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I was working on a an old worksheet problem here. It asks

Let $S=\{\tan(k):k=1,2,\dots\}$. Find the set of limit points of $S$ on the real line.

The answer is $(-\infty,\infty)$. Intuitively I feel that if we keep evaluating tangent at positive integer points, they will be so scattered over the real line that we could always construct some subsequence converging to any real number. How can this be made rigorous to get this purported conclusion? Thanks.

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    I don't have the energy to type out a full answer, but I'd start by thinking "what would it mean if the conclusion was false?" and then trying to show why that doesn't happen (roughly speaking, trying to show that there aren't any points that are "missed")2012-09-17

3 Answers 3

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For each real number $x$ there is a unique integer $n_x$ such that $n_x\pi\le x<(n_x+1)\pi$; let $\hat x=x-n_x\pi\in[0,\pi)$, and observe that $\hat x$ is the unique element of $[0,\pi)$ such that $\tan\hat x=\tan x$. Thus, $\{\tan k:k\in\Bbb N\}=\{\tan\hat k:k\in\Bbb N\}$. Let $D=\{\hat k:k\in\Bbb N\}$. It suffices to show that $D$ is dense in $[0,\pi)$: the tangent function is continuous and maps $[0,\pi)$ onto $\Bbb R$, so $\tan[D]=\{\tan\hat k:k\in\Bbb N\}$ must then be dense in $\Bbb R$.

Note that for any $x,y\in\Bbb R$, $\hat x=\hat y$ iff $\frac{x}{\pi}-\frac{y}{\pi}\in\Bbb Z$. Thus, instead of showing that $D$ is dense in $[0,\pi)$, we can scale everything by a factor of $1/\pi$ and show that $D_0=\{\hat k/\pi:k\in\Bbb N\}$ is dense in $[0,1)$.

This is a nice application of the pigeonhole principle. Let $n$ be a positive integer, and divide $[0,1)$ into the $n$ subintervals $\left[\frac{k}n,\frac{k+1}n\right)$ for $k=0,\dots,n-1$. Two of the $n+1$ numbers $\frac{\hat k}{\pi}$ for $k=0,\dots,n$ must belong to the same one of these subintervals; say $\frac{\hat k}{\pi},\frac{\hat\ell}{\pi}\in\left[\frac{i}n,\frac{i+1}n\right)\;,$ where $0\le k<\ell\le n$ and $0\le i. Then $0<\left|\frac{\hat\ell}{\pi}-\frac{\hat k}{\pi}\right|<\frac1n\;.$ Let $m=\ell-k$; then $\dfrac{\hat m}{\pi}\in\left[0,\dfrac1n\right)$ if $\hat\ell-\hat k>0$, and $\dfrac{\hat m}{\pi}\in\left[1-\dfrac1n,1\right)$ if $\hat\ell-\hat k<0$.

In the first case let $N$ be the smallest positive integer such that $\dfrac{N\hat m}{\pi}>1$, and in the second let $N$ be the smallest positive integer such that $N\left(1-\dfrac{\hat m}{\pi}\right)>1$. Then every point of $[0,1)$ is within $1/n$ of one of the multiples $\dfrac{\widehat{jm}}{\pi}$ for $j=1,\dots,N-1$. Thus, every $x\in[0,1)$ is within $1/n$ of some element of $D_0$, and since $n$ was arbitrary, $D_0$ is dense in $[0,1)$.

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    @joriki: You’re right: I tried to make it simpler than it actually is.2012-09-19
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For convenience's sake we use $2\pi$ instead of $\pi$. You need to prove the images of the $a\rightarrow a\pmod{2\pi}, a\in \mathbb{N}$ is equidistributed. This can be done by Weyl's criterion that modified with $\pmod{2\pi}$ instead of $\pmod{1}$. We need to prove that $ \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n-1}_{0}e^{2ilx_{j}}=0, \forall l\in \mathbb{Z} $ with $x_{j}$ be $j$'s image by the quotient map. The summation is a geometric series with $e^{2il}, e^{2i2l},e^{2i3l}$, etc. So we have $\sum^{n-1}_{0}e^{2ilx_{j}}=\frac{1-e^{2iln}}{1-e^{2il}}$ which can be bounded by $\left|\frac{1-e^{2iln}}{1-e^{2il}}\right|\le \frac{2}{|1-e^{2il}|}$ which is finite since $l$ is fixed. Therefore the limit $ \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n-1}_{0}e^{2ilx_{j}}=0, \forall l\in \mathbb{Z} $must be 0, and the original sequence be equidisributed.

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    Updated. Thanks for pointing out.2012-09-17
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The function $\tan(x)$ is continuous and $2\pi$-periodic. Suppose I want to find a sequence that converges to the real number $y$. Let $y = \tan(x)$ for some real number $x$. Well, we know that the multiples of an irrational number are equidistributed modulo $1$. Thus, we pick some natural number $N_1$ so that $x + 2N_1\pi$ is really close to $0$ modulo $1$. This means that $x + 2N_1\pi$ is really close to some integer, say $M_1$. Then $\tan(x+2N_1\pi) = \tan(x) = y$ is very close to $\tan(M_1)$ by continuity, so $\tan(M_1)$ is really close to $y$. To get the next integer $M_2$, pick $N_1$ so that $x + 2N_2\pi$ is even closer to $0$ modulo $1$, which is to say that $M_2$ is even closer to $x + 2N_2\pi$ then $M_1$ was close to $x + 2N_1\pi$. By continuity, this means the approximation $\tan(M_2)$ should be even close to $y$ than $\tan(M_1)$. Repeat.

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    @Sasha: yeah, I guess my proof cause a misleading effect.2012-09-17