Exam (elaborations) TEST BANK FOR Principles of Mathematical Analysis

Exam (elaborations) TEST BANK FOR Principles of Mathematical Analysis Similarly, it follows from Definition 1.12(M2), (M3), (M4) and (M5) that xy = x implies that y = 1. (c) Similarly, it follows from Definition 1.12(M2), (M3), (M4) and (M5) that xy = 1 implies that y = 1 x . (d) Since 1 x ∈ F, we have 1 1 x ∈ F such that 1 x · 1 1 x = 1. Now we have 1 x · 1 1 x = 1 x ·x(= 1), then Proposition 1.15(a) implies that 1 1 x = x. This completes the proof of the problem.  1.2 Properties of supremums and infimums Rudin Chapter 1 Exercise 4. Problem 1.4 Proof. Since E ⊂ S, the definitions give α ≤ x and x ≤ β for all x ∈ E. Thus Definition 1.5(ii) implies that α ≤ β. This completes the proof of the problem.  Rudin Chapter 1 Exercise 5. Problem 1.5 Proof. Theorem 1.19 says that R is an ordered set with the least-upper-bound property. Since A is a non-empty subset of R and A is bounded below, inf A exists in R by Definition 1.10. Furthermore, −A is a non-empty subset of R. Let y be a lower bound of A, i.e. y ≤ x for all x ∈ A. Then we have −x ≤ −y for all x ∈ A. Thus −y is an upper bound of −A and sup(−A) exists in R by Definition 1.10. Let α = inf A and β = sup(−A). By definition, we have y ≤ β for all y ∈ −A, where y = −x for some x ∈ A. It implies that x = −y ≥ −β for all x ∈ A, so −β is a lower bound of A and then −β ≤ α. Similarly, we have α ≤ x for all x ∈ A so that −α ≥ −x for all x ∈ A. It implies that −α is an upper bound of −A, so β ≤ −α and then −β ≥ α. Hence we have α = −β, i.e. inf A = − sup(−A). This completes the proof of the problem.  1.3 An index law and the logarithm Rudin Chapter 1 Exercise 6. Problem 1.6 Proof. (a) Since b m > 0 and n ∈ N, Theorem 1.21 implies that there exists one and only one real y such that y n = b m. Similarly, there exists one and only one real z such that z q = b p . We have y nq = (y n ) q = (b m) q = b mq = b pn = (b p ) n = (z q ) n = z qn which implies that y = z, i.e., (b m) 1 n = (b p ) 1 q . (b) Let b r = b m n and b s = b p q . Without loss of generality, we may assume that n and q are positive. Then the corollary of Theorem 1.21 implies that b r+s = b mq+np nq = (b mq+np) 1 nq = (b mq × b np) 1 nq = (b mq) 1 nq × (b np) 1 nq = b m n × b p q = b r × b s . 3 1.3. An index law and the logarithm (c) By definition, B(r) = {b t |t ∈ Q, t ≤ r}, where r ∈ Q. It is clear that b r ∈ B(r), so it is a nonempty subset of R. Since b > 1, we have b t ≤ b r for all t ≤ r so that b r is an upper bound of B(r). Therefore, Theorem 1.19 and Definition 1.10 show that sup B(r) exists in R. Now we show that b r = sup B(r). If 0 < γ < br , then γ is obviously not an upper bound of B(r) because b r ∈ B(r). By Definition 1.8, we have b r = sup B(r). (d) By part (c), we know that b x , by and b x+y are all well-defined in R. By definition, we have B(x) = {b r | r ∈ Q, r ≤ x}, B(y) = {b s | s ∈ Q, s ≤ y}, B(x + y) = {b t |t ∈ Q, t ≤ x + y}. Before continuing the proof, we need to show several results: For every real x and y, we define B(x, y) = B(x) × B(y) = {b r × b s | r, s ∈ Q, r ≤ x, s ≤ y}. Then we have b x × b y = sup B(x, y). Lemma 1.1 Proof of Lemma 1.1. By definition, b x and b y are upper bounds of B(x) and B(y) respectively, so we have b