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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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Please note $\phi$ is different from $\emptyset$! See en.wikipedia.org/wiki/Empty_set#Notation –  wj32 Nov 1 '12 at 10:03
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Not really. Try to show that if the equivalence classes are not disjoint, they must coincide. –  Michael Greinecker Nov 1 '12 at 10:04
    
@wj32: Sorry, I'm not familiar with the notation and how I'd do it in LaTeX. –  Transcedental Number Nov 1 '12 at 10:06
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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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Well, if there exists a z element in the intersection of the equivalence classes, isn't it automatically equivalent to every other element in the intersection? –  Transcedental Number Nov 1 '12 at 10:26
    
@TranscedentalNumber: That's not important here. The key thing is that $\class{a}\cap\class{b}$ is not empty, so we can choose some $z$ in the intersection. After you do that, the transitive property tells you that $a \sim b$. –  wj32 Nov 1 '12 at 10:27
    
Ok, we know that the transitive property should hold for some $z$ in this non-empty set because this is an intersection of equivalence classes right? –  Transcedental Number Nov 1 '12 at 10:33
    
@TranscedentalNumber: Yes, that's right. –  wj32 Nov 1 '12 at 10:34
    
Ok now, if $a$~$b$ and if they are arbitrary respective elements of $[a]$ and $[b]$ does this imply that they are equal already, again by the properties of the equivalence relation? –  Transcedental Number Nov 1 '12 at 10:39
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