# Is this strengthening of the definition of weakly Shelah cardinals equivalent to being weakly Shelah?

The MathOverflow question A weak (?) form of Shelah cardinals 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 $$V_{j(f)(\kappa)} \subset M$$.

I would like to add a requirement that $$j(f) \upharpoonright \kappa = f$$ but as a comment by Sean Cox on this question made me realize, it is not clear that that definition is equivalent to Trevor Wilson's definition. Are the definitions equivalent?

My strengthening is not equivalent because a cardinal is weakly Shelah iff it is Woodin and $$\Sigma_3$$-0-extendible ($$\Sigma_3$$-0-extendible cardinals could also be called $$\Sigma_3$$-otherworldly), while my strengthening is equivalent to being Woodin and weakly superstrong.
A cardinal is weakly Shelah iff it is Woodin and $$\Sigma_3$$-0-extendible (proof inspired by comment by Trevor Wilson). Suppose that $$\kappa$$ is Woodin and $$\Sigma_3$$-0-extendible, that is there exist $$\theta$$ such that $$V_\kappa \prec_3 V_\theta$$. Since $$\kappa$$ is a Woodin cardinal, for every $$f: \kappa \to \kappa$$ there exists $$\nu$$ such that $$f"\nu \subset \nu$$ (that is for every $$\alpha \lt \nu$$, $$f(\alpha) \lt \nu$$) and $$\nu$$ is $$\lt \kappa$$-$$f$$-strong, meaning that for every $$\alpha \lt \kappa$$ there is an extender for an elementary embedding $$j: V \to M$$ with critical point $$\nu$$ such that $$V_{f(\alpha)} \subset M$$ and $$j(f\upharpoonleft\nu)\upharpoonleft(\alpha+1)=f\upharpoonleft(\alpha+1)$$. In particular, it holds in $$V_\kappa$$ that for every $$\alpha$$ there is an extender for an elementary embedding $$j: V \to M$$ with critical point $$\nu$$ such that $$V_{j(f)(\alpha)} \subset M$$. This is a $$\Pi_3$$ formula with parameter $$f\upharpoonleft\nu$$, so it also holds in $$V_\theta$$. Thus, in $$V_\theta$$, there is an extender for an elementary embedding $$j: V \to M$$ with critical point $$\nu$$ such that $$V_{j(f)(\kappa)} \subset M$$, as in the definition of weakly Shelah cardinals. Since this works for every $$f: \kappa \to \kappa$$, $$\kappa$$ is weakly Shelah.
Conversely, suppose that $$\kappa$$ is weakly Shelah. To prove that $$\kappa$$ is Woodin, I first have to prove that it is inaccessible, that is regular and a strong limit cardinal. If $$\kappa$$ is not regular, that is there is a function $$f: \lambda \to \kappa$$, with $$\lambda \lt \kappa$$, that enumerates a cofinal sequence, then we can assume without loss of generality that $$f$$ is strictly increasing and $$f(0) \ge \lambda$$, and $$f$$ can be extended to a function $$\kappa \to \kappa$$ by defining $$f(\alpha)$$ arbitrarily for $$\alpha \gt \lambda$$. This extended function has no closure points less than $$\kappa$$, so $$\kappa$$ can't be a weakly Shelah cardinal. Now suppose that $$\kappa$$ is not a strong limit cardinal, meaning that there is an ordinal $$\lambda \lt \kappa$$ and a surjective function $$\pi: V_\lambda \to \kappa$$. Define $$f(\alpha)=\lambda+\alpha$$. This $$f$$ has no closure points below $$\lambda \cdot \omega$$, and there are no measurable cardinals between $$\lambda$$ and $$\kappa$$, so no ordinal less than $$\kappa$$ can be a critical point as in the definition of a weakly Shelah cardinal.
To prove that $$\kappa$$ is a Woodin cardinal given that it is weakly Shelah, I have to prove for every function $$f: \kappa \to \kappa$$ that there exists a $$\lt \kappa$$-$$f$$-strong cardinal $$\nu$$ such that $$f"\nu \subset \nu$$. Define $$g$$ by $$g(\alpha)=$$ the least $$\beta \gt \alpha$$ such that $$\langle V_\beta, \in, f \rangle \prec \langle V_\kappa, \in, f \rangle$$ (since $$\kappa$$ is inaccessible, it is a limit of such cardinals $$\beta$$). Since $$\kappa$$ is weakly Shelah, there exist a cardinal $$\nu$$ such that $$g"\nu \subset \nu$$ and an elementary embedding $$j: V \to M$$ with critical point $$\nu$$ such that $$V_{j(g)(\kappa)} \subset M$$. By elementarity of $$j$$, $$\langle V_{j(g)(\kappa)}, \in, j(f) \rangle \prec \langle V_{j(\nu)}, \in, j(f) \rangle$$. Thus $$j(f)(\alpha) \lt j(g)(\kappa)$$ for every $$\alpha \lt j(g)(\kappa)$$, so $$V_{j(g)(\kappa)} \vDash \text {\nu is f-strong}$$ since $$V_{j(g)(\kappa)}$$ sees enough extenders witnessing this. Thus $$\nu$$ is a limit of cardinals that are $$\lt \nu$$-$$f$$-strong, and since $$\nu$$ is a limit of cardinals $$\beta$$ such that $$\langle V_\beta, \in, f \rangle \prec \langle V_\kappa, \in, f \rangle$$, so that $$\langle V_\nu, \in, f \rangle \prec \langle V_\kappa, \in, f \rangle$$, those are $$\lt \kappa$$-$$f$$-strong.
