Solution Suppose x y and z are elements of a field and x 6 0 The axioms. y 1y x y xy xz x z 1z z, proving part a Part b follows from part a by setting z 1 and part c. follows from part a by setting z 1 x Since,replacing x with 1 x in part c gives part d. 1 4 Let E be a nonempty subset of an ordered set suppose is a lower bound. of E and is an upper bound of E Prove that, Solution Since E is nonempty there exists a point x E Note that is a. lower bound of E and is an upper bound of E by the definition of supremum. and infimum Thus x by the definition of lower bounds and x by the. definition of upper bounds and therefore by Definition 1 5 ii. 1 5 Let A be a nonempty set of real numbers which is bounded below Let. A be the set of all numbers x where x A Prove that. inf A sup A, Solution By the least upper bound property inf A exists By the definition. of infimum x for all x A and if then there exists a y A such. that y Therefore Proposition 1 18 a implies x for all x A. and if then there exists a y A such that y Thus is an. upper bound of A and if then is not an upper bound of A. By the definition of supremum it follows that sup A and therefore. inf A sup A,1 6 Fix b 1, a If m n p q are integers n 0 q 0 and r m n p q prove that. Hence it makes sense to define br bm,b Prove that br s br bs if r and s are rational. c If x is real define B x to be the set of all numbers bt where t is rational. and t x Prove that,br sup B r,when r is rational Hence it makes sense to define. bx sup B x,for every real x,d Prove that bx y bx by for all real x and y. Solution Fix b 1, a If m n p q then mq nq np nq and therefore mq np For simplic. ity let bm and bp Then,1 nq 1 nq 1 nq 1 nq,nq bmq bnp nq. 1 n 1 q 1 n,so bm bp Thus br bm is well defined since any two. representations of r yield the same value, b Suppose r m n and s p q for some integers m n p q with n 0 and. q 0 Then r s mq np nq and by part a and the corollary to. Theorem 1 21 we have,br bs bm n bp q bmq nq bnp nq bmq bnp. bmq np b mq np nq br s, c Suppose r p q for some integers p and q 0 and fix bs B r Then. s is rational and s r so br s 1 since b 1 Multiplying through by. bs yields br bs so br is an upper bound of B r, Since r r we have br B r Thus if br then is not an upper. bound of B r Hence br sup B r, The definition bx sup B x for every real x makes sense because the. equation br sup B r holds for every rational r by the argument above. d Suppose x and y are real r and s are rational r x and s y Then. r s x y so,br bs br s bx y, Thus taking the supremum over all rationals r x and s y gives. bx by bx y, Now suppose t is rational and t x y Then there exist rational numbers. r and s such that t r s r x and s y Thus,bt br s br bs bx by. and taking the supremum over all rationals t x y gives bx y bx by. Hence bx y bx by, 1 8 Prove that no order can be defined in the complex field that turns it into. an ordered field, Solution By Definition 1 27 we have i 6 0 and by Theorem 1 28 and Propo. sition 1 18 a d we have i2 1 0 This contradicts Proposition 1 18 d. and therefore no order can be defined in the complex field that turns it into an. ordered field, 1 9 Suppose z a bi w c di Define z w if a c and also if a c. but b d Prove that this turns the set of all complex numbers into an ordered. set Does this ordered set have the least upper bound property. Solution The real numbers form an ordered set so one and only one of the. following holds a c a c a c Similarly either b d b d or b d. There are 5 possible cases,1 If a c then z w,2 If a c then z w. 3 If a c and b d then z w,4 If a c and b d then z w. 5 If a c and b d then z w, Thus one and only one of z w z w and z w holds so Definition 1 5 i. Let z1 z2 z3 be complex numbers satisfying z1 z2 and z2 z3 Then one. and only one of the following holds,1 Re z1 Re z2,2 Re z1 Re z2 and Im z1 Im z2. and similarly one and only one of the following holds. 1 Re z2 Re z3,2 Re z2 Re z3 and Im z2 Im z3,This yields 2 possible cases. 1 If Re z1 Re z2 or if Re z2 Re z3 then Re z1 Re z3. 