Mapping Image of Union

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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:

$\displaystyle f \left({\bigcup \mathbb S}\right) = \bigcup_{X \in \mathbb S} f \left({X}\right)$


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

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

$\blacksquare$


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