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Let $\kappa$ be a cardinal and $\alpha$ any ordinal less than $\kappa$. Is it true that $\operatorname{ZF} \vdash L \cap P(L_\alpha) \subseteq L_\kappa$, where $L$ denotes the constructible universe?

I wrote this down in some notes a long time ago but I'm no longer certain that it's true, and I can't find a reference for it. A proof or disproof, or a reference, would be much appreciated. A much weaker statement appears in Jech's book, Set Theory.

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Yes, this follows from the condensation principle. It is clearly true for finite cardinals. For any infinite ordinal $\alpha$, consider a subset $A\subseteq L_\alpha$ with $A\in L$. Since $A\in L_\beta$ for some ordinal $\beta$, we can find an elementary substructure $X\prec L_\beta$ containing every element of $L_\alpha$ and $A$ itself, with $X$ of size $|\alpha|$. By condensation, $X$ is isomorphic to some $L_\gamma$, and since this has size $\alpha$, it must be that $\gamma<|\alpha|^+$. The isomorphism will fix every element of $L_\alpha$ and hence also $A$ itself. So $A$ appears in $L$ before the next cardinal.

This is the method Gödel used to prove that the GCH holds in $L$. For any infinite cardinal $\kappa$, all the subsets of $\kappa$ appear before $\kappa^+$, but since $L_{\kappa^+}$ has size $\kappa^+$, it means that $P(\kappa)$ has size $\kappa^+$ in $L$, so the GCH holds.

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    $\begingroup$ Since the question asks about provability in ZF, not ZFC, it seems worth noting that the absence of AC does no harm to this argument, because $L$ is well-orderable even without AC. $\endgroup$ Commented Sep 2 at 16:30

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