How to show that two equivalence classes are either equal or have an empty intersection? For $x \in X$, let $[x]$ be the set $[x] = \{a \in X | \ x \sim a\}$.  
Show that given two elements $x,y \in X$, either
a) $[x]=[y]$ or
b) $[x] \cap [y] = \varnothing$.
How I started it is, if $[x] \cap [y]$ is not empty, then $[x]=[y]$, but then I am kind of lost. 

Lemma If $u \in [v]$ then $[u] = [v]$ proof: If $u \in [v]$ then by definition of equivalence class we have $v\sim u$, by transitivity of equivalence relations we have $\forall a \in X$, $\ v \sim a$ iff $\ u \sim a$. Hence we conclude $[v] = \{a \in X | \ v \sim a\} = \{a \in X | \ u \sim a\} = [u]$.
Corollary Suppose $u \in [x] \cap [y]$ then $[u] = [x]$ and $[u] = [y]$ so $[x] = [y]$.
