# Image of Singleton under Relation

## Theorem

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

Then the image of an element of $S$ is equal to the image of a singleton containing that element, the singleton being a subset of $S$:

$\forall s \in S: \mathcal R \left({s}\right) = \mathcal R \left[{\left\{{s}\right\}}\right]$

## Proof

We have the definitions:

 $\displaystyle \mathcal R \left({s}\right)$ $=$ $\displaystyle \left\{ {t \in T: \left({s, t}\right) \in \mathcal R}\right\}$ Definition of Image of Element under Relation $\displaystyle \mathcal R \left[{\left\{ {s}\right\} }\right]$ $=$ $\displaystyle \left\{ {t \in T: \exists s \in \left\{ {s}\right\}: \left({s, t}\right) \in \mathcal R}\right\}$ Definition of Image of Subset under Relation

 $\displaystyle t$ $\in$ $\displaystyle \mathcal R \left({s}\right)$ $\displaystyle \implies \ \$ $\displaystyle \left({s, t}\right)$ $\in$ $\displaystyle \mathcal R$ $\displaystyle \implies \ \$ $\, \displaystyle \exists s \in \left\{ {s}\right\}: \,$ $\displaystyle \left({s, t}\right)$ $\in$ $\displaystyle \mathcal R$ since $s \in \left\{ {s}\right\}$ by definition of Singleton $\displaystyle \implies \ \$ $\displaystyle t$ $\in$ $\displaystyle \mathcal R \left[{\left\{ {s}\right\} }\right]$ $\displaystyle \implies \ \$ $\displaystyle R \left({s}\right)$ $\subseteq$ $\displaystyle \mathcal R \left[{\left\{ {s}\right\} }\right]$ Definition of Subset

 $\displaystyle t$ $\in$ $\displaystyle \mathcal R \left[{\left\{ {s}\right\} }\right]$ $\displaystyle \implies \ \$ $\, \displaystyle \exists s \in \left\{ {s}\right\}: \,$ $\displaystyle \left({s, t}\right)$ $\in$ $\displaystyle \mathcal R$ $\displaystyle \implies \ \$ $\displaystyle \left({s, t}\right)$ $\in$ $\displaystyle \mathcal R$ since $r \in \left\{ {s}\right\} \implies r = s$ by Singleton Equality $\displaystyle \implies \ \$ $\displaystyle t$ $\in$ $\displaystyle \mathcal R \left({s}\right)$ $\displaystyle \implies \ \$ $\displaystyle \mathcal R \left[{\left\{ {s}\right\} }\right]$ $\subseteq$ $\displaystyle R \left({s}\right)$

Finally:

$\mathcal R \left({s}\right) = \mathcal R \left[{\left\{{s}\right\}}\right]$

by definition of set equality.

$\blacksquare$