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The following sentence was at inaccessible cardinal:
A weakly inaccessible cardinal may be strengthened to produce a (generally larger) strongly inaccessible cardinal.
I have removed it after complaints by a set theorist, and in light of the below discussion box, which I have copied here and removed.
Mike: What does that last sentence mean? It seems obviously false to me in the absence of CH.
Toby: It means that if a weakly inaccessible cardinal exists, then a strongly inaccessible cardinal exists, but I couldn't find the formula for it. Something like $\beth_\kappa$ is strongly inaccessible if $\kappa$ is weakly inaccessible (note that $\aleph_\kappa = \kappa$ then), but I couldn't verify that (or check how it holds up in the absence of choice).
Mike: I don’t believe that. Suppose that the smallest weakly inaccessible is not strongly inaccessible, and let $\kappa$ be the smallest strongly inaccessible. Then $V_\kappa$ is a model of set theory in which there are weakly inaccessibles but not strong ones. I’m almost certain there is no reason for the smallest weakly inaccessible to be strongly inaccessible.
JCMcKeown: Surely $\beth_\kappa$ has cofinality at most $\kappa$, so it can’t be regular. Maybe the strengthening involves some forcing or other change of universe? E.g., you can forcibly shift $2^\lambda = \lambda^+$ for $\lambda \lt \kappa$, and then by weak inaccessibility, etc… I think. Don’t trust me. —- (some days later) More than that: since the ordinals are well ordered, if there is any strongly inaccessible cardinal greater than $\kappa$, then there is a least one, say $\theta$. Then $V_\theta$ is a universe with a weakly inaccessible cardinal and no greater strongly inaccessible cardinal. Ih! Mike said that already… So whatever construction will have to work the other way around: if there is a weakly inaccessible cardinal that isn’t strongly inaccessible, and if furthermore a weakly inaccessible cardinal implies a strongly inaccessible cardinal, then the strongly inaccessible cardinal implied must be less than $\kappa$. And that sounds really weird.
I was also supplied with a AC-free proof that weakly inaccessibles are $\aleph_{(-)}$-fixed points:
First, a quick induction shows that $\alpha\leq\omega_\alpha$ is always true. If $\kappa$ is a limit cardinal, then the set of cardinals below $\kappa$ is unbounded; but since it’s also regular there are $\kappa$ of them. So $\omega_\kappa\leq\kappa\leq\omega_\kappa$, and equality ensues.
I can edit this into the page if desired.
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