Derivative of Union is Union of Derivatives

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $A$, $B$ be subsets of $S$.

Then
 * $\paren {A \cup B}' = A' \cup B\,'$

where
 * $A'$ denotes the derivative of $A$.

Derivative of Union subset of Union of Derivatives
It will be shown that:
 * $\paren {A \cup B}' \subseteq A' \cup B\,'$

Let $x \in \paren {A \cup B}'$.

By definition of set derivative:
 * $x$ is an accumulation point of $A \cup B$.

Then by definition of accumulation point of set:
 * $(1): \quad x \in \paren {\paren {A \cup B} \setminus \set x}^-$

where $A^-$ denotes the closure of $A$.

By Set Difference is Right Distributive over Union:
 * $\paren {A \cup B} \setminus \set x = \paren {A \setminus \set x} \cup \paren {B \setminus \set x}$

Then by Closure of Finite Union equals Union of Closures:
 * $\paren {\paren {A \cup B} \setminus \set x}^- = \paren {A \setminus \set x}^- \cup \paren {B \setminus \set x}^-$

Then by $(1)$ and definition of set union:
 * $x \in \paren {A \setminus \set x}^-$ or $x \in \paren {B \setminus \set x}^-$

Then by definition of accumulation point of set:
 * $x$ is an accumulation point of $A$ or $x$ is an accumulation point of $B$.

Then by definition of set derivative:
 * $x \in A'$ or $x \in B\,'$

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

Union of Derivatives subset of Derivative of Union
By Set is Subset of Union:
 * $A \subseteq A \cup B$ and $B \subseteq A \cup B$

Then by Derivative of Subset is Subset of Derivative:
 * $A' \subseteq \paren {A \cup B}'$ and $B\,' \subseteq \paren {A \cup B}'$

Hence by Union of Subsets is Subset:
 * $A' \cup B\,' \subseteq \paren {A \cup B}'$