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I'm aware of this other post, but I've approached the problem in a different manner, and there's a step I consider dobtful: Let $p$ in $[-1,1]$. There exists $q$ in $\mathbb{R}$ such that $cos(q)=p$. Now let $\epsilon>0$ be given. Since $cos(x)$ is continuous, exists $\delta$ such that if $|x-q|<\delta$, then $|cos x-p|<\epsilon$. Since $cos x$ is periodic, the $\delta$ is the same for all the numbers $y=q+2k\pi$, $k \in \mathbb{Z}$. The numbers $y$ can be written as $k(\dfrac{q}{k}+2\pi)$=$kr$. Now suppose $q$ irrational(rational is simpler) and hence $r$ irrational. There are infinitely many naturals at a distance less than $\delta$ to a multiple of $r$, and therefore there are infinitely many naturals with $|cos n-p|<\epsilon$. We get an infinite set $X_1$ of naturals which acompplish that inequality. Doing the same, but now taking $\dfrac{\epsilon}{2}$, we get an infinite set $X_2$ of naturals which accompplishes $|cos (n)-p|<\dfrac{\epsilon}{2}$.[Now comes the doubtful step] Doing this for all the $\dfrac{\epsilon}{i}$ ,$i$ in $\mathbb{N}$, we take $X=\cup^{\infty}_{i}X_i$, and the subsequence whose indexes are all the elements of $X$, converges to $p$.

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Diego
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  • so you used that the set of limits points of the sequence $u_n = n r - \lfloor nr \rfloor$ is $[0;1]$ ? and your doubtful argument makes sense for constructing a subsequence of $u_n = \cos(x_n)$ converging to $p$, and the sequence will be such that $x_n \in \cup_{i=n}^\infty X_i$ (or simply $x_n \in X_n$ since $X_{n+1} \subset X_n$) – reuns Feb 25 '16 at 04:11
  • The problem I've got is to repeat the process for every $\dfrac{\epsilon}{i}$, when they are an infinite number of steps. – Diego Feb 25 '16 at 04:16
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    that's a proof by induction, and if you prefer, take the minimum of $X_n \cap \mathbb{N}$ to be deterministic (so that you don't need the axiom of choice). – reuns Feb 25 '16 at 04:17
  • the algorithm simply becomes : $x_i = \min_{n \in \mathbb{N}} |cos(n) - p| < \frac{\epsilon}{i}$ (since you proved that minimum always exists, everything is ok) – reuns Feb 25 '16 at 04:19
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    and don't forget that if $r$ is rational, the limit points of $nr - \lfloor nr \rfloor$ are not $[0;1]$ at all $\implies$ this requires its own proof (similar to thisone) – reuns Feb 25 '16 at 04:21
  • Thinking it better, if it were q rational, then r still would be irrational, so I guess the proof is the same? – Diego Feb 25 '16 at 04:47

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