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Why the proof of Heine-Borel theorem doesn't work for open sets.

  1. Apr 29, 2007 #1
    Hi all,

    I am wondering why the "creeping along" proof of Heine-Borel theorem doesn't work for open subsets. I have adapted Spivak's proof in "Calculus on Manifolds" here, but I can't seem to find where I am wrong:

    The theorem for open sets would be:
    The open interval (a, b) is compact.


    Let O be an open cover of (a,b).
    Let A = {x : a <= x <=b and (a,x) is covered by some finite number of open sets in O}.
    Then A is clearly bounded above by b.
    We want to show that b \in A, which is done by proving that \alpha, the supremum of A, is in A and that \alpha = b.

    Since O is a cover, \alpha \in U for some U in O. Then all points in some interval to the left of \alpha are also in U. Since \alpha is the least upper bound of A, there are x_i and x_j in this interval such that x_i, x_j \in A and x_i < x_j. Thus (a, x_j) is covered by some finite number of open sets of O, while (x_i, \alpha) is covered by the single set U. Hence (a,\alpha) is covered by a finite number of open sets of O, and \alpha \in A.

    Now, suppose instead that \alpha < b. Then there is a point y between \alpha and b such that (\alpha, y) is a subset of U. Since \alpha \in A, the interval (a, \alpha) is covered by finitely many open sets of O, while (\alpha, y) is covered by U. Hence y \in A, contradicting the fact that \alpha is an upper bound of A.

    I would be grateful to anyone who finds mistakes in the above paragraphs.
  2. jcsd
  3. Apr 29, 2007 #2
    You can't invoke the least upper bound of A unless A is nonempty. Unfortunately, you cannot prove A is never empty, since this is not the case. A simple counter-example will suffice to show that.
    Last edited: Apr 29, 2007
  4. Apr 29, 2007 #3


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    the clim that alpha is in A is not valid, because O is not a cover of [a,b], but only of (a,b). but perhaps you mean, if alpha is not b then alpha is in (a,b), so in some U. this is not a mistake, merely a fillable gap.

    moo of doom has pinpointed the fatal error. for the open cover (1/n,1) of (0,1) , the set A is empty.
  5. Apr 29, 2007 #4
    At this point the normal proof will assert "and the set A is non-empty" by exhibiting the element "a", but it does not make sense that [/itex] a\in A[/itex] in your case.

    Also, you can easily find a counter example to the theorem in the case of open sets. Consider the open cover of (0,1) given by all open intervals of the form (1/n , 1). This is an open cover because e > 0 => there exist n such that 1/n < e.

    Now, suppose A is a finite subset of this open cover, then there is a largest N such that (1/N , 1) is in A. Then (0 , 1/N) does not interesect any of the elements of A, and so A is not a cover of (0,1).
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