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

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 $$\alpha > \omega$$.)

• It’s $\omega_1$. Aug 1, 2020 at 5:46
• I suspected so. Do you have a reference for a theorem that implies this? Aug 1, 2020 at 5:55
• If $0^\sharp$ exists, then every cardinal is inaccessible in L. Aug 1, 2020 at 6:04
• 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. Aug 1, 2020 at 6:08
• 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. Aug 1, 2020 at 14:22

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.

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