Union of Derivatives is Subset of Derivative of Union

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

Let:
 * $\mathcal F \subseteq \mathcal P \left({S}\right)$ be a set of subsets of $S$

where $\mathcal P \left({S}\right)$ denotes the power set of $S$.

Then:
 * $\displaystyle \bigcup_{A \mathop \in \mathcal F} A' \subseteq \left({\bigcup_{A \mathop \in \mathcal F} A}\right)'$

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

Proof
Let $\displaystyle x \in \bigcup_{A \mathop \in \mathcal F} A'$.

Then by definition of union there exists $A \in \mathcal F$ such that:
 * $(1): \quad x \in A'$

By Set is Subset of Union:
 * $\displaystyle A \subseteq \bigcup_{A \mathop \in \mathcal F} A$

Then by Derivative of Subset is Subset of Derivative:
 * $\displaystyle A' \subseteq \left({\bigcup_{A \mathop \in \mathcal F} A}\right)'$

Hence by $(1)$ the result:
 * $\displaystyle x \in \left({\bigcup_{A \mathop \in \mathcal F} A}\right)'$

follows by definition of subset.