Open Set minus Closed Set is Open

Theorem
Let $\left(X,\tau\right)$ be a topological space.

For $A \in X$ denote by $\complement_X \left({A}\right)$ the relative complement of $A$ in $X$.

Let $O \in \tau$ and $\complement_X \left(C\right) \in \tau$.

Then $O \setminus C \in \tau$ and $\complement_X \left(C \setminus O\right) \in \tau$.

Proof
By definition of Set Difference as Intersection with Relative Complement: $O \setminus C = O \cap \complement_X \left(C\right)$.

Since $\tau$ is a topology: $O, \complement_X \left(C\right) \in \tau \implies O \cap \complement_X \left(C\right) \in \tau \implies O \setminus C \in \tau$

The other statement follows mutatis mutandis.