# Making all cardinals countable and its HOD

Suppose that $G$ is $Col(\omega, <Ord)$-generic over $V$ and let $W=HOD^{V[G]}$. Is $CH$ true in $W$? In general, what can we say about the behaviour of the power function in $W$?

Update. Are the models $HOD^{V^{Col(\omega, <ord)}}$ and $gHOD$ different, where $gHOD$ is as defined in Hamkins answer.

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Perhaps a good test case would be to collapse an inaccessible $\kappa$ like that, and consider $V_{\omega_1}^{V[G]}$. –  Asaf Karagila Aug 19 '13 at 9:39
There is also a paper by Roguski called "Hartogs's numbers and axiom of power-set" which is impossible to find online (and the library at HUJI, where the paper can be found, is closed this week). I don't recall the exact construction the author uses, but it might be relevant. –  Asaf Karagila Aug 19 '13 at 9:42
@MonroeEskew It seems a natural problem. Under appropriate assumptions, we would expect an initial segment of $\mathsf{HOD}$ (certainly beyond $\mathsf{HOD}\cap V_{\omega_1}$) to be "core like", in particular to satisfy $\mathsf{GCH}$. This is a natural test question to try to clarify what those additional assumptions ought to require. –  Andres Caicedo Aug 19 '13 at 18:56
@EverettPiper Here is a weak reason: The model $\mathsf{HOD}$ of natural models of determinacy certainly has this property. This was shown by Steel (and extended by Woodin) for $L(\mathbb R)$, and has since been established for much larger models; in fact, the possibility of such a description of $\mathsf{HOD}$ is key in recent developments in descriptive inner model theory. Under very large cardinals, we expect "higher order instances" of results that hold under determinacy to be true. –  Andres Caicedo Aug 22 '13 at 2:44
@EverettPiper Another reason is that there is a very compelling picture of the universe as a (class) forcing extension of its $\mathsf{HOD}$, and the ultimate $L$ analysis would suggest the same ought to hold under appropriate assumptions replacing $\mathsf{HOD}$ with "ultimate $L$". But, in fact, since core-like models are ordinal definable, we expect that (ultimately), this largest core model ought to coincide with $\mathsf{HOD}$ itself. I'm sure there are more serious and sophisticated reasons as well. –  Andres Caicedo Aug 22 '13 at 2:47


I believe that it was Peter Koepke who first proposed that we should investigate this model, including the question of GCH in it.

Let me confess at the outset that I am troubled by the very definition of the model as $\text{HOD}^{V[G]}$. The problem is that I don't know a sensible way to define $\text{HOD}$ inside $V[G]$, which, lacking power set, lacks the usual von Neumann $V_\theta$ hierarchy that is central to the usual definition of $\text{HOD}$. That is, one usually defines $\text{HOD}$ in a $\text{ZFC}$ context by saying $x\in\text{HOD}$ if it is ordinal definable in some structure $\langle V_\theta,{\in}\rangle$. This is formalizable in set theory, and it works precisely because the $V_\theta$ sets themselves are ordinal definable. (One cannot say, naively, that $x$ is ordinal definable just in case there is a definition of $x$ using ordinal parameters, since this uses a universal truth predicate, which we cannot have on account of Tarski's theorem on the non-definability of truth.) I am not aware of any theory of $\text{HOD}$ that may be undertaken in a $\text{ZFC}^-$ context. So at the very least, the question needs clarification as to what $\text{HOD}^{V[G]}$ is supposed to mean.

But let us suppose that this issue can be overcome, and that we have a definable way to speak of $W=\text{HOD}^{V[G]}$, so that it is a definable class in $V[G]$ consisting of the ordinal definable elements there (and certainly, over some models $V$ this assumption is legitimate). In this case, let me show why $W$ is an inner model of $\text{ZF}$ contained in $V$. That is, although in $V[G]$, every set has become countable and so ZF of course fails there, by moving to the HOD of $V[G]$, you only take the hereditarily ordinal-definable part of $V[G]$, and I claim this must in fact be contained in $V$. To see this, note that the forcing is almost homogeneous, and every condition has an automorphic image compatible with any other. It follows that any statement $\varphi(\check x)$ in the forcing language involving only parameters from the ground model is forced either by every condition or by none, since the automorphisms fix the names of the parameters of the statement. It follows that any if all elements of a hereditarily ordinal definable element of $V[G]$ are in $V$, then we can determine these elements in $V$ via the forcing relation. (Note: this argument relies on the forcing lemmas for the class forcing to make every set countable; but this is OK, since this forcing is an iteration of set-sized forcing and hence every set is contained in a set-sized complete subforcing notion.) Thus, $W\subset V$, and it is a definable transitive class containing all the ordinals.

