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 $\alpha > \omega$.)



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


