A family of sets closed under difference is closed under intersection I was propposed to prove that a family of sets that is closed under difference is also closed under finite intersection.
At my understading, this is easy to prove provided my family of sets is a ring. Indeed, given $A, B$ on the family,
$$A\cap B = (A\cup B)- \{(A-B)\cup(B-A)\}.$$
Since $A, B$ lie on the family, that is closed under difference, $A\cap B$ lies on the family if, and only if, $A\cup B$ also belongs to the family.
The problem is that I do not have this information. How to proceed?
 A: Suppose $A$ and $B$ are both elements of our family of sets which is closed under set difference.  It follows that $A\setminus B$ is also an element of our family.
It follows then that $A\setminus (A\setminus B)=A\cap B$ is an element of our family and so our family is also closed under intersection.

 For proving that set identity, remember that $A$ can be partitioned as $A=(A\cap B)\cup (A\setminus B)$, that is to say given an arbitrary element of $A$, it is either also in $B$ or it is not in $B$.


Note, the given conditions in the problem do not mention the existence of a universal set, nor do they mention that the universal set must be in our collection.  As such, it does not follow that our collection must also be closed under union.  For example, $\{\{1\},\{2\},\emptyset\}$ is a perfectly valid collection of sets closed under set difference and intersection despite that the set $\{1,2\}$ is missing.
Also, although being closed under difference implies being closed under intersection, the converse is not true.  $\{\{1\}\}$ is a perfectly valid family closed under intersection, but it is not closed under difference since the empty set is missing.
It is only after you additionally include as a hypothesis that $\emptyset$ and $X$ (our universal set) are elements of our family as well alongside the original hypothesis that the family is closed under differences that we get that the set is closed under complements and unions too.
