Derivative of Union is Union of Derivatives

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

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

Then
 * $\operatorname{Der} \left({A \cup B}\right) = \operatorname{Der} A \cup \operatorname{Der} B$

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

Proof
First inclusion: $\operatorname{Der} \left({A \cup B}\right) \subseteq \operatorname{Der} A \cup \operatorname{Der} B$

Let $x \in \operatorname{Der} \left({A \cup B}\right)$

Then x is an accumulation point of $A \cup B$ by Definition:Set derivative Then (1): $x \in \operatorname{Cl} \left({\left({A \cup B}\right) \setminus \{x\}}\right)$ by Definition:Accumulation Point of a Set.

$\left({A \cup B}\right) \setminus \{x\} = \left({A \setminus \{x\}}\right) \cup \left({B \setminus \{x\}}\right)$ by Union Distributes over Difference.

Then $\operatorname{Cl} \left({\left({A \cup B}\right) \setminus \{x\}}\right) = \operatorname{Cl} \left({A \setminus \{x\}}\right) \cup \operatorname{Cl} \left({B \setminus \{x\}}\right)$ by Closure of Union is Union of Closures.

Then $x \in \operatorname{Cl} \left({A \setminus \{x\}}\right)$ or $x \in \operatorname{Cl} \left({B \setminus \{x\}}\right)$ by (1), Definition:Set Union.

Then $x$ is an accumulation point of $A$ or $x$ is an accumulation point of $B$ by Definition:Accumulation Point of a Set.

Then $x \in \operatorname{Der} A$ or $x \in \operatorname{Der} B$ by Definition:Set derivative.

Hence $x \in \operatorname{Der} A \cup \operatorname{Der} B$ by Definition:Set Union.

Second inclusion:

$A \subseteq A \cup B$ and $B \subseteq A \cup B$ by Set is Subset of Union.

Then $\operatorname{Der} A \subseteq \operatorname{Der} (A \cup B)$ and $\operatorname{Der} B \subseteq \operatorname{Der} (A \cup B)$ by Derivative of Subset is Subset of Derivative.

Hence $\operatorname{Der} A \cup \operatorname{Der} B \subseteq \operatorname{Der} (A \cup B)$ by Union of Subsets is Subset.