How a Path Rules Out a Separation
Connectedness and path connectedness describe different features of a space. Connectedness rules out a split into two disjoint, nonempty relatively open sets. Path connectedness supplies a continuous route within the space between each pair of points. The key link is that a path has the interval \([0,1]\) as its domain, and intervals in \(\mathbb{R}\) are connected. If a space were separated, a path joining points on opposite sides would pull that separation back to a separation of the interval.
We use the definitions of connectedness and path connectedness established earlier. In particular, a path in a subset \(A\) is a continuous map \(\gamma:[0,1]\to A\), and a separation of \(A\) is a pair of disjoint, nonempty relatively open subsets whose union is \(A\). The connectedness of intervals in \(\mathbb{R}\) will provide the contradiction.
Proof. If \(A=\varnothing\), then \(A\) is connected: it cannot be written as a union of two nonempty sets. Now suppose \(A\ne\varnothing\), and assume for contradiction that \(A\) is disconnected. Then there are disjoint, nonempty, relatively open sets \(U,V\subseteq A\) such that \(A=U\cup V\). Choose \(a\in U\) and \(b\in V\). Since \(A\) is path connected, there is a continuous map \(\gamma:[0,1]\to A\) with \(\gamma(0)=a\) and \(\gamma(1)=b\).
Consider the inverse images of the two sets under \(\gamma\). They are disjoint because \(U\) and \(V\) are disjoint. They cover \([0,1]\) because every value of \(\gamma\) lies in \(A=U\cup V\). Each inverse image is relatively open in \([0,1]\): \(\gamma\) is continuous as a map into \(A\), and \(U,V\) are relatively open there. Both inverse images are nonempty, since \(0\in\gamma^{-1}(U)\) and \(1\in\gamma^{-1}(V)\). Thus they would separate \([0,1]\), contradicting the fact that every interval in \(\mathbb{R}\) is connected. Therefore \(A\) is connected. \(\square\)
Take two disjoint, nonempty relatively open sets covering the path-connected set.
Path connectedness gives a path whose endpoints lie in the two different sets.
Their inverse images would separate \([0,1]\), which is impossible because intervals are connected.
The proof depends on the whole path staying in \(A\), not merely on its endpoints belonging to \(A\). That condition ensures the inverse images cover the entire interval. It also depends on continuity: without it, the inverse images of relatively open sets need not be relatively open, so they would not necessarily form a separation.
Applying the Implication
Worked Example: A Graph Over an Interval
Let \(E=\{(t,t^2-2t): -1\leq t\leq 2\}\subseteq\mathbb{R}^2\). We show that \(E\) is connected by first constructing paths within it. Choose any two points of \(E\), say \((s,s^2-2s)\) and \((t,t^2-2t)\), where \(s,t\in[-1,2]\). Define \(\gamma(r)=(u(r),u(r)^2-2u(r))\) for \(r\in[0,1]\), where \(u(r)=(1-r)s+rt\).
For each \(r\in[0,1]\), \(u(r)\) lies between \(s\) and \(t\), so \(u(r)\in[-1,2]\) and \(\gamma(r)\in E\). The coordinate functions are continuous, and \(\gamma(0)=(s,s^2-2s)\) while \(\gamma(1)=(t,t^2-2t)\). Hence \(E\) is path connected. The Path Connected Implies Connected Theorem now shows that \(E\) is connected.
Worked Example: Two Coordinate Rays Joined at the Origin
Consider \(E=\{(x,0):x\geq0\}\cup\{(0,y):y\geq0\}\). This set is not convex: the segment joining \((1,0)\) to \((0,1)\) contains \((1/2,1/2)\), which is not in \(E\). Nevertheless, any two points \(p,q\in E\) can be joined by a path in \(E\). Define \(\gamma(r)=(1-2r)p\) for \(0\leq r\leq1/2\), and \(\gamma(r)=(2r-1)q\) for \(1/2\leq r\leq1\).
On the first half of the interval, \(\gamma\) moves from \(p\) toward the origin along the ray containing \(p\); on the second half, it moves from the origin toward \(q\) along the ray containing \(q\). Both formulas give \((0,0)\) at \(r=1/2\), so the pieces join continuously. The scalar factors are between \(0\) and \(1\) on their respective halves, so every value remains in \(E\). Also, \(\gamma(0)=p\) and \(\gamma(1)=q\). Thus \(E\) is path connected and, by the theorem, connected.
Worked Example: A Spiral Arc
Define \(\gamma:[0,1]\to\mathbb{R}^2\) by \(\gamma(t)=(e^{-t}\cos(4\pi t),e^{-t}\sin(4\pi t))\), and let \(S=\gamma[[0,1]]\) be its image. The coordinate functions are continuous, so \(\gamma\) is continuous. Moreover, the distance of \(\gamma(t)\) from the origin is \(e^{-t}>0\), so the curve never reaches the origin. Its endpoints are \(\gamma(0)=(1,0)\) and \(\gamma(1)=(e^{-1},0)\).
The interval \([0,1]\) is path connected: the linear path between any two of its points stays in the interval. By the Continuous Images of Path-Connected Sets Theorem, \(S\) is path connected. The Path Connected Implies Connected Theorem then gives that \(S\) is connected. This argument does not require a direct description of every possible separation of the spiral image.
Path Components Fit Inside Connected Components
The implication also clarifies how the two kinds of components relate. A path component groups the points that can be joined to a given point by paths. A connected component groups the points lying in the maximal connected subset containing that point. Since every path-connected set is connected, each path component must lie within a connected component. This says that paths cannot join points from different connected components; it does not say that every pair of points in a connected component can be joined by a path.
Proof. By the Path Components Partition Theorem, \(P_X(x)\) is path connected and contains \(x\). The Path Connected Implies Connected Theorem therefore shows that \(P_X(x)\) is connected. By the Construction of Components Theorem, \(C_X(x)\) is the maximal connected subset of \(X\) containing \(x\). Since \(P_X(x)\) is a connected subset containing \(x\), maximality gives \(P_X(x)\subseteq C_X(x)\). \(\square\)
Worked Example: A Separation Prevents Paths Across It
Let \(E=\{-3,4\}\subseteq\mathbb{R}\). The singleton sets \(\{-3\}\) and \(\{4\}\) are disjoint, nonempty, and relatively open in \(E\); for example, a ball of radius \(1\) about either point meets \(E\) only at that point. They cover \(E\), so \(E\) is disconnected. The contrapositive of the Path Connected Implies Connected Theorem shows that \(E\) is not path connected. More generally, if a subset is known to be disconnected, no path-connected subset can equal it.
The Direction of the Result
The theorem establishes one implication: path connectedness guarantees connectedness. A common pitfall is to treat the two conditions as interchangeable. Connectedness excludes a separation, but that exclusion does not itself construct a continuous path between arbitrary points. The theorem therefore cannot be used in reverse without a separate result and additional hypotheses or examples.
When using the proof, keep track of where each condition enters. The endpoints on opposite sides make both inverse images nonempty. The fact that the path stays in the set makes those inverse images cover \([0,1]\). Continuity makes them relatively open. Finally, the connectedness of the interval rules out the resulting separation. This sequence of steps is the mechanism behind the implication, and it applies regardless of the geometry of the paths or the ambient metric space.
Check Your Understanding
Answer these questions using the theorem and its proof.
- Why must both inverse images in the separation argument be nonempty?
- Where does the proof use the requirement that a path stays inside the subset?
- Why are the inverse images of relatively open sets relatively open in \([0,1]\)?
- What contradiction follows from assuming a path-connected set is disconnected?
- What containment between path components and connected components follows from the theorem?