A lower bound for X is an element z F such that

Math 316, Intro to Analysis Completeness. Definition 1 (Upper bounds). Let F be an ordered field. For a subset X F an upper bound for X is an element y F such that A lower bound for X is an element z F such that A set is called bounded above if it has an upper bound. It is called bounded below if it has a lower bound. A set is called bounded if it has an upper and a lower bound. Let F be an ordered field and a, b F. Which of the following sets are bounded above / below? Give an upper / lower bound if one exists. On your own time, prove that the unbounded sets are unbounded. [a, b] = {x F : a x b}. (, b] = {x F : x b}. {x F : x 2 1}. {x F : x 2 1}. F. As a warm up, let s prove the following theorem Theorem 2. Let X F. Let X = { x : x X} be the set of additive inverses of elements of X. Let u F. Then u is an upper bound for X if and only if u is a lower bound for X. Definition 3. Let F be an ordered field and X F. supremum or least upper bound if (1) s is an upper bound for X and (2) and if u is any upper bound for X then An element s F is called the For you: Fill in the blacks below to define the infemum of a set X. Definition 4. Let F be an ordered field and X F. An element s F is called the infemum or greatest lower bound if (1) (2) The use of the word the needs justification: Theorem 5 (uniqueness of the supremum if it exists). Let F be an ordered field and X F. If a and b each satisfy the properties of the supremum of X then a = b. If a and b each satisfy the properties of the infemum of X then a = b. Suppose that each of a and b is a supremum for X. We will proceed in two steps: Claim 1 a b. 1

2 Claim 2 b a. For an ordered field F and a F what do you think that the suprema of these sets are? Do they have one? {x : x a} {x : x < a} {x : x 2 < a} There are other equivalent definitions of the supremum. The following theorem is a first appearance of a concept based an an arbitrarily small ɛ > 0. Theorem 6. Let F be an ordered field, s F and X F. s is the supremum of X if and only if (1) s is an upper bound for X. (2) For all ɛ > 0 there is an x X such that s ɛ < x. Let F be a fields, s F and X F. Step 1: Assume that s is the supremum. Prove that (1) s is an upper bound for X and that (2) Since s is the supremum we know that (a) and that (b) How can we deduce (1) from (a)? Now we need to deduce (2). Consider any ɛ > 0. Then s ɛ s. Thus, s ɛ cannot be an upper bound on X. Why? Negate the definition of what it means to be an upper bound. Thus we see that there must exist some x X such that Since ɛ > 0 was arbitrary, we see property (2). This completes half of the proof. Now suppose that s is an upper bound for X and (2) for all ɛ > 0 there is an x X such that s ɛ < x. We must show that s is the supremum. That is we have to show that (a) and that (b) How can we deduce (a) from (1)? We need to show (b). For the sake of contradiction, let u be an upper bound for X and u s. Let ɛ = s u. The ɛ > 0 Why?. Moreover, u = s ɛ Why?. What does (1) tell you? Does it contradict the assumption that u is an upper bound?

3 Write a sentence completing the proof. The existence of suprema Completeness. Definition 7. An ordered field F is called complete if every set which is nonempty and bounded above has a supremum. Now we can completely summarize the properties of R. Axiom. R is a complete ordered field. In some sense (which I might discuss later) R is the only complete ordered field. What about infema? Theorem 8. Let X be a subset of R which is bounded below. Prove that X has an infemum. The Archemedian property: The natural numbers are unbounded. The Natural numbers sit inside of R. They satisfy the properties that (1) 1 N and (2) If n N then n + 1 N. Theorem 9 (Archemedian property version 1). N R has no upper bound Proof. Suppose that N is bounded. Then by completeness, N has a supremum, let s call it s. Is it possible that s 1 is an upper bound? Why? Since s 1 is not an upper bound for N, there exists an element n N such that n We can add 1 to both sides of this inequality (why? Cite a theorem see that n + 1 s. Is n + 1 a natural number? Why? s 1. ) to But then s is not even an upper bound. This contradicts the fact that s was defined to be the supremum.

