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The Gandy-Harrington topology on $\omega^\omega$ is the topology generated by all lightface $\Sigma^1_1$ sets; that is, all sets which are continuous-in-the-usual-sense images of $\omega^\omega$.

Although this topology is definitely less nice than the standard one - for example, it is non-metrizable - it satisfies the strong Choquet property: this is the statement that player I has a winning strategy in a certain topological game. From the strong Choquet property we can deduce many results whose proof would follow easily from metrizability, so in some sense the Gandy-Harrington topology is "close enough" to metrizable. One particularly nice result we can show is: if $A$ is a nonempty lightface $\Sigma^1_1$ subset of $\omega^\omega$, then $A$ has an element $x$ such that $\mathcal{O}^x\equiv_T\mathcal{O}\oplus x$.

A natural question now is to ask about topologies generated by lightface sets further up the projective hierarchy; e.g., let $\tau_k$ be the topology on $\omega^\omega$ generated by lightface $\Sigma^1_k$ sets. And similarly, we can define $\tau_\Gamma$ for any pointclass $\Gamma$ (although we're probably only interested in, say, Spector pointclasses). Unfortunately, for $k>1$ the strong Choquet property fails badly in $\tau_k$!

My question is:

Is there a weakening of the strong Choquet property that holds for $\tau_2$? (Or more generally for $\tau_\Gamma$ for nice enough pointclass $\Gamma$.)

Relatedly, for a real $x$ let $\mathcal{O}_2^x$ be the set of $x$-computable codes for nonempty $\Sigma^1_2$ sets, and let $\mathcal{O}_2=\mathcal{O}_2^\emptyset$.

Is it the case that every lightface $\Sigma^1_2$ subset of $\omega^\omega$ contains an $x$ with $\mathcal{O}_2^x\equiv_T\mathcal{O}_2\oplus x$?

I've tagged "set-theory" and "inner-model-theory" because it seems reasonable to me that answers might depend on structural hypotheses about the universe; feel free to remove either tag if this seems bonkers.

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  • $\begingroup$ I could be mistaken but I don't think this generalization can be true. For example if one want to generalize Kleene's basis theorem for $\Sigma^1_1$ sets then one needs to use non-trivial $\Pi^1_{2n+1}$ singletons instead of complete $\Pi^1_{2n+1}$ sets of integers. Basically the naive generalisation of Kleene's basis theorem is provably false. So maybe the same phenomenon happens here. Maybe the question you mean to ask is for example, if every $\Sigma^1_3$ subset of $\mathbb{R}$ contains an $x$ with $y^0_3(x)$ Turing equivalent to $y^0_3 \oplus x$. Would this follow from your definition? $\endgroup$ – Calico Jack Rackham Jun 16 '15 at 22:40
  • $\begingroup$ ($y^0_3$ is the least nontrivial $\Pi^1_3$ singleton). I meant to also add that the clue is as you stated, that the strong Choquet property fails for $k>1$. Anyway your generalization may require boldface determinacy of $\Delta^1_2$ for example for the $\Sigma^1_3$ level. $\endgroup$ – Calico Jack Rackham Jun 16 '15 at 22:41
  • $\begingroup$ What Carlo said is probably related to Kechris-Martin-Solovay "introduction to Q-theory", Lemma14.9(i). $\endgroup$ – Yizheng Zhu Jun 17 '15 at 10:10
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Related to the second question: Every non-empty $\Sigma^1_2$ subset of $\omega^\omega$ has a $\Delta^1_2$ element $x$, namely its $L$-least member. The complete $\Sigma^1_2$ subset of $\omega$ relative to this $x$ is $\Sigma^1_2$, so $\mathcal{O}_2^x$ is recursively isomorpic to $\mathcal{O}_2$.

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  • $\begingroup$ I'm probably just tired, but: why is $\mathcal{O}^x_2$ $\Sigma^1_2$? Since $x$ is $\Delta^1_2$ the best I can see is $\Sigma^1_2(\Delta^1_2)=\Sigma^1_3$. $\endgroup$ – Noah Schweber Jun 17 '15 at 4:16
  • $\begingroup$ Say that $x$ is defined as $\{n:\psi(n)\}=\{n:\phi(n)\}$, where $\psi$ is $\Sigma^1_2$ and $\phi$ is $\Pi^1_2$. Then, $\endgroup$ – Theodore Slaman Jun 17 '15 at 5:45
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    $\begingroup$ Sorry, I didn't realize that I would run out of time on the comment. Try again: We may assume that $x$ is an element of $2^\omega$. If $x$ is $\Delta^1_2$, then the statement about $z$ that $z$ is equal to $x$ is $\Sigma^1_2(z)$: for every $n$, $n\in z$ implies $n\in x$ and $n\not\in z$ implies $n\not\in x$. Both $n\in x$ and $n\not\in x$ are $\Sigma^1_2$ statements about $n$ and $\Sigma^1_2$ statements are closed under universal number quantifiers. Then, for $\theta\in\Sigma^1_2$, $\theta(x)$ iff $\exists z(z=x \wedge \theta(z))$. $\endgroup$ – Theodore Slaman Jun 17 '15 at 5:54
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As mentioned by Ted, the appropriate generalization is for odd levels only. In Hjorth "Variations of the Martin-Solovay tree" and "Some applications of coarse inner model theory", the following folklore result that generalizes Harrington's proof of Silver's dichotomy is stated:

Assume boldface $\bf\Delta^1_2$ determinacy. If $E$ is a thin $\Pi^1_3$ equivalence relation on $\mathbb{R}$, then for every $x$, there is a $\Delta^1_3(<u_\omega)$ set $A$ such that $x \in A \subseteq [x]_E$.

As defined in these papers, for a fixed $\alpha<u_\omega$, $A\subseteq \mathbb{R}$ is $\Sigma^1_3(\alpha)$ means there is $B\subseteq \mathbb{R}^2$ such that $x\in A$ iff $(x,y)\in B$ for some sharp code $y$ for $\alpha$. $\Sigma^1_3(<u_\omega)$ means $\Sigma^1_3(\alpha)$ for some $\alpha<u_\omega$. $\Delta$-pointclasses are defined in the obvious way.

The correct generalization of the Gandy-Harrington topology appears to be generated by $\Sigma^1_3(<u_\omega)$ sets. Also as a corollary, every thin $\Pi^1_3$ equivalence relation is $\Delta^1_3$ reducible to a $\Pi^1_3$ equivalence relation on a $\Pi^1_3$ subset of $u_\omega$ (in the sharp codes); every thin $\Delta^1_3$ equivalence is $\Delta^1_3$ reducible to equality on $u_\omega$.

I haven't read the proof of this generalization. So it would be nice if someone can expand a bit more. I also don't know the proof of the basis theorem at lower level with this approach, even though it seems to be straightforward.

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  • $\begingroup$ There is another clone of the GHT, the OD topology in the Solovay model, which I used long ago to prove a Glimm-Effros type dichotomy in the said model. There also exist $\kappa$-Souslin generalizations of the GHT, where the non-emptiness condition works in the form that the $\kappa$-Souslin set considered is non-empty at least in some collapse generic extension of the universe. $\endgroup$ – Vladimir Kanovei Jul 9 '16 at 21:08

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