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$\def\class#1{\mathopen{[\![}#1\mathclose{]\!]}}$Proposition: If $\sim$ is an equivalence relation on $A$ and $a,b\in A$, then either $\class a \cap \class b = \emptyset$ or $\class a = \class b$.

$\class a$, $\class b$ are the respective equivalence classes of $a$ and $b$.

I want to verify these two cases, this is how I've thought it should go:

For the first case where the intersection of the two equivalence classes is the empty set; it happens when $a$ and $b$ are elements of $A$ that are not related by the equivalence relation.

For the latter case; it's true when $a$ and $b$ are related by the equivalence relation and thus the sets: $\class a = \{x\in A \mid x \sim a\}$ and $\class b = \{x \in A \mid x \sim b\}$ are equal due to the symmetric properties of the equivalence relation.

This is probably farfetched, but I want to know if this is even near a answer to the problem.

If this turns out to be completely wrong, then, I'd appreciate it if somebody could enlighten me.

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    @wj32: Sorry, I'm not familiar with the notation and how I'd do it in LaTeX.2012-11-01

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Suppose that $\class{a}\cap\class{b}\ne\emptyset$, i.e. there exists a $z \in \class{a}\cap\class{b}$. How might we "connect" $a$ and $b$ using $z$? (Think transitive.) You can then use this to prove that the two sets $\class{a}$ and $\class{b}$ are equal.

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    Oh, yeah, thanks. I just want to verify if I'm at least at the right track. This helped me$a$lot.2012-11-01