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Assume AD+DC. Assume $\kappa$ is an infinite cardinal and $N$ is a (set or class) transitive model of ZFC containing $\kappa$.

Is it true that for all $\alpha<\kappa$, $N$ thinks that the power set of $\alpha$ has size less than $\kappa$?

This seems like obvious, at least for $\kappa<\Theta$, but I don't find a reference or an idea yet.

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  • $\begingroup$ So... every cardinal is strongly inaccessible in any inner model of $\sf ZFC$? $\endgroup$
    – Asaf Karagila
    May 3, 2015 at 17:31
  • $\begingroup$ If so, every regular cardinal is strongly inaccessible in any inner model of đť–ąđť–Ąđť–˘ $\endgroup$ May 3, 2015 at 17:33

1 Answer 1

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Some partial answers:

(1) Under the (possibly) additional assumption of $\mathsf{AD}^+$ we can prove the weaker conclusion that $\alpha^{+N} < \alpha^+$ for every infinite ordinal $\alpha < \Theta$:

  • Assume toward a contradiction that $\alpha^{+N} = \alpha^+$ and use $\mathsf{AC}$ in $N$ to take a sequence of functions $(f_\xi : \xi < \alpha^+)$ such that each function $f_\xi$ is a surjection from $\alpha$ onto $\xi$. This shows that $\alpha^+$ is regular in $V$; otherwise we could combine cofinally many of these surjections to get a surjection from $\alpha \times \alpha$ to $\alpha^+$, a contradiction.

  • By a theorem of Woodin every uncountable regular cardinal less than $\Theta$ is measurable (as stated by Koellner and Woodin, Large cardinals from determinacy, p. 12,) so $\alpha^+$ is measurable. But then we can take an ultrapower of the sequence $(f_\xi : \xi < \alpha^+)$ to again get a surjection from $\alpha$ to $\alpha^+$, a contradiction.

(2) Under the stronger assumption of $\mathsf{AD} + V = L(\mathbb{R})$ we have $(2^\alpha)^N < \alpha^+$ for every infinite ordinal $\alpha < \Theta$, which implies the desired conclusion. This follows directly from Steel, An outline of inner model theory, Theorem 8.26, which says that for every infinite cardinal $\alpha < \Theta$, every wellordered family of subsets of $\alpha$ has cardinality at most $\alpha$. Steel calls this result the "boldface $\mathsf{GCH}$ for $L(\mathbb{R})$" and proves it from $\mathsf{GCH}$ in $\text{HOD}_x$ where $x$ is a real, which can be understood as a fine-structural model.

Note that the theorem that every uncountable regular cardinal less than $\Theta$ is measurable was first proved under the hypothesis $\mathsf{AD} + V = L(\mathbb{R})$ also using fine structure (see Steel, Theorem 8.27) before being generalized to the $\mathsf{AD}^+$ setting by Woodin. I don't know whether we should expect a similar generalization of the boldface $\mathsf{GCH}$.

(3) For cardinals $\kappa = \alpha^+ > \Theta$, even the weaker statement $\alpha^{+N} < \alpha^+$ can fail. Consider the case that $V = L(\mathcal{P}(\mathbb{R}))$, $\mathsf{AD}^+ + \mathsf{AD}_\mathbb{R}$ holds, and $N = \text{HOD}$. We can add $\mathcal{P}(\mathbb{R})$ back to $\text{HOD}$ by a Vopěnka-type forcing of cardinality $\Theta$; this argument is due to Woodin and is written up in Section 4.3.4.2 of Nam Trang's thesis. This forcing doesn't collapse cardinals above $\Theta$, so $\alpha^{+N} = \alpha^+ = \kappa$.

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  • $\begingroup$ If you work in $V=L(\mathcal P(\Bbb R))$ which does not satisfy choice; I don't see how you can add $\mathcal P(\Bbb R)$ into $\rm HOD$ without either adding new sets to $V$ or violating the axiom of choice. $\endgroup$
    – Asaf Karagila
    May 4, 2015 at 4:19
  • $\begingroup$ @Asaf You are right; I was being imprecise. The generic filter does not live in $V$, but the extension of $\text{HOD}$ by this generic filter is precisely $V$. $\endgroup$ May 4, 2015 at 5:24
  • $\begingroup$ But any extension of $\rm HOD$ by a generic filter will satisfy choice, so it couldn't possibly be $V$. $\endgroup$
    – Asaf Karagila
    May 4, 2015 at 5:34
  • $\begingroup$ @Asaf Oops, today I am not very good at saying things that are literally true. $V$ is a symmetric submodel of the generic extension of $\text{HOD}$. $\endgroup$ May 4, 2015 at 5:36
  • $\begingroup$ Okay, I am willing to buy that. Although I don't see an immediate proof. $\endgroup$
    – Asaf Karagila
    May 4, 2015 at 5:38

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