# Image of Domain of Relation is Image Set

## Theorem

Let $S$ and $T$ be sets.

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

The image of the domain of $\mathcal R$ is the image set of $\mathcal R$:

$\mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right] = \operatorname{Im} \left ({\mathcal R}\right)$

where $\operatorname{Im} \left ({\mathcal R}\right)$ is the image of $\mathcal R$.

## Proof

Let $y \in \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]$.

 $\displaystyle y$ $\in$ $\displaystyle \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle \exists x \in \operatorname{Dom} \left({\mathcal R}\right): \ \$ $\displaystyle \left({x, y}\right)$ $\in$ $\displaystyle \mathcal R$ $\quad$ Definition of Image of Subset under Relation $\quad$ $\displaystyle \implies \ \$ $\displaystyle y$ $\in$ $\displaystyle \operatorname{Im} \left ({\mathcal R}\right)$ $\quad$ Definition of Image of Relation $\quad$ $\displaystyle \implies \ \$ $\displaystyle \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]$ $\subseteq$ $\displaystyle \operatorname{Im} \left ({\mathcal R}\right)$ $\quad$ Definition of Subset $\quad$

Let $y \in \operatorname{Im} \left ({\mathcal R}\right)$.

 $\displaystyle y$ $\in$ $\displaystyle \operatorname{Im} \left ({\mathcal R}\right)$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle \exists x \in S: \ \$ $\displaystyle \left({x, y}\right)$ $\in$ $\displaystyle \mathcal R$ $\quad$ Definition of Image of Relation $\quad$ $\displaystyle \implies \ \$ $\displaystyle y$ $\in$ $\displaystyle \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]$ $\quad$ Definition of Domain of Relation $\quad$ $\displaystyle \implies \ \$ $\displaystyle \operatorname{Im} \left ({\mathcal R}\right)$ $\subseteq$ $\displaystyle \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]$ $\quad$ Definition of Subset $\quad$

The result follows by definition of set equality.

$\blacksquare$