Closure of Derivative is Derivative in T1 Space

Theorem
Let $T = \left({S, \tau}\right)$ be a $T_1$ topological space.

Let $A$ be a subset of $S$.

Then
 * $A'^- = A'$

where
 * $A'$ denotes the derivative of $A$
 * $A^-$ denotes the closure of $A$.

Proof
So:
 * $A'^- \subseteq A'$

Then by definition of closure:
 * $A' \subseteq A'^-$

Hence the result by definition of set equality.