**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](https://mathoverflow.net/questions/243064/what-is-the-smallest-density-of-a-metrizable-space-without-countable-separation).


**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.



Does a continuous map from $\kappa^\omega$ to $[0,1]^\omega$ have a non-scattered fiber?
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.


