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This question seems really hard, I'm trying to prove that the set of the adherent values of the sequence $x_n=\cos (n)$ is the closed interval $[-1,1]$, i.e., every point of this interval is a limit of a subsequence of $x_n$, and also the limit of any subsequence of $(x_n)$ is in $[-1,1]$

It's obvious that every adherent values is in $[-1,1]$, I'm having troubles to prove the converse, i.e., a point in $[-1,1]$ is an adherent value of $(x_n)$.

I need help

Thanks a lot

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up vote 6 down vote accepted

One way to do this:

  • The set $\def\ZZ{\mathbb Z}A=\ZZ+2\pi\ZZ$ is dense in $\mathbb R$. This can be done by mimicking my answer here

  • As a consequence, the set $B=\{\exp(in):n\in\mathbb Z\}$ is dense in the unit circle $S^1\subseteq\mathbb C$. Indeed, $B$ is the image of $A$ under the map $t\in\mathbb R\mapsto\exp(i t)\in S^1$, which is continuous surjective, and the image of a dense set under a surjective map is dense.

  • The set $C=\{\cos n:n\in\mathbb N\}$ is dense in $[-1,1]$. Indeed, the function $a+bi\in S^1\mapsto a\in[-1,1]$ is continuous, surjective, and maps the set $B$ to the set $C$.

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After you have that $A$ is dense, wouldn't it suffice to say that $C$ is the image of $A$ through $\cos$, and since $A$ is dense, and $\cos$ is continuous and surjective onto $[-1,1]$, $C$ is dense? The set $B$ seems slightly artificial to me. – Jakub Konieczny Feb 18 '13 at 6:51
Heh, Indeed. I tend to think of density in the circle as the more fundamental result: that's why I think of this this way. – Mariano Suárez-Alvarez Feb 18 '13 at 6:53

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