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?



Neighborhoods with proper multiplication
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.


