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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 oneone 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 oneone function f from A into B and a oneone function g from B into C. From Exercise
11 of the exercises on functions we see that g f is a oneone 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.
 Fall '08
 STAFF
 Math, Calculus

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