Two Notions of Component
Connectedness asks whether a space can be separated into two disjoint, nonempty open pieces. Path-connectedness asks whether any two points can be joined by a continuous map from the interval. The first condition is weaker: a path-connected space is connected, but a connected space need not be path-connected. The component statements in this tutorial describe precisely how the two notions fit together.
Let \((X,\mathcal{T})\) be a topological space. A separation of \(X\) is a pair of disjoint, nonempty open subsets whose union is \(X\). A space is connected if it has no separation. A subset is connected when it is connected in the subspace topology. A path from \(x\) to \(y\) is a continuous map \(\gamma:I\to X\), where \(I=[0,1]\) has its standard topology, with \(\gamma(0)=x\) and \(\gamma(1)=y\). A subset is path-connected if every pair of its points can be joined by a path taking values in that subset.
We will show that each point lies in exactly one connected component and exactly one path component. If \(X=\varnothing\), there are no points and hence no components; the statements about a component containing a point are then vacuous.
Connected Components Contain Path Components
Proof. If \(y,z\in P(x)\), 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 \(z\). Reversal and concatenation preserve continuity, so this gives a path from \(y\) to \(z\) whose image lies in \(P(x)\). Thus \(P(x)\) is path-connected.
We use the standard result that a path-connected space is connected. Its reason is that a separation of a path-connected space would, by choosing a point from each side and taking a path between them, give a separation of \(I\) by the two preimages. The interval \(I\) is connected, so this is impossible. Therefore \(P(x)\) is connected.
For completeness, a connected subset containing \(x\) lies in a maximal connected subset containing \(x\). Indeed, take the union \(C(x)\) of all connected subsets of \(X\) containing \(x\). This union is connected: if it had a separation, \(x\) would lie on one side, and each connected subset in the union, since it contains \(x\), would have to lie entirely on that side. The other side could then contain no point of the union, a contradiction. The union is maximal because it includes every connected subset containing \(x\). Since \(P(x)\) is connected and contains \(x\), it is among the subsets in this union, so \(P(x)\subseteq C(x)\). \(\square\)
The same union argument shows that two connected components that intersect must be equal: their union is connected, so maximality forces each to be the whole union. Consequently, the connected component containing any given point is unique. Path components also partition \(X\): paths can be reversed and concatenated, so being joined by a path is an equivalence relation.
Proof. Let \(C\) be a connected component and let \(P(x)\) be a path component that meets \(C\). Choose \(y\in P(x)\cap C\). The path component \(P(y)\) equals \(P(x)\), since paths can be reversed and concatenated. Also \(P(y)\) is contained in the connected component containing \(y\), by the proposition. That component is \(C\), by uniqueness of the connected component containing \(y\). Hence \(P(x)\subseteq C\). Every point of \(C\) lies in its own path component, which meets \(C\). These two containments prove the stated union. \(\square\)
The theorem does not say that the union has only one path component. It says that path components cannot cross from one connected component into another, while a single connected component may contain several distinct path components.
Worked Example: The Real Line
Give \(\mathbb{R}\) its standard topology. We determine its connected and path components.
For any \(a,b\in\mathbb{R}\), define \(\gamma:I\to\mathbb{R}\) by \(\gamma(t)=(1-t)a+tb\). This is continuous, \(\gamma(0)=a\), and \(\gamma(1)=b\). Thus \(\mathbb{R}\) is path-connected, and its only path component is \(\mathbb{R}\). By the result that path-connected spaces are connected, it is connected as well, so its only connected component is \(\mathbb{R}\).
Check: every point of \(\mathbb{R}\) is joined to \(0\) by the path \(\gamma(t)=(1-t)0+tb=tb\), where \(b\) is that point. So the computed path component does contain every real number. Here the two types of component agree.
Worked Example: Two Separated Intervals
Let \(X=[0,1]\cup[2,3]\) have the subspace topology inherited from the standard topology on \(\mathbb{R}\). Set \(A=[0,1]\) and \(B=[2,3]\). In \(X\), \(A=X\cap(-1,1.5)\) and \(B=X\cap(1.5,4)\), so both are open in \(X\). Each is also the complement in \(X\) of the other.
No connected subset \(D\) of \(X\) can meet both \(A\) and \(B\). If it did, \(D\cap A\) and \(D\cap B\) would be disjoint, nonempty, relatively open subsets of \(D\) covering \(D\), a separation. On the other hand, \(A\) is connected because it is an interval, and \(B\) is connected for the same reason. Therefore \(A\) and \(B\) are exactly the connected components of \(X\).
