Two Notions of a Component
A space can split into pieces in more than one way, depending on what kinds of subsets are allowed to count as a single piece. Connected components group points that belong to the same connected subset. Path components group points that can be joined by a path. The word maximal is important in both definitions: a component cannot be enlarged while retaining the relevant property.
Throughout, \((X,\mathcal{T})\) is a topological space. A subset \(C\subseteq X\), with the subspace topology, is connected if it cannot be written as the union of two disjoint, nonempty sets that are open in \(C\). A subset \(P\subseteq X\) is path-connected if for every \(x,y\in P\) there is a continuous map \(\gamma:I\to P\) with \(\gamma(0)=x\) and \(\gamma(1)=y\), where \(I=[0,1]\) has its usual topology.
Maximal does not mean “largest by number of points.” It means that no strictly larger subset of the specified kind contains the subset. In particular, the definition does not assume that \(X\) itself is connected or path-connected.
Connected Components Exist and Are Unique
Proof. Let \(\mathcal{C}_x\) be the collection of connected subsets of \(X\) that contain \(x\), and put \(C_x=\bigcup_{C\in\mathcal{C}_x}C\). The collection is nonempty because the singleton \(\{x\}\) is connected: it cannot be divided into two nonempty subsets. We first prove that \(C_x\) is connected. Suppose to the contrary that \(C_x=U\cup V\), where \(U\) and \(V\) are disjoint, nonempty, and open in \(C_x\). Since \(x\in C_x\), assume \(x\in U\), after switching the names if necessary. For every \(C\in\mathcal{C}_x\), the sets \(U\cap C\) and \(V\cap C\) are disjoint and open in \(C\), and their union is \(C\). Also \(U\cap C\) contains \(x\). If \(V\cap C\) were nonempty, these sets would disconnect \(C\). Thus \(V\cap C=\varnothing\) for every \(C\in\mathcal{C}_x\), which implies \(V=\varnothing\), a contradiction. Therefore \(C_x\) is connected.
Every connected set containing \(x\) is included in \(C_x\) by its definition. Hence \(C_x\) is maximal connected and is a connected component. To see uniqueness, suppose \(D\) is another connected component containing \(x\). Then \(D\subseteq C_x\), since \(D\in\mathcal{C}_x\). Since \(C_x\) is connected and contains \(D\), maximality of \(D\) gives \(C_x\subseteq D\). Thus \(C_x=D\). \(\square\)
This proof also establishes a useful maximality test: any connected subset containing \(x\) lies inside the connected component of \(x\). Consequently, two connected components that meet must be equal. Indeed, if they meet at \(x\), each is contained in the component \(C_x\), and maximality forces each to equal \(C_x\).
Path Components and Their Maximality
Paths give a direct way to construct path components. Say \(x\sim y\) if there is a path in \(X\) from \(x\) to \(y\). Constant paths give \(x\sim x\); reversing a path gives \(y\sim x\); and concatenating paths gives transitivity. Thus \(\sim\) is an equivalence relation. Its equivalence classes partition \(X\).
Proof. Fix \(x\in X\), and let \(P_x=\{y\in X:x\sim y\}\). If \(y,z\in P_x\), choose paths from \(x\) to \(y\) and from \(x\) to \(z\). Reverse the first path and concatenate it with the second. The result is a path from \(y\) to \(z\), so \(P_x\) is path-connected. If a path-connected set \(A\) contains \(P_x\), then it contains \(x\); for each \(y\in A\), path-connectedness gives a path from \(x\) to \(y\), so \(y\in P_x\). Therefore \(A\subseteq P_x\), and \(P_x\) is maximal path-connected. Conversely, if \(A\) is a maximal path-connected set and \(x\in A\), every point of \(A\) is joined to \(x\) by a path in \(A\), so \(A\subseteq P_x\). Since \(P_x\) is path-connected, maximality gives \(A=P_x\). \(\square\)
Equivalence classes are either equal or disjoint. Thus every point belongs to exactly one path component, and two path components that intersect are equal. The construction is parallel to the connected-component construction, but it uses the relation “joined by a path” rather than the larger condition “together in a connected subset.”
Worked Examples
Worked Example: Components of the Real Line
Give \(\mathbb{R}\) its standard topology. Find its connected components and path components.
The real line is connected, so its only connected component is \(\mathbb{R}\). One way to check connectedness is the standard connectedness theorem for intervals: every interval in \(\mathbb{R}\) is connected, and \(\mathbb{R}\) is an interval.
