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# 45 - a least element Proof To prove part 1 suppose that a...

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5.2 Analysing Partial Orders The shape of the partial orders ( N , ) and ( Z , ) are different from each other. The number 0 is the smallest element with respect to the natural numbers, but not with respect to the integers. D EFINITION 5.6 (A NALYSING P ARTIAL O RDERS ) Let ( A, ) be a partial order. 1. An element a A is minimal iff b A. ( b a b = a ) . 2. An element a A is least iff b A. a b . 3. An element a A is maximal iff b A. ( a b a = b ) . 4. An element a A is greatest iff b A. b a . In example 5.4, the least (and minimal) element is 1 , the maximal elements are 12 and 18 , and there is no greatest element. In example 5.5, the least (and minimal) element is , and the greatest (and maximal) element is { 1 , 2 , 3 } . With the usual partial order ( N , ) , the least element is 0 , and there is no maximal element. P ROPOSITION 5.7 Let ( A, ) be a partial order. 1. If a is a least element, then a is a minimal element. 2. If a is a least element, then it is unique. 3. If A is finite and non-empty, then ( A, ) must have a minimal element. 4. If ( A, ) is a total order, where A is finite and non-empty, then it has
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Unformatted text preview: a least element. Proof To prove part 1, suppose that a ∈ A is least and assume for contra-diction that b < a for some b ∈ A such that b ± = a . But a ≤ b by deFnition of a being least. This contradicts anti-symmetry. 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 anti-symmetry, 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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