Open Set minus Closed Set is Open

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

For $A \subseteq S$ denote by $\relcomp S A$ the relative complement of $A$ in $S$.

Let $U \in \tau$ and $\relcomp S V \in \tau$.

Then:
 * $U \setminus V \in \tau$

and:
 * $\relcomp S {V \setminus U} \in \tau$

Proof
From Set Difference as Intersection with Relative Complement:
 * $U \setminus V = U \cap \relcomp S V$

Since $\tau$ is a topology:
 * $U, \relcomp S V \in \tau \implies U \cap \relcomp S V \in \tau \implies U \setminus V \in \tau$

The other statement follows mutatis mutandis.