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.
$A$ is open and $B$ is closed. Let $x \in A \setminus B$. Then there is some $r&gt;0$ such that $B(x,r) \subseteq A$ as $A$ is open and $x \in A$. 

$x \notin B$ so $x$ is not a limit point of $B$ (as $B$ is closed it would contain all its limit points), so there is soem $s&gt;0$ such that $B(x,s) \cap B =\emptyset$. 

But then $B(x, \min(r,s)) \subseteq A\setminus B$ showing $x$ to be an interior point of $A \setminus B$, so $A \setminus B$ is open.

Now suppose $x$ is a limit point of $B\setminus A$. Then $x$ is also a limit point of the larger set $B$ and as $B$ is closed, $x \in B$. Suppose that $x \in A$, then we'd have $r&gt;0$ such that $B(x,r) \subseteq A$ and then $B(x,r) \cap (B \setminus A)=\emptyset$, and $x$ wouldn't be a limit point of $B \setminus A$. So, in fact, we must have that $x \notin A$, which together with $x \in B$ implies $x \in B\setminus A$, which thus contains all its limit points and is thus closed.
If we assume A/B is not open, then there must be a point $x$ in $A \setminus B$ for which all neighbourhoods intersect $(A \setminus B)^c$. $(A \setminus B)^c$ is the set of all points which either lie in $B$, or lie not in $A$ (or both). 

Pick a point in $A \setminus B$. It lies in $A$, so at least one of its neighbourhoods lies completely in $A$. It lies in $B^c$. There has to be at least one of its neighbourhoods which lies completely in $B^c$ (i.e. does not intersect $B$), as $B$ is its own closure.

Take the intersection of it with the first neighbourhood to provide a counterexample for our assumption that $A \setminus B$ is not open.

A similar idea could be used for the $B \setminus A$ case.