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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

__. Since r is arbitrary in R+, we then__

**sup S − inf S ≤ b − a + r ≤ diam(S) + r**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?