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

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1 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 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 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 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. The proof depends on the following facts (which are homework) Proposition 13. (a) Every bounded subset of N is finite (b) If X R is nonempty and finite, then the supremum of X is an element of X. Since x < y, it follows that y x there is an n N such that n (y x) (1) n y < n x By version 1 of the Archemedes principle 1. Rearranging this inequality we see that Case 1: Assume that n y > 1. Let W = {j N : j < n y}. By assumption 1 W so that W is not empty. Since W is a bounded subset of N ( is an upper bound), W is finite by part (a) of Proposition 13. Let m be the supremum of W. By Proposition 13 part (b), m W so (2) m n y. Since m + 1 > m, and (3) n y m + 1 it follows that m + 1 / W and

5 By equation (3) and by (1), n y m 1 < n y n x so that m > n x. Combining this with equation (2) we see that n x m n y Dividing by n > 0, we see that n x y m This completes the proof 5

6 6 homework These exercises will take you through a proof of proposition 13 Problem 1 Prove that every bounded subset X N is finite. Here is how to begin: Let X be a bounded subset. Let s R be the supremum of X. Use the Archemedian property (which one?) to find an n N greater than s. Next prove that X {m N : m N}. You can now appeal to the following facts: Facts: (1) For all n N, {m N : m n} is a finite set. (2) Subsets of finite sets are finite. Problem 2 Prove that if X R is finite and nonempty then the supremum of X is an element of X. Prove this by inducting on the number of elements of X. See me if you need more of a hint.