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I'm learning about Real Analysis, specifically accumulation points of a sequence, and need some help with this problem:

Show that the set of accumulation points of the sequence $a_n = \sin{n}$ is the closed interval $[-1, 1]$.

My thoughts:

Obviously we have $-1 \leq \sin{n} \leq 1 \forall n \in \mathbb{N}$.

The set $a_n$ will be of the form $a_n = \{\sin1, \sin2, \sin3, ...\}$.

Given $l \in [-1, 1]$ we want to show that there exists a subsequence $\{a_{n_k}\}$ such that $a_{n_k} \rightarrow l$ as $k \rightarrow \infty$.

This problem is different from the problems I'm used to solve in the sense that instead of finding the accumulation points of a sequence I need to prove that the accumulation points of a sequence is a closed interval and I don't know how to approach this.

glpsx
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  • Hint: you want to make an argument using the fact that the period of sin is not rational and n is. – Jürgen Sukumaran Jan 21 '16 at 18:52
  • Your second thought: Why do you think the set of accumulation points is that set, when you're trying to show the set of accumulation points is all of $[-1,1]?$ – zhw. Jan 21 '16 at 19:32
  • This is an obvious mistake on my part. Thank you for noticing me. – glpsx Jan 21 '16 at 19:41

2 Answers2

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Consider the bounded sequence $$n\mapsto z_n:=e^{in}\in{\mathbb C}\qquad(n\in{\mathbb N}_{\geq0})\ .$$ By Bolzano's theorem the $z_n$ have an accumulation point $\zeta\in S^1$. It follows that for any $\epsilon>0$ there are natural numbers $p<q$ with $|z_q-z_p|<\epsilon$. Put $r:=q-p$, and consider the subsequence $$k\mapsto w_k:=z_{k r}=e^{i kr}\quad(k\geq0)\ .$$ From $w_{k+1}-w_k=e^{ik r}(e^{ir}-1)$ it follows that $|w_{k+1}-w_k|=|e^{iq}-e^{ip}|$; hence consecutive $w_k$ have a distance $<\epsilon$. From this we can conclude that the set $Z:=\{z_n\>|\>n\geq0\}$ is in fact dense on $S^1$.

Now let an arbitrary $y\in[{-1},1]$, as well as an $\epsilon>0$, be given. There is an $x\geq0$ with $z:=(x,y)\in S^1$, and there is a $z_n\in Z$ with $|z-z_n|<\epsilon$. It follows that $$|y-\sin n|\leq|z-e^{in}|=|z-z_n|<\epsilon\ .$$ As $\epsilon>0$ was arbitrary we can conclude that $y$ is an accumulation point of the sequence $(\sin n)_{n\geq0}$.

Christian Blatter
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  • This is a nice answer! (Could you have also used the Pigeonhole Principle to conclude that for any $\epsilon>0$ there are natural numbers $p<q$ with $|z_q-z_p|<\epsilon$?) – user84413 Mar 13 '16 at 16:31
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Hint: This is equivalent to show that the set of rest of the division of $n\in N$ by $2\pi$ is dense in $[0,2\pi]$. This follows from the fact that the subgroup generated by $n\in N$ and $2\pi$ is dense in $R$.

Tsemo Aristide
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