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Both of the commonly studied $\sigma$-ideals (meager sets and null sets) in Polish spaces with a natural measure (i.e. $\mathbb{R}$, $[0,1]$, $[0,1]^\omega$, $2^{\omega}$, etc.) have the nice property that every Borel set is equivalent to a Borel set of uniformly bounded rank modulo the ideal, specifically for any Borel set $A$,

  • there is an open set $U$ such that $A \Delta U$ is meager, and
  • there is a $G_\delta$ set $B$ such that $A \Delta B$ is null,

where $\Delta$ is the symmetric difference.

A notable result in topological dimension theory is the fact that the Hilbert cube, $[0,1]^\omega$, is strongly infinite dimensional, meaning that it cannot be covered by a countable collection of zero-dimensional sets (a set is zero-dimensional if it has a basis of clopen sets in the induced subspace topology). Note that a union of zero-dimensional sets is not in general itself zero-dimensional. In fact, a metrizable space has topological dimension $n$ if and only if it can be written as a union of $n+1$ zero-dimensional subsets.

This implies that the collection of zero-dimensional subsets of $[0,1]^\omega$ generates a non-trivial $\sigma$-ideal, which doesn't seem to have a standard name. What is interesting about this $\sigma$-ideal is that it is orthogonal to both the meager set $\sigma$-ideal and the null set $\sigma$-ideal in the sense that it contains $([0,1] \setminus \mathbb{Q})^{\omega}$, which is both comeager and full measure.

Question: Let $\mathcal{Z}$ be the $\sigma$-ideal of subsets of $[0,1]^\omega$ generated by zero-dimensional subsets. Does there exist a countable ordinal $\alpha$ such that for any Borel set $A \subseteq [0,1]^\omega$ there exists a Borel set $B \subseteq [0,1]^\omega$ with Borel rank less than $\alpha$ such that $A\Delta B \in \mathcal{Z}$? If such an $\alpha$ exists, what is the optimal $\alpha$?

A possibly relevant result (EDIT: due to Smirnov) is that (with no set theoretic assumptions) every separable metric space is the union of $\aleph_1$ zero-dimensional subsets. I can't remember who showed this originally. I also can't remember if separability is necessary, but that doesn't matter for this question.

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  • $\begingroup$ To get off the ground, can we find a closed set $E$ that is not of the form $E = U \Delta B$ for $U$ open and $B \in \mathcal{Z}$? $\endgroup$ Commented Feb 29, 2020 at 18:54
  • $\begingroup$ I think $\{0\} \times [0,1]^\omega$ should have that property? $\endgroup$ Commented Feb 29, 2020 at 20:18
  • $\begingroup$ The $\sigma$-ideal $\mathcal Z$ has Borel base consisting of $G_{\delta\sigma}$-sets. I do not know if this will help to answer the Question. $\endgroup$ Commented Mar 1, 2020 at 20:03
  • $\begingroup$ I believe that the answer to the Question should be negative: just take a set $H$ of high Borel class in $[0,1]$ and consider the set $A=H\times[0,1]^{\mathbb N}$ in $[0,1]^{\omega}$. How to find a Borel subset $B$ of low class with $A\Delta B\in\mathcal Z$? $\endgroup$ Commented Mar 1, 2020 at 20:09
  • $\begingroup$ The fact that each separable metrizable space is the union of $\aleph_1$ is an old result of Smirnov, cited as Problem 5.1.B in Engelking's "Theory of dimensions finite and infinite''. For more information on the $\sigma$-ideals on the Hilbert cube, see doi.org/10.1007/s11856-015-1235-z $\endgroup$ Commented Mar 1, 2020 at 20:18

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