# Compactness of subsets

1. Aug 24, 2012

### bedi

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.

2. Aug 24, 2012

### jgens

This is pretty much nonsense.

3. Aug 24, 2012

### micromass

Staff Emeritus
OK. Are you working in topological spaces, metric spaces, $\mathbb{R}^n$, $\mathbb{R}$?

4. Aug 24, 2012

### bedi

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

5. Aug 24, 2012

### micromass

Staff Emeritus
OK, here is a criticism of the proof:

What is A? What is B?

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

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

Where did you that the subset of the compact set is closed?

6. Aug 24, 2012

### bedi

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

7. Aug 24, 2012

### Bacle2

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.

8. Aug 24, 2012

### micromass

Staff Emeritus
The result is actually always true, even without Hausdorff condition. The "converse" that every compact set is closed uses Hausdorff though...

9. Aug 24, 2012