r ≤ b x and b s ≤ b y for every b r ∈ B(x) and b s ∈ B(y). Therefore, we have b r × b s ≤ b x × b y for every b r × b s ∈ B(x, y). In other words, b x × b y is an upper bound of B(x, y). Let 0 < α < bx × b y . Then we have α bx < by . We define the number p = 1 2 ( α bx + b y ). It is obvious from this definition that α b x < p < by . By α b x < p, we have α p < bx and so there exists b r ∈ B(x) such that α p < br . (1.1) Similarly, the inequality p < by implies that there exists b s ∈ B(y) such that p < bs . (1.2) Now inequalities (1.1) and (1.2) show that α < br × b s for some b r × b s ∈ B(x, y). Hence α is not an upper bound of B(x, y) and we have b x × b y = sup B(x, y), completing the proof of the lemma.  Let S be a set of positive real numbers and bounded above and S −1 = {x −1 | x ∈ S}. Then we have sup S = 1 inf S−1 . Lemma 1.2 Proof of Lemma 1.2. Suppose that α is an upper bound of S, i.e., 0 < x ≤ α for all x ∈ S. Then we have 0 < α−1 ≤ x −1 for all x −1 ∈ S −1 . Hence the result follows directly from the definitions of the least upper bound and the greatest lower bound.  Chapter 1. The Real and Complex Number Systems 4 For every real x, we have b −x = 1 b x . Lemma 1.3 Proof of Lemma 1.3. We have two facts: – Fact 1. If b > 1, then b 1 n > 1 for every positive integer n. Otherwise, 0 < b 1 n < 1 implies that 0 < b = (b 1 n ) n < 1 n = 1 by Theorem 1.21, a contradiction. – Fact 2. If m and n are positive integers such that n > m, then b 1 m > b 1 n . Otherwise, it follows from Fact 1 that 1 < b 1 m < b 1 n and so it implies that b = (b 1 m ) m < (b 1 n ) m < (b 1 n ) n = b, a contradiction. Let r and s be rational. Define A(x) = {b s | s ∈ Q, s ≥ x}. We next want to prove that sup B(x) = inf A(x). In fact, it is clear that sup B(x) ≤ inf A(x) by definitions. Suppose that D = inf A(x) − sup B(x). Assume that D > 0. By Fact 2 above, b 1 n − 1 is decreasing as n is increasing, so there exists a positive integer N such that b x (b 1 n − 1) < D for all n ≥ N. By Theorem 1.20(b), we see that there exist r, s ∈ Q such that x − 1 2n < s < x and x < r < x + 1 2n . Thus we have r − s < x + 1 2n − (x − 1 2n ) = 1 n and this implies that D < br − b s = b s (b r−s − 1) ≤ b x (b r−s − 1) < D, a contradiction. Hence we must have sup B(x) = inf A(x). Define B′ (−x) = {b −t |t ∈ Q, t ≤ −x}. Let s = −t. Then we have b −t = b s and the inequality t ≤ −x is equivalent to s ≥ x. Thus we have B′ (−x) = A(x). Now it follows from Lemma 1.2 and the above facts that b −x = sup B(−x) = 1 inf B′(−x) = 1 inf A(x) = 1 sup B(x) = 1 b x , completing the proof of the lemma.  We can continue our proof of the problem. Let b r and b s be any elements of B(x) and B(y) respectively. Then we have r and s are rational numbers such that r ≤ x and s ≤ y. Since the sum t = r + s is also rational and t ≤ x + y, part (b) implies that b r × b s = b r+s ≤ b x+y . Thus b x+y is an upper bound of the set B(x, y) and Lemma 1.1 implies that b x × b y ≤ b x+y . Assume that b x b y < bx+y . We deduce from Lemma 1.3 that b y = (b −x b x )b y = b −x (b x b y ) < b−x b x+y ≤ b −x+(x+y) = b y , a contradiction. Hence we have the desired result that b x b y = b x+y . This completes the proof of the problem.  Rudin Chapter 1 Exercise 7. Problem 1.7 Proof. 