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Let us accept the von Neumann cardinal assignment for this question. Furthermore, given a cardinal number $\kappa$, let us write $2^\kappa$ for the unique cardinal number isomorphic to the powerset of $\kappa$ (so in particular, we're doing cardinal exponentiation, not ordinal).

Also, write $C$ for the subclass of the ordinals consisting of all cardinal numbers. Then we have the following observation. $$\forall \kappa \in C :\mathrm{ord}(\kappa \cap C) \leq \kappa.$$

Question. Does ZFC prove the following? For all $\kappa \in C$, the following are equivalent.

  1. $\mathrm{ord}(\kappa \cap C) = \kappa$
  2. $\kappa$ is weakly inaccessible

I think I have the backward direction. (Is it correct?)

$(\leftarrow)$ Proof. Suppose $\kappa$ is a weakly inaccessible cardinal number. Then since $\kappa$ is a limit cardinal, hence $\kappa \cap C$ is cofinal in $\kappa$. Hence $|\kappa \cap C| = \kappa,$ since $\kappa$ is regular. Therefore $\mathrm{ord}(\kappa \cap C) \geq \kappa.$ Hence by the observation preceding the question, it follows that $\mathrm{ord}(\kappa \cap C) = \kappa.$

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No, there are singular cardinals $\kappa$ such that $\kappa\cap C$ has order-type $\kappa$. The first such $\kappa$ is the supremum of the sequence (of length just $\omega$) defined inductively by $\xi(0)=\aleph_0$ and $\xi(n+1)=\aleph_{\xi(n)}$.

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  • $\begingroup$ Something that always bothered me, and I never sat down to think it through, but maybe you have a quick answer instead. Why is this the first fixed point? How can we be sure that we didn't skip over the first in one of the steps? $\endgroup$
    – Asaf Karagila
    Apr 28, 2014 at 3:01
  • $\begingroup$ @AsafKaragila, I know your comment is directed at Andreas, but here's a partial answer. Suppose that some $\kappa$ satisfies $\xi(n) \leq \kappa < \xi(n+1)$. Then$ \aleph_{\xi(n)} \leq \aleph_\kappa \leq \aleph_{\xi(n+1)}$. So $\xi(n+1) \leq \aleph_\kappa \leq \xi(n+2)$. Hence $\kappa < \aleph_\kappa$. So no "skipping over" can occur. That being said, I cannot see why no $n \in \omega$ can satisfy "$\xi(n)$ is a fixed point of $\aleph$." $\endgroup$ Apr 28, 2014 at 3:29
  • $\begingroup$ @user18921: Yes, you're right. As to the question you ask, this is simply by induction, if $\xi(n)=\aleph_{\xi(n)}$ then $n>0$ (since $\xi(0)=\aleph_0$), so we have that $\xi(n)=\aleph_{\xi(n-1)}=\aleph_{\xi(n)}$, and from here you can get contradiction in several ways. $\endgroup$
    – Asaf Karagila
    Apr 28, 2014 at 5:10

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