a1notes - Math 486 Homework 1 Notes Paul Shafer...

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Math 486 Homework 1 Notes Paul Shafer pshafer@math.cornell.edu February 5, 2008 Problem I.1.8a We are to prove that ( N n ,< L ) is a well order. It is reasonably clear that ( N n ,< L ) is a linear order, so we need to prove the following: every nonempty A N n has a < L -least element. This is stronger than saying N n has a < L -least element. It is easy to give an example of a linear order with a least element that is not a well order. Consider the set {-∞} ∪ Z with Z ordered in the usual way and -∞ < z for all z Z . Then -∞ is least, but the order is not a well order because the subset Z has no least element. Given A N n nonempty, we can find the least element of A exactly as you would in the dictionary. Let A 1 = { x 1 | ∃ x 2 ,x 3 ,...,x n < x 1 ,x 2 ,x 3 ,...,x n > A } = the set of 1 st coordinates of A a 1 = the least element of A 1 A 2 = { x 2 | ∃ x 3 ,...,x n < a 1 ,x 2 ,x 3 ,...,x n > A 1 } = the set of 2 nd coordinates of A 1 a 2 = the least element of
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a1notes - Math 486 Homework 1 Notes Paul Shafer...

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