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Work in ZF. Let $\lambda$ be strongly inaccessible. Let $\kappa$ be such that for every $\alpha\lt\lambda$, there is some $j: V_\lambda\prec V_\lambda$, with critical point $\kappa$, such that $j(\kappa)\gt\alpha$.

In this setting, can we define a sequence $A=\langle j_\alpha: V_\lambda\prec V_\lambda\mid \alpha\lt\lambda\rangle$, such that every $j_\alpha$ is a non-trivial elementary embedding with critical point $\kappa$, and $j_\alpha(\kappa)\gt\alpha$?

(Note: $A$ needs to be a sequence, not just the set of all embeddings.)

Motivation: I was doing research on large cardinals without choice, and was attempting to use a similar collection of embedding to $A$ to give a first order definition of cardinals that were $n$-huge.

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  • $\begingroup$ Can you show in ZF that the set $\{ j(cr(j)) | j : V_\lambda \to V_\lambda$ is elementary$\}$ is cofinal in $\lambda$, assuming it’s nonempty? $\endgroup$
    – Monroe Eskew
    Oct 30, 2020 at 8:20
  • $\begingroup$ That is the definition of the problem. For any $\alpha$, we have that there is some $j: V_\lambda\prec V_\lambda$ with $j(crit(j))\gt\alpha$. $\endgroup$
    – Master
    Oct 30, 2020 at 16:23
  • 2
    $\begingroup$ If $\lambda$ is the least inaccessible for which there is such an elementary embedding, then that set is not cofinal since $\lambda$ is not a limit of inaccessibles. $\endgroup$
    – Gabe Goldberg
    Oct 30, 2020 at 17:26
  • $\begingroup$ You're right. I assumed he was talking about this specific problem. $\endgroup$
    – Master
    Oct 30, 2020 at 18:00
  • $\begingroup$ May we assume AC in $V_\kappa$ (and therefore also in $V_\lambda$)? $\endgroup$
    – Joel David Hamkins
    Oct 31, 2020 at 21:25

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