# 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.

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Small note: it should be safe to assume that $a=0$. Only the relative shift matters in a topological group. – dfeuer Jul 20 '13 at 20:49
Small note 2: for the reals, you should be able to get some good info from minima and maxima. – dfeuer Jul 20 '13 at 20:51
By considering the Lebesgue measure, we see $\mu(A' \cup B')\le \mu(A \cup B)$ and $\mu(A' \cap B')\le \mu(A \cap B)$. Now $\mu(A)+\mu(B)=\mu(A')+\mu(B')=\mu(A' \cup B')+\mu(A' \cap B')\le \mu(A \cup B)+ \mu(A \cap B)=\mu(A)+\mu(B)$ implies that we must have $\mu(A' \cup B')= \mu(A \cup B)$ and $\mu(A' \cap B')= \mu(A \cap B)$ which yield $\mu(((A \cup B)+t_1)\setminus (A' \cup B'))=0$ and $\mu(((A \cap B)+t_2)\setminus (A' \cap B'))=0$ – Dominik Jul 23 '13 at 19:18
@Dominik: yes, i know. – Alexey Jul 23 '13 at 22:45

## 1 Answer

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

$$t_2(A\cap B)\subset t_2A=aA\subset aA\cap bB=A'\cap B'.$$

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Good observation. This in fact can be easily deduced from the case $A\cap B = \varnothing$. (I have not checked your proof, for me it was easier to apply me previous method.) The case of $A\cap B = \varnothing$ with $B=\varnothing$ implies that if $A'\subset A$ or $A'\supset A$, then $A' = A$. Now, if $A\subset B$, then $A'\cup B' \subset t_1(A\cup B)=t_1B$ implies $B'=t_1B$, $A'\subset B'$, $A'=t_2A$, $A'\cup B'=B'=t_1B=t_1(A\cup B)$, $A'\cap B'=A'=t_2A=t_2(A\cap B)$. The case $A'\subset B'$ can be handled symmetrically: $A'=t_2A$ and hence $A\subset B$. – Alexey Jul 26 '13 at 11:51
@Alexey Yes, it seems that both proofs are based on the lemma: if $A$ is a compact subset of a (Hausdorff) topological group $\Gamma$ and $tA\subset A$ for some $t\in\Gamma$ then $tA=A$. – Alex Ravsky Jul 26 '13 at 13:39