HW 7 - MATH 290 HOMEWORK 7 SOLUTIONS The following problems...

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MATH 290 – HOMEWORK 7 – SOLUTIONS The following problems are based on Exercise 24 of Section 3.4, pp. 168-169 in the book. Let A be a set with partial order R . For each a A , let S a = { x A : xRa } . Let F = { S a : a A } . Then F is a collection of subsets of A , and F is partially ordered by set inclusion (that is, “ ” is a partial order on F ). 1. Prove that for every a, b A , aRb if and only if S a S b . Solution: Let a, b A . (= ) Suppose that aRb and let x S a . This means that xRa . Since also aRb , the transitivity of R implies that xRb and hence that x S b . Therefore S a S b . ( =) Suppose that S a S b . Since R is reflexive, aRa and hence a S a . Since S a S b , a S b which means that aRb . 2. Prove that for every B A , x A is the least upper bound (that is, the supremum) of B if and only if S x is the least upper bound (supremum) of the set { S b : b B } (Note that { S b : b B } is a subset of
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HW 7 - MATH 290 HOMEWORK 7 SOLUTIONS The following problems...

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