Question. Let $\kappa>\mathfrak c$ be a cardinal endowed with the discrete topology and $f:\kappa^\omega\to[0,1]^\omega$ be a continuous map. Is there a point $y\in[0,1]^\omega$ whose preimage $f^{-1}(y)$ is not scattered?

Let us recall that a topological space $X$ is scattered if each non-empty subspace of $X$ contains an isolated point.

An affirmative answer to this question would give an affirmative answer to this problem.

Remark. If $\kappa<\kappa^\omega$, then the answer to the above question is affirmative. Indeed, assuming that every preimage $f^{-1}(y)$, $y\in[0,1]^\omega$, is scattered, we conclude that $|f^{-1}(y)|\le w(f^{-1}(y))\le w(\kappa^\omega)=\kappa$ and hence $|\kappa^\omega|\le\sum_{y\in[0,1]^\omega}|f^{-1}(y)|\le\mathfrak c\cdot \kappa=\kappa<|\kappa^\omega|$, which is a desired contradiction. So, actually, the question concerns cardinals $\kappa=\kappa^\omega>\mathfrak c$. The smallest such cardinal is $\mathfrak c^+$, the successor of the continuum.


1 Answer 1


The continuity of $f$ is not needed. Indeed, suppose to the contrary that $\kappa^\omega$ is a union of the family $\{F_\alpha:\alpha<\frak c\}$ of fibers of $f$. Let $\alpha<\frak c$ be any index. Since the fiber $F_\alpha$ is scattered, there is an injective enumeration $F_\alpha=\{f_{\alpha,\beta}:\beta<\beta_\alpha\}$ such that for each $\beta<\beta_\alpha$, $f_{\alpha,\beta}=(f_{\alpha,\beta,n})_{n\in\omega}$ is an isolated point of the space $F_{\alpha,\beta}=F_\alpha\setminus \{f_{\alpha,\gamma}:\gamma<\beta\}$. Therefore there exists a natural number $m=m(f_{\alpha,\beta})>0$ such that $(f_{\alpha,\beta,n})_{n\in m}\ne(f_n)_{n\in m}$ for each $f=(f_n)_{n\in\omega}\in F_{\alpha,\beta}\setminus\{ f_{\alpha,\beta}\}$. It follows that for each natural $m>0$ the projection $\pi_m:\kappa^\omega\to \kappa^m$ is injective on the set $F_{\alpha,m}=\{f\in F_{\alpha}: m(f)=m\}$. Since $\kappa>\frak c$, by induction we can construct a sequence $g=(g_m)_{m\in\omega}$ of elements of $\kappa$ such that $g|m\not\in\pi_m(F_{\alpha,m-1})$ for each $\alpha<\frak c$ and $m>1$. It follows that $g$ does not belong to any $F_{\alpha,m}$, a contradiction.

  • $\begingroup$ Ну все, тепер ідеально! Congratulations! Треба ще згадати до чого мені та проблема була потрібна. $\endgroup$ Feb 21, 2020 at 12:39

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