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Theorem about bounded sets.

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Theorem: If S is any bounded set in n space, and d>0 is given, then it is possible to choose a finite set of points pi in S such that every point p existing in S is within a distance d of at least one of the points p1, p2, ..., pm.

    Prove this theorem assuming that the set S is both closed and bounded.

    Prove this theorem, assuming only that S is bounded. [The difficulty lies in showing that the points pi can be chosen in S itself.


    3. The attempt at a solution

    Let S be a bounded set in n-space. By definition, there exists an M such that |p|< M for all p E S and S is a subset of B(0, M). Take po and p E S. ...
     
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  3. Oct 8, 2009 #2

    Dick

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    Use compactness?
     
  4. Oct 8, 2009 #3

    HallsofIvy

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    Well, that's the problem, Dick. It is easy if S is closed but suppose S is not closed?

    You could, in that case, look at the closure of S but then you run into the problem that some of the finite number of points you get are boundary points of S that are not in S itself.
     
  5. Oct 8, 2009 #4

    Dick

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    Oh yeah. Pick a point p0 in S. If B(p0,d) doesn't cover S, pick a point p1 outside the ball. If the union of B(p0,d)UB(p1,d) doesn't cover S, pick a point p2 outside the union. Continue. Doesn't that make the balls B(pk,d/2) disjoint? What can you conclude from that?
     
    Last edited: Oct 8, 2009
  6. Oct 8, 2009 #5

    Dick

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    The problem doesn't say that the set is bounded by d. It just says that it's bounded. d is a given number. Do what you did before and just say that there is an M such that |p|<M for all p in S.
     
  7. Oct 8, 2009 #6
    I guess I don't know where to go after that though? Take p and poES. Say the distance between them is some d >0. Show that the distance between them and another point pm is also d? But how?
     
  8. Oct 8, 2009 #7

    Dick

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    Did you read post 4? Did it make sense?
     
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