A question about the Ordinal Definable elements of Power Sets

Question

Assume that we are working in a set theory T, formalized in the language of ZFC, whose axioms---in addition to those of ZFC---include also the negation of "V=OD".

For any set X, let CARD(X) denote the cardinal number of X, let ORDEF(X) denote the set of ordinal definable elements of X and let P(X) denote the the set of all subsets of X. As usual, N denotes the set of all non-negative integers and R denotes the set of all real numbers.

It is known that T+"CARD(ORDEF(P(N)))=CARD(N)" is consistent if ZFC is.

I would like to know whether this phenomenon---that the consistency of ZFC implies the consistency of T+"CARD(ORDEF(P(X)))=CARD(X)"---is true for many sets X? In particular, is it true for X=R?

My reason for being interested in this question is that, for many uncountable sets X, it becomes difficult to work in P(X) because so many sets---most of which we do not care about---belong to P(X).

If CARD(ORDEF(P(X)))= CARD(X) and if no inconsistency is introduced, we could work in ORDEF(P(X)) instead of in P(X). Most of the sets belonging to P(X) in which we are interested should still belong to ORDEF(P(X)) and we would be working in a set is no greater than that of X.

Just consider how few of the sets belonging to P(R) are of any interest to anybody. Most of them are literally indescribable in any conceivable language.

Answer 1

$ \newcommand\N{\mathbb{N}} \newcommand\R{\mathbb{R}} \newcommand\ZFC{\text{ZFC}} \newcommand\HOD{\text{HOD}} \newcommand\OD{\text{OD}} $For your initial remark, yes, it is relatively consistent with ZFC that there are only countably many ordinal-definable subsets of $\N$. To see this, suppose that $V$ is any model of $\ZFC$, and then go to the forcing extension $V[G]$ obtained by collapsing the continuum of $V$ to be countable in the extension. This forcing is weakly homogeneous, and it follows from this that $\HOD^{V[G]}\subset V$. In particular, every ordinal-definable set of natural numbers is in $V$, and in $V[G]$, this collection of sets was made countable. So $V[G]$ is a model of $\ZFC+V\neq\OD+$ there are only countably many $\OD$ reals.

Now, for your question, the point is that a similar argument works quite generally. If $X$ is any set in $V$, we can simply collapse $2^{|X|}$ to $|X|$ by forcing, and in the resulting extension $V[G]$, since the forcing is weakly homogeneous, all the ordinal-definable subsets of $X$ will be in $V$, and because of the collapse, this collection has the same size as $X$.

In the case of the reals $X=\R$, the collapse forcing of $2^\frak{c}$ to $\frak{c}$ is countably closed, and does not add any new reals. So in the resulting forcing extension $V[G]$, we would have the same reals $\R$ as in $V$, but now there would be only continuum many ordinal-definable sets of reals.

But lastly, let me remark that I don't really agree with the perspective that we necessarily have a much better grasp of the ordinal definable sets of reals than the "arbitrary" sets of reals. Indeed, any given set can be made ordinal definable in a forcing extension; we can even make them all ordinal definable: starting in any model $V$ of $\ZFC$, there is a forcing extension $V[G]$ in which every set, including all the original arbitrary sets in $V$, becomes ordinal definable in the extension (and many new sets as well).

Answer 2

A marginal comment. ORDEF(P(N)) can be countable while ORDEF(P(R)) cannot since for any countable ordinal $\alpha$ there is a non-empty OD set of all reals which code $\alpha$. Is there an UNcountable set $X$ such that ORDEF(X) is countable (or just empty?)