Local Path-Connectedness and Open Subspaces
Local path-connectedness is a condition about the neighborhoods available near each point: however small an open neighborhood is chosen, it contains a smaller open neighborhood in which any two points can be joined by a path. This property is preserved when we pass to an open subspace. The openness assumption is essential, and the rational numbers with their subspace topology provide a useful contrast.
A path in a space \(X\) is a continuous map \(\gamma:I\to X\), where \(I=[0,1]\). It joins \(x\) to \(y\) when \(\gamma(0)=x\) and \(\gamma(1)=y\). A subset \(P\subseteq X\) is path-connected if every pair of its points can be joined by a path whose image lies in \(P\). We use the following neighborhood formulation of local path-connectedness.
If \(U\subseteq X\), its subspace topology consists of the sets \(U\cap V\), where \(V\) is open in \(X\). Thus an open set in \(U\) need not be open in \(X\). When \(U\) itself is open in \(X\), however, every open set in \(U\) is open in \(X\) as well: it has the form \(U\cap V\), an intersection of two open sets of \(X\). This is the key point in the inheritance result.
Proof. Let \(x\in U\), and let \(O\) be an open neighborhood of \(x\) in \(U\). By the definition of the subspace topology, \(O=U\cap V\) for some open set \(V\) in \(X\). Since \(U\) and \(V\) are both open in \(X\), \(O\) is open in \(X\). It contains \(x\), so local path-connectedness of \(X\) gives a path-connected open set \(W\) in \(X\) with \(x\in W\subseteq O\). In particular, \(W\subseteq U\), and \(W\) is open in \(U\) because it is open in \(X\). Its paths in \(X\) also give paths in \(U\): their images lie in \(W\subseteq U\), and a map into \(U\) is continuous when its composite with the inclusion \(U\to X\) is continuous. Therefore \(O\) contains a path-connected open neighborhood of \(x\) in \(U\). Since \(x\) and \(O\) were arbitrary, \(U\) is locally path-connected. \(\square\)
The argument uses openness twice: it makes the chosen neighborhood \(O\) open in \(X\), and it ensures that the smaller open set \(W\) is an open neighborhood in the subspace. The theorem does not say that every subspace of a locally path-connected space is locally path-connected.
Why Euclidean Open Sets Have the Property
Equip \(\mathbb{R}^n\), for \(n\geq 1\), with its standard Euclidean topology. Its open balls are path-connected. To verify this, take a ball \(B(c,r)\) and points \(u,v\in B(c,r)\). Define the straight-line path \(\gamma(t)=(1-t)u+tv\), for \(t\in I\). For every \(t\in I\), the triangle inequality gives
The strict inequality holds because \(\|u-c\|<r\), \(\|v-c\|<r\), and the coefficients \(1-t\) and \(t\) are nonnegative and sum to \(1\). Thus \(\gamma(t)\in B(c,r)\) for all \(t\). The formula is continuous, and \(\gamma(0)=u\), \(\gamma(1)=v\), proving that the ball is path-connected. Every open neighborhood of a point in \(\mathbb{R}^n\) contains a ball centered at that point. Hence \(\mathbb{R}^n\) is locally path-connected.
Proof. The Euclidean-space argument shows that \(\mathbb{R}^n\) is locally path-connected. The set \(G\) is open in \(\mathbb{R}^n\), so the theorem applies with \(X=\mathbb{R}^n\) and \(U=G\). \(\square\)
Worked Example: A Neighborhood in the Punctured Plane
Let \(G=\mathbb{R}^2\setminus\{(0,0)\}\), with the subspace topology inherited from standard \(\mathbb{R}^2\). Consider \(p=(3,4)\) and the open neighborhood \(O=B(p,1)\cap G\). Since \(\|p\|=5\), every \(q\in B(p,1)\) satisfies, by the reverse triangle inequality, \(\|q\|\geq \|p\|-\|q-p\|>5-1=4\). Thus \(B(p,1)\) does not contain the origin, so \(O=B(p,1)\).
The smaller ball \(W=B(p,1/2)\) contains \(p\) and lies in \(O\), because \(\|q-p\|<1/2\) implies \(\|q-p\|<1\). If \(u,v\in W\), their straight-line path stays in \(W\): the estimate from the preceding argument gives \(\|(1-t)u+tv-p\|\leq(1-t)\|u-p\|+t\|v-p\|<1/2\). Therefore \(W\) is a path-connected open neighborhood of \(p\) in \(G\) contained in \(O\). This verifies the local condition in this particular neighborhood; the corollary gives it at every point of \(G\).
Check: The radius estimate also ensures that the entire path lies in \(G\), not merely in \(\mathbb{R}^2\): every point \(z\in W\) has \(\|z\|\geq 5-\|z-p\|>5-1/2=9/2>0\).
Worked Example: An Open Annulus
In standard \(\mathbb{R}^2\), consider \(A=\{x\in\mathbb{R}^2:1<\|x\|<3\}\). First check that \(A\) is open. For \(p\in A\), set \(r=\min(\|p\|-1,3-\|p\|)/2\). Both numbers inside the minimum are positive, so \(r>0\). If \(\|q-p\|<r\), then the reverse triangle inequality and triangle inequality give
Here \(r\leq(\|p\|-1)/2\) gives \(\|p\|-r>1\), and \(r\leq(3-\|p\|)/2\) gives \(\|p\|+r<3\). Thus \(B(p,r)\subseteq A\), proving that \(A\) is open. The corollary now says that \(A\) is locally path-connected.
