# 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 $\mathcal R$:

- $\forall A \subseteq \Dom {\mathcal R}: \mathcal R \sqbrk A \subseteq T$

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

- $\forall A \in \powerset S: \map {\mathcal R^\to} A \in \powerset T$

### Corollary 1

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

The image of $\mathcal R$ is a subset of the codomain of $\mathcal R$:

- $\Img {\mathcal R} \subseteq T$

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

\(\displaystyle A\) | \(\subseteq\) | \(\displaystyle \Dom {\mathcal R}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \mathcal R \sqbrk A\) | \(\subseteq\) | \(\displaystyle \Img {\mathcal R}\) | $\quad$ Image of Subset is Subset of Image | $\quad$ | ||||||||

\(\displaystyle \) | \(\subseteq\) | \(\displaystyle T\) | $\quad$ | $\quad$ |

$\blacksquare$