Suppose that the set $A$ is open and the set $B$ is closed. Then $A \setminus B$ is open.
I can show this for two cases:
If $A \cap B=\emptyset$, then $A \setminus B=A$, which is obviously open.
If $A \subseteq B$, $A \setminus B=\emptyset$, which is open by definition
How do I show this for the general case, ie where $A \cap B \neq \emptyset$ and $A \not\subseteq B$?
Since $A\setminus B=A\cap(X\setminus B)$, where $X$ is the whole space, and since $A$ and $X\setminus B$ are open, $A\setminus B$ is open.