Image of Preimage under Relation is Subset

Theorem

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

Then:

$B \subseteq T \implies \paren {\RR \circ \RR^{-1} } \sqbrk B \subseteq B$

where:

$\RR \sqbrk B$ denotes the image of $B$ under $\RR$

$\RR^{-1} \sqbrk B$ denotes the preimage of $B$ under $\RR$

$\RR \circ \RR^{-1}$ denotes composition of $\RR$ and $\RR^{-1}$.

Let $B \subseteq T$.

Then:

$\ds y$

$\in$

$\ds \paren {\RR \circ \RR^{-1} } \sqbrk B$

$\ds \leadsto \ \ $

$\ds y$

$\in$

$\ds \RR \sqbrk {\RR^{-1} \sqbrk B}$

$\ds \leadsto \ \ $

$\ds \exists x \in \RR^{-1} \sqbrk B: \, $

$\ds \tuple {x, y}$

$\in$

$\ds \RR$

$\ds \leadsto \ \ $

$\ds y$

$\in$

$\ds B$

So by definition of subset:

$B \subseteq T \implies \paren {\RR \circ \RR^{-1} }\sqbrk B \subseteq B$

$\blacksquare$