For a concrete point, take \(p=(2,0)\), for which \(\|p\|=2\) and \(r=\min(1,1)/2=1/2\). The ball \(B(p,1/2)\) lies in \(A\): for each \(q\) in that ball, \(\|q\|>2-1/2=3/2>1\) and \(\|q\|<2+1/2=5/2<3\). Straight-line paths between points of this ball stay in the ball, so it is a path-connected neighborhood of \(p\) inside the annulus.
Check: At \(p=(2,0)\), the computed radius is positive and less than both distances from \(\|p\|=2\) to the boundary values \(1\) and \(3\). The inequalities above therefore verify both required strict bounds.
Worked Example: A Disconnected Open Subset of the Line
Let \(G=(-2,-1)\cup(1,3)\subseteq\mathbb{R}\), with the subspace topology from the standard topology. It is open in \(\mathbb{R}\), so the corollary shows that it is locally path-connected. We can check the neighborhood condition directly at \(x=5/2\). The interval \(W=(2,3)\) contains \(5/2\) and is contained in \(G\). It is open in \(G\), since it is open in \(\mathbb{R}\) and \(W=W\cap G\).
For any \(u,v\in(2,3)\), the path \(\gamma(t)=(1-t)u+tv\) lies strictly between \(2\) and \(3\): both endpoints are strictly between these bounds, and their weighted average with nonnegative weights summing to \(1\) remains strictly between them. Thus \(W\) is path-connected and is a suitable neighborhood of \(5/2\). The same reasoning works near every point of either interval. Local path-connectedness is a local property; it does not require that the whole open set be path-connected.
Check: For the chosen point, \(5/2\in(2,3)\subseteq(1,3)\subseteq G\), and the interval path never leaves \((2,3)\). This verifies the neighborhood and path conditions without requiring a path between points in different intervals.
Why the Openness Hypothesis Matters
One tempting but invalid extension is to replace “open subspace” in the theorem with “any subspace.” The issue is not that paths in the larger space cease to exist; it is that an open neighborhood in the subspace may fail to contain a smaller path-connected open neighborhood there. The rational numbers give an explicit failure.
Proof. First, no path in \(\mathbb{Q}\) can join two distinct points. Suppose a path \(\gamma:I\to\mathbb{Q}\) joined distinct rationals \(a<b\). Composing with the inclusion \(\mathbb{Q}\to\mathbb{R}\) gives a continuous real-valued function on \(I\). By the intermediate value theorem, its image contains every real number between \(a\) and \(b\), including an irrational number. That contradicts \(\gamma(I)\subseteq\mathbb{Q}\). Hence every path-connected subset of \(\mathbb{Q}\) contains at most one point.
Now consider the open neighborhood \(O=\mathbb{Q}\cap(-1,1)\) of \(0\) in \(\mathbb{Q}\). Any path-connected open neighborhood \(W\) of \(0\) contained in \(O\) would have to be the singleton \(\{0\}\), by the preceding argument. But \(\{0\}\) is not open in \(\mathbb{Q}\): every open neighborhood of \(0\) in \(\mathbb{Q}\) has the form \(\mathbb{Q}\cap V\) for some real open neighborhood \(V\) of \(0\), and \(V\) contains a real interval \((-\delta,\delta)\) for some \(\delta>0\). Choose \(N\in\mathbb{N}\) with \(1/N<\delta\). Then \(1/N\in\mathbb{Q}\cap V\) and \(1/N\ne0\). So no such neighborhood is the singleton \(\{0\}\). The neighborhood \(O\) therefore contains no path-connected open neighborhood of \(0\) in \(\mathbb{Q}\), proving the claim. \(\square\)
Common pitfall. The incorrect argument is: “\((-1,1)\) is a path-connected neighborhood of \(0\) in \(\mathbb{R}\), so \(\mathbb{Q}\cap(-1,1)\) is a path-connected neighborhood of \(0\) in \(\mathbb{Q}\).” The conclusion does not follow: the intersection contains \(0\) and \(1/2\), but no path in \(\mathbb{Q}\) can join them. Such a path, viewed as real-valued, would have to take the irrational value \(\sqrt{2}/4\), which lies strictly between \(0\) and \(1/2\). The theorem avoids this error by requiring the subspace to be open, so that the neighborhoods to which local path-connectedness is applied are open in the larger space.
Check Your Understanding
Use the neighborhood definition, and keep track of which topology each open set belongs to.
- In the inheritance theorem, where is the assumption that \(U\) is open used to show that \(O=U\cap V\) is open in \(X\)?
- Prove that every open interval in standard \(\mathbb{R}\) is path-connected using a straight-line path.
- For an open ball \(B(c,r)\subseteq\mathbb{R}^n\), give the path joining \(u,v\in B(c,r)\) and the inequality that keeps it in the ball.
- Why does the argument for \(\mathbb{Q}\) rule out a path between any two distinct rationals?
- Does the theorem imply that an open subset of a locally path-connected space is path-connected? Explain the difference between the local conclusion and this global claim.