Closed subsets of $\mathbb{C}^*$ proper for multiplication Let $S_1$ and $S_2$ be two proper closed subsets of $\mathbb{C}^*$. Let's denote by $\overline{S_1}$ and $\overline{S_2}$ their closure in $\mathbb{C}_{\infty}.$ (Alexandrov compactification) 
Theorem : If $S_1 \times S_2$ is proper for the complex multiplication (i.e. : $S_1 \cap \frac{K}{S_2}$ is compact for each compact $K \subset \mathbb{C}^*$) then $(0,\infty) \notin \overline{S_1} \times \overline{S_2}.$
I must say I'm not sure of the validity of this theorem. Actually I try to find a counter-example (with discrete closed subsets, etc...) without success. However I can't find any proof of this theorem. 
Any help (proof or counter-example) will be greatly appreciated.
 A: Let
$$S_1 = \biggl\{ \frac{1}{(2m+1)!} : m \in \mathbb{N}\biggr\}\quad\text{and}\quad S_2 = \{ (2n)! : n \in \mathbb{N}\}.$$
Then $S_1$ and $S_2$ are closed subsets of $\mathbb{C}^{\ast}$, $0 \in \overline{S_1}$ and $\infty \in \overline{S_2}$ (closures taken in $\mathbb{C}_{\infty}$), and yet for every compact $K \subset \mathbb{C}^{\ast}$ the set $S_1 \cap \frac{K}{S_2}$ is finite, hence compact.
For, any compact subset of $\mathbb{C}^{\ast}$ is contained in a closed annulus $A_R := \{ z : R^{-1} \leqslant \lvert z\rvert \leqslant R\}$, and
$$\frac{1}{R} \leqslant \frac{(2n)!}{(2m+1)!} \implies (m < n) \lor (2m+1 \leqslant R),\tag{1}$$
while
$$\frac{(2n)!}{(2m+1)!} \leqslant R \implies (m \geqslant n) \lor (2n \leqslant R).\tag{2}$$
Thus for $m > \frac{R-1}{2}$, we can by $(1)$ only have $\frac{(2n)!}{(2m+1)!} \in A_R$ if $n > m$, but then $2n > 2m + 1 > R$ and then $(2)$ shows $\frac{(2n)!}{(2m+1)!} \notin A_R$, so
$$\operatorname{card} \biggl(S_1 \cap \frac{A_R}{S_2}\biggr) \leqslant \frac{R+1}{2}.$$
