Exercise 2.4.9 (Connectedness defines an equivalence relation)

Proof.

1.

First, let us show that $\sim$ is an equivalence relation.

$x\sim x$ : Take $E=\{x\}\subseteq X.E$ is certainly connected, and both $x$ and $x$ are contained in $E$. Thus $x\sim x$.

$x\sim y\Rightarrow y\sim x$ : Clearly true by definition of $\sim$.

$x\sim y,y\sim z\Rightarrow x\sim z:$ If $x\sim y$ then there is some connected set $E\subseteq X$ such that $x,y\in E$. And if $y\sim z$ then there is some connected set $F\subseteq X$ such that $y,z\in F$.

Since $E\cap F\ne \varnothing$ we can use the result from Exercise 2.4.6 to conclude that $E\cup F$ is connected. But $x,z\in E\cup F$. Hence $x\sim z$.

Thus $\sim$ is an equivalence relation on $X$.

2.

Now let us show that the equivalence classes are connected.

Let $[x]$ be an equivalence class. Now suppose (for sake of contradiction) that $[x]$ is disconnected. That is, there are two non-empty, disjoint, open sets $V,W\subseteq [x]$ such that $V\cup W=[x]$. If $[x]$ has only one element then it is clear that $[x]$ is connected. So suppose that $[x]$ has at least two elements, $x,y\in [x]$. And without loss of generality, let $x\in V$, and $y\in W.x\sim y$, so by definition there is some connected set $E\subseteq X$ with $x,y\in E$. We also know that $E\subseteq [x]$ since if $x\in E$ connected, then for all $y\in E$ there is a connected set, $E$, that contains $x$ and $y$. So $E\subseteq [x]$.

But, $E=E\cap [x]=E\cap (W\cup V)=(E\cap W)\cup (E\cap V)$ And since $E\cap V\subseteq E\subset [x]$, and $V$ is open in $[x]$ we can use Proposition $12.3.4$ to conclude that $E\cap V$ is open in $E$. The same argument gives us that $E\cap W$ is open in $E$.

But, $E=(E\cap W)\cup (E\cap V),(E\cap W)\cap (E\cap V)=E\cap V\cap W=\varnothing$ and $x\in E\cap V$, and $y\in E\cap W$, so these sets are non-empty, and from above they are open in $E$. Thus, by definition, $E$ is disconnected, a contradiction.

Thus $[x]$ is connected.

3.

Now let us show that $[x]$ is closed.

First, note that $[x]$ is the largest connected set that contains $x$ by the following argument: If there were a larger connected set containing $x$, say $A$, then there would be some element $y\in A\setminus [x]$. But $A$ is connected, so we have to have $y\sim x$ which gives us $y\in [x]$.

From Exercise 2.4.8, we know that $\overline{[x]}$ is a connected set. But $x$ is always in $[x]$ and hence is always in $\overline{[x]}$ thus $\overline{[x]}\subseteq [x]\subseteq \overline{[x]}$. Thus $[x]=\overline{[x]}$, and therefore $[x]$ is closed.