# Boolean algebra of ambiguous Borel class

Suppose $$X$$, $$Y$$ are uncountable compact metric spaces and $$\Delta^0_\xi(X)$$, $$\Delta^0_\xi(Y)$$ ($$2\le\xi\le\omega_1$$) are the Boolean algebras of Borel sets of ambiguous class $$\xi$$. So for $$\xi=2$$ these are the simultaneously $$\mathbf{F}_\sigma$$ and $$\mathbf{G}_\delta$$ sets of $$X$$ and $$Y$$. When are $$\Delta^0_2(X)$$ and $$\Delta^0_2(Y)$$ isomorphic as Boolean algebras? Since each singleton $$\{x\}$$ is $$\Delta^0_2$$, the Boolean algebras are totally atomic, so any Boolean isomorphism must be realized by a point bijection (Borel isomorphism) between $$X$$ and $$Y$$. What if $$X=[0,1]$$ and $$Y=2^\omega$$? This must be well-known; a reference would be appreciated.

• A Borel isomorphism is easy to construct from between $2^\omega$ and $[0,1]$. Use binary coding to get a Borel isomorphism between $2^\omega$ minus a countable subset and $[0,1]$. Then it is clear that for $X$ uncountable, $X$ is Borel-isomorphic to $X$ minus any countable subset.
– YCor
Commented Jul 20, 2022 at 12:52
• The Borel isomorphisms described by @YCor are in fact quite low in the Borel hierarchy and therefore provide isomorphisms between $\Delta^0_\xi(X)$ and $\Delta^0_\xi(Y)$ once $\xi$ is above a certain small value (probably $2$, but I haven't checked that). Commented Jul 20, 2022 at 13:23

## 1 Answer

This is a very interesting question whose answer depends on dimension properties of the spaces $$X,Y$$.

First we introduce a suitable terminology. A function $$f:X\to Y$$ between topological spaces is called

$$\bullet$$ $$\mathcal B$$-measurable where $$\mathcal B$$ is a Borel class if $$\{f^{-1}[B]:B\in\mathcal B(Y)\}\subseteq \mathcal B(X)$$;

$$\bullet$$ $$\sigma$$-continuous (resp. $$\bar\sigma$$-continuous) if there exists a countable (closed) cover $$\mathcal C$$ of $$X$$ such that for every $$C\in\mathcal C$$ the restriction $$f{\restriction}_C$$ is continuous;

$$\bullet$$ a $$\sigma$$-homeomorphism (resp. $$\bar\sigma$$-homeomorphism) if $$f$$ is bijective and the functions $$f$$ and $$f^{-1}$$ are $$\sigma$$-continuous (resp. $$\bar\sigma$$-continuous).

Topological spaces $$X,Y$$ are defined to be $$\sigma$$-homeomorphic (resp. $$\bar\sigma$$-homeomorphic) if there exists a $$\sigma$$-homeomorphism (resp. $$\bar\sigma$$-homeomorphism) $$h:X\to Y$$.

The following theorem was proved by Jayne and Rogers in J. Math. Pure Appl. 61 (2) (1982), 177-205.

Theorem (Jayne-Rogers): A function $$f:X\to Y$$ between analytic spaces if $$\bar\sigma$$-continuous if and only if $$f$$ is $$\Pi^0_2$$-measurable.

This theorem was completed by the following deep theorem that follows from Theorem 1.1 and Lemnma 6.1 in this paper of Pawlikowski and Sabok.

Theorem (Pawlikowski-Sabok): Let $$n\in\mathbb N$$. Any $$\Pi^0_n$$-measurable function between analytic spaces is $$\sigma$$-continuous.

Now let us return to the Booleans algebras $$\Delta^0_\xi(X)$$. Since the $$\Pi^0_2$$-measurability is equivalent to the $$\Delta^0_2$$-measurability, the Jayne-Rogers Theorem implies

Theorem $$\Delta^0_2$$. For analytic spaces $$X,Y$$ the Boolean algebras $$\Delta^0_2(X)$$ and $$\Delta^2(Y)$$ are isomorphic if and only if the spaces $$X,Y$$ are $$\bar\sigma$$-homeomorphic.

Since $$\bar\sigma$$-homeomorphisms preserve the dimension of metrizable separable spaces (see Corollary 3.10 of this paper), Theorem $$\Delta^0_2$$ implies

Corollary. If for analytic spaces $$X,Y$$ the Boolean algebras $$\Delta^0_2(X)$$ and $$\Delta^0_2(Y)$$ are isomorphic, then $$\dim(X)=\dim(Y)$$.

