Two Kinds of Components
A space can be connected even when some pairs of points cannot be joined by a path. The distinction is between connected components, which are defined using connected subsets, and path components, which are defined using paths. Local path connectedness supplies the key link: it makes each path component open. Since path components partition the space, a connected component cannot contain points from two different ones.
A subset \(C\) of a topological space \((X,\mathcal{T})\) is connected if it cannot be written as the union of two disjoint, nonempty sets open in the subspace topology on \(C\). A path from \(x\) to \(y\) in \(X\) is a continuous map \(\gamma:I\to X\), where \(I=[0,1]\), such that \(\gamma(0)=x\) and \(\gamma(1)=y\). A subset is path-connected if every pair of its points can be joined by a path whose image lies in that subset.
Path components really do form a partition. Every point is joined to itself by a constant path. Reversing a path joins its endpoints in the other order, and concatenating two paths with a shared endpoint gives a path joining the remaining endpoints. For concatenation, reparameterize the first path on \([0,1/2]\) and the second on \([1/2,1]\); the two formulas agree at \(1/2\), so the resulting map is continuous. Thus “can be joined by a path” is an equivalence relation, and its equivalence classes are the path components.
Why Path-Connected Sets Are Connected
We will need the fact that a path-connected subset is connected. The argument reduces to connectedness of the interval \(I\). Here is a brief justification of that interval fact. If \(I=A\cup B\) were a separation, choose points in the two sets and, after reversing their order or the names of the sets if necessary, take \(a<b\) with \(a\in A\) and \(b\in B\). Let \(c=\sup(A\cap[a,b])\). If \(c\in A\) and \(c<b\), relative openness of \(A\) gives points of \(A\cap[a,b]\) greater than \(c\), contradicting the supremum; \(c=b\) is impossible because \(b\in B\). If \(c\in B\), relative openness of \(B\) and the definition of supremum give a point of \(A\) arbitrarily close to \(c\) from below inside a neighborhood contained in \(B\), also a contradiction. Therefore \(I\) is connected.
Proof. Let \(A\subseteq X\) be path-connected, and suppose for contradiction that \(A=E\cup F\) is a separation: \(E\) and \(F\) are disjoint, nonempty, and open in \(A\). Choose \(e\in E\) and \(f\in F\). A path \(\gamma:I\to A\) joins \(e\) to \(f\). The sets \(\gamma^{-1}(E)\) and \(\gamma^{-1}(F)\) are disjoint, nonempty, relatively open in \(I\), and cover \(I\), since \(\gamma\) is continuous and \(E,F\) are open in \(A\). This contradicts connectedness of \(I\). Hence \(A\) is connected. \(\square\)
Connected components exist as well. For each \(x\in X\), take the union of all connected subsets containing \(x\). This union is connected: if it had a separation, \(x\) would lie on one side, and each connected set in the union, since it contains \(x\), would have to lie wholly on that same side. The other side would then be empty. The union is maximal because any connected set containing it also contains \(x\), and so was already included in the union. This construction applies to every \(x\); if \(X=\varnothing\), there are no components.
The Component Theorem
Proof. We first show that each path component is open. Let \(P\) be the path component of \(x\), and take any \(y\in P\). Local path connectedness, applied to the open set \(X\), gives an open path-connected neighborhood \(V\) of \(y\). For any \(z\in V\), there is a path in \(V\) from \(y\) to \(z\). Since \(y\in P\), there is also a path from \(x\) to \(y\); concatenating the paths shows that \(z\in P\). Thus \(V\subseteq P\). Every point of \(P\) therefore has an open neighborhood contained in \(P\), proving that \(P\) is open.
Now let \(P\) be a path component. It is path-connected: any two points in it can be joined by reversing a path to a fixed point and concatenating with another path from that point. The proposition shows that \(P\) is connected. Therefore \(P\subseteq C(x)\) for any \(x\in P\), by maximality of the connected component.
For the reverse inclusion, suppose a connected component \(C(x)\) met both \(P\) and \(X\setminus P\). The set \(C(x)\cap P\) would be nonempty and open in \(C(x)\), because \(P\) is open in \(X\). The set \(C(x)\cap(X\setminus P)\) would also be open in \(C(x)\): the complement of \(P\) is the union of all the other path components, each of which is open. These two sets are disjoint, nonempty, and cover \(C(x)\), contradicting its connectedness. Hence \(C(x)\subseteq P\). Combining the inclusions gives \(C(x)=P\). Since \(x\) was arbitrary, the two kinds of components coincide. \(\square\)
The proof uses both parts of the local hypothesis in a specific way. A path-connected neighborhood of \(y\) lets us extend a path that reaches \(y\) to every nearby point; the neighborhood’s openness then makes the whole path component open. Once those open sets partition the space, connectedness rules out a connected component meeting more than one of them.
Worked Example: Two Intervals in the Real Line
Let \(X=(-2,-1)\cup(1,3)\) with the subspace topology inherited from \(\mathbb{R}\). We find its path and connected components.
