Image of Union under Mapping

Theorem
Let $f: S \to T$ be a mapping.

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

Then:
 * $f \left({A \cup B}\right) = f \left({A}\right) \cup f \left({B}\right)$

General Theorem
Let $f: S \to T$ be a mapping.

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

Let $\mathbb S \subseteq \mathcal P \left({S}\right)$.

Then:
 * $f \left({\bigcup \mathbb S}\right) = \bigcup_{X \in \, \mathbb S} f \left({X}\right)$

Proof
As $f$, being a mapping, is also a relation, we can apply Image of Union:


 * $\mathcal R \left({A \cup B}\right) = \mathcal R \left({A}\right) \cup \mathcal R \left({B}\right)$

and


 * $\mathcal R \left({\bigcup \mathbb S}\right) = \bigcup_{X \in \, \mathbb S} \mathcal R \left({X}\right)$

Also see

 * Mapping Preimage of Union
 * Mapping Image of Intersection
 * Mapping Preimage of Intersection