# 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.

• Unless I'm mistaken, $(0,\infty)\in \overline{S_1}\times \overline{S_2}$ means exactly $0\in S_1$ and $S_2$ is not bounded. And $S_1\times S_2$ being proper means that for every $r\ge 0$, $S_1\cap \left\{\frac{r}{z}\mid z\in S_2\right\}$ is compact. So your theorem says that if $0\in \overline{S_1\cap \left\{\frac{1}{z}\mid z\in \overline{S_2}\right\}}$, then it's not compact... Whereas it probably should be $\infty\in \overline{S_1\cap \left\{\frac{1}{z}\mid z\in \overline{S_2}\right\}}$. Are you sure it's not $(\infty,0)$ instead of $(0,\infty)$? – xavierm02 Feb 12 '16 at 12:27
• $S_1 \subset \mathbb{C}^*$. Hence $0 \in S_1$ has no sens. $0 \in \overline{S_1}$ only says there is a convergent sequence $x_n \in S_1 \to 0.$ – C. Dubussy Feb 12 '16 at 12:33
• Yeah. I edited and added an overline. But I'm starting to think that you can never have $0\in\overline{S_1}$: $S_1$ is closed so if it has a sequence converging to $0$, it contains $0$ but it can't because it's a subset of $\mathbb C^*$. And so $\overline{S_1}$ can't contain $0$... – xavierm02 Feb 12 '16 at 12:36
• It is a closed subset of $\mathbb{C}^*$ for the topology of $\mathbb{C}^*$ . Take for exemple $S_1 = \{\frac{1}{n} : n \in \mathbb{N}\}$. This a closed subset of $\mathbb{C}^*$ and $0 \in \overline{S_1}.$ – C. Dubussy Feb 12 '16 at 12:38

## 1 Answer

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

$$\operatorname{card} \biggl(S_1 \cap \frac{A_R}{S_2}\biggr) \leqslant \frac{R+1}{2}.$$

• Nice counter-example, thank you ! – C. Dubussy Feb 12 '16 at 13:26