sin(n) subsequence limits set

We were given a challenge by our calculus professor and I’ve been stuck on it for a while now.

Show that the set of subsequence limits of $A_n=\sin(n)$ is $[-1, 1]$ (another way to phrase this would be: $\forall r\in [-1,1]$ there exists a subsequence of $A_n=\sin(n)$ that converges to $r$).

What would be a good way to start?

Solutions Collecting From Web of "sin(n) subsequence limits set"

Following Qiaochu hint, I’ll try to elaborate a bit.

Note: This is a complete rewrite of the proof to fix a flaw pointed out by Qiaochu and make it overall clearer.

First some notation. Let $S^1 = [0, 2\pi)$ with 0 and $2\pi$ identified. For two points $a, b \in S^1$ we’ll denote by $a \oplus b = a + b \mod 2\pi$ (and likewise for $\ominus$) and let $A_n = \{n \mod 2\pi\}_{n=1}^{\infty}$. Also denote by $U_{\varepsilon}(p)$ a punctured $\varepsilon$-neighborhood of $p \in S^1$.

Our strategy will be to show that $\{A_n\}$ has at least one limit point in $S^1$. From this we will conclude that 0 is also a limit point and finally we will use this fact to show that every point in $S_1$ is a limit point of $A_n$ (this means that $\{A_n\}$ is dense in S^1). Having established this, we will use continuity of $\sin$ to finally resolve the problem.


Proof

As a preparation we will note the relation $A_{n+m} = A_n \oplus A_m \quad (1)$. As a corollary of this we have that the sequence $\{A_n\}$ is injective (for if not, there would exist $n, m \in \mathbb{N}, k \in \mathbb{Z} \quad n>m$ such that $A_n = A_m$ and so $n – m = 2 k \pi$, a contradiction with irrationality of $\pi$). This implies simple but crucial fact that the image of the sequence contains infinitely many points.

Now, to establish the density of ${A_n}$ in $S^1$ we will first show that there exists at least one limit point $p \in S^1$. This is established by a standard argument: consider intervals $I = [0, \pi]$ and $J = [\pi, 2\pi]$ and take the one which contains infinitely many points of $\{A_n\}$ (if both do, take the “bottom” one, i.e. $I$). Call this interval $I_0$. Now again divide it in two intervals of same length and let $I_1$ be the one with infinitely many points. Continue in this way to obtain an infinite sequance of intervals $I_0 \supset I_1 \supset \cdots$. Then the set $K = \cap_{n=0}^{\infty} I_n$ is non-empty (this should be covered in standard calculus course, I hope) and any point $p \in K$ is surely a limit point (by construction of $\{I_n\}$).

Thanks to the above we now know that for each $\varepsilon > 0$ there exist infinitely many points in $U_\varepsilon(p)$, i.e. infinitely many $n, m \quad n > m$ such that $|A_n – A_m – 2k\pi| < 2\varepsilon$ for some $k \in \mathbb{Z}$. But from (1) we get that $|A_{n-m} – 2k\pi| < 2\varepsilon$. Because we identify 0 and $2\pi$ in $S^1$ we can see that these differences are concentrated around 0. Therefore we have shown that 0 is also a limit point.

Now we will show that every $p \in S^1$ is a limit point. Suppose we are given $\varepsilon > 0$. We just take any $A_k$ from $U_{\varepsilon}(0)$ and note that for $l = \lfloor {p \over A_k} \rfloor$ we have $A_{lk} \in U_{\varepsilon}(p)$.

To conclude the answer we will show that any $r \in [-1, 1]$ is indeed a limit point of $\{\sin(n)\}$. Take $s = \arcsin (r) \mod 2\pi$. Then by the above, we know that $s$ is a limit point of $\{A_n\}$ so there is a subsequence $\{A_{n_k}\} \to s$ and observe by continuity of $\sin$ that $\{\sin(A_{n_k})\} = \{\sin(n_k)\} \to sin(s) = r$.

Suppose you want to find positive integers $n, k$ such that $|n – 2 \pi k – \arcsin r| < \frac{1}{m}$ for some large positive integer $m$ and $r \in [-1, 1]$. Then I claim that some value of $n$ less than $2\pi m$ works. This is the pattern I was trying to get you to see, although I wasn’t doing a very good job of it (I should’ve asked you to plot $k$ instead of $n$).

Unfortunately I think this is a little hard to prove except when $r = 0$; fortunately, using the second half of Marek’s argument you can prove the same result with a weaker bound on $n$ for general $r$ using the case of $r = 0$, so I would encourage you to try that case first. To do this, try plotting the fractional parts of the numbers $\frac{n}{2\pi}$ in $[0, 1)$.