Corollary. The Boolean algebras $$\Delta^0_2(X)$$ and $$\Delta^0_2(Y)$$ of the spaces $$X=[0,1]$$ and $$Y=2^\omega$$ are not isomorphic.

On the other hand, Pawlikowski-Sabok Theorem implies

Theorem $$\sigma$$. Let $$X,Y$$ be analytic spaces. If for some $$n\in\mathbb N$$ the Boolean algebras $$\Delta^0_n(X)$$ and $$\Delta^0_n(Y)$$ are isomorphic, then the spaces $$X,Y$$ is $$\sigma$$-homeomorphic.

Proof. Since $$\Delta^0_n(X)\cong\Delta^0_n(Y)$$, there exists a bijective function $$f:X\to Y$$ such that the functions $$f,f^{-1}$$ are $$\Delta^0_n$$-measurable. By Theorem 22.21 in Kechris' book, any set $$A\in\Sigma^0_{n}(Y)$$ can be written as the union $$\bigcup_{n\in\omega}A_n$$ of pairwise disjoint sets $$A_n\in\Delta^0_n(Y)$$. Then $$f^{-1}[A]=\bigcup_{n\in\omega}f^{-1}[A_n]$$ is the countable union of sets $$f^{-1}[A_n]\in \Delta^0_n(X)\subseteq\Sigma^0_n(X)$$ and hence $$f^{-1}[A]\in\Sigma^0_n(X)$$. This means that the function $$f$$ is $$\Sigma^0_n$$-measurable and also $$\Pi^0_n$$-measurable. By Pawlikowski-Sabok Theorem, $$f$$ is $$\sigma$$-continuous.

By analogy we can prove that $$f^{-1}$$ is $$\sigma$$-continuous, which means that $$f$$ is a $$\sigma$$-homeomorphism between $$X$$ and $$Y$$. $$\quad\square$$

Theorem $$\sigma$$ will help us to prove the following characterization

Theorem $$\Delta^0_n$$. For Polish spaces $$X,Y$$ the following conditions are equivalent:

1. The Boolean algebras $$\Delta^0_3(X)$$ and $$\Delta^0_3(Y)$$ are isomorphic;
2. The Boolean algebras $$\Delta^0_n(X)$$ and $$\Delta^0_n(Y)$$ are isomorphic for all finite $$n\ge 3$$;
3. The Boolean algebras $$\Delta^0_n(X)$$ and $$\Delta^0_n(Y)$$ are isomorphic for some finite $$n\ge 3$$;
4. The spaces $$X,Y$$ are $$\sigma$$-homeomorphic.

Proof. $$(1)\Rightarrow(2)$$ Assuming that the Boolean algebras $$\Delta^0_3(X)$$ and $$\Delta^0_3(Y)$$ are isomorphic, find a bijection $$f:X\to Y$$ such that the maps $$f$$ and $$f^{-1}$$ are $$\Delta^0_3$$-measurable. Repeating the argument of the proof of Theorem $$\sigma$$, we can show that the maps $$f,f^{-1}$$ are $$\Sigma^0_3$$-measurable and $$\Pi^0_3$$-measurable. Next, by induction, it can be shown that $$f$$ and $$f^{-1}$$ are $$\Sigma^0_n$$-measurable and $$\Pi^0_n$$-measurable for all finite $$n\ge 3$$ and hence $$f,f^{-1}$$ are $$\Delta^0_n$$-measurable, witnessing that the Boolean algebras $$\Delta^0_n(X)$$ and $$\Delta^0_n(Y)$$ are isomorphic for all finite $$n\ge 3$$.

The implication $$(2)\Rightarrow(3)$$ is trivial and $$(3)\Rightarrow(4)$$ follows from Theorem $$\sigma$$.

$$(4)\Rightarrow(1)$$ Assume that the spaces $$X,Y$$ are $$\sigma$$-homeomorphic and fix a $$\sigma$$-homeomorphism $$h:X\to Y$$.

Claim. There are disjoint countable covers $$(X_n)_{n\in\omega}$$ and $$(Y_n)_{n\in\omega}$$ of the spaces $$X$$ and $$Y$$, respectively, such that for every $$n\in\omega$$ the restriction $$h{\restriction}_{X_n}:X_n\to Y_n$$ is a homeomorphism.

