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Compact sets and coverings

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that S = [0,1) is not compact by giving an closed cover of S that has no finite subcover.

    2. Relevant equations



    3. The attempt at a solution

    I know that S is not compact because it is [STRIKE]an open[/STRIKE] not a closed set even though it is bounded.

    But I am completely lost on the open cover part.

    I understand an open cover is a union of open sets where S is a subset of the union...

    But I appear to missing something very fundamental. If I picked (-1, 2) for the cover that is an open set and S is a subset of it's "trivial" union. Why doesn't that work?

    <edit> OK, I understand it needs to work for every cover, that's why,
    but is (-1,2) a cover? <end edit>

    If instead I had [0,1], that is closed and bounded so it is compact.
    What sort of sets would go into a cover for it?

    Does anyone know some extremely elementary references for this?
     
    Last edited: Oct 30, 2009
  2. jcsd
  3. Oct 29, 2009 #2

    Dick

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    (-1,2) is a cover. And it has a finite subcover. No problems there. Really big hint. What about (-1,1-1/n) for all n. Can you tell me why i) it is a cover and ii) why it has no finite subcover? And S ISN'T open.
     
  4. Oct 30, 2009 #3
    Thanks, I believe I understand.

    [STRIKE]I said it backwards ... S is closed and bounded on the Reals so it is compact.[/STRIKE]

    The union of the open intervals U = (-1, 1-1/n) for all positive n, covers S.

    That is, S is a subset of U.

    But if one takes a finite subset of u, then the finite union F is the interval (-1, 1-1/f) where f is the last interval in F.

    Since 1/f < 1 for every positive integer f S is not a subset of F.

    Does that sound good?
     
    Last edited: Oct 30, 2009
  5. Oct 30, 2009 #4

    Dick

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    S=[0,1) isn't compact either. You just proved it! It's bounded but neither open nor closed. It's 'closed' at one end and 'open' at the other. You understand the point perfectly. You could work on phrasing it a little better. It sounds like you mean U to be the set of intervals (-1,1-1/n). Then S is contained in the union of the U, not S is contained in U. So it's a cover. And a finite subcover would a set {(-1,1-1/n1),(-1,1-1/n2),...(-1,1-1/nk)}. Defining f=max(nk), then the union of that set is just (-1,1-1/f) and just as you say, 1/f<1.
     
    Last edited: Oct 30, 2009
  6. Oct 30, 2009 #5
    Thanks
     
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