From Local Paths to Open Path Components
A path component is defined by whether points can be joined by paths, a condition that may involve the whole space. Local path-connectedness supplies a way to test membership locally: near each point, there are open neighborhoods in which any two points can be joined. The key fact is that a path-connected neighborhood of a point lies entirely inside that point’s path component.
Throughout, \(X\) has topology \(\mathcal{T}\), and \(U\subseteq X\) has the subspace topology unless stated otherwise. A path in a space \(Y\) from \(y_0\) to \(y_1\) is a continuous map \(\gamma:I\to Y\), where \(I=[0,1]\), with \(\gamma(0)=y_0\) and \(\gamma(1)=y_1\). A space is path-connected if each pair of its points can be joined by a path.
The topology matters here. A set open in \(U\) is open in \(X\) when \(U\) itself is open in \(X\); this will let us conclude that the path components are open in \(X\), not just in \(U\).
Why Paths Define Equivalence Classes
For completeness, the relation used in the definition really is an equivalence relation. This also clarifies why it makes sense to speak of a point’s path component.
Proof. Every point \(y\in Y\) is joined to itself by the constant path \(\gamma(t)=y\), so the relation is reflexive. If \(\gamma\) joins \(y_0\) to \(y_1\), then \(t\mapsto\gamma(1-t)\) is a continuous path from \(y_1\) to \(y_0\), so the relation is symmetric. For transitivity, suppose \(\gamma\) joins \(y_0\) to \(y_1\) and \(\eta\) joins \(y_1\) to \(y_2\). Define \[ \lambda(t)= \begin{cases} \gamma(2t),&0\le t\le \tfrac12,\\ \eta(2t-1),&\tfrac12\le t\le 1. \end{cases} \] At \(t=\tfrac12\), both formulas give \(y_1\). Each formula is continuous on its half of \(I\); the two continuous pieces agree at their common endpoint, so the pasting property gives a continuous map on \(I\). Also \(\lambda(0)=y_0\) and \(\lambda(1)=y_2\). Thus the relation is transitive. Equivalence classes of an equivalence relation partition the set. \(\square\)
The Openness Theorem
Proof. If \(U=\varnothing\), it has no path components, so the claim holds. Otherwise let \(P\) be any path component of \(U\), and fix \(x\in P\). Since \(U\) is an open neighborhood of \(x\) in \(X\), local path-connectedness gives a path-connected open set \(V\) with \(x\in V\subseteq U\). For any \(y\in V\), path-connectedness gives a path in \(V\) from \(x\) to \(y\). Because \(V\subseteq U\), this is also a path in \(U\). Thus \(y\) is in the same path component of \(U\) as \(x\), so \(y\in P\). We have shown \(x\in V\subseteq P\). This works for every \(x\in P\), so every point of \(P\) has an open neighborhood in \(X\) contained in \(P\). Therefore \(P\) is open in \(X\). \(\square\)
The argument uses both hypotheses. Local path-connectedness supplies the path-connected open neighborhood; openness of \(U\) lets us use \(U\) as the neighborhood to which that local property is applied. Once such a neighborhood is found, its paths stay in \(U\), which is essential because the path component is taken inside \(U\).
Worked Example: Two Path Components in the Real Line
Give \(\mathbb{R}\) its standard topology and let \(U=(-2,-1)\cup(1,3)\). We identify the path components and check their openness. The intervals are open in \(\mathbb{R}\), so \(U\) is open. The real line is locally path-connected: around any point in an open neighborhood, one can choose a smaller open interval, and the straight-line path between two points of that interval stays in the interval.
Take \(a,b\in(-2,-1)\). The formula \(\gamma(t)=(1-t)a+tb\) defines a continuous path. Since \(0\le t\le1\), \(\gamma(t)\) lies between \(a\) and \(b\), and both endpoints lie strictly between \(-2\) and \(-1\). Hence \(\gamma(t)\in(-2,-1)\) for every \(t\). So this interval is path-connected. The same formula shows that \((1,3)\) is path-connected.
No path in \(U\) can join a point \(a\in(-2,-1)\) to a point \(b\in(1,3)\). If \(\gamma:I\to U\) were such a path, then as a real-valued continuous function it would have \(\gamma(0)=a<0\) and \(\gamma(1)=b>0\). The intermediate value theorem would give some \(t\in I\) with \(\gamma(t)=0\), contradicting \(0\notin U\). Thus the path components are exactly \((-2,-1)\) and \((1,3)\). Their endpoints are excluded, as indicated by the strict interval notation, and both components are open in \(\mathbb{R}\), as the theorem predicts.
Check: the paths within either interval stay in that interval, while a path between intervals would have to take the value \(0\). These two facts verify both that the listed sets are path-connected and that they are distinct path components.
Worked Example: A Path-Connected Open Disk
Give \(\mathbb{R}^2\) its Euclidean topology and let \(U=\{x\in\mathbb{R}^2:\|x\|<2\}\), the open disk of radius \(2\) centered at the origin. We show directly that \(U\) is one path component and exhibit path-connected open neighborhoods inside it.
