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James E Hanson
James E Hanson
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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 can't remember who showed this originally. I also can't remember if separability is necessary, but that doesn't matter for this question.

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 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.

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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James E Hanson
James E Hanson
  • 13.1k
  • 3
  • 37
  • 72

Borel rank collapse in Hilbert cube modulo $\sigma$-ideal generated by zero-dimensional sets

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 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.