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$\mathbb{R^{2}}$ is a metric space with distance function $d((x_{1},y_{1}),(x_{2},y_{2})) = \sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2}}$. Want to show that a subset of $\mathbb{R^{2}}$ is compact iff closed and bounded.

To be honest, I didn't get the whole metric space at all and been working on this for over 5-6 hours with no clue what I'm going to do. I know Heine-Borel theorem for $\mathbb{R}$, but I'm clueless about whole metric space stuff. Like how can I write an open cover for this subset or any metric space($\mathbb{R^{2}}$ or $\mathbb{R^{}}$ doesn't matter) at all? I know that the open ball definition etc but completely lost.

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    $\begingroup$ There is a typo in the defined metric. ⇒ is true for all metric space but ⇐ is not true in general and the proof of ⇐ will be highly similar to the proof on the real line…just take a decreasing sequence of rectangles instead of intervals in this case. So did you study the proof of that on the real line? $\endgroup$ – Li Chun Min Apr 11 '17 at 0:12
  • $\begingroup$ Yes I did, also thanks for the hints. $\endgroup$ – dankmemer Apr 11 '17 at 0:19
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    $\begingroup$ If it is not bounded then it is easily seen not to be compact (consider the cover of open disks with radius $n=1,2,3...$). If not close a similar argument works. So, compact implies closed and bounded. $\endgroup$ – Mirko Apr 11 '17 at 3:14
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Since the subset is bounded it can be enclosed in a closed square. By Heine Borel the edges are compact. Thus the subset is included in a compact square for the square is the product of compact spaces, hence compact. Finally, since the subset is a closed subset of a compact space, it is compact.

Simularily, a closed and bounded subset of R^d is compact. Metric spaces whose closed and bounded subsets are compact are called Borel compact spaces. A closed subspace of a Borel compact space is Borel compact. The plain with an open disk removed, for example.

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