TITLE: partition and equivalence relation's classes
QUESTION [5 upvotes]: My question concerns the definitions more than semantics.
That is, a family of sets $P$ is a partition of $X$ if the following conditions hold:

$P$ doesn't contain the empty set;
union of all $P$'s sets gives $X$;
elements of $P$ are pairwise disjoint.

Now for an equivalence relation $R$ over $X$, its classes are defined as:

$\forall x\in X,$ $c_R(x)=\{y \mid (x,y) \in R\}$

My question is: why is the family of equivalence classes over $R$ a partition of $X$?
Why it shouldn't be: because we can have a relation $R$ and a set $X$ such that: $ \exists x,y \in X,  c_R(x)\cap c_R(y)\ne \emptyset$ which violates the condition 3. in the definition of a partition.
EDIT
I am convinced that every two classes of an equivalence relation are either disjoint or equal. But I still have a problem with the definition of equivalence classes:
So since we have a class for each element, we can have equal classes and so non pairwise disjoint.
unless family word in the family of equivalence classes over a relation $R$ on a set $X$ is a partition of $X$ refers to unique classes.

REPLY [2 votes]: Let $R$ be an equivalence relation on non empty set $X$. Let's denote class of $p\in X$ by $cl(p)$. For any $x,y \in X$. We can have only two cases:  
Case (1): $(x,y)\notin R$ 
If $a\in cl(x)\cap cl (y) $, then $ (x,a)\in R, (a,y)\in R$ and hence $(x,y)\in R$ (as R is equivalence relation). This is a contradiction and hence $ cl(x)\cap cl(y)=\phi$ 
Case (2): $(x,y)\in R$ 
For any $b\in cl(x)$, we have $(b,x)\in R$ and also $(x,y)\in R$ and hence by transitivity of $R$, we have $(b,y)\in R$ and hence $b\in cl(y)$. Thus $cl(x)\in cl (y)$. Similarly, $cl(y)\in cl(x)$ and hence $cl(x)=cl(y)$ 

Note that any two equivalent classes of an equivalent relation are
either equal or disjoint.

Proof: Let's consider two equivalent classes $cl(x)$ and $cl(y)$ such that $cl(x)\cap cl(y)\ne \phi$. Hence, $\exists a\in X$ such that $a\in cl(x)\cap cl(y)$. This implies that $(x,a)\in R$ and $(a,y)\in R\implies (x,y)\in R$ (by transitivity of $R$) and hence by case (2) above, $cl(x)=cl(y)$. Thus two equivalence classes of $R$ are either equal or disjoint! 
Now for any $x\in X$, we have $x\in cl (x)\subseteq X$. Hence, $cl(x)$ is non empty. [Condition 1 is satisfied] 
Any two classes of $R$ are either pairwise disjoint or equal. From here, you can show that they form a partition. Can you take it from here?