Why is $\{\sin(n): n\in \mathbb{N}\}$ not a closed set in $\mathbb{R}$?
I am considering $\mathbb{R}$ with the usual topology but I have no real intuition for finding the limit points of this set.
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Because it's dense in $[-1, 1]$. – Jakobian May 11 '21 at 15:20
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5How do we know that? – Partey5 May 11 '21 at 15:21
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See my answer here for example. There should be elementary methods as well. – Jakobian May 11 '21 at 15:24
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$[-1, 1]$ is closed but also dense in $[-1, 1]$, in case someone misunderstands the hint by Jakobian. – Improve May 11 '21 at 15:25
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@Improve The set is countable so if it is dense in $[-1,1]$ then any point in that interval not in the set cannot have a neighborhood around it not in the set. We still must prove that the set is dense in that interval but if it is, the result follows. – John Douma May 11 '21 at 15:35
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Take a look at http://users.auth.gr/~siskakis/sin(n).pdf It is interesting to note that a Google search of this yields some really poor answers including an accepted answer on this site. – John Douma May 11 '21 at 16:14
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A good way to understand this set is to imagine the set of points ${(\cos(n),\sin(n)): n \in \mathbb{N}}$. One can prove these are dense on the unit circle with the pigeonhole principle and the irrationality of $\pi$. – TomKern May 11 '21 at 16:21