To see that $W\models\text{ZF}$, one gets the easier axioms without much difficulty. Beyond this, we can see that $W\cap V_\alpha\in W$, by induction on $\alpha$. If true at $\alpha$, then $W\cap V_\alpha$ is definable in $V[G]$ as the elements of HOD of rank less than $\alpha$, and so its power set is also definable in $V[G]$, and is a set there since it is subset of $V_{\alpha+1}$. If true below a limit $\lambda$, then $W\cap V_\lambda$ is definable in $V[G]$ as the sets in HOD of rank less than $\lambda$. Thus, $W$ is almost universal and this gives the collection axiom, and so $W\models\text{ZF}$.

I don't see yet why we should expect the axiom of choice always to hold in $W$, and this is related to my issues with the definition of $W$. Ordinarily, one get AC in HOD by defining in ZFC a well-ordering of HOD, so that $x\lt y$ just in case $x$ is coded by a set of ordinals definable in $V_\theta$ for a smaller $\theta$ than $y$ is, or for the same $\theta$ but with a smaller definition, or with the same $\theta$ and the same definition, but with smaller parameters. So, this argument will continue to work in the context of this question, provided that we define $\text{HOD}^{V[G]}$ by reference to some kind of ordinal-definable hierarchy in $V[G]$ inside of which we may run the definitions. So solving the definition issue mentioned above may also show how to get AC in $W$. But in general, I am unsure.

Meanwhile, let me mention another approach to a similar model, about which more progress has been made. Namely, Gunter Fuchs defined the generic HOD, denoted $\text{gHOD}$, and we give some of the basic facts in our paper Set-theoretic geology. The generic HOD is the intersection of the HOD's of all the set-forcing extensions of $V$. $$\text{gHOD}\ \ = \ \ \bigcap_\theta \text{HOD}^{V^{\text{Coll}(\omega,\theta)}}$$ This model is similar to yours, in the sense that we are collapsing more and more sets to be countable, and only taking the sets in the HODs of the models along the way (whereas you take the HOD of the final model at the end, which is not a ZFC model). The nice thing about the generic HOD is that it is an inner model $V$ satisfying $\text{ZFC}$ and containing all ordinals. One of the theorems of set-theoretic geology is that every model of set theory arises as the $\text{gHOD}$ of another model of set theory, a class forcing extension.

Theorem. (Fuchs,Hamkins,Reitz Set-theoretic geology) Every model of set theory $V$ is the generic HOD, as well as the HOD, the mantle and the generic mantle, of another model $V[G]$, obtained by class forcing over $V$. $$V=\text{gHOD}^{V[G]}=\text{HOD}^{V[G]}=\text{M}^{V[G]}=\text{gM}^{V[G]}.$$

(The mantle of a model of set theory is the intersection of all the ground models of that model, and the generic mantle is the intersection of the ground models of all the set-forcing extensions of the model.)

Thus, for the generic HOD version of your question, the answer is that one cannot make any general assertions about what is true there, since it could be an arbitrary model of ZFC.

Update. Regarding your updated question, yes, in general we should expect $\gHOD$ to be very different from $\HOD^{V[G]}$. For example, for example, if you start with a model having the continuum coding axiom, which asserts that every set of ordinals arises as a block in the $\text{GCH}$ pattern, then $\gHOD=V$, since every set in $V$ continues to be coded high up in the $\GCH$ pattern in any set forcing extenion $V[g]$. Thus, all such sets remain in the $\HOD^{V[g]}$ of any set forcing extension and thus remain in $\gHOD$. But if $V$ as obtained by forcing the continuum coding axiom over some deeper model $V_0$, then the $\HOD^{V[G]}$ will be contained in $V_0$, and hence much smaller than $V=\gHOD$. I guess the homogeneity arguments show that $\HOD^{V[G]}\subset\gHOD$.

(But once again, I don't think we have actually defined $\HOD^{V[G]}$, and so I wouldn't want to state many theorems about it until we have a proper definition.)

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I'm surprised that you don't make use of \newcommand in such posts to define \HOD or \ZF commands in your answer. I always found that it saves a lot of time (and then it is also useful for the comments too!). –  Asaf Karagila Aug 19 '13 at 22:31
Maybe I'll start to do that... –  Joel David Hamkins Aug 19 '13 at 22:51
Then again, it will cut into my lead as the chattiest: data.stackexchange.com/mathoverflow/query/5153/…. :-) –  Joel David Hamkins Aug 19 '13 at 22:54
I think that you're far enough into the lead to relax and allow yourself the comfort of custom commands! ;-) –  Asaf Karagila Aug 19 '13 at 22:57
Dear Prof. Hamkins, thanks so much –  Mohammad Golshani Oct 28 '13 at 15:05