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to restate part (ii). Lemma 1.3.8. Assume s ER is an upper bound for a set A CR. Then, sup A if and only if, for every choice of e> 0, there exists an element a Є A satisfying s - € < a. 18 Chapter 1. The Real Numbers Proof. Here is a short rephrasing of the lemma: Given that s is an upper bound, s is the least upper bound if and only if any number smaller than s is not an upper bound. Putting it this way almost qualifies as a proof, but we will expand on what exactly is being said in each direction. () For the forward direction, we assume s = sup A and consider s-e, where € > 0 has been arbitrarily chosen. Because s-e 0 is chosen, s-e is no longer an upper bound for A. Notice that what this implies is that if b is any number less than s, then b is not an upper bound. (Just let e = s-b.) To prove that s = sup A, we must verify part (ii) of Definition 1.3.2. (Read it again.) Because we have just argued that any number smaller than s cannot be an upper bound, it follows that if b is some other upper bound for A, then s sup B. (b) A finite set that contains its infimum but not its supremum. (c) A bounded subset of Q that contains its supremum but not its infimum. Exercise 1.3.3. (a) Let A be nonempty and bounded below, and define B = {bЄR: b is a lower bound for A}. Show that sup B = inf A. (b) Use (a) to explain why there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness. Exercise 1.3.4. Let A1, A2, A3,... be a collection of nonempty sets, each of which is bounded above.

to restate part (ii). Lemma 1.3.8. Assume s ER is an upper bound for a set A CR. Then, sup A if and only if, for every choice of e> 0, there exists an element a Є A satisfying s - € < a. 18 Chapter 1. The Real Numbers Proof. Here is a short rephrasing of the lemma: Given that s is an upper bound, s is the least upper bound if and only if any number smaller than s is not an upper bound. Putting it this way almost qualifies as a proof, but we will expand on what exactly is being said in each direction. () For the forward direction, we assume s = sup A and consider s-e, where € > 0 has been arbitrarily chosen. Because s-e 0 is chosen, s-e is no longer an upper bound for A. Notice that what this implies is that if b is any number less than s, then b is not an upper bound. (Just let e = s-b.) To prove that s = sup A, we must verify part (ii) of Definition 1.3.2. (Read it again.) Because we have just argued that any number smaller than s cannot be an upper bound, it follows that if b is some other upper bound for A, then s sup B. (b) A finite set that contains its infimum but not its supremum. (c) A bounded subset of Q that contains its supremum but not its infimum. Exercise 1.3.3. (a) Let A be nonempty and bounded below, and define B = {bЄR: b is a lower bound for A}. Show that sup B = inf A. (b) Use (a) to explain why there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness. Exercise 1.3.4. Let A1, A2, A3,... be a collection of nonempty sets, each of which is bounded above.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.4: Binary Operations
Problem 9E: 9. The definition of an even integer was stated in Section 1.2. Prove or disprove that the set of...
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Please answer Exercise 1.3.1

to restate part (ii). Lemma 1.3.8. Assume s ER is an upper bound for a set A CR. Then, sup A if and only if, for every choice of e> 0, there exists an element a Є A satisfying s - € < a.
Transcribed Image Text:to restate part (ii). Lemma 1.3.8. Assume s ER is an upper bound for a set A CR. Then, sup A if and only if, for every choice of e> 0, there exists an element a Є A satisfying s - € < a.
