Are equivalence classes closed under intersection? How do I prove whether the set of equivalence classes for some equivalence relation is closed under intersection; that is the intersection of equivalence classes is an equivalence class? 
If you're asking why... I'm trying to prove that Every two equivalence classes $[x]$ and $[y]$ are either equal or disjoint but that problem I can solve. I just need to know whether countable intersection of equivalence classes is an equivalence class.
 A: Your goal is two show that if $[x]$ and $[y]$ are two classes, then they are the same or disjoint. This is easy to prove directly and does not necessitate us to consider arbitrary or countable intersections of classes at all:
Suppose $z \in [x] \cap [y]$ so that $[x]$ and $[y]$ are not disjoint.
Then in fact $[x] = [y]$: 
Pick $u \in [x]$, so $u \sim x$ (where $\sim$ is the equivalence relation).
We also have $z \in [x]$ (so $z \sim x$ and also $x \sim z$ by symmetry) and $z \in [y]$, so $z \sim y$.
So $u \sim x \sim z \sim y$, so the properties of an equivalence relation tell us that $u \in [y]$. As $u$ was arbitrary $[x] \subseteq [y]$.
The proof that $[y] \subseteq [x]$ is entirely symmetric. So the classes are equal when they're not disjoint.
In words: If there is some $z$ equivalent to both $x$ and $y$, then everything equivalent to $x$ is equivalent to $y$ and vice versa.
Another way to see it: If you know that $x \sim y$ iff $[x] = [y]$ then
$z \in [x] \cap [y]$ implies $[x] = [z] = [y]$ quite directly.
All of this implies that the intersection of two classes already need not be a class, whenever there are two non-equivalent points $x \not \sim y$, so that $[x] \neq [y]$ and $[x] \cap [y] = \emptyset$ which is not a class.
A: If I understood correctly, yes, the intersection of classes is a class. It will either be empty if you are intersecting inequivalent classes or , if not, it will agree with the first class in your intersection, which will coincide, (intersection being non-empty) with all classes being intersected.. 
A: If they are either equal or disjoint, then that means the (2-ary) intersection is either one set or the nullset. It is easily shown that this is also true for a k-ary intersection. 
A: Not exactly sure what you are asking...
But if you are saying $R_1$ is an equivalence relation
and $R_2$ is an equivalence relation, is $R_1$ and $R_2$ an equivalence relation.
It is.  To prove it, show that it is reflexive, symmetric and transitive.
A partition is an equivalence relation.  And a sup-partition is still a partition.  So, pictorally. 

A: Let $A$ be any set with equivalence relations $R_1$ and $R_2$ defined on it.
• $R_1 \cap R_2:$
Reflexive:
Since $R_1$ and $R_2$ are  both  reflexive,  then $$\forall x \in A:(x,x) \in R_1 \wedge (x,x) \in R_2.$$
Therefore, $(x,x) \in R_1 \cap R_2,$ so it is reflexive.
Symmetric:
Let  $(x, y)\in R_1 \cap R_2.$ Then $(x,y) \in R_1$ and $(x,y) \in R_2.$ Since both $R_1$ and $R_2$ are symmetric, we have $(y,x) \in R_1$ and $(y,x) \in R_2.$ Therefore, $(y,x) \in R_1 \cap R_2,$ and the intersection is symmetric.
Transitive:
Let $(x, y),(y,z) \in R_1 \cap R_2.$ Then, since $R_1$ and $R_2$ are both transitive, we have $(x,z) \in R_1$ and $(x,z) \in R_2.$ So, $(x,z) \in R_1 \cap R_2$ and the intersection is transitive.
This shows that $R_1 \cap R_2$ is an equivalence relation. That is, equivalence relations on a set $A$ are closed under intersection.
