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

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Corollary to Image of Subset under Relation is Subset of Image

Let $S$ and $T$ be sets.

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


Let $C, D \subseteq T$.


Then:

$C \subseteq D \implies \RR^{-1} \sqbrk C \subseteq \RR^{-1} \sqbrk D$

where $\RR^{-1} \sqbrk C$ is the preimage of $C$ under $\RR$.


Proof

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

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

$\blacksquare$