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!
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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.
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@Gerry Myerson Ah, very true. Thanks for pointing it out. – EuYu Aug 28 '12 at 00:22
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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 00:49
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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 00:50
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The link you posted earlier said that cos(n) is dense in [-1,1]. – Ted Johnson Aug 28 '12 at 00:55
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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 01:00
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Minor nitpickery: while it still follows easily from density, you might want to note that you can find a 'sufficiently large' $n$ satisfying your condition; in particular, given $\epsilon$ and some $n_0$, you can find $n\gt n_0$ s.t. $\cos(n)\in(x-\epsilon,x+\epsilon)$ - this lets you produce an increasing sequence of $n$s. – Steven Stadnicki Aug 28 '12 at 01:09
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@Steven Stadnicki I've added the corrections. Thank you. – EuYu Aug 28 '12 at 01:18