25 It should also be noted that a set may have neither a maximum nor a minimum element. Consider for example the set A =(0,1). If a ∈ (0,1), then 0 < a < 1 and therefore there are elements b1 and b2 in (0,1) such that 0 < b1 < a < b2 < 1. This shows that a is neither a maximum nor a minimum element of A. The following proposition is convenient in working with suprema. Proposition 1.5.1 Let A be a nonempty subset of R that is bounded above. Then α = supA if and only if (1’) x ≤ α for all x ∈ A, (2’) For any ε > 0, there exists a ∈ A such that α − ε < a. Proof: Suppose frst that α = supA. Then clearly (1’) holds (since this is identical to condition (1) in the defnition of supremum). Now let ε > 0. Since α − ε < α, condition (2) in the defnition of supremum implies that α − ε is not an upper bound of A. Therefore, there must exist an element a of A such that α − ε < a as desired. Conversely, suppose conditions (1’) and (2’) hold. Then all we need to show is that condition (2) in the defnition of supremum holds. Let M be an upper bound of A and assume, by way of contradiction, that M < α. Set ε = α − M. Note that ε > 0. By condition (2’), there is a ∈ A such that a > α − ε = α − (α − M)= M. This contradicts the fact that M is an upper bound. The conclusion now follows. □ The following is an axiom of the real numbers and is called the completeness axiom. The Completeness Axiom. Every nonempty subset A of R that is bounded above has a least upper bound. That is, supA exists and is a real number. This axiom distinguishes the real numbers from all other ordered felds and it is crucial in the proofs of the central theorems of analysis. There is a corresponding defnition for the infmum of a set. Defnition 1.5.3 Let A be a nonempty subset of R that is bounded below. We call a number β a greatest lower bound or infmum of A, denoted by β = infA, if the following conditions hold: (1) x ≥ β for all x ∈ A (that is, β is a lower bound of A). (2) If L is a lower bound of A, then β ≥ L (this means β is largest among all lower bounds). Using the completeness axiom, we can prove that if a nonempty set is bounded below, then its infmum exists (see Exercise 1.5.5). ■ Example 1.5.2 (a) inf(0,3]= inf[0,3]= 0. (b) inf{3,5,7,8,10} = 3. (−1)n (c) inf : n ∈ N = −1. n 1 (d) inf{1 + : n ∈ N} = 1. n (e) inf{x2 : −2 < x < 1,x ∈ R} = 0.
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