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Well i've been having problems trying to prove that $\mathscr{P}(A) \subseteq\mathscr{P}(\mathscr{P}(A)) $ if $ A \subseteq \mathscr{P}(A)$ What i need is to get a proof by using quantifiers

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    Pick an element of the set on the left. By definition, it is a subset of $A$. By hypothesis, it is a subset of ${\mathcal P}(A)$. By definition, it belongs to the set on the right. Once you follow this, if you really need to "use quantifiers", all you need to do is to write all these statements formally.2012-02-29

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Before I actually do the problem, we should at least think about what we are really trying to prove here. The problem can be translated into words to mean :

If $A$ is a transitive set, then $\mathcal{P}(A)$ is a transitive set. (assuming that our universe contains only sets)

Anyway, here is the proof:

Let $x\in \mathcal{P}(A)$, then by definition of the power set $x \subseteq A$. But by assumption $A\subseteq \mathcal{P}(A)$ thus we see that $x\subseteq A \subseteq \mathcal{P}(A)$ and so $x\in \mathcal{P}(\mathcal{P}(A))$

I'm not sure how detailed "proof by quantifiers" is perhaps something like

$\forall x \in \mathcal{P}(A) (x\subseteq A) \land (A \subseteq \mathcal{P}(A)) \implies \forall x\in \mathcal{P}(A) (x\subseteq \mathcal{P}(A))$

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    @Mayhem You do need the $\forall$ is you want to do it by quantifiers. The definition of $A\subseteq B$ is that $(\forall x\in A)\Rightarrow x\in B$, and you can think of $x$ as a dummy variable (like when you consider the integral $\int_0^a x^2 dx$ $x$ is bound which is the technical name)2012-03-01
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$X \subseteq Y \implies \mathscr{P}(X) \subseteq \mathscr{P}(Y)$

take $x \in \mathscr{P}(X)$ so $x \subseteq X$ and since $X \subseteq Y$ then $x \subseteq Y$ so finaly $x \in \mathscr{P}(Y)$