Image of Set Difference under Relation/Corollary 2

Corollary to Image of Set Difference under Relation

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

By definition of the image of $\RR$:

$\Img \RR = \RR \sqbrk S$

So, when $B = S$ in Image of Set Difference under Relation: Corollary 1:

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

Hence:

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

means exactly the same thing as:

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

that is:

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

$\blacksquare$