Solution verification: $A\sim B\implies \mathscr P(A)\sim \mathscr P(B)$. Prove that if $A$~$B$ then $\mathscr{P}(A)$~$\mathscr{P}(B)$, where ~ shows equivalence between sets.
Proof: Since $A$~$B$ we define a function $f:A\to B$ such that f is one-to-one and onto by the formula $f(a)=a+1$. Define  $g:\mathscr{P}(A)\to\mathscr{P}(B)$ for arbitrary $k,n\in\mathbb{Z^+}$ such that $n=|A|$, 
$g(\{a_1,...,a_k\})= 
\begin{cases} \{a_1 +1,...,a_k+1\}, &\text{$1\le k\le 2^{n-1}$}\\
\{a_k+1\}, &\text{$n=1$}\\
\end{cases}$
It's easy to check that $g$ is one-to-one and onto.
 A: Your proof is wrong, and has several key problems.


*

*Nowhere it says that $A$ and $B$ are finite sets of integers, they might be infinite sets which contain nothing that resembles integers at all.

*The function $f$ from $A$ to $B$ is given, and it is not defined by you.

*More to the previous point, functions are not necessarily given by a formula, it's not the early 19th century anymore. Functions are sets of ordered pairs with certain properties.

*Even more to the last two points, assuming that $f(a)=a+1$ seems to be an awfully specific formula, so given $A$, $B$ is uniquely determined by that function.

*"It's easy to check that $g$ is a bijection", that right there is -50% of the grade, had I been grading that problem. This check is exactly what you are asked to prove. It is true that it is easy to check, and if you were a Ph.D. student and the course was on a much harder topic, I would have overlooked that. But this sort of problem, including your "proof" indicate that you are far from the point where you should overlook that part of the proof.
Finally, the definition of $g$, is fairly correct. You need to adjust it to the general and arbitrary case. Let me give you a hint, then.

HINT: If $X\subseteq A$ then $\{f(x)\mid x\in X\}$ is a subset of $B$.

