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$B ∈ \{\mathcal{P} (A) | A ∈ F\}$; where $\mathcal{P}(A)$ is the power set of $A$ and $F$ means Family of sets

the book gave the following steps:

$$∃A ∈ F(B = \mathcal{P} (A))$$

$$∀x(x ∈ B \iff x ⊆ A)$$

$$∃A ∈ F ∀x(x ∈ B \iff ∀y(y ∈ x → y ∈ A))$$

Basically, i do not understand $B=\mathcal{P}(A)$. The way i reasoned through this is if $B$ is an element of the set of unique subsets of the set $A$ then $B$ is a subset of $A$ where $A$ is a set element of $F$. Now if $B$ is a subset of $A$ then every element in $B$ must be in $A$, but the converse does not necessarily hold true. Where in my line of reasoning was i mistaken? Thanks!

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    But $B$ **is** a set of all subsets of the set $A$, not an element of this set. – 2017-01-30
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    why is there a B∈{...}? don't the ∈ indicate B is an element of P(A)? – 2017-01-30
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    $B$ belongs to the set consisting of all the power sets of elements of $F$, and is equal to one of these power sets, namely, $\mathcal P(A)$. – 2017-01-30
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    Read it aloud: $B$ is an element of the set of all the $\mathcal{P}(A)$ such that $A$ belongs to $F$. So, $B$ is one of the $\mathcal{P}(A)$s. – 2017-01-30
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    dammit i was reading it wrong all along lol – 2017-01-30

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