To the set 0 1 s and deduce that cantors inequality

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Unformatted text preview: t a successor. We know that whenever x  a we have x  a and since the equation x   a does not occur we must have x   a whenever x  a. Now suppose that x   a whenever x  a. This condition tells us that there can’t exist a member x of S for which x   a and so a fails to be a successor. 9. If S is a well ordered set and a S, then we saw that a is a limit member of S if a is not the least member of S and a has no predecessor. Prove that if a well ordered set S has a limit member then S has a nonempty subset E that has no limit member and no largest member. Suppose that a well ordered set S has a limit member. We define q to be the least limit member of S and we define E  x S x  q . We see at once that E is nonempty and has no limit member. Now given any member x of E, it follows from Exercise 8 and the fact that x  q that x  E and so E has no largest member. 10. Give an example of a totally ordered set S that is not well ordered even though every member of S has a successor. The system Z of integers has the desired properties. Exercises on Zorn’s Lemma 1. a. A set S of real numbers is said to be linearly independent over the set Q of rational numbers if for every positive integer n and every choice of members x 1 , , x n of the set S and rational numbers r 1 , , r n , the condition r1x1    rnxn  0 cannot hold unless all of the numbers r 1 , , r n are zero. Prove that if is the family of subsets of R that are linearly independent over Q and , is partially ordered by the relation , then has a maximal member. We need to show that every chain in the partially ordered set has an upper bound. Suppose that is a chain in . Each member of is linearly independent over the set Q of rational numbers and if A and B are any two members of then either A B or B A. We define U  Þ and we observe that U is a set of numbers and A U whenever A . All we need to do is show that U is linearly independent. Suppose that n is a positive integer and that x 1 , x 2 , , x n belong to . We see easily that some member A of contains all of the numbers x 1 , x 2 , , x n . Since A is linearly independent, no linear combination r1x1    rnxn with rational coefficients can be zero unless all of the coefficients are zero. This completes the proof that U and so the chain has an upper bound in . 69 b. Prove that if H is a maximal member of the family that was defined in part a then every real number can be expressed uniquely in the form r 1 x 1    r n x n . 2. An additive subgroup of the system R of real numbers is defined to be a nonempty subset G of R with the property that whenever x and y belong to G, then so do the numbers x  y and x y. a. Which of the following subsets of R are additive subgroups of R? These simple exercises have appeared before. i. The set 0 . ii. The set 1 . iii. The set 1, 0, 1 . iv. The set Z of all integers. v. The set Q of all rational numbers. vi. The set R Q of all irrational numbers. vii. The empty set . b. Given that G is an...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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