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If the zero-dimensional set $X$ is not closed, then the answer is "no". To construct a suitable example, take any open bounded neighborhood $U\subset\mathbb R^2$ of zero, whose boundary $\partial U$ …

As John Samples noted in his comment, Dranishnikov's Theory of cohomological dimension implies the positive answer to this problem for compact (even $\sigma$-compact) metrizable spaces. Namely, accord …

A topological space $X$ is called totally disconnected if for any distinct points $x,y\in X$ there exists a clopen set $U\subseteq X$ such that $x\in U$ and $y\notin U$. In 1973 Roman Pol proved that …

As asked Dusan Repovs (who is an expert in the theory of topological manifolds), and he sent me the following answer: This is indeed best possible result, since whenever a product of two spaces is a …

Problem. Does a compact metric space of finite packing dimension admit an equi-Hölder embedding into a Hilbert space? A map $f:X\to Y$ between metric spaces $(X,d_X)$, $(Y,d_Y)$ is called equi-Hölder …

For every $n\in\mathbb N$ consider the space $$X_n=\{x\in\mathbb R^\omega:\lvert x^{-1}(0)\rvert\ge n\}.$$ Theorem. For any positive integer numbers $n<m$, the spaces $X_n$ and $X_m$ are not homeomorp …

The answer to both questions is affirmative. Theorem 1. The complete Erdos space $\mathfrak E_c$ has a self-homeomorphism whose every orbit is dense in $\mathfrak E_c$. Proof. We use a known result …

A topological space $X$ is called $\bullet$ totally disconnected if for any distinct points $x,y\in X$ there exists a clopen set $U\subseteq X$ such that $x\in U$ and $y\notin U$; $\bullet$ hereditari …

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{\ …

Let $\omega_{cf}$ be the countable space $\omega=\{0,1,2,3,\dots\}$ endowed with the cofinite topology $$\tau_{cf}=\{\emptyset\}\cup\{U\subseteq\omega:\omega\setminus U\mbox{ is finite}\}.$$ It is eas …

The Hilbert cube can be written as the union of two punctiform spaces. Just take any Bernstein set $X\subset[0,1]^\omega$ and observe that compact subsets in $X$ and $Y=[0,1]^\omega\setminus X$ are at …

Such an enumeration does not exist. To derive a contradiction, take any family of Cantor sets $(C_\alpha)_{\alpha\in A}$ in $[0,1]$, indexed by an uncountable subset $A\subset[0,1]$. Let $\mathcal …

This question was motivated by this MO-question asking about the example of a hereditarily disconnected metrizable separable space, which is not the union of countably many totally disconnected subspa …

This question (in a bit different form) I leaned from Olena Karlova. Question. Let $f:X\to Y$ be a bijective continuous map between metrizable separable spaces such that for every open set $U\subset …

This question is motivated by this problem of Dominic van der Zypen. Problem. Let $X=\prod_{i=1}^nX_i$ be the Tychonoff product of linearly ordered compact Hausdorff spaces $X_1,\dots,X_n$. Is it tru …