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

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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

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

$\blacksquare$


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