Paths Without a Metric
A path is a continuous way to move from one point to another, with the parameter ranging over the interval \([0,1]\). No distance is needed: continuity is understood using the topologies on the interval and on the space. This lets us discuss paths in spaces that are not metric spaces, and it makes path connectedness a property of topology itself.
Earlier in this course, path connectedness and path components were introduced for metric spaces. The same definitions work in arbitrary topological spaces. We will also examine a product-space result: paths in a product are precisely paths whose coordinate functions are continuous. That observation gives a useful description of both path connected products and their path components.
With this definition, the empty space is path connected vacuously, since it has no pair of points to test. A singleton is path connected by its constant path. For a subset \(A\subseteq X\), path connectedness refers to \(A\) with its subspace topology; a path in \(A\) is equivalently a continuous map into \(X\) whose image lies in \(A\).
Reversing and Joining Paths
Two elementary constructions will be useful. If \(\gamma\) is a path from \(x\) to \(y\), define its reverse by \(\bar\gamma(t)=\gamma(1-t)\). The map \(t\mapsto 1-t\) is continuous, so \(\bar\gamma\) is continuous, and \(\bar\gamma(0)=y\), \(\bar\gamma(1)=x\).
If \(\gamma\) goes from \(x\) to \(y\) and \(\eta\) goes from \(y\) to \(z\), they can be joined into one path:
At the joining parameter, both formulas give \(y\): the first gives \(\gamma(1)=y\), and the second gives \(\eta(0)=y\). The resulting map is continuous by the pasting lemma, since the two formulas are continuous on the closed intervals \([0,\frac12]\) and \([\frac12,1]\) and agree on their intersection. Its endpoints are \(x\) and \(z\). Thus paths can be reversed and concatenated without requiring any metric structure.
Worked Example: A Path Along the Unit Circle
Let \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\), with the subspace topology. A path from \((1,0)\) to \((0,1)\) is \(\gamma(t)=(\cos(\pi t/2),\sin(\pi t/2))\). Its coordinate functions are continuous, so the map into \(\mathbb{R}^2\) is continuous; since its image lies in \(S^1\), it is also continuous as a map into the subspace \(S^1\).
For every \(t\in[0,1]\), \(\cos^2(\pi t/2)+\sin^2(\pi t/2)=1\), so \(\gamma(t)\in S^1\). At the endpoints, \(\gamma(0)=(\cos 0,\sin 0)=(1,0)\) and \(\gamma(1)=(\cos(\pi/2),\sin(\pi/2))=(0,1)\). More generally, any two points of \(S^1\) can be written as \((\cos\alpha,\sin\alpha)\) and \((\cos\beta,\sin\beta)\) for suitable angles. Replacing \(\pi t/2\) by \((1-t)\alpha+t\beta\) gives a path between them.
Worked Example: An Indiscrete Space Is Path Connected
Let \(X=\{a,b\}\) have the indiscrete topology \(\{\varnothing,X\}\). Define \(\gamma:[0,1]\to X\) by \(\gamma(t)=a\) when \(t<1\), and \(\gamma(1)=b\). The only open sets in \(X\) are \(\varnothing\) and \(X\), whose preimages are respectively \(\varnothing\) and \([0,1]\); both are open in \([0,1]\). Therefore \(\gamma\) is continuous, and it joins \(a\) to \(b\). The space is path connected.
The formula for a map alone does not determine whether it is continuous: continuity depends on the topology of the target. Here even a map that changes value at one parameter is continuous because the indiscrete topology has so few open sets to test.
Worked Example: A Discrete Two-Point Space Is Not Path Connected
Give \(D=\{a,b\}\) the discrete topology. Suppose a path \(\gamma:[0,1]\to D\) joined \(a\) to \(b\). The set \(\{a\}\) is both open and closed in \(D\), so continuity would make \(\gamma^{-1}(\{a\})\) both open and closed in \([0,1]\). This preimage contains \(0\) but not \(1\), so it is nonempty and is not all of \([0,1]\). That contradicts connectedness of the interval \([0,1]\). Hence no such path exists, and \(D\) is not path connected.
This example shows why a path needs more than its endpoints: the topology can prevent any continuous map from changing between them.
Paths in Product Spaces
For a product \(X=\prod_{i\in I}X_i\) with the product topology, a map into \(X\) is continuous if and only if each of its coordinate maps is continuous, as established in the continuity theorem for product spaces. Consequently, a path in a product is exactly a collection of coordinate paths with a common parameter \(t\in[0,1]\).
Proof. First suppose every factor is path connected. Take any \(x=(x_i)_{i\in I}\) and \(y=(y_i)_{i\in I}\) in \(X\). For each \(i\), choose a path \(\gamma_i:[0,1]\to X_i\) from \(x_i\) to \(y_i\). Define \(\gamma:[0,1]\to X\) by \(\gamma(t)=(\gamma_i(t))_{i\in I}\). Every coordinate map of \(\gamma\) is continuous, so the product continuity theorem shows that \(\gamma\) is continuous. Also \(\gamma(0)=x\) and \(\gamma(1)=y\). Thus \(X\) is path connected.
