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.

When you talk about the equivalence classes of a relation on a set:

*

*if you violate (1), i.e. $\emptyset \in P$, you have an invalid relation

*if you violate (2), either the union of class elements has something which is not in the original set (which means you have an invalid relation since it is defined on things outside the set of interest) or the set has something not in the union of equivalence classes (which invalidates the relation being an equivalence relation)

*if you violate (3), 2 equivalence classes contain the same item $x$, then your equivalence classes just collide into the same class because the equivalence relation is transitive, and if $a \ne x$ is in the first class and $b \ne x$ in the second, since $R$ is transitive you have $(a,x)$ and $(x,b)$ imply $(a,b)$ and classes collapse.

