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Question on If a set is unbounded, then it cannot be compact

  1. Jan 1, 2007 #1
    question on "If a set is unbounded, then it cannot be compact"

    Hello,

    I am not a mathematician so wanted to understand by picturizing and got stuck in between.

    While trying to understand the proof as given in Wikipedia
    http://en.wikipedia.org/wiki/Heine-Borel_theorem

    I was not sure if I understood this statement very clearly.

    It says the following
    If a set is unbounded, then it cannot be compact
    Why? Because one can always come up with an infinite cover, whose elements have an upper finite bound to their size, i.e. the elements of the cover are not allowed to grow in size without bound.

    Please confirm my understanding here
    an example of an unbounded set is R
    then how is it that u can cover R with any (a,b) which are bounded ie how can u cover while |a| < some x and |b| < some y.

    The second problem refers to this particular statement

    I was trying to understand the theorem and so got confused with this line

    http://www.du.edu/~etuttle/math/heinebo.htm

    Now, the theorem says that in any such case, the closed interval a < x < b can be covered by a finite number of finite intervals. If we choose δ0 to be the smallest of the finite number of values of δ(ci) at the points ci, then we have that |f(x) - f(c)| < ε, for any ε > 0, whenever |x - c| < δ0, independently of c.

    So if I am understanding correctly it means there are
    |x1-c| < delta1 |f(x1) -f(c)| < epsilon1
    |x2-c| < delta2 |f(x2) -f(c)| < epsilon2

    surely |x1| < |x2| if delta1< delta2 and is epsilon1 <= epsilon2 ?
    If the above is true then how is |f(x) -f(c)| < any epsilon greater than 0.
    I can understand that it will be true for any epsion greater than equal to the smallest epsilon in that set but not sure how for all cases greater than 0 it is true.
     
  2. jcsd
  3. Jan 1, 2007 #2

    matt grime

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    (a,b) is a subinterval. You don't cover R with subintervals where the elements in the interval are all bounded above. It is the lengths of the intervals that is the issue.

    The intervals ( n-1,n+1) with n in N are an infinite cover and the lengths of all the intervals are 2.





    What particular statement? That page contains several statements, and you've introduced an f from nowhere.
     
  4. Jan 2, 2007 #3

    HallsofIvy

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    You don't neet to assume you are in the real numbers (and so talk about intervals). It is easy to prove that in any metric space, where we can define "bounded", it is true that any compact set is bounded (and so unbounded sets cannot be compact).

    Proof: Let A be a compact set in a metric space and let p be any point in the set. For every integer n, Define Un to be the open "n" ball centered on p- the set of all points whose distance from p is less than n. Since any point in the set has some distance from p and there exist an integer larger than that distance, this is an open cover for the set A. Since A is compact, there exist a finite subcover and, so, a largest such integer, N. We now have that the distance from p to any point in A is less than N and so the distance between any two points in A is less than 2N
     
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