Proving $\displaystyle\limsup_{n\to\infty}\cos{n}=1$ using $\{a+b\pi|a,b\in\mathbb{Z}\}$ is dense

Question

I have seen What is $\limsup\limits_{n\to\infty} \cos (n)$, when $n$ is a natural number? and got this question. How can I proceed to prove $\displaystyle\limsup_{n\to\infty}\cos{n}=1$ If you know that $\{a+b\pi|a,b\in\mathbb{Z}\}$ is dense in $\mathbb{R}$?

My idea:

If we could prove that $\sup\{\cos{k}|k\geq n\}=1$ for any $n$, then we are done. So I tried to prove this.

Fix $n$.

For any $\delta>0$, there exists some $a,b\in \mathbb{Z}$ such that $-\frac{\delta}{2}-\frac{n}{2}<a+b\pi<\frac{\delta}{2}-\frac{n}{2}$ by density. This means that $|2a+n+2b\pi|<\delta$.

then by the continuity of $\cos{x}$, we have that $\forall \epsilon>0,\exists a\in\mathbb{Z}, |\cos{(2a+n)}-1|<\epsilon,$ which means that $1-\epsilon<\cos{(2a+n)}$. If $a\geq 0$ then we might have $\sup\{\cos{k}|k\geq n\}=1$ but it is possible that $a<0$. So this approach is not good.

I tried using $-\frac{\delta}{2}+\frac{n}{2}<a+b\pi<\frac{\delta}{2}-\frac{n}{2}$ But this only makes sense when $n<\delta$. Now I am lost.

Answer 1

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.

Answer 2

Just remember for the rest of your life: geometry is the most important thing in all the mathematics. :) Vladimir Arnold once said: "A good person draws a picture, a bad one does not".

Now, draw a circle. Then you know that $cos(n)$ is a point on a circle. Cosine takes $1$ if it's at the point of $2\pi k$ for $k \in ℤ$. The thing is that if you start at some point on the circle, say $cos(0)$, and then go for every $cos(n)$ with $n\in ℤ_+$, will cosine take same points at the circle? No. It will go for a finite set of same points only if you take points that's are multiples of $2\pi$. Otherwise, $cos(n)$ will always pick different points and more over, you can prove that for any arc of any length there will be infinite points taken by $cos(n)$.

So, you can take any arc centered at the point $0$ (i.e. where the axes $y=0$ and $x=1$) and get there points taken by $cos(n)$. It satisfies the definition of supremum. It's irrelevant from what $n$ to start of course, because the reasoning above applies nevertheless. So cosine and sine are dense on $[-1, 1]$ for $cos(n)$ where $n\inℤ$!

Although I did not give you an answer in analytic notation. Please consider what I said on the drawing! It will really improve your understanding of the problem. (and this exercise will help you in the future while learning differential equations and measures).