# The axiom $I_0$ in the absence of $AC$

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.

• (+1) Could you please explain more about your interesting remark? Is it a long proof or a short and fairly straightforward one? – Morteza Azad May 23 '16 at 8:03
• 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. – Master Jul 26 at 22:38

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 $$L(V_{δ+1})$$.
• 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. – YCor Oct 3 at 9:03