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This possibly a very basic descriptive set-theory question; if it is too basic for MO, feel free to migrate.

Throughout we work in ZF+AD. My question is:

If $A$ is an uncountable OD set of reals, need $A$ have an OD perfect subset?

Motivation: The Solovay sequence is given by:

  • $\theta_0$ is the least ordinal onto which $\mathbb{R}$ doesn't surject via an OD function;

  • $\theta_\lambda=\sup\{\theta_\alpha: \alpha<\lambda\}$ for $\lambda$ limit; and

  • $\theta_{\alpha+1}$ is the least ordinal onto which $\mathcal{P}(\theta_\alpha)$ doesn't surject via an OD function.

Ignore the later stages of the hierarchy; if the answer to my question is no, then we get a potentially interesting way to compare the "size" of two uncountable sets of reals under AD (which of course have cardinality continuum): let $\theta_0(X)$ be the least ordinal onto which $X$ doesn't surject in an OD way. If the answer to my question is "no," we might have $\theta_0(X)<\theta_0$ for some uncountable $X$, even assuming AD. This looks very interesting to me.

. . . which is why I think the answer to my question will be "yes" :P. But oh well.

(To keep going and define $\theta_\alpha(X)$, we need $X$ to be "ordinal-independent": that is, we want a "$\kappa$-version" $X_\kappa\subseteq 2^\kappa$ of $X$ for all (or enough) $\kappa$s. Then we can talk about things like the rate at which the Solovay sequence grows, for a given such $X$; but I think I'm getting well ahead of myself, so I'll stop for now.)

A note on tags: I used the inner-model-theory tag because of the importance of the Solovay sequence in descriptive inner model theory, and more generally the hope that inner model theory might be relevant to this question. This might be incorrect on my part; if so, feel free to remove that tag.

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  • $\begingroup$ A guess: let X be the OD set of all Cohen generic reals over L. It would be strange to get an OD surjection X onto even $\omega_2^L$ - ? $\endgroup$
    – Vladimir Kanovei
    Commented Jul 9, 2016 at 20:30
  • $\begingroup$ @VladimirKanovei That seems promising!Maybe add it to my other question mathoverflow.net/questions/243461/… if you can resolve it further? $\endgroup$
    – Noah Schweber
    Commented Jul 9, 2016 at 20:37
  • $\begingroup$ I am thinking more about the choiceless Solovay model, where my conjecture would look like a sure fact. $\endgroup$
    – Vladimir Kanovei
    Commented Jul 9, 2016 at 21:01
  • $\begingroup$ @VladimirKanovei See my answer to mathoverflow.net/questions/243461/…. $\endgroup$
    – Noah Schweber
    Commented Jul 14, 2016 at 0:32

1 Answer 1

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Let $A$ be the set of those reals that are not OD. Then $A$ is OD (since I've just defined it), and it's uncountable (since its complement, being a well-orderable set of reals, must be countable under AD). But any perfect OD set $P$ has an element that is OD and thus outside $A$, namely the first element of $P$ (in the usual ordering of the reals, or, if your reals are $2^\omega$, in the lexicographic order), which exists as $P$ is compact. That first element is OD because it's defined from the OD set $P$.

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  • $\begingroup$ Very nice! I forgot that "ordinal definable" was ordinal definable :P. Given this, do you know if it's possible for an uncountable OD set $X$ to have $\theta_0(X)<\theta_0$? $\endgroup$
    – Noah Schweber
    Commented Jun 30, 2016 at 16:50
  • $\begingroup$ EDIT: "Possible" should of course be "possible under AD." $\endgroup$
    – Noah Schweber
    Commented Jun 30, 2016 at 16:59
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    $\begingroup$ I like the answer a lot. Meanwhile, let me point out that Andreas's set has a perfect subset definable from a real, in fact, from any non-OD real $z$. If you fix $z$, consider the set of reals whose even-bits form a copy of $z$. $\endgroup$
    – Joel David Hamkins
    Commented Jun 30, 2016 at 23:25
  • $\begingroup$ I've asked a new question based on my comments/motivation here: mathoverflow.net/questions/243461/…. $\endgroup$
    – Noah Schweber
    Commented Jul 1, 2016 at 13:44

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