It is well-known that if $AC$ holds and if $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ is a non-trivial elementary embedding with $crit(j) < \lambda,$ then $\lambda$ has countable cofinality (and in fact it is the least fixed point of $j$ above $crit(j)$).

Question. Is $AC$ needed to show that $\lambda$ has countable cofinality.

In other words, is it possible to show, just working in $ZF$ that, $cf(\lambda)=\omega.$

Remark. I can show that if we can prove the result without AC, then there are no Reinhardt cardinals in ZF.

  • 2
    $\begingroup$ (+1) Could you please explain more about your interesting remark? Is it a long proof or a short and fairly straightforward one? $\endgroup$ May 23, 2016 at 8:03
  • $\begingroup$ I'm not sure what the proof is for Reinhardt cardinals, but if $\kappa$ is super Reinhardt as witnessed by (j), then $V_\kappa\prec V$ and let $\delta$ be the least inaccessible above $\lambda=lim_{n\rightarrow\omega}j^n(\kappa)$. Then $V_\delta\prec V$ and so $j\restriction V_\delta$ extends to an $I-1$ embedding, which is therefore an $I0$ embedding with regular target. $\endgroup$
    – Master
    Jul 26, 2019 at 22:38

1 Answer 1


It is consistent that $AC$ fails and there exists a non-trivial elementary embedding $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ with $crit(j) < \lambda,$ and $\lambda$ has uncountable cofinality. See BERKELEY CARDINALS AND THE STRUCTURE OF $L(V_{δ+1})$.

  • $\begingroup$ Literally the question asks whether ($\lambda$ has countable cofinality) implies AC. The "in other words" is rather another question, as it indicates you're asking whether (ZF and $\lambda$ does not have countable cofinality) is consistent), which is distinct. So it seems that "AC is needed" is not what is proved there. $\endgroup$
    – YCor
    Oct 3, 2019 at 9:03
  • $\begingroup$ I don't think asking if AC is needed for a statement is the same as saying that if that statement implies AC. $\endgroup$ Oct 7, 2019 at 8:14
  • $\begingroup$ Formally speaking, it is the same. In practice, you're unfortunately right. $\endgroup$
    – YCor
    Oct 7, 2019 at 8:21

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