Vtyjkyuk

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 $exists A in Fforall y(y in x to y in A)$."

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: $$bigcup F = 1,2,3,4cup2,3,4,5cup3,4,5,6=1,2,3,4,5,6$$

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 $xin mathcalP(A)$ ? 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?

I do not completely understand your question/s, but I hope this helps you.

Let $mathcalF$ be a family of sets. Example: $mathcalF=a,b, c,d$. It has 2 elements.

Then $Ain mathcalF$ is a family of sets of sets. For our example, $Ain mathcalF= emptyset, a, b,a,b, emptyset, c,d ,c,d$ (please parse this carefully). It still has 2 elements.

Then, $bigcup Ain mathcalF$ will just become a set of sets. As you already know, union shatters the "family" by "concatenating" all "family members". For our example, $bigcup Ain mathcalF= emptyset, a, b,a,b,c,d,c,d$. Now 8 elements.

Now, from our example, if $xin bigcup Ain mathcalF$, 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 $mathcalP(A)$ (Understand that the set $c$ is different from the element $c$).