Image of Union under Mapping

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

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

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

Generalized Theorem
Let $$S_i \subseteq S: i \in I = \N_n$$.

Then:
 * $$f \left({\bigcup_{i \in I} S_i}\right) = \bigcup_{i \in I} f \left({S_i}\right)$$.

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


 * $$\mathcal{R} \left({\bigcup_{i \in I} S_i}\right) = \bigcup_{i \in I} \mathcal{R} \left({S_i}\right)$$