Joining Points by Continuous Paths
Connectedness says that a set cannot be split into separated pieces. Path connectedness asks a more direct question: can any two points in the set be joined by a continuous path that stays inside it? For subsets of the real line, this question has a particularly simple answer. The path-connected sets are exactly the intervals, so they are also exactly the connected sets. The key is that a continuous path cannot move between two real numbers without taking every value between them.
The parameter \(t\in[0,1]\) records progress along the path: \(t=0\) is the starting point and \(t=1\) is the endpoint. A path need not move steadily in one direction. It may turn back or revisit points. What matters is that it is continuous and that every value it takes belongs to \(E\).
Under this definition, the empty set is path connected because it has no pair of points for which the required condition could fail. A singleton is path connected as well: the constant function \(\gamma(t)=x\) joins its only point \(x\) to itself. These cases agree with the usual convention that the empty set and singletons are connected.
Paths Force Intervals
Suppose a path joins \(x\) to \(y\) in \(E\). As a continuous image of \([0,1]\), its image \(\gamma([0,1])\) is connected, by the Continuous Images of Connected Sets theorem. It contains both endpoints. The Characterization of Connected Subsets of \(\mathbb{R}\) then says that this image contains every real number between \(x\) and \(y\). Since the image lies in \(E\), those intermediate numbers lie in \(E\) too.
Proof. If \(E\) is empty, it is connected. Otherwise, let \(x,y\in E\). Since \(E\) is path connected, there is a continuous \(\gamma:[0,1]\to E\) with \(\gamma(0)=x\) and \(\gamma(1)=y\). The interval \([0,1]\) is connected, so the Continuous Images of Connected Sets theorem implies that \(\gamma([0,1])\) is connected. It contains \(x\) and \(y\); by the Characterization of Connected Subsets of \(\mathbb{R}\), it contains the interval between them. As \(\gamma([0,1])\subseteq E\), that interval is contained in \(E\). Thus \(E\) is an interval, and the same characterization implies that \(E\) is connected. \(\square\)
The proof uses two facts in sequence: a continuous image of an interval is connected, and a connected subset of \(\mathbb{R}\) contains every point between any two of its points. This argument also explains why a path cannot get around a missing real number: if its endpoints lie on opposite sides of the missing number, its image would have to include that number.
Worked Example: A Path Inside a Closed Interval
Let \(E=[-3,2]\), and choose \(x=-2\) and \(y=1\). Define \(\gamma(t)=(1-t)x+ty=-2+3t\) for \(t\in[0,1]\). At the endpoints, \(\gamma(0)=-2=x\) and \(\gamma(1)=1=y\). Also, \(0\leq t\leq1\) gives \(-2\leq -2+3t\leq1\), so \(\gamma(t)\in[-3,2]\) throughout. The function is continuous, and therefore it is a path in \(E\) from \(-2\) to \(1\).
This construction works for any two points \(x,y\) in any interval: use \(\gamma(t)=(1-t)x+ty\). For each \(t\in[0,1]\), the value \((1-t)x+ty\) lies between \(x\) and \(y\), so it belongs to the interval.
Intervals Admit Paths
The converse follows from the simplest possible route between two real numbers: the straight-line interpolation between them. An interval contains every point between any two of its points, so this interpolation stays inside the set. Together with the preceding theorem, this gives the full characterization.
Proof. If \(E\) is path connected, the preceding theorem shows that it is an interval. Conversely, suppose \(E\) is an interval. If \(E\) is empty, it is path connected by the definition. If \(E\) is nonempty, take any \(x,y\in E\) and define \(\gamma(t)=(1-t)x+ty\) for \(t\in[0,1]\). This function is continuous, and \(\gamma(0)=x\), \(\gamma(1)=y\). For every \(t\in[0,1]\), \(\gamma(t)\) lies between \(x\) and \(y\), so it belongs to \(E\). Thus \(\gamma\) is a path in \(E\) joining \(x\) to \(y\), and \(E\) is path connected.
Earlier in this course, the Characterization of Connected Subsets of \(\mathbb{R}\) established that connected sets in the real line are exactly the intervals. Combining that result with the equivalence just proved gives the connectedness formulation. \(\square\)
Worked Example: A Path-Connected Set With Open Endpoints
Consider \(E=(1,5)\). Given any \(x,y\in E\), define \(\gamma(t)=(1-t)x+ty\). The value \(\gamma(t)\) lies between \(x\) and \(y\), and both endpoints lie strictly between \(1\) and \(5\). Consequently \(1<\gamma(t)<5\) for every \(t\in[0,1]\). The path stays in \(E\), so \(E\) is path connected.
The same proof applies to unbounded intervals such as \((4,\infty)\) and \(\mathbb{R}\). No boundedness or inclusion of endpoints is needed: the interpolation between two points of an interval remains between those points.
