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.


$ \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).

  • $\begingroup$ Your first line misses "consistent" before "relatively with". $\endgroup$ – Asaf Karagila Dec 31 '13 at 0:08
  • $\begingroup$ I think the source of that perspective may be that ordinals somehow 'feel' much tamer than other sets - we have a clean well-ordering on them, etc - even though of course in practice they can be remarkably complicated themselves. $\endgroup$ – Steven Stadnicki Dec 31 '13 at 0:31
  • $\begingroup$ @Joel: Many thanks for answering all my questions and providing background that goes far beyond my rudimentary knowledge of forcing. My feeling about ordinal definability stems from my preference for set theories in which many undefinable sets exist (particularly undefinable real numbers). Intuitively, the languages in which most set theories are formalized contain only countably many formulae that can be used to define these sets. So I would like the class OD to be as small as possible. It seems to me rather counter-intuitive for all sets to be definable-which is what occurs with V=OD. $\endgroup$ – Garabed Gulbenkian Jan 1 '14 at 16:21
  • $\begingroup$ Oh, I like your questions. Concerning ordinal-definability, one may think of V=OD as the principle that there is a definable well-ordering of the universe. In this case, every set becomes ordinal-definable, since any set $x$ is the $\alpha^{th}$ set in that well-ordering for some ordinal $\alpha$. So the ordinal parameter in the definition is simply pointing at a particular index in the enumeration of all sets. Much weirder models arise with the pointwise definable models, in which every set is definable without any parameters at all. $\endgroup$ – Joel David Hamkins Jan 1 '14 at 16:30

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?)

  • $\begingroup$ Nice observation. For your question, if $X\subset V-\text{OD}$, then $\text{OrDef}(X)=\emptyset$, and if $V\neq\text{HOD}$, there are uncountable such $X$. Perhaps you want $X$ itself to be OD, or to allow $X$ as a parameter in the definitions? $\endgroup$ – Joel David Hamkins Mar 22 '15 at 22:00
  • $\begingroup$ You can have $X$ itself definable, if you let $X$ consist of the minimal rank sets not in $\text{OD}$. If there are any non-OD sets at all, then this is an uncountable definable set with only countably many (actually zero) ordinal definable elements. $\endgroup$ – Joel David Hamkins Mar 22 '15 at 23:04

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