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), then0<a<1 and therefore there are elements b1 and b2 in (0,1) such that 0<b1 <a<b2 <1. This shows that ais neither a maximum nor a minimum element of A. The following proposition is convenient in working with suprema. Proposition 1.5.1 Let Abe a nonempty subset of Rthat is bounded above. Thenα=supAif and only if (1’) x ≤α for all x ∈A, (2’) For anyε >0, there exists a∈Asuch that α−ε <a. Proof: Suppose first that α=supA. Then clearly (1’) holds (since this is identical to condition (1) in the definition of supremum). Now let ε >0. Since α−ε <α, condition (2) in the definition of supremum implies that α−ε is not an upper bound of A. Therefore, there must exist an element a of Asuch 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 definition of supremum holds. Let Mbe 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 Mis 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 Aof Rthat is bounded above has a least upper bound. That is, supAexists and is a real number. This axiom distinguishes the real numbers from all other ordered fields and it is crucial in the proofs of the central theorems of analysis. There is a corresponding definition for the infimum of a set. Definition 1.5.3 Let Abe a nonempty subset of Rthat is bounded below. We call a number β a greatest lower bound or infimumof A, denoted byβ =infA, if the following conditions hold: (1) x ≥β for all x ∈A(that is, β is a lower bound of A). (2) If Lis 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 infimum 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. (c) inf (−1)n n : n∈N =−1. (d) inf{1+ 1 n : n∈N}=1. (e) inf{x2 : −2<x <1,x ∈R}=0.
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