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A subset $A\subseteq \mathbb{R}$ is said to be a $Q$-set if every subset $B\subseteq A$ is $F_\sigma$ wrt the subspace topology on $A$. For example $\mathbb{Q}$ is a $Q$-set. The first time I have seen this definition is in: Balogh, Zoltán, There is a $Q$-set space in ZFC, Proc. Am. Math. Soc. 113, No. 2, 557-561 (1991).
Now my questions are:

  1. Can we prove in $\mathtt{ZF}+\mathtt{DC}$, where $\mathtt{DC}$ stands for dependent choice, that every co-analytic $Q$-set is countable?
  2. Is the descriptive complexity (with connections with set-theoretic hypothesis) of these sets been studied in any paper/book/thesis?

Thanks!

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I do not know about full references but if you haven't seen it yet there is a nice discussion of this topic in Miller's https://people.math.wisc.edu/~miller/res/dstfor.pdf. (Incidentally this may be the most fun math book to read). See in particular sections 2-5.

In particular, Theorem 5.1 states that Martin's Axiom implies every second countable Hausdorff space of cardinality less than the continuum is a Q-set. If $V=L$ there are $\Pi^1_1$ sets $A \subseteq \mathbb R$ of size $\aleph_1$ so that in any $\aleph_1$-preserving forcing remain uncountable and no new elements are added. Forcing martin's Axiom over $L$ with ccc forcing (so the standard way) will therefore give a consistent example of an $\aleph_1$-sized co-analytic Q-set.

Also Corollary 3.2 can be used to answer your first question. This corollary states that any second countable space which contains a perfect set is not Q (actually it uses a weaker hypothesis to make a stronger conclusion). Staring at the proof of the relevant results towards proving this corollary you need to convince yourself that DC is enough. Given this, if every set $A \subseteq \mathbb R$ contains a perfect set (for instance in the Solovay model) then no uncountable set of reals can be a Q-set.

Putting these two results together also we get (modulo an inaccessible) that the existence of a projective uncountable Q-set is independent of ZFC.

I do not know if anyone has thought about these definability issues further, though I would check the references of the Miller book to start.

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  • $\begingroup$ Thanks for the answer and for the reference. I have a question which may be a bit silly: once we force Martin's Axiom, how can know that the set $A$ is still $\Pi_1^1$ in the forcing extension? Thanks $\endgroup$
    – Lorenzo
    Apr 23, 2022 at 10:20
  • $\begingroup$ There are two ways of interpreting this question. A $\Pi^1_1$ set is a set defined by a formula of the form $\forall x \in \mathbb R \varphi(x, y)$ with $\varphi$ arithmetic. Of course in any forcing extension this formula still exists and so it re-evaluates to a $\Pi^1_1$ set. However, when we force, two (non mutually exclusive) things can happen - new reals $x \notin V$ can be forced to satisfy this formula and/or the set of x satisfying the formula can become countable. If we force MA with ccc forcing over $L$ then of course the latter does not happen. The point is that in L we can... [1/3] $\endgroup$
    – Corey Bacal Switzer
    Apr 25, 2022 at 12:58
  • $\begingroup$ ...in L we can find a $\Pi^1_1$ set $A$ so that $\mathsf{ZFC} \vdash$``$A \subseteq L \cap \mathbb R$" i.e. it is provable that no non-constructible reals satisfy the defining formula for $A$. There are many examples of such. A particularly obvious one is the set of reals coding an $L_\alpha$ for $\alpha < \omega_1$ (for some fixed coding that makes it constructible). [2/3] $\endgroup$
    – Corey Bacal Switzer
    Apr 25, 2022 at 13:02
  • $\begingroup$ Note that in L none of these sets are Q however - CH implies that there are no uncountable Q sets (in fact $2^{\aleph_0} < 2^{\aleph_1}$ suffices). The point is that for any subset $B$ of any such $A$ there is a ccc forcing to make $B$ relatively $G_\delta$ in $A$ (and this persists upwards) so under MA every such set is relatively $G_\delta$. [3/3] $\endgroup$
    – Corey Bacal Switzer
    Apr 25, 2022 at 13:02

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