*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.)




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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.



Ordinal-definable witnesses to the perfect set property?
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 $P$.


