1

Show that if $r\in\mathbb{R}\backslash \mathbb{Q}$, then $\{e^{i2\pi r n}\}_{n\in\mathbb{N}}$ have at least one limit point

I've been sitting with this problem for at while now, but can't figure it out. I've tried to rewrite it to its real form, because I know that a bounded reel sequence have at least one limit point, but that doesn't do it for me

njlieta
  • 411

3 Answers3

2

Hint: The unit circle (or even the closed unit disc) is compact. And if $r$ is irrational, then the $e^{i2\pi rn}$s are all different ...

hmakholm left over Monica
  • 286,031
1

Hint: The set $R$ of the remainders of the division of $2\pi nr, n\in N$ by $2\pi$ is dense in $[0,2\pi]$. Remark that $R$ is $2\pi$ $\{$ $rn-\lfloor rn \rfloor$, $n\in N\}$.

Tsemo Aristide
  • 87,475
1

Your approach works fine, just apply it twice.

The real part of the sequence is the sequence $(\cos (2\pi r n))_n$ this is clearly bounded and thus has a convergent subsequence $(\cos (2\pi r n_k))_k$.

The imaginary part of the respective subsequence of the original sequence is $\sin (2\pi r n_k)_k$. This is again bounded and you find a convergent subsequence $\sin (2\pi r n_{k_l})_l$. Of course $(\cos (2\pi r n_{k_l}))_l$ is also convergent, and you are done—the complex sequence converges as real and imaginary part converge.

quid
  • 42,135
  • But the sequence of all $1$ is convergent, yet the set ${1}$ has no limit point, so shouldn't you also have that almost every term of your sequence is different from the limit? – Hetebrij Jun 08 '16 at 20:46
  • I took the notation to mean the sequence not the set. I would never denote a set like this. But if it is the set, one can remark in addition that not two terms differ by a multiple of $2\pi$, i.e. they are all different, for if $2\pi rn = 2\pi rm + 2\pi k $ with integral $k$ it would follow that $r$ is rational. – quid Jun 08 '16 at 20:51