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Compact Sets

  1. Oct 3, 2011 #1
    The problem statement, all variables and given/known data

    Show that every compact set must be closed.

    I am looking for a simple proof.

    This is supposed to be Intro Analysis proof.

    Relevant equations

    Any compact set must be bounded.

    3. The attempt at a solution

    Suppose A is not closed, so let a be an accumulation point of A such that a is in A etc. (I think I can finish this proof, but I do not like it.)

    Does anyone know of a simpler proof.
     
  2. jcsd
  3. Oct 3, 2011 #2
    I don't understand: if you are working in R^n (which I imagin you may be doing, since you mention an analysis book), then Heine-Borel is the resul. If not, the result is true only if you have a compact subset of a Hausdorff space, where you show that the complement in the space of the compact set is open, by finding a 'hood (neighborhood) of a point x in the complement that is disjoint from the compact space, by using Hausdorffness (so you can choose disjoint pairs of open sets containing x, disjoint from finitely-many open subsets of the cover. This is the main idea ). So this last one should take care of the case of R^n, or for any Hausdorff space.
     
  4. Oct 3, 2011 #3

    micromass

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    Yes, what is the context you are working in?? Are you working in [itex]\mathbb{R}^n[/itex]?? In a metric space? A topological space? A Hausdorff space??

    And how did you define compact??
     
  5. Oct 3, 2011 #4
    The section in the book is called Topology of the Real Number System. We have not covered metric spaces, topological spaces, Hausdorff space, etc.

    A set A subset R is compact if every open cover C of A contains a finite subcover F of A.

    And by the way can anyone explain what does a cover, a subcover, etc. really represent? I understand the definitions but what is really meant by something like C covers A?
     
  6. Oct 3, 2011 #5
    C covers A if the union of the open sets contains A; a subcover of C is a subcollection of the subsets of C, so you want a finite subcollection that contains C.

    An example, maybe somewhat-trivial, is that of a convergent sequence {xn} in R, with limit x; take a collection of open intervals (an,bn)--take it, please! (sorry, prepping for my 2nd job)-- with an:= xn-0.1, (any small number will do ), and bn:=xn+0.1, and a 'hood (neighborhood) (x-0.1,x+0.1) of the limit point x . Then the union of the intervals (a,n,bn) and (x-0.1,x+0.1)covers {xn}, together with a 'hood , trivially, since each member of {xn} is in the n-th interval. Now, you can find a subcover ( you, and you alone!), by using (x-0.1,x+0.1) , which , by convergence of {xn}, will contain all-but-finitely many of the elements of the sequence, so that you can use finitely-many intervals (an,bn) to cover all the remaining elements.
     
  7. Oct 4, 2011 #6
    It may be good too, for contrast, to consider a cover of the integers in the real line, by
    open intervals (n-1/4,n+1/4), and see if you can find a finite subcollection of these open sets that covers the integers. For a more advanced problem, show that a half-open interval [a,b) in the real line is not compact. I'm pretty sure you will eventually see Heine-Borel theorem which tells you that, for Rn, the compact sets are always those that are close and bounded.
     
  8. Oct 4, 2011 #7

    HallsofIvy

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    If you are working in a general vector space or general metric space, use the fact that a set is closed if its complement is open. To show that the complement of a set is open, show that any point NOT in the set (and so in the original compact set) has an open neighborhood that does not intersect the intended closed set.
     
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