More Heine-Borel

1. Apr 29, 2007

jostpuur

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 $$\mathbb{R}^n$$ 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 $$[-R,R]^n$$, and then start splitting this box into smaller pieces, but how could you replace this procedure with something in an arbitrary metric space?

2. Apr 29, 2007

Hurkyl

Staff Emeritus
I'm pretty sure no. Try looking at function spaces.

3. Apr 29, 2007

jostpuur

Oh, well. No wonder I didn't understand how to extend this proof into general metric spaces...

Now when you mentioned function spaces, I just remembered, that I do know the Riesz's theorem of non-compactness. Just couldn't put pieces together. Ok, sorry for bothering!

4. Apr 29, 2007

HallsofIvy

You don't need function spaces. The Heine-Borel theorem, for the real numbers, is equivalent to the "least upper bound property". That means that it is NOT true for the set of rational numbers with the "usual" topology.

In particular, {x| x2<= 2} is both closed and bounded as a subset of the rational numbers but is not compact. (The crucial point in the proof is that there is no rational number, x, such that x2= 2.)

5. Apr 29, 2007