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11 Homework 03 Solutions Theory of Analysis Solutions Assignment 2 Problem 1: Let S be an ordered set. Let A ⊂ S be a nonempty subset that is bounded above. Suppose sup(A) exists and sup(A) 62 A. Show that A contains a countably infinite subset. Solution 1: Since A is nonempty, let x1 be any element of A. Now assume an has been defined to be an element of A. If an were the largest element of A, then it would be sup(A), and sup(A) 2 A which is against the hypothesis, so there must elements of A larger than an. Choose one and call it an+1. The subset fan : n 2 Zg is a countably infinite subset of A. Problem 2: Let S be a set of real numbers with supremum s, and let k be a positive real number. Define kS = fkx : x 2 Sg, that is, the set obtained by multiplying each element of S by k. Prove that kS has supremum ks. Solution 2: Since s ≥ x for any x 2 S, we can multiply by the positive number k to find ks ≥ kx. Thus ks is an upper bound for kS, and by completeness kS has a least upper bound, t. Since t is the least upper bound, ks ≥ t. But if ks > t then s > t=k and thus t=k will not be an upper bound for S|there is some x 2 S with x > t=k. But then kx > t and t is not an upper bound for kS, even though we know it is. So ks = t and the proof is complete. Problem 3: Given a real number y, define a subset Qy of the rational numbers as Qy = fq 2 Q : q < yg. That is, Qy is the set of rational numbers strictly less than real number y (which might itself be rational or irrational. Consider Qy to be a subset of R. (a). In R show that sup(Qy) = y. (b). Let A be a subset of rationals that is bounded above, that whenever x 2 A and t < x then t is also 2 A, and A does not contain its supremum. Such a set A is called a Dedekind cut. Show that, for every Dedekind cut A, there is a real number y such that A = Qy. (c). Show, in fact, that there is a bijection between R and Dedekind cuts in the rationals. Solution 3: (a). In R, y is certainly an upper bound for Qy. Let u = sup(Qy). If y 6= u then by the density theorem there is a rational number q strictly between u and y|that is u < q < y. Since q < y, q 2 Qy by definition. But then u ≥ q since u is the supremum of Qy, contradicting the fact that u < q. So y = u. Theory of Analysis Solutions Assignment 2 (b). Since A is bounded above, as a set of real numbers it must have a supremum y. We show that A = Qy. Let q be any rational number q < y. Choose = y−q. Then by the approximation property there must be an element x 2 A greater than y − = q. Since A does not contain its supremum y, x 6= y. By hypothesis, since q < x then q 2 A. On the other hand, since y is the supremum of A and A does not contain its supremum, no rational number ≥ y is in A. So A consists exactly of all rational numbers less than y, which is the definition of Qy. (c). Claim: sup(A) is a bijection from Dedekind cuts to real numbers. Obviously each cut has a supremum, so this is a function from cuts to real numbers. We show it is surjective and injective. Let A 6= B be two cuts in the rational numbers. Without loss of generality, let q be a rational number in A that is not in B. No rational number larger than q can be in B or by the definition of a Dedekind cut q would be in B. Thus, all numbers in B are smaller than q and q is an upper bound for B. Thus sup(B) ≤ q. But since q 2 A there must be numbers larger than q in A since A is a Dedekind cut and thus does not contain its supremum. Thus q is not an upper bound for A, and sup(A) > q. Therefore different cuts have different suprema, and the supremum is injective. On the other hand, given a real number y, Qy is a Dedekind cut as shown in part (b), and part (a) shows that sup(Qy) = y, so the function is surjective. Problem 4: Prove Dirichlet's Approximation Theorem. Given any positive real number x and any integer N > 1, prove that there exist integers h and k with 0 < k ≤ N such that jkx − hj < 1=N. That is, there is some relatively small multiple of x that is close to an integer. Hint: Consider the numbers tx − btxc as t varies from 0 upto N. Show that some pair of these must differ by at most 1=N. Solution 4: Consider the numbers tx − btxc as t varies from 0 to N. Each of these N + 1 numbers is a real number that is ≥ 0 and < 1. Consider splitting the interval into evenly- sized pieces [0; 1=N); [1=N; 2=N);::: [(N − 1)=N; 1). There are N of these sub-intervals of width 1=N but N + 1 numbers we are considering, so by the pigeonhole principle, at least one of the sub-intervals contains at least two of the numbers. In particular, there are two choices of t, say a and b, so that 0 ≤ j(ax − baxc) − (bx − bbxc)j < 1=N. Then let h be the integer jbaxc − bbxcj and k the integer a − b. By construction jhx − kj < 1=N and the proof is complete. Problem 5: If x is irrational, prove that there are infinitely many rational numbers h=k with k > 0 for which jx − h=kj < 1=k2. That is, there are a lot of rationals that approximate x so closely that the error is less than 1=k2 where k is the denominator of the rational number. Theory of Analysis Solutions Assignment 2 Hint: assume there are only a finite number of such approximations, h1=k1; h2=k2; : : : ; hp=kp. Let δ be the smallest of the number x − hi=ki and apply the previous problem with any integer N > 1/δ. Solution 5: As indicated in the hint, assume that there are only finitely many pairs, (h1; k1); (h2; k2);:::; (hn; kn) that satisfy the relationship. Choose the pair (hi; ki) for which δ = jx − hi=kij is as small as possible. Since x is irrational and hi=ki is rational, δ cannot be zero. By the Archimedean property we can choose an integer N larger than 1/δ and apply the previous problem to find h and k so that jkx − hj < 1=N. Then, dividing through by k, we find jx − h=kj < 1=(kN) < 1=k2 since k < N. So we have found a new approximation h=k with an error less than 1=k2, and for this approximation jkx − hj < 1=N < δ while all the previous approximations had jkix − hij ≥ δ so this new approximation was not on our previous list. So we left one out of our list|and this applies to any finite set, so the set of approximations must be infinite. Problem 6: Prove that there is no way to define an order relation < on the complex numbers that is compatible with the axioms as we have given them for real numbers. Solution 6: First, it is clear that i 6= 0 so either i > 0 or i < 0. If i > 0 then by closure i · i = −1 > 0 which we know is false. But if i < 0 then −i > 0 and (−i) · (−i) = −1 would be positive, which is still false..
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