# The union of powersets is contained in the powerset of union

Prove or Disprove: For any family of sets $\{A_n\}_{n\in\mathbb N}$

$$\bigcup_{n=1}^\infty\mathcal P \left({A_n}\right)\subseteq \mathcal P \left({\bigcup_{n=1}^\infty A_n}\right)$$

How do I approach proving this? I know how to unpack the definition of powersets ($\mathcal P \left({A}\right) = \{x | x \subseteq A\}$) but I'm not sure what else I can do. I've done powerset proofs before but none involving indexed family of sets.

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Like one approaches many such problems. show that if the set $w$ is an element of the left-hand side, it is an element of the right-hand side. – André Nicolas Oct 25 '12 at 4:06

The first thing to try when you’re faced with proving an inclusion like this is to show that each element of the lefthand side is an element of the righthand side. That is, begin by letting $x$ be an arbitrary element of $\bigcup\limits_{n\in\Bbb N}\wp(A_n)$. Then start applying definitions. In order for $x$ to belong to a union, it must belong to one of the sets whose union is being taken, so if $x\in\bigcup\limits_{n\in\Bbb N}\wp(A_n)$, then $x\in\wp(A_k)$ for some particular $k\in\Bbb N$. Is that enough to ensure that $x\in\wp\left(\bigcup\limits_{n\in\Bbb N}A_n\right)$?
Added: Yes, but there’s still some work to be done. Since $x\in\wp(A_k)$, $x\subseteq A_k$. But $A_k\subseteq\bigcup\limits_{n\in\Bbb N}A_n$ (why?), so $x\subseteq\bigcup\limits_{n\in\Bbb N}A_n$, and hence $x\in\wp\left(\bigcup\limits_{n\in\Bbb N}A_n\right)$. The amount of justification that you’ll need for these steps depends on what you’ve already proved; for instance, you may need to justify the fact that if $x\subseteq y\subseteq z$, then $x\subseteq z$.
I don't think so, because all we've proven is that $x\in\wp(A_n)$ for some $n\in\Bbb N$. Don't we need to prove it for all n in the set of natural numbers? I guess my main confusion with this proof is how the union is affecting it. Edit: Actually, I take that back. I just looked over the definition of Union of Indexed Family of Sets and I think that we DID prove it. – bill Oct 25 '12 at 4:54