For \(x,y\in U\), define \(\gamma(t)=(1-t)x+ty\). This is continuous, and the triangle inequality gives \[ \|\gamma(t)\|\le (1-t)\|x\|+t\|y\|. \] Both \(\|x\|\) and \(\|y\|\) are strictly less than \(2\). Since \(1-t\) and \(t\) are nonnegative and sum to \(1\), the right side is strictly less than \(2\). Hence \(\gamma(t)\in U\) for every \(t\in I\). Every pair of points in \(U\) is joined by a path in \(U\), so \(U\) itself is its only path component.
To see the local neighborhood mechanism, fix \(x\in U\) and set \(r=(2-\|x\|)/2\). Since \(\|x\|<2\), we have \(r>0\). If \(y\in B(x,r)\), then \[ \|y\|\le\|x\|+\|y-x\|<\|x\|+r=\frac{\|x\|+2}{2}<2. \] Thus \(B(x,r)\subseteq U\). This ball is open, and it is path-connected: the straight-line path between any two of its points stays in the ball by the triangle inequality. In particular, these are path-connected open neighborhoods of the points of \(U\), and the theorem’s local argument applies.
Check: \(r\) is positive for every \(x\in U\), and the displayed bound gives \(\|y\|<2\), not merely \(\|y\|\le2\). The strict inequality ensures that the whole ball lies in the open disk.
Why the Hypothesis Cannot Be Dropped
Worked Example: The Rationals with the Subspace Topology
Give \(X=\mathbb{Q}\) the subspace topology inherited from the standard topology on \(\mathbb{R}\), and take \(U=X\), which is open in \(X\). We show that every path component is a singleton and that these singletons are not open.
Suppose a path \(\gamma:I\to\mathbb{Q}\) had two distinct values \(a\) and \(b\). The inclusion \(\mathbb{Q}\to\mathbb{R}\) is continuous, so \(\gamma\), viewed as a map into \(\mathbb{R}\), is continuous. The interval \(I\) is connected, and a continuous image of a connected space is connected. A connected subset of \(\mathbb{R}\) containing \(a<b\) contains every real number between them. But \(a,b\in\mathbb{Q}\), and \(a+(b-a)/\sqrt{2}\) is irrational and lies strictly between \(a\) and \(b\). This contradicts \(\gamma(I)\subseteq\mathbb{Q}\). Therefore every path in \(\mathbb{Q}\) is constant, and the path component of each \(q\in\mathbb{Q}\) is \(\{q\}\).
No singleton \(\{q\}\) is open in \(\mathbb{Q}\). Indeed, every open neighborhood of \(q\) in the subspace topology contains a set of the form \((q-\varepsilon,q+\varepsilon)\cap\mathbb{Q}\) for some \(\varepsilon>0\). By density of \(\mathbb{Q}\) in \(\mathbb{R}\), this set contains a rational other than \(q\). Thus the path components of \(U\) are not open.
This example also verifies that \(\mathbb{Q}\) is not locally path-connected: no open neighborhood of any \(q\) can be path-connected, because every such neighborhood contains another rational, and no path joins distinct rationals. The openness theorem does not apply.
A Common Mistake
A tempting but invalid step is: “For each point \(x\) of a path component \(P\), choose an open neighborhood \(V\subseteq U\); then \(V\subseteq P\).” Openness alone says nothing about whether points of \(V\) can be joined to \(x\) by paths in \(U\).
For a concrete failure, take \(U=\mathbb{Q}\) as above, \(P=\{0\}\), and \(V=(-1,1)\cap\mathbb{Q}\). The set \(V\) is an open neighborhood of \(0\) in \(U\), but \(1/2\in V\) and \(1/2\notin P\). The erroneous inclusion \(V\subseteq P\) fails. The correction is precisely the extra local hypothesis: choose \(V\) to be both open and path-connected. Then every point of \(V\) can be joined to \(x\) by a path staying in \(V\subseteq U\), which does imply \(V\subseteq P\).
The result is useful because it upgrades a local property of the space to a global structural fact about an open subset: its path components form an open partition. The proof does not claim that every path component in every space is open. It identifies the specific neighborhood condition that guarantees openness, and the rational example shows what can go wrong when that condition is absent.
Check Your Understanding
Use the definitions of path component, local path-connectedness, and the subspace topology to answer the following.
- In the standard topology on \(\mathbb{R}\), what are the path components of \(U=(-3,-2)\cup(0,1)\)? Explain why no path joins points from different intervals.
- In the standard topology on \(\mathbb{R}^2\), why does the straight-line path between two points of an open disk stay inside that disk?
- Where in the openness theorem’s proof is the assumption that \(U\) is open used?
- Why does an open neighborhood of \(0\) in \(\mathbb{Q}\) contain a point other than \(0\)?
- State the exact property of the neighborhood \(V\) that lets the proof conclude \(V\subseteq P\).