1873_solutions

See similarly that f e we conclude that e f 6 prove

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Unformatted text preview: m members and B has n members, then the set A B has m n members. (See also this exercise.) This exercise is identical to the earlier one and the solution was given to the earlier one in the instructor’s manual. 4. Given A ß U and any set B, prove that A B ß Solution: Using the fact that A ß function  from AB to UB UB. U we choose a one-one function h from A to U. We now define a as follows: B hf f h U A For each member f of the set A B we define  f to be the function h f : B U. In other words, f x hfx whenever x B. To see that  is one-one, suppose that f 1 and f 2 are different members of the st A B . f 2 x . Since h is one-one we have Choose x B such that f 1 x h f1 x h f2 x , 62 in other words,  f 1 x  f 2 x and therefore the functions  f 1 and  f 2 are unequal to each other. V, prove that A B ß 5. Given that A and B are nonempty sets and that B ß Solution: Using the fact that B is nonempty and B ß now define a function from A B into A V as follows: AV. V we choose a function h from V onto B. We B h f fh V A For each member f of the set A B we define  f to be the function f h. In other words f x fhx whenever x V. To see that the function  is one-one, suppose that f 1 and f 2 are different members of the f 2 y . Using the fact that the function h is onto the set A B and choose a member y of B such that f 1 y set B, choose a member x of V such that h x  y. Thus  f1 x  f1 h x  f1 y f2 y  f2 h x   f2 x and so the functions  f 1 and  f 2 are different from each other. ,Aß 6. Combine this exercise and this exercise to show that if A U and B ß 7. Use this exercise and the equivalence theorem to give a quick proof that if A AB ß UV. V, then A B ß UV. , A ß U and B ß V, then 8. Given any sets A, B and C, prove that AB C ßA BC . Solution: We begin by observing that a typical member of the set A B C is a function f from C to A B . For such a function f we know that, whenever x C, the function f x is a function from B to A. Thus if f A B C then for every member x of the set C and every member y of the set B we have fx y A. Now we take a similar look at the set A . A typical member of this set is a function g from B Thus if g A B C then whenever x C and y B we have g y, x A. BC C into A. With these thoughts in mind we define a function  from A B C to A B C as follows: Given any member f of the set A B C , the function  f is the function from B C to A defined by  f y, x  f x y whenever x C and y B. We need to show that the function  is one-one and that the range of  is the entire set A B C . To see that  is one-one, suppose that f 1 and f 2 are different members of the set A B C . Choose a member x of C such that f 1 x f 2 x . Using the fact that f 1 x and f 2 x are different functions from B to A we now choose a member y of B such that f1 x y f2 x y and we observe that  f 1 y, x  f 2 y, x which tells us that  f 1  f2 . Finally, to see that  is onto the set A B C , suppose that g A B C . We need to find a membe...
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