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 $A$.

$\ds A'^-$

$=$

$\ds A' \cup A$

$\ds $

$\subseteq$

$\ds A' \cup A'$

$A \subseteq A'$ by Derivative of Derivative is Subset of Derivative in T1 Space

$\ds $

$=$

$\ds A'$

So:

$A'^- \subseteq A'$

Then by definition of closure:

$A' \subseteq A'^-$

Hence the result by definition of set equality.

$\blacksquare$