> 
> Suppose $\{A\_{i} | i \in I\}$ is an indexed family of sets. Prove
>  that $\bigcup\_{i\in I} \mathscr P(A\_{i}) \subseteq \mathscr P
> (\bigcup\_{i \in I}A\_{i})$
> 
> 
> 

My attempt:

Let's denote $\{A\_{i} | i \in I\}$ as $\mathcal F$.

Let $x \in \bigcup\_{i\in I} \mathscr P(A\_{i})$. It implies that there is at least one set, call it $B$, such that $B \in \mathcal F$ and $x \subseteq B$. $\bigcup\_{i \in I}A\_{i}$ (or $\bigcup \mathcal F$) is the set containing all elements of all sets in $\mathcal F$. Since $B \in \mathcal F$, it follows that $x \subseteq \bigcup\_{i \in I}A\_{i}$. And by definition of a power set, we can conclude that $x \in \mathscr P (\bigcup\_{i \in I}A\_{i})$. Since $x$ was arbitrary, we conclude that $\forall x \bigl(x \in \bigcup\_{i\in I} \mathscr P(A\_{i}) \implies x \in \mathscr P (\bigcup\_{i \in I}A\_{i})\bigr)$, or in other words, $\bigcup\_{i\in I} \mathscr P(A\_{i}) \subseteq \mathscr P (\bigcup\_{i \in I}A\_{i})$. $\Box$

Is it correct?