6
$\begingroup$

I try to prove $L_{SO}=\mathrm{HOD}$, where $L_{SO}$ is second-order constructible universe which has similar definition with $L$ but it uses second-order definability rather than the first-order definability, and I found the answer in MO. Also, I found the referred article in the answer which is written by Myhill and Scott.

The proof of $L_{SO}=\mathrm{HOD}$ in the answer mentioned previoisly and the article uses axiom of choice. (To be precise, it uses trichotomy for cardinals and they use it to prove $\mathrm{HOD}\subset L_{SO}$). My question is: using the axiom of choice is essential to prove $\mathrm{HOD}\subset L_{SO}$? Thanks for any information or clarification.

$\endgroup$
4
  • $\begingroup$ Earlier this year one of the talks in the Friday set theory seminar was (amongst other things) about this construction, and later we had a discussion one whether or not the axiom of choice is used there. I can't recall the answer or semi-answer that we arrived at, though. $\endgroup$
    – Asaf Karagila
    Feb 19, 2015 at 15:10
  • $\begingroup$ The proof is definitely using AC in a fundamental way, since when a set is in HOD, then the proof argues that it is definable in some $V_\theta$, and then you wait for the $L_{SO}$ hierarchy to grow large enough that you can have a predicate coding all of $V_\theta$ in order to define $x$ at that stage. But if $V_\theta$ is not well-orderable, there can be no such stage where $V_\theta$ is coded like that. $\endgroup$ Feb 19, 2015 at 15:23
  • 1
    $\begingroup$ @Joel: That much is clear, but the interesting question is whether or not the proof can be modified slightly to accommodate this. For example, if you could ensure that instead of $V_\alpha$, you could take some elementary submodel of $V_\alpha$ which can be encoded by ordinals and that it calculates $A$ correctly. $\endgroup$
    – Asaf Karagila
    Feb 19, 2015 at 15:25
  • $\begingroup$ (Of course the existence of elementary submodels of this type generally requires $\sf DC$ and $\sf AC_\kappa$ for every aleph number $\kappa$.) $\endgroup$
    – Asaf Karagila
    Feb 19, 2015 at 15:27

1 Answer 1

8
$\begingroup$

The equality $L_{SO}=HOD$ can not be proved just in $ZF$. This is proved in the paper ``The consistency of the theory $ZF+L^1\neq HOD$'' by Szczepaniak.

Here $L^1$ refers to what you named $L_{SO}$. The idea of the proof is as follows:

$(1)$ If two models of $ZF$ have the same sets of ordinals, then they have the same classes $L^1,$

$(2)$ There are models $N_1 \subset N_2$ with the same sets of ordinals, such that there is a real $a\in N_1$ such that $a\notin HOD^{N_1}$ but $a\in HOD^{N_2}.$

Now the result follows from $(1)$ and $(2)$. The paper can be find here

$\endgroup$
4
  • $\begingroup$ Could you say something about the proof? I imagine some kind of higher symmetric inner model of a model where $\text{HOD}^{\text{HOD}}\neq\text{HOD}$. For example, add a Cohen real, and then code it into the GCH pattern, but take a symmetric inner model of that model. So the idea is that the Cohen real will be in $\text{HOD}$, but not in $L_{SO}$. $\endgroup$ Feb 20, 2015 at 13:54
  • $\begingroup$ Do you have a reference for the paper? $\endgroup$
    – Asaf Karagila
    Feb 20, 2015 at 15:18
  • $\begingroup$ @Joel: I would actually expect something slightly more clever than that. For example force over Cohen's first model in a way which doesn't add sets of ordinals, but encodes one of the generic reals into $\sf HOD$. But I'd much rather see the paper first. $\endgroup$
    – Asaf Karagila
    Feb 20, 2015 at 15:21
  • $\begingroup$ Mohammad, thanks for the reference! I've looked it up. It's a nice idea which sort of combines both mine and Joel's comments. $\endgroup$
    – Asaf Karagila
    Feb 22, 2015 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.