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I know that $A\subseteq B\iff (\forall x)(x\in A\to x\in B)$. But how do I expand the powerset that way?

$$\mathcal{P}(A)\subseteq \mathcal{P}(B)\iff(\forall X)(X\in \mathcal{P}(A)\to X\in\mathcal{P}(B))$$

Now... how do I expand $X\in\mathcal{P}(A)$?

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$\mathcal{P}(A)\subseteq \mathcal{P}(B)\iff A\subseteq B$ – Dominic Michaelis Apr 7 '13 at 19:44
$x \in \mathcal{P}(A)$ just means that $x \subset A$. – Tom Oldfield Apr 7 '13 at 19:44
@DominicMichaelis I need to prove that fact, but I don't know how to go about it. – agent154 Apr 7 '13 at 19:45
$X\in\wp(A)$ iff $\forall x(x\in X\to x\in A)$. – Brian M. Scott Apr 7 '13 at 19:46
up vote 3 down vote accepted

First, the definition of powerset: $P(A) = \{ X; X \subseteq A\}$

So: $$X \in P(A) \iff X \subseteq A \iff \forall y:(y \in X \implies y \in A)$$

Therefore: $$P(A) \subseteq P(B) \iff \forall X : (X \subseteq P(A) \implies X \subseteq P(B)) \iff $$ $$ \forall X : (\forall y: (y \in X \implies y \in A) \implies \forall y: (y \in X \implies y \in B))$$

The negation is as follows: $$ \exists X :(\forall y: (y \in X \implies y \in A) \land \exists y : (y \in X \land y \notin B))$$

Using the rules: $\neg\forall X : \phi(x) \sim \exists X : \neg \phi(x) $ and $\neg(p \implies q) \sim p \land \neg q$.

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$P(A) = \{ X; X \subseteq A\}$ – suissidle Apr 7 '13 at 20:26
of course, my bad – jureslak Apr 8 '13 at 16:53

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