ИСТИНА

Several examples concerning the covering dimension $\dim_0$ in the sense of Kat\v etov of topological groups are presented. Recall that, given a topological space $X$, $\dim_0 X$ is the least integer $n\ge -1$ such that any finite cozero cover of $X$ has a finite cozero refinement of order $n$, provided that such an integer exists. If it does not exist, then $\dim_0 X = \infty$. For normal spaces, this dimension coincides with the covering dimension $\dim$ in the sense of \v Cech, whose definition coincides with that of $\dim_0$ in which ``cozero'' is everywhere replaced by ``open.'' In all of the examples ``$\dim_0$'' can be replaced by $\dim$. A space $X$ with $\dim_0(X)=0$ is said to be \emph{strongly zero-dimensional}. All examples are based on the following one. \begin{uexample} There exist spaces $C_1$ and $C_2$ with the following properties: \begin{enumerate} \item $\dim_0 C_i=0$ for $i=1,2$; \item $\dim_0 (C_1\times C_2)>0$; \item $C_1^n$ is Lindel\"of for each $n\in \mathbb N$; \item the underlying set of the space $C_1$ is the Cantor set $C$, its topology is finer than that of $C$, and it has a base consisting of sets closed in $C$; \item $C_2$ is second-countable; in fact, $C_2$ is a subspace of the Cantor set $C$. \end{enumerate} \end{uexample} Both spaces $C_1$ and $C_2$ are retracts of certain Abelian topological groups $G_1$ and $G_2$, respectively. Therefore, $C_1\times C_2$ is a retract of (and hence $C$-embedded in) $G_1\times G_2$, which implies that $\dim_0(G_1\times G_2)>0$. The groups $ G_1$ and $G_2$ have the following properties: \begin{enumerate} \item $G_1^n$ is Lindel\"of for every $n\in \mathbb N$; \item $G_1$ is topologically isomorphic to a closed subgroup of a group $M_1$ being a product of zero-dimensional second-countable topological groups; \item $G_2$ is second-countable; \item $\dim_0(G_1)= \dim_0(G_2)=0$; \item $\dim_0(G_1\times G_2)>0$. \end{enumerate} This gives the following examples, which answer old questions of Arkhangel'skii, Shakhmatov, Tkachenko, and Zambakhidze. \begin{theorem} There exist two strongly zero-dimensional Abelian topological groups whose product has positive dimension. \end{theorem} \begin{theorem} There exists a strongly zero-dimensional Abelian topological group containing a closed subgroup of positive dimension. \end{theorem} Recall that a topological group $G$ is said to be \emph{$\mathbb R$-factorizable} if, for every continuous function $f\colon G\to \mathbb R$, there exists a continuous epimorphism $h\colon G \to H$ onto a second-countable topological group $H$ and a continuous function $g\colon H\to \mathbb R$ such that $f = g \circ h$. It is known that any Lindel\"of group is $\mathbb R$-factorizable and that $\dim_0 H\le \dim_0 G$ for any topological group $G$ and any $\mathbb R$-factorizable subgroup $H$ of $G$. This implies the following result. \begin{theorem} There exists an $\mathbb R$-factorizable topological group $G$ and a second-countable topological group $H$ such that the product $G\times H$ is not $\mathbb R$-factorizable. \end{theorem}