Find connected closed subgroup of $(\mathbb C,+)$

Question

I am asked to show that the only connected closed subgroup of $(\mathbb C,+)$ are $\{0\}$, $\mathbb C$, or a line passing through the origin. Since the subgroup is connected, $0$ is a limit point. It suffices to show that there must be a sequence of complex numbers $z_n$ on a line approaching $0$, since then by closedness the entire line is contained in the subgroup and we can quotient it out. But I couldn't prove the existence of such a sequence. Can anyone help?

Source: This is given as exercise 14 (iv)(b) in this blog post.

Answer 1

Suppose $G\subset\mathbb{C}$ is a nontrivial proper closed connected subgroup; we wish to show $G$ is a line. As you say, by considering the quotient, it suffices to show that $G$ contains a line. Since $0$ is not isolated in $G$, there is a sequence $(g_n)$ of nonzero elements of $G$ approaching $0$. Let $h_n=g_n/|g_n|$. Since the unit circle is compact, we can pass to a subsequence and assume that $(h_n)$ converges to some $h$.

I now claim that $G$ contains the line $\mathbb{R}h$. Indeed, given $\epsilon>0$, choose $n$ such that $|h_n-h|<\epsilon$ and $|g_n|<\epsilon$. Note then that for any $t\in\mathbb{R}$ there is $m\in\mathbb{Z}$ such that $m|g_n|$ is within $\epsilon$ of $t$. Thus $|mg_n-th|\leq |mg_n-th_n|+|th_n-th|\leq \epsilon+t\epsilon$. For fixed $t$, this can be made arbitrarily small by choosing $\epsilon$ to be sufficiently small, and thus $th$ can be approximated arbitrarily closely by elements of the form $mg_n$ which are in $G$.

Answer 2

Partial answer

Let’s start with an observation. For any $0 \neq c \in \mathbb C$, let $L_c$ be the line passing through the origin and $c$. If $G$ is an additive connected closed subgroup of $\mathbb C$, then $G_c =G \cap L_c$ is isomorphic (as a group) to a closed subgroup of $\mathbb R$. Hence $G_c$ is either equal to $\{0\}$, to the whole line $L_c$ or to $c^\prime \mathbb Z$ where $c^\prime \in L_c$.

Also, $\{0\}$ and lines passing through the origin are the only connected closed subgroups of $(\mathbb C, +)$ included in a line. Let $G$ be a connected closed subgroup of $(\mathbb C, +)$ not included in a line. We have to prove that $G = \mathbb C$. By hypothesis, it exist $0,a,b \in G$ not aligned. If $G_a= G \cap L_a =L_a$ and $G_b = G \cap L_b= L_b$ we are done. So let suppose that $G_a = a \mathbb Z$.

We have to prove that this can’t happen… and I’m stuck for the moment.