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We have the following theorem:

Suppose $\omega_1^L=\omega_1$ then there exists a $\Pi_1^1$ subset of reals without the perfect set property

Moreover, under the same hypotheses, we can prove actually the existence of a "greatest" $\Pi_1^1$ set without the perfect set property, one which contains (as a subset) every $\Pi_1^1$ set without the perfect set property.
My question is the following (we are working in $\mathtt{ZF}+\mathtt{DC}$, where $\mathtt{DC}$ stands for Dependent Choice):

  • Suppose $\omega_1^L=\omega_1$, can we prove the existence of a subset $X$ of reals without the perfect set property and which is not a $Q$-set, i.e. such that there exists a $A\subseteq X$ with $A$ not $F_\sigma$ wrt the subspace topology on $X$?

Ideas?
Thanks!

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  • $\begingroup$ I'm confused by your claim that there can exist a "greatest" $\Pi^1_1$ set without the perfect set property. Given any $\Pi^1_1$ set without the perfect set property, you can add countably many points to it, and it will still be a $\Pi^1_1$ set without the perfect set property. Right? $\endgroup$
    – Will Brian
    Commented Apr 20, 2022 at 15:57
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    $\begingroup$ @WillBrian I think the issue is lightface vs. boldface - adding a single point to a $\Pi^1_1$ set may indeed make it no longer $\Pi^1_1$, although it will of course still be ${\bf \Pi^1_1}$. Similarly, if $\omega^\omega\cap L$ is countable then it is the largest countable lightface $\Sigma^1_2$ set. And there are various other results along these lines. $\endgroup$
    – Noah Schweber
    Commented Apr 20, 2022 at 18:03
  • $\begingroup$ @NoahSchweber: Oh right! Thanks -- that makes sense now. $\endgroup$
    – Will Brian
    Commented Apr 20, 2022 at 18:20
  • $\begingroup$ In fact, the hypothesis is not necessary for the existence of a largest $\Pi^1_1$ set without a perfect subset. Provably in ZF+DC, the set $\{x\in \mathbb{R} \mid x\in L_{\omega_1^x} \}$ is such a set. $\endgroup$
    – Jason Zesheng Chen
    Commented Apr 20, 2022 at 18:57
  • $\begingroup$ @喻良 the link sends me back to this question... is the question's link correct? $\endgroup$
    – Lorenzo
    Commented Apr 22, 2022 at 9:47

1 Answer 1

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If $\kappa$ is such that $2^\kappa > 2^{\aleph_0}$ then there are no $\kappa$-sized Q sets. In particular, no set of reals of size continuum is Q. This is because for any set of reals viewed as a topological space with the subspace topology, there are only continuum many Borel sets so if your space has more than continuum many subsets it cannot be Q. Therefore in fact ZFC proves there is a non-Q set without the perfect set property, without any hypothesis on the $\omega_1$ of L. See this paper for more info: https://arxiv.org/pdf/1611.08152.pdf

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  • $\begingroup$ Thanks for the answer and the reference, but I already knew that in ZFC we have the existence of such a set. In my question instead I ask if we can prove its existence assuming DC and $\omega_1^L=\omega_1$ $\endgroup$
    – Lorenzo
    Commented Apr 25, 2022 at 14:42
  • $\begingroup$ Oh I see, sorry for my misunderstanding $\endgroup$
    – Corey Bacal Switzer
    Commented Apr 25, 2022 at 15:03

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