5 1.4. Properties of the complex field (a) Since b > 1, it is obvious that b n − 1 = (b − 1)(b n−1 + b n−2 + · · · + 1) ≥ (b − 1)(1 + 1 + · · · + 1) = n(b − 1) for any positive integer n. (b) It is obvious that b 1 n > 1; otherwise, we have b = (b 1 n ) n < 1 which is impossible. The result follows by replacing b by the real number b 1 n in part (a) and Problem 1.6(a). (c) If t > 1 and n > b−1 t−1 , then part (b) implies that b − 1 ≥ n(b 1 n − 1) > b − 1 t − 1 × (b 1 n − 1) and so b 1 n < t. (d) Let w be a number such that b w < y. Let t = y · b −w. It is easily to check that t > 1. If n is sufficiently large enough, then we have n > b−1 t−1 . Hence it follows from parts (c) and (b) that b 1 n < t = y · b −w and thus b w+ 1 n < y for sufficiently large n. (e) Let w be a number such that b w > y. Let t = y −1 · b w. It is obvious that t > 1. If n is sufficiently large enough, then we have n > b−1 t−1 . Hence it follows from part (c) and then part (b) that n(y −1 b w − 1) > b − 1 ≥ n(b 1 n − 1) and thus b w− 1 n > y for sufficiently large n. (f) We have A = {w ∈ R | b w < y}. Since x is the least upper bound of A, we have w ≤ x for all w ∈ A. If b x < y, then part (d) implies that b x+ 1 n < y for sufficiently large n and so x + 1 n ∈ A. Therefore, we have x + 1 n ≤ x and then 1 n ≤ 0, a contradiction. Similarly, if b x > y, then x /∈ A and so w < x for all w ∈ A. Now part (e) implies that b x− 1 n > y for sufficiently large n, so we have w < x − 1 n < x for some sufficiently large n. This means that x − 1 n is an upper bound of A, contradicting to the fact that x is the least upper bound of A. Hence, we must have b x = y. (g) The uniqueness of x follows from the uniqueness of the least upper bound of the set A. We finish the proof of the problem.  1.4 Properties of the complex field Rudin Chapter 1 Exercise 8. Problem 1.8 Proof. Assume that i > 0. Then we have i·i > i·0 which implies that −1 > 0, a contradiction. Similarly, the case i < 0 is impossible.  Rudin Chapter 1 Exercise 9. Problem 1.9 Chapter 1. The Real and Complex Number Systems 6 Proof. We check Definition 1.5 in this case. Let z = a + bi, w = c + di ∈ C. If a 6= c, then we have either z < w or z > w. If a = c and b 6= d, then we have either z < w or z > w. If a = c and b = d, then we have z = w. Therefore, this relation satisfies Definition 1.5(i). Let z = a + bi, w = c + di and q = e + f i be complex numbers such that z < w and w < q. Since z < w, we have either a < c or a = c and b < d. Similarly, since w < q, we have either c < e or c = e and d < f. Combining the above two results, we get either a < e or a = e and b < f. This proves Definition 1.5(ii). Hence this turns C into an ordered set. Let S = {z = a + bi | b ∈ R, a < 0} ⊂ C. Since −1 ∈ S, it is clear that S is not empty. We will show that sup S does not exist in C. Assume that w = c + di was the least upper bound of S for some c, d ∈ R. That is, z ≤ w for all z ∈ S. If c > 0, then the definition of the dictionary order implies that z < ζ = 0 + bi < w for all z ∈ S, contradicting to the fact that w = sup S. If c < 0, then we have c < c 2 so that ζ = c 2 + di ∈ S and w < ζ, contradicting to the fact that w = sup S. Hence this order set does not have the least-upper-bound property, completing the proof of the problem.  