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Suppose we have an infinite compact (Hausdorff) group $G$, and a subgroup $H\leq G$ which is meagre.

Can $H$ always be covered by a countable family of nowhere dense sets $H_n$ such that $H_n^2$ is still nowhere dense, for each $n$?

Clearly, we can assume that $H$ is dense in $G$. I believe it holds when $H$ is (cointained in) a meagre $F_\sigma$ subgroup, because then we can just take for $H_n$ the closed nowhere dense sets which add up to $H$, and then use the fact that a meager closed set is nowhere dense.

On the other hand, just taking for $H_n$ a family of closed nowhere dense sets covering $H$ is not good enough -- for instance if any $H_n=H_n^{-1}$ is not null, its square will have nonempty interior by Steinhaus theorem.

I don't have much experience with abstract compact groups, so I have hard time even imaging $H,G$ which do not satisfy the assumptions of the previous paragraph...

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  • $\begingroup$ Do you know the answer when G is the additive group of reals? $\endgroup$
    – Ashutosh
    Commented May 28, 2015 at 14:51
  • $\begingroup$ @Ashutosh: The additive group of reals is not compact. If anything, a base example would be the circle group (or more generally, compact metrisable groups). And no, I don't know the answer even in this case. $\endgroup$
    – tomasz
    Commented May 28, 2015 at 20:44
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    $\begingroup$ By reals i meant $2^{\omega}$. Here's a test situation to ponder: add $\aleph_1$ random reals and let H be the group generated by them. Can you write H as a countable union of sets whose self-sums are nowhere dense? $\endgroup$
    – Ashutosh
    Commented May 28, 2015 at 22:17
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    $\begingroup$ I am glad to inform you that an international group of mathematicians is trying to answer your question. :-) We obtained negative answers under some additional axiomatic assumptions (for instance, Martin Axiom) and we are trying to obtain a negative answer in ZFC. Also we are trying to obtain a positive answer when the group $H$ is Borel. $\endgroup$
    – Alex Ravsky
    Commented Jun 2, 2015 at 19:13
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    $\begingroup$ @AlexRavsky: Wow, thanks. Consistency of the negative answer is enough, as far as I am concerned. I might be interested in the special case where $H$ is analytic in the sense that it can be obtained from compact sets through the Souslin operation. Maybe this will prove a better assumption than Borelness? Borel sets tend to be rather odd in non-second-countable spaces. $\endgroup$
    – tomasz
    Commented Jun 2, 2015 at 22:40

1 Answer 1

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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 analytic subgroup $H$ of a Polish locally compact group $G$ can be covered by countably many closed sets of Haar measure zero.

Trying to generalize this result to non-locally compact groups, Laczkovich proved that any non-open analytic subgroup of a Polish group $G$ belongs to the sigma-ideal generated by the family $\mathcal F$ consisting of closed sets $A$ such that any non-empty open subspace of $A$ contains two relatively open non-empty sets $U,V$ with nowhere dense sum $U+V$ and difference $U-V$. Truly speaking, this result of Laczkovich is not quite satisfactory as each closed subset of $G$ containing a dense set of isolated points belongs to the family $\mathcal F$. Consequently, the $\sigma$-ideal generated by the family $\mathcal F$ coincides with the ideal of meager subsets of $G$.

But we can ask another problem: can each non-open analytic subgroup of a Polish Abelian group be covered by countably many closed Haar null subsets?

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  • $\begingroup$ Thanks, this answers my question completely. :) (If not in the direction I had hoped for.) $\endgroup$
    – tomasz
    Commented Aug 26, 2015 at 2:15
  • $\begingroup$ Actually, I withdraw my comment: the group of reals is not compact, just locally compact, so it does not really answer my question. Can this be easily fixed? $\endgroup$
    – tomasz
    Commented Aug 26, 2015 at 12:21
  • $\begingroup$ To obtain a "compact" example, take the quotient of the real line by the cyclic subgroup $C$ generated by a point n the Borel subgroup $H$ constructed by Laczkovich. Then $H/C$ with a Borel subgrop of the compact group $R/C$ which does not admit the desired representation. $\endgroup$
    – Taras Banakh
    Commented Aug 26, 2015 at 21:36
  • $\begingroup$ Right, of course. Well, you might as well take a quotient by ${\bf Z}$. I thought about that, but then took a glance at the paper and was mildly distressed with the use of convex hulls at the very beginning -- they would have been useless in the circle group. But of course this is locally just taking a countable union, so it preserves both borelness and meagreness of the subgroup, and if we had a "bad" (actually good) cover of the quotient, we could just pull it back to the reals, since meagreness is a local property. So thanks again. :) $\endgroup$
    – tomasz
    Commented Aug 27, 2015 at 9:17

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