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A simple example of seqeuntial topological groups $G,H$ with non-seqeuntial product $G\times H$: $G=\mathbb R^\omega$ and $H=\mathbb R^\infty=\lim \mathbb R^n$ be the direct limit of finite-dimensional Euclidean spaces. The group $G$ is metrizable and $H$ is a sequential $k_\omega$-space. The non-sequentiality of $G\times H$ follows from a result [Banakh, ...


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If a topological group is contractible, then it is locally contractible (using the group operation produce a contraction which does not move the unit of the group). By a classical result of [A. Gleason, R. Palais, On a class of transformation groups, Amer. J. Math. 79 (1957), 631–648], a locally path-finite finite-dimensional topological group is a Lie group ...


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It seems that the answer to this problem is affirmative for compact groups. Indeed, if the union $E$ of $H$-cosets is measurable in $G$, then for every $\epsilon>0$ we can find two compact subsets $A\subset E$ and $B\subset G\setminus B$ such that $\mu(A)+\mu(B)>1-\epsilon$. It follows that $q(A)\subset q(E)\subset q(G)\setminus q(E)$ and the ...


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This problem was been answered in negative by M.Laczkovich (http://www.ams.org/journals/proc/1998-126-06/S0002-9939-98-04241-5/S0002-9939-98-04241-5.pdf). He constructed a proper Borel subgroup $H$ of the real line which cannot be covered by countably many sets $H_i$ with nowhere dense sums $H_i+H_i$. On the other hand, Laczkovich proved that each non-open ...


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It is contractible, according to Kuiper himself who also writes: Putnam and Wintner [5, 6] proved with the help of spectral resolutions that $U_{\mathbf R}$ and hence $GL_{\mathbf R}$ are connected.


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The answer is No. In fact, if the lattice of compatible topologies of some group $G$ is $M_n$ (=the lattice of height three with $n$ atoms), then $n$ must be $1, 2$ or $p+1$ for some prime $p$. Here is why. If $G$ is a group and $N$ is a normal subgroup of $G$, let $T_N$ denote the topology on $G$ whose open sets are the unions of cosets of $N$. Claim 1. ...



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