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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.

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  • $\begingroup$ Strengthening the question. Is there a condensation from $D^{\aleph_0}$ onto $N^N$? $\endgroup$ – Alexander Osipov Jun 13 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 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]. $\endgroup$ – Alexander Osipov Jun 13 at 13:03
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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.

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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 $(\aleph_1\times(\omega^\omega\sqcup\aleph_1))^\omega\cong \aleph_1^\omega$.

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  • $\begingroup$ Taras, Thanks!!! $\endgroup$ – Alexander Osipov Jun 14 at 4:53

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