From Local Paths to a Path Across the Space
A space can have paths near every point without having a path between every pair of points: the paths might remain confined to separate regions. The key extra condition is connectedness. We will show that in a locally path-connected space, each region accessible by paths is open. These regions partition the space, so connectedness rules out having more than one.
A path in a space \(X\) is a continuous map \(\gamma:I\to X\), where \(I=[0,1]\) has its usual topology. Its initial and final points are \(\gamma(0)\) and \(\gamma(1)\). A subset \(A\subseteq X\) is path-connected if for every \(x,y\in A\) there is a path in \(A\) from \(x\) to \(y\). A space is connected if it cannot be written as the union of two disjoint, nonempty open sets.
For \(x,y\in X\), write \(x\sim y\) when there is a path in \(X\) from \(x\) to \(y\). The equivalence classes of this relation are called the path components of \(X\). The relation is reflexive because the constant path at \(x\) joins \(x\) to itself. It is symmetric because reversing a path \(\gamma\), by using \(t\mapsto\gamma(1-t)\), reverses its endpoints. It is transitive because two paths with a matching endpoint can be concatenated: traverse the first on \([0,1/2]\) and the second on \([1/2,1]\), with each parameter rescaled linearly. The resulting map is continuous by the pasting lemma for two closed subintervals of \(I\). Thus path components form a disjoint partition of \(X\).
Why Local Path Connectivity Makes Path Components Open
Proof. Let \(P\) be a path component and let \(x\in P\). By local path connectivity, there is an open path-connected set \(V\) with \(x\in V\). For every \(y\in V\), path connectivity gives a path in \(V\), and hence in \(X\), from \(x\) to \(y\). Thus \(y\) belongs to the same path component as \(x\), so \(V\subseteq P\). Every point \(x\in P\) therefore has an open neighborhood contained in \(P\), which proves that \(P\) is open. \(\square\)
The argument needs only one path-connected neighborhood of each point, contained in its path component. Local path connectivity supplies exactly that. Once the path components are known to be open, their partition has a useful additional feature: the complement of any one component is a union of other open components, and is therefore open too.
Proof. If \(X=\varnothing\), path connectivity holds vacuously, because there are no pairs of points to join. Suppose \(X\ne\varnothing\). Its path components partition \(X\), and the proposition shows that every component is open. If there were at least two components, choose one of them, \(P\). Then \(P\) is nonempty and open. The union \(Q\) of all the other components is also open, is nonempty, and is disjoint from \(P\). Since the components partition \(X\), \(P\cup Q=X\). This contradicts connectedness. Hence \(X\) has exactly one path component. Every pair of points in \(X\) lies in that component, so a path joins them. Therefore \(X\) is path-connected. \(\square\)
In this proof, openness is the link between the local hypothesis and the global conclusion. Connectedness by itself does not say that arbitrary regions determined by paths cannot exist. It rules out a partition into disjoint nonempty open sets; local path connectivity ensures that the path components provide just such open sets if there is more than one.
Worked Examples
Worked Example: The Closed Unit Interval
Consider \(I=[0,1]\) with the subspace topology inherited from \(\mathbb{R}\). The interval is connected by the standard connectedness theorem for real intervals. We verify local path connectivity and then exhibit paths explicitly.
Let \(x\in I\) and let \(U\) be an open neighborhood of \(x\) in \(I\). By the subspace topology, there is an open set \(O\subseteq\mathbb{R}\) with \(U=O\cap I\). Since \(x\in O\), there is \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq O\). The set \(V=(x-\varepsilon,x+\varepsilon)\cap I\) is open in \(I\), contains \(x\), and lies in \(U\). It is an interval, possibly including 0 or 1. For any \(a,b\in V\), the map \(\gamma(t)=(1-t)a+tb\) takes values between \(a\) and \(b\), so it lies in \(V\) and joins \(a\) to \(b\). Thus \(V\) is path-connected, and \(I\) is locally path-connected.
For any \(a,b\in I\), the same formula gives a path from \(a\) to \(b\). Indeed, \(\gamma(0)=(1-0)a+0b=a\), \(\gamma(1)=(1-1)a+1b=b\), and every value is in \(I\). This direct check agrees with the theorem. If \(a=b\), the formula is the constant path \(\gamma(t)=a\), so the coincident-endpoint case is included.
Check: For \(a=1/4\), \(b=3/4\), and \(t=1/2\), the path gives \(\gamma(1/2)=(1/2)(1/4)+(1/2)(3/4)=1/8+3/8=1/2\), which lies in \(I\). Its endpoint values are \(1/4\) and \(3/4\), as required.
Worked Example: Removing a Point from the Real Line
Let \(X=\mathbb{R}\setminus\{0\}\) with the subspace topology from the standard topology on \(\mathbb{R}\). This space is locally path-connected but not connected, and it is not path-connected.
