Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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
1  
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
add comment

1 Answer 1

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

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

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.