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Let $X$ be a connected space and let $\mathcal{R}$ be an equivalence relation on $X$ such that for each $x \in X$, there exists an open set $O_x$ containing $x$ such that $O_x \subseteq [x]$. Prove that $\mathcal{R}$ has only one (distinct) equivalence class.

Lost for ideas on this one... any help would be appreciated.

user56031
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1 Answers1

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Fix any equivalence class $[x]$. Then since for all $y\in[x]$ there exists a neighborhood $O_y\ni y$ such that $O_y\subseteq[y] = [x]$, $[x]$ is open. Can you proceed from here?

Tai
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  • Thanks. I've thought for a long time on your answer and I really have no idea :( Firstly, why is $[x]$ open? I'm a beginner to all of this and the definition I'm exposed to for open is that a subset $A$ of a metric space $X$ is open if for all $x \in A$ there exists an open ball centered on $x$ that is contained in $A$. Where are the open balls here? Also how does the metric space $X$ being connected come into play? – user56031 Jul 27 '17 at 10:35
  • No problem! We showed that for all $y\in[x]$, there exists an open set $O_y$ that contains $y$ that is contained in $[x]$. Since $O_y$ is open, there must be an open ball centered at $y$ that is contained in $O_y$, which is in turn contained in $[x]$. So, there is an open ball centered at $y$ that is contained in $[x]$.

    Now, we finish off with the fact that $X$ must be connected. If $X$ had more than one equivalence class, then $X$ is disjoint union of open sets, since all the equivalence classes are disjoint. That contradicts connectedness.

     – Tai Jul 27 '17 at 15:22