Is there a model of $\mathsf{ZFC}$ such that for every cardinal $\beta > \aleph_0$ there is a cardinal $\alpha < \beta$ such that $|{\cal P}(\alpha)| >\beta$?
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4Take the limit of $\aleph_0,2^{\aleph_0},2^{2^{\aleph_0}},\dots$ – Wojowu Nov 17 '19 at 09:54
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There cannot be such a model. Let for instance $\beta=\beth_\omega$, the supremum of $\beth_0=\aleph_0,\beth_1=2^{\beth_0},\beth_2=2^{\beth_1},\dots$, for any $\alpha<\beta$ we have $\alpha\leq\beth_n$ for some $n$, and hence $|\mathcal P(\alpha)|\leq\beth_{n+1}<\beta$.
The exact same argument shows that this proposition necessarily fails for all $\beta$ which are strong limit cardinals (i.e. $\kappa<\beta\implies 2^\kappa<\beta$). However, at least relative to large cardinals, it is consistent that those are the only points of failure - by a result of Woodin (apparently unpublished, discussed here) it is consistent with ZFC that for all $\alpha$ we have $2^{\aleph_\alpha}=\aleph_{\alpha+2}$. This implies that the strong limit cardinals are precisely $\aleph_\alpha$ with $\alpha$ limit. Hence if we take any infinite cardinal which is not a strong limit, it is of the form $\aleph_{\beta+1}$. Then we have $\aleph_\beta<\aleph_{\beta+1}$ but $2^{\aleph_\beta}>\aleph_{\beta+1}$.
Wojowu
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