TITLE: Property of set exclusion set.
QUESTION [6 upvotes]: Let $T$ have the property that for all sets $A, B \in T$ we have that $(A\backslash B) \in T$. 
How can I prove that $\forall A,B \in T, A\cap B \in T$?
I was thinking I should start with both expressions:
$(A\backslash B) \in T$. 
$(B\backslash A) \in T$. 
and show that$ (A \cup B)\backslash((A\backslash B)\cup (B \backslash A)=A\cap B \in T$.
I'm not sure how to show that the final part is in that set. It doesn't say anything about unions.

REPLY [10 votes]: Hint: $A\cap B=A\setminus (A\setminus B)$

REPLY [6 votes]: Write $A\cap B = A\setminus(A\setminus B)$.