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I am trying to show that the set of numbers of the form $e^{2\pi ina}$, where $a$ is irrational and $n\in\mathbb Z$ , is dense in the unit circle.

Solution: Noting that elements of the unit circle can be written as $z=e^{2\pi i\theta}$ with $\theta\in\mathbb R$, and using the standard topology of $\mathbb C$;

\begin{align*} \left|e^{2\pi ian}-e^{2\pi i\theta}\right| &= \big|[\cos(2\pi an)-\cos(2\pi\theta)]-i[\sin(2\pi an)-\sin(2\pi\theta)]\big|\\ &= \left|-2\sin[\pi(na-\theta)]\sin[\pi(na+\theta)]-2i\sin[\pi(na-\theta)\cos[\pi(na+\theta)]\right|\\ &=2|\sin[\pi(na-\theta)]||\sin[\pi(na+\theta)]+i\cos[\pi(na+\theta)]|\\ &=2|\sin[\pi(na-\theta)]|\leq 2\pi|na-\theta|<2\pi\epsilon_1=\epsilon. \end{align*}

I have used the following: $$\cos(P)-\cos(Q)=-2\sin\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right),~ \sin(P)-\sin(Q) = 2\cos\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right)$$ and $|\sin x|\leq |x|$.

The last inequality follows from the fact that there is always a real number ($\theta$) arbitrarily close to any rational number ($na$).

This last argument makes me feel like I have cheated (a bit). Is there a way to complete this exercise without the prior knowledge of the denseness of the rationals on the real line? Furthermore, please recommend alternative approaches, if there are any.

Thato Tsabone
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2 Answers2

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Your proof is wrong. For fixed $a$ and $\theta$ you can't make $na$ arbitrarily close to $\theta$ in general. Also the statement of the theorem is wrong since that set is dense only if $a$ is irrational. This problem can be solved in several ways, one of them being that you map the circle homeomorphically and homomorphically to $[0,1]$ $mod(1)$. This is already done by the exponential map you introduced. Then you reduced the problem of studying the circle to studying subsets of R. Now you should be able to prove that any additive subgroup of R is either dense (corresponding to when $a$ is irrational) or it is a discrete group having least positive element (corresponding to when $a$ is rational). Since homeomorphisms preserve all topological properties (density included) then you are done. If you need help with specifics or alternative approaches you can take a look at similar questions such as this one.

Leonid
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  • Thanks for the useful comment. I made a mistake yes: $a$ is irrational. – Thato Tsabone Jul 08 '22 at 22:26
  • Would it have been been better if I had chosen $\theta$ to be a real number in the neighbourhood of $na$ a priori? – Thato Tsabone Jul 09 '22 at 08:50
  • That's not the way to go because you are trying to prove that given any $\theta$ you can make $e^{2\pi ian}$ arbitrarily close to $e^{2\pi i\theta}$ for some $n \in \mathbb{N}$. So to be more concrete, your job is the following: Given a random $\theta$ and $\epsilon > 0$ you must find an $n \in \mathbb{N}$ s.t. $e^{2\pi ian}$ is within an $\epsilon$ distance of $e^{2\pi i\theta}$ – Leonid Jul 09 '22 at 11:30
  • I see…. Thanks a lot! – Thato Tsabone Jul 10 '22 at 07:24
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As Leonid observed, your proof is incorrect. Since you asked for the alternative approach, this is my try.

  • Let $a_n:=a\cdot n-\lfloor a\cdot n\rfloor$. Then $e^{2\pi i\cdot a\cdot n} = e^{2\pi i\cdot a_n}$.
  • It sufficies to show that $\{a_n:n\in\Bbb N\}$ is dense in $[0,1]$. Indeed, for any $z\in S$ (unit circle) there exists $\theta\in [0,1]$ such that $z=e^{2\pi i \theta}$. If $a_n$ is close to $\theta$ then $e^{2\pi i\cdot a_n}$ is close to $z$.
  • Observe that irrationality of $a$ implies that $a_m\neq a_n$ for $m\neq n$.
  • Take any $N\in\Bbb N$ (big number). Consider $N$ boxes $[0,1/N],[1/N,2/N],\ldots,[1-1/N,1]$ covering $[0,1]$.
  • One of the boxes contains infinitely many numbers $a_n$. Particularly, there are $m,n\in\Bbb N$, $m<n$, such that $|a_m-a_n|\leq 1/N$.
  • Assume $0<a_n-a_m\leq 1/N$ (the other case is similar, as we consider indices from $\Bbb Z$, not only $\Bbb N$). Then $0<a_{k}\leq 1/N$ for $k:=n-m$.
  • Observe that $a_{2k}=2a_{k}$, $a_{3k}=3a_{k}$ and generally $a_{jk}=ja_{k}$ as long as $ja_{k}<1$. These points form an arithmetic progression with the difference $d=a_{k}\leq 1/N$.
  • Therefore for any $\theta\in[0,1]$ there exists $j$ such that $|a_{jk}-\theta|\leq 1/N$.
  • Since $N$ can be arbitrarily large, the set $\{a_n:n\in\Bbb N\}$ is dense in $[0,1]$.
Mateo
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