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 wellknown; a reference would be appreciated.

2$\begingroup$ 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 Borelisomorphic to $X$ minus any countable subset. $\endgroup$– YCorCommented Jul 20, 2022 at 12:52

1$\begingroup$ 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). $\endgroup$– Andreas BlassCommented 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), 177205.
Theorem (JayneRogers): 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 (PawlikowskiSabok): 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 JayneRogers 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, PawlikowskiSabok 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 PawlikowskiSabok 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:
 The Boolean algebras $\Delta^0_3(X)$ and $\Delta^0_3(Y)$ are isomorphic;
 The Boolean algebras $\Delta^0_n(X)$ and $\Delta^0_n(Y)$ are isomorphic for all finite $n\ge 3$;
 The Boolean algebras $\Delta^0_n(X)$ and $\Delta^0_n(Y)$ are isomorphic for some finite $n\ge 3$;
 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 countabledimensional if it can be written as the countable union of zerodimensional spaces. It is wellknown that each finitedimensional separable metrizable space is countabledimensional. On the other hand, the Hilbert cube $[0,1]^\omega$ is not countabledimensional.
The countabledimensionality is preserved by $\sigma$homeomorphisms.
Theorem cd. Assume that $X,Y$ are $\sigma$homeomorphic spaces. If the space $X$ is countabledimensional, then the space $Y$ is countabledimensional, 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 countabledimensional, then it has a disjoint cover $\{Z_k\}_{k\in\omega}$ by zerodimensional subspaces. Then $\{h[X_{n,m}\cap Z_k]:n,m,k\in\omega\}$ is a countable cover of $Y$ by zerodimensional subspaces, witnessing that the space $Y$ is countabledimensional. $\quad\square$
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 countabledimensional 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 countabledimensional 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:
 $X_n\cap X_m=\emptyset$;
 $X_{2n}$ is a singleton;
 $X_{2n+1}$ is homeomorphic to $\omega^\omega$.
Proof. By definition, the countabledimensional space $X$ can be written as the union $X=\bigcup_{n\in\omega}X_n$ of zerodimensional spaces. Using Lavrentiev Theorem 3.9 in Kechris' book, we can enlarge each $X_n$ to a zerodimensional $G_\delta$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 countabledimensional Polish space $X$ admits a countable cover $\mathcal X=\{X_n\}_{n\in\omega}$ by pairwise disjoint zerodimensional 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 CantorBendixson 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 zerodimensional 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 AleksandrovUryson 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 zerodimensional 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 zerodimensional 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 $$\Delta^0_\omega(X)\cong\Delta^0_\omega(P)\cong\Delta^0_\omega(Z)\cong\Delta^0_\omega(Y).\quad\square$$

$\begingroup$ 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

1$\begingroup$ @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 finitedimensional 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 zerodimensional 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

1$\begingroup$ @FredDashiell I added to my answer the proof of the fact that for any uncountable countabledimensional Polish spaces $X,Y$ the Boolean algebras $\Delta^0_3(X)$ and $\Delta^0_3(Y)$ are isomorphic. The proof essentially uses the countabledimensionality 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 $\bigwedge_{a\in S_1,b\in S_2}\{ba\}=0$ then there exists $c$ between $S_1$ and $S_2$. $\endgroup$ Commented Jul 21, 2022 at 23:59