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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.

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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Jul 20, 2022 at 12:52
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    $\begingroup$ 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). $\endgroup$ Commented Jul 20, 2022 at 13:23

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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 $$\Delta^0_\omega(X)\cong\Delta^0_\omega(P)\cong\Delta^0_\omega(Z)\cong\Delta^0_\omega(Y).\quad\square$$

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  • $\begingroup$ 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. $\endgroup$ Commented Jul 20, 2022 at 19:50
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    $\begingroup$ @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. $\endgroup$ Commented Jul 20, 2022 at 21:05
  • $\begingroup$ @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? $\endgroup$ Commented Jul 20, 2022 at 21:18
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    $\begingroup$ @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. $\endgroup$ Commented Jul 21, 2022 at 6:50
  • $\begingroup$ 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 $S_2$. $\endgroup$ Commented Jul 21, 2022 at 23:59

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