8
$\begingroup$

It has been discussed already whether a countable OD set necessarily contains an OD element. See e.g. A question about ordinal definable real numbers . A negative answer was obtained in Archive for Mathematical Logic 2015, 54, 5, 711-723, on the base of a very non-homogeneous forcing notion, in fact a clone of Jensen's minimal-$\Pi^1_2$-singleton forcing. In the opposite direction, one would believe that typical homogeneous-forcing models (Cohen, random, Solovay etc) provide a positive answer. And indeed it holds in a simple Cohen extension $L[a]$ that

(1) a countable OD set of reals necessarily contains an OD element (see arxiv1)

Is the following stronger claim still true in Cohen's $L[a]$?

(2) a countable OD set of sets of reals (or weaker but in fact equivalent, an OD partition of an OD subset $D$ of $\mathbb R$ into countably many pieces) necessarily contains an OD element

The key step would be to prove (2) restricted to the domain $$ D=\{x\in\mathbb R\cap L[a]:x\text{ is Cohen and }L[x]=L[a]\}. $$ And now, one may want to get a counterexample of the following kind. Let $G$ be the group of Borel category-preserving bijections of the reals, coded in $L$. Suppose $G$ is presented in the form $G=C\times U$, where $C$ is a countable subgroup and $U$ a subgroup. Then $D$ is invariant under the action of $G$, but one may hope that there are countably many of $U$-orbits in $D$ (as $C$ countable), none of which is OD.

To conclude, is (2) true in Cohen's $L[a]$?

$\endgroup$
  • $\begingroup$ It seems to me that every OD partition of $\mathbb{R}$, regardless of the number of pieces, contain an OD piece, namely, the piece of the partition containing $0$. Have I misunderstood the question? $\endgroup$ – Joel David Hamkins Jul 24 '16 at 22:38
  • $\begingroup$ My oversight, thanks. Should be: an OD partition of an OD subset of $\mathbb R$ into countably many pieces necessarily contains an OD piece $\endgroup$ – Vladimir Kanovei Jul 25 '16 at 6:11
  • $\begingroup$ If you're partitioning an OD subset $A$ of $\mathbb R$, can't you just let $a$ be the first element of $A$ in the standard well-ordering of OD, and then use Joel's comment with $a$ in place of $0$? $\endgroup$ – Andreas Blass Jul 25 '16 at 17:28
  • $\begingroup$ Andreas, we don't know that $A$ has any OD elements. For example, perhaps $A$ is the set of non-OD reals. $\endgroup$ – Joel David Hamkins Jul 25 '16 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.