# Image is Subset of Codomain

## Theorem

Let $\RR = S \times T$ be a relation.

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

$\forall A \subseteq \Dom \RR: \RR \sqbrk A \subseteq T$

In the notation of direct image mappings, this can be written as:

$\forall A \in \powerset S: \map {\RR^\to} A \in \powerset T$

### Corollary 1

Let $\RR = S \times T$ be a relation.

The image of $\RR$ is a subset of the codomain of $\RR$:

$\Img \RR \subseteq T$

These results also hold for mappings:

### Corollary 2

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 $f$:

$\forall A \subseteq S: f \sqbrk A \subseteq T$

### Corollary 3

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

The image of $f$ is a subset of the codomain of $f$:

$\Img f \subseteq T$

## Proof

 $\ds A$ $\subseteq$ $\ds \Dom \RR$ $\ds \leadsto \ \$ $\ds \RR \sqbrk A$ $\subseteq$ $\ds \Img \RR$ Image of Subset is Subset of Image $\ds$ $\subseteq$ $\ds T$

$\blacksquare$