Preimage of Intersection under Relation
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Theorem
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $C$ and $D$ be subsets of $T$.
Then:
- $\RR^{-1} \sqbrk {C \cap D} \subseteq \RR^{-1} \sqbrk C \cap \RR^{-1} \sqbrk D$
General Result
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\map \PP T$ be the power set of $T$.
Let $\mathbb T \subseteq \map \PP T$.
Then:
- $\ds \RR^{-1} \sqbrk{\bigcap \mathbb T} \subseteq \bigcap_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$
Family of Sets
Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $\RR \subseteq S \times T$ be a relation.
Then:
- $\ds \RR^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} \subseteq \bigcap_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where $\ds \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\family {T_i}_{i \mathop \in I}$.
Also see
Proof
This follows from Image of Intersection under Relation, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules.
$\blacksquare$