540hw3ans - HW 3 1 1. (a) Let ( A; & ) be a...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW 3 1 1. (a) Let ( A; & ) be a well-ordered set, B ¡ A and f : A onto ! B be an order isomorphism of ( A; & ) onto ( B; & ) . Prove that for every a 2 A , a & f ( a ) . (b) Let ( A; & ) be a well-ordered set and a 2 A . Recall that the set I a = f x 2 A j x ¢ a g is called an initial segment of A . Use part (a) to prove that ( A; & ) cannot be order isomorphic to one of its initial segments. Proof. (a) Contrary to the claim, suppose that there is a 2 A such that f ( a ) ¢ a . Then the set A = f a 2 A j f ( a ) ¢ a g is not empty because a 2 A . Because ( A; & ) is a well-ordered set and A 6 = ? , the set A contains the smallest element a 2 A , i.e., a & a 8 a 2 A : Now set b = f ( a ) . Because b = f ( a ) ¢ a and because f preserves the order, we have f ( b ) ¢ f ( a ) = b , i.e., b 2 A and b ¢ a , a contradiction because a is the smallest element of A . (b) Contrary to the claim, suppose that there is a 2 A such that ( A; & ) is order isomorphic to the initial segment I a . Then 8 x 2 A f ( x ) ¢ a; in particular, f ( a ) ¢ a , a contradiction with the result of part (a). 2. An ideal in a partial ordered set ( X; & ) is a set I ¡ X such that ( x 2 I ) ^ [( y 2 X ) ^ ( y & x )] ) y 2 I . Prove that every ideal I in a well-ordered set ( X; & ) is either an initial segment of ( X; & ) or X . Proof. If I = X , then we are done. Let I 6 = X . Let X n I 6 = ? . Set b = inf ( X n I ) . We claim that I = I b . Indeed, if x 2 I , then x & b ; 1 Note to the grader. Please grade the following problems: #1, 2, #3, 4, 5 (20 pts for each problem) 1 otherwise, b & x . Because I is an ideal, b 2 I , a contradiction. So, I ¡ I b . Now we show that I b n I = ? . Contrary to the claim, suppose that there is y 2 I b n I . Then y & b because y 2 I b and y < b because y 2 X n I and b is the smallest element in X n A . Hence, because y & b and y < b , we arrive at a contradiction. So, I...
View Full Document

This note was uploaded on 02/16/2012 for the course MATH 540 taught by Professor Ruan during the Spring '08 term at University of Illinois, Urbana Champaign.

Page1 / 5

540hw3ans - HW 3 1 1. (a) Let ( A; & ) be a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online