# Possible inconsistency of weakly Shelah cardinals (I hope not)

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?

• 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. Commented Aug 31, 2021 at 20:03
• @GabeGoldberg Why don't you post that as an answer? Commented Aug 31, 2021 at 21:05
• 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 $j(f)$. Commented Sep 1, 2021 at 16:56
• @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. Commented Sep 1, 2021 at 17:36
• @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. Commented Sep 3, 2021 at 1:19

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.