partition and equivalence relation's classes 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.

Every $x$ is in its own $c_R(x)$, which is a member of the set of classes. So that takes care of 1 and 2. It uses reflexivity of $R$.

If $c_R(x)=c_R(y)$ as sets, this is equivalent to $(x,y) \in R$.

Proof: Suppose $c_R(x)=c_R(y)$. Then in particular $y \in c_R(y)$ as $(y,y) \in R$ so $y \in c_R(x)$ which by definition means $(x,y) \in R$.
Suppose $(x,y) \in R$. Let $z \in c_R(x)$ so that $(x,z) \in R$. We also have that $(z,x) \in R$ by symmetry, and then $(z,x),(x,y) \in R$ allows us to say that $(z,y) \in R$ so that $z \in c_R(y)$, and we've shown $c_R(x) \subseteq c_R(y)$ and the reverse inclusion is similar.
Also, if two classes $c_R(x), c_R(y)$ intersect, they are equal; this follows from the previous: if $z$ is in the intersection, $(x,z) \in R, (z,y) \in R$ and so $(x,y) \in R$ and so equal classes..