Conversely, suppose \(X\) is path connected. Since \(X\) is nonempty, fix \(z=(z_i)_{i\in I}\in X\). For any fixed index \(j\) and any \(u,v\in X_j\), form points \(x,y\in X\) by setting \(x_j=u\), \(y_j=v\), and \(x_i=y_i=z_i\) for \(i\ne j\). A path \(\gamma\) in \(X\) from \(x\) to \(y\) exists. Its \(j\)-th coordinate, \(t\mapsto p_j(\gamma(t))\), is continuous because the projection \(p_j:X\to X_j\) is continuous. This coordinate path joins \(u\) to \(v\), so \(X_j\) is path connected. Since \(j\) was arbitrary, every factor is path connected. \(\square\)
The nonempty hypothesis matters for the converse. If one factor is empty, the product is empty and is path connected under the vacuous definition above, even if other factors are not path connected. If the index set is empty, the product is a singleton, so the theorem holds as well: there are no factors to check.
Worked Example: A Product of a Line and a Circle
Both \(\mathbb{R}\) and \(S^1\) are path connected. For \(\mathbb{R}\), the map \(\gamma(t)=(1-t)r+ts\) joins any real numbers \(r\) and \(s\). Path connectedness of \(S^1\) was verified in the circle example. The product theorem therefore shows that \(\mathbb{R}\times S^1\), with the product topology, is path connected.
For explicit endpoints \((0,(1,0))\) and \((2,(0,1))\), one path is \(\Gamma(t)=(2t,(\cos(\pi t/2),\sin(\pi t/2)))\). Its first coordinate is continuous, and its second coordinate is continuous, so the product continuity theorem gives continuity of \(\Gamma\). Substitution gives \(\Gamma(0)=(0,(1,0))\) and \(\Gamma(1)=(2,(0,1))\).
Path Components of Products
For \(x\in X\), write \(P_X(x)\) for the path component of \(x\): the set of all points of \(X\) that can be joined to \(x\) by a path. Path components and their partition property were studied earlier for metric spaces. The coordinate description below gives an additional product-space conclusion.
Proof. Suppose \(y\in P_X(x)\). There is a path \(\gamma:[0,1]\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=y\). For each \(i\), the coordinate map \(p_i\circ\gamma\) is continuous and joins \(x_i\) to \(y_i\). Therefore \(y_i\in P_{X_i}(x_i)\) for every \(i\), which proves \(P_X(x)\subseteq\prod_{i\in I}P_{X_i}(x_i)\).
For the reverse inclusion, take \(y\in\prod_{i\in I}P_{X_i}(x_i)\). For each \(i\), choose a path \(\gamma_i\) in \(X_i\) from \(x_i\) to \(y_i\). Define \(\gamma(t)=(\gamma_i(t))_{i\in I}\). Since all coordinate maps are continuous, the product continuity theorem makes \(\gamma\) continuous. Its endpoints are \(x\) and \(y\), so \(y\in P_X(x)\). Thus \(\prod_{i\in I}P_{X_i}(x_i)\subseteq P_X(x)\), proving equality. \(\square\)
Worked Example: Path Components of \(\mathbb{R}\times\mathbb{Q}\)
Give \(\mathbb{R}\) its usual topology and \(\mathbb{Q}\) the subspace topology inherited from \(\mathbb{R}\). Every two real numbers are joined by a straight-line path. In contrast, every path in \(\mathbb{Q}\) is constant. Indeed, a path \(\gamma:[0,1]\to\mathbb{Q}\), viewed as a map into \(\mathbb{R}\), has connected image by the theorem on continuous images of connected spaces, since \([0,1]\) is connected. A connected subset of \(\mathbb{R}\) is an interval. If the image contained distinct rationals \(q<r\), it would contain every real number between them, including an irrational number, which is impossible for an image lying in \(\mathbb{Q}\).
It follows that the path component of \(q\in\mathbb{Q}\) is \(\{q\}\). Applying the product theorem, the path component of \((r,q)\in\mathbb{R}\times\mathbb{Q}\) is \(\mathbb{R}\times\{q\}\). This set is path connected: for any \(s,t\in\mathbb{R}\), the map \(\gamma(u)=((1-u)s+ut,q)\) joins \((s,q)\) to \((t,q)\). The product has many path components, one for each rational coordinate.
What Path Connectedness Tells Us
Path connectedness is stronger than connectedness. Earlier in the course, the topologist’s sine curve was shown to be connected but not path connected. In the other direction, the theorem “Path Connected Implies Connected” established earlier for metric spaces follows from the same basic idea in the topological setting: a path is a continuous image of the connected interval, and a separation of its image would contradict connectedness of that interval. In particular, a path-connected space cannot be divided into separated pieces, but connectedness alone does not guarantee that points can be joined by paths.
The product results explain how this stronger property behaves coordinate by coordinate. To connect two points in a product, each coordinate must be able to move between its corresponding endpoints. Conversely, choosing such coordinate paths produces a path in the product topology. The same reasoning identifies the path component of a point as the product of the path components of its coordinates.
A common pitfall is to assume that a continuous path can be constructed by drawing a plausible curve without checking continuity in the space’s topology. In a product, checking each coordinate is sufficient; in an unusual topology, as the indiscrete example illustrates, continuity may behave very differently from intuition based on distance. The endpoint conditions and continuity of the map must both be verified.
Check Your Understanding
Use the definitions and product results to answer the following questions.
- Why is a path required to be continuous as a map from the usual interval \([0,1]\), rather than merely to have specified endpoints?
- How does the pasting lemma justify joining two paths that meet at a common endpoint?
- Why is the nonempty-product hypothesis needed for the converse direction of the path-connected product theorem?
- Describe the path component of \((r,q)\) in \(\mathbb{R}\times\mathbb{Q}\).
- Give one reason a connected space might fail to be path connected.