Rubin's Theorem 2.34 says that "Compact subsets of metric spaces are closed." But I'm wondering should it also be true that, compact subsets of metric spaces are closed and bounded at the same time? The closeed-ness comes from Rudin's proof, and the bounded-ness comes from the fact that for a compact subset $E$, we can take the finite open cover and than this open cover is what bounds $E$? Discussions are welcomed! Thank you!!!
Real Analysis – Are Compact Subsets of Metric Spaces Closed and Bounded?
compactnessmetric-spacesreal-analysis
Best Answer
Yes, this is correct. To be a bit more precise, if $E$ is compact, you can fix any point $x$ and consider the open balls $B_n(x)$ for each $n\in\mathbb{N}$. These are an open cover of $E$, so there is a finite subcover. It follows that there exists an $N$ such that $d(x,y)<N$ for all $y\in E$, so $E$ is bounded.