# Preimage of Intersection under Mapping/Family of Sets/Proof 2

## Theorem

Let $S$ and $T$ be sets.

Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.

Let $f: S \to T$ be a mapping.

Then:

$\displaystyle f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$

where:

$\displaystyle \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\family {T_i}_{i \mathop \in I}$.
$f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.

## Proof

 $\displaystyle x$ $\in$ $\displaystyle f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i}$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \map f x$ $\in$ $\displaystyle \bigcap_{i \mathop \in I} T_i$ $\displaystyle \leadstoandfrom \ \$ $\, \displaystyle \forall i \in I: \,$ $\displaystyle \map f x$ $\in$ $\displaystyle T_i$ $\displaystyle \leadstoandfrom \ \$ $\, \displaystyle \forall i \in I: \,$ $\displaystyle x$ $\in$ $\displaystyle f^{-1} \sqbrk {T_i}$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle x$ $\in$ $\displaystyle \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$

$\blacksquare$