Rudin Chapter 1 Exercise 10. Problem 1.10 Proof. If v ≥ 0, then we have z 2 = a 2 + 2abi − b 2 = u + 2 |w| 2 − u 2 4 1 2 i = u + (|w| 2 − u 2 ) 1 2 i = u + (v 2 ) 1 2 i = u + vi = w. If v ≤ 0, then we have z 2 = a 2 − 2abi − b 2 = u − (|w| 2 − u 2 ) 1 2 i = u − (v 2 ) 1 2 i = u − (−v)i = u + vi = w. By the above results, we see that every non-zero complex number w has two complex square roots z, z which are defined as in the question. However, when w = 0, then u = v = 0 which imply that a = b = 0 and thus z = z = 0. Hence, the complex number 0 has only one complex square root which is 0 itself, completing the proof of the problem.  Rudin Chapter 1 Exercise 11. Problem 1.11 Proof. Let |z| = r > 0. We define z = rw, where w = z r . Then it is easily to see that this expression satisfies the required conditions. Assume r1w1 = r2w2, where r1 > 0, r2 > 0 and |w1| = |w2| = 1. Then we have r1 r2 = w1 w2 which leads to r1 r2 = | w1 w2 | = |w1| |w2| = 1. Hence we have r1 = r2 and then w1 = w2. This completes the proof of the problem.  Rudin Chapter 1 Exercise 12. Problem 1.12 Proof. This result follows from induction and Theorem 1.33(e).  Rudin Chapter 1 Exercise 13. Problem 1.13 7 1.5. Properties of Euclidean spaces Proof. Since |x| = |x − y + y| ≤ |x − y| + |y| by Theorem 1.33(e), we have |x| − |y| ≤ |x − y|. Similarly, since |y| = |y − x + x| ≤ |y − x| + |x| by Theorem 1.33(e), we have −|x−y| ≤ |x| −|y|. Hence these two results together imply that the desired result, completing the proof of the problem.  Rudin Chapter 1 Exercise 14. Problem 1.14 Proof. It follows from Definition 1.32 that |1 + z| 2 + |1 − z| 2 = (1 + z)(1 + z) + (1 − z)(1 − z) = 2(1 + |z| 2 ) holds. This completes the proof of the problem.  Rudin Chapter 1 Exercise 15. Problem 1.15 Proof. By the proof of the Schwarz inequality (Theorem 1.35), the equality holds if and only if each term in the sum Xn j=1 |Baj − Cbj | 2 is zero, i.e., |Baj − Cbj | 2 = 0 (Baj − Cbj )(Baj − Cbj ) = 0 B 2 |aj| 2 + C 2 |bj| 2 − 2BCRe (aj bj ) = 0. (1.3) By Theorem 1.33(d), we have Re (aj bj) ≤ |aj bj | and so the relation (1.3) implies that 0 ≥ B 2 |aj | 2 + C 2 |bj | 2 − 2BC|aj ||bj | = (B|aj | − C|bj |) 2 . Hence we have the equality holds if and only if B|aj | = C|bj | if and only if |aj | |bj | is a constant for j = 1, 2, . . . , n. This completes the proof of the problem.  1.5 Properties of Euclidean spaces Rudin Chapter 1 Exercise 16. Problem 1.16 Proof. Let m be the mid-point of x and y and u = z − m = (u1, u2, . . . , uk). Geometrically, these conditions say that the three vectors x, y and u form an isosceles triangle with sides r, r and d. In other words, u must satisfy the equations u · (x − y) = 0 and |u| 2 = r 2 − d 2 4 . (1.4) Chapter 1. The Real and Complex Number Systems 8 (a) Since |x − y| = d > 0, we have from Theorem 1.37(b) that x and y are distinct and then we may assume without loss of generality that x1 6= y1. If u2, u3, . . . , uk are arbitrary, then we define u1 = −1 x1 − y1 [u2(x2 − y2) + · · · + uk(xk − yk)]. Then it is clear that this vector u = (u1, u2, . . . , uk) satisfies the equation u · (x − y) = 0. Since u2, u3, . . . , uk are arbitrary, there are infinitely many u1 and then u. Next, suppose that u is a vector satisfying the equation u·(x−y) = 0. Since 2r > d, we must have r 2 − d 2 