Show that $\{e^{in}: n\in\Bbb N\}$ is Dense in the Unit Circle

Question

This problem gave me fits when I was in grad school. Looking back at it now, it still escapes me. The problem is from Conway's Functions of One Complex Variable. I'm looking for a proof from basic principles, so no big theorems please.

I remember that the Pigeonhole principle was involved, but I don't seem to be able to formulate it in the proper way.

Let $S$ be the unit cirlce, and $T=\{e^{in}:n\in\Bbb N\}$. For a fixed $k$, $S$ can be paritioned into $$A_{m,k}=\left\{e^{i\theta}:\frac{2\pi i(m-1)}{k}\le\theta\le\frac{2\pi im}{k},m=0,\dots k-1\right\}$$. Now infinitely many members of $T$ must belong to $A_{m,k}$. Then $A_{m,k}$ can be partitioned in a similar manner, and in this way we can create a sequence of sets $S_n$ such that $S_{n+1}\subset S_n$ each of which contains infinitely many points of $T$.

So, $\cap_{n=1}^{\infty}S_n$ is not empty, and since $\rm diam\, S_n\to 0$, then intersection is a singleton, which is a limit point of $T$. But this doesn't help, we already know $T$ has limit points in $S$ because it is infinite and $S$ is compact.

I rember the Pigeonhole princple being involved, but I can't seem to make it work. I'll show my thoughts on that. — Tim Raczkowski, Dec 18 '14 at 15:23
My impression is this at least has to use that pi is irrational, which is maybe a big theorem?? For example, e^(i * pi*n/20) is not dense in the unit circle. — CJD, Dec 18 '14 at 15:29
Does the equidistribution theorem count as a big theorem? I seem to recall that there's a proof relying on Fourier series, or something similar. — Jonathan Y., Dec 18 '14 at 15:50
@JonathanY. Yes, that would be a big theorem. The proof only requires basic principle of general topology. — Tim Raczkowski, Dec 18 '14 at 15:52

score 9 · Accepted Answer · edited Apr 13 '17 at 12:21

9

This is identical to showing that the fractional parts of $\dfrac{n}{2\pi}$ are dense in $[0,1)$, which is true since $\dfrac{1}{2\pi}$ is irrational (see e.g. this question).

edited Apr 13 '17 at 12:21

Community

1

answered Dec 18 '14 at 15:59

TonyK

64,559

score 0 · Answer 2 · answered Dec 04 '19 at 21:33

It is not hard to prove that $e^{in}$ is not periodic, in fact, that it never takes the same value twice. Given $\epsilon>0$, consider the partition of the circle into the $A_{m,k}$ arcs, with $k$ such that $\frac{2\pi}{k}<\epsilon$. Consider now the set of $k+1$ points $\{e^{in}: n = 0, 1, ... k\}$; at least one of the arcs contains two of the points in the set, say $e^{in_1}$ and $e^{in_2}$. Then, $d(e^{in_1}, e^{in_2})<\epsilon$, then, the metric being rotation invariant, $d(e^{i(n_2-n_1)}, 1)<\epsilon$. Let $r:=n_2-n_1$. Now, each two consecutive points of the form $e^{i r s}$, $s=0, 1, 2\; ...\;$, are $\epsilon$-close.

Show that $\{e^{in}: n\in\Bbb N\}$ is Dense in the Unit Circle

2 Answers2