# Preimage of Union under Relation

## Theorem

Let $S$ and $T$ be sets.

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

Let $T_1$ and $T_2$ be subsets of $T$.

Then:

$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$

### General Result

Let $S$ and $T$ be sets.

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

Let $\powerset T$ be the power set of $T$.

Let $\mathbb T \subseteq \powerset T$.

Then:

$\ds \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$

where $\RR^{-1} \sqbrk X$ denotes the preimage of $X$ under $\RR$.

### 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:

$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$

where:

$\ds \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \mathop \in I}$
$\RR^{-1} \left[{T_i}\right]$ denotes the preimage of $T_i$ under $\RR$.

## Proof

We have that $\RR^{-1}$ is a relation.

The result follows from Image of Union under Relation.

$\blacksquare$