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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.

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  • $\begingroup$ 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 ♦
    Commented 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. $\endgroup$
    – Joel David Hamkins
    Commented Feb 19, 2015 at 15:23
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    $\begingroup$ @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 ♦
    Commented 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 ♦
    Commented Feb 19, 2015 at 15:27

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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

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  • $\begingroup$ 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 $L_{SO}$. $\endgroup$
    – Joel David Hamkins
    Commented Feb 20, 2015 at 13:54
  • $\begingroup$ Do you have a reference for the paper? $\endgroup$
    – Asaf Karagila ♦
    Commented 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 ♦
    Commented 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 ♦
    Commented Feb 22, 2015 at 19:23

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