smallest ordinal $\alpha$ such that $L \cap P(L_\alpha)$ is uncountable

Question

Let $V$ denote the von Neumann universe and $L$ Gödel's constructible universe. For any set $X$, let $P(X)$ denote the power set of $X$.

Assume that $0^\sharp$ exists (and ZFC).

What is the smallest ordinal $\alpha$ such that $L \cap P(L_{\alpha})$ is uncountable? (If $V = L$, then $\alpha = \omega$, but if $0^\sharp$ exists, then $\alpha > \omega$.)

Answer 1

Really, this was answered in the comments; I'm putting this answer down to move this off the unanswered queue. I've made this CW and will delete it if one of the original commenters adds their own answer.

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We have in $L$, for each (infinite) $\alpha$, the following bijections:

• $f_\alpha:\alpha\rightarrow L_\alpha$.

• $g_\alpha: \mathcal{P}(L_\alpha)^L=\mathcal{P}(L_\alpha)\cap L\rightarrow L_{(\vert\alpha\vert^+)^L}$.

Hence $\vert\mathcal{P}(L_\alpha)^L\vert=\vert(\vert\alpha\vert^+)^L\vert$. Now assuming $0^\sharp$ we have that $\omega_1^V$ is a limit cardinal in $L$, so for each $\alpha<\omega_1^V$ we have $\vert\mathcal{P}(L_\alpha)^L\vert=\aleph_0$.

So the answer to your question is $\omega_1^V$.

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Note that all this requires is that $\omega_1^V$ be a limit cardinal in $L$. More generally, let $\kappa$ be the supremum of the $L$-cardinals whose $L$-successor is $<\omega_1^V$; then the $\kappa$th level of $L$ is the first whose $L$-powerset is truly uncountable.