# A metric space in $\mathbb{R^{2}}$ is compact iff closed and bounded

$\mathbb{R^{2}}$ 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.

• 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? – Li Chun Min Apr 11 '17 at 0:12
• Yes I did, also thanks for the hints. – dankmemer Apr 11 '17 at 0:19
• 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. – Mirko Apr 11 '17 at 3:14