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The following question was originally asked here, by C. Dubussy: https://math.stackexchange.com/questions/1802111/neighbourhoods-with-proper-multiplication

Assume we have two closed subsets $F$ and $G$ of $\mathbb{C}^*$ which are proper for the multiplication, i.e. $$KF^{-1}\cap G$$ is a compact of $\mathbb{C}^*$ when $K$ is a compact of $\mathbb{C}^*$. The aim here is to find a closed neighbourhood $N_F$ of $F$ and $N_G$ of $G$ such that $N_F$ and $N_G$ remain proper for the multiplication.

Here is my try. For each connected component $i$ of $\mathbb{C}^*\backslash F$ we choose a $z_i$ and $r_i$ such that $\overline{D(z_i,r_i)}$ is included in that component. We set $$N_F = \bigcap_{i\in CC(\mathbb{C}^*\backslash F)} \mathbb{C}^*\backslash D(z_i,r_i).$$ This is clearly a closed neighborhood of $F$ and we can do the same for $G$. But with that construction I can't prove that $N_F$ and $N_G$ are proper for the multiplication. I guess I have some lattitude in the choice of $z_i$ and $r_i$ (I can take the disks "big" as possible) but even with that, it seems complicated to me.

It's possible the only problem are the non-bounded components of $\mathbb{C}^*\backslash F$? Maybe the property is false, and we can find a counterexample?

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  • $\begingroup$ Btw, it suffices to consider only annuli as $K$, thus reducing the problem to $\mathbb R^+$ instead of $\mathbb C^*$. $\endgroup$ Commented Jun 8, 2016 at 10:41
  • $\begingroup$ I see my question interested you math.stackexchange.com/questions/1802111/… ;) $\endgroup$
    – C. Dubussy
    Commented Jun 8, 2016 at 11:22

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Do I misunderstand something? Let $B$ be some small closed ball centered at $1$. Then $BF$ and $BG$ are closed neighborhoods of $F$ and $G$, respectively. Now, $K(BF)^{-1}\cap (BG)$ is a closed subset of $A=B((KB^{-2})F^{-1}\cap G)$; indeed, if $z=kb_1^{-1}f^{-1}=b_2g$, then $z=b_2\cdot k(b_1b_2)^{-1}f^{-1}=b_2g$. Since $KB^{-2}$ is compact, so is $A$ and hence any its close subset.

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