Subsequential limit of sequence

I'm trying to determine all subsequential limit points of the following sequence

X_n = cos(n)


Not sure how to decompose this into subsequences.

Anyone know how to tackle this problem?

Thanks!

-

1 Answer

It is rather well known that the image of the integers under sine and cosine is dense in $[-1, 1]$. For a reference, see the related question here.

edit in response to op's comment

With the knowledge that $\cos(n)$ is dense in $[-1, 1]$, we shall show that any $x\in[-1,1]$ is a sub-sequential limit point. Indeed, given any $\epsilon > 0$ and some $n_0 > 0$, we can find $n > n_0$ such that $$\cos(n) \in (x - \epsilon, x + \epsilon)$$ so we can iteratively pick a sub-sequence that lies within this $\epsilon$-neighborhood. This shows that any point is a sub-sequential limit point of the sequence.

-
@Gerry Myerson Ah, very true. Thanks for pointing it out. –  EuYu Aug 28 '12 at 0:22
Does that mean all subsequential limit points fall between [0,1]? I'm having trouble relating density to subsequential limits. –  Ted Johnson Aug 28 '12 at 0:49
It means that every point in $[0,1]$ is a sub-sequential limit point. I will make an edit to make this more explicit. –  EuYu Aug 28 '12 at 0:50
The link you posted earlier said that cos(n) is dense in [-1,1]. –  Ted Johnson Aug 28 '12 at 0:55
My mistake, it should be $[-1, 1]$. For some reason I was interpreting the sequence as positive. The argument still holds though. Every point in $[-1, 1]$ is a limit point of the sequence. –  EuYu Aug 28 '12 at 1:00