Question about constructible universe and $V = L$ I'm having trouble with the following problem:
Assume $V = L$, For each $\alpha$, let $\beta(\alpha)$ be the least such that $L_{\beta(\alpha)+1}$ contains a surjection from $\omega$ to $\alpha$. Let $X_\alpha = \{ A \subseteq \alpha : A \in L_{\beta(\alpha)} \}$. Show that for any $A \subseteq \omega_1$, there is a club C such that $A \cap \alpha \in X_\alpha$ for all $\alpha \in C$
and was wondering if anyone could help or give a hint?
 A: Consider any $A\subseteq\omega_1$. Then $A\in L_{\omega_2}$ (by the proof that $L$ satisfies GCH). Now let $(M\in,A)$ be a countable elementary submodel of $(L_{\omega_2},\in,A)$, and let $\pi:M\to N$ be the transitive collapsing map. Then $N=L_\gamma$ for some countable $\gamma$, and $\pi(\omega_1)=\alpha$ for some $\alpha<\gamma$. It's well-known that all ordinals $<\alpha$ are in $M$ and are fixed by $\pi$, and these are the only countable ordinals in $M$. So the construction of $\pi$ gives us that $\pi(A)=A\cap\alpha$. In particular, $A\cap\alpha\in L_\gamma$.
Note that there is no surjection $\omega\to\omega_1$ in $M$ (or anywhere else), and so, since $\pi$ is an isomorphism, there is no surjection in $N$ from $\omega$ to $\alpha$. Thus $\beta(\alpha)\geq\gamma$ and therefore $A\cap \alpha\in L_{\beta(\alpha)}$. 
So, given $A$, we have one $\alpha$ with $A\cap \alpha\in L_{\beta(\alpha)}$.  But we need a whole club of such $\alpha$'s. Fortunately, we can get a suitable $\alpha$ from any countable elementary submodel $M$ of $(L_{\omega_2},\in,A)$. So build a continuous elementary chain of such submodels (where "continuous" means taking unions at limit stages) of length $\omega_1$, gradually covering more and more of the countable ordinals. The $\alpha$'s associated to these submodels form a continuous increasing sequence of countable ordinals. (To prove the continuity, use again that $\alpha$ equals the set of countable ordinals in $M$.) So they constitute the desired club.
A: For $A\subseteq\omega_1$, let $C$ be the set of countable $\alpha$ such that there is some countable $M\prec L_{\omega_2}$ with $A\in M$, such that if $\pi$ is the Mostowski collapse of $M$, then $\pi(\omega_1)=\alpha$.
Exercise. $C$ is a club in $\omega_1$.
Claim. If $\alpha\in C$, then $A\cap\alpha\in X_\alpha$.
Fix $\alpha\in C$ and $M$ witnessing that. To see that $A\cap\alpha\in X_\alpha$, note that if $\pi(M)=L_\beta$, then $\beta\leq\beta(\alpha)$, since $L_\beta\models\alpha=\omega_1$.
It is enough to show that $A\cap M=A\cap\alpha$, and therefore $\pi(A)=A\cap\alpha$. If we show that, then $A\cap\alpha\in L_\beta$, and therefore $A\cap\alpha\in L_{\beta(\alpha)}$, and therefore $A\in X_\alpha$.
Exercise. $A\cap M=A\cap\alpha$. (Hint: if $\xi\in A\cap M$, then $M$ knows that $\xi$ is countable.)