Proof. Find countable covers $$\{A_n\}_{n\in\omega}$$ and $$\{ B_n\}_{n\in\omega}$$ of $$X$$ and $$Y$$, respectively, such that for every $$n\in\omega$$ the restrictions $$h{\restriction}_{A_n}$$ and $$h^{-1}{\restriction}_{B_n}$$ are continuous. For every $$n,m\in\omega$$ consider the sets $$A_{n,m}=A_n\cap h^{-1}[B_m]$$ and $$B_{n,m}=h[A_n]\cap B_m$$, and observe that the map $$h_{n,m}=h{\restriction}_{A_{n,m}}:A_{n,m}\to B_{n,m}$$ is a homeomorphism. Finally, put $$X_n=A_{2^i(2j+1)-1}$$ and $$Y_n=B_{2^i(2j+1)-1}$$ where $$i,j\in\omega$$ are unique numbers such that $$n=2^i(2j+1)-1$$. $$\quad\square$$

Let $$(X_n)_{n\in\omega}$$ and $$(Y_n)_{n\in\omega}$$ be disjoint covers of $$X$$ and $$Y$$ from the Claim. By the Lavrentiev Theorem 3.9 in Kechris' book, for every $$n\in\omega$$, the homeomorphism $$h{\restriction}_{X_n}:X_n\to Y_n$$ can be extended to a homeomorphism $$\tilde h_n:\tilde X_n\to\tilde Y_n$$ of some $$G_\delta$$-sets $$\tilde X_n$$ and $$\tilde Y_n$$ in the Polish spaces $$X$$ and $$Y$$, respectively.

By Theorem 22.16 in Kechris' book, the class $$\Sigma^0_3$$ has the generalized reduction property, so we can find disjoint covers $$(\check X_n)_{n\in\omega}$$ and $$(\check Y_n)_{n\in\omega}$$ of the Polish spaces $$X,Y$$ by $$G_{\delta\sigma}$$-sets such that $$\check X_{n}\subseteq\tilde X_{n}$$ and $$\check Y_{n}\subseteq \tilde Y_{n}$$.

For every $$n,m\in\omega$$, consider the $$G_{\delta\sigma}$$-sets $$X_{n,m}=\check X_n\cap \tilde h_n^{-1}[\check Y_m]$$ and $$Y_{n,m}=\tilde h_{\check X_n}\cap \check Y_{m}$$ in $$X$$ and $$Y$$, respectively, and observe that $$\tilde h_n{\restriction}_{X_{n,m}}:X_{n,m}\to Y_{n,m}$$ is a homeomorphism. It is easy to see that $$(X_{n,m})_{n,m\in\omega}$$ and $$(Y_{n,m})_{n,m\in\omega}$$ are disjoint covers of $$X$$ and $$Y$$ by $$G_{\delta\sigma}$$-sets. Let $$f:X\to Y$$ be a unique bijective map such that $$f{\restriction}_{X_{n,m}}=\tilde h_n{\restriction}_{X_{n,m}}$$. It is easy to see that $$\{f[A]:A\in\Sigma^0_3(X)\}=\Sigma^0_3(Y)$$ and hence $$f$$ induces an isomorphism of the Boolean algebras $$\Delta^0_3(X)$$ and $$\Delta^0_3(Y)$$. $$\quad\square$$

A topological space is countable-dimensional if it can be written as the countable union of zero-dimensional spaces. It is well-known that each finite-dimensional separable metrizable space is countable-dimensional. On the other hand, the Hilbert cube $$[0,1]^\omega$$ is not countable-dimensional.

The countable-dimensionality is preserved by $$\sigma$$-homeomorphisms.

Theorem cd. Assume that $$X,Y$$ are $$\sigma$$-homeomorphic spaces. If the space $$X$$ is countable-dimensional, then the space $$Y$$ is countable-dimensional, too.

Proof. Let $$h:X\to Y$$ be a $$\sigma$$-homeomorphism. Then there exist countable disjoint covers $$\{X_n\}_{n\in\omega}$$ and $$\{Y_n\}_{n\in\omega}$$ of $$X$$ and $$Y$$, respectively, such that for every $$n\in\omega$$ the restrictions $$h{\restriction}_{X_n}$$ and $$h^{-1}{\restriction}_{Y_n}$$ are continuous. For every $$n,m\in\omega$$ consider the sets $$X_{n,m}=X_n\cap h^{-1}[Y_m]$$ and $$Y_{n,m}=h[X_n]\cap Y_m$$, and observe that $$h{\restriction}_{X_{n,m}}:X_{n,m}\to Y_{n,m}$$ is a homeomorphism. If the space $$X$$ is countable-dimensional, then it has a disjoint cover $$\{Z_k\}_{k\in\omega}$$ by zero-dimensional subspaces. Then $$\{h[X_{n,m}\cap Z_k]:n,m,k\in\omega\}$$ is a countable cover of $$Y$$ by zero-dimensional subspaces, witnessing that the space $$Y$$ is countable-dimensional. $$\quad\square$$

