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