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It is known that if there is a measurable cardinal then every $\Pi_1^1$ set has the perfect set property (i.e it is either countable or contains a copy of $2^{\omega}$). Also if we have $\Pi_1^1$-determinacy (or in other words $0^{\sharp}$) then we get that $\Sigma_2^1$ has the perfect set property. Note the result that $\Sigma_1^1$ has the perfect set property is a $ZFC$ result.

Is there a reason as to why we need stronger infinity axioms to prove the perfect set property for $\Pi$ classes in comparison to what we need to prove the perfect set property for $\Sigma$ classes? This is weird because we have $\Pi_1^1 \subseteq \Sigma_2^1$. Do we actually need less than a measurable to prove the PSP for $\Pi_1^1$?

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  • $\begingroup$ Someone please change occurences of "prefect" to "perfect". Thank you in advance. Gerhard "Ask Me About System Design" Paseman, 2011.10.22 $\endgroup$ – Gerhard Paseman Oct 23 '11 at 3:24
  • $\begingroup$ Note that $2^\omega$ is not the only perfect set. Lightface $\Pi^1_1$-determinacy is equivalent to the existence of $0^\sharp$; you need all sharps for boldface $\boldsymbol{\Pi}^1_1$-determinacy. Why does $\Pi^1_1 \subseteq \Sigma^1_2$ make this weird? $\endgroup$ – François G. Dorais Oct 23 '11 at 3:25
  • $\begingroup$ @Gerard:changed. $\endgroup$ – Rachid Atmai Oct 23 '11 at 3:44
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Solovay showed that the following are equivalent:

  1. $\boldsymbol{\Sigma}^1_2$ sets have the perfect set property
  2. $\boldsymbol{\Pi}^1_1$ sets have the perfect set property
  3. $\aleph_1^{L[a]} < \aleph_1$ for every real $a$

You only need an inaccessible to force (3).

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  • $\begingroup$ Francois, maybe this should be asked as a full-fledged question, but have such "nice" characterizations been obtained further up in the projective hierarachy? My knowledge of descriptive set theory tends to peter out around the $\mathbf{\Sigma}^1_3$ level... $\endgroup$ – Todd Eisworth Oct 23 '11 at 3:43
  • $\begingroup$ @Todd: That's what I was asking about. What happens on the projective hierarchy and why do we need more to prove the PSP for $\Pi$ classes in comparison to proving it for $\Sigma$ classes? $\endgroup$ – Rachid Atmai Oct 23 '11 at 3:47
  • $\begingroup$ Todd, I don't think so but I must admit I suffer from the same type of nearsightedness you do. There is a great paper by Brendle and Löwe that has a bunch of these characterizations; I don't think they do large subscripts, but that's where I'd check first. $\endgroup$ – François G. Dorais Oct 23 '11 at 3:58
  • $\begingroup$ @alephomega: The only way I know to prove PSP for large subscripts is using determinacy and the unfolding trick. Unfolding immediately gives the next $\Sigma$ pointclass up the ladder. I have no idea how you could avoid the extra step and just have PSP for $\Pi$ pointclass and not the next $\Sigma$ pointclass. $\endgroup$ – François G. Dorais Oct 23 '11 at 4:02

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