How to Analyze $x \in \bigcup\{P(A) | A \in F\}$ logically? This is from Velleman's How to Prove It. 
$F$ is a family of sets. $P(A)$ is the power set of $A$. 
The book says: "This time we start by writing out the deﬁnition of union. According to this deﬁnition, the statement means that x is an element of at least one of the sets $P(A)$, for $A \in F$. In other words, $\exists A \in F(x \in P (A))$. Inserting our analysis of the statement $x \in P(A)$ from Example 2.3.3, we get $\exists A \in F\forall y(y \in x \to y \in A)$." 
I understand that $x$, if it belongs to the union of $P(A)$, must also exist inside of a set in $P(A)$, with $A$ belonging to $F$. But what I don't understand is how $\exists A \in F(x \in P (A))$ suffices the English form. The union definition means to essentially concatenate the subsets of all $P(A)$ from the $A \in F$ into one large non-redundant set. 
This explanation of union on the previous page was very helpful: $$\bigcup F =\{ 1,2,3,4\}\cup\{2,3,4,5\}\cup\{3,4,5,6\}=\{1,2,3,4,5,6\}$$
But then why does he make $x \in P (A)$ when the items in the union of $P(A)$ are all integers? Wouldn't doing $x \in P(A)$ imply that $x$ is a set? In addition, belonging to the union of a set would mean that, ultimately, only $A$ matters because the elements of the sets in $P(A)$ all belong to $A$?
Edit: this is not a duplicate of the other question because I have tangential, more general questions about sets. 
Edit 2: one more question. How can $x\in \mathcal{P}(A)$ ? Isn't $x$ an integer? 
Edit 3: in the linked thread, many commenters noted that $x \in \mathcal P(A)$ rather than $x \in A$. But how is this true? I understand this from the logic/definition point of view. But if you take the union of a family of sets and write them out in their elements, then it would have to be literally anything other than a set. How does this work? 
 A: I do not completely understand your question/s, but I hope this helps you.
Let $\mathcal{F}$ be a family of sets. Example: $\mathcal{F}=\{\{a,b\}, \{c,d\}\}$. It has 2 elements.
Then $\{\mathcal{P}(A)|A\in \mathcal{F}\}$ is a family of sets of sets. For our example, $\{\mathcal{P}(A)|A\in \mathcal{F}\}=\{ \{\emptyset, \{a\}, \{b\},\{a,b\}\},\{ \emptyset, \{c\},\{d\} ,\{c,d\}\}\}$ (please parse this carefully). It still has 2 elements.
Then, $\bigcup \{\mathcal{P}(A)|A\in \mathcal{F}\}$ will just become a set of sets. As you already know, union shatters the "family" by "concatenating" all "family members". For our example, $\bigcup \{\mathcal{P}(A)|A\in \mathcal{F}\}=\{ \emptyset, \{a\}, \{b\},\{a,b\},\{c\},\{d\},\{c,d\}\}$. Now 8 elements.
Now, from our example, if $x\in \bigcup \{\mathcal{P}(A)|A\in \mathcal{F}\}$, then $x$ should be a set! $x$ is not an element of some $A$, but a subset of it. Hence, $x$ should be an element of $\mathcal{P}(A)$ (Understand that the set $\{c\}$ is different from the element $c$).
