What is the supremum of $A=\{\cos{n},n\in\Bbb{N}\}$ I would like to find the supremum of the following set:
$$A=\{\cos{n},n\in\Bbb{N}\}$$
Is it $1$ or something smaller?
One way to show that the supremum is $1$ (If it were), is to prove that for any $\epsilon>0$, there are positive integers $p,q$ such as $\vert2\pi q-p\vert<\epsilon$  , but I have not managed to do that...
Any suggestions-solutions, to this problem? Thanks in advance.
 A: Well the answer would be $1$ if it were true that for any $\epsilon > 0$ we can find $n_\epsilon$ so that $|\cos n_\epsilon -1| < \epsilon$.
Now each natural $n$ is congruent $\pmod {2\pi}$ to some value $n': -\pi \le n' < \pi$, or in other words $n = n' + 2k\pi$ for some integer $k$.  And we want to show that $n'$ can, for values of $n$, be arbitrarily close to $0$.  ($n'$ can't actually every equal $0$ because that would mean $n = 2k\pi$ and $\pi$ is irrational so that can't happen [unless $n$ and $k$ equal $0$]).
If we go the other way, though for any $d > 0$ where $d$ is pretty dang small, can we choose $d$ small enough that $\cos (\pm d) =\cos d > 1- \epsilon$?  Well the answer to that has to be, yes.  $\cos$ is a continuous function, so for any $\epsilon > o$ there is a $\delta$ so that $|x-0| <\delta$ (or in other words $-d < x < d$) then $|\cos x - \cos 0|=|\cos x - 1| < \epsilon $ (or in other words $1-\epsilon < \cos x$.
So Let $\delta$ be so that $-\delta < x < \delta \implies 1-\epsilon < \cos x$.
Now we need to find an $n$ so that $|2k\pi - n| < \delta$ for some integer $k$.
Well for that to happen we need $n-\delta < 2k\pi < n+ \delta$ or $\frac n{2k}-\frac {\delta}{2k} < \pi < \frac n{2k}+\frac {\delta}{2k}$ which is clearly possible by the archimedian principal.
So, yep.  Supremum is $1$.
