Path-Connectedness Near a Point
A space can fail to be path-connected as a whole while still having a well-behaved path-connected neighborhood around each of its points. Local path-connectedness records precisely this kind of near-point behavior. The condition is about finding a suitable smaller neighborhood inside each given neighborhood; it does not require every neighborhood to be path-connected.
Throughout, a neighborhood means an open set containing the point in question. For a subset \(A\) of a topological space \(X\), paths in \(A\) are understood using the subspace topology on \(A\). The interval \(I=[0,1]\) has the subspace topology inherited from \(\mathbb{R}\) with its standard topology.
The distinction between “at \(x\)” and “in \(X\)” is important: the first condition concerns one point, while the second requires the condition at all points. Also, the chosen set \(V\) must be open in \(X\), not merely open as a subset of \(U\). Since \(U\) itself is open in \(X\), a set open in \(U\) is open in \(X\), but the definition directly asks for an open neighborhood in the whole space.
Intervals in the Real Line
Worked Example: Intervals and Their Endpoints
Let \(J\subseteq\mathbb{R}\) be any interval with at least two points, equipped with the subspace topology. This includes open, closed, and half-open intervals, as well as unbounded intervals. We verify that \(J\) is locally path-connected, including at its endpoints when they belong to \(J\).
Fix \(x\in J\), and let \(U\) be an open neighborhood of \(x\) in \(J\). By the definition of the subspace topology, there is an open set \(G\subseteq\mathbb{R}\) such that \(U=J\cap G\). Since \(x\in G\) and \(G\) is open in the standard topology, there is \(r>0\) such that \((x-r,x+r)\subseteq G\). Set \(V=J\cap(x-r,x+r)\). This set contains \(x\), is contained in \(U\), and is open in \(J\), so it is an open neighborhood of \(x\) in \(J\).
To check path-connectedness, take any \(y,z\in V\). For \(t\in I\), define \(\gamma(t)=(1-t)y+tz\). Because \(J\) is an interval, every point between \(y\) and \(z\) belongs to \(J\). Also, since \(y,z\in(x-r,x+r)\), each convex combination \((1-t)y+tz\) lies between \(y\) and \(z\), hence also in \((x-r,x+r)\). Thus \(\gamma(t)\in V\) for every \(t\in I\). The map is continuous, and \(\gamma(0)=y\), \(\gamma(1)=z\). Therefore \(V\) is path-connected.
This argument does not require \(x\) to be an interior point of \(J\) in \(\mathbb{R}\). For example, if \(J=[0,1]\) and \(x=0\), the neighborhood \(V=[0,1]\cap(-r,r)\) works just as well. The endpoints are handled by taking neighborhoods relative to the interval.
Euclidean Open Sets
Worked Example: A Neighborhood Inside a Euclidean Open Set
Let \(n\geq 1\), and let \(\Omega\) be an open subset of \(\mathbb{R}^n\), with the subspace topology (which agrees with its inherited Euclidean topology). We show directly that \(\Omega\) is locally path-connected.
Fix \(x\in\Omega\) and an open neighborhood \(U\) of \(x\) in \(\Omega\). There is an open set \(G\subseteq\mathbb{R}^n\) with \(U=\Omega\cap G\). Since \(x\in\Omega\) and \(\Omega\) is open, choose \(r_1>0\) with \(B(x,r_1)\subseteq\Omega\). Since \(x\in G\) and \(G\) is open, choose \(r_2>0\) with \(B(x,r_2)\subseteq G\). Let \(r=\min(r_1,r_2)\). Then \(r>0\), and the open ball \(V=B(x,r)\) satisfies \(x\in V\subseteq\Omega\cap G=U\). It is open in \(\Omega\), because it is open in \(\mathbb{R}^n\) and contained in \(\Omega\).
For \(y,z\in V\), define \(\gamma(t)=(1-t)y+tz\). The Euclidean norm satisfies \(\|\gamma(t)-x\|=\|(1-t)(y-x)+t(z-x)\|\leq(1-t)\|y-x\|+t\|z-x\|\). Since \(y,z\in B(x,r)\), we have \(\|y-x\|<r\) and \(\|z-x\|<r\). Therefore \(\|\gamma(t)-x\|<(1-t)r+tr=r\), so \(\gamma(t)\in V\) for all \(t\in I\). This continuous path joins \(y\) to \(z\) inside \(V\), proving that \(V\) is path-connected.
The openness of \(\Omega\) matters in this argument: it lets us choose a ball around \(x\) that stays inside \(\Omega\). The conclusion concerns open subsets of Euclidean space, not arbitrary subspaces.
Isolated Points and the Pointwise Condition
Worked Example: An Isolated Point in a Discrete Space
Let \(X=\{p,q\}\) with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\). In particular, \(\{p\}\) and \(\{q\}\) are open. We check local path-connectedness at \(p\).
