# $\sin(n)$ 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?

• Pick r. What are the values of x such that sin x = r? How can you get n to be close to one of these values? These are the obvious questions you should be asking yourself. Nov 7 '10 at 20:32
• @daniel.jackson maybe you want to delete the 0 from your image. That way your cardinality grows with your reputation. Nov 7 '10 at 22:43
• @Qiaochu: not sure this is what you meant, but for x = arcsin(r) + 2πk? Nov 8 '10 at 9:15
• @daniel.jackson: yes. How can you get n to be close to one of these numbers? Nov 8 '10 at 10:22
• @Qiaochu: not really sure how. Nov 8 '10 at 15:45

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 $\{\sin(A_{n_k})\} = \{\sin(n_k)\} \to sin(s) = r$.

• You don't need Bolzano-Weierstrass, and the last line of the argument is wrong. Nov 7 '10 at 22:32
• Thank you for pointing that out. I now see that it suffices to use the injectivity of the sequence to obtain the desired approximation property. Nov 7 '10 at 23:22
• Thanks for the detailed answer, unfortunately I don't think I've learned the required material to understand your solution. Nov 8 '10 at 9:16
• @Marek: The last line of the argument is still wrong. Nov 8 '10 at 10:23
• I am aware of that Qiaochu. But I wonder what I should do? Rewrite the solution completely? Or post new answer with correct solution? I am not sure what the policy is, being new to the site. Nov 8 '10 at 11:32

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 $[0, 1)$.

• This works because $1/2\pi$ is irrational. More generally, the fractional parts of the sequence $n\alpha$ is dense in the unit interval for any irrational $\alpha$, and is not dense for any rational $\alpha$. So, at the very least, any proof should mention irrationality. Nov 9 '10 at 1:21