§ 25. Components and Local Connectedness¶
Definition 25.1 : Components
Given \(X\), define an equivalence relation on \(X\) by setting \(x \sim y\) if there is a connected subspace of \(X\) containing both \(x\) and \(y\). The equivalence classes are called the components (or the "connected components") of \(X\).
Theorem 25.0
Given \(X\), define a relation on \(X\) by setting \(x \sim y\) if there is a connected subspace of \(X\) containing both \(x\) and \(y\). The defined relation is indeed, an equivalence relation.
Proof
Symmetry and reflexivity of the relation are obvious. Transitivity follows by noting that if \(A\) is a connected subspace containing \(x\) and \(y\), and if \(B\) is a connected subspace containing \(y\) and \(z\), then \(A \cup B\) is a subspace containing \(x\) and \(z\) that is connected because \(A\) and \(B\) have the point \(y\) in common.
Theorem 25.1
The components of \(X\) are connected disjoint subspaces of \(X\) whose union is \(X\), such that each nonempty connected subspace of \(X\) intersects only one of them.
Proof
Being equivalence classes, the components of \(X\) are disjoint and their union is \(X\). Each connected subspace \(A\) of \(X\) intersects only one of them. For if \(A\) intersects the components \(C_{1}\) and \(C_{2}\) of \(X\), say in points \(x_{1}\) and \(x_{2}\), respectively, then \(x_{1} \sim x_{2}\) by definition; this cannot happen unless \(C_{1}=C_{2}\).
To show the component \(C\) is connected, choose a point \(x_{0}\) of \(C\). For each point \(x\) of \(C\), we know that \(x_{0} \sim x\), so there is a connected subspace \(A_{x}\) containing \(x_{0}\) and \(x\). By the result just proved, \(A_{x} \subset C\). Therefore,
Since the subspaces \(A_{x}\) are connected and have the point \(x_{0}\) in common, their union is connected.
Definition 25.2 : Path Components
We define another equivalence relation on the space \(X\) by defining \(x \sim y\) if there is a path in \(X\) from \(x\) to \(y\). The equivalence classes are called the path components of \(X\).
Theorem 25.1.1
Let's define a relation on the space \(X\) by defining \(x \sim y\) if there is a path in \(X\) from \(x\) to \(y\). The defined relation is indeed, an equivalence relation.
Proof
Let us show this is an equivalence relation. First we note that if there exists a path \(f:[a, b] \rightarrow X\) from \(x\) to \(y\) whose domain is the interval \([a, b]\), then there is also a path \(g\) from \(x\) to \(y\) having the closed interval \([c, d]\) as its domain. (This follows from the fact that any two closed intervals in \(\mathbb{R}\) are homeomorphic.)
Now the fact that \(x \sim x\) for each \(x\) in \(X\) follows from the existence of the constant path \(f:[a, b] \rightarrow X\) defined by the equation \(f(t)=x\) for all \(t\).
Symmetry follows from the fact that if \(f:[0,1] \rightarrow X\) is a path from \(x\) to \(y\), then the "reverse path" \(g:[0,1] \rightarrow X\) defined by \(g(t)=f(1-t)\) is a path from \(y\) to \(x\).
Finally, transitivity is proved as follows: Let \(f:[0,1] \rightarrow X\) be a path from \(x\) to \(y\), and let \(g:[1,2] \rightarrow X\) be a path from \(y\) to \(z\).
We can "paste \(f\) and \(g\) together" to get a path \(h:[0,2] \rightarrow X\) from \(x\) to \(z\); the path \(h\) will be continuous by the "pasting lemma," Theorem 18.3.
Theorem 25.2
The path components of \(X\) are path-connected disjoint subspaces of \(X\) whose union is \(X\), such that each nonempty path-connected subspace of \(X\) intersects only one of them.
Theorem 25.2.1
Each component of a space \(X\) is closed in \(X\), since the closure of a connected subspace of \(X\) is connected.
If \(X\) has only finitely many components, then each component is also open in \(X\), since its complement is a finite union of closed sets. But in general the components of \(X\) need not be open in \(X\).
One can say even less about the path components of \(X\), for they need be neither open nor closed in \(X\).
Example
The "topologist's sine curve" \(\bar{S}\) of the preceding section is a space that has a single component (since it is connected) and two path components. One path component is the curve \(S\) and the other is the vertical interval \(V=0 \times[-1,1]\). Note that \(S\) is open in \(\bar{S}\) but not closed, while \(V\) is closed but not open.
If one forms a space from \(\bar{S}\) by deleting all points of \(V\) having rational second coordinate, one obtains a space that has only one component but uncountably many path components.
Definition 25.3 : Locally Connected / Locally Path Connected
A space \(X\) is said to be locally connected at \(x\) if for every neighborhood \(U\) of \(x\), there is a connected neighborhood \(V\) of \(x\) contained in \(U\). If \(X\) is locally connected at each of its points, it is said simply to be locally connected.
Similarly, a space \(X\) is said to be locally path connected at \(x\) if for every neighborhood \(U\) of \(x\), there is a path-connected neighborhood \(V\) of \(x\) contained in \(U\). If \(X\) is locally path connected at each of its points, then it is said to be locally path connected.
Example
Each interval and each ray in the real line is both connected and locally connected. The subspace \([-1,0) \cup(0,1]\) of \(\mathbb{R}\) is not connected, but it is locally connected. The topologist's sine curve is connected but not locally connected. The rationals \(\mathbb{Q}\) are neither connected nor locally connected.
Theorem 25.3
A space \(X\) is locally connected if and only if for every open set \(U\) of \(X\), each component of \(U\) is open in \(X\).
Proof
Suppose that \(X\) is locally connected; let \(U\) be an open set in \(X\); let \(C\) be a component of \(U\). If \(x\) is a point of \(C\), we can choose a connected neighborhood \(V\) of \(x\) such that \(V \subset U\). Since \(V\) is connected, it must lie entirely in the component \(C\) of \(U\). Therefore, \(C\) is open in \(X\).
Conversely, suppose that components of open sets in \(X\) are open. Given a point \(x\) of \(X\) and a neighborhood \(U\) of \(x\), let \(C\) be the component of \(U\) containing \(x\). Now \(C\) is connected; since it is open in \(X\) by hypothesis, \(X\) is locally connected at \(x\).
Theorem 25.4
A space \(X\) is locally path connected if and only if for every open set \(U\) of \(X\), each path component of \(U\) is open in \(X\).
Theorem 25.5
If \(X\) is a topological space, each path component of \(X\) lies in a component of \(X\).
If \(X\) is locally path connected, then the components and the path components of \(X\) are the same.
Proof
Let \(C\) be a component of \(X\); let \(x\) be a point of \(C\); let \(P\) be the path component of \(X\) containing \(x\). Since \(P\) is connected, \(P \subset C\). We wish to show that if \(X\) is locally path connected, \(P=C\). Suppose that \(P \subsetneq C\). Let \(Q\) denote the union of all the path components of \(X\) that are different from \(P\) and intersect \(C\); each of them necessarily lies in \(C\), so that
Because \(X\) is locally path connected, each path component of \(X\) is open in \(X\). Therefore, \(P\) (which is a path component) and \(Q\) (which is a union of path components) are open in \(X\), so they constitute a separation of \(C\). This contradicts the fact that \(C\) is connected.