Connected components of a topological space Let X be a topological space. Define a binary relation $\sim$ in $X$ as follows: $x \sim y$ if there exists a connected subspace $C$ included in $X$ such that $x,y$ belong to $C$. Show the following.
(i) $\sim$ is an equivalence relation.
(ii) Each equivalence class is a maximal connected subspace of $X$. These equivalence
classes are called the connected components of $X$.
(iii) Each connected component is a closed subset of $X$. To this end, show that the closure
of a connected set is connected.
 A: Hint: (i) I guess you're ok with $x \sim x$ and $x\sim y \Rightarrow  y \sim x$. For transitivity, recall that the union of two connected sets with nonempty intersection is also a connected set.
(ii) Use the same fact of (i) (possibly with infinite elements) to check that the equivalence classes are connected. If C is a connected set in $X$, note that any two points in $C$ are equivalent, so they all must be contained in an equivalence class.
(iii) Closure of a connected subset of $\mathbb{R}$ is connected?
A: $\textbf{Hint:}$ (i) is straightforward. 
(ii) If $A$ is an equivalence class and $A \subseteq B$ where $B$ is connected, show that $B \subseteq A$ (note that $\forall x \in B$, $\forall a \in A$ we have $x$~  $a$).
(iii) If $A$ is a connected component, note that $A$ is dense in $cl(A)$ and apply (ii) to get $A=cl(A)$. 
A: You can prove the following: If $A$ is connected in $X$, then $A\subseteq B\subseteq \bar A$ implies $B$ is connected. Argue that if $B$ is not connected, then neither is $A$.
Thus, the closure of a connected set is connected. In particular, $\overline{\operatorname{Cmp}(a)}\ni a$ is connected, so $\overline{\operatorname{Cmp}(a)}\subseteq {\operatorname{Cmp}(a)}$ and the reverse inclusion always holds, so  $$\overline{\operatorname{Cmp}(a)}={\operatorname{Cmp}(a)}$$ 
and the component is closed.
