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While I have seen and can understand a similar conversion in a different context, I can't seem to wrap my head around how this change works:

$B \in \{\mathscr P(A) | A \in \mathscr F\} \equiv \exists A \in \mathscr F(B = \mathscr P(A))$

In a simpler context, I can see that

$x \in \{n^2 | n \in \mathbb N\} \equiv \exists n \in \mathbb N(x = n^2)$

Because the former would be described as "x is a member of the set of all n-squared such that n is a member of the set of natural numbers" while the latter would be "There exists a number n in the set of natural numbers for which x is equal to n-squared." Obviously if x is a member of the set of square numbers, it makes sense that there must be a number in the given universe of discourse that can be squared to get x.

However, the equivalence I'm struggling with is a hang-up for me. If the set B is a member of the power set of A wherein A is a member of family F, I don't see how it follows that there must be a value of the set A for which B is equivalent to the power set of A. There could be, but I see no logically sound reason to assume there is one.

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  • $\begingroup$ I'm parsing the first formula as "$B$ is the powerset of $A$ for some $A \in \mathscr{F}$." I was initially reading it as $B$ being merely an element of some power set, which sounds like what you were/are thinking. $\endgroup$
    – pjs36
    Commented Dec 13, 2015 at 5:22

3 Answers 3

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The set $\{P(A) \mid A \in \mathscr{F}\}$ is the set of powersets of $A$, where $A$ runs over all sets in the family $\mathscr{F}$. So if $B$ is an element of this set, it must be the powerset of some $A \in \mathscr{F}$, not just any old element in one of those power sets, that is, not just any generic subset of some $A \in \mathscr{F}$.

I initially had the latter interpretation as well. There's an extra layer of "nesting" here that's tricky.

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    $\begingroup$ So, "B is a member of the set of $\mathbf all$ powersets of set A, such that A is $\mathbf any$ element of the family $\mathscr F$" would be a more accurate representation than what I was thinking of? That makes a lot of sense. Thank you. $\endgroup$
    – user242007
    Commented Dec 13, 2015 at 23:03
  • $\begingroup$ Yes, that's more like it! $\endgroup$
    – pjs36
    Commented Dec 14, 2015 at 0:24
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We can start from your second example, rewriting it as :

$X=\{x \mid ∃n \in \mathbb N \ (x=n^2) \}$.

The formula inside the set-builder operator $\{ \ \mid \ \}$ can be thought as a sort of "procedure" : as long as $n$ "spans" the set $\mathbb N$, we get the output of the "equation" $x=n^2$ and "throw in" the result $x$ into the set $X$.

In the same way, for :

$\mathscr X = \{ B \mid ∃A \in \mathscr F \ (B = \mathscr P(A)) \}$

we have a family of sets : $\mathscr F$ (in place of $\mathbb N$). As long as $A$ spans it ($A$ is a set), we get the corresponding power-set $\mathscr P(A)$ (obviously, it exists, because $A$ is a set; and also $\mathscr P(A)$ is a set) and "throw it in" into the family $\mathscr X$ of power-sets defined by formula.

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The thing is, writing $$ \{\mathscr{P}(A)|A\in\mathscr{F}\} $$ is a succinct way of writing $$ \left\{E\in\mathscr{P}\left(\mathscr{P}\left(\bigcup\mathscr{F}\right)\right)\bigg|\exists A\in\mathscr{F},E=\mathscr{P}(A)\right\}. $$ As you may well be aware of, when defining a set by describing the properties that its members have, you need to take those members in a set (here $\mathscr{P}(\mathscr{P}(\mathscr{\bigcup F}))$ was such a set). In general, one should always write something like $\{x\in X|\phi(x)\}$. This last set distinguishes the elements of $X$ that have the property $\phi(x)$. If you don't take the members from a set $X$, then you can get a contradiction, e.g. Russell's Paradox.

So,

If the set B is a member of the power set of A wherein A is a member of family F

would really be

If the set $B$ is a member of the family of power sets of sets in $\mathscr{F}$

So, the RHS of your equivalence just says that $B$ is one of those power sets.

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  • $\begingroup$ Initially I thought I understood this explanation, but now I'm wondering, how is it that $\mathscr P(A)$ is equivalent to $E \in \mathscr P(\mathscr P(\mathscr F))$? My understanding is that the sets within a power-set are not just made of all possible individual elements, but also all combinations. So, for example, if I had $x \in F$ where F was {1, 2, 3}, then x could be 1, 2, or 3, but $\mathscr P(F)$ could be $\emptyset$, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}. So wouldn't $\mathscr P(\mathscr F)$ be different from A? $\endgroup$
    – user242007
    Commented Dec 13, 2015 at 4:11
  • $\begingroup$ @user242007 I think you might be right that $\mathscr{P}(\mathscr{P}(\mathscr{F}))$ doesn't work. What about $\mathscr{P}(\mathscr{P}(\mathscr{\bigcup\mathscr{F}}))$? $\bigcup\mathscr{F}$ contains all the elements of all the $A$ in $\mathscr{F}$, so for a fixed $A$, $\mathscr{P}(\mathscr{\bigcup\mathscr{F}})$ contains all the subsets of $A$ and $\mathscr{P}(\mathscr{P}(\mathscr{\bigcup F}))$ contains the set of all those subsets, that is, $\mathscr{P}(A)$. $\endgroup$
    – Guest
    Commented Dec 13, 2015 at 4:28
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    $\begingroup$ Minor LaTeX remark, you can (and should) use \left and \right modifiers with \{\}. $\endgroup$
    – Asaf Karagila
    Commented Dec 13, 2015 at 12:19

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