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 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. Then α=supAif and only if (1’) x ≤α for all x ∈A, (2’) For any ε >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 <aas 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 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 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.
RkJQdWJsaXNoZXIy NTc4NTAz