A is an open set and B is a closed set. Prove that A\B is open and B\A is closed. The problem consists of using the epsilon neighborhood definition to prove that A\B is open and B\A is closed. 
I'm thinking of using a proof by contradiction and assuming that A\B is not open and vice versa for B\A. Any help would be greatly appreciated.
Sorry, forgot to mention that we can't use the fact that a set is open iff its compliment is closed and vice versa.

Let $x\in A\setminus B$. Denote by $U_\epsilon$ the $\epsilon$-neighborhood of $x$. If for any $n$ there was a point $x_n\in B\cap U_{1/n}$, then $x_n\to x$. But $B$ is closed and so $x\in B$. Contradiction! Hence, there exists $n$ such that $U_{1/n}$ is completely contained in $A\setminus B$.
