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Define a pointclass to be:

  • boldface inductive-like if it is $\mathbb{R}$-parameterized, has the scale property, and is closed under $\wedge$, $\vee$, $\forall^\mathbb{R}$, $\exists^\mathbb{R}$, and preimages by continuous functions, and

  • lightface inductive-like if it is $\omega$-parameterized, has the scale property, and is closed under $\wedge$, $\vee$, $\forall^\mathbb{R}$, $\exists^\mathbb{R}$, and preimages by continuous functions.

Given a lightface inductive-like pointclass $\Gamma$ we can define the boldface inductive-like pointclass ${\bf \Gamma} = \bigcup_{x \in \mathbb{R}} \Gamma(x)$ as usual (this does not have much to do with the particulars of "inductive-like.")

My question is, can we do the reverse? That is, given a boldface inductive-like pointclass ${\bf \Gamma}$ can we find a lightface inductive-like pointclass $\Gamma$ such that ${\bf \Gamma} = \bigcup_{x \in \mathbb{R}} \Gamma(x)$? (We wouldn't expect uniqueness; if $\Gamma$ works then so does $\Gamma(z)$ for any real $z$.)

I have heard that Howard Becker proved something like this, but I can't find it and I don't know what kind of pointclasses it was for (probably more general than "inductive-like," which is merely the case I happen to be interested in at the moment.)

For a motivating example, consider the case where $\mathsf{AD}$ holds and $\bf \Gamma$ is the pointclass of $\kappa$-Suslin sets. If $\bf \Gamma$ is closed under $\forall^\mathbb{R}$ then it is boldface inductive-like, but as far as I know there is no canonical choice of lightface pointclass that corresponds to it.

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  • $\begingroup$ Let $\bf \Gamma$ be any inductive-like pointclass (say this is on some product space $X$). Let $U \subset \mathbb{R} \times X$ be a universal set for $\bf \Gamma$ subsets of $X$. Define a lightface pointclass: $\Sigma_1^1 (U)$, the least lightface pointclass containing $U$ and closed under $\cup$, $\cap$, integer quantification and $\exists^{\mathbb{R}}$. Can it be that $\bf \Gamma$ = $\cup_{x \in \mathbb{R}} \forall^{\mathbb{R}} \Sigma_1^1 (U)(x)$? Or maybe $\bf \Gamma$ = $\cup_{x \in \mathbb{R}} \exists^{\mathbb{R}} (\Sigma_1^1 (U) \wedge \Pi_1^1 (U))(x)$. These are random guesses. $\endgroup$
    – Rachid Atmai
    Commented Feb 27, 2014 at 1:47
  • $\begingroup$ @CarloVonSchnitzel I think in the definition of $\Sigma^1_1(U)$ one should also require closure under preimages by recursive functions. If so, then it looks to me like ${\bf \Gamma} = \bigcup_{x\in \mathbb{R}} \Gamma(x)$ where $\Gamma = \Sigma^1_1(U)$. Probably $\Sigma^1_1(U)$ is $\omega$-parameterized (although at the moment I have only tracked down the corresponding result that ${\bf \Sigma}^1_1(U)$ is $\mathbb{R}$-parameterized.) However, it is not clear to me whether $\Sigma^1_1(U)$ will have the scale property. $\endgroup$
    – Trevor Wilson
    Commented Feb 27, 2014 at 2:24
  • $\begingroup$ I should have said $\Gamma = \forall^\mathbb{R} \Sigma^1_1(U)$ in my last comment, I think, in order to ensure that $\Gamma$ is closed under $\forall^\mathbb{R}$. However the scale property still seems trickier to obtain. $\endgroup$
    – Trevor Wilson
    Commented Feb 27, 2014 at 2:33
  • $\begingroup$ Assuming some amount of determinacy probably, if $\Sigma_1^1(U)$ has the scale property then $\forall^{\mathbb{R}} \Sigma_1^1(U)$ will have the scale property since $\forall^{\mathbb{R}} \Sigma_1^1(U) = \Game \Sigma_1^1(U)$ and the game quantifier propagates scales (using some local determinacy, if I recall correctly this is in chapter 6 of Moschovakis). So maybe using the scale on the universal set $U$ (as $\bf \Gamma$ is inductive-like) one can build a scale for any $\Sigma_1^1(U)$ set. I trying to write this down rigorously. (Also maybe $\Sigma_1^1(U)$ is $\omega$-parametrized). $\endgroup$
    – Rachid Atmai
    Commented Feb 27, 2014 at 3:33
  • $\begingroup$ @CarloVonSchnitzel Yes, it does sound like a reasonable idea although it seems like one might need an effective way of finding an index of a $\bf \Gamma$-scale on a given $\bf \Gamma$ set. $\endgroup$
    – Trevor Wilson
    Commented Feb 27, 2014 at 3:53

1 Answer 1

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See Lemma 3.4 and preceding paragraph of my paper "A characterization of jump operators", JSL vol. 53, 1988.

Howard Becker

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