Theorems cd and $$\Delta^0_n$$ imply

Corollary. For every $$n\in\mathbb N$$ the Booleans algebras $$\Delta^0_n(\{0,1\}^\omega)$$ and $$\Delta^0_n([0,1]^\omega)$$ are not isomorphic.

On the other hand, we have

Theorem $$\Delta^0_3$$. Any countable-dimensional uncountable Polish spaces $$X,Y$$ are $$\sigma$$-homeomorphic. Consequently, the Boolean algebras $$\Delta^0_3(X)$$ and $$\Delta^0_3(Y)$$ are isomorphic.

Theorem $$\Delta^0_3$$ follows from Theorem $$\Delta^0_n$$ and

Lemma $$\sigma$$. Each countable-dimensional uncountable Polish space $$X$$ admits a cover $$\{X_n\}_{n\in\omega}$$ such that for any distinct numbers $$n,m\in\omega$$ the following conditions hold:

1. $$X_n\cap X_m=\emptyset$$;
2. $$X_{2n}$$ is a singleton;
3. $$X_{2n+1}$$ is homeomorphic to $$\omega^\omega$$.

Proof. By definition, the countable-dimensional space $$X$$ can be written as the union $$X=\bigcup_{n\in\omega}X_n$$ of zero-dimensional spaces. Using Lavrentiev Theorem 3.9 in Kechris' book, we can enlarge each $$X_n$$ to a zero-dimensional $$G_\delta$$-set and assume that $$X_n$$ is a $$G_\delta$$-set in $$X$$. By Theorem 22.16 in Kechris' book, the class $$\mathbf\Sigma^0_3$$ of absolute $$G_{\delta\sigma}$$-sets has the generalized reduction property, which allows us to find a sequence $$(X_n')_{n\in\omega}$$ of pairwise disjoint $$G_{\delta\sigma}$$-sets such that $$X=\bigcup_{n\in\omega}X_n'$$ and $$X_n'\subseteq X_n$$ for every $$n\in\omega$$. Each $$G_{\delta\sigma}$$-set $$X_n'$$ can be written as a disjoint union of $$G_\delta$$-sets. This without loss of generality, we can assume that the $$G_\delta$$-sets $$X_n$$, $$n\in\omega$$, are pairwise disjoint. This shows that the countable-dimensional Polish space $$X$$ admits a countable cover $$\mathcal X=\{X_n\}_{n\in\omega}$$ by pairwise disjoint zero-dimensional Polish subspaces. Since $$X$$ is uncountable, one of the $$G_\delta$$-sets, say $$X_0$$ is uncountable. Then $$X_0$$ contains a family $$\{X_{0,n}\}_{n\in\omega}$$ of pairwise disjoint subsets, homeomorphic to the Cantor cube $$2^\omega$$. Replacing the cover $$\mathcal X$$ by $$\{X_0\setminus\bigcup_{n\in\omega}X_{0,n},X_{0,n}:n\in\omega\}\cup\{X_n:n\ge 1\}$$, we can assume that the cover $$\mathcal X$$ contains infinitely many sets homeomorphic to the Cantor cube $$2^\omega$$. By Cantor-Bendixson Theorem, each Polish space $$P$$ can be written as the disjoint union $$C\cup D$$ of an open countable subspace $$C$$ and a closed crowded (= without isolated points) subspace $$D$$. Then we can assume that each set in $$\mathcal X$$ is either countable or crowded. Moreover, replacing each countable set in $$\mathcal X$$ by the union of singetons, we can assume that each countable set in $$\mathcal X$$ is a singleton. Therefore, $$X$$ has a countable disjoint cover $$\mathcal X$$ whose elements are either singletons or crowded zero-dimensional Polish spaces. Let $$\mathcal X_1=\{C\in\mathcal X:|C|=1\}$$ and $$\mathcal X_c=\mathcal X\setminus\mathcal X_1$$. For each crowded space $$C\in\mathcal X_c$$ choose a countable dense set $$D_C$$ in $$C$$ and observe that the space $$C\setminus D_C$$ is Polish, crowded and nowhere locally compact. By Aleksandrov-Uryson Theorem 7.7 in Kechris' book, the space $$C\setminus D_C$$ is homeomorphic to the Baire space $$\omega^\omega$$. Now we see that the disjoint countable cover $$\mathcal X_1\cup \bigcup_{C\in\mathcal X_c}\{C\setminus D_C,\{x\}:x\in D_C\}$$ of $$X$$ consists of infinitely many singletons and infinitely many sets homeomorphic to $$\omega^\omega$$. $$quad\square$$

