# Image of Subset under Relation is Subset of Image/Corollary 1

## Corollary to Image of Subset under Relation is Subset of Image

Let $S$ and $T$ be sets.

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

Let $C, D \subseteq T$.

Then:

$C \subseteq D \implies \mathcal R^{-1} \left[{C}\right] \subseteq \mathcal R^{-1} \left[{D}\right]$

where $\mathcal R^{-1} \left[{C}\right]$ is the preimage of $C$ under $\mathcal R$.

## Proof

We have that $\mathcal R^{-1}$ is itself a relation, by definition of inverse relation.

The result follows directly from Image of Subset under Relation is Subset of Image.

$\blacksquare$