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Consider two closed compacts $A$ and $B$ in a topological group $\Gamma$. Let $A'$ be a left translation of $A$ and $B'$ a left translation of $B$:

  • $A' = aA$,
  • $B' = bB$.

Suppose it is known that $A'\cup B'$ is contained in a translation of $A\cup B$, and $A'\cap B'$ is contained in a translation of $A\cap B$:

  • $A'\cup B'\subset t_1(A\cup B)$,
  • $A'\cap B'\subset t_2(A\cap B)$.

Is it always true in this case that $A'\cup B' = t_1(A\cup B)$ and $A'\cap B' = t_2(A\cap B)$?

I cannot prove it even under the additional assumptions that $\Gamma = \mathbb R$ and $A\cap B$ consists of a single element.

I have asked this question on Math.StackExchange first, but it seems sufficiently hard to be posted on MO (in a slightly different form).

Easy cases

The case $A\cap B = \varnothing$ is fairly easy (using minimal covers by left translates of a given open set).

Cases when $A\subset B$ or $A'\subset B'$ are also easy and can be deduced from the fact that if $A'\subset A$, then $A' = A$ (this is the case of $B = \varnothing$).

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  • $\begingroup$ Compacts are always closed, or I miss something? $\endgroup$
    – Anton Klyachko
    Commented Jul 30, 2013 at 14:17
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    $\begingroup$ Only if the group is Hausdorff :). For me personally, restricting to closed compacts is more natural than restricting to Hausdorff topologies, at least it sounds less complicated.. $\endgroup$
    – Alexey Muranov
    Commented Jul 30, 2013 at 16:17
  • $\begingroup$ Can you maybe give an example of an ordinary group where this property does not hold? ($A,B$ being arbitrary subsets). $\endgroup$
    – Gerrit Begher
    Commented Aug 5, 2013 at 8:23
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    $\begingroup$ @GarlefWegart: in an ordinary group, and ordinary subset can be translated strictly inside itself, like $\mathbb{N}$ in $\mathbb{Z}$, or like the set of positive integer powers of $e^i$ in the "complex unit circle" multiplicative group. It follows that the property from the question does not hold. $\endgroup$
    – Alexey Muranov
    Commented Aug 5, 2013 at 10:39
  • $\begingroup$ I would restrict to Hausdorff topologies: Doesn't your reply to Garlef show that the property doesn't hold, if we endow a group (like $\mathbb{Z}$) with the indiscrete topology? (That, of course, depends on what you mean with Compact. I prefer to require quasi-compact and Hausdorff, but to me it seems you don't.) $\endgroup$
    – Ben
    Commented Aug 5, 2013 at 20:25

1 Answer 1

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This is not an answer to Muranov's problem. Below I present some partial results which can be interesting or helpful for researchers that will try to attack this problem in future. The proofs of the following statements can be found in this paper.

Theorem 1. Let $A,B$ be compact subsets of a topological group $G$ such that $aA\cup bB\subset A\cup B$ and $aA\cap bB\subset c(A\cap B)$ for some elements $a,b,c\in G$. The equalities $aA\cup bB=A\cup B$ and $aA\cap bB=c(A\cap B)$ hold if either the subgroup $H_3$ generated by the set $\{a,b,c\}$ is discrete or for some set $T\subset \{a,b,c\}$ with $\{a,b\}\subset T$, $\{a,c\}\subset T$ or $\{b,c\}\subset T$ the subgroup $H_2$ generated by $T$ is discrete and closed in $G$ and $H_2$ is normal in $H_3$.

This Theorem implies

Corollary 1. Let $A,B$ be compact subsets of a topological group $G$ such that $aA\cup bB\subset A\cup B$ and $aA\cap bB=\emptyset$ for some elements $a,b\in G$. The equalities $aA\cup bB=A\cup B$ and $A\cap B=\emptyset$ hold if either the subgroup $H_2$ generated by the set $\{a,b\}$ is discrete or for some non-empty set $T\subset\{a,b\}$ the subgroup $H_1$ generated by $T$ is discrete and closed in $G$ and $H_1$ is normal in $H_2$.

These results (and the original problem of Muranov) motivate the following definitions.

Definition. A topological group $G$ is called

  • Muranov if for any compact subsets $A,B\subset G$ and points $a,b,c\in G$ the inclusions $aA\cup bB\subset A\cup B$ and $aA\cap bB\subset c(A\cap B)$ imply the equalities $aA\cup bB=A\cup B$ and $aA\cap bB=c(A\cap B)$;
  • weakly Muranov if for any compact subsets $A,B\subset G$ and points $a,b\in G$ with $aA\cup bB\subset A\cup B$ and $aA\cap bB=\emptyset$ we get the equalities $aA\cup bB=A\cup B$ and $A\cap B=\emptyset$.

Theorem 1 and Corollary 1 imply

Corollary 2. An (abelian) topological group $G$ is

  • Muranov if each 3-generated (resp. 2-generated) subgroup of $G$ is discrete;
  • weakly Muranov if each 2-generated (resp. 1-generated) subgroup of $G$ is discrete.

Corollary 3. For every $n\in\mathbb N$ the topological group $\mathbb Q^n$ is Muranov and $\mathbb R^n$ is weakly Muranov.

Corollary 4. Each locally finite topological group is Muranov.

However the original problem of Muranov remains wide open:

Problem. Is the real line $\mathbb R$ Muranov? Is the circle $\mathbb R/\mathbb Z$ weakly Muranov?

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  • $\begingroup$ Yes, of course. $\endgroup$
    – Alexey Muranov
    Commented Oct 3, 2015 at 10:58
  • $\begingroup$ Alexey, I am thinking about the answer in case of rational numbers $a,b$ and $\Gamma=\mathbb R$. It seems that if all numbers $a,b,t_1,t_2$ are rational, then we can reduce the problem to the case of the group $\mathbb Z$ and apply the counting measure. $\endgroup$
    – Taras Banakh
    Commented Oct 3, 2015 at 11:14
  • $\begingroup$ I have just observed that the same answer was given on Stackexchange by Dominik math.stackexchange.com/questions/448277/… $\endgroup$
    – Taras Banakh
    Commented Oct 3, 2015 at 16:19
  • $\begingroup$ Well, this does not answer my question. $\endgroup$
    – Alexey Muranov
    Commented Oct 3, 2015 at 21:38
  • $\begingroup$ But at least for some non-discrete topological groups (like $\mathbb Q^n$) we have the affirmative answer. The case of the group $\Gamma=\mathbb R$ is indeed very interesting. One can try to understand what is going on with finitely generated subgroups of $\mathbb R$ (like $\mathbb Z+\sqrt{2}\mathbb Z$). $\endgroup$
    – Taras Banakh
    Commented Oct 3, 2015 at 22:41

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