# Compactness of subsets

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.

## Answers and Replies

jgens
Gold Member
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.

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:

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

Bacle2
Science Advisor
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.

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...

Bacle2
Science Advisor
Ah, yea, I misread it, my bad....