Hi I am reading about matrix lie groups from the book by Krishtopher Tapp. The following problem is one from the exercises: I want to prove the following:
Let $\lambda \in \mathbb{R}$ be an irrational multiple of $2\pi$. Define $$G : = \{ (e^{\lambda ti}) \mid t \in \mathbb{Z}\} \subset U (1) \subset GL_{1}(\mathbb{C})$$ Prove that $G$ is a subgroup of $GL_{1}(\mathbb{C})$, but not a matrix group. Prove that $G$ is dense in $U(1)$. Proof. For $t_{1}, t_{2} \in \mathbb{Z}$, we have $$e^{\lambda i t_{1}}\cdot e^{-\lambda i t_{2}} = e^{\lambda i (t_{1} - t_{2})} \in G$$ By one-step subgroup test $G \leq GL(n,\mathbb{C})$
In order to proceed with the proof let us first prove the following lemma.
Lemma. Given $\beta \in \mathbb{R}$ and any integer $N$ there exists $n,m$ such that $1 \leq n \leq N$ $$|n\beta - m | < \frac{1}{N}$$ Proof. Let $f(x) = x - \lfloor x \rfloor$. Consider the following sequence of $N + 1$ numbers {$f(0), f(\beta), f(2 \beta), f(3 \beta), \dots , f(N \beta)$ }. By definition of $f$ all these $N+1$ integers lie in $[0,1)$. Consider the following subdivision of the interval $[0,1)$ into $N$ sub intervals: $[0, \frac{1}{N})$, $[\frac{1}{N}, \frac{2}{N})$, $[\frac{2}{N}, \frac{3}{N})$, \dots $[\frac{N-1}{N}, 1)$. By pigeonhole principle there exist atleast 2 numbers, from the above constructed sequence of $N+1$ numbers that lie in the same sub interval. Hence there exist $0\leq i<j \leq N $ such that $$| f(i \beta) - f(j \beta) | < \frac{1}{N}$$ So we take $n = i - j$ and $m = \lfloor i \beta \rfloor - \lfloor j \beta \rfloor$.
Since $U(1) = \mathbb{S}^{1}$ any element of $U(1)$ can be represented as $e^{i \theta}$ for $\theta \in [0,2\pi)$. To prove that $G$ is dense in $U(1)$ we need to show any $\epsilon$- ball around $e^{i \theta}$ for some $\theta \in [0,2\pi)$ contains a point of the set $G$.\ Observe that for $t_{1}, t_{2} \in \mathbb{R}$, we have $$| e^{it_{1}} - e^{it_{2}} | < | t_{1} - t_{2} |$$ Since the straight line distance between two points on the unit circle is less than the arc along the circle joining those two points.
Now fix $\epsilon \in \mathbb{R}$. We can chose a natural number $N$ such that $\frac{1}{N} < \epsilon$ (Archimedean Property). Let $\alpha = \frac{\lambda}{2\pi}$. Applying previous lemma on $\alpha$ and $N$ we conclude that there exists $t,m \in \mathbb{N}$ and $1 \leq t \leq N$ $$|t \alpha - m | < \frac{1}{N}$$ Observe that $e^{2 \pi i \alpha t} \in G$. $$|e^{2 \pi i \alpha t} - 1| < | e^{2 \pi i \alpha t} - e^{2\pi i m}| <2\pi(t \alpha - m) < 2\pi \epsilon$$ Thus arbitrary $\epsilon$-ball around $1 \in U(1)$ contains a point of $G$. But I cannot prove why every epsilon ball around any point in the circle should contain a point of the set $G$. Can someone give me a rigorous proof explaining why this is the case? I know there are a lot of proofs of this particular problem in stack exchange, but not a single one contains a rigorous proof of this particular point.
BTW... If We have proved $G$ is dense in $U(1)$, then the rest of the proof is straightforward. We would have $\bar{G} = U(1)$. If $G$ was a matrix group then it would have to be closed in $GL_{1}(\mathbb{C})$ then $\bar{G} = G$. Hence $G = U(1)$ which is absurd. Hence although $G$ is a subgroup $GL_{1}(\mathbb{C})$ it is not a matrix group
Thanks in Advance
