CH imples the existence of a function I was studying an article and the author stated that CH implies that there exists a function from $\omega_1 \setminus \omega$ onto the set of all countable subsets of $\omega_1$ such that for each $\omega\leq \alpha<\omega_1$, $F(\alpha)\subset \alpha$. He stated that it's a trivial consequence, but I can't see why. Can someone tell me how to prove the existence of this function?
 A: Let $A=[\omega,\ \omega_1)=\{\alpha:\omega\le\alpha\lt\omega_1\}$ and let $B=\{X\subseteq\omega_1:X\text{ is countable }\}$. Using CH it is trivial to construct an injective function $g:B\to A$ with $X\subset g(X)$ for every $X\in B$; just list the elements of $B$ in a transfinite sequence of length $\omega_1$ and use transfinite recursion. Let $F$ be the inverse function of $g$, extended by setting $F(\alpha)=\emptyset$ whenever $\alpha$ is not in the range of $g$.
A: Here is a more 'elementary' solution: Let $M \prec H(\omega_2)$ be of size $\aleph_1$ and closed under countable sequences. Such a structure exists by $\mathsf{CH}$. Further, suppose that $M = \bigcup_{\alpha < \omega_1} M_\alpha$, where $\langle M_\alpha\rangle_{\alpha < \omega_1}$ is a continuous increasing sequence of countable elementary substructures of $H(\omega_2)$ (that is, $M_\alpha \in M_{\alpha+1}$, and if $\alpha$ is a limit ordinal, $M_\alpha = \bigcup_{\beta < \alpha} M_\beta$). 
Let $\delta_\alpha = \mathrm{sup}(M_\alpha \cap \omega_1)$. Note that $M_{\alpha+1}\setminus M_\alpha$ is countable, and that $\delta_{\alpha +1} > \delta_{\alpha} + \omega$, and that if $s\in M_\alpha$ is a countable subset of $\omega_1$, then $s \subset \delta_\alpha$. Also, by our choice of $M$, each countable subset of $\omega_1$ occurs in some $M_\alpha$. 
So now, define a function $F$ with domain $\bigcup_{\alpha < \omega_1}[\delta_{\alpha+1}, \delta_{\alpha+1}+\omega)$ satisfying the properties you require, where $F$ restricted to $[\delta_{\alpha+1}, \delta_{\alpha+1}+\omega)$ simply enumerates the countable subsets of $\omega_1$ in $M_{\alpha+1} \setminus M_\alpha$. By our observations, the range of this function is the set of all the countable subsets of $\omega_1$, so we can extend $F$ in any well-behaved way to $\omega_1 \setminus \omega$ to obtain a function as required. 
