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Unclear step in diam(S) = supS − inf S proof

  1. Jan 4, 2012 #1
    Suppose S is a subset of R. Then diam(S) = supS − inf S.

    Proof: First, if S = ∅, then inf S = ∞, supS = −∞ and diam(S) = sup∅ = −∞,
    so the proposition is true by definition (B.7.1). Second, if sup S = ∞, pick
    b ∈ S. Then, for each p ∈ R+, there exists some a ∈ S with a > b + p, so that
    diam(S) > p. Since p is arbitrary in R+, we have diam(S) = ∞ = supS−inf S.
    By a similar argument, the proposition is true if inf S = −∞.
    Finally, suppose that inf S and sup S are both real. Let r ∈ R+. Then there
    exist a, b ∈ S such that a − r/2 < inf S ≤ a ≤ b ≤ sup S ≤ b + r/2. So
    sup S − inf S ≤ b − a + r ≤ diam(S) + r. Since r is arbitrary in R+, we then
    have sup S − inf S ≤ diam(S). But, for all x, y ∈ S, we have inf S ≤ x ≤ sup S
    and inf S ≤ y ≤ sup S, so that |y − x| ≤ sup S − inf S. Since x and y are
    arbitrary in S, this yields diam(S) ≤ sup S − inf S. The two inequalities then
    lead to the desired conclusion.


    How does choosing small arbitrarily positive r, mean that the S − inf S ≤ diam(S)? I would have thought that, if the sign was < in sup S − inf S≤ diam(S) + r then removing a small positive number would lead to ≤?

    Can some one please explain the reasoning behind arbitrarily small r?
     
  2. jcsd
  3. Jan 4, 2012 #2

    mathman

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    Science Advisor
    Gold Member

    It is simple if you use proof by contradiction.
    If sup S - inf S > diam(S), then sup S - inf S = diam(S) + c, where c > 0. Choose r with 0 < r < c and you get a contradiction.
     
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