Derivative of Union is Union of Derivatives

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

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

Then
 * $\left({A \cup B}\right)' = A' \cup B\,'$

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

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

Let $x \in \left({A \cup B}\right)'$.

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

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

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

By Union Distributes over Difference:
 * $\left({A \cup B}\right) \setminus \left\{ {x}\right\} = \left({A \setminus \left\{ {x}\right\} }\right) \cup \left({B \setminus \left\{ {x}\right\} }\right)$

Then by Closure of Finite Union equals Union of Closures:
 * $\left({\left({A \cup B}\right) \setminus \left\{ {x}\right\} }\right)^- = \left({A \setminus \left\{ {x}\right\} }\right)^- \cup \left({B \setminus \left\{ {x}\right\} }\right)^-$

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

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:
 * $A' \subseteq \left({A \cup B}\right)'$ and $B\,' \subseteq \left({A \cup B}\right)'$

Hence by Union of Subsets is Subset:
 * $A' \cup B\,' \subseteq \left({A \cup B}\right)'$