Closure Equals Union with Derivative

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

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

Then:
 * $A^- = A \cup A'$

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

Closure Subset of Union
It is to be proved that:
 * $A^- \subseteq A \cup A'$

Let $x \in A^-$.

In the case where $x \in A$ then $x \in A \cup A'$ by definition of set union.

Let:
 * $(1): \quad x \notin A$

From Characterization of Derivative by Open Sets, to prove $x \in A'$ it is enough to show that:
 * for every open set $U$ of $T$:
 * if $x \in U$
 * then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$.

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

Let $x \in U$.

Then by Condition for Point being in Closure:
 * $A \cap U \ne \varnothing$

Then by definition of empty set:
 * $\exists y \in S: y \in A \cap U$

By definition of set intersection:
 * $y \in A$ and $y \in U$

But as $x \notin A$ it follows by definition of set intersection that:
 * $x \notin A \cap U$

So by $(1)$:
 * $x \ne y$

Thus $y$ fulfils the conditions of the hypothesis, and so:
 * $x \in A'$

Hence by definition of set union:
 * $x \in A \cup A'$

Thus in all cases:
 * $x \in A^- \implies x \in A \cup A'$

and so:
 * $A^- \subseteq A \cup A'$

Union Subset of Closure
It is to be proved that:
 * $A \cup A' \subseteq A^-$

By Set is Subset of its Topological Closure:
 * $A \subseteq A^-$

By Derivative is Included in Closure:
 * $A' \subseteq A^-$

Hence by Union of Subsets is Subset:
 * $A \cup A' \subseteq A^-$

Hence by definition of set equality:
 * $A^- = A \cup A'$