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How can I use compactness?

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Let K be a compact subset of a metric space M, and let {U_α}_α∈I be an open cover of
    K. Show that there is a positive real number δ with the property that for every x ∈ K there is
    some α ∈ A with B_δ (x) ⊆ U_α.

    2. Relevant equations

    n/a

    3. The attempt at a solution

    Since U_α are open, every point of them is a an interior point. For any positive δ, B_δ (x) [itex]\subset[/itex] U_α, for some α?

    I'm lost here, anything I think seems trivial...
     
  2. jcsd
  3. Aug 30, 2012 #2

    Dick

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    Define f(x) for x in K to be the diameter of the largest ball around x that is contained in one of the U's. Can you tell me why f(x) is positive? Now define the set A_n to be the set of all x such that f(x)>1/n. Do the A_n cover K?
     
  4. Aug 30, 2012 #3
    f(x) is positive because it's a distance?

    I can't decide if A_n cover K because K may be the set of all integers and n may be too large to cover the gaps between the integers.

    Actually I'm still not sure what's wrong with my proof
     
  5. Aug 30, 2012 #4
    Oh I was totally wrong, but I still don't understand your argument
     
  6. Aug 30, 2012 #5

    Bacle2

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    I think Dick was going for finding a finite subcover of the cover of all such balls,all

    of which have positive diameter. Re A_n covering: is the diameter of any given U_n

    larger than 1/n for given n? (to be very formal, this is the Archimedean Property.)
     
  7. Aug 30, 2012 #6
    I've started to learn these kind of things a month ago but I still don't get the Archimedean property and can't use it in proofs. Maybe mathematics is unfortunately not my thing afterall...
     
  8. Aug 30, 2012 #7

    Bacle2

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    Don't take it so hard, we all have problems at some point. Besides, I think it was overkill

    on my part, sorry. Try to see why the diameter is larger than 1/n for some n.
     
  9. Aug 30, 2012 #8

    HallsofIvy

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    The infinite sequence 1/n, for n a positive integer, goes to 0 so it is NOT true that there is a positive lower bound. However, since the set is compact, there exist a finite subcover, giving a finite subset of {1/n} and every finite set of numbers has a smallest number.
     
  10. Aug 30, 2012 #9

    Dick

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    For a given x, f(x)>0 because x is contained in SOME U_a. So there is a ball of some diameter d>0 centered on x and contained in U_a (definition of an open set). So f(x) must be greater than or equal to d>0. Now think about how the sets A_n are related to each other. Which is larger A_2, or A_3?
     
  11. Aug 30, 2012 #10
    Would not it be easier to just say that since f(x) is never zero on the compact set, it reaches a positive minimum there?
     
  12. Aug 30, 2012 #11

    Dick

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    If you have that theorem, that would be a nice way to prove it.
     
  13. Aug 30, 2012 #12
    How do we know that such a maximum diameter exists? If K is a compact subset of the real line, and one of the U's is the entire space, then there is no such maximum, right? (maybe not, I don't know, but clarification would be helpful for me).

    Can we just exclude this case (generalized to metric spaces), because the theorem is clear for the case when {Ua} contains such a set in which any ball around x is contained (I think, because any real-valued delta will do the job)?
     
    Last edited: Aug 30, 2012
  14. Aug 30, 2012 #13
    Exactly.
     
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