Showing Closed Subsets of Compact Sets are Compact

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Discussion Overview

The discussion revolves around the proof that closed subsets of compact sets are compact, specifically within the context of metric spaces. Participants are examining the validity of a proposed proof and addressing various aspects of compactness and closed sets.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant attempts to prove that closed subsets of compact sets are compact but expresses uncertainty about their proof's correctness.
  • Another participant critiques the proof, questioning the definitions of sets A and B and the choice of open cover.
  • Concerns are raised about the use of the maximum function on elements that are not real numbers.
  • A participant notes the importance of the closed nature of the subset in the proof but acknowledges a lack of clarity on how to incorporate this aspect.
  • Some participants mention that the result holds in general for closed subsets of compact Hausdorff spaces, with metric spaces being a specific case.
  • There is a discussion about the implications of the Hausdorff condition, with one participant stating that the result is true even without it, while another points out that the converse requires the Hausdorff condition.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the initial proof and the definitions used. There is no consensus on the correctness of the proof or the best approach to demonstrate the compactness of closed subsets.

Contextual Notes

Participants note missing definitions and assumptions, particularly regarding the sets involved and the nature of the open cover. The discussion highlights the need for clarity in the proof's structure and the implications of compactness in different topological contexts.

bedi
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I'm trying to show that "closed subsets of compact sets are compact". I think I proved (or didn't) that every subset of a compact set is compact, which may be wrong. Here is what I've done so far, please correct me.

q in A, q not in B, p in B implies p in A. Let {V_a} an open cover of A where V_a = N_r (p) if r = max(p,q). By compactness {V_a} has finite subcover {V_a_i}. Due to our choice of V_a, B can also covered by that finite subcover of {V_a}. Hence B is compact.
 
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bedi said:
q in A, q not in B, p in B implies p in A. Let {V_a} an open cover of A where V_a = N_r (p) if r = max(p,q). By compactness {V_a} has finite subcover {V_a_i}. Due to our choice of V_a, B can also covered by that finite subcover of {V_a}. Hence B is compact.

This is pretty much nonsense.
 
bedi said:
I'm trying to show that "closed subsets of compact sets are compact".

OK. Are you working in topological spaces, metric spaces, \mathbb{R}^n, \mathbb{R}?
 
I'm working in some metric space. It seems like I picked a really bad way to show that. Although I've seen rudin's solution and it does make sense to me, I just wanted to know if my proof could be corrected somehow
 
OK, here is a criticism of the proof:

bedi said:
q in A, q not in B, p in B implies p in A.

What is A? What is B?

Let {V_a} an open cover of A where V_a = N_r (p)

What is N_r(p)? Why can we choose V_a=N_r(p)?

if r = max(p,q).

p and q are not real numbers, how can you take the max?

By compactness {V_a} has finite subcover {V_a_i}. Due to our choice of V_a, B can also covered by that finite subcover of {V_a}. Hence B is compact.

Where did you that the subset of the compact set is closed?
 
Oops I forgot to mention a lot of things... But I still can't figure out how to use that the subset is closed. I'll be happy with rudin's proof. Thanks
 
Just to note an implicit point I think Micromass brought up. Your result is true in

general for closed subsets of compact Hausdorff spaces; metric spaces are Hausdorff,

so Hausdorff condition is not explicitly used.
 
Bacle2 said:
Just to note an implicit point I think Micromass brought up. Your result is true in

general for closed subsets of compact Hausdorff spaces; metric spaces are Hausdorff,

so Hausdorff condition is not explicitly used.

The result is actually always true, even without Hausdorff condition. The "converse" that every compact set is closed uses Hausdorff though...
 
Ah, yea, I misread it, my bad...
 

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