Can compacts on a real line behave "paradoxically" with respect to unions, intersections, and translation? What about other topological groups? I have the following question i cannot answer myself.
Consider two compacts $A$ and $B$ on the real line $\mathbb R$. Let $A'$ be a translation of $A$ and $B'$ a translation of $B$:


*

*$A' = A + a$,

*$B' = B + b$.


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 $A\cap B$:


*

*$A'\cup B'\subset (A\cup B) + t_1$,

*$A'\cap B'\subset (A\cap B) + t_2$.


Is it always true in this case that $A'\cup B' = (A\cup B) + t_1$ and $A'\cap B' = (A\cap B) + t_2$?
I am in fact interested in the natural generalization of this question to closed compacts in arbitrary topological groups.
I can only prove it in the case $A\cap B = \varnothing$, which is fairly easy. I cannot prove it even for the real line, even if i assume that $A\cap B$ consists of a single point.

I have posted this question in a slightly modified form on MathOverflow.
 A: Such problems are quite specific and have no standard approach. I am going to pose yours at the divertissement at our seminar “Topology & Applications” – maybe the collective mind will be able to move farther than  I did. 
I can handle the case $A\subset B\subset\Gamma$, where $\Gamma$ is an arbitrary Hausdorff topological group. We have $bB\subset t_1B$. We shall deal similarly to the Swelling Lemma proof. Put $$T=\{x\in\Gamma: b^{-1}t_1B\subset xB\}.$$ We show that $T$ is a subsemigroup of $\Gamma$. Let $x,y\in T$. Then $b^{-1}t_1B\subset xB\subset xb^{-1}t_1B\subset xyB$. Since $T$ is closed, $T$ is a compact semigroup. Therefore $T$ contains an idempotent $e$ (this is a well-known and easily provable fact in topological algebra), which should be the unit of $G$. Then $b^{-1}t_1B\subset eB=B$ and $bB=t_1B$. Hence 
$$bB\subset A'\cup B'=aA\cup bB\subset t_1(A\cup B)=t_1B=bB,$$
and all the inclusions are equalities. Now we have $aA=aA\cap bB\subset t_2(A\cap B)=t_2A$. Similarly to the above we can prove that $aA=t_2A$. Hence 
$$t_2(A\cap B)\subset t_2A=aA\subset aA\cap bB=A'\cap B'.$$
