# Homework Help: About compactness in topology

1. Oct 22, 2011

### edcvfr

1. The problem statement, all variables and given/known data
E is a compact set, F is a closed set. Prove that intersection of E and F is compact

2. Relevant equations

3. The attempt at a solution
On Hausdoff space (the most general space I can work this out), compact set is closed. So E is closed. So intersection of E and F is closed. That is a closed subset of compact set E so intersection of E and F is closed.

My problem is i can't generalize the proof to general topological space. If E is not in Hausdoff space, it is not necessarily closed. I don't know if I was going in the right direction or not. Please help me with it. Thank you very much.

2. Oct 22, 2011

### Dick

Go straight to the definition. Pick any open cover of F. Can you think of a way of extending that to an open cover of E?

3. Oct 22, 2011

### edcvfr

F is closed. Call {Gi} class of sets such that F$\subset$$\bigcup$iGi and {Hi} class of sets such that E$\subset$$\bigcup$iHi. Because E is compact, H has finite number of subcovers. If we union G and H, we have open cover that covers both E, F and definitely E$\cap$F. However, because F is not compact (I read that a closed set is not necessarily compact, though it must be compact in Hausdoff space), G may have infinite number of subcovers. Union of G and H may have infinite subcovers so the intersection can't be proved to be compact.

4. Oct 22, 2011

### canis89

I'm sorry, but, if you want to proof the compactness of the intersection between E and F, should you start from picking any open cover of that intersection, instead of F? That is, if you want to go straight by definition like Dick said.

Anyway, once you pick such open cover, try to augment it with some set such that the union, say M, equals(or contains, which implies equality) the topological space, say, X. From there, deduce that E is a subset of some open cover which is a subset of this M. Then, use the compactness of E until you complete the proof.

Sorry, if it's not clear. I'm currently writing it with my phone.

5. Oct 23, 2011

### edcvfr

Sorry canis, I'm afraid I don't get your idea. Can you please elaborate on that? I apologize for my stupidity.

6. Oct 23, 2011

### canis89

Alright, let me be more straightforward then. If you want to proof that the intersection of E and F, say, A, is compact by using the definition of compact set, then, you could start by picking any open cover of A. Then, (this is what I intend to say back then), union this open cover with the complement of A (which is the union between the complements of E and F). See if you can continue from there.

7. Oct 23, 2011

### edcvfr

Alright, I tried to think by your hints and get some results. I'm not sure my explanation is correct. Please correct it for me:
Call G the finite set of open subcover of E, A' be an open cover of A. If we union A' and complement of F and call it M (open because F is closed), we still have an open cover of A. Moreover, M also cover E and can be reduced to a finite number of subcover because E is compact. After all M cover A and can be reduced so A is compact.
Sorry I can't figure out the meaning of union between A' and complement of E.

8. Oct 23, 2011

### canis89

Yes, you're right. That's the idea. Forget about the union of A' and the complement of E. It's redundant. You only have to union A' with F complement. Now, can you write your argument more specific?

9. Oct 24, 2011

### edcvfr

could it be more specific? To me it is comprehensive enough.

10. Oct 24, 2011

### canis89

Oh, I just want to make sure that when you reduce M, the result must contain only the finite subfamily of A' (it must not contain F complement). But it is straightforward since $A=E\cap F$ cannot be a subset of F complement.

11. Oct 24, 2011

### edcvfr

Thank you very much for helping me with this problem.