Special case of the Schroeder-Bernstein Theorem I have tried to work through the following, but I cannot do so on my own (or at least not in a way that I have convinced myself is correct). I know that it is elementary, but there's something to learn here that I am not picking up!

In the special case of the Schroeder-Bernstein Theorem where $A=B=\mathbb{N}$, define f as an injection A to B and g is an injection B to A, each mapping $n\mapsto 2n$.  f and g n are maps between P(A) and P(B). U is defined as a subset of A that maps f(U)' surjectively onto U'. Prove that:
$$U = \{l2^k \mid \text{$l$ is odd and $k$ is even}\}.$$

Any help would be appreciated. Note: This would technically be homework, though I am not enrolled in any course. This is my text: http://books.google.com/books?id=8KtRMofBKc0C&pg=PA118&source=gbs_toc_r&cad=4#v=onepage&q&f=false (see pages 118 and 120).
EDIT: During the proof of Schroeder-Bernstein: if $A$ and $B$ are sets, $f:A\rightarrow B$ and $g:B\rightarrow A$ are functions, we define $w:\mathscr{P}(A)\ni X\mapsto A\setminus g\left(B\setminus f(A)\right)\in\mathscr{P}(A)$ and $U=\bigcup\left\{X\in\mathscr{P}(A):X\subseteq \omega(X)\right\}$.
 A: In this case, $A=B=\mathbb{N}$ and $g=f:\mathbb{N}\rightarrow\mathbb{N}$ are given by $g(n)=f(n)=2n$. Also, $\omega:\mathscr{P}(\mathbb{N})\rightarrow\mathscr{P}(\mathbb{N})$ is given by
$$\omega(X)=\mathbb{N}\setminus g\left(\mathbb{N}\setminus f(X)\right)$$
and, finally, $U=\bigcup\left\{X\in\mathscr{P}(\mathbb{N}):X\subseteq \omega(X)\right\}$.
First, notice that, given $X\subseteq\mathbb{N}$, we have $f(X)=\left\{2x:x\in X\right\}$, then, since $g$ is injective,
$$g(\mathbb{N}\setminus f(X))=g(\mathbb{N})\setminus g(f(X))=\left\{2n:n\in\mathbb{N}\right\}\setminus\left\{4x:x\in X\right\},$$
therefore,
$$\omega(X)=\mathbb{N}\setminus g\left(\mathbb{N}\setminus f(X)\right)=\left\{\text{odd numbers}\right\}\cup\left\{4x:x\in X\right\}\qquad (*).$$
Let $V=\left\{2^k l:k\text{ even}, l\text{ odd}\right\}$. The exercise is to show that $U=V$.


*

*Let's show that $U\subseteq V$.


Let $x\in U$. Then, there exists $X\in\mathscr{P}(\mathbb{N})$ such that
$$x\in X\subseteq \omega(X)=\left\{\text{odd numbers}\right\}\cup\left\{4x:x\in X\right\}.$$
If $x$ is odd, then $x=2^0 x$, and we have $x\in V$.
If not, then $x=4x_1$, for some $x_1\in X$ (notice that this implies that $x\geq 4$). Again, if $x_1$ is odd, then $x=2^2x_1\in V$.
If not, then $x_1=4x_2$, for some $x_2\in X$, hence $x=2^4 x_2$ (notice that this imples that $x\geq 16$). Again, $x_2$ is either odd or not...
You can continue with this reasoning, and at some point you will find an odd $x_n$ such that $x=2^{2n}x_n$ (since $x<2^{2k}$ for some $k\in\mathbb{N}$).
Therefore, $x=2^{2n}x_n\in V$.
This shows that $U\subseteq V$.


*

*Let's show that $V\subseteq U$.


By the definitions of $U$, it suffices to show that $V\subseteq \omega(V)$.
Let $x=2^{2n}m\in V$, with $n,m\in\mathbb{N}$. By $(*)$, we have
$$\omega(V)=\left\{\text{odd numbers}\right\}\cup\left\{2^{k+2}l:k\text{ even},l\text{ odd}\right\},\quad\text{that is,}$$
$$\omega(V)=\left\{2^0l:l\text{ odd}\right\}\cup\left\{2^{k}l:k\text{ even},k\geq 2,l\text{ odd}\right\}
=\left\{2^k l:k\text{ even}, l\text{ odd}\right\},$$
since every even number is either $0$ or $\geq 2$.
Then $V\subseteq\omega(V)$, hence $V\subseteq U$.
Remark: In this book, $\mathbb{N}$ is defined as not containing $0$. But
in this exercise we use that $0$ is an even number. If you can's assume that $0$
exists, then we would have proved that $U=V\cup\left\{\text{odd numbers}\right\}$. But I don't see this as a big problem.
