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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?

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    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.2010-11-07
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    @daniel.jackson maybe you want to delete the 0 from your image. That way your cardinality grows with your reputation.2010-11-07
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    @Qiaochu: not sure this is what you meant, but for `x = arcsin(r) + 2πk`?2010-11-08
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    @daniel.jackson: yes. How can you get n to be close to one of these numbers?2010-11-08
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    @Qiaochu: not really sure how.2010-11-08
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    @daniel.jackson: at this point I would encourage you to do some experimentation, for example with a calculator. Pick some value of r and compute the numbers of arcsin(r) + 2 pi k. See if you can come up with any conjectures about when these numbers are close to integers.2010-11-08
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    @Qiaochu: OK, here's what I tried http://pastebin.com/raw.php?i=AZsdWjGW2010-11-08
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    @daniel.jackson: cool. What result did you get?2010-11-08
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    @Qiaochu: well as I predicted, the closer I want to get to an integer, the further I have to go in the naturals. But I have no idea how this helps, I mean it seems obvious that at some point I'll get close enough since I can go to infinity.2010-11-08
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    @daniel.jackson: if you can predict how much further you have to go, you'll have a proof. It might help to write down a list of the first n that works with various choices of error parameter.2010-11-08
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    @Qiaochu: here are some numbers http://pastebin.com/raw.php?i=pnhSL1rr - am I missing something obvious here?2010-11-08
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    @daniel.jackson: try printing 1/epsilon instead of epsilon, and use more values of epsilon.2010-11-08
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    @Qiaochu: you mean for smaller values of epsilon? This is how the test app currently looks like: http://pastebin.com/raw.php?i=YVX9fbc62010-11-08
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    @daniel.jackson: you won't see it until you print 1/epsilon instead of epsilon.2010-11-08
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    @Qiaochu: like this http://pastebin.com/raw.php?i=aV0UbNUY ?2010-11-08
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    @daniel.jackson: you could just be printing m, but sure. What kind of output do you get?2010-11-08
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    @Qiaochu: there are certain values of n that repeat for many many epsilons2010-11-08
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    @daniel.jackson: this is actually a red herring, and it will disappear if you use larger values of m. But do you notice anything about the relationship between m and n?2010-11-08
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    @Qiaochu: I'm afraid not, this sucks. Here's a more compact output, it only prints when n changes http://pastebin.com/raw.php?i=NnXpJFZP2010-11-08
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    See also: http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-12015-01-18

2 Answers 2

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

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    You don't need Bolzano-Weierstrass, and the last line of the argument is wrong.2010-11-07
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    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.2010-11-07
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    Thanks for the detailed answer, unfortunately I don't think I've learned the required material to understand your solution.2010-11-08
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    @Marek: The last line of the argument is still wrong.2010-11-08
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    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.2010-11-08
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    @daniel.jackson: The argument (flawed but that can be fixed) is just about epsilon-delta gymnastics and properties of sequencies, which I thought you should understand, being a calculus student. If this is not so, please let me know, and I'll try to explain it more clearly (also, I'll of course fix the problem Qiaochu mentioned).2010-11-08
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    @Marek: We haven't learned what's a dense sequence.2010-11-08
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    @Marek: rewriting the solution is fine.2010-11-08
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    @daniel.jackson: sorry for that. I used the term just as the terminology, you don't need to know any properties of dense sets. But just that you know, a set $A \subset B$ is dense in $B$, if points of $B$ can be approximated by points of $A$ to arbitrary precision (informally speaking). E.g. $\mathbb{Q}$ is dense in $\mathbb(R)$. This is esentially also what your problem is about.2010-11-08
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    @Marek: 1) the step where you claim that A_{lk} is in U_e(p) is still wrong, or at least I do not understand how you are justifying it. In any case as I have said, you do not need to use compactness at all; the argument is one or two lines. 2) You may not be aware of this, but in America (which I am guessing is where daniel.jackson is from) "calculus" is not the same as "real analysis." We don't generally talk about compactness or the finite intersection property in our calculus courses, so even the parts of this argument which are valid aren't likely to be helpful to the OP at all.2010-11-08
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    1) Well, by homomorphism property we have A_{lk} = A_k + ... + A_k = lA_k (if you assume epsilon << p and A_k > 0 but you can always take such A_k), so this should be completely obvious. One of us must be missing something. By the way I realize that there might be (a lot) simpler proof, but this is the only thing I came up with.2010-11-08
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    @Qiaochu: 2) Very interesting, that never occurred to me. But I assume that the most important points of topology aren't omitted (at least in disguised form). One doesn't have to mention normed vector spaces to discuss properties of |.| norm, etc. So in the same vein, intersection of intervals could be covered even without any mention of compactness, right? Just wondering :-)2010-11-08
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    @Marek: 1) all that tells you is that A_{lk} is in U_{le}(0). 2) No, calculus students generally don't learn any topology. All most calculus students get is the intermediate value theorem.2010-11-08
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    1) That too, but that's not important. Consider p = 2.17, e = 0.2 and A_k = 0.1. Then l = 21, A_kl = 2.1 and its surely in U_e(p). The same construction works in general. 2) I see. Thank you for pointing that out, I'll try to keep it in mind.2010-11-08
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    @Marek: ah! My apologies.2010-11-08
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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 $[0, 1)$.

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    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.2010-11-09