How do you go from $x\in\cup\{\mathcal{P}(A)|A\in\mathcal{F}\}$ to $\exists A\in\mathcal{F}(x\in \mathcal{P}(A))$? These are statement 4 of example 2.3.6 and its solution from section
2.3 in Daniel J. Velleman's "How to Prove It - A Structured Approach"
(great book), where the author asks the reader to analyze the logical
form of several statements. On the solution to this particular statement,
i.e.:
$x\in\cup\{\mathcal{P}(A)|A\in\mathcal{F}\}$
he argues that, according to the definition of union given earlier:
$\cup\mathcal{F}=\{x|\exists A\in\mathcal{F}(x\in A)\}=\{x|\exists A(A\in\mathcal{F}\wedge x\in A)\}$
the statement means that "... $x$ is an element of at least one
of the sets $\mathcal{P}(A)$, for $A\in\mathcal{F}$. In other words,
$\exists A\in\mathcal{F}(x\in\mathcal{P}(A))$."
Intuitively it makes sense, but I can't write it down formally. If
I state that $x\in\cup\mathcal{F}$ is true, I know that $\exists A\in\mathcal{F}(x\in A)$.
But I get lost when I try to replace $\mathcal{F}$ with $\mathcal{P}(A)$,
and can't figure out the rest.
 A: This is the same as other answers in essence, written down differently (except that personally I don't like this way of writing quantifiers).
For me, these kind of problems are just about carefully unpacking of the definitions, and then using logic to simplify.
So I would calculate as follows:
\begin{align}
& x\in\cup\{\mathcal{P}(A)|A\in\mathcal{F}\} \\
\equiv & \qquad \text{"Velleman's definition of $\;\cup\;$, but using $\;B\;$ to prevent confusion"} \\
& x\in\{x|\exists B (B\in\{\mathcal{P}(A)|A\in\mathcal{F}\} \land x\in B)\} \\
\equiv & \qquad \text{"definition of set builder notation"} \\
& \exists B (B\in\{\mathcal{P}(A)|A\in\mathcal{F}\} \land x\in B) \\
\equiv & \qquad \text{"definition of (a more complex version of) set builder notation"} \\
& \exists B (\exists A \in \mathcal{F} (B = \mathcal{P}(A)) \land x\in B) \\
\equiv & \qquad \text{"logic: pull $\;x\in B\;$, which does not use $\;A\;$, into $\;\exists A\;$"} \\
& \exists B \exists A \in \mathcal{F} (B = \mathcal{P}(A) \land x\in B) \\
\equiv & \qquad \text{"logic: exchange $\;\exists\;$ quantifiers"} \\
& \exists A \in \mathcal{F} \exists B (B = \mathcal{P}(A) \land x\in B) \\
\equiv & \qquad \text{"logic: one-point rule"} \\
& \exists A \in \mathcal{F} (x\in \mathcal{P}(A)) \\
\end{align}
The OP seems gone, but this may still help someone...
A: From $x\in\cup\{\mathcal{P}(A)|A\in\mathcal{F}\}$ you can conclude that there is an $A\in \mathcal{F}$ such that $x\in \mathcal{P}(A)$. The $x$ is a subset of $A$ and (not necessarily) an element of $A$. 
A: Suppose $x \in \cup \{ \cal{P}(A) | A \in \cal{F} \}$. Since $x$ is in the union of a set, that means there is an element of the set of which $x$ is an element. That is, there is an element of $\{ \cal{P}(A) | A \in \cal{F} \}$ for which $x$ is an element. Every element of this set has the form $\cal{P}(A)$ for some $A \in \cal{F}$, so there must be an $A \in \cal{F}$ such that $x \in \cal{P}(A)$, in symbols $\exists A \in \cal{F}(x \in \cal{P}(A))$.
A: Define $\mathcal{G} = \{P(A) : A \in \mathcal{F}\}$. Then
$x \in \bigcup \{P(A) : A \in \mathcal{F}\}$ 
is equivalent to
$x \in \bigcup \mathcal{G}$
By definition of union, it is equivalent to 
$(\exists B \in \mathcal{G})(x \in B)$
Recall that $(\exists B \in \mathcal{G}$ if and only if there $(\exists A \in \mathcal{F})(B = \mathcal{P}(A))$. So replacing $B$ by $\mathcal{P}(A)$, you get
$(\exists A \in \mathcal{F})(x \in \mathcal{P}(A))$
A: Something in what you wrote looks a little suspicious: if I'm understanding correctly your notation, $\,\mathcal F\,$ is a set of sets and, thus, its elements are sets, whereas that $\,x\,$ there seems to be an element of some set in $\,\mathcal F\,$...
I think the problem is: by $\,x\in\cup\{P(A)\;|\;A\in\mathcal F\}\,$ , we have that x in an element of $\,P(A)\,$ , for some $\,A\in\mathcal F\,$ , so $\,x\,$ is a subset of some such $\,A\,$ .
Yet later we have that $\,x\in A\,$ , for some $\,A\in\mathcal F\,$...and this already messes all up, since above we saw $\,x\,$ is a subset of some set, yet now it is an element of some set...so unless some, or all, the elements in $\,\mathcal F\,$ are sets that contain both elements and the sets determined by them, we've reached a dead-end here.
Now, what you say the author says is correct: the statement $\,x\in\cup\{P(A)\;|\;A\in\mathcal F\}\,$ means "there exists some set $\,A\,$ in $\,\mathcal F\,$ s.t. $\,x\subset A\,$ , but I can't make any sense of $\,\cup\mathcal F\,$ ...
