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Let $j\colon V \to M$ be a superstrong elementary embedding (i.e. $M$ is transitive, $j\neq id$, and $V_{j(\kappa)} \subseteq M$ where $\kappa$ is the critical point of $j$).

Is $j(\kappa)$ necessarily inaccessible in $V$?

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  • $\begingroup$ ברוך הבא לעולם התחתון! ;) $\endgroup$
    – Asaf Karagila
    Jan 29 '18 at 19:46
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    $\begingroup$ Well, it's clearly worldly. But I don't see why $j(\kappa)$ will be regular... $\endgroup$
    – Asaf Karagila
    Jan 30 '18 at 9:14
  • $\begingroup$ @AsafKaragila: Gitik claims that the answer is negative and Cantor's attic claims that the answer is positive. I'm trying to figure out which one is correct. $\endgroup$
    – Yair Hayut
    Jan 30 '18 at 10:19
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    $\begingroup$ Who edited that on Cantor's Attic? My money is on Moti being right. But you can get another opinion from Menachem... $\endgroup$
    – Asaf Karagila
    Jan 30 '18 at 10:20
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The answer seems to be negative. In fact, more is true:

Claim: Let $j\colon V\to M$ be a superstrong embedding with critical point $\kappa$ and target $\lambda$ (i.e. $j(\kappa) = \lambda$). If $\lambda$ is regular then there is a club $C \subseteq \lambda$ such that for all $\mu \in C$, there is a superstrong embedding with critical point $\kappa$ and target $\mu$. In particular, the first superstrong cardinal has a singular target.

Proof: Let $E = \langle E_\alpha \mid \alpha < \lambda\rangle$ be an extender that witnesses the superstrength of $\kappa$ with target $\lambda$. For every $x\in V_\lambda$ there is a finite sequence $\eta \in \lambda^{<\omega}$ and a function $f\colon \kappa\to V_\kappa$, such that $j_E(f)(\eta) = x$.

Let $C$ be the set of all $\beta < \lambda$, such that:

  • For all $\eta \in \beta^{<\omega}$ and $f\colon \kappa \to \kappa$, $j_E(f)(\eta) < \beta$.
  • For all $x\in V_\beta$ there is $\eta \in \beta^{<\omega}$ and $f\colon \kappa \to V_\kappa$ such that $j_E(f)(\eta) = x$.

$C$ is a club as the set of closure points of $2^\kappa < \lambda$ many functions.

I claim that for every $\beta\in C$ the elementary embedding using the extender $E \restriction \beta$ is a superstrong embedding with target $\beta$. Indeed, every $x\in V_\beta$ is represented in this extender by a function from $\kappa$ to $V_\kappa$ and a generator $\eta\in\beta^{<\omega}$. Clearly, $j_{E\restriction \beta}(\kappa) \geq \beta$. If $j_{E\restriction \beta}(\kappa) > \beta$ then there is a function $f\colon \kappa \to \kappa$ and $\eta \in \beta^{<\omega}$ such that $j_{E\restriction \beta}(f)(\eta) = \beta$. But $j_E(f)(\eta) \geq j_{E\restriction \beta}(f)(\eta) = \beta$, which contradicts the assumption that $\beta \in C$. Thus, we conclude that $\beta = j_{E\restriction \beta}(\kappa)$, as wanted.

Edit: Apparently, for any superstrong cardinal $\kappa$, the minimal target has cofinality between $\kappa^{+}$ and $2^\kappa$ (so under GCH - it is always $\kappa^{+}$). This follows from Theorem 3.3 in this paper of Perlmutter. The idea is that if $j\colon V \to M$ is a superstrong embedding with critical point $\kappa$ and $\theta = \sup_{f\colon \kappa \to \kappa} j(f)(\kappa)$, then there is a superstrong embedding with critical point $\kappa$ and target $\theta$. If the cofinality of $j(\kappa)$ is below $\kappa$ or above $2^\kappa$, it is possible to deduce that $\theta < j(\kappa)$.

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    $\begingroup$ Can you push this to check that for any superstrong $\kappa$ the least target $\lambda_\kappa$ always has cofinality $\omega$? $\endgroup$ Jan 30 '18 at 18:51
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    $\begingroup$ I don't know. This proof only shows that the cofinality of the $\lambda_\kappa$ is at most $2^\kappa$. I don't see how to modify the argument in order to get a club in those cases. $\endgroup$
    – Yair Hayut
    Feb 1 '18 at 7:59
  • $\begingroup$ Isn't it true that $M$ thinks that $j(\kappa)$ is superstrong, and applying the same extender gives us $N$ such that $(V_{j(j(\kappa))})^M\subseteq N$? $\endgroup$
    – Asaf Karagila
    Feb 4 '18 at 4:26
  • $\begingroup$ This is true. But I don't see how it helps us to obtain a superstrong embedding with singular target, since at each point in the iteration the closure is only with respect to the previous model and not the initial one (which can see the real cofinality of the targets). $\endgroup$
    – Yair Hayut
    Feb 4 '18 at 6:56
  • $\begingroup$ That's an interesting theorem. It's odd the the "least" will be something in that interval. Is it at least consistent to get something above $\kappa^+$ (e.g. if $2^\kappa=\kappa^{+++}$ or so?) $\endgroup$
    – Asaf Karagila
    Feb 5 '18 at 15:53

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