Compact Projections to $S^1$ Let $S^{1} \subset \mathbb{R}^2$ be the circle, and let $\pi_i : S^1 \to \mathbb{R}$ the natural projections. Let $\pi_{1}^{-1} ( [0,\frac{1}{2}]) \subset S^1$. 
Is $\pi_{1}^{-1} ( [0,\frac{1}{2}])$ compact?
I think it is, but I don't know how to prove it.
This question comes from the fact that most probably I have misunderstood a comment I received in another question. Anyway, this question is particularly important for me, because I have quite some problems with proving that a specific space is compact.
Any feedback or help is most welcome.
Thank you for your time.
 A: Some important prerequisites:


*


*$\mathbb{R}^2$ is Hausdorff and Complete.


* $S^1$ is closed in $\mathbb{R}^2$.


*A closed subset of a complete space is also a complete space (and thus Hausdorff). 


* A closed subset of a compact Hausdorff space is compact.


* $\pi:S^1\to \mathbb{R}$ is continuous.


*The preimage of a closed set under a continuous function is closed. 



If you are comfortable will all of these facts, the rest is as follows:
Notice that $S^1$ is closed and bounded in $\mathbb{R}^2$, so it is complete, and thus compact and Hausdorff. Since $\pi$ is continuous and $\left[0,\frac{1}{2}\right]$ is closed, $\pi^{-1}\left(\left[0,\frac{1}{2}\right]\right)$ is closed. But this means that $\pi^{-1}\left(\left[0,\frac{1}{2}\right]\right)$ is a closed subset of a compact Hausdorff space, so $\pi^{-1}\left(\left[0,\frac{1}{2}\right]\right)$ is itself compact.
A: $S^{1}$ is compact and $\pi_{1}$ is a continuous function on $S^{1}$, and $[0,\frac{1}{2}]$ is a closed subset of the range of $\pi_{1}$. Use the fact that preimages of closed sets under continuous maps are closed, and closed subsets of compact sets in the Euclidean topology are compact.
Note that in particular $\pi_{1}^{-1}(F)$ is a compact subset of $S^{1}$ for any closed set $F\subseteq\mathbb{R}$. So the choice of $[0,\frac{1}{2}]$ plays no particular role in this case.
