Elementary set theory notation verification Reading Velleman's "How To Prove It" I came across the following expression:
$$ x \in\bigcup\{\mathscr P(A)\mid A\in \mathcal F\} $$ such that $\mathcal F$ is a family of sets, $A$ is a set, and $\mathscr P(A)$ is the power set of $A$. Now according to Velleman, it holds: $$ x \in\bigcup\{\mathscr P(A)\mid A\in \mathcal F\} \iff \exists A \in \mathcal F(x\in \mathscr P(A)).$$ The last term means "there is the set A which is an element of F, such that x is an element of the power set of A". Now, it seems to me, however, that the following equality also: $$ \exists A \in \mathcal F(x\in \mathscr P(A)) \iff x \in\{\mathscr P(A)\mid A\in \mathcal F\}.$$ Which implies:$$ x \in\bigcup\{\mathscr P(A)\mid A\in \mathcal F\} \iff x \in\{\mathscr P(A)\mid A\in \mathcal F\}.$$ Is that correct? Thanks in advance.
 A: No, not quite.
$x\in\{\mathscr P(A)\mid A\in\mathcal F\}$ if and only if there is some $A\in\cal F$, such that $x=\mathscr P(A)$. 
It might be the case that $x\in\mathscr P(A')$ for a different $A'\in\cal F$. But just from the given information we cannot infer this. And since under the assumption that $\mathcal F=\{A\}$ we can show this to be false (meaning we can show that $\mathscr P(A)\notin\mathscr P(A)$), the equivalence you suggested is not true.
A: You're wrong:
$\exists A \in \mathcal F(x\in \mathscr P(A))$ means in plain English: $x$ is a subset of some $A$ in family $\mathcal F$, i.e. a member  of the powerset of some $A$ in family $\mathcal F$.
while $\,x \in\{\mathscr P(A)\mid A\in \mathcal F\}$ is the powerset of some $A$ in family $\mathcal F$.
A: Let us simply analyze what the statement $x \in \bigcup \{ \mathscr P (A) \mid A \in \mathcal{F} \}$ means. For $x$ to be in a union means exactly that $x$ is in at least one of the members of the union. So we can rewrite this as
$$ \exists A \in \mathcal F : x \in \mathscr P(A)$$
(which is what I think you meant to write). However, your last equivalence is not true. Let's consider an example; take $\mathcal F = \{\emptyset, \{a\} \}.$ Now $\mathscr P(\emptyset) = \{ \emptyset \}$ and $\mathscr P(\{a\}) = \{ \emptyset , \{a\}\}.$ Thus
\begin{align} x \in \bigcup \{ \mathscr P (A) \mid A \in \mathcal{F} \} &\iff \exists A \in \mathcal F : x \in \mathscr P(A) \\
&\iff x \in \{\emptyset\} \text{ or } x \in \{ \emptyset , \{a\}\} \\
&\iff x = \emptyset \text{ or } x = \{ a \}.
\end{align}
On the other hand, \begin{align} x \in \{ \mathscr P(A) \mid A \in \mathcal F\}
&\iff  x \in \{ \{ \emptyset \}, \{ \emptyset , \{a\}\}\} \\
&\iff x = \{ \emptyset \} \text{ or } x = \{ \emptyset , \{a\}\}.
\end{align}
In other words: in the first case, $x$ is an element of some $\mathscr P(A)$, whereas in the second case, $x$ is equal to some $\mathscr P(A).$