Write \(L=(-2,-1)\) and \(R=(1,3)\). If \(a,b\in L\), the formula \(\gamma(t)=(1-t)a+tb\) defines a path in \(L\): for every \(t\in I\), \(\gamma(t)\) lies between \(a\) and \(b\), both of which are strictly between \(-2\) and \(-1\). The same formula gives a path in \(R\) when \(a,b\in R\). Thus each interval is path-connected.
No path in \(X\) can join a point \(a\in L\) to a point \(b\in R\). Such a path, regarded as a real-valued continuous function, would take the value \(0\) by the intermediate value theorem, because \(a<0<b\). But \(0\notin X\). Thus the path components are exactly \(L\) and \(R\).
Each point of \(L\) has a small open interval contained in \(L\), and each point of \(R\) has a small open interval contained in \(R\). These intervals are path-connected, so \(X\) is locally path-connected. To check connected components directly, any subset meeting both \(L\) and \(R\) is separated by its intersections with \(L\) and \(R\), which are disjoint, nonempty, and relatively open. Since \(L\) and \(R\) themselves are connected, they are exactly the connected components as well.
Check: the two listed sets are disjoint and their union is \(X\); the path argument rules out joining across them, and the separation argument rules out a connected subset crossing between them. Both component partitions therefore agree.
Worked Example: A Finite Discrete Space
Let \(X=\{a,b,c\}\) have the discrete topology, in which every subset is open. We determine its two types of components.
Each singleton is open and path-connected, since the constant path joins its only point to itself. There is no path from \(a\) to \(b\): if \(\gamma:I\to X\) were such a path, its image would be a connected subset of \(X\), because the continuous image of a connected space is connected. To verify that image fact, a separation of the image would pull back under \(\gamma\) to a separation of \(I\). But every subset of a discrete space with at least two points is disconnected: choose one point in it, and separate that singleton from the remaining points; both pieces are open in the subspace. Thus the image of \(\gamma\) must be a singleton, contradicting \(\gamma(0)=a\) and \(\gamma(1)=b\).
The same argument applies to every distinct pair, so the path components are \(\{a\},\{b\},\{c\}\). These singletons are also connected. No connected subset can contain two points, by the discrete-space separation just described, so the connected components are precisely the same three singletons.
Check: each singleton is open, and its complement is open. Hence it is both open and closed, and no connected subset containing it can also contain another point.
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This space is not discrete. We show that it is nevertheless locally path-connected and has one component of each kind.
Define \(\gamma:I\to X\) by \(\gamma(t)=a\) for \(0\le t<1/2\), and \(\gamma(t)=b\) for \(1/2\le t\le1\). The preimages of the three open sets are \(\gamma^{-1}(\varnothing)=\varnothing\), \(\gamma^{-1}(\{a\})=[0,1/2)\), and \(\gamma^{-1}(X)=I\). Each is open in the subspace topology on \(I\), so \(\gamma\) is continuous. It joins \(a\) to \(b\); therefore \(X\) is path-connected. By the proposition, it is connected too.
For local path connectedness at \(a\), the open neighborhood \(\{a\}\) is path-connected. At \(b\), the only open neighborhood is \(X\), and \(X\) is path-connected by the path just constructed. These neighborhoods verify the definition at both points. Hence the path component and connected component of either point are both all of \(X\).
Check: continuity requires checking preimages of open sets, not whether the path looks continuous in the usual real-line sense. The only nonempty proper open set in this topology has preimage \([0,1/2)\), which is open in \(I\), so the stated path is continuous.
A Common Pitfall: Unions Need a Shared Point
A useful way to construct a connected component is to take the union of all connected sets containing one fixed point. The shared point matters. It is not valid to conclude that an arbitrary union of connected sets is connected.
Common mistake. In \(\mathbb{R}\), the sets \(A=[0,1]\) and \(B=[2,3]\) are each connected, so one might incorrectly infer that \(A\cup B\) is connected. But \(A\cup B\) is the disjoint union of the nonempty relatively open sets \(A\) and \(B\): for example, \(A=(A\cup B)\cap(-1,1.5)\) and \(B=(A\cup B)\cap(1.5,4)\). Thus \(A\cup B\) is disconnected. The component construction in the proof avoids this error because every set in the union contains the same point \(x\).
The theorem’s conclusion also has a precise scope: local path connectedness is what makes the path components open in the proof. Path-connected subsets are always connected, but that fact alone does not show that a path component is a maximal connected subset. The open partition is the additional step that rules out a larger connected component.
Check Your Understanding
Use the definitions and the component theorem to answer the following.
- Why does every path component contain an open neighborhood of each of its points when \(X\) is locally path-connected?
- Explain why a connected component cannot meet two different open path components.
- In the discrete topology on a set with at least two points, what are the path components and connected components?
- For the Sierpinski topology \(\{\varnothing,\{a\},X\}\), which open-set preimages verify continuity of the path in the example?
- Give two connected subsets of \(\mathbb{R}\) whose union is disconnected, and exhibit a separation of that union.