Each of \(A\) and \(B\) is path-connected. For \(u,v\) in either interval, \(\gamma(t)=(1-t)u+tv\) stays between \(u\) and \(v\), hence stays in that interval, and joins \(u\) to \(v\). A path in \(X\) cannot join a point of \(A\) to a point of \(B\): the continuous image of \(I\) would be connected and meet both sets, which was just shown impossible. Thus the path components are also \(A\) and \(B\).
Check: the path formula has endpoints \(\gamma(0)=u\) and \(\gamma(1)=v\). If \(u,v\in[0,1]\), every value lies in \([0,1]\); if \(u,v\in[2,3]\), every value lies in \([2,3]\). The computed paths respect the subspace \(X\).
A Connected Component with Many Path Components
To see why the containment can be strict, consider an uncountable set \(X=\mathbb{R}\) with the countable-complement topology: the open sets are \(\varnothing\) and all subsets whose complement in \(X\) is countable. This is a topology: arbitrary unions of nonempty open sets have complements contained in the countable complement of any one member, and finite intersections of nonempty open sets have complements that are finite unions of countable sets.
Worked Example: The Countable-Complement Topology
First, \(X\) is connected. If \(U,V\) were disjoint, nonempty open sets covering \(X\), then both \(X\setminus U\) and \(X\setminus V\) would be countable. Since \(U\cap V=\varnothing\) and \(U\cup V=X\), we would have \(X=(X\setminus U)\cup(X\setminus V)\), making \(X\) countable. This contradicts the uncountability of \(\mathbb{R}\). Thus \(X\) has no separation, and its connected component is all of \(X\).
Now let \(f:I\to X\) be continuous. We show that \(f\) must be constant. Write \(Y=f(I)\) for its image. Suppose first that \(Y\) is countable and contains at least two points. Choose \(y\in Y\). Since \(Y\setminus\{y\}\) is countable, \(X\setminus(Y\setminus\{y\})\) is open. Its preimage under \(f\) is \(f^{-1}(\{y\})\), so this fiber is open in \(I\). Also \(X\setminus\{y\}\) is open, so \(I\setminus f^{-1}(\{y\})\) is open. The fiber is therefore both open and closed, nonempty, and proper. That contradicts the connectedness of \(I\).
Suppose instead that \(Y\) is uncountable. The set \(D=\mathbb{Q}\cap I\) is a countable dense subset of \(I\), and \(C=f(D)\) is countable. The set \(X\setminus C\) is open, so \(f^{-1}(C)\) is closed in \(I\). It contains \(D\), and a closed set containing a dense subset is all of \(I\). Hence \(f(I)\subseteq C\), contradicting that \(Y\) is uncountable. Both possibilities for a nonconstant \(f\) lead to contradictions; thus every continuous path \(I\to X\) is constant.
It follows that every path component is a singleton: a path can join a point only to itself. In contrast, the connected component is all of \(X\). For example, the path component of \(0\) is \(\{0\}\), and \(\{0\}\) is properly contained in the connected component \(X=\mathbb{R}\).
Check: the two cases cover every possible image \(Y\): it is either countable or uncountable. In the countable case, the assumption that \(Y\) has at least two points makes the fiber nonempty and proper; in the uncountable case, the countability of \(f(D)\) contradicts the conclusion that it contains the whole image.
A Common Mistake
Common pitfall: concluding that a connected component must itself be a path component. For the countable-complement topology just examined, that would give the false calculation \(C(0)=P(0)=\{0\}\). The corrected calculation is \(P(0)=\{0\}\), because every path is constant, while \(C(0)=\mathbb{R}\), because the whole space is connected. The valid general conclusion goes in only one direction: every path component is contained in a connected component.
This distinction matters whenever connectedness is used to describe the large-scale pieces of a space. Paths provide a particularly concrete way of moving between points, but the absence of paths does not by itself separate a space into disconnected pieces. A connected component may consist of many path components, as the example demonstrates.
Check Your Understanding
For each question, distinguish containment in a connected component from equality with a path component.
- Explain why the path component \(P(x)\) is path-connected, using reversal and concatenation of paths.
- Why can two connected components not intersect without being equal?
- In \(X=[0,1]\cup[2,3]\) with the subspace topology, identify the connected components and path components.
- For the countable-complement topology on \(\mathbb{R}\), why does a countable nontrivial path image contradict connectedness of \(I\)?
- State the conclusion that always holds between path components and connected components, and explain why its converse can fail.