Paths Between Points
Connectedness asks whether a space can be separated into two disjoint, nonempty relatively open sets. Path connectedness asks a more direct question: can any two points be joined by a continuous journey that stays inside the space? This extra structure is useful when studying how spaces map into one another and how their points group into path components.
A path in a metric space \(X\) is a continuous map from the interval \([0,1]\) into \(X\). Its starting point is the image of \(0\), and its endpoint is the image of \(1\). Continuity is understood with respect to the usual metric on \([0,1]\) and the given metric on \(X\).
A path in \(A\) is also a path in \(X\), but the requirement that its values stay in \(A\) matters: path connectedness is a property of the set with its subspace metric. Reversing a path exchanges its endpoints, and paths can be concatenated when the endpoint of one agrees with the starting point of the next. These operations will let us organize points according to which ones can be joined.
Basic Examples and Constructions
Worked Example: Convex Sets Are Path Connected
A subset \(A\subseteq\mathbb{R}^n\) is convex if, whenever \(a,b\in A\) and \(t\in[0,1]\), the point \((1-t)a+tb\) belongs to \(A\). If \(A\) is nonempty and convex, take any \(a,b\in A\) and define \(\gamma(t)=(1-t)a+tb\). Each \(\gamma(t)\) lies in \(A\) by convexity. The map is continuous, since for \(s,t\in[0,1]\), \[ \|\gamma(t)-\gamma(s)\|=|t-s|\,\|b-a\|. \] Also, \(\gamma(0)=a\) and \(\gamma(1)=b\). Thus \(A\) is path connected. The empty set is path connected by the vacuous convention as well.
For example, the filled triangle with vertices \((0,0),(2,0),(0,1)\) is convex: it is the set of points \((x,y)\) satisfying \(x\geq0\), \(y\geq0\), and \(x/2+y\leq1\), and these linear inequalities are preserved under convex combinations. The straight segment between any two points of the triangle therefore supplies a path in it.
Worked Example: Paths in the Punctured Plane
Let \(A=\mathbb{R}^2\setminus\{(0,0)\}\). We show that \(A\) is path connected, even though it is not convex: the segment joining \((1,0)\) and \((-1,0)\) passes through the removed origin.
Take arbitrary \(x,y\in A\). Write their polar forms as \(x=r_x(\cos\theta_x,\sin\theta_x)\) and \(y=r_y(\cos\theta_y,\sin\theta_y)\), where \(r_x,r_y>0\). Define \[ \gamma(t)=\big((1-t)r_x+tr_y\big) \big(\cos((1-t)\theta_x+t\theta_y),\, \sin((1-t)\theta_x+t\theta_y)\big). \] The formula is continuous in \(t\). Its radius is positive because \((1-t)r_x+tr_y\geq\min(r_x,r_y)>0\) for \(0\leq t\leq1\). Hence \(\gamma(t)\neq(0,0)\) throughout. At \(t=0\), the formula gives \(x\), and at \(t=1\), it gives \(y\). Thus \(\gamma\) is a path in \(A\) from \(x\) to \(y\).
This example shows why requiring a path to stay in the set is important. The direct line segment is not always suitable, but another continuous route may still join the points.
Continuous Images of Path-Connected Sets
A continuous map carries each path in its domain to a path in its image. The construction is composition: if \(\gamma\) joins two points and \(f\) is continuous, then \(f\circ\gamma\) joins their images. This gives a preservation theorem parallel to the continuous-image result for connected spaces, without requiring any claim here about the relationship between path connectedness and connectedness.
Proof. If \(A=\varnothing\), then \(f[A]=\varnothing\), which is path connected by convention. Otherwise, take any \(u,v\in f[A]\). By the definition of image, there are \(a,b\in A\) such that \(f(a)=u\) and \(f(b)=v\). Since \(A\) is path connected, there is a continuous path \(\gamma:[0,1]\to A\) with \(\gamma(0)=a\) and \(\gamma(1)=b\). Regard \(\gamma\) as a map into \(X\). The composition \(f\circ\gamma:[0,1]\to Y\) is continuous, and its values lie in \(f[A]\). Moreover, \((f\circ\gamma)(0)=u\) and \((f\circ\gamma)(1)=v\). Thus any two points of \(f[A]\) can be joined by a path in \(f[A]\), as required. \(\square\)
Worked Example: The Circle as a Continuous Image
Consider the map \(\gamma:[0,1]\to\mathbb{R}^2\) defined by \(\gamma(t)=(\cos(2\pi t),\sin(2\pi t))\). It is continuous, and its image is the unit circle \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\): every value of \(\gamma\) satisfies \(\cos^2(2\pi t)+\sin^2(2\pi t)=1\), and as \(2\pi t\) ranges over \([0,2\pi]\), the parametrization traverses the full circle.
The interval \([0,1]\) is path connected: for \(s,t\in[0,1]\), the map \(\eta(r)=(1-r)s+rt\), \(0\leq r\leq1\), stays in \([0,1]\) and joins \(s\) to \(t\). Applying the Continuous Images of Path-Connected Sets Theorem to \(\gamma\) shows that \(S^1\) is path connected. The theorem guarantees paths between any two points of the circle by composing \(\gamma\) with paths in its domain.
