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...
View
Full Document
 Spring '09
 Koskesh
 Math, Integers, Natural Numbers, Order theory, Greatest element, Partially ordered set, Maximal element, L O R D E R

Click to edit the document details