Prove $P(A) \subset P(B)$ if and only if $A \subset B$ I have to prove the proposition: Prove $P(A) \subset P(B)$  if and only if  $A \subset B$
Here's what I wrote:
($\impliedby$) If   $A \subset B$ with $A \in P(A)$ and $B \in P(B)$ then
$\Rightarrow A \subset B \in P(B)$
$\Rightarrow A \in P(B)$ and
$\Rightarrow A \in P(A)$ so
$\Rightarrow P(A) \subset P(B)$
What do you think?
 A: In order to show that $P(A)\subset P(B)$, you should take $C\in P(A)$ and show that $C\in P(B)$.
But if $C\subset A$ and $A\subset B$, it comes $C\subset B$, so $C\in P(B)$
For the other way, if you assume that $P(A)\subset P(B)$, you have $A\in P(A)$, so $A\in P(B)$, i.e. $A\subset B$.
A: If $X \in P(A) \to X \subseteq A\to X \subseteq B\to X \in P(B)\to P(A) \subseteq P(B)$ .If $P(A) \subseteq P(B)$, and $x \in A \to \{x\} \in P(A) \to \{x\} \in P(B) \to x \in B \to A \subseteq B$. Thus $P(A) \subseteq P(B) \iff A \subseteq B$
A: Looks like you have the right idea in mind but as others have pointed out, your logic is flawed in the last few lines. Moreover you'll want to abide by the standard way to show that one set is a subset of another. $C$ is a subset of $D$ iff for each $x$ in $C$ you can show $x \in D$. Let $C = P(A)$ and $D = P(B)$. To show that $P(A) \subset P(B)$ follows from the assumption that $A \subset B$, you'll want to pick an arbitrary element $x \in P(A)$ and show it is contained in $P(B)$. But what does $x$ look like? Well by definition $x$ is some subset of $A$. By assumption $x$ is also a subset of $B$. Hence $x \in P(B)$ and you are done. Does this make sense? If so, then try to complete the proof in the other direction; suppose that $P(A) \subset  P(B)$ from the start to conclude that $A \subset B$. Once you show this your proof will be complete.
