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# 540hw3ans - HW 31 onto 1(a Let(A be a well-ordered set B A...

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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 ² 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 0 2 A such that f ( a 0 ) ² a 0 . Then the set A = f a 2 A j f ( a ) ² a g is not empty because a 0 2 A . Because ( A; ° ) is a well-ordered set and A 6 = ? , the set A contains the smallest element a 0 2 A , i.e., a 0 ° a 8 a 2 A : Now set b 0 = f ( a 0 ) . Because b = f ( a 0 ) ² a 0 and because f preserves the order, we have f ( b ) ² f ( a 0 ) = b 0 , i.e., b 0 2 A and b 0 ² a 0 , a contradiction because a 0 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 ) ² a; in particular, f ( a ) ² 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

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otherwise, b ² 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 ² 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 ² b and y < b , we arrive at a contradiction. So, I b n I = ? .
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