# Image of Set Difference under Relation

## Theorem

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

Let $A$ and $B$ be subsets of $S$.

Then:

$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$

where:

$\setminus$ denotes set difference
$\RR \sqbrk A$ denotes image of $A$ under $\RR$.

### Corollary 1

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

Let $A \subseteq B \subseteq S$.

Then:

$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp B A}$

where:

$\RR \sqbrk B$ denotes the image of $B$ under $\RR$
$\complement$ (in this context) denotes relative complement.

### Corollary 2

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

Let $A$ be a subset of $S$.

Then:

$\relcomp {\Img \RR} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp S A}$

where:

$\Img \RR$ denotes the image of $\RR$
$\RR \sqbrk A$ denotes the image of $A$ under $\RR$.

## Proof

 $\ds y$ $\in$ $\ds \RR \sqbrk A \setminus \RR \sqbrk B$ $\ds \leadsto \ \$ $\ds \exists x \in A: x \notin B: \,$ $\ds \tuple {x, y}$ $\in$ $\ds \RR$ Definition of Image of Subset under Relation $\ds \leadsto \ \$ $\ds \exists x \in A \setminus B: \,$ $\ds \tuple {x, y}$ $\in$ $\ds \RR$ Definition of Set Difference $\ds \leadsto \ \$ $\ds y$ $\in$ $\ds \RR \sqbrk {A \setminus B}$ Definition of Image of Subset under Relation

$\blacksquare$

## Also see

Note that equality does not hold in general.

See Difference of Images under Mapping not necessarily equal to Image of Difference for an example of a mapping (which is of course a relation) for which it does not.