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.
$\blacksquare$