\). The sets \(A\cap G\) and \(B\cap G\) would be disjoint, nonempty, relatively open in \(G\), and cover \(G\), contradicting its connectedness. Therefore \(T=\overline{G}\) is connected.
Proof that \(T\) is not path connected. Suppose a path \(\gamma:[0,1]\to T\) joined \((0,0)\) to \((1,\sin 1)\). Write its coordinates as \(\gamma(t)=(u(t),v(t))\). Both coordinate functions are continuous; \(u(0)=0\) and \(u(1)=1\). Whenever \(u(t)>0\), the point \(\gamma(t)\) must lie on the graph \(G\), so \(v(t)=\sin(1/u(t))\).
There is an interval of times immediately before \(1\) on which \(u(t)>0\), by continuity and \(u(1)=1\). Let \(a\) be the left endpoint of the interval of positive values that reaches \(1\). Then \(0\leq a<1\), \(u(a)=0\), and \(u(t)>0\) for \(a<t\leq1\). Choose \(t_*\) with \(a<t_*<1\); in particular, \(u(t_*)>0\).
For each sufficiently large integer \(n\), consider the two positive numbers \(\displaystyle r_n^+=1/(\pi/2+2\pi n)\) and \(\displaystyle r_n^-=1/(3\pi/2+2\pi n)\). Both tend to \(0\), and \(\sin(1/r_n^+)=1\), while \(\sin(1/r_n^-)=-1\). For large \(n\), each number is less than \(u(t_*)\). Since \(u\) is continuous on \([a,t_*]\), with \(u(a)=0\), the Intermediate Value Theorem gives times \(s_n^+,s_n^-\in(a,t_*)\) such that \(u(s_n^+)=r_n^+\) and \(u(s_n^-)=r_n^-\).
These times approach \(a\). Indeed, for any \(\delta>0\) with \(a+\delta\leq t_*\), \(u\) is positive on the compact interval \([a+\delta,t_*]\). By the Extreme Value Theorem it has a positive minimum there. Once \(r_n^+\) and \(r_n^-\) are smaller than that minimum, neither corresponding time can lie in \([a+\delta,t_*]\). Thus \(s_n^+\to a\) and \(s_n^-\to a\). But the graph identity gives \(v(s_n^+)=1\) and \(v(s_n^-)=-1\) for every such \(n\). Continuity of \(v\) at \(a\) would require both sequences of values to converge to \(v(a)\), which is impossible. Therefore no such path exists, and \(T\) is not path connected. \(\square\)
Worked Example: Why the Oscillation Blocks a Path
The obstruction in the proof is not merely that the graph oscillates near \(x=0\). Oscillation alone does not prove that a path is impossible; the path might, in principle, approach the vertical segment in some other way. The coordinate argument rules this out. Once a path moves from first coordinate \(0\) to first coordinate \(1\), it must pass through positive first-coordinate values approaching \(0\). At each such value \(x\), the second coordinate is forced to be \(\sin(1/x)\).
For example, \(x=1/(\pi/2+2\pi n)\) forces the second coordinate to equal \(1\), while \(x=1/(3\pi/2+2\pi n)\) forces it to equal \(-1\). Both sequences of first coordinates tend to \(0\). A continuous path approaching a time with first coordinate \(0\) would have a second coordinate tending to a single value there, not alternating between \(1\) and \(-1\). The contradiction depends on both the exact graph equation and continuity of the path.
What the Comparison Tells Us
Path connectedness is a stronger condition than connectedness: the theorem proves that every path-connected set is connected, while \(T\) shows that the reverse implication fails. In \(\mathbb{R}\), the two conditions coincide because connected subsets are intervals and straight-line paths join any two points of an interval. The planar example shows that this one-dimensional equivalence should not be treated as a general principle.
The same distinction matters when discussing components. A path component groups points that can be joined by paths; a connected component is a maximal connected subset. Every path component is path connected, hence connected, so it lies inside a connected component. But one connected component can contain more than one path component: the topologist’s sine curve is connected even though the points \((0,0)\) and \((1,\sin1)\) cannot be joined by a path.
A useful proof check is to ask what the argument actually supplies. A separation argument tests whether a set can be split into relatively open pieces. A path argument requires a continuous map from \([0,1]\) whose entire image stays in the set. Establishing the first does not automatically construct the second. In later work with metric spaces, this distinction will remain important: connectedness is about separations, whereas path connectedness is about joining points continuously.
Check Your Understanding
Use the definitions and arguments in this tutorial to answer the following questions.
- In the proof that path connectedness implies connectedness, why do the preimages of the two pieces of a separation give a contradiction?
- What feature of the open disk makes the straight-line path between any two of its points stay inside the disk?
- Why is the vertical segment \(\{0\}\times[-1,1]\) contained in the closure of the graph \(G\)?
- In the proof that \(T\) is not path connected, which two sequences force incompatible values for the second coordinate?
- Why does connectedness of \(T\) not imply that \((0,0)\) and \((1,\sin1)\) can be joined by a path?