Local Path-Connectedness as a Basis Condition
A space can be path-connected as a whole without having small path-connected neighborhoods around each of its points. Local path-connectedness captures this local requirement: near any point, one can find path-connected open sets that fit inside any prescribed open neighborhood. The open-set condition matters. A path-connected subset inside an open set need not contain any open neighborhood of the point.
Throughout, a path is a continuous map \(\gamma:I\to X\), where \(I=[0,1]\). A subset \(A\) of a topological space \((X,\mathcal{T})\) is path-connected if it is nonempty and, for every \(a,b\in A\), there is a path \(\gamma:I\to A\) with \(\gamma(0)=a\) and \(\gamma(1)=b\). A subset \(N\subseteq X\) is a neighborhood of \(x\) if it contains an open set containing \(x\). A neighborhood basis at \(x\) is a collection of neighborhoods of \(x\) such that every open set containing \(x\) contains a member of the collection.
The Path-Connected Open-Set Criterion
Proof. Suppose first that \(X\) is locally path-connected. Fix \(x\in X\), and let \(U\) be any open set containing \(x\). By the definition, there is a path-connected open set \(V\) with \(x\in V\subseteq U\). Thus every open neighborhood of \(x\) contains a path-connected open neighborhood of \(x\). The collection of all such sets \(V\) is therefore a neighborhood basis at \(x\).
Conversely, suppose that for each \(x\in X\), the path-connected open sets containing \(x\) form a neighborhood basis at \(x\). Given any open \(U\) containing \(x\), the basis property supplies one of those sets, say \(V\), with \(x\in V\subseteq U\). By its membership in the collection, \(V\) is both open and path-connected. This is precisely the defining condition for \(X\) to be locally path-connected.
If \(X\) is empty, both conditions hold vacuously: there are no points \(x\) at which a neighborhood condition must be checked. This covers that degenerate case as well. \(\square\)
The theorem packages the local condition into a collection of sets that can be used directly. To test local path-connectedness at \(x\), take an arbitrary open neighborhood \(U\) and find a path-connected open \(V\) with \(x\in V\subseteq U\). It is not enough to find a path-connected subset of \(U\) that contains \(x\): unless that subset is itself a neighborhood, it may be too small to give the required local basis.
Worked Example: The Real Line
Give \(\mathbb{R}\) its standard topology. We verify the criterion at an arbitrary \(x\in\mathbb{R}\). Let \(U\) be open and contain \(x\). By openness, there is \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). Set \(V=(x-\varepsilon,x+\varepsilon)\). It is open, contains \(x\), and is contained in \(U\).
To verify path-connectedness, take any \(a,b\in V\) and define \(\gamma(t)=(1-t)a+tb\) for \(t\in I\). This is continuous, \(\gamma(0)=a\), and \(\gamma(1)=b\). Since \(0\leq t\leq1\), the value \((1-t)a+tb\) lies between \(a\) and \(b\); both endpoints lie strictly between \(x-\varepsilon\) and \(x+\varepsilon\), so every value of \(\gamma\) lies in \(V\). Thus \(V\) is path-connected and the criterion holds at every \(x\).
Check: the path has the required endpoints because \(\gamma(0)=(1-0)a+0b=a\) and \(\gamma(1)=(1-1)a+1b=b\). Hence \(\mathbb{R}\) is locally path-connected.
Worked Example: A Closed Interval at an Endpoint
Give \([0,1]\) the subspace topology inherited from \(\mathbb{R}\). We check the criterion at \(x=0\), where neighborhoods are relative to \([0,1]\). Take the specific open neighborhood \(U=[0,1]\cap(-0.2,0.4)=[0,0.4)\). The set \(V=[0,1]\cap(-0.1,0.2)=[0,0.2)\) is relatively open, contains \(0\), and is contained in \(U\).
For \(a,b\in V\), define \(\gamma(t)=(1-t)a+tb\). The endpoints are \(a\) and \(b\), and \(0\leq a,b<0.2\). For \(t\in I\), \(\gamma(t)\) lies between \(a\) and \(b\), so \(0\leq\gamma(t)<0.2\); therefore \(\gamma(t)\in V\). This verifies path-connectedness of this particular neighborhood, including the endpoint \(0\).
More generally, for any \(x\in[0,1]\) and any relatively open neighborhood \(U\) of \(x\), the subspace topology gives an open set \(O\subseteq\mathbb{R}\) with \(U=O\cap[0,1]\). Choose \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq O\), and set \(V=(x-\varepsilon,x+\varepsilon)\cap[0,1]\). This is relatively open, contains \(x\), and is contained in \(U\). For \(a,b\in V\), the same linear path \((1-t)a+tb\) stays in \([0,1]\), since both endpoints lie there, and stays in \((x-\varepsilon,x+\varepsilon)\), since it lies between \(a\) and \(b\). Thus it stays in \(V\). The criterion holds at every point of \([0,1]\).
Check: for the chosen neighborhood, \(V=[0,0.2)\subseteq[0,0.4)=U\), and the path values satisfy \(0\leq\gamma(t)<0.2\). In particular, no separate argument excluding the endpoint \(0\) is needed.
Worked Example: A Discrete Space
Let \(X=\{a,b,c\}\) with the discrete topology, in which every subset is open. Fix any \(x\in X\) and any open set \(U\) containing \(x\). The singleton \(V=\{x\}\) is open and satisfies \(x\in V\subseteq U\).
The singleton is path-connected: its only pair of points is \(x,x\), and the constant path \(\gamma(t)=x\) joins them. It is continuous, and both endpoint values equal \(x\). Thus every open neighborhood of every point contains a path-connected open neighborhood, so this discrete space is locally path-connected. The argument works for any discrete space, not just this three-point example.
