This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 317 HW #2 Solutions 1. Exercise 1.3.3. (a) Let A be bounded below, and define B = { b ∈ R : b is a lower bound for A } . Show that sup B = inf A . Proof. First, note that B is bounded above by every element of A , so, by the Axiom of Completeness, B has a supremum. Moreover, inf A ∈ B since inf A is a lower bound for A and B is the set of all lower bounds for A . Therefore, since sup B is an upper bound for B , it must be the case that sup B ≥ inf A . If it were the case that sup B > inf A , then := sup B inf A > , so, by Lemma 1.3.7, there exists b ∈ B such that b > sup B = sup B (sup B inf A ) = inf A. But b ∈ B means that b is a lower bound for A , so this means that inf A is not the greatest lower bound for A . This is plainly impossible, so it must not be the case that sup B > inf A . The only remaining possibility is that sup B = inf A , as desired. Note: By definition of the infimum, inf A ≥ b for all b ∈ B , since B is precisely the set of lower bounds of A . If you had Exercise 1.3.7 in hand, this would complete the proof (and, really, the bulk of the above proof is essentially the proof of Exercise 1.3.7). [Added 09.12.10] Here’s a better proof, borrowed (with minor modifications) from Zach: Proof. If b ∈ B , then b is a lower bound for A , meaning that b ≤ a for all a ∈ A . Therefore, every element of A is an upper bound for B , so B is bounded above and thus, by the Axiom of Completeness, has a least upper bound sup B . We want to see that sup B satisfies the two conditions for being the greatest lower bound of A . i. If sup B is not a lower bound for A , then there exists a ∈ A such that a < sup B ....
View
Full
Document
This note was uploaded on 09/08/2011 for the course MAT 472 taught by Professor Spielberg during the Spring '06 term at ASU.
 Spring '06
 Spielberg
 Math

Click to edit the document details