B is an element of some power set of A such that A is an element of F.

Question

$$B \in \{\,\mathcal P(A) \mid A \in F\,\}$$

I can't quite figure this out. My textbook says this statement is equivalent to

$$\exists A \in F\ \forall x\ (x \in B \leftrightarrow \forall y\ (y \in x \rightarrow y \in A))$$

How do you derive the last statement from the first? And how would you read this in a natural language?

Answer 1

The notation $B\in\{\,\mathcal P(A)\mid A\in F\,\}$ is equivalent to $$ \exists A\in F(B=\mathcal P(A))$$ Also, $B= \mathcal P(A)$ is equivalent to $$\forall x(x\in B\leftrightarrow x\subseteq A)$$ and $x\subseteq A$ is equivalent to $$\forall y(y\in x\to y\in A).$$ Combining all this you obtain the textbook expression.

Answer 2

Let $F$ be a family of sets and fix a particular set $B$. Then the statement says that:

$B$ is the power set of one of the sets in $F$.

Indeed: \begin{align*} B \in \{\mathcal P(A) \mid A \in F\} &\iff \exists A \in F \text{ such that } B = \mathcal P(A) \\ &\iff \exists A \in F \text{ such that } \forall x ~ [x\in B \leftrightarrow x\in \mathcal P(A)] \\ &\iff \exists A \in F \text{ such that } \forall x ~ [x\in B \leftrightarrow x \subseteq A] \\ &\iff \exists A \in F \text{ such that } \forall x ~ [x\in B \leftrightarrow \forall y ~ [y \in x \rightarrow y \in A]] \\ \end{align*}

Answer 3

I think the notation might be throwing you off. It may be easier to start with an example:

Let $$F = \{\mathbb{Z},\mathbb{Q}, \mathbb{R}\} \tag{1}$$ then $A$ is any one of the integers, rationals, or reals considered as sets. The power set of any of these sets is the set of all subsets them.

So, here the statement $$B \in \{\,\mathcal P(A) \mid A \in F\,\},$$ is saying $B$ is one such power set. For example, $\mathcal P (\mathbb{Z})$.

For your second statement, the part to focus on is $$x \in B \leftrightarrow \forall y\ (y \in x \rightarrow y \in A) \tag{2}.$$ You should take some time to convince yourself, that when in the scope of this part of the statement, the set $F$ becomes irrelevant. This statement is saying that membership of $x$ in $B$ is equivalent to claiming that any element, $y$, of $x$ must guarantee that it is also an element of $A$. Here you should remember that this says the implication is true, not that any element is claimed to be a member of $A$ as it might be the null set for example.

This fits in well, and is true, since $B$ is a power set. Working with the definition of a power set, you will see that this statement follows very easily. Of course, though I have mentioned to the contrary, it is technically correct to carry through all of the quantifiers in and before $(2)$.