2 If Re z1 Re z2 Re z3 then Im z1 Im z2 Im z3, Thus z1 z3 in all cases so Definition 1 5 ii holds. Hence the set of all complex numbers is an ordered set. Now let E in n Z Then E is a nonempty subset of C which is. bounded above by 1 Suppose i sup E Then 0 since is an upper. bound But if 0 then 2 is a smaller upper bound of E contradicting. that i sup E Thus 0 and so i sup E But for any there. exists an integer n contradicting that i is an upper bound of E Hence. sup E does not exist so the least upper bound property does not hold. Real Analysis,Math 131AH,Rudin Chapter 2,Dominique Abdi. 2 1 Prove that the empty set is a subset of every set. Solution Assume the contrary that there is a set E such that the empty set is. not a subset of E Then there is an element x such that x E but this. contradicts that the empty set is empty Hence E, 2 2 A complex number z is said to be algebraic if there are integers a0 an. not all zero such that,a0 z n a1 z n 1 an 1 z an 0. Prove that the set of all algebraic numbers is countable. Solution Let p be a polynomial of degree n with integer coefficients and let. Zp z C p z 0, By the fundamental theorem of algebra Zp has at most n elements. Now define the set,Pn Z ak z ak Z an 6 0, consisting of all polynomials of degree n with integer coefficients and let. En a0 a1 an ak Z an 6 0, Example 2 5 can be easily modified to show that the set of non zero integers is. countable Thus by Example 2 5 and Theorem 2 13 En is countable Define. f En Pn Z f a0 a1 an ak z k, Every polynomial of degree n is uniquely determined by its n 1 coefficients. so f is one to one and onto Hence Pn Z is countable and by Theorem 2 12. consisting of all polynomials with integer coefficients is countable. Since the algebraic numbers are defined to be the roots of polynomials with. integer coefficients the set A of all algebraic numbers can be written as. a union of a countable collection of finite hence at most countable sets By. the corollary to Theorem 2 12 A is at most countable But for any n Z. is a polynomial with integer coefficients for which z n is a root so Z A. We have shown that A is an at most countable set which contains an infinite. subset Hence A is infinite and at most countable and thus A is countable. 2 3 Prove that there exist real numbers which are not algebraic. Solution Assume that every real number is algebraic Then R is countable by. Exercise 2 This contradicts the corollary to Theorem 2 43 which states that R. is uncountable, 2 4 Is the set of all irrational real numbers countable. Solution Assume that the set Qc of all irrational real numbers is countable. Since R Q Qc and Q is countable by the corollary to Theorem 2 13 it. follows that R is countable by the corollary to Theorem 2 12 This contradicts. the corollary to Theorem 2 43 which states that R is uncountable. 2 5 Construct a bounded set of real numbers with exactly three limit points. Solution Define the sets,1 n 1 2n 1,E0 n N E1 n Z E2 n N. and E E0 E1 E2 Then E is bounded since E 0 2, We now show that E 0 0 1 2 Fix 0 and choose N such that N 1. Then 0 1 n for all n N and in particular,1 n 1 2n 1. for all n N This shows that an open interval of arbitrarily small radius 0. centered at 0 1 or 2 intersects E at infinitely many points so 0 1 2 E 0. Now suppose x E 0 If,0 1 2 If x 0 then 2x 0 E so x. x 0 then there are three possible cases,Case 1 If 0 x 1 then choose k N such that. and define to be the smallest non zero element of the set. Case 2 If 1 x 2 then choose k N such that, and define to be the smallest non zero element of the set. Case 3 If x 2 then define x 2, In all three cases we have x x E by the choice of so x. Hence E 0 1 2, 2 6 Let E 0 be the set of all limit points of a set E Prove that E 0 is closed. Prove that E and E have the same limit points Recall that E E E 0 Do. E and E 0 always have the same limit points, Solution By Definition 2 26 and Theorem 2 27 b to prove that E 0 is closed. it suffices to show that E 0 contains its limit points Let x be a limit point of E 0. and let Nr x be a neighborhood of x Then there exists a point y Nr x E 0. with x 6 y Let r0 r d x y and note that Nr0 y E contains a point. z E since y is a limit point of E Moreover x 6 z since d x z 0 The. triangle inequality gives,d x z d x y d y z d x y r0 r. so z Nr x Thus x is a limit point of E so therefore E 0 0 E 0 and hence. E 0 is closed, We now show that E and E have the same limit points If x E 0 and r 0. then there exists a point y Nr x E with x 6 y Since E E we have. y Nr x E and therefore x E 0 Conversely suppose that x E 0 and. r 0 Then there exists a point y Nr x E with x 6 y If y E then. y E 0 since E E E 0 Setting r0 r d x y then there exists a z E. with z 6 y and z 6 x since d x z 0 In either case Nr x contains a point. of E distinct from x so x E 0 Thus E 0 E 0, The sets E and E 0 need not have the same limit points Consider the set. E defined in the solution to Exercise 2 5 We showed that E 0 0 1 2 If. x E 0 then x 1 x 1 E 0 x and if x,E 0 then for,min x y y E 0. we have x x E 0 Thus E 0 0 and therefore E 0 6 E 0 0. 