Is there a condensation (continuous bijective mapping) from $D^{\aleph_0}$ onto a metrizable compact space ?

$D$ - discrete space of cardinality $\aleph_1$.

CH implies it is a positive answer. In general, I don’t know the answer.

  • $\begingroup$ Strengthening the question. Is there a condensation from $D^{\aleph_0}$ onto $N^N$? $\endgroup$ Jun 13, 2020 at 12:12
  • $\begingroup$ What's $N$? also, is it implicit that there exists a condensation of $\aleph_0^{\aleph_0}$ onto a metrizable compact space? $\endgroup$
    – YCor
    Jun 13, 2020 at 12:48
  • $\begingroup$ $N$ is a space of natural numbers. We can assume that $N^N$ is a space of irrational numbers. Sierpinski proved that irrational numbers admits a condensation onto [0,1]. $\endgroup$ Jun 13, 2020 at 13:03

2 Answers 2


If $|D| < \aleph_\omega$, then there is a condensation from $D^\omega$ onto $\omega^\omega$ (the Baire space) if and only if there is a partition of $\omega^\omega$ into exactly $|D|$ Borel sets.

As far as I know, this theorem was first proved by me and Arnie Miller in

"Partitions of $2^\omega$ and completely ultrametrizable spaces," Topology and its Applications 184 (2015), pp. 61-71.

See Theorem 3.9. I still don't know what happens for $|D| \geq \aleph_\omega$.

It is a theorem of Hausdorff that $\omega^\omega$ can be partitioned into $\aleph_1$ Borel sets, regardless of whether $\mathsf{CH}$ holds or not. This, together with the theorem quoted above, provides a positive answer to your question.

It is consistent with any allowed value of $\mathfrak{c}$ that there is a partition of $\omega^\omega$ into $\kappa$ Borel sets for every $\kappa \leq \mathfrak{c}$. (This is Theorem 3.11 in the linked paper.) It is also consistent with any allowed value of $\mathfrak{c}$ that the only partitions of $\omega^\omega$ into Borel sets have size $\leq \aleph_0$, $\aleph_1$, and $\mathfrak{c}$ (see Corollary 3.16 in the linked paper), or size $\leq \aleph_0$, $\aleph_1$, $\kappa$, and $\mathfrak{c}$ for any particular $\kappa$ between $\aleph_1$ and $\mathfrak{c}$ (see Proposition 3.17 and the comments following). Generally, though, it's not known how to get some prescribed list of sizes and not others.


The answer is affirmative and can be derived from

Theorem (Banakh, Plichko). The Hilbert space $\ell_2(\aleph_1)$ condenses onto the Hilbert cube.

By the way, this theorem is related to Problem 1 from the Scottish Book.

In order to answer the original problem of Alexander Osipov, it suffices to construct a condensation of $\aleph_1^\omega$ onto the Hilbert space $\ell_2(\aleph_1)$. This can be done as follows. Using the Torunczyk's characterization of the Hilbert space topology, one can prove that $\ell_2(\aleph_1)$ is homeomorphic to the countable power of the hedgehog $$H(\aleph_1)=\bigcup_{\alpha\in\aleph_1}[0,1]\cdot e_\alpha\subset \ell_2(\aleph_1)$$ where $(e_\alpha)_{\alpha\in\aleph_1}$ is the standard orthonormal basis of the Hilbert space.

Next, using the observation that $[0,1]$ is the union of irrationals and rational, one can show that $H(\aleph_1)$ is a continuous bijective image of $\aleph_1\times (\omega^\omega\sqcup \omega)$ and then $H(\aleph_1)^\omega$ is a continuous bijective image of $(\aleph_1\times(\omega^\omega\sqcup\aleph_1))^\omega\cong \aleph_1^\omega$.

  • $\begingroup$ Taras, Thanks!!! $\endgroup$ Jun 14, 2020 at 4:53

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