When Points Cannot Be Joined
A path-connected set must contain the interval between any two of its points. This gives a quick way to rule out paths: find a missing point strictly between the proposed endpoints. The argument does not depend on whether a candidate path is straight, curved in some imagined sense, or repeatedly changes direction. Every continuous map into \(\mathbb{R}\) still has the intermediate-value property.
Worked Example: No Path Across a Missing Point
Let \(E=\mathbb{R}\setminus\{0\}\). There is no path in \(E\) from \(-2\) to \(3\). If \(\gamma:[0,1]\to E\) were such a path, then \(\gamma\) would be continuous, \(\gamma(0)=-2\), and \(\gamma(1)=3\). By the Intermediate Value Theorem, \(\gamma(t)=0\) for some \(t\in[0,1]\). This contradicts the requirement that every value of \(\gamma\) lie in \(E\).
Points on the same side of zero can be joined: the straight-line path between two positive numbers stays positive, and the straight-line path between two negative numbers stays negative. Thus the two intervals \((-\infty,0)\) and \((0,\infty)\) are the path-connected pieces of \(E\).
Worked Example: The Rational Numbers Have No Nonconstant Paths
Let \(E=\mathbb{Q}\), and suppose \(q,r\in\mathbb{Q}\) with \(q<r\). If a path in \(\mathbb{Q}\) joined \(q\) to \(r\), its image would be a connected subset of \(\mathbb{R}\) containing \(q\) and \(r\). It would therefore contain every point of \([q,r]\). But \(q+(r-q)/\sqrt{2}\) lies strictly between \(q\) and \(r\) and is irrational, so it does not belong to \(\mathbb{Q}\). This is a contradiction.
The expression is irrational because \(r-q\) is a nonzero rational number and \(1/\sqrt{2}\) is irrational; multiplying a nonzero rational by an irrational number remains irrational. It lies strictly between \(q\) and \(r\) because \(0<1/\sqrt{2}<1\). Hence no two distinct rational numbers can be joined by a path in \(\mathbb{Q}\). Each singleton is path connected, so the path-connected pieces of \(\mathbb{Q}\) are singletons.
Path Components on the Real Line
For any set, one can group together points that can be joined by paths. The resulting maximal path-connected subsets are called path components. In \(\mathbb{R}\), these pieces agree exactly with the connected components. The reason is specific and simple: a connected component is an interval, and any two points in an interval can be joined by straight-line interpolation.
Proof. Path equivalence is an equivalence relation. The constant path joins each point to itself. If \(\gamma\) joins \(x\) to \(y\), then \(t\mapsto\gamma(1-t)\) joins \(y\) to \(x\). If \(\alpha\) joins \(x\) to \(y\) and \(\beta\) joins \(y\) to \(z\), define \(\gamma(t)=\alpha(2t)\) for \(0\leq t\leq 1/2\), and \(\gamma(t)=\beta(2t-1)\) for \(1/2\leq t\leq1\). Both formulas give \(y\) at \(t=1/2\), so the pieces join continuously. This path joins \(x\) to \(z\), proving transitivity.
Each path component is path connected: if \(u\) and \(v\) belong to the same equivalence class, there is a path from \(u\) to \(v\). By the Path Connected Sets in \(\mathbb{R}\) Are Connected theorem, each path component is connected.
Now fix \(x\in E\), let \(P\) be its path component, and let \(C_x\) be its connected component. Because \(P\) is connected and contains \(x\), the Maximality Characterization of Components gives \(P\subseteq C_x\). Conversely, \(C_x\) is connected, so it is an interval by the Characterization of Connected Subsets of \(\mathbb{R}\). For each \(y\in C_x\), the straight-line path from \(x\) to \(y\) stays in \(C_x\), and hence in \(E\). Thus \(y\in P\), which gives \(C_x\subseteq P\). Therefore \(P=C_x\). Since every point belongs to its connected component, this proves that the path components and connected components coincide. \(\square\)
This result is useful when identifying the pieces of a subset of the real line: either connectedness or path connectedness gives the same decomposition. Keep the setting in mind, however. The proof that intervals are path connected uses the real-line interpolation \((1-t)x+ty\). The equivalence established here is a feature of subsets of \(\mathbb{R}\), not a definition of connectedness itself.
Try the straight-line formula \(\gamma(t)=(1-t)x+ty\), and verify that its values remain in the set.
Check whether the interval between the proposed endpoints contains a point missing from the set.
On the real line, find the connected components; they are exactly the path components.
Check Your Understanding
Use the definition of a path and the interval characterization to answer the following questions.
- Why does a path from \(x\) to \(y\) in a subset of \(\mathbb{R}\) force that set to contain every point between \(x\) and \(y\)?
- Write a path joining two given points \(x,y\) of an interval, and explain why it stays in the interval.
- Why can no path in \(\mathbb{R}\setminus\{0\}\) join a negative point to a positive point?
- Which part of the argument shows that the path components of a subset of \(\mathbb{R}\) are connected components?
- What path-connected pieces does \(\mathbb{Q}\) have, and why can no two distinct rational numbers be joined within \(\mathbb{Q}\)?