1873_solutions

S and therefore c ny x n z and x s 3 a nonempty family

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Unformatted text preview: and E 2 belongs to . then  Z Exercises on Choice Dependent Properties of Sets 1. Given that A and B are well ordered sets and that every initial segment of A is strictly subequivalent to B and every initial segment of B is strictly subequivalent to A, prove that A and B are order isomorphic. We know that if A and B are not order isomorphic to one another then either A is order isomorphic to an initial segment of B or B is order isomorphic to an initial segment of A. But both of these options are impossible because no initial segment of B can be equivalent to A and so initial segment of A can be equivalent to B. 2. Given that S is an infinite set and that A is strictly subequivalent to S, prove that S A ß S. We observe first that if S A were subequivalent to A then it would follow from the preceding theorem on unions that S  A Þ S A ß A, which we know to be false. It therefore follows from the theorem on comparision of sets that A ß S A and we can apply the union theorem again to yield S  A Þ S A ß S A. 3. Suppose that S n is a sequence of sets, that for each n, the set S n is strictly subequivalent to S n1 , and that S  Sn. n Z Prove that there does not exist a set E such that p E ß S. This exercise is related to an earlier exercise on applications of the equivalence theorem. If we had S ß p E then we would have  SZ ß 0, 1 E Z ß 0, 1  Now since the set S is infinite, so is the set E and so Z ß  SZ ß 0, 1 E Z ß 0, 1 E Z E Z . E and the theorem on products gives us ß 0, 1 E ßS which we know to be false. 4. Prove that there is a well ordered set S such that 1 is strictly subequivalent to S and such that for every x S, we have P S, x ß 1 . Prove that any set of this type is cardinally ordered and that any two sets of this type are order isomorphic. Choose a set T such that 1 is strictly subequivalent to T. For example, we chould take T  p 1 . We now assign a well order to the set T. If for every member x of T we have P S, x ß 1 then we take S  T. Otherwise we define y to be the least member of T for which the segment P S, y fails to be subequivalent to 1 and we define S  P S, y . It is clear that S has the desired properties. Finally, if S 1 and S 2 are two sets with the specified properties then, neither of the sets can be order isomorphic with an initial segment of the other, it follows from the uniqueness theorem for well ordered sets that S 1 and S 2 are order isomorphic to one another. 71 5. Prove that if we choose any one set of the type described in the preceding exercise and call it 2 then for any set S, the set 1 will be strictly subequivalent to S if and only if 2 ß S. It is clear that if 2 ß S then 1 must be strictly subequivalent to S. Now suppose that 1 is strictly subequivalent to S. We assign a well order to S. Since S cannot be order isomorphic to any initial segment in 2 we conclude that either 2 is order isomorphic to S or S. 2 is order isomorphic to an initial segment of S. In either case we have 2...
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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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