1873_solutions

1873_solutions

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Unformatted text preview: P n is countable. Hint: This exercise follows very simply from Exercise 2. 5. Prove that if P is the set of all polynomials with rational coefficients then P is countable. Hint: Use the fact that if P n is defined as in Exercise 4 for each n then P Ý  Pn. n1 6. As indicated earlier, a number x is said to be algebraic if there exists a nonzero polynomial f with integer coefficients such that f x  0. Prove that the set of algebraic numbers is countable. Solution: The set F of polynomials with integer coefficients is countable by Exercise 5. Now for every member f of the set F we know that the set of solutions of the equation f x  0 is finite; and therefore countable. Therefore the set  x R fx 0 fF is countable. 7. A real number that is not algebraic is said to be transcendental. Complete the details of Cantor’s proof outlined earlier that there exist transcendental numbers. The work has been done. The set of algebraic numbers is countable and the set R of all real numbers is uncountable. Therefore there must exist numbers that are not algebraic. In fact, we can do better. If T is the set of transcendental numbers then, since R T is countable we have R ß T. 8. Prove that the set of all functions from Z  into 0, 1 is uncountable. The proof given here is modelled on the diagonal method given earlier, We define F to be the set of all functions from Z  into 0, 1 and, to obtain a contradiction suppose that the members of F can be arranged in a sequence f 1 , f 2 , f 3 . For each n we define g n  1 fn n .  We see that g is a function from Z into 0, 1 and that g is not equal to any of the functions f n . This is the desired contradiction. 9. Prove that if S is the set of all functions from Z  into 0, 1 and p Z  is the family of all subsets of Z  , then S ß p Z  . Deduce that p Z  is uncountable. Hint: For every subset E of Z  we define a function  E from Z  to 0, 1 by the equation 1 if x E 0 if x E x  Z E Now show that the function  defined in this way is a one-one function from p Z  onto S. Exercises on Subequivalence 60 1. Given sets A, B and C satisfying A ß B and B ß C, prove that A ß C. Choose a one-one function f from A into B and a one-one function g from B into C. From Exercise 11 of the exercises on functions we see that g f is a one-one function from A into C. 2. Given that A is strictly subequivalent to B and that B ß C. C, explain why A must be strictly subequivalent to Solution: The assertion we want to prove is obvious if A is empty. Suppose now that A . Using the fact that that B ß C we choose a function f from C onto B. Now to obtain a contradiction, suppose that A fails to be strictly subequivalent to C and choose a function g from A onto C. The composition f g is a function from A onto B, contradicting our assumption that A is strictly subequivalent to B. 3. Given that a set A is countable and that B is uncountable, explain why A is strictly subequivalent to B. Since B is an infinite set we have Z  ß B. Sinc...
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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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