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Compactness Proof w/o Heine-Borel

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove directly (i.e., without using the Heine-Borel theorem) that if K [tex]\subseteq[/tex] Rd is compact and F [tex]\subseteq[/tex]K is closed, then F is compact.

    2. Relevant equations

    Definition of a Compact Set: A set K is said to be compact if, whenever it is contained in the union of a collection G={Ga} of open sets, then it is also contained in the union of some finite number of the sets in G.


    3. The attempt at a solution

    I think that I need to be more detailed in this, so any help would be greatly appreciated. :-D

    If K is compact, then there exists a finite subcover for every open cover of K. If F [tex]\subseteq[/tex] K, then every open cover of K also covers F. And because K is compact, every open subcover of K is once again a cover for F.

    Therefore F is compact.

    It seems to me that there has to be more to this proof. Help?
     
  2. jcsd
  3. Feb 18, 2008 #2

    morphism

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    We want to look at open covers of F. While it's true that every open cover of K is an open cover of F, it isn't necessarily true that these are all possible open covers of F.
     
  4. Feb 18, 2008 #3
    So, you're saying that I need to pay attention to the open covers that are not open covers of K, even though they're open covers of F, because they are contained in K? That makes sense, I'm just not sure how to approach that case.

    Maybe... Because F is closed, for every open cover, there must exist a finite open subcover because it can get infinitely close to F without being equal to or contained in F?

    I'm not sure how else to word it.
    Thanks.
     
  5. Feb 18, 2008 #4

    StatusX

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    Take an open cover of F, say [itex]O_\alpha[/itex], which we can assume are open subsets of K. You want to show this has a finite subcover.

    Now, this won't necessarily be an open cover of K, but you can get one by adding more sets, specifically, ones that cover the rest of K. Call these [itex]P_\beta[/itex]. Then you can use the fact that K is compact to say that this collection of all O's and P's has a finite subcollection [itex]O_1,..,O_m,P_1,...,P_n[/itex] that covers K. This then clearly covers F as well.

    However, remember that you need to find a subset of the O's alone that covers F, ie, you can't use any of the P's to cover part of F. One way to guarantee you won't need any P's to cover F is to pick each P to lie outside of F. Hint: you'll only need one P, and constructing it will use the fact that F is closed.
     
  6. Feb 19, 2008 #5
    "However, remember that you need to find a subset of the O's alone that covers F, ie, you can't use any of the P's to cover part of F. One way to guarantee you won't need any P's to cover F is to pick each P to lie outside of F. Hint: you'll only need one P, and constructing it will use the fact that F is closed."


    This is what I was attempting to do. Unfortunately, my trouble is in wording this type of thing. How do I use the fact that F is closed to write a rigid proof? I want to say that because F is closed, all covers and subcovers must be open, and for any given cover, there exists another cover that is even smaller than the first (a subcover), because F is closed. Is there a better way to phrase this?
     
  7. Feb 19, 2008 #6
    I think you may be over thinking the problem. You have an open cover of F. So you've covered everything K except for the stuff outside of F. What's the "simplest" open set that covers that stuff? Note that up till now you haven't used ANY of the assumptions you've made.
     
  8. Feb 19, 2008 #7
    I think that that comment right there might have helped. :-D

    Because F [tex]\subseteq[/tex] K, the open covers of K also cover F. Let's take a look at the covers of F that do not also cover K. One such cover is the complement of F within K, which can be written as K\F. This is open because it is the complement of a closed set. Because this is an open covering of F within K with finite subcovers, we know that F is compact.

    Yes? And thank you.
     
  9. Feb 19, 2008 #8
    I think you have the right idea, but you seem to be stating it poorly. For one, K - F certainly isn't an open cover of K: It doesn't cover F. But an open cover of F along with K - F (or just F' if you want) is an open cover of K.

    Also, from a purely aesthetic point of view, it's probably better to just start with an open cover of F (either in the relative topology or in the topology of the ambient space) and extend it as indicated. That way you don't have to deal with cases.
     
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