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Help one question about analysis

  1. Oct 9, 2007 #1
    The set A={1/n : n in N} is not compact.
    A) prove this by explicitly finding a family of open sets which covers A but has no finite subfamily whcih also covers A.
    B) Find another family of open sets which covers A and does have a finite subfamily which cobers A.
  2. jcsd
  3. Oct 9, 2007 #2
    i have no idea on how it works.....
    i am learning by myself right now....so could some one help me out?...thank you very much...
  4. Oct 9, 2007 #3
    I assume you are working in the standard topology for R. Then just take the family of open sets defined by

    {U_n = (1/n,infinity) | n in N}.

    This does not have a finite subcover. To see this notice that U_n+1 contains U_n for all n. Then any finite subcollection of these sets will have a maximal lement U_M where M is in N. Clear the point 1/(M+1) is not in any of the sets in the finite subcollection.

    B. Take the collection {(-infinty, infinity)}.

    Try to extend this argument for the set {(x,y) | |x| + |y| < 1} and to be able to say any open set in R or R^2 is not compact.
  5. Oct 10, 2007 #4


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    Here's another example:
    The distance from 1/n to 1/(n+1) is 1/(n(n+1)) (the distance from 1/n to 1/(n-1) is larger). Take as your open sets each Un the neighborhood of 1/n with radius 1/(n(n+1)). Each of those contains exactly one point in {1/n} and so no finite subset will cover all of {1/n}.

    For B, just take { (0, 2)}. That is a single set that covers all of {1/n} by itself. Of course, that tells us nothing about the compactness. {1/n} is "still" compact because not every open cover has a finite subcover.
  6. Oct 11, 2007 #5
    is still NOT compact is what halls of ivy means.
  7. Oct 11, 2007 #6


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    Oops, thanks! My fingers don't always connect to my brain.
  8. Oct 11, 2007 #7


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    Homework Helper

    here is a clue. if you throw in 0, it becomes compact.
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