1873_solutions

Numbers is countable and the set r of all real

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Unformatted text preview: e A ß Z  ß B we have A ß B. Since A is countable and B is uncountable it is impossible to have A ß B. 4. Given that a set S is finite and that x S, explain why the set S x must be strictly subequivalent to S. Since S x S we know that S x is subequivalent to S. We also know from Exercise 2 of the x are not equivalent to each other. exercises on finite sets that the sets S and S 5. Given that A ß B prove that p A ß pB. Hint: Using the fact that A ß B we choose a one-one function f from A to B. We now define a function  from p A to p B as follows: E fE for every subset E of A. Show that the function  defined in this way is one-one. To see that  is one-one, suppose that E and F are subsets of A and that  E   F , in other words f E  f F . Given any member x of the set E we know that f x f F . Choose u F such that f x  f u . Since f is one-one we have x  u and so x F. Therefore E F and we see similarly that F E. We conclude that E  F. 6. Prove that R ß pQ. Hint: Consider the function f : R p Q defined by fx  r for every x Q rx R. 7. Give an example of two nonempty sets A and B such that the set A Þ B fails to be subequivalent to the set A B. The set 0 Þ 1 fails to be subequivalent to 0 1. 8. Given that each of two given sets A and B has more than one member, prove that A Þ B ß A B. Choose two different members x and y of A and choose two different members u and v of B. We define A1  A u and B Þ y, v x, u B1  x Since A ß A 1 and B ß B 1 and since A 1 B 1  it is easy to see that A Þ B ß A 1 Þ B 1 . As a matter of fact, if we define f t, u  t for all t A and define f x, t  t for all t B u and define f y, v  u then f is a function from A 1 Þ B 1 onto A Þ B. Since A 1 Þ B 1 A B we conclude that A Þ B ß A B. Exercises on The Equivalence Theorem 61 The exercises in this subsection explore some of the harder properties of set equivalence that can be deduced with the help of the equivalence theorem. We also need a little notation: Given any two sets A and B, the set A B is defined to be the set of all functions from B to A. Thus AB  f f : B A . 1. Prove that for any set A we have A 0, 1 Hint: We define a function  from 0, 1 A ßpA. to p A as follows: f  x A fx 0 for every member f of the set 0, 1 A . Prove that the function  is one-one and onto the set p A . To see that  is one-one, suppose that f and g belong to 0, 1 A and that  f   g . In other words x A fx 0  x A gx 0 . Given any x A, if f x  0 then g x  0  f x and if f x 0 then f x  1 and g x 0 which gives us g x  1  f x . Therefore f  g. Finally, to see that  is onto the set p A , suppose that B is any subset of A. We define 0 if x B 1 if x fx  A B and observe that  f  B. 2. Prove that if S is the set defined in the proof of this theorem then S ß R. Solution: Since S R we know that S ß R. On the other hand,  S ß 0, 1 Z ß p Z  ß R and so the result follows from the equivalence theorem. 3. Prove that if A and B are finite sets and A and...
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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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