- #1
jostpuur
- 2,116
- 19
The other thread about heine-borel theorem just reminded me of something that has been unclear to me. I understand how you can prove, that a closed and bounded subset of [tex]\mathbb{R}^n[/tex] is compact, but isn't this true also for an arbitrary metric space? The proof I've read relies on the fact that we can first put the subset in a box [tex][-R,R]^n[/tex], and then start splitting this box into smaller pieces, but how could you replace this procedure with something in an arbitrary metric space?