1873_solutions

That b solution using the fact that b is nonempty and

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Unformatted text preview: r f of the set A B C such that g   f . Given any member x of the set C we define f x to be the function from B to A 63 whose value at each member y of the set B is g y, x . The function f from C to A B that we have defined clearly satisfies the condition g   f . Z 9. Show that R ß 0, 1  and then use this exercise to show that R Z ß R. Solution: We already know that R ß p Z  ß 0, 1 Z . Therefore  RZ ß 10. Prove that R Z 0, 1 Z ß 0, 1 Z Z Z ß 0, 1 ß R, R ß R. Solution: We know that R ß whenever x, y theorem that R R R. On the other hand, if we define f x, y  x, y, 0, 0, 0,   R R then we can see that R R ß R Z ß R and so it follows from the equivalence R ß R. 11. Prove that R 3 ß 0, 1 . We know that R ß R 3 . On the other hand, if we define f x, y, z  x, y, z, 0, 0,   3 whenever x, y, z R then we can see that R 3 ß R Z ß R and so it follows from the equivalence theorem that R 3 ß R. 12. Prove that p R Z ß pR R ßpR. Solution: On the one hand we know that p R ß Z pR and on the other other hand we know that pR Z ß R pR ß 0, 1 R R ß 0, 1 RR ß 0, 1 R ßpR and so the result follows from the equivalence theorem. 13. Prove that R R ß p R . On the one hand, p R ß 0, 1 RR ß 0, 1 R ß Z R R and on the other hand R ß 0, 1 Z R ß 0, 1 RR ß 0, 1 R ßpR and so the result follows from the equivalence theorem. 14. Prove that if A ß R and A has more than one member then A R ß p R . Since A has more than one member we have p R ß 0, 1 R ß A R . On the other hand, AR ß RR ß p R and so the result follows from the equivalence theorem. 15. Given that S is a set with more than one member and that S S ß S, prove that S S ß p S shall show in a later theorem that the condition S S ß S is satisfied by every infinite set. Since S has more than one member we have p S ß 0, 1 S ß S S ß p S S . On the other hand, pS S ß 0, 1 S S ß 0, 1 SS ß 0, 1 S S ß p S . We ßpS and so the result follows from the equivalence theorem. 16. Given that S S ß S, that A ß S and that A has more than one member, prove that A S ß p S . Since A has more than one member we have 64 p S ß 0, 1 S ß AS. On the other hand, AS ß SS ß p S by Exercise 15. So the result follows from the equivalence theorem. 17. The preceding exercises show that when A ß B and the set A has more than one member and the set B satisfies the condition B B ß B, then we have A B ß p B . Now what happens when B is the smaller set? In other words, what happens when B is strictly subequivalent to A? In this case, Exercises this one, this one, this one and this one seem to suggest that we should have A B ß A. In this exercise we see that this statement is false even if A is uncountable and B is countable: a. Suppose that S n is a sequence of sets, that S  Sn n Z  and that, for each n, the set S n is strictly subequivalent to S. Prove that the sets S and S Z are not  equivalent to each other. Show, in fact, that there is no function from S onto S Z .  Solutio...
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