Path Components
Being joinable by a path defines an equivalence relation on the points of a nonempty metric space. Reflexivity comes from the constant path; symmetry comes from reversing a path; transitivity comes from concatenating paths. The equivalence classes are called path components. They collect exactly the points that can be joined to one another by paths.
Proof. For \(x\in X\), the constant path at \(x\) shows that \(x\in P_X(x)\). Hence the path components cover \(X\). If \(y,z\in P_X(x)\), there is a path from \(x\) to \(y\) and a path from \(x\) to \(z\). Reverse the first path to obtain a path from \(y\) to \(x\), then concatenate it with the path from \(x\) to \(z\). The resulting path joins \(y\) to \(z\), so \(P_X(x)\) is path connected.
If \(P_X(x)\) and \(P_X(y)\) intersect at \(z\), then paths join \(x\) to \(z\) and \(y\) to \(z\). Reversing the second and concatenating shows that \(x\) and \(y\) are joined by a path. For any \(w\in P_X(x)\), concatenate a path from \(y\) to \(x\) with a path from \(x\) to \(w\); this shows \(w\in P_X(y)\). Thus \(P_X(x)\subseteq P_X(y)\). Interchanging \(x\) and \(y\) gives the reverse inclusion, so the two path components are equal.
Now let \(E\subseteq X\) be nonempty and path connected, and choose \(e\in E\). For each \(w\in E\), path connectedness gives a path in \(E\), hence in \(X\), from \(e\) to \(w\). Therefore \(w\in P_X(e)\), and \(E\subseteq P_X(e)\). Since each path component is itself path connected, this containment also proves maximality: no strictly larger path-connected subset can contain a whole path component. If \(X=\varnothing\), there are no equivalence classes; the empty set is path connected, but the assertion about containing every path-connected subset in a class is intentionally stated only for nonempty subsets. \(\square\)
Worked Example: The Path Components of the Punctured Real Line
Let \(X=\mathbb{R}\setminus\{0\}\). The sets \((-\infty,0)\) and \((0,\infty)\) are each path connected: the linear path \(\gamma(t)=(1-t)a+tb\) stays strictly negative if \(a,b<0\), and stays strictly positive if \(a,b>0\).
There cannot be a path in \(X\) from a negative point \(a\) to a positive point \(b\). If \(\gamma:[0,1]\to X\) were such a path, then it would also be a continuous real-valued function with \(\gamma(0)=a<0<b=\gamma(1)\). The Intermediate Value Property on a Connected Domain, established earlier, would imply that \(\gamma(t)=0\) for some \(t\in[0,1]\), contradicting that all values of \(\gamma\) lie in \(X\). Therefore the two displayed sets are exactly the path components of \(X\).
Homeomorphisms Preserve Path Components
A continuous map can send distinct path components into the same path component, just as it can identify distinct points. If the map is a homeomorphism, however, paths can be transported in both directions. This ensures that path components correspond exactly.
Proof. If \(y\in P_X(x)\), take a path \(\gamma\) in \(X\) from \(x\) to \(y\). The composition \(f\circ\gamma\) is a path in \(Y\) from \(f(x)\) to \(f(y)\). Thus \(f(y)\in P_Y(f(x))\), proving \(f[P_X(x)]\subseteq P_Y(f(x))\).
For the reverse inclusion, take \(z\in P_Y(f(x))\). There is a path \(\alpha:[0,1]\to Y\) from \(f(x)\) to \(z\). Since \(f^{-1}\) is continuous, \(f^{-1}\circ\alpha\) is a path in \(X\) from \(x\) to \(f^{-1}(z)\). Hence \(f^{-1}(z)\in P_X(x)\), so \(z\in f[P_X(x)]\). This proves equality.
Every path component of \(X\) has the form \(P_X(x)\), and the equality shows its image is a path component of \(Y\). Since \(f\) is onto, every path component of \(Y\) is obtained this way. Injectivity ensures that distinct path components cannot have the same image: if their images were equal, applying \(f^{-1}\) would make the original sets equal. Therefore the correspondence is one-to-one and onto. \(\square\)
What Path Connectedness Does and Does Not Say
A path gives more information than a statement that two points lie in the same connected piece: it supplies a continuous map with specified endpoints. Convexity is a particularly convenient sufficient condition, since straight segments provide paths. But convexity is not necessary, as the punctured plane and the circle illustrate.
The path components partition any nonempty space, but one should not assume that each path component is open or closed without additional hypotheses. Nor should path connectedness be confused with the existence of a shortest path or a path of a particular geometric form. The definition requires only a continuous map from \([0,1]\); it imposes no length or smoothness condition.
When applying these results, check the set in which the path is required to lie and the endpoints it must have. For a continuous image, compose a path with the given map. For path components, use the relation of being joined by a path, remembering that the containment statement for path-connected subsets requires a nonempty subset. A homeomorphism preserves these classes because both it and its inverse carry paths to paths.
Check Your Understanding
Use the definitions and theorems in this tutorial to answer the following questions.
- Why does reversing a path establish symmetry of the relation “joined by a path”?
- How does convexity guarantee that the straight-line path between two points stays in the set?
- Why does a continuous image of a path-connected set remain path connected?
- Why is the statement that every path-connected subset lies in one path component restricted to nonempty subsets?
- Where is continuity of the inverse used when proving that a homeomorphism maps path components onto path components?