0
$\begingroup$

I just want to understand the embedding behind Reinhardt's cardinals. We have an elementary embedding $j: V \to V$. Let the background theory be $\sf MK - Choice$. We know that $V$ itself is a class stage of the cumulative hierarchy, i.e. there is a class ordinal $\kappa$ such that $V=V_\kappa$, but this means that the range of $j$, being structure preserving, would also be a stage of the cumulative hierarchy, this would be $V_{j(\kappa)}$. So, either $j(\kappa)= \kappa$, by then $j$ would be an automorphism, but $V$ is a well-founded transitive class model of $\sf ZF$, and I think it can't admit automorphisms over it. So $j(\kappa) \neq \kappa$, but this leads to $j(\kappa) < \kappa$, and so $V_{j(\kappa)} \in V$, but this would be inconsistent since $j$ is freely used in Replacement and Separation.

Where is the flaw in the above argument?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

An elementary embedding is not an automorphism, although an automorphism would be an elementary embedding. But an elementary embedding is not requiring to be surjective.

An elementary embedding just means that the images satisfy the same properties of their origin, in the codomain space that is. In the case of an elementary embedding $V\to V$ it just means that $x$ and $j(x)$ have the same first order properties. And indeed, if $\kappa<j(\kappa)$, then there is no reason to expect that some $\alpha$ is mapped to $\kappa$ at all. We do not expect that in the case of measurable cardinal, or even weaker theories, like the transfer principle from the reals to the hyperreals.

However, much like any other function that the universe is closed under, it will have fixed points. Many of them. So having $j(x)=x$ also tells you very little about the general behaviour of $j$. So, again, it is not enough to deduce that it is surjective just because we found some $\lambda$ such that $j(\lambda)=\lambda$.

Your main problem is overusing $\kappa$. $V$ is just $V$, if you really must insist on thinking of it as a stage in the von Neumann hierarchy, then it is $V_{\rm Ord}$. The elementary embedding is defined on sets, so $j(\mathrm{Ord})$ is not a set, and while we can define $j$ on class as $j(X)=\bigcup_{\alpha\in\rm Ord}j(X\cap V_\alpha)$, there is no reason to expect $j$ to behave exactly as it does with sets.

Improve this answer
$\endgroup$
17
  • $\begingroup$ if an elementary embedding is also surjective wouldn't that make it an automorphism? $\endgroup$
    – Zuhair Al-Johar
    Commented Dec 21, 2022 at 9:44
  • 1
    $\begingroup$ Yes, but why would it be surjective, unless it is trivial? $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2022 at 9:55
  • 1
    $\begingroup$ Why would it be an elementary embedding? $j^+(\kappa)=j[\kappa]$ is not an ordinal if $j$ is non-trivial. This may preserve some modicum of elementarity with regards to the second-order theory of $V_\kappa$, but it is no longer a fully elementary embedding. There is a whole theory on when and how these embeddings can be lifted and by how much. It is complicated, it is difficult, and it is full of subtlety which you are currently sweeping off the table. I am not an expert on those things either, but Gabe and Farmer both have papers on this and both are around often. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2022 at 10:17
  • 1
    $\begingroup$ Either the critical point of $j$ is below $\kappa$, in which case $j[\kappa]\neq\kappa$, or it is trivial. Those are the only two options. Choose one of them, then go back to my previous comments. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2022 at 15:10
  • 1
    $\begingroup$ No. If $j$ is non-trivial, there is some $\alpha<\kappa$ such that $j(\alpha)>\alpha$. Then the first such $\alpha$ will not be in $j[\kappa]$. Full stop. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2022 at 15:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .