S2 - 92.305 Homework 2 Solutions September 25, 2009...

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Unformatted text preview: 92.305 Homework 2 Solutions September 25, 2009 Exercise 1.3.2. (a) Write a formal definition in the style of Definition 1.3.2 for the infimum or greatest lower bound of a set. Answer: A number x ∈ R is the infimum or greatest lower bound of a set A ⊆ R of real numbers if • it is a lower bound for A , i.e. for all a ∈ A one has x ≤ a , and • it is maximal with this property, i.e. for all lower bounds l for A , one has l ≤ x . (b) Now, state and prove a version of Lemma 1.3.7 for greatest lower bounds. Answer: Assume s ∈ R is a lower bound for a set A ⊆ R . Then s = inf A if and only if, for every choice of ǫ > 0, there exists an element a ∈ A satisfying a < s + ǫ . Proof: Assume first that s = inf A , and suppose ǫ > 0 is given. Since s is maximal among lower bounds, it follows that s + ǫ is not a lower bound. In other words, there must exist a ∈ A satisfying a < s + ǫ . This shows ⇒ . For ⇐ , assume that for every ǫ > 0 there exists a ∈ A with a < s + ǫ . Given any number s ′ with s ′ > s , apply our hypothesis to the choice ǫ = s ′ − s > 0 to produce an a ∈ A with a < s + ǫ = s + ( s ′ − s ) = s ′ . The existence of such an a implies that s ′ cannot be a lower bound for A . Summarizing, no s ′ with s ′ > s can be a lower bound for A , so every lower bound l for A satisfies l ≤ s . The latter claim is precisely what is required for s = inf A to hold. squaresolid Exercise 1.3.3. (a) Let A be bounded below and nonempty and define B = { b ∈ R : b is a lower bound for A } . Show that sup B = inf A . Proof: (Although the problem doesn’t mention it, we point out that B is nonempty and bounded above. That B negationslash = ∅ is exactly the claim that A is bounded below. On the other hand, any a ∈ A and b ∈ B are related by the inequality b ≤ a because b is a lower bound for A . Otherwise said, reciprocally, any a ∈ A is an upper bound for B . That B is bounded above thus follows from the nonemptiness of A . Only now that 1 we know that B is nonempty and bounded above, we can appeal to the Completeness Axiom to say that sup B exists in the first place!) We show that the number x = sup B satisfies the definition of an infimum of A : that it is a lower bound, and maximal among lower bounds. Assume, towards a contradiction, thatis a lower bound, and maximal among lower bounds....
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.

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S2 - 92.305 Homework 2 Solutions September 25, 2009...

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