Prove that $\{e^{k\alpha \pi i}: k\in \mathbb N \}$ is dense in $S^1 $ by using(or modifying the proof of) the result when $\mathbb N$ is $\mathbb Z$

Question

Let $\alpha$ be an irrational number. There is a well-known result saying that $\{e^{k\alpha \pi i}: k\in \mathbb Z \}$ is dense in $S^1 $. In the proof of this result, the Pigeonhole Principle can't guarantee that we can always choose $i>j$. So I wonder how we can prove that $\{e^{k\alpha \pi i}: k\in \mathbb N \}$ is dense in $S^1 $ by using that result or modifying its proof($\mathbb N$ is the set of natural numbers). Thanks in advance.

Answer 1

Proof assuming the $\mathbb Z$ result: Suppose the result for $\mathbb Z$ holds, which implies the result for $\mathbb Z\setminus \{0\}$ holds.

Claim: There exists a sequence of positive integers $k_j$ such that $e^{k_j\alpha \pi i} \to 1.$

Proof: There is a sequence $e^{k_j\alpha \pi i} \to 1,$ where each $k_j\in \mathbb Z\setminus \{0\}.$ If infinitely many of these $k_j$ are positive, we're done. If not, note $e^{-k_j\alpha \pi i} \to 1$ as well, and infinitely many $-k_j$ are positive, giving the claim.

Now suppose $e^{it}\in S^1.$ There is a sequence $e^{k_j\alpha \pi i} \to 1$ where each $k_j\in \mathbb Z\setminus \{0\}.$ By the claim we can choose $k_j'>|k_j|$ such that $e^{k_j'\alpha\pi i} \to 1.$ We then have $e^{(k_j+k_j')\alpha\pi i} \to \zeta$ and each $k_j+k_j'>0.$ This gives the result.

Brief proof without using the $\mathbb Z$ result: Because the circle is compact, the sequence of distinct points $e^{k\alpha \pi i},k\in \mathbb N,$ has a limit point $\zeta$ on the circle. Thus there is a subsequence $0< k_1 < k_2 < \cdots $ such that $e^{k_j\alpha\pi i} \to \zeta.$ We then have $e^{(k_{j+1}-k_j)\alpha\pi i} \to 1.$ Thus for any $\epsilon>0,$ there exists $k_\epsilon\in \mathbb N$ such that $0<|1-e^{k_\epsilon\alpha\pi i}|<\epsilon.$ The set $A_\epsilon = \{e^{\alpha k_\epsilon k \pi i}:k\in \mathbb N\}$ then has the property that every point on the circle is within $\epsilon$ of some point in $A_\epsilon.$ Since $\epsilon>0$ is arbitrary, we have the desired density.