Let \(U\) be any open neighborhood of \(p\). Then \(V=\{p\}\) is open, contains \(p\), and is contained in \(U\). It is path-connected: the only pair of points to join is \(p\) to \(p\), and the constant map \(\gamma(t)=p\) is a continuous path in \(V\). Thus \(X\) is locally path-connected at \(p\). The same argument, with the constant path at \(q\), proves local path-connectedness at \(q\).
However, \(X\) is not path-connected. Any map \(f:I\to X\) that takes both values would have \(f^{-1}(\{p\})\) and \(f^{-1}(\{q\})\) as nonempty disjoint open sets whose union is \(I\), contradicting the connectedness of \(I\). This example separates the local condition from the global one: every point has a path-connected open neighborhood, but the whole space is not path-connected.
The same singleton argument works in any topological space at any isolated point. Here “isolated” means that \(\{x\}\) is open in \(X\). For every open neighborhood \(U\) of such a point, \(V=\{x\}\) is an open path-connected neighborhood contained in \(U\). In particular, every discrete space is locally path-connected, regardless of its cardinality.
Two Results About the Local Condition
Proof. Let \(U\) be an arbitrary open neighborhood of \(x\). Local path-connectedness gives a path-connected open set \(V\) with \(x\in V\subseteq U\). A standard result from the earlier material on path-connected spaces is that every path-connected space is connected. Apply it to \(V\) with its subspace topology: \(V\) is connected. Thus \(V\) is a connected open neighborhood of \(x\) contained in \(U\), as required. \(\square\)
This implication explains the relation between the local condition here and local connectedness: local path-connectedness at a point is stronger, because it supplies path-connected neighborhoods, which in particular are connected neighborhoods. The reverse implication is not asserted by this argument; a connected neighborhood need not be path-connected.
Proof. A homeomorphism is a bijection that is continuous and whose inverse is continuous. These conditions imply that \(f\) maps open sets to open sets: for an open set \(W\subseteq X\), \(f(W)=(f^{-1})^{-1}(W)\), which is open in \(Y\) by continuity of \(f^{-1}\). Also, the restriction of \(f\) to any subset \(V\subseteq X\) is a homeomorphism from \(V\) to \(f(V)\) with their subspace topologies, so it preserves path-connectedness by composing paths with \(f\) or \(f^{-1}\).
Suppose first that \(X\) is locally path-connected at \(x\), and let \(O\) be any open neighborhood of \(f(x)\) in \(Y\). The set \(f^{-1}(O)\) is open in \(X\) and contains \(x\). Choose a path-connected open neighborhood \(V\) of \(x\) with \(V\subseteq f^{-1}(O)\). Then \(f(V)\) is open in \(Y\), contains \(f(x)\), and lies in \(O\). Since \(f|_V\) is a homeomorphism and \(V\) is path-connected, \(f(V)\) is path-connected. This proves local path-connectedness at \(f(x)\).
Conversely, apply the direction just proved to the homeomorphism \(f^{-1}:Y\to X\) and the point \(f(x)\). It follows that local path-connectedness at \(f(x)\) implies local path-connectedness at \(x\). Hence the two conditions are equivalent. \(\square\)
A Common Misreading
Common pitfall: “Locally path-connected at \(x\)” does not mean that every open neighborhood of \(x\) is path-connected. For a concrete example, take \(X=\mathbb{R}\) with its standard topology and \(x=0\). The open set \(U=(-1,1)\cup(2,3)\) is a neighborhood of \(0\), but it is not path-connected: no continuous path in \(U\) can join \(0\) to \(5/2\), since such a path would have to take the value \(3/2\) by the intermediate value theorem, and \(3/2\notin U\).
The incorrect argument is to test this one \(U\) and conclude that \(\mathbb{R}\) is not locally path-connected at \(0\). The definition asks instead for a path-connected open neighborhood inside \(U\). The set \(V=(-1,1)\) is open, contains \(0\), and satisfies \(V\subseteq U\); it is path-connected because the linear path between any two of its points remains in the interval. Indeed, the interval argument above proves that \(\mathbb{R}\) is locally path-connected at every point.
When checking the condition, keep its quantifiers in order: begin with an arbitrary open neighborhood \(U\) of the point, then find a path-connected open neighborhood \(V\) with \(x\in V\subseteq U\). A single path-connected neighborhood is not enough to establish the condition unless it can be found inside every such \(U\).
Check Your Understanding
For each question, pay attention to whether the topology is the standard topology or a subspace topology.
- State the definition of local path-connectedness at a point, including the required relationship between \(V\) and an arbitrary neighborhood \(U\).
- Why does the interval argument still work at \(0\in[0,1]\), even though \(0\) is not an interior point of \([0,1]\) in \(\mathbb{R}\)?
- Show that every point of a discrete space has an open path-connected neighborhood contained in each of its open neighborhoods.
- In the Euclidean open-set argument, identify where openness of \(\Omega\subseteq\mathbb{R}^n\) is used.
- Give an open neighborhood of \(0\) in \(\mathbb{R}\) that is not path-connected, and find a path-connected open neighborhood of \(0\) contained in it.