Prove that: Given an equivalence relation $∼$ on a set $X$, the equivalence classes of $X$ form a partition of $X$.
Well, I first define $O_x ={\{y:x∼y}\}$. If $O_x ∩ O_y \neq ∅$, so there is some $z\in O_x ∩ O_y$, implies $x∼z$ and $y∼z$ for every $x$ and $y$; and (because of definition of equivalence) thus $x∼y$ for every $x$ and $y$. But how to prove that it implies both of $O_x \subset O_y$ and $O_y \subset O_x$ (to prove $O_y = O_x$)?
Thank you.
EDIT By the question above I mean that why $x∼y \implies O_x \subset O_y$
Clearly every element of $X$ is in some equivalence class, namely its own, so you need to show that distinct equivalence classes are disjoint.
If $O_x\cap O_y\neq\emptyset$ like you say, let $z\in O_x\cap O_y$. Pick $a\in O_x$. Then $a\sim x$, and $x\sim z$, so by transitivity, $a\sim z$. But $z\sim y$, so again by transitivity, $a\sim y$. Thus $a\in O_y$, so $O_x\subset O_y$. That $O_y\subset O_x$ holds by a symmetric argument, so $O_x=O_y$.