What are the limit points of $\tan(\mathbb{N})$ in $\mathbb{R}$?

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.

Solution 1:

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 $0\le i<n$. 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 $[0,1)$.

Solution 2:

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.