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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.

Community
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Kolmin
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    Well, the first question is to identify $\pi_1^{-1}([0,1])$. Have you done so? – Kevin Carlson Mar 19 '15 at 20:23
  • I have deleted the [tag:self-learning] tag, as this question is not about the process of self-learning. –  Mar 19 '15 at 20:31
  • While it would be helpful for intuition to identify the given preimage, it's probably not neccessary in this scenario. Are you able to assume that the projection is continuous? – rnrstopstraffic Mar 19 '15 at 20:31
  • @rnrstopstraffic I think any student who takes more than half a second to identify the given preimage should avoid using any general facts for this problem. – Kevin Carlson Mar 19 '15 at 20:52
  • @MikeMiller I agree that it's valuable to identify the preimage, but we're going to use general facts at some point, no? Trying to prove that this particular set is compact using first principles would involve particular versions of the proofs of said general facts. – rnrstopstraffic Mar 19 '15 at 20:59
  • @MikeMiller: I have used the self-learning tag as it has been used in the past, which is not the strict meaning of the tag, but as a signal that the OP is a self-learner. You can find some posts on the meta part of the site concerning the issue. – Kolmin Mar 19 '15 at 20:59
  • I have changed the original question, because I realized that it did not really tackle the issue I was interested in, namely that from the preimage of a prejection of a compact set we get two discconnected subsets of $S^1 \subset \mathbb{R}^2$. – Kolmin Mar 19 '15 at 21:01
  • The idea remains the same; the connected or not, the preimage is closed. – rnrstopstraffic Mar 19 '15 at 21:11

2 Answers2

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Some important prerequisites:

  1. $\mathbb{R}^2$ is Hausdorff and Complete.
  2. $S^1$ is closed in $\mathbb{R}^2$.
  3. A closed subset of a complete space is also a complete space (and thus Hausdorff).
  4. A closed subset of a compact Hausdorff space is compact.
  5. $\pi:S^1\to \mathbb{R}$ is continuous.
  6. 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.

rnrstopstraffic
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$S^{1}$ 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.

T. Eskin
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