Denote by $$\theta$$ the weakly Shelah witnessing ordinal of $$\kappa$$, that is the supremum of the least $$j(f)(\kappa)$$ as in the definition of a weakly Shelah cardinal as $$f$$ ranges over the functions $$f: \kappa \to \kappa$$. To prove that $$V_\kappa \prec_3 V_\theta$$, I have to prove for every $$\Pi_3$$ formula $$\forall x \exists y \phi(x, y)$$ and every parameter combination $$\vec{z}$$ that $$V_\kappa \vDash \phi$$ implies $$V_\theta \vDash \phi$$. Every $$\Pi_3$$ formula $$\forall x \exists y \phi(\vec{z})$$ is equivalent over ZFC to a formula of the form $$\forall \alpha \exists \beta (\beta \ge \alpha \wedge \text{\beta is a \beth fixed point} \wedge V_\beta \vDash( \forall x \in V_\alpha \exists y \phi(x, y, \vec{z})))$$ (note that formulas of the form $$V_\beta \vDash \psi$$ are absolute between V and ranks $$V_\eta$$ for $$\beth$$ fixed points $$\eta \gt \beta$$; also note that $$\kappa$$ is inaccessible and thus $$V_\kappa$$ satisfies ZFC and, in particular, $$\kappa$$ is a limit of $$\beth$$ fixed points). By the assumption on $$\theta$$, for every $$\gamma \lt \theta$$ there is a function $$f: \kappa \to \kappa$$ and an elementary embedding $$j: V \to M$$ with critical point less than $$\kappa$$ such that $$V_{j(f)(\kappa)} \subset M$$ and $$j(f)(\kappa) \ge \gamma$$. This $$j$$ can be chosen so that $$j(f)(\kappa)$$ is minimal (that is there's no elementary embedding $$i$$ satisfying the same conditions such that $$i(f)(\kappa) \lt j(f)(\kappa)$$). If $$V_\kappa \vDash \forall x \exists y \phi(\vec{z})$$, define $$g$$ by $$g(\alpha)=$$ the least $$\beth$$ fixed point $$\beta$$ such that $$V_\beta \vDash \forall x \in V_{f(\alpha)} \exists y \phi(x, y, \vec{z})$$. Since $$\kappa$$ is weakly Shelah with witnessing ordinal $$\theta$$, there is an elementary embedding $$i: V \to N$$ with critical point less than $$\kappa$$ such that $$V_{i(g)(\kappa)} \subset N$$. By definition of $$g$$ we have $$i(g)(\kappa) \gt i(f)(\kappa)$$ and by minimality of $$j(f)(\kappa)$$ we have $$i(f)(\kappa) \ge j(f)(\kappa)$$. By the definition of $$g$$ and elementarity of $$i$$ we have $$V_{i(g)(\kappa)} \vDash \forall x \in V_{i(f)(\kappa)} \exists y \phi(x, y, \vec{z})$$; together with the fact that $$\gamma \le j(f)(\kappa) \le i(f)(\kappa) \le i(g)(\kappa) \le \theta$$, this implies that $$V_\theta \vDash \forall x \in V_\gamma \exists y \in V_{i(g)(\kappa)} \phi(x, y, \vec{z})$$. Hence we have $$V_\theta \vDash \forall \gamma \exists \beta \forall x \in V_\gamma \exists y \in V_\beta \phi(x, y, \vec{z})$$
If $$\kappa$$ satisfies my strengthening of the definition of weakly Shelah cardinals, it is weakly superstrong. Suppose that for every function $$f: \kappa \to \kappa$$ there exists an elementary embedding $$j: V \to M$$ with critical point less than $$\kappa$$ such that $$V_{j(f)(\kappa)} \subset M$$ and $$j(f) \upharpoonleft \kappa = f$$. Fix an $$A \subseteq V_\kappa$$ and define $$f: \kappa \to \kappa$$ by $$f(\alpha)=$$ the least $$\beta \gt \alpha$$ such that $$\langle V_\beta, \in, A \rangle \prec \langle V_\kappa, \in, A \rangle$$; this function is total because $$\kappa$$ is inaccessible, as proved above. By the assumption on $$\kappa$$, there exists an elementary embedding $$j: V \to M$$ such that $$V_{j(f)(\kappa)} \subset M$$ and $$j(f) \upharpoonleft \kappa = f$$; I will denote the critical point of $$j$$, which is less than $$\kappa$$, by $$\nu$$. Since $$\kappa$$ is a limit of cardinals $$\beta$$ such that $$\langle V_\beta, \in, A \rangle \prec \langle V_{j(\nu)}, \in, j(A) \rangle$$, we have $$\langle V_\kappa, \in, A \rangle \prec \langle V_{j(\nu)}, \in, j(A) \rangle$$ by the Tarski-Vaught test, and we have $$\langle V_{j(f)(\kappa)}, \in, j(A) \rangle \prec \langle V_{j(\nu)}, \in, j(A) \rangle$$ by the definition of $$f$$. Thus we have $$\langle V_\kappa, \in, A \rangle \prec \langle V_{j(f)(\kappa)}, \in, j(A) \rangle$$, again by the Tarski-Vaught test. Since this works for every $$A \subseteq V_\kappa$$, $$\kappa$$ is strongly 1-uplifting, which is equivalent to being weakly superstrong by theorem 5 of the paper that defined weakly superstrong cardinals. Conversely, if $$\kappa$$ is Woodin and weakly superstrong, it is weakly Shelah, as I've proved in another MathOverflow answer.