18 Chapter 1. The Real Numbers Proof. Here is a short rephrasing of the lemma: Given that s is an upper bound, s is the least upper bound if and only if any number smaller than s is not an upper bound. Putting it this way almost qualifies as a proof, but we will expand on what exactly is being said in each direction. () For the forward direction, we assume s = sup A and consider s-e, where € > 0 has been arbitrarily chosen. Because s-e<s, part (ii) of Definition 1.3.2 implies that s-e is not an upper bound for A. If this is the case, then there must be some element a Є A for which s-e <a (because otherwise 8 - € would be an upper bound). This proves the lemma in one direction. (<) Conversely, assume s is an upper bound with the property that no matter how € > 0 is chosen, s-e is no longer an upper bound for A. Notice that what this implies is that if b is any number less than s, then b is not an upper bound. (Just let e = s-b.) To prove that s = sup A, we must verify part (ii) of Definition 1.3.2. (Read it again.) Because we have just argued that any number smaller than s cannot be an upper bound, it follows that if b is some other upper bound for A, then s <b. It is certainly the case that all of our conclusions to this point about least upper bounds have analogous versions for greatest lower bounds. The Axiom of Completeness does not explicitly assert that a nonempty set bounded below has an infimum, but this is because we do not need to assume this fact as part of the axiom. Using the Axiom of Completeness, there are several ways to prove that greatest lower bounds exist for nonempty bounded sets. One such proof is explored in Exercise 1.3.3. Exercises D9D Exercise 1.3.1. (a) Write a formal definition in the style of Definition 1.3.2 for the infimum or greatest lower bound of a set. (b) Now, state and prove a version of Lemma 1.3.8 for greatest lower bounds. Exercise 1.3.2. Give an example of each of the following, or state that the request is impossible. (a) A set B with inf B > sup B. (b) A finite set that contains its infimum but not its supremum. (c) A bounded subset of Q that contains its supremum but not its infimum. Exercise 1.3.3. (a) Let A be nonempty and bounded below, and define B = {bЄR: b is a lower bound for A}. Show that sup B = inf A. (b) Use (a) to explain why there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness. Exercise 1.3.4. Let A1, A2, A3,... be a collection of nonempty sets, each of which is bounded above.
Transcribed Image Text:18 Chapter 1. The Real Numbers Proof. Here is a short rephrasing of the lemma: Given that s is an upper bound, s is the least upper bound if and only if any number smaller than s is not an upper bound. Putting it this way almost qualifies as a proof, but we will expand on what exactly is being said in each direction. () For the forward direction, we assume s = sup A and consider s-e, where € > 0 has been arbitrarily chosen. Because s-e<s, part (ii) of Definition 1.3.2 implies that s-e is not an upper bound for A. If this is the case, then there must be some element a Є A for which s-e <a (because otherwise 8 - € would be an upper bound). This proves the lemma in one direction. (<) Conversely, assume s is an upper bound with the property that no matter how € > 0 is chosen, s-e is no longer an upper bound for A. Notice that what this implies is that if b is any number less than s, then b is not an upper bound. (Just let e = s-b.) To prove that s = sup A, we must verify part (ii) of Definition 1.3.2. (Read it again.) Because we have just argued that any number smaller than s cannot be an upper bound, it follows that if b is some other upper bound for A, then s <b. It is certainly the case that all of our conclusions to this point about least upper bounds have analogous versions for greatest lower bounds. The Axiom of Completeness does not explicitly assert that a nonempty set bounded below has an infimum, but this is because we do not need to assume this fact as part of the axiom. Using the Axiom of Completeness, there are several ways to prove that greatest lower bounds exist for nonempty bounded sets. One such proof is explored in Exercise 1.3.3. Exercises D9D Exercise 1.3.1. (a) Write a formal definition in the style of Definition 1.3.2 for the infimum or greatest lower bound of a set. (b) Now, state and prove a version of Lemma 1.3.8 for greatest lower bounds. Exercise 1.3.2. Give an example of each of the following, or state that the request is impossible. (a) A set B with inf B > sup B. (b) A finite set that contains its infimum but not its supremum. (c) A bounded subset of Q that contains its supremum but not its infimum. Exercise 1.3.3. (a) Let A be nonempty and bounded below, and define B = {bЄR: b is a lower bound for A}. Show that sup B = inf A. (b) Use (a) to explain why there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness. Exercise 1.3.4. Let A1, A2, A3,... be a collection of nonempty sets, each of which is bounded above.
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