Is the set $U=\{ x \in \mathbb{R} : \sin (x) = \frac{1}{2} \}$ closed or open set? I tried to simplify the set definition by solving for different values of $x$'s and ended up with this set. 
$U = \{ ... , -\pi-\frac{\pi}{6} , \frac{\pi}{6} , \pi-\frac{\pi}{6}, ... \}$
I know that a set is said to be closed if all of its cluster points belong to the set. 
I don't know how to go ahead with the problem. 
Thanks for the help in advance 
Edit 1: From the comments although I don't know exactly why the set is closed but yes the set U is closed. Then π/6 is a cluster point i.e every neighborhood of π/6 has atleast one element in U leaving π/6 . How is that true?
$\sin(x): \mathbb{R} \to [-1,1]$ is a continuous function. Now since $\{\frac{1}{2} \}$ is closed in $[-1,1]$ and $U=\sin^{-1}\left(\frac{1}{2}\right)$ we get that $U$ is closed in $\mathbb{R}$.
Take any arbitrary $x$=nπ + $(-1)^n$$\frac{π}{6}$.
Now let $\epsilon$ &lt; min {[(n+1)π + $(-1)^{(n+1)}$$\frac{π}{6}$] - [nπ + $(-1)^n$$\frac{π}{6}$], [nπ + $(-1)^n$$\frac{π}{6}$] - [(n-1)π + $(-1)^{(n-1)}$$\frac{π}{6}$] }.
Although it looks a bit complex but it's not at all complex. If you see the format of any three consecutive numbers in U, you'll understand what I have written in that minimum part. Now let's say $x_1$, $x$, $x_2$∈U such that $x_1$ &lt; $x$ &lt; $x_2$. Now what I have done is I have taken $\epsilon$&lt;min{$x$-$x_1$, $x_2$-$x$}.(You can take equality but &lt; makes the argument stronger and more explicit.) Then the neighbourhood N($x$,$\epsilon$) contains no point of U.This shows that there exists a neighbourhood of $x$ such that it doesn't contain any point of U. Hence, x is not a limit point. (Remember the definition of limit point). Since x is arbitrary, none of the points of U is a limit point. Hence, derived set U$'$ = {$\phi$} $\subset$ U and therefore, U is a closed set.