Given two subsets $A,B$ of the Cantor cube $2^\omega$ we write $A\le\_W B$ (resp. $A\le\_H B$) if there is a continuous (and injective) function $f:2^\omega\to 2^\omega$ such that $f^{-1}(B)=A$. The relation $\le\_W $ is the well-known (and well-studied) Wadge reducibility relation. The relation $\le\_H$ is called *the Hurewicz reducibility* to make associations with some known Hurewicz-type tests (proved by Hurewicz, Louveau, Saint Raymond and others). 


**Problem 1.** *What is known about the relation of Hurewicz reducibility between Borel subsets of the Cantor cube?
Is it equivalent to the Wadge reducibility for sufficiently complex Borel subsets of $2^\omega$?*


More precisely: *Assume that $A,B\subset 2^\omega$ are two Borel subsets of complexity (say) $A,B\notin \Sigma^0\_3\cap \Pi^0\_3$. Does $A\le\_W B$ imply $A\le\_HB$?*


By a result of Louveau and Saint Raymond (1987) this implication is true if $A,B$ belong to the class $\Gamma\setminus \check\Gamma=\{C\in\Gamma:2^\omega\setminus C\notin\Gamma\}$ for some class $\Gamma$ of Borel subsets of $2^\omega$ which is closed under intersections with $G\_\delta$-sets and unions with $F\_\sigma$-sets in $2^\omega$. So, for any $\xi\ge 3$ and any subsets $A,B\in\Pi^0\_\xi\setminus \Sigma^0\_\xi$ we get $A\le\_H B$ and $B\le\_H A$.


Let us say that two topological spaces $X,Y$ are *Hurewicz equivalent* if each of them admits a closed topological embedding into the other.
It is clear that two subsets $A,B\subset 2^\omega$ are Hurewicz equivalent if $A\le\_H B$ and $B\le \_H A$. The converse is not true: the Cantor cube contains two homeomorphic dense $G\_\delta$-sets $A,B$ such that $A\not\le\_H B$.


**Problem 2.** Assume that $A,B\subset 2^\omega$ are two Borel subsets of sifficiently hight Borel complexity (say $A,B\notin \Sigma^0\_3\cap \Pi^0\_3$). Assume that $A,B$ are Hurewicz equivalent. Is $A\le\_H B$?


**Problem 3.** Classify Borel subsets of $2^\omega$ up to Hurewicz equivalence. Is it true that for each Borel subset $B\subset 2^\omega$ of sufficiently hight Borel complexity (say $B\notin\Pi^0\_2\cup \Sigma^0\_2$) the Wadge class $W(A)=\{B\subset 2^\omega:B\le\_W A,\;A\le\_W B\}$ of $A$ contains only finitely many spaces (better, a unique space) that are pairwise Hurewicz non-equivalent?



Hurewicz versus Wadge hierarchy of zero-dimensional Borel sets?
After some thoughts I realized that the answers to Problems 1 and 3 are negative. Namely, each limit Wadge class contains infinitely many Hurewicz non-equivalent spaces. Indeed, take a sequence $(U\_n)\_{n\in\omega}$ of pairwise disjoint clopen subsets of the Cantor cube $2^\omega$ that converge to some point $x\_\infty$. In each set $U\_n$ choose a Borel subset $A\_n$ so that the sequence $(A\_n)\_{n\in\omega}$ is strictly increasing in the Warge hierarchy. Finally, consider the Borel set $A=\{x\_\infty\}\cup\bigcup\_{n\in\omega}A\_n$. It can be shown that for every number $n=\{0,\dots,n-1\}\in\mathbb N$ endowed with the discrete topology the space $A\times n$ is Wadge equivalent to $A$. Nonetheless, for $m>n$ the space $A\times m$ admits no closed embedding into $A\times n$. So, the spaces $A\times n$, $n\in\mathbb N$, are pairwise Hurewicz non-equivalent.


