Image is Subset of Codomain

Theorem

Let $\mathcal R = S \times T$ be a relation.

For all subsets $A$ of the domain of $\mathcal R$, the image of $A$ is a subset of the codomain of $\mathcal R$:

$\forall A \subseteq \operatorname{Dom} \left ({\mathcal R}\right): \mathcal R \left({A}\right) \subseteq T$

In the language of induced mappings, this can be written as:

$\forall A \in \mathcal P \left({S}\right): f_\mathcal R \left({A}\right) \in \mathcal P \left({T}\right)$

Corollary

This also holds for mappings:

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

For all subsets $A$ of the domain $S$, the image of $A$ is a subset of the codomain of $f$:

$\forall A \subseteq S: f \left({A}\right) \subseteq T$

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle A\)	\(\subseteq\)	\(\displaystyle \operatorname{Dom} \left ({\mathcal R}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \mathcal R \left({A}\right)\)	\(\subseteq\)	\(\displaystyle \operatorname{Im} \left ({\mathcal R}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Subset of Image
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof of Corollary

As a mapping is by definition also a relation, the result follows immediately.

$\blacksquare$