4 Theorem 10 (Archemedian property version 2). If a and b are real numbers and 0 < a then there is an n N such that a n > b Proof. Case 1: If b 0 then since a 1 = a > 0 b we can take n = 1 to and conclude that a n > b. So assume that b > 0. and for the sake of contradiction assume that for all n N a n b. Multiplying this inequality by 1 a > 0 we see that for all n N But this means that is an upper bound on N. This contradicts the first version of the Archemedian property. Corollary 11 (Archemedian property version 3). The infemum of {1, 1/2, 1/3, 1/4,... } = {n 1 : n N} is 0 Proof. We ll do this one together. First let s show that 0 is a lower bound. (this is easy.) Second We ll show that if ɛ > 0 then ɛ is not a lower bound. We will use contradiction. Theorem 12 (Archemedian property version 4, the density of Q). For any real numbers with x < y there is a q Q such that x q < y. Proof. We first deal with the case that x and y have opposite sign. If x is not positive and y is positive then x 0 < y so that is the desired rational number. There are now two cases either (1) 0 x < y or (2) x < y 0. The proof depends on the following facts (which are homework) Proposition 13. Let W R be a nonempty subset of R which is bounded above. Suppose that w 1 w 2 1 for all distinct w 1, w 2 W. Then sup(w ) W Proposition 14. For any distinct m, n N, m n 1. We will assume that we are in case (1). Namely that 0 x < y. Since x < y, it follows that y x 0. By version 1 of the Archemedes principle there is an n 1 N such that (1) n 1 (y x) 1. Similarly, since y > 0, there exists some n 2 N such that (2) n 2 y 1.

Let n = max(n 1, n 2 ) be the bigger of the two. Then n n 1 and n n 2 so that equations (1) and (2) give us that (3) n (y x) n 1 (y x) 1 and n y n 2 y 1 So, we assume that n y > 1. Let W = {j N : j < n y}. By the inequality 1 W so that W is not empty. W is bounded above since is an upper bound. By completeness then W has a supremum. Let m be the supremum of W. Since W N, Proposition 14 implies that W satisfies the hypotheses of Proposition 13. Thus, m W so (4) m n y. Since m + 1 > m, and (5) n y m + 1 it follows that m + 1 / W and By equation (5) and by (??), n y m 1 < n y n x so that m > n x. Combining this with equation (4) we see that n x m n y Dividing by n > 0, we see that x n m This completes the proof Finally, we have to deal with the case that x < y 0. By setting x = y and y = x we see that 0 x y so that the case of the proof we have already proven implies that there exists some rational number q with x < q < y. For you: Finish the proof Use the rational number between x and y to get a rational number between x and y. y 5

6 homework This exercise will take you through a proof of proposition 13 Proposition (Proposition 13). Let W R be a nonempty subset of R which is bounded above. Suppose that w 1 w 2 1 for all distinct w 1, w 2 W. Then sup(w ) W Proof. Let W be a nonempty subset of R which is bounded above. Suppose also that for all distinct w 1, w 2 W that w 1 w 2 1. Since W is nonempty and bounded, it has a supremum, let s = sup(w ). We make use of the reformulation given by Theorem 6 to conclude that ( ) ( ) For all ɛ > 0 and Let ɛ 1 = 1 3 then ( ) implies that there exists some w 1 W such that. Now s / W so that s w 1. Accordign to ( ) s w 1 so that we conclude s w 1. Thus, if we set ɛ 2 = s w 1, then ɛ 2 0. Employing ( ) again, we see that there is some w 2 W such that. Claim 1: s w 1 1 3 your writeup should include a proof Claim 2: s w 2 1 3 your writeup should include a proof Claim 3: w 1 < w 2. your writeup should include a proof

7 Now, Claim 1, Claim 2 and the triangle inequality imply that w 1 w 2 = (w 1 s)+(s w 2 ) + + < 1 However, by Claim 3, w 1 and w 2 are distinct so that by assumption, w 1 w 2. We thus, have a contradiction. Our assumption that s / W must be false and so s W. This completes the proof.