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Unformatted text preview: a least element. Proof To prove part 1, suppose that a A is least and assume for contradiction that b < a for some b A such that b = a . But a b by deFnition of a being least. This contradicts antisymmetry. To prove part 2, suppose that a and b are both least elements. By deFnition of least element, we have a b and b a . By antisymmetry, it follows that a = b . To prove part 3, pick any a A . If a is not minimal we can pick a 1 < a . If a 1 is not minimal, we can pick a 2 < a 1 . In this way, we get a decreasing chain a > a 1 > a 2 > ... . All the elements of the chain must be different, by construction. Since A is a Fnite set, we must Fnd a minimal element at some point. [Notice that in this proof we not only show the existence of minimal elements, but also how to Fnd one.] Part 4 is left as an exercise. 45...
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This note was uploaded on 03/02/2010 for the course MATH Math2009 taught by Professor Koskesh during the Spring '09 term at SUNY Empire State.
 Spring '09
 Koskesh
 Math, Integers, Natural Numbers

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