A tree coding all of $L_\alpha$ within $L_\alpha\cap\mathcal{P}(\omega)$ This question is about a sentence in the paper Gaps in the Constructible Universe by Marek & Srebrny. A countable ordinal is called a gap ordinal iff $(L_{\alpha+1}\smallsetminus L_{\alpha})\cap \mathcal{P}(\omega)=\emptyset$; that is, no reals are added in going from $L_\alpha$ to $L_{\alpha+1}$. Lemma 2.4 in the paper states:

Let $\alpha$ be a gap ordinal. If $X\in \mathcal{P}(\omega)\cap L_{\alpha}$ and $X$ is a real well-ordering, then the type of $X$ is less than $\alpha$.

The authors proved this by contradiction: assume not. We can construct a tree coding all of $L_{\alpha}$ within $L_{\alpha}\cap \mathcal{P}(\omega)$. Now, we apply Cantor's diagonal procedure to obtain a new set of natural numbers.
My question is about the italicized sentence in the paragraph above: what does this mean? More specifically, how do we construct such a tree?
 A: The point is that $X$ is a well-order, thus we may use recursion along $X$, but we do not know that the result of the recursion has to be a member of $L_\alpha$. Fortunately, we know that $L_\alpha$ for a gap ordinal $\alpha$ satisfies the complete arithmetical
schema of choice. (Lemma 2.2 of Marek and Srebrny:)

Lemma. If $\alpha$ is a gap ordinal, and $\phi(x,y)$ is a set-theoretic formula, then $L_\alpha$ satisfies
$$\forall n\in\omega\exists A\subseteq \omega \phi(n,A)\to \exists A\subseteq\omega \forall n\in\omega \phi(n,A^{(n)}).$$

Let $X\in L_\alpha\cap\mathcal{P}(\omega)$ be a well-order. I assume that $X$ is a well-order over $\omega$. Moreover, we may assume that the ordertype of $X$ is $>\alpha$.
Now consider the following formula $\phi(n, A)$ given by the conjunction of

*

*$A$ is a function whose domain is of the form $\{m\in\omega\mid m\le_X n\}$,


*For any $k$,

*

*if $k$ is the least element of $X$, then $A(k)=\langle(\varnothing,\varnothing)\rangle$,

*if $k$ is a $X$-successor of some $m\in\omega$, then $A(k)=(M,E)$, where $(M,E)=\operatorname{Def}(A(m))$, and

*if not, then $A(k)$ is the direct limit of all $A(m)$, $m<_Xk$.



By the lemma, there is $A$ such that $\phi(n,A^{(n)})$ for all $n\in\omega$. Especially, if $n$ is the $\alpha$th element of $X$, then $A^{(n)}$ codes an isomorphic copy of $\langle L_\xi \mid \xi\le\alpha\rangle$.
