Closure Equals Union with Derivative

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

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

Then
 * $\operatorname{Cl} A = A \cup \operatorname{Der} A$

where
 * $\operatorname{Der} A$ denotes the derivative of $A$,
 * $\operatorname{Cl} A$ denotes the closure of $A$.

Proof
First inclusion: $\operatorname{Cl} A \subseteq A \cup \operatorname{Der} A$.


 * Let $x \in \operatorname{Cl} A$.


 * In case when $x \in A$ then $x \in A \cup \operatorname{Der} A$ by Definition:Set Union.


 * In case when (1): $x \not\in A$ to prove $x \in \operatorname{Der} A$ according to Characterization of Derivative by Open Sets it is enough to show that
 * for every an open subset $U$ of $T$ if $x \in U$ there exists a point $y$ of $T$ such that $y \in A \cap U$ & $x \ne y$.


 * Let $U$ be an open subset of $T$.
 * Assume $x \in U$.
 * Then $A \cap U \neq \varnothing$ by Condition for Point being in Closure.
 * Then there exists a point $y$ of $T$ such that
 * $y \in A$ and $y \in U$ by Definition:Set Intersection, Definition:Empty Set.
 * Hence $y \in A \cap U$ and $x \ne y$ by (1), Definition:Set Intersection.
 * Hence $x \in A \cup \operatorname{Der} A$ by Definition:Set Union.

Second inclusion:

$A \subseteq \operatorname{Cl} A$ by Set is Subset of its Topological Closure.

$\operatorname{Der} A \subseteq \operatorname{Cl} A$ by Derivative is Included in Closure. Hence $A \cup \operatorname{Der} A \subseteq \operatorname{Cl} A$ by Union of Subsets is Subset