Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice) is consistent, does it remain consistent when the following statement is added to it as a new axiom?

"There exists a denumerably infinite and ordinal definable set of real numbers, not all of whose elements are ordinal definable"

If the answer to the above question is negative, then it must be provable in ZFC that every denumerably infinite and ordinal definable set of real numbers is hereditarily ordinal definable. This is because every real number can be regarded as a set of finite ordinal numbers and every finite ordinal number is ordinal definable. Garabed Gulbenkian

share|improve this question
If you adjoin a Cohen real to L, in the resulting model are the set of L-Cohen reals countable? I am not sure, but if so then I think Joel David Hamkin's answer here mathoverflow.net/questions/10413/… gives an example showing the consistency of your statement –  Justin Palumbo Mar 10 '10 at 0:19
@Justin: If r is Cohen over L then so is r+s for every constructible real s. Since the constructible reals are uncountable in L[r], there are uncountably many Cohen reals in L[r]. –  François G. Dorais Mar 10 '10 at 15:28
I wonder if the question I am asking is actually an open problem, although I would be surprised if it were since so much is now known about ordinal definability. Perhaps the following statement-which yields a negative answer to my question-would be considered of some interest and even be well known, if it were provable in ZFC. Is it? "If S is any ordinal definable set of real numbers which contains at least one real number among its elements that is not ordinal definable, then S is uncountable." –  Garabed Gulbenkian Mar 18 '10 at 20:44
+1. This is a fascinating, outstanding question. –  Joel David Hamkins Mar 22 '10 at 22:38
Ali Enayat posted this on FOM: cs.nyu.edu/pipermail/fom/2010-July/014944.html –  Kaveh Jul 24 '10 at 22:51
add comment

1 Answer 1

(I am replacing prior nonsense with a completely different suggestion. I am also turning this into CW so details can be added by somebody with time (which, sadly, most likely won't be me). Comments prior to Feb. 9, 2011, refer to said prior nonsense.)

Start with $V=L$ and force to add a Matthias real $s$. Let $W$ be the resulting extension. Let $A$ be the set of reals $r$ that are Matthias generic over $L$ and such that $L[r]=W$. I strongly suspect that a real $r$ is in $A$ iff it differs from $s$ finitely often, and so $A$ is countable, ordinal definable, and lacks any ordinal definable members.

(I have briefly discussed this idea with other set theorists, but we did not elaborate any details.)

From Andreas Blass: The following began as a comment, but Andres suggested adding it to his answer, for improved visibility. As it stands, with ordinary Mathias forcing, this won't work, because if $r\subset\omega$ is a Mathias real then so is the result of shifting it to the right (or left) by 1, and it still generates the same model. Instead of a simple shift, you could apply any strictly monotone function from $L$. But suppose you did Mathias forcing with respect to the constructibly-first non-principal ultrafilter on $\omega$ in $L$. That would avoid this problem. (Note that Joel David Hamkins's comment also depends on the fact that Prikry forcing is with respect to an ultrafilter in the ground model.)

share|improve this answer
Andres, homogeneous forcing doesn't necessarily preserve OD. (Although you are likely thinking of the fact that it doesn't enlarge OD.) So I don't see why X should be OD-definable after the collapse. Indeed, it cannot be, since the same extension could have arisen by absorbing additional Cohen forcing into the collapse forcing. –  Joel David Hamkins Jun 6 '10 at 18:18
About the Laver theorem: the history is that Laver proved this theorem, and Woodin later observed a similar fact independently. But this theorem doesn't give you OD preservation. It is possible to have models of V=HOD whose HOD's drop in further extensions by homogeneous forcing. –  Joel David Hamkins Jun 6 '10 at 18:22
To explain, for homogeneous forcing what you have is $HOD^{V[G]}\subset HOD^V$, rather than equality. If your model is $V[c][G]$, where $c$ is the Cohen real and $G$ is the collapse forcing, then we can reconstitute it as $V[c][G]=V[c'][G']$, where $c'=c\oplus d$ is built from an additional Cohen real $d$ added by $G$, and the $g'$ is the corresponding quotient forcing, which is still collapse forcing. Any proposed definition of $X$ cannot distinguish between the reals of $V[c]$ and those of $V[c']$. So $X$ is not ordinal definable in $V[c][G]$. –  Joel David Hamkins Jun 6 '10 at 18:33
Is the problem then that this $V_{\delta+1}$ isn't OD in the extension? If it were you could define the ground model, and thus the levels of the cumulative hierarchy of the ground model, and thus the ground model OD sets. Is that right? –  Justin Palumbo Jun 6 '10 at 18:41
You can see that $HOD^{V[G]}$ can differ from $HOD^V$ by first adding a Cohen real $c$, then coding it into the continuum function of $V[c][H]$, and then collapsing cardinals to $V[c][H][g]$. The combined forcing $H+g$ is equivalent to collapsing forcing, as is the whole forcing $c+H+g$, so HOD of the final model is contained in $V$. Thus, $c$ went from definable in $V[c][H]$ to nondefinable in $V[c][H][g]$, even though it was homogeneous forcing. –  Joel David Hamkins Jun 6 '10 at 18:43
show 6 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.