Open Set Less Closed Set is Open

Theorem
Let $T$ be a topological space.

Let $U$ be an open subset of $T$.

Let $D$ be a closed subset of $T$.

Then
 * $U \setminus D$ is open.

Proof
By definition of closed set
 * $\complement_T\left({D}\right)$ is open

where $\complement_T\left({D}\right) = T \setminus D$ denotes the complement of $D$ relative to $T$.

By Set Difference as Intersection with Relative Complement:
 * $U \setminus D = U \cap \complement_T\left({D}\right)$

By definition of topological space
 * intersection of two open sets is an open set.

Thus $U \setminus D$ is open.