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?

  • 1
    $\begingroup$ Another idea to consider: This equivalence relation collapses each connected component of $X$ to a point. $\endgroup$ – 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 $M$.

  • $\begingroup$ I will look at this later. But i think this will help me. Thanks! $\endgroup$ – Marm Jun 24 '14 at 7:55
  • $\begingroup$ One question: how u know what the sets q(U) and q(V) are Open sets? Out map q in general is not an open map, hence does not map open sets to open sets. $\endgroup$ – Marm Jun 24 '14 at 12:06
  • $\begingroup$ 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. $\endgroup$ – Fabio Lucchini Jun 24 '14 at 13:39
  • $\begingroup$ Ah, i really have forgotton this. Thanks :) $\endgroup$ – Marm Jun 24 '14 at 14:41
  • $\begingroup$ But $U$ is not open, it is just open in $q^{-1}(M)$. $\endgroup$ – user87690 Dec 24 '15 at 14:27

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