Set Theory – Baire Subset of Reals That is Not Suslin Measurable

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EDIT: The definition of a Suslin measurable set I wrote here is incorrect. It should be that $\mathcal{S}$ contains the field (or algebra) of open subsets of ${}^\omega\omega$ (or, in other words, it contains all open and closed subsets of ${}^\omega\omega$). I do not intend to re-ask this question, so the question and answers here refer to the unchanged definition below.

Recall that a Suslin scheme is a set of subsets of reals ${}^\omega\omega$ of the form:
$$
\langle X_s : s \in {}^{<\omega}\omega\rangle
$$
and that the Suslin operation $\mathcal{A}$ is an operation that takes a Suslin scheme $\mathcal{X} := \langle X_s : s \in {}^{<\omega}\omega\rangle$ and yield:
$$
\mathcal{A}(\mathcal{X}) := \bigcap_{a \in {}^\omega\omega}\bigcup_{n \in \omega} X_{a\upharpoonright n}
$$
The set of all Suslin measurable sets, call it $\mathcal{S}$, is the smallest set of subsets of reals such that:

  • $\mathcal{S}$ contains all open subsets of ${}^\omega\omega$.
  • $\mathcal{S}$ is closed under the Suslin operation (i.e. if $\mathcal{X}$ is a Suslin scheme in which $X_s \in \mathcal{S}$ for all $s$, then $\mathcal{A}(\mathcal{X}) \in \mathcal{S}$).

A result of Nikodym says that the set of Baire subsets of reals is closed under the Suslin operation (Corollary 4.8 of Todorcevic's Introduction to Ramsey spaces). Thus, every Suslin measurable subset of reals has the Baire property. The questions I have are:

  1. Can we prove, in $\mathsf{ZFC}$ and $\mathsf{ZF}$, that there exists a Baire subset of reals that is not Suslin measurable?

  2. If it is not provable in $\mathsf{ZF}$, is there a well-known model of set theory in which every subset of reals is Suslin measurable?

(Side question: There seems to be very little literature that discusses Suslin measurable sets. Is there another term for such sets?)

EDIT: To clarify, a subset $X$ of real is Baire (or has the Baire property) if $X = U \, \triangle \, M$, where $U$ is open and $M$ is meagre.

Best Answer

While Gabe's answer is deleted ("One shouldn't try to work in ZF at 5am"), let me work in ZFC.

(a) The usual middle-thirds Cantor set $C$ is nowhere dense in $\mathbb R$. It has cardinal $\mathfrak c = 2^{\aleph_0}$. So every subset of $C$ is again nowhere dense in $\mathbb R$. Every subset of $C$ has the property of Baire. There are at least $2^{\mathfrak c}$ sets with the property of Baire. [In fact, there are exactly $2^{\mathfrak c}$ sets with the property of Baire.]

(b) Start with any family $\mathscr U$ of sets, and define $$ \mathcal A(\mathscr U) = \{ \mathcal A(\mathcal X) : \mathcal{X} = \langle X_s : s \in {}^{<\omega}\omega\rangle , X_s \in \mathscr U \text{ for all } s \in {}^{<\omega}\omega\} . $$ The Suslin operation is idempotent. That is, if $\mathscr U$ is any family of sets, then $\mathcal A(\mathcal A(\mathscr U)) = \mathcal A(\mathscr U)$.
The family of "Suslin measurable sets" (a.k.a. coanalytic sets) is $\mathcal A(\mathscr G)$, where $\mathcal G$ is the family of all open subsets of $\mathbb R$.

(c) Let $\mathscr G_0 = \{(a,b) : a,b\in\mathbb Q, a<b\}$. Then $\mathcal A(\mathscr G) = \mathcal A(\mathscr G_0)$. Since $\mathscr G_0$ is countable and ${}^{<\omega}\omega$ is countable, there are at most $\aleph_0 ^{\aleph_0} = \mathfrak c$ Souslin schemes in $\mathscr G_0$. We conclude that $\mathcal A(\mathscr G_0)$ has cardinal at most $\mathfrak c$. So there are at most $\mathfrak c$ Suslin measurable sets in $\mathbb R$. [In fact, there are exactly $\mathfrak c$ Suslin measurable sets.]

(d) Conclude there is a set with the property of Baire that is not Suslin measurable.

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