If $V=L$, is every $V_\alpha$ an $L_\beta$? I guess that amounts to if there is a continuous $f$ with $\mathbb{P}(\mathbb{L}_{f(\alpha)}) \cap \mathbb{L} = \mathbb{L}_{f(\alpha+1)}$
I seem to remember reading that it is, but I forget where or when I read it, why it's true, what f is, and I can't find it now.
Thanks for any info.

The answer is no.


*

*It is a theorem of $ZFC$ that $V\models\alpha>\omega\rightarrow |L_\alpha|=|\alpha|$, therefore it is consistent with $V=L$ as well, so $L\models |L_\alpha|=|\alpha|$ for infinite $\alpha$.

*On the other hand, $V\models V_{\alpha+1}=\mathcal P(V_\alpha)$, which for $\alpha>0$ is not the case that $|V_\alpha|=|\alpha|$, but rather much larger. (Of course for inaccessible cardinals there is an equality but this requires a consistency strength greater than $Con(ZFC)$).

*Now consider the following case: $L\models 2^\omega=\omega_1$, since $L_{\omega+1}$ is countable it cannot have all the subsets of $\omega$. On the other hand $\mathcal P(\omega)\subseteq V_{\omega+1}$ so clearly $L_{\omega+1}\neq V_{\omega+1}$.

*And finally $\beta$ is the set of all ordinals in both $L_\beta$ and $V_\beta$. If for some $\beta$ we had $V_{\omega+1}=L_\beta$ then $\beta=\omega+1$ but we have seen that this is not the case.

