Stuck because of possible error in exercise of "How to Prove It" by Velleman I am self-studying "How to Prove It" by Velleman, and I believe there must be an error on exercise 3.3 #14.
I'll show the question here, and where I think the error is, and then I'd love to find out if you believe I'm correct that there is an error, or if not, where am I missing the boat?
The exercise:
"Suppose $ \{A_i \mid i \in I \} $ is an indexed family of sets. Prove that $ \bigcup_{i \in I} \mathbb{P}(A_i) \subseteq \mathbb{P}(\bigcup_{i \in I} A_i) $."
Where I believe the problem to be:
It seems to me that the left-hand side of the subset symbol would be a "set" (in other words, "flat" if you will), while the right-hand side would be a "set of sets." To me, it appears that the definition of subset makes it impossible for a "set" to be a subset of a set-of-sets. 
Is the book wrong, or am I? (I suspect I am wrong, but after approaching it from several different angles, I simply DO NOT SEE WHERE).
Thanks.
 A: Suppose $A_1=\{1,2\}$ and $A_2=\{2,3\}$.  Then $\mathbb P(A_1) = \{\varnothing,\{1\},\{2\},\{1,2\}\}$ and $\mathbb P(A_2) = \{\varnothing,\{2\},\{3\},\{2,3\}\}$, so
$$
\mathbb P(A_1)\cup\mathbb P(A_2) = \{\varnothing,\{1\},\{2\},\{3\},\{1,2\},\{2,3\}\}
$$
and that is a subset of
$$
\mathbb P(A_1\cup A_2) = \{ \varnothing,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\} \}.
$$
Of course this doesn't prove the result, but it shows how your objection to the conclusion misses something.
A: The exercise is correct. Actually, I do not understand your problem. To say that $A\subseteq B$ means that any element of $A$ is an element of $B$. It is completely irrelevant whether the elements of $A$ are apples, numbers, or sets. (This is important: In mathematics, we keep track with the notation of the "type" of a set, and have elements, sets, collections of sets, families of collections, etc. But the distinction is purely there for notational convenience. In reality, there are no "types", and the elements of a set may be themselves sets whose elements are sets whose elements are sets ... Although it may appear difficult to keep track of all this at a given point, the moral is that probably you do not need to. For example, in the case at hand, this plays no role, and the nature of the actual members of the $A_i$ is not important.)
Anyway, the two things to remember are that 


*

*If $A\subseteq B$, then also $\mathcal P(A)\subseteq \mathcal P(B)$, and

*If for each $i$ in an index set $I$ we have a set $B_i$, and it happens that each $B_i$ is a subset of $C$, then also $\bigcup_{i\in I}B_i\subseteq C$. 


These are the two statements you need to understand, and verify, and the exercise follows immediately from them. If either statement gives you difficulties, make sure that the difficulties are mathematical, and not a result of your current intuition (and, of course, feel free to ask for further clarifications). But again, check that the problem is not that "it seems that something should not work". After all, if you are just learning a new subject, most likely your intuitions are not fully developed yet and may suggest erroneous conclusions (as apparently it is the case here).
A: When viewing mathematics as founded on set theory, the idea is that "everything is a set". Sets of sets are still sets.
It is entirely reasonable to worry about hierarchies, though. Various work on foundations has, indeed, used systems in which "things", and "sets of things", and "sets of sets of things" were different.
But the usual use of set theory does not attempt this distinction, avoiding paradoxes by restricting set-formation in other ways.
A: When you have a doubt, go back to the definitions and try hiding what seems problematic. Set $B_i=\mathbb{P}(A_i)$.
An element of $\bigcup_{i \in I} B_i$ is an object, let's call it $X$, such that $X\in B_j$, for some $j\in I$. Don't worry about the case of the letters, they are just arbitrary labels. Well, $X\in B_j$ means $X\in\mathbb{P}(A_j)$, so $X\subseteq A_j$.
Good! Now we recall that $A_j\subseteq\bigcup_{i\in I}A_i$, so, by transitivity, $X\subseteq\bigcup_{i\in I}A_i$, which means
$$
X\in\mathbb{P}\Bigl(\bigcup_{i\in I}A_i\Bigr)
$$
Done! We have proved
$$
\bigcup_{i\in I}\mathbb{P}(A_i)
\subseteq
\mathbb{P}\Bigl(\bigcup_{i\in I}A_i\Bigr).
$$
By the way, the inclusion is strict, in general. Indeed, in order the equality holds, any subset $Y$ of $\bigcup_{i\in I}A_i$ should be contained in some $\mathbb{P}(A_j)$, in particular
$$
\bigcup_{i\in I}A_i\subseteq A_j
$$
for some $j\in I$. This implies $A_i\subseteq A_j$ for all $i\in I$. So, if equality holds, one of the sets of the family contains all the others; the converse is almost obvious.
