Can closed compacts in a topological group behave "paradoxically" with respect to unions, intersections, and one-sided translations? 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 $B = \varnothing$).
 A: 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?
