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 $\alpha > \omega$.)

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    $\begingroup$ It’s $\omega_1$. $\endgroup$ Aug 1, 2020 at 5:46
  • $\begingroup$ I suspected so. Do you have a reference for a theorem that implies this? $\endgroup$ Aug 1, 2020 at 5:55
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    $\begingroup$ If $0^\sharp$ exists, then every cardinal is inaccessible in L. $\endgroup$ Aug 1, 2020 at 6:04
  • $\begingroup$ Ah, so $L \cap V_\alpha$ is countable for all countable $\alpha$, too, since $L \cap V_\alpha$ is the $V_\alpha$ of $L$. Thank you. $\endgroup$ Aug 1, 2020 at 6:08
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    $\begingroup$ In ZFC alone, the $\alpha$ in the title can be described as: (1) If genuine $\omega_1$ is a successor cardinal of $L$, then $\alpha$ is its immediate predecessor cardinal of $L$. (2) If genuine $\omega_1$ is a limit (and therefore inaccessible) cardinal of $L$, then it is equal to $\alpha$. The additional hypothesis that $0^\#$ exists implies that case (2) occurs. $\endgroup$ Aug 1, 2020 at 14:22

1 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.

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 $\omega_1^V$.

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.

  • $\begingroup$ OK, except I think you meant $P(L_\alpha)^L = L \cap P(L_\alpha)$, not $P(L_\alpha)^L = P(\alpha) \cap L$. $\endgroup$ Aug 3, 2020 at 0:30
  • $\begingroup$ @JesseElliott Whoops, quite right - fixed! $\endgroup$ Aug 3, 2020 at 0:36

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