For any \(x,y\in\mathbb{R}\), define \(\gamma(t)=(1-t)x+ty\) for \(t\in I\). This map is continuous, \(\gamma(0)=x\), and \(\gamma(1)=y\). Its values lie in \(\mathbb{R}\), so \(\mathbb{R}\) is path-connected. Since no subset of \(\mathbb{R}\) can be larger than \(\mathbb{R}\), its only path component is also \(\mathbb{R}\).
Check: If \(x=2\), \(y=5\), and \(t=1/4\), the formula gives \(\gamma(1/4)=(3/4)2+(1/4)5=11/4\). The endpoint values are \(\gamma(0)=2\) and \(\gamma(1)=5\), as required.
Worked Example: Components of the Rational Numbers
Give \(\mathbb{Q}\) the subspace topology inherited from the standard topology on \(\mathbb{R}\). Show that both its connected components and its path components are singletons.
Take distinct \(a,b\in\mathbb{Q}\), with \(a<b\). Choose an irrational \(c\) with \(a<c<b\). The sets \(\mathbb{Q}\cap(-\infty,c)\) and \(\mathbb{Q}\cap(c,\infty)\) are disjoint, nonempty, open in \(\mathbb{Q}\), and together cover \(\mathbb{Q}\). Their intersections with any subset \(A\subseteq\mathbb{Q}\) containing \(a\) and \(b\) give a separation of \(A\): both intersections are nonempty, disjoint, open in \(A\), and cover \(A\). Thus no connected subset of \(\mathbb{Q}\) contains two distinct points. Every singleton is connected, so the connected components are exactly the singletons.
Now suppose a path \(\gamma:I\to\mathbb{Q}\) joined distinct rationals \(a\) and \(b\). Composing with the inclusion \(\mathbb{Q}\subseteq\mathbb{R}\) gives a continuous real-valued function on \(I\). The intermediate value theorem says it takes every real value between \(a\) and \(b\), including an irrational number. That contradicts the fact that every value of \(\gamma\) is rational. Hence no two distinct points of \(\mathbb{Q}\) lie in the same path component, and the path components are also singletons.
Check: The argument uses strict inequalities \(a<c<b\), so \(a\) and \(b\) lie in opposite open sets. A singleton cannot be split into two nonempty subsets, and its constant path verifies path-connectedness.
Worked Example: A Connected Space with Different Components
Consider the standard topologist’s sine curve \(S=\{(x,\sin(1/x)):0<x\leq 1\}\cup(\{0\}\times[-1,1])\), with the subspace topology inherited from the standard topology on \(\mathbb{R}^2\). This is a standard example of a connected space that is not path-connected. Its connected component is all of \(S\), while it has more than one path component: a path component cannot be all of \(S\), because that would make \(S\) path-connected.
This example shows why the two definitions must be kept separate. The connected-component question asks whether the entire set can be split by a separation. The path-component question asks whether every pair of points can be joined by a continuous path staying in the set. For this particular curve, verifying connectedness and determining its path components requires a detailed analysis of oscillation near \(x=0\); the example is included here to mark the distinction, not as a shortcut proof of those facts.
A Common Pitfall: Maximal Is Not Maximum
A common error is to assume that every component is the unique largest connected subset of \(X\). The definitions guarantee maximal subsets, and the theorem guarantees that connected components partition \(X\); neither definition says that one component must contain all the others. For example, in the discrete topology on \(X=\{a,b\}\), every subset is open. The set \(\{a,b\}\) is disconnected, since \(\{a\}\) and \(\{b\}\) separate it. The connected components are \(\{a\}\) and \(\{b\}\), not a single largest connected subset.
The same example checks the path notion directly. A continuous map from the connected interval \(I\) into this discrete two-point space cannot take both values: the preimages of \(\{a\}\) and \(\{b\}\) would be disjoint, nonempty, open sets covering \(I\). Thus the path components are also \(\{a\}\) and \(\{b\}\). The correct conclusion is that the components are maximal pieces of the specified kind, not that there is one maximum piece.
Check Your Understanding
For each question, distinguish carefully between connectedness and path-connectedness.
- Explain why a singleton is both connected and path-connected.
- In the discrete topology on a set \(X\), what are the connected components and path components?
- Why do two connected components that intersect have to be equal?
- State the equivalence relation whose classes are the path components.
- What does “maximal” mean in the definitions of connected component and path component, and how does it differ from “maximum”?