Heine-Borel Theorem: Proving Compactness in Metric Spaces

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Discussion Overview

The discussion centers on the Heine-Borel theorem and its implications for compactness in metric spaces. Participants explore the conditions under which closed and bounded subsets are compact, particularly in the context of arbitrary metric spaces versus \(\mathbb{R}^n\).

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about extending the proof of the Heine-Borel theorem from \(\mathbb{R}^n\) to arbitrary metric spaces, noting the reliance on enclosing subsets in boxes.
  • Another participant asserts that the Heine-Borel theorem does not hold in arbitrary metric spaces, suggesting that function spaces provide a counterexample.
  • A participant recalls Riesz's theorem of non-compactness, indicating a recognition of the limitations of the Heine-Borel theorem in broader contexts.
  • It is noted that the Heine-Borel theorem for real numbers is linked to the least upper bound property, and that this property does not hold for the rational numbers, providing an example of a closed and bounded set that is not compact.
  • Another participant states that a metric space is compact if and only if it is complete and totally bounded, introducing a different perspective on compactness.
  • A further comment highlights that in \(\mathbb{R}^n\), closed subsets are complete, and questions what property of \(\mathbb{R}^n\) ensures that every bounded subset is totally bounded.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the Heine-Borel theorem to arbitrary metric spaces, with some asserting it does not hold while others provide conditions under which compactness can be determined. The discussion remains unresolved regarding the extension of the theorem beyond \(\mathbb{R}^n\).

Contextual Notes

Participants reference specific properties of metric spaces, such as completeness and total boundedness, and the implications of these properties on compactness. There is an acknowledgment of the limitations of the Heine-Borel theorem in certain topological contexts, particularly with respect to the rational numbers.

jostpuur
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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?
 
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jostpuur said:
but isn't this true also for an arbitrary metric space?
I'm pretty sure no. Try looking at function spaces.
 
Hurkyl said:
I'm pretty sure no. Try looking at function spaces.

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. :rolleyes: Ok, sorry for bothering!
 
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.)
 
In general a metric space is compact if and only if it is complete and totally bounded.
 
deadwolfes remark means compactness follows if every sequence has a cauchy subsequence, and every cauchy sequence converges.

notice R^n is already complete, so any closed subset is also complete. what property does R^n have causing every bounded subset to be totally bounded?
 
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