Suppose $A\subseteq \mathscr P (A)$. Prove that $ \mathscr P (A)\subseteq \mathscr P ( \mathscr P (A))$ This is Velleman's exercise 3.3.4. Suppose $A\subseteq \mathscr P (A)$. Prove that $ \mathscr P (A)\subseteq \mathscr P ( \mathscr P (A))$. I started reexpressing the terms in their equivalent forms to discover relations I did not believe are possible. $A\subseteq \mathscr P (A)$ is equivalent to $\forall x(x\in A \rightarrow x \subseteq A)$. Is there such an x at all? Anyway, $ \mathscr P (A)\subseteq \mathscr P ( \mathscr P (A))$ is equivalent to $ \forall x (x\subseteq A \rightarrow x \subseteq \mathscr P (A))$. Setting x arbitrary I have:


*

*Givens: $x\in A \rightarrow x \subseteq A$

*Goal:  $ x\subseteq A \rightarrow x \subseteq \mathscr P (A)$.


Two question:


*

*How do I proceed with the proof? 

*Can someone please show an example where these relations actually hold? 


Thanks in advance.
 A: Your result is an immediate consequence of the following proposition.
Proposition. Suppose $X\subseteq Y$. Then $\mathscr P(X)\subseteq\mathscr P(Y)$.
Proof. Let $E\in\mathscr P(X)$. Then $E\subseteq X\subseteq Y$ so that $E\subseteq Y$. Hence $E\in\mathscr P(Y)$. This proves $\mathscr P(X)\subseteq\mathscr P(Y)$. $\Box$
Do you see how your problem is now a corollary?
A: In general, if $A\subseteq B$, then $\mathscr P (A)\subseteq \mathscr P (B)$ because every subset of $A$ is a subset of $B$. 
More formally, if $a\in \mathscr P (A)$, we need to show that $a\in \mathscr P (B)$. But this is trivial, since if $x\in a$, then $x\in B$ which implies that $a\subseteq B$ which is the same as $a\in \mathscr P (B)$. 
Now take $B=\mathscr P (A)$ to obtain your claim. 
For an example, take $A=\emptyset$. Then, $\mathscr P (A)=\left \{ \emptyset \right \}$, $\mathscr P(\mathscr P(A))=\left \{ \emptyset,\left \{ \emptyset \right \} \right \}$ and then $\mathscr P (A)\subset \mathscr P ( \mathscr P (A))$.
A: $X\subset Y$ implies every element of X is an element of Y, so subsets of X are subsets of Y, so $\mathcal{P}(X)\subset\mathcal{P}(Y)$. Finally, for $Y=\mathcal{P}(X)$ you have $\mathcal{P}(X)\subset\mathcal{P}(\mathcal{P}(X))$.
