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$\newcommand{\w}{\omega}\newcommand{\A}{\mathcal A}\newcommand{\F}{\mathcal F}\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal J}\newcommand{\M}{\mathcal M}\newcommand{\N}{\mathcal N}\newcommand{\x …

$\newcommand{\IR}{\mathbb R}$ $\newcommand{\IT}{\mathbb T}$ $\newcommand{\w}{\omega}$ $\newcommand{\e}{\varepsilon}$ Taras Banakh and me proceed a long quest answering a question of ougao at Mathemati …

For any infinite group $G$ we can easily construct a Vitali subset $V$ of $G$. Indeed, pick an arbitrary countable infinite subgroup $H$ of $G$ and let $V$ be a subset of $G$ which intersects each rig …

This is a draft proof of an affirmative answer to Problem 3. Proposition. For any $n\in\mathbb N$, $\varepsilon>0$ and vectors $x_1,\dots,x_n\in\mathbb R^{2n}$ there exists a linear transformation $ …

Yes. The paper [KKM] contains more general results. In particular, by Corollary 1 each regular semitopological group which is a cover semi-complete Baire space is a topological group. It remains to …

There is a huge number of such topologies. Let $G$ be any discrete abelian group. By $\widehat G$ we denote the family (in fact, a group) of its characters, that is of homomorphisms from $G$ to the un …

Clearly, there are examples for the second question. Each Hausdorff abelian paratopological group (that is, a group endowed with a topology making the multiplication continuous) which is not a topolog …

How if we have a neighborhood $U$ of $1$, with $\overline{U^{-1}}$ compact? It seems the following. The answer is positive, because in this case $G$ is a (locally compact) topological group. S …

It seems the following. In general the answer is no, because compactness does not imply sequential compactness. Let $\Bbb T=\{z\in\Bbb C:|z|=1\}$ be the unit circle endowed with the standard topology. …

As I already told at the workshop, Taras Banakh and me solved this problem. But, surprisingly and converse to that I told at the workshop, the answer is affirmative. This is why is good to write a com …