# Preimage of Intersection under Relation

## 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 $\family {T_i}_{i \mathop \in I}$.

## 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.

$\blacksquare$