2 7 Let A1 A2 A3 be subsets of a metric space,a If Bn i 1 Ai prove that B n i 1 Ai for n 1 2 3. b If B i 1 Ai prove that B i 1 Ai, Show by an example that the inclusion can be proper. a Fix x i 1 Ai Then x Ai Ai A0i for some i If x Ai then. x B n since Ai Bn B n If x A0i then for every r 0 there is a. Sn y Ai Bn with x 6 y such that d x y r Thus x Bn B n. so i 1 Ai B n,i 1 Ai Then,that Nri x Ai c for each. Thus there exist r1 rn 0 such T Tn i Let,r min r1 rn Then Nr x i 1 Nri x so Nr x i 1 Ai c. In particular Nr x contains no points of any Ai and therefore no points. of Bn If Nr x contains a point y Bn0 then letting r0 r d x y we. see that Nr0 y contains a point z Bn by the definition of limit points. and z Nr0 y Nr x by the triangle inequality contradicting that. Nr x Bn Thus,x Bnc Bn0 c Bn Bn0 c B n c, so therefore B n i 1 Ai and by the result proved above B n i 1 Ai. b Fix x i 1 Ai Then x Ai Ai A0i for some i If x Ai then. x B since Ai B B If x A0i then for every r 0 there is a. pointSy Ai B with x 6 y such that d x y r Thus x B 0 B so. For all i N let Ai 1 i Then Ai Ai since finite sets are closed by. the corollary to Theorem 2 20 However B 1 i i N and therefore. B i N 0 1 i i N Ai,and the inclusion is proper since 0. 2 8 Is every point of every open set E R2 a limit point of E Answer the. same question for closed sets in R2, Solution Yes every point of every open set E R2 a limit point of E Let. E R2 be open and fix x E Then x is an interior point of E so there. exists an r 0 such that Nr x E Let Ns x be an arbitrary neighborhood. of x If s r then Nr x Ns x so Nr x Ns x E If s r then. Ns x Nr x E so Ns x Ns x E In either case Ns x E is itself a. neighborhood of x in R2 and therefore contains infinitely many points and in. particular some point y 6 x Thus x is a limit point of E. The same result does not hold for closed sets in R2 A set consisting of a. single point x R2 is closed since its complement is open but finite sets have. no limit points by the corollary to Theorem 2 20, 2 9 Let E denote the set of all interior points of a set E. a Prove that E is always open,b Prove that E is open if and only if E E. c If G E and G is open prove that G E, d Prove that the complement of E is the closure of the complement of E. e Do E and E always have the same interiors,f Do E and E always have the same closures. a Fix x E Then there exists an r 0 such that Nr x E If d x y r. and r0 r d x y then d y z r0 implies,d x z d x y d y z r. so y E This shows that Nr x E so x is an interior point of E. Thus E is open, b If E E then E is open by part a If E is open then every point of. E is an interior point so E E Since every interior point of a set must. be contained in the set the reverse inclusion holds as well so E E. c Suppose G E and G is open and fix x G Then x is an interior point. of G so there exists an r 0 such that Nr x G E Thus x is an. interior point of E so G E, d Fix x E c Then x is not an interior point of E so every neighborhood. Nr x intersects E c Thus x is in E c or x is a limit point of E c so. x E c E 0 c E, Now suppose x E c Then x E so there exists an r 0 such that. Nr x E Thus x E and x is not a limit point of E c so. x E E 0 c E c c E 0 c E E 0 c E c, e No E and E do not always have the same interiors For example note. that Q because every neighborhood of a rational number contains. irrational numbers However Q R so Q R since R is open. f No E and E do not always have the same closures For example note. that Q because every neighborhood of a rational number contains. irrational numbers Therefore Q since the empty set is closed. 2 10 Let X be an infinite set For p X and q X define. 1 if p 6 q, Prove that this is a metric Which subsets of the resulting metric space are. open Which are closed Which are compact, Solution Parts a and b of Definition 2 15 follow immediately from the def. inition of d If p q then part c is trivial If p 6 q then for every r X. either p 6 r or q 6 r Thus d p r d q r 1 and since d p q 1 part c. Real Analysis Math 131AH Rudin Chapter 1 Dominique Abdi 1 1 If ris rational r6 0 and xis irrational prove that r xand rxare irrational Solution

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