0
$\begingroup$

I propose the following large cardinal axiom:

There exists a proper class of cardinals $\lambda$, such that for each $\lambda$, there exists a rank-into-rank embedding $j: V_\lambda \rightarrow V_\lambda$.

Question. Is the above consistent with ZFC? If so, is it equivalent to any of the rank into rank axioms stronger than I3?

Motivation. If you had a property like the above, then maybe one could try "stitching together" the embeddings to get a non-trivial elementary embedding from $V \to V$, which is inconsistent with ZFC. But I don't see if this is guaranteed to happen or not.

Improve this question
$\endgroup$
4
  • $\begingroup$ One issue re: building an embedding $V\rightarrow V$ out of smaller elementary embeddings via some limit process is that even if we do get such an embedding, it might be the identity. For example, it is consistent as far as we know so far that there is a sequence $\langle j_\eta, \lambda_\eta\rangle_{\eta\in \mathsf{Ord}}$ such that $(i)$ $\eta<\alpha\implies \lambda_\eta<\lambda_\alpha$, $(ii)$ $j_\eta$ is a nontrivial elementary embedding $V_{\lambda_\eta}\rightarrow V_{\lambda_\eta}$, but $(iii)$ $\forall\alpha\exists\beta\forall\gamma>\beta(j_\gamma(\alpha)=\alpha)$ (limit-triviality). $\endgroup$
    – Noah Schweber
    Commented Feb 24, 2021 at 0:55
  • $\begingroup$ The point is that while each of the $j_\eta$s are individually nontrivial, if we try to paste them together (at least, in the most obvious way) we just get the identity map on $V$. In order to push back against the Kunen inconsistency you'll need your rank-into-rank embeddings to "cohere" somehow, which (even before hitting a contradiction) will result in principles a lot stronger than a proper class of $\mathsf{I3}$s (which is what you've proposed). $\endgroup$
    – Noah Schweber
    Commented Feb 24, 2021 at 0:56
  • 4
    $\begingroup$ The consistency follows from $I_2(V_\lambda)$, which implies $V_\lambda$ satisfies your principle. $\endgroup$
    – Gabe Goldberg
    Commented Feb 24, 2021 at 2:25
  • $\begingroup$ See Laver's "Implications between strong large cardinal axioms." $\endgroup$
    – Gabe Goldberg
    Commented Feb 24, 2021 at 15:35

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .