Suppose that $M$ is a transitive class, denote by $\mathrm{HOD}(M)$ the class of all those sets which are hereditarily definable from ordinals and parameters in $M$.

Some trivial examples include $\mathrm{HOD}(V)=V$, and $\rm HOD(Ord)=HOD$. Note that this model might not satisfy the axiom of choice, for example if $M=V$ and $V$ does not satisfy the axiom of choice.

Is it possible for $M$ to be an inner model of $\sf ZFC$ such that $M$ is not definable (with parameters) in $\mathrm{HOD}(M)$?

Alternatively, we can consider $L(M)$ as the smallest model of $\sf ZF$ such that $M\cap V_\alpha\in L(M)$ for all $\alpha$. Is $M$ always a class of $L(M)$?


Yes, this can happen.

Let's start in $L$ and first perform Easton forcing to add a Cohen subset to every regular cardinal, giving rise to the forcing extension $L[G]$. Next, over $L[G]$ we force to code every set into the GCH pattern, for example, by iterating the forcing that forces GCH or its negation with a lottery sum at each such cardinal stage. In the resulting forcing extension $V=L[G][H]$, we therefore have $\newcommand\HOD{\text{HOD}}\HOD^V=V$, since every set is coded.

Next, over $V$, force to add a $V$-generic class $A$ of the regular cardinals. This forcing adds no sets, and $A$ is not a definable class in $V$, but we may consider it in GBC. Let $M=L[G\upharpoonright A]$ be the inner model arising by taking only the sets of $G$ at coordinates in $A$. So this is a transitive inner model of $V$, but it is not first-order definable in $V$, since $A$ is generic over $V$.

But meanwhile, $\text{HOD}(M)$ as computed in $V$ will be the same as $\HOD$, since it is already all of $V$ and so the extra parameters of $M$ didn't add anything.

So we have an inner model $M$ that is not definable in $\HOD(M)$, as desired.

Let's now modify the example to produce a model $V$ of ZFC with a definable inner model $M$, such that $M$ is not definable in $\HOD(M)$. Start in $L$ as before, and this time, add the class $A$ first, over $L$. In particular, every initial segment of $A$ is in $L$. Now, over the GBC model $L[A]$, force to add Cohen sets at every uncountable regular cardinal, forming $L[A][G]$. Finally, using product forcing rather than an iteration, we may force to code every set in $L[G]$, and also the classes $G$ and $A$, into the GCH pattern, producing an extension $V=L[A][G][H]$ in which $\HOD=L[G]$. (For example, the methods of my paper G. Fuchs, J. D. Hamkins, J. Reitz, Set-theoretic geology show how to do this, while also controlling the mantle and the generic $\HOD$.)

Let $M=L[G\upharpoonright A]$. This is a transitive inner model of ZFC. It is definable in $V$, since the class $G\upharpoonright A$ is definable there. Further, since every initial segment of $A$ is in $L$, it follows that $L[G\upharpoonright A]\subset L[G]$, which is the $\HOD$ of $V$, and so $\HOD(M)=\HOD=L[G]$ in $V$. But $A$ is not definable in $L[G]$, since $A$ and $G$ are mutually generic. So $M$ is not definable in $L[G]$.

Which is to say, $M$ is a definable transitive inner model of $V$, but it is not definable in $\HOD(M)$, as you desired.

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  • $\begingroup$ I think one should also be able to make a slightly better example, where $M$ is definable in $V$, but not in HOD(M), which you would probably prefer. $\endgroup$ – Joel David Hamkins Sep 4 '16 at 20:57
  • $\begingroup$ I updated with such an improved example, where M is now definable in V. $\endgroup$ – Joel David Hamkins Sep 4 '16 at 21:20
  • $\begingroup$ Thanks Joel. Your updated answer is what I had meant, as clearly the idea was for $M$ to be definable in $V$ (I generally reserve the term "class" for "definable class", but I guess I should have been clearer in my phrasing). $\endgroup$ – Asaf Karagila Sep 4 '16 at 21:26
  • $\begingroup$ (I'll give it a closer reading tomorrow morning and see if I have any follow up questions. Thanks for the answer!) $\endgroup$ – Asaf Karagila Sep 4 '16 at 21:33
  • $\begingroup$ For the record, I would encourage you henceforth to say "definable class" when that is what you mean, since it is a comparatively small minority usage to assume every class is definable, and I recall a few other questions on MO that were misunderstood more seriously because of that misunderstanding. $\endgroup$ – Joel David Hamkins Sep 4 '16 at 21:34

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