Some question about proving $\displaystyle\limsup_{n\to\infty}|\cos{n}|=1$ by using density of $\{a+b\pi|a,b\in\mathbb{Z}\}$ I have seen Proving $\displaystyle\limsup_{n\to\infty}\cos{n}=1$ using $\{a+b\pi|a,b\in\mathbb{Z}\}$ is dense and got this question.
Hagen von Eitzen gave the solution as following:
Pick an integer $n$. By density of $\Bbb Z+\pi\Bbb Z$, there exist $a_n,b_n\in\Bbb Z$ with $\frac 1{n+1}<a_n+b_n\pi<\frac1 n$. If $a_m=a_n$, then $|b_n\pi-b_m\pi|<1$, which implies $b_n=b_m$ and ultimately $n=m$. We conclude that $|a_n|\to \infty$.
As $$\cos|2a_n|=\cos 2a_n=\cos(2a_n+2\pi b_n)>\cos\frac 2n\to 1, $$
the desired result follows.
I was wondering why  $|a_n|\to \infty.$ Could someone give more details about it? –Moreover, is  $|a_n|$ increasing to $\infty?$
 A: Hagen von Eitzen shows that $a_m=a_n$ implies $m=n$, so the integers $a_1,a_2,a_3,...\;$are distinct.

Hence for any fixed positive integer $N$, we must have $|a_n| \ge N$ for all but at most $2N-1$ values of $n$.

It follows that $|a_n|$ approaches infinity as $n$ approaches infinity.

However it's not automatic that the sequence $(|a_n|)$ is an increasing sequence.
 
A: To facilitate beginners, I assume as little as possible and make the proof self-contained.
Proposition 1: Let $\alpha\in(0,1)$ be an irrational number. Let $A=\{m+n\alpha\mid m,n\in\mathbb{Z}\}$.
We have that $\inf\{x\mid x\in A\mbox{ and }x>0\}=0$.
Proof of Proposition 1: We remark that $A\cap(0,1)\cap\mathbb{Q}=\emptyset$.
For, if there exists $x\in A\cap(0,1)\cap\mathbb{Q},$we may write
$x=m+n\alpha$ for some $m,n\in\mathbb{Z}$. If $n\neq0$, then $\alpha=(x-m)/n\in\mathbb{Q}$,
which is a contradiction. If $n=0$, then $x=m$, contradicting to
$x\in(0,1)$. We will need this fact at later time.
Prove by contradiction. Suppose the contrary that $\inf\{x\mid x\in A\mbox{ and }x>0\}=\beta>0$.
Note that $\alpha\in A$ and $\alpha>0$, so $\beta\leq\alpha<1$.
Consider two cases.
Case 1: $\beta$ is rational. Suppose that $\beta=\frac{m}{n}$ for
some $m,n\in\mathbb{N}$. Choose $x\in A$ such that $0\leq x-\beta<\frac{1}{2n^{2}}$,
then $0\leq nx-m<\frac{1}{2n}$. Since $A$ is a subgroup of $(\mathbb{R},+$),
$nx-m\in A$. If $nx-m>0$, it contradicts to the fact that $\beta$
is the infimum of the set $\{x\mid x\in A\mbox{ and }x>0\}$. Therefore
$nx-m=0$ and hence $x=\frac{m}{n}\in\mathbb{Q}$, contradicting to
$A\cap(0,1)\cap\mathbb{Q}=\emptyset$.
Case 2: $\beta$ is irrational. Let $k=\max\{k\mid k\beta<1,k\in\mathbb{N}\}$,
then $k\beta<1<(k+1)\beta$. We have that $0<1-k\beta\leq\frac{1}{2}\beta$
or $0<(k+1)\beta-1\leq\frac{1}{2}\beta$. Consider two sub-cases.
Case 2.1: Suppose that $0<1-k\beta\leq\frac{1}{2}\beta$. Choose $x\in A$
such that $0\leq x-\beta<\frac{1}{2k}(1-k\beta)$. Then, $0\leq kx-k\beta<\frac{1}{2}(1-k\beta)$
and hence $1-kx>1-[k\beta+\frac{1}{2}(1-k\beta)]>0$. That is, $0<1-kx\leq1-k\beta\leq\frac{\beta}{2}$.
Note that $1-kx\in A$ and $1-kx>0$. This contradicts to the fact
that $\beta$ is the infimum of the set $\{x\mid x\in A\mbox{ and }x>0\}$.
Case 2.2: Suppose that $0<(k+1)\beta-1\leq\frac{1}{2}\beta$. Choose
$x\in A$ such that $0\leq x-\beta<\frac{1}{2(k+1)}\left[(k+1)\beta-1\right]$,
then $0\leq(k+1)x-(k+1)\beta<\frac{1}{2}\left[(k+1)\beta-1\right]$.
It follows that $0<(k+1)x-1<\frac{3}{2}\left[(k+1)\beta-1\right]\leq\frac{3}{4}\beta$.
Note that $(k+1)x-1\in A$ and $(k+1)x-1>0$. This contradicts to
the fact that $\beta$ is the infimum of the set $\{x\mid x\in A\mbox{ and }x>0\}$.

Corollary 2: For any irrational $\alpha>0$, $\inf\{m+n\alpha\mid m,n\in\mathbb{Z}\mbox{ and }m+n\alpha>0\}=0$.
Proof: Let $\beta=\alpha-[\alpha]$, where $[\alpha]$ denotes the
largest integer that is not greater than $\alpha$, then $\beta$
is an irrational and $\beta\in(0,1)$. Note that $\{m+n\alpha\mid m,n\in\mathbb{Z}\mbox{ and }m+n\alpha>0\}=\{m+n\beta\mid m,n\in\mathbb{Z}\mbox{ and }m+n\beta>0\}$.
The result follows from Proposition 1.

Proposition 3: For any $\alpha>0$ such that $\frac{\alpha}{\pi}$
is irrational, $\limsup_{n\rightarrow\infty}\cos(n\alpha)=1$.
Proof of Prop 3: Let $\varepsilon>0$. Choose $\delta\in(0,1)$ such
that $|\cos(x)-1|<\varepsilon$ whenever $x\in(-\delta,\delta)$.
Let $N\in\mathbb{N}$ be arbitrary. Since $\frac{\alpha}{2\pi}$ is
irrational, there exist $m,n\in\mathbb{Z}$ such that $0<m+n(\frac{\alpha}{2\pi})<\frac{\delta}{4N\pi}$.
If $n=0$, we have $0<m<\frac{\delta}{4N\pi}<1$ which is impossible.
Therefore $n\neq0$. If $n>0$, we have $0<mN\cdot2\pi+(nN)\alpha<\frac{\delta}{2}$.
It follows that $\cos((nN)\alpha)=\cos\left(mN\cdot2\pi+(nN)\alpha\right)\in(1-\varepsilon,1]$.
Note that $nN\geq N$. Hence $\sup_{k\geq N}\cos(k\alpha)>1-\varepsilon$.
If $n<0,$ we have $-\frac{\delta}{2}<-mN\cdot2\pi+(-nN)\alpha<0$,
so $\cos\left((-nN)\alpha\right)=\cos\left(-mN\cdot2\pi+(-nN)\alpha\right)\in(1-\varepsilon,1]$.
Note that $-nN\geq N$. Therefore, we also have $\sup_{k\geq N}\cos(k\alpha)>1-\varepsilon$.
This shows that $\limsup_{n\rightarrow\infty}\cos(n\alpha)=1$.