Check: the chosen \(V\) is open by discreteness, and it is contained in \(U\) because \(x\in U\). Its constant path remains in \(V\) for every \(t\in I\).
A Space Where the Criterion Fails
Worked Example: The Rational Numbers
Give \(\mathbb{Q}\) the subspace topology inherited from the standard topology on \(\mathbb{R}\). We show that it is not locally path-connected at \(0\). First note that every path-connected subset of \(\mathbb{Q}\) contains at most one point.
Indeed, suppose a path \(\gamma:I\to\mathbb{Q}\) had distinct endpoint values \(a\) and \(b\). The inclusion \(i:\mathbb{Q}\to\mathbb{R}\) is continuous: for every open \(O\subseteq\mathbb{R}\), its preimage is \(O\cap\mathbb{Q}\), open in the subspace topology. Therefore \(i\circ\gamma:I\to\mathbb{R}\) is a continuous real-valued function. The intermediate value theorem says that it takes every value between \(a\) and \(b\). There is an irrational number strictly between any two distinct real numbers, so \(i\circ\gamma\) would take an irrational value, contradicting that every value of \(\gamma\) belongs to \(\mathbb{Q}\). Thus no path in \(\mathbb{Q}\) has distinct endpoint values.
Now consider any neighborhood \(N\) of \(0\) in \(\mathbb{Q}\), not merely a particular one chosen in advance. By the definition of neighborhood, \(N\) contains an open neighborhood \(V\) of \(0\) in \(\mathbb{Q}\). There is an open \(O\subseteq\mathbb{R}\) with \(V=O\cap\mathbb{Q}\), and since \(0\in O\), some \(\delta>0\) satisfies \((-\delta,\delta)\subseteq O\). Density of \(\mathbb{Q}\) in \(\mathbb{R}\) gives a rational \(s\) with \(0<s<\delta\). Hence \(s\in V\subseteq N\) and \(s\ne0\). So \(N\) contains two distinct points, \(0\) and \(s\), and cannot be path-connected. In particular, there is no path-connected open neighborhood of \(0\) inside the open neighborhood \(\mathbb{Q}\cap(-1,1)\). The criterion fails at \(0\).
Check: the inequality \(0<s<\delta\) implies \(s\in(-\delta,\delta)\subseteq O\); since \(s\in\mathbb{Q}\), it follows that \(s\in O\cap\mathbb{Q}=V\). The argument applies to every candidate neighborhood \(N\), so it rules out all path-connected neighborhoods of \(0\), not just one selected subset.
Homeomorphisms Preserve the Local Criterion
Proof. Suppose \(X\) is locally path-connected, and let \(y\in Y\) lie in an open set \(W\). Write \(y=f(x)\), using surjectivity. Since \(f\) is continuous, \(f^{-1}(W)\) is open and contains \(x\). By local path-connectedness, there is a path-connected open \(V\) with \(x\in V\subseteq f^{-1}(W)\). A homeomorphism maps open sets to open sets, so \(f(V)\) is open; it contains \(y\) and is contained in \(W\). It is path-connected as well: for any \(u,v\in f(V)\), choose \(a,b\in V\) with \(f(a)=u\) and \(f(b)=v\), and take a path \(\gamma:I\to V\) from \(a\) to \(b\). The composite \(f\circ\gamma\) is a path in \(f(V)\) from \(u\) to \(v\). This verifies the criterion in \(Y\). Applying the same argument to the homeomorphism \(f^{-1}:Y\to X\) proves the reverse implication. \(\square\)
This proposition lets us check local path-connectedness in whichever of two homeomorphic spaces is easier to describe. The essential test remains local: for each point and each open neighborhood, find a path-connected open neighborhood contained within it.
A Common Mistake
In the rational example, a tempting argument is to choose one rational \(r\in\mathbb{Q}\cap(-1,1)\), note that no path-connected subset of \(\mathbb{Q}\) contains both \(0\) and \(r\), and conclude that there is no path-connected neighborhood of \(0\) inside \(\mathbb{Q}\cap(-1,1)\). That conclusion does not follow: the proposed neighborhood might not contain the chosen \(r\). In set terms, \(N\subseteq U\) and \(r\in U\) do not imply \(r\in N\). For example, \(N=\{0\}\), \(U=\mathbb{Q}\cap(-1,1)\), and \(r=1/2\) satisfy \(N\subseteq U\) and \(r\in U\), but \(r\notin N\).
The correction is to begin with an arbitrary candidate neighborhood \(N\). Its open core \(V\) around \(0\) contains some rational \(s\ne0\), by density and the fact that \(V\) is open in \(\mathbb{Q}\). This point \(s\) is guaranteed to belong to \(N\), because \(V\subseteq N\). Since no path in \(\mathbb{Q}\) can join \(0\) to \(s\), the candidate \(N\) is not path-connected. Controlling the point chosen relative to the candidate set is what completes the argument.
Check Your Understanding
Use the criterion and justify each claim about openness and paths.
- Give \(\mathbb{R}\) its standard topology. For an open neighborhood \(U\) of \(x\), explain how to choose an open interval contained in \(U\), and show that it is path-connected.
- Give \(\mathbb{Z}\) the subspace topology inherited from \(\mathbb{R}\). Is it locally path-connected? Verify the criterion at an arbitrary integer.
- Why does the fact that a path-connected subset \(N\) lies inside an open set \(U\) fail to establish that \(N\) is a neighborhood of a point?
- In the rational example, identify where the intermediate value theorem is used and why it contradicts a path taking values only in \(\mathbb{Q}\).
- Use the homeomorphism proposition to explain why any space homeomorphic to \([0,1]\) is locally path-connected.