Let $G$ be a subgroup of the multiplicative group $\mathbb{C}^*$. such that $g \in G$ where $|g| = 1$ then is the set bounded and closed subset of $\mathbb{C}$? If so can we use the heine-borel theorem in this case to show that it is compact (As $\mathbb{C} \cong \mathbb{R}^2$)
I tried to show that any subset of $\mathbb{C}$ which is bounded will have $|g| = 1$.
Assume that the set $G$ is bounded and there is a $g \in G$ such that $|g| > 1$ . Let $|g| = 1+\epsilon$ where $\epsilon > 0 $ . Then as $G$ is a subgroup of $G$ then we can conclude that $g^n \in G$ for all $n \in \mathbb{Z}$.
Then $|g|^n = (1+\epsilon)^n$ .
Now we know that $|(a+ib)| = \sqrt{(a^2 + b^2)}$ which defines a norm on $\mathbb{C}$.
We know that in a normed linear space any set is bounded iff $||x|| \le M$ for some $M \in \mathbb{R}$.
Then assume that there exists a $M >0 $ such that $|g| \le M$ for all $g \in G$.
Now we know that by the archimedian property of $\mathbb{R}$ there exists an $N_1 \in \mathbb{N}$ such that $N_1 . \epsilon > M$ .
let $|g|^{N_1} = (1+\epsilon)^{N_1} > 1 + N_1.\epsilon > M $ which is a contradiction.
Is the set $G$ closed? If so can we use the heine-borel theorem to conclude that it is compact?(by assuming $\mathbb{C} \cong \mathbb{R}^2$?)
The answer is negative. Take, for instance, $G=\left\{e^{i\theta}\,\middle|\,\theta\in\Bbb Q\right\}$. Then $G$ is not a closed subset of $S^1$, and therefore it is not compact.