[Munkres Topology] 25. Components and Local Connectedness

 This article is one of the posts in the Textbook Commentary Project.

Definition. 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 of $X$.

Check. Symmetry and reflexivity of the relation are obious. Transitivity follows by noting that if $A$ is a connected subspace containing $x$ and $y$, and if $B$ is a connected subspace contining $y$ 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 conneted subspace $A$ of $X$ intersects only one of them: 

[Only one] 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 $C_1=C_2$.

[Connceted] 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, $$C=\bigcup_{x\in C} A_x.$$ Since the subspaces $A_x$ are connected and have the point $x_0$ in common, their union is connected.

Definition. We define another equivalence realsion 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 $X$.

Check. First we note that if there exsits a path $f:[a,b]\rightarrow X$ from $x$ to $y$ whos 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 pathe from $x$ to $y$. 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.

Definition. 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.

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 $U$. $\forall x\in C$, 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 is 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 with 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 lie in $C$, so that $$C=P\cup Q.$$ 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.