A function $f:X\to Y$ between topological spaces is called

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

$\bullet$ a $\sigma$-homeomorphism if $f$ is bijective and the maps $f$ and $f^{-1}$ are $\sigma$-continuous.

Two topological spaces $X,Y$ are called $\sigma$-homeomorphic if there exists a $\sigma$-homeomorphism $h:X\to Y$.

By Theorem $\Delta^0_3$ from this MO-post, any two countable-dimensional uncountable Polish spaces are $\sigma$-homeomorphic. We recall that a topological space is countable-dimensional if it is the countable union of zero-dimensional spaces.

By a famous result of Pol, there exists a totally disconnected Polish space $P$ which is not countable-dimensional. Since the countable-dimensionality is preserved by $\sigma$-homeomorphisms, Pol's space $P$ is not $\sigma$-homeomorphic to the Cantor cube $2^\omega$. By the same reason, the Hilbert cube $[0,1]^\omega$ is not $\sigma$-homeomorphic to the Cantor cube. Since $[0,1]^\omega$ is not a countable union of totally disconnected spaces, the Hilbert cube $[0,1]^\omega$ is not $\sigma$-homeomorphic to Pol's space $P$.

Therefore, we have three spaces: $2^\omega$, $[0,1]^\omega$, $P$ which are not pairwise $\sigma$-homeomorphic.

Problem. Are there infinitely (countably, continuum) many uncountable Polish spaces which are not pairwise $\sigma$-homeomorphic?

Remark. Repeating the argument of the proof of Theorem $\Delta^0_n$ in this MO-post, it is possible to show that two Polish space $X,Y$ are $\sigma$-homeomorphic if and only if there exist countable partitions $\{X_n\}_{n\in\omega}$ and $\{Y_n\}_{n\in\omega}$ of the spaces $X,Y$ such that $X_n$ and $Y_n$ are homeomorphic Polish spaces for every $n\in\omega$. This characterization should imply that the maximal number of pairwise non-$\sigma$-homeomorphic Polish spaces belongs to the set $\omega\cup\{\omega,\aleph_1,\mathfrak c\}$.

  • $\begingroup$ @Anonymous Any countable-dimensional uncountable spaces are $\sigma$-homeomorphic, is particular $[0,1]^n$ and $[0,1]^m$ are $\sigma$-homeomorphic. $\endgroup$ Jul 23, 2022 at 12:44
  • $\begingroup$ Oh, right. In order to avoid trivialities (like finite spaces of different cardinalities) you probably are interested in spaces of cardinality $\frak c$. $\endgroup$
    – Anonymous
    Jul 23, 2022 at 13:07
  • $\begingroup$ @Anonymous Thank you for this comment. Indeed, I am interested in uncountable Polish spaces. $\endgroup$ Jul 23, 2022 at 13:28

1 Answer 1


There are indeed continuum many. See:

Kihara, T., & Pauly, A. (2022). Point Degree Spectra of Represented Spaces. Forum of Mathematics, Sigma, 10, E31. doi:10.1017/fms.2022.7


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