A Mathoverflow question by Trevor Wilson defines weakly Shelah cardinals as follows:

A cardinal $\kappa$ is weakly Shelah if for all $f : \kappa \to \kappa$ there is some $\alpha < \kappa$ that is closed under $f$ and there is some elementary embedding $j : V \to M$ (where $M$ is a transitive class) such that $\operatorname{crit}(j) = \alpha$ and $j(\alpha) > \kappa$ and $V_{j(f)(\kappa)} \subset M$.

Edit: Because of the issue noted by Sean Cox in the comments, a better definition also requires that $j(f)(\beta)=f(\beta)$ whenever $\beta \lt \kappa$.

If $f$ is the function enumerating ordinals $\beta$ such that $V_\beta \prec V_\kappa$ (and there exists $\kappa$-many such ordinals since $\kappa$ is inaccessible so $V_\kappa \vDash KM$ and the argument in the blog post Kelley–Morse set theory implies Con(ZFC) and much more by Joel David Hamkins applies) but are not limits of such ordinals, then $V_{\beta_1} \prec V_{\beta_2}$ whenever $\beta_1$ and $\beta_2$ are in the range of $f$ or limits of ordinals in the range of $f$. By elementarity of $j$, $V_\kappa \prec V_{j(f)(\kappa)}$. Thus $\kappa$ is 0-extendible (defined by Bagaria et al. 2015, Superstrong and other large cardinals are never Laver indestructible, also known as otherworldly).

However, a comment by Trevor Wilson says that every $\Sigma_3$-reflecting Woodin cardinal is weakly Shelah:

On further thought, it seems like $\kappa$ Woodin and $\Sigma_3$-reflecting implies κ weakly Shelah: given $f:\kappa \to \kappa$, use Woodinness to get some $\alpha \lt \kappa$ that is $\lt \kappa$-$f$-strong. Then the desired conclusion (on existence of $j$) holds for cofinally many $\bar\kappa \lt \kappa$ in place of $\kappa$. Formulating this in terms of extenders, $\Sigma_3$-reflection implies the desired conclusion holds for cofinally many $\bar\kappa \in Ord$ in place of $\kappa$, and therefore for $\kappa$ itself (since $f$ is increasing.)

If $V_\kappa \prec V_{j(f)(\kappa)}$ and $\kappa$ is Woodin, then $V_{j(f)(\kappa)} \vDash \text{"$\kappa$ is $\Sigma_3$-reflecting and Woodin"}$. That would mean that there cannot be a least $\lambda$ such that $V_\lambda \vDash \text{"$\kappa$ is weakly Shelah"}$ and the existence of $\Sigma_3$-reflecting Woodin cardinal is inconsistent. Where does my argument or Trevor Wilson's go wrong?

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    $\begingroup$ Interesting... maybe the problem is in Trevor's argument, which seems to require the function $f : \kappa\to \kappa$ as a parameter in the $\Sigma_3$-reflection argument. $\endgroup$ Aug 31, 2021 at 20:03
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    $\begingroup$ @GabeGoldberg Why don't you post that as an answer? $\endgroup$ Aug 31, 2021 at 21:05
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    $\begingroup$ In your argument, why is $V_\kappa \prec V_{j(f)(\kappa)}$? I don't see why it follows by elementarity, because I don't see why $\kappa$ is a limit of ordinals in the range of $j(f)$. $\endgroup$
    – Sean Cox
    Sep 1, 2021 at 16:56
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    $\begingroup$ @SeanCox I was assuming that $j(f) \upharpoonright \kappa =f$ but that's not part of the definition. I think that's a flaw of the definition. $\endgroup$ Sep 1, 2021 at 17:36
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    $\begingroup$ @ArvidSamuelsson It seems to me that there are now two separate reasons this is not an inconsistency: (1) I don't follow Trevor's proof that $\Sigma_3$-reflecting Woodins are weakly Shelah, and (2) I don't see how to patch your proof that they need not be. Experience suggests one of these assertions is provable ("large cardinals are linearly ordered"), and I don't want to end this discussion before figuring out which it is. Of course one can change the definition, but that won't make the original question go away. $\endgroup$ Sep 3, 2021 at 1:19

1 Answer 1


This question was answered in the comments:

As Gabe Goldberg pointed out, Trevor Wilson's argument that $\Sigma_3$-reflecting Woodin cardinals are weakly Shelah fails because the reflection argument would have to use the function $f$ as a parameter, and $\kappa$ being $\Sigma_3$-reflecting is not enough for that.

Additionally, as Sean Cox pointed out, we can't be sure that $V_\kappa \prec V_{j(f)(\kappa)}$ since it is possible that there is some $\beta \lt \kappa$ such that $j(f)(\beta) \gt \kappa$. I assumed that $j(f) \upharpoonleft \kappa = f$ but I don't know how to prove that from the definition.


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