$\def\class#1{\mathopen{[\![}#1\mathclose{]\!]}}$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.