I was reading this and I got stuck on a step in the proof of Theorem 1.3.18 (page 25).
In proving the implication (4) $\Rightarrow $ (5) it states:
Since the cyclic group generated by $T$ has no element of small argument (otherwhise $T$ will have a finite order) one can construct a sequence $\{T^{a_n} : a_n \in \mathbb{Z}\}$ such that $T^{a_n} \to \mathrm{Id}$ where $n\to \infty$.
Now, I cannot see why the implication: $T$ has element of smallest argument implies $T$ having finite order.
I attempt to justify the existence of such sequence $(a_n)$ by showing that $\liminf (|e^{2in\pi \theta}-1|)=0.$ But I have no clue of how this would happen.
Can you suggest me a solution please? Thanks in advance.
Suppose that the cyclic group $\langle T \rangle$ that is generated by $T$ has an element of smallest argument, call that argument $\theta \in (0,2\pi)$.
Let $k \ge 1$ be the least exponent such that $(k-1)\theta < 2\pi$. It follows that $$k\theta - 2 \pi < \theta $$ It also follows that $$k\theta \ge 2\pi $$ If the last inequality is strict, i.e. if $k\theta > 2 \pi$, then we have $$0 < k\theta - 2 \pi < \theta $$ and therefore $k\theta - 2 \pi$ is an argument of $T$ that is smaller than the smallest argument $\theta$, a contradiction.
Therefore $k\theta = 2 \pi$. Choosing $m$ such that the argument of $T^m$ is equal to $\theta$, it follows that $T^{mk}$ is the identity and so $\langle T \rangle$ is finite.