Our final theorem shows that the finite ordinal $$n$$ in Theorem $$\Delta^0_n$$ cannot be replaced by $$\omega$$.

Theorem $$\Delta^0_\omega$$. For any uncountable Polish spaces $$X,Y$$ the Boolean algebras $$\Delta^0_\omega(X)$$ and $$\Delta^0_\omega(Y)$$ are isomorphic.

Proof. By Theorem 22.21 in Kechris' book, there exists a continuous bijective map $$h:Z\to Y$$ from a zero-dimensional Polish space $$Z$$ such that $$Z$$ has a countable base $$\mathcal B$$ of the topology such that $$h[B]\in\Delta^0_2(Y)$$ for any set $$B\in\mathcal B$$. This property implies that for every $$n\in\mathbb N$$ and $$A\in\Sigma^0_n(Z)$$ we have $$h[A]\in\Sigma^0_{n+1}(Y)$$. Consequently, $$h$$ induces an isomorphism of the Boolean algebras $$\Delta^0_\omega(Z)$$ and $$\Delta^0_\omega(Y)$$.

By analogy, we can find an uncountable zero-dimensional Polish space $$P$$ such that the Boolean algebras $$\Delta^0_\omega(P)$$ and $$\Delta^0_\omega(X)$$ are isomorphic (denoted by $$\Delta^0_\omega(P)\cong\Delta^0_\omega(X)$$). By Theorem $$\Delta^0_3$$, the Boolean algebras $$\Delta^0_3(P)$$ and $$\Delta^0_3(Z)$$ are isomorphic, which implies that $$\Delta^0_\omega(P)\cong\Delta^0_\omega(Z)$$ and hence $$\Delta^0_\omega(X)\cong\Delta^0_\omega(P)\cong\Delta^0_\omega(Z)\cong\Delta^0_\omega(Y).\quad\square$$

• This certainly answers the posted question. Thanks. But it suggests the following: What is the least $\xi$ such that $\Delta^0_\xi(X)$ and $\Delta^0_\xi(Y)$ are isomorphic as Boolean algebras for all uncountable compact metric $X$ and $Y$. I believe the Borel isomorphism between $X$ and $Y$ mentioned above would suggest that $\xi=\omega$ would be an upper bound. Commented Jul 20, 2022 at 19:50
• @FredDashiell I expect that this $\xi$ can be equal to 3, at least it is certainly 3 for $[0,1]$ and $2^\omega$. In any case, very good question. Commented Jul 20, 2022 at 21:05
• @FredDashiell For any uncountable finite-dimensional compact metrizable space $X$ the Boolean algebra $\Delta^0_3(X)$ is isomorphic to $\Delta^0_3(2^\omega)$. The reason: $X$ can be written as the disjoint union of countably many zero-dimensional Polish spaces. So, the crucial question is whether the Boolean algebras $\Delta^0_3(2^\omega)$ and $\Delta^0_3([0,1]^\omega)$ are isomorphic? Commented Jul 20, 2022 at 21:18
• @FredDashiell I added to my answer the proof of the fact that for any uncountable countable-dimensional Polish spaces $X,Y$ the Boolean algebras $\Delta^0_3(X)$ and $\Delta^0_3(Y)$ are isomorphic. The proof essentially uses the countable-dimensionality and doe not work for Hilbert cube. Commented Jul 21, 2022 at 6:50
• Very nice! A rather complete picture. We know some facts about the Stone spaces of the Boolean algebras $\Delta^0_\xi(X)$ but not in sufficient detail to distinguish them for different spaces $X$ and the same $\xi$. These Boolean algebras all satisfy a weak countable interpolation property: if $S_1$ and $S_2$ are countable subsets with $a\le b$ for all $a\in S_1$ and $b\in S_2$, AND $\bigwedge_{a\in S_1,b\in S_2}\{b-a\}=0$ then there exists $c$ between $S_1$ and $S_2$. Commented Jul 21, 2022 at 23:59