# Universal $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set

Does anyone know of a reference for the fact that if $\lambda$ is a limit of Woodin cardinals, then the pointclass $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ is $\omega$-parameterized? By this I mean that there is a $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $U \subset \omega \times \omega^\omega$ that is universal in the sense that every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $A \subset \omega^\omega$ is equal to the section $U_n$ for some $n < \omega$.

As far as I am concerned it would be enough to have a complete $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $U \subset \omega^\omega$ such that every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $A \subset \omega^\omega$ is equal to the pre-image $f^{-1}(C)$ by some recursive function $f$.

In either case, this seems not entirely trivial to me. Suppose that $A \in (\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$, say $$x \in A \iff \exists B \in \text{Hom}_{\mathord{<}\lambda}\,(HC; \in, A) \models \varphi[x].$$ It seems like we need to bound the complexity of the formulas $\varphi$ we are considering. One way to do this is to show that if there is a set $B \in \text{Hom}_{\mathord{<}\lambda}$ with $(HC; \in, B) \models \varphi[x]$ then there is also a set $B' \in \text{Hom}_{\mathord{<}\lambda}$ with $(HC; \in, B') \models \psi[x]$, where $\psi$ is the Skolem normal form of $\varphi$. We can get $B'$ using the fact that $\text{Hom}_{\mathord{<}\lambda}$ is projectively closed and every $\text{Hom}_{\mathord{<}\lambda}$ set admits a $\text{Hom}_{\mathord{<}\lambda}$ uniformization.

This might be more familiar in the equivalent context of $\mathsf{AD}^+$ where we would talk about $\Sigma^2_1$ rather than $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$, but I don't recall having seen any argument for it anywhere. Is there a published reference for this? Or perhaps an obvious argument that is simpler than the one I started to sketch above?

By the way, my motivation for asking this is the following. If $\lambda$ is a limit of Woodin cardinals and the derived model at $\lambda$ satisfies $\theta_0 < \Theta$ (or equivalently, every $\Pi^2_1$ set is Suslin) then every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set becomes $\lambda$-universally Baire in some $\mathord{<}\lambda$-generic extension $V[g]$. (More precisely, its re-interpretation in $V[g]$ becomes $\lambda$-universally Baire.) To get a single $\mathord{<}\lambda$-generic extension $V[g]$ in which every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set is universally Baire, it seems like the simplest way would be to consider a universal set.

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I don't know if this answer will address your specific problem directly, but under AD, Wadge's Lemma implies that every non-selfdual pointclass has a universal set. There is an argument in Jackson's article in the Handbook (p.1761): basically if you have a $\Gamma$-complete set $A$, then define for every product space $X=X_0 \times X_1 \times ...\times X_n$ some sets $U_X \subseteq \omega^{\omega} \times X$ by $U_x(y, x_0,...,x_n) \leftrightarrow f_y(<x_0,...,x_n>) \in A$ where $y \in \mathbb{R}$ code real-valued Lipschitz continuous functions on $\mathbb{R}$, i.e $y$ computes the output value of $f_y(a_0,...,a_n)$ given the input $(a_0,...,a_n)$ by $(y(<a_0>),...,y(<a_0,...,a_n>))$. Then the sets $U_x$ are in $\Gamma$ because we assume $\Gamma$ is at least adequate (so we get recursive substitution) and whenever you take some set $B$ in $\Gamma$ then by Wadge's Lemma, it is Lipschitz reducible to $A$ so for some real $y$, $(U_X)_y=B$.
Now since $\Sigma^2_1$ is obviously adequate, and it is nonselfdual, then it has a universal set. I guess that makes it a Spector pointclass (it is normed, even has the scale property).
Ok, this might be the better route, to forget about the syntactic notion and to use general pointclass arguments in the derived model. However, I probably should have clarified that I want a lightface $\Sigma^2_1$ subset of $\omega \times \mathbb{R}$ that is universal for subsets of $\mathbb{R}$. Does this argument give such a thing? It would also be enough for my purposes to have a lightface $\Sigma^2_1$ subset of $\mathbb{R}$ from which every other lightface $\Sigma^2_1$ subset of $\mathbb{R}$ can be obtained via recursive substitution. – Trevor Wilson Aug 20 '13 at 1:15
Actually that's a good question. I think in this case instead of a general space $X$ like in the above argument, we need to only look at type 1 spaces, because Wadge's Lemma is formulated for type 1 space (see Moschovakis section 7). So I guess we need to restrict to sets of reals here instead of subsets of a general space $X$ and then the argument would be OK. – Carlo Von Schnitzel Aug 20 '13 at 1:34
I wasn't concerned about the difference between $X$ and $\mathbb{R}$ (by which I mean $\omega^\omega$) so much as I was about the difference between $\mathbb{R}$ and $\omega$. That is to say, to reduce another lightface set to this universal (or just complete) lightface set, how do we know we don't need a non-recursive real as a parameter? – Trevor Wilson Aug 20 '13 at 1:38