# Proof: The quotient space is totally disconnected

I want to prove the following:

Let $X$ be a topological space. Remark: $x \sim y \iff$ There is a connected component which contains $x$ and $y$. And now I want to show that the quotient space $X/\sim$ is totally disconnected.

My way so far:

We suppose the opposite: $X/\sim$ is not totally disconnected. So there is a connected component $M\subset X/\sim$ which contains $2$ elements, named $[x]$ and $[y]$. $M$ is closed, hence $q^{-1}(M)$ is also closed in $X$ and contains $[x]$ and $[y]$. $[x]$ and $[y]$ are disjoint, hence $q^{-1}(M)$ is not connected, since it contains at least $2$ connected components. So there exists two open (not empty and disjoint) sets $U,V \subset$ $q^{-1}(M)$ with $U \cup V = q^{-1}(M)$.

But from now I don't know how to get a contradiction. Does someone have an idea to complete this proof?

• Another idea to consider: This equivalence relation collapses each connected component of $X$ to a point. – Ayman Hourieh Jun 23 '14 at 20:34

Note that $q^{-1}(q(U))=U$, for if $u\in q^{-1}(q(U))$ then $q(u)=q(u')$ for some $u'\in U$. Hence $u$ and $u'$ live in the same connected component $C=q^{-1}(\{u\})\subseteq q^{-1}(M)$. Since $C=(C\cap U)\cup(C\cap V)$ and $C\cap U\neq\emptyset$, it follows $C\cap V=\emptyset$, thus $C\subseteq U$.
Consequently $M=q(U)\cup q(V)$ and $q(U)$, $q(V)$ are nonempty, open, disjoint subsets of $M$.
• Recall the caracteristic property of quotient topology: a sub set $S$ of $X\diagup\sim$ is open if and only if $q^{-1}(S)$ is open. In our case, $q^{-1}(q(U))=U$ is open, hence $q(U)$ is open. – Fabio Lucchini Jun 24 '14 at 13:39
• But $U$ is not open, it is just open in $q^{-1}(M)$. – user87690 Dec 24 '15 at 14:27