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Let S be a set and $\{S_\alpha\}$ be nonempty subsets such that S = $\bigcup_{\alpha} $ $S_\alpha$ and $S_\alpha \cap S_\beta$ =$\emptyset $ if $\alpha \neq \beta $ Define an equivalence relation on S in such a way that the $S_\alpha$ are precisely all the equivalence classes.

I don't understand what this is asking. What am I supposed to do?

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Given two elements of $S$, you have to decide if they are equivalent or not. So, what good equivalence relation would you use? Under this relation all the equivalent elements should be on the same subset. –  Sigur Oct 16 '12 at 22:57
    
@Sigur I would say that if 2 classes aren't disjoint, then they must be equal. But how do I put that into math/equivalence-y notation? –  user39794 Oct 16 '12 at 23:01

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What about this: given $x,y\in S$, they are equivalent if and only if there exists $\alpha$ such that $x,y\in S_\alpha$. Now you can prove that this is an equivalence relations. Note that it's fundamental to have the union equal to $S$.

  1. $x\sim x$, for all $x\in S$;
  2. if $x\sim y$ then $y\sim x$, for all $x,y\in S$;
  3. if $x\sim y$ and $y\sim z$ then $x\sim z$, for all $x,y,z\in S$.
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Why is it fundamental to have the union equal to S though? I'm sorry if these are stupid questions I'm just new to the whole equivalence relation thing. –  user39794 Oct 16 '12 at 23:03
    
If you want to define an equivalence relation on $S$ you must have $x$ equivalent to $x$, for all $x\in S$. Did you follow? –  Sigur Oct 16 '12 at 23:06
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Yeah I get it now, thank you very much! –  user39794 Oct 16 '12 at 23:11

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