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

Proof of previously announced result

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edited Oct 31, 2023 at 15:36

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To do: prove that ifIf $\kappa$ 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 elsewhereanother MathOverflow answer.

To do: prove that if $\kappa$ satisfies my strengthening of the definition of weakly Shelah cardinals, it is weakly superstrong. Conversely, if $\kappa$ is Woodin and weakly superstrong, it is weakly Shelah, as I've proved elsewhere.

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.

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edited Oct 31, 2023 at 0:08

LSpice

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My strengthening is not equivalent because a cardinal is weakly Shelah iff it is Woodin and $\Sigma_3$-0-extendible $\Sigma_3$-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.

To do: prove that if $\kappa$ satisfies my strengthening of the definition of weakly Shelah cardinals, it is weakly superstrong weakly superstrong. Conversely, if $\kappa$ is Woodin and weakly superstrong, it is weakly Shelah, as I've proved elsewhere.

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created Oct 30, 2023 at 20:08

Arvid Samuelsson

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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 $V_\theta \vDash \forall \gamma \exists \beta \forall x \in V_\gamma \exists y \in V_\beta \phi(x, y, \vec{z})$

To do: prove that if $\kappa$ satisfies my strengthening of the definition of weakly Shelah cardinals, it is weakly superstrong. Conversely, if $\kappa$ is Woodin and weakly superstrong, it is weakly Shelah, as I've proved elsewhere.

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