To check local path connectivity, take \(x\in X\) and an open neighborhood \(U\) of \(x\) in \(X\). There is an open set \(O\subseteq\mathbb{R}\) with \(U=O\cap X\). Choose \(\varepsilon>0\) small enough that \((x-\varepsilon,x+\varepsilon)\subseteq O\) and \(\varepsilon<|x|\). Then \(V=(x-\varepsilon,x+\varepsilon)\) lies in \(X\), is open in \(X\), and is contained in \(U\). It is an interval, so the linear path between any two of its points stays in \(V\). Hence \(X\) is locally path-connected.
The sets \(P=(-\infty,0)\) and \(Q=(0,\infty)\) are disjoint, nonempty, open in \(X\), and have union \(X\). Therefore \(X\) is disconnected. They are also path-connected: for two points in either interval, every value of their linear path lies between those points and hence remains in that interval.
No path in \(X\) can join \(-1\) to \(1\). If \(\gamma:I\to X\) did so, its composite with the inclusion \(X\subseteq\mathbb{R}\) would be a continuous real-valued function taking values \(-1\) and \(1\) at the endpoints. The intermediate value theorem would give \(t\in I\) with \(\gamma(t)=0\), contradicting \(\gamma(t)\in X\). Thus \(X\) has more than one path component, consistent with the theorem: it fails the connectedness hypothesis.
Worked Example: A Connected Space That Is Not Locally Path-Connected
Consider the topologist’s sine curve \(S=\{(x,\sin(1/x)):0<x\leq 1\}\cup(\{0\}\times[-1,1])\) with the subspace topology inherited from the Euclidean plane. This example illustrates why local path connectivity is also needed. We verify that \(S\) is connected but not path-connected.
Set \(G=\{(x,\sin(1/x)):0<x\leq1\}\). The map \(x\mapsto(x,\sin(1/x))\) is continuous on the connected interval \((0,1]\), so its image \(G\) is connected. Here we use the standard fact that continuous images of connected spaces are connected: a separation of the image would pull back to a separation of the domain. The closure of \(G\) in the plane is exactly \(S\). For positive first coordinate, continuity identifies the only possible closure point at coordinate \(x\) as \((x,\sin(1/x))\); at first coordinate 0, every limit has second coordinate in \([-1,1]\). Conversely, for each \(y\in[-1,1]\), the oscillations of \(\sin(1/x)\) give a sequence \(x_n\to0\) with \(\sin(1/x_n)=y\), so \((0,y)\) is in the closure.
The closure of a connected set is connected. To see this, suppose its closure were the union of disjoint nonempty relatively open sets \(A\) and \(B\). Since \(G\) is dense in its closure, each nonempty relatively open set meets \(G\); thus \(A\cap G\) and \(B\cap G\) would separate \(G\), contradicting its connectedness. It follows that \(S=\overline{G}\) is connected.
Now suppose a path \(\gamma(t)=(u(t),v(t))\) in \(S\) joined a point of the vertical segment to a point of \(G\), with the vertical point at \(t=0\). The function \(u\) is continuous, \(u(0)=0\), and \(u(1)>0\). The part of the set \(\{t\in I:u(t)>0\}\) containing 1 has the form \((a,1]\) for some \(a\in[0,1)\), and continuity gives \(u(a)=0\). Choose \(s_n\in(a,1]\) with \(s_n\to a\). For each \(n\), choose \(c_n\) with \(0<c_n<u(s_n)\), \(c_n<1/n\), and \(\sin(1/c_n)=(-1)^n\); such values exist because the sine function takes each of \(1\) and \(-1\) at arbitrarily large arguments. By the intermediate value theorem, there is \(t_n\in[a,s_n]\) with \(u(t_n)=c_n\). Since \(t_n\to a\), continuity of \(v\) would imply \(v(t_n)\to v(a)\). But \(v(t_n)=\sin(1/c_n)=(-1)^n\), which does not converge. This contradiction rules out such a path. So \(S\) is not path-connected.
The theorem now shows that \(S\) cannot be locally path-connected: if it were, its connectedness would force path connectivity. This example shows that connectedness alone does not give the conclusion.
A Common Pitfall
Common pitfall. It is not enough to know that every point has some path-connected neighborhood. To prove that a path component is open, the neighborhood must be contained in that point’s path component. Local path connectivity provides an open path-connected neighborhood inside any chosen open neighborhood. Without that containment, the argument that the entire component is open has not been established.
There is also a distinction between the two global hypotheses. In \(\mathbb{R}\setminus\{0\}\), local path connectivity holds but connectedness fails, and there are two path components. In the topologist’s sine curve, connectedness holds but local path connectivity fails, and there is more than one path component. The theorem uses both conditions for different steps: local path connectivity makes the path components open, and connectedness forces the open partition to have only one member.
Check Your Understanding
In each question, attend separately to openness, connectedness, and the existence of paths.
- Explain why the path components of a locally path-connected space are not only open but also closed.
- Give the two path components of \(\mathbb{R}\setminus\{0\}\) and state which hypothesis of the theorem fails.
- In the theorem’s proof, why is the union of all path components other than \(P\) open?
- Show directly that any two points in \((0,1)\), with its subspace topology from \(\mathbb{R}\), can be joined by a path in \((0,1)\).
- Why does the theorem not imply that every connected space is path-connected?