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1.2 Supremum and Infimum Integers, Rational Numbers and Real Numbers Completeness of Real Numbers Supremum and Infimum Characterizations of Sup and of Inf Calculus and Elementary Analysis 1.2.1 Integers, Rational and Real Numbers 0,1,2,3, Notations Natural Numbers 0, 1, 2, 3, p p q Theorem Notation ,q Integers This condition means that p and q have no common factors. , q 0, gcd p, q 1Rational Numbers There are no rational numbers r such that r2=2. Real Numbers Real numbers consist of rational and of irrational numbers. 2017/5/24 Calculus and Elementary Analysis Integers, Rational and Real Numbers Proposition Proof There are not rational numbers r such that r2=2. Assume the contrary. Then there are positive integers p and q such that p2/q2=2 and p and q do not have common factors. This means that p2=2q2. Hence the area of a square with side length p is twice the area of the square with side length q. Furthermore p and q are smallest such numbers since they do have common factors. Graphically: p G q 2017/5/24 B Now p and q are integers so that B=p2=2q2=2G. I.e. the area B of the large brown square is twice the area G of the green square. Calculus and Elementary Analysis Integers, Rational and Real Numbers There are not rational numbers r such that r2=2. Theorem Proof (cont’d) p Furthermore, p and q are smallest integers such that the area B of the large brown square is twice the area G of the smaller green square, B=2G. q Now move a copy of the green square to p the upper right hand corner of the larger square. The intersection I of the two copies of the green square is the square I with the area I. A I A p-q q The two squares marked by A in the picture have the same area A. Furthermore, by the assumptions, I=2A. This is impossible, since B and G were the smallest squares with integer side lengths such that B=2G. 2017/5/24 Calculus and Elementary Analysis 1.2.2 Upper and Lower Bounds Definition Let A be a non-empty set of real numbers. We say that a number M is an upper bound for the set A if a A : a M. A number m is an lower bound for the set A if a A : a m. A set A need not have neither upper nor lower bounds. The set A is bounded from above if A has a finite upper bound. The set A is bounded from below if A has a finite lower bound. The set A is bounded if it has finite upper and lower bounds. 2017/5/24 Calculus and Elementary Analysis 1.2.3 Bounds of Integers, Rationals and Reals Observations 1 2 Any non-empty bounded set of integers has the largest (and the smallest) element which is also an integer. Consider the set of rational numbers r such that r2 ≤ 2. This set is clearly bounded both from the above and from below. Observe that the set of rational upper bounds for the elements of this set does not have a smallest element. This follows from the previous considerations showing that 2 is not rational. This means that the set of rational numbers is not complete in the sense that the set of rational upper bounds of a bounded set does not necessarily have the smallest rational upper bound. 2017/5/24 Calculus and Elementary Analysis 1.2.4 Supremum Let A , A , be bounded from above. Completeness of Real Numbers The set A has finite upper bounds. An important completeness property of the set of real numbers is that the set A has a unique smallest upper bound. Definition The smallest upper bound of the set A is called the supremum of the set A. Notation sup(A) = the supremum of the set A. Example 1 3 7 Let A , , , 1 2 n n , n 0 . Then sup A 1. 2 4 8 2017/5/24 Calculus and Elementary Analysis Infimum Let A , A , be bounded from below. The set A has finite lower bounds. As in the case of upper bounds, the set of real numbers is complete in the sense that the set A has a unique largest lower bound. Definition The largest lower bound of the set A is called the infimum of the set A. Notation inf(A) = the infimum of the set A. Example 3 5 9 Let A , , , 1 2 n n , n 0 . Then inf A 1. 2 4 8 2017/5/24 Calculus and Elementary Analysis 1.2.5 Characterization of the Supremum (1) Let A , A , be bounded from above. This is equavalent to saying that sup A . Theorem Proof 1) a A : s a and s sup A 2) 0 : a A such that s a Assume that s sup A . Then since s is an upper bound of A, the condition 1) is trivially satisfied. To prove the condition 2) assume the contrary, i.e. assume that there is an 0 such that there are no elements a A such that s a . If this is the case, then also s is an upper bound for the set A. 2 But this is impossible, since s is the smallest upper bound for A. 2017/5/24 Calculus and Elementary Analysis Characterization of the Supremum (2) Theorem For a non-empty set A: 1) a A : s a, and s sup A 2) 0 : a A such that s- a s. Proof Cont’d To prove the converse, assume that the number s satisfies the conditions 1) and 2). We have to show that s sup A . By the condition 1), s is an upper bound for the set A. If s sup A , then s sup A , and s sup A 0. By the condition 2), a A such that s- a s. By the definition of the number this means that a sup A . This is impossible, hence s sup A . 2017/5/24 Calculus and Elementary Analysis 1.2.6 Characterization of the Infimum Theorem For a non-empty set A: 1) a A : a m, and m inf A 2) 0 : a A such that m<a<m+ . The proof of this result is a repetition of the argument the previous proof for the supremum. 2017/5/24 Calculus and Elementary Analysis 1.2.7 Usage of the Characterizations Example Assume that sup A , and let 2A 2a | a A. Claim sup 2A 2sup A. Proof of the Claim 1 Let m sup A . Then a A : m a. Hence a A : 2m 2a sup 2 A 2m 2sup A . 2 Let 0. Apply the above characterization for sup A replacing by to conclude: a A such that a- 2 implying sup 2 A 2m 2sup A . 1 and 2017/5/24 2 2 a m. Then m- 2a m. sup 2A 2sup A. Calculus and Elementary Analysis