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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.

dmtri
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    Apply Dirichlet's approximation theorem (https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem) to $\alpha=2\pi$ and $N>\frac{1}{\varepsilon}$. –  Jan 03 '20 at 19:56
  • @StinkingBishop, Thanks a lot for the ref, it was easy and I missed it... – dmtri Jan 03 '20 at 20:17
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    Or even better, use Kronecker's approximation theorem, like here. – rtybase Jan 03 '20 at 21:00

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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 $1$.

fleablood
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  • Ah....Dirichlet's approximation theorem... that's what I wanted for the final step. I was sort of hand waving it into existence but it sure seemed like it'd have to work. – fleablood Jan 04 '20 at 06:44
  • Oh! Kronecker's Approximation Theorem. I knew someone had to have ironed out the details somehow.... – fleablood Jan 04 '20 at 06:45