4 > 0. If |u| 2 6= r 2 − d 2 4 , then we consider the vector u defined by u = q r 2 − d2 4 |u| u. Thus it is easy to check that u satisfies both equations (1.4). In fact, this proves that there are infinitely many u satisfying both equations (1.4) and hence there are infinitely many z such that |z − x| = |z − y| = r. (b) If 2r = d, then we have r 2 − d 2 4 = 0 and so |u| 2 = 0. It follows from Theorem 1.37(b) that u = 0, i.e., z = m. This prove the uniqueness of such z. (c) If 2r < d, then we have r 2 − d 2 4 < 0 and it is clear that there is no u such that |u| 2 < 0. Hence there is no such z in this case. When k = 1 or 2, the results will be different from those when k ≥ 3 and the analysis is given as follows: • Case (i): k = 2. If 2r > d, then u = q r 2 − d2 4 |u| u are the only vectors satisfying the equations (1.4), where u = ±  − x2 − y2 x1 − y1 u2, u2  . If 2r = d, then u = 0 is the only vector such that u · (x − y) = 0 and |u| 2 = 0. If 2r < d, then there is no such u. Hence, there are two such z if 2r > d, exactly one such z if 2r = d and no such z if 2r < d. • Case (ii): k = 1. Then there is no point u satisfying both u(x1 − y1) = 0 and |u| 2 = r 2 − d 2 4 if 2r 6= d. However, u = 0 is the only point such that u(x1 − y1) = 0 and |u| 2 = 0 if 2r = d. Hence, there is no such z if 2r 6= d and only one such z if 2r = d. This completes the proof of the problem.  Rudin Chapter 1 Exercise 17. Problem 1.17 9 1.6. A supplement to the proof of Theorem 1.19 Proof. Let x = (x1, x2, . . . , xk) and y = (y1, y2, . . . , yk). Then we have |x + y| 2 + |x − y| 2 = X k j=1 [(xj + yj ) 2 + (xj − yj ) 2 ] = 2X k j=1 x 2 j + 2X k j=1 y 2 j = 2|x| 2 + 2|y| 2 . Suppose that x and y are sides of a parallelogram. Then x + y and x − y are the diagonals of the parallelogram and the above result can be interpreted as follows: the sum of the squares of the lengths of the diagonals (left-hand side) is double to the sum of the squares of the lengths of the sides (right-hand side). This completes the proof of the problem.  Rudin Chapter 1 Exercise 18. Problem 1.18 Proof. We define x = (x1, x2, . . . , xk), where xj ∈ R for j = 1, 2, . . . , k. If x1 = x2 = · · · = xk = 0, then the element y = (1, 0, . . . , 0) satisfies the requirements that y 6= 0 and x · y = 0. Otherwise, without loss of generality, we may assume that x1 6= 0. If we define y =  − x2 + x3 + · · · + xk x1 , 1, . . . , 1  , then we still have y 6= 0 and x · y = 0. Let x be a non-zero real number. If y 6= 0, then Proposition 1.16(b) implies that x · y 6= 0. Hence this is not true if k = 1, finishing the proof of the problem.  Rudin Chapter 1 Exercise 19. Problem 1.19 Proof. Let 3c = 4b − a and 3r = 2|b − a|. Since |x| 2 = x · x (Definition 1.36), we have the following relations: |x − a| = 2|x − b| ⇔ |x| 2 − 2a · x + |a| 2 = 4|x| 2 − 8b · x + 4|b| 2 ⇔ 3|x| 2 − 8b · x + 2a · x + 4|b| 2 − |a| 2 = 0 ⇔ 9|x| 2 − 24b · x + 6a · x + 12|b| 2 − 3|a| 2 = 0 ⇔ (3x − 4b + a) · (3x − 4b + a) = 4(b − a)(b − a) ⇔  x − 4 3 b + 1 3 a  ·  x − 4 3 b + 1 3 a  = 4 9 (b − a)(b − a) ⇔     x − 4 3 b + 1 3 a     = 2 3 |b − a| ⇔ |x − c| = r, completing the proof of the problem.  1.6 A supplement to the proof of Theorem 1.19 Rudin Chapter 1 Exercise 20. Problem 1.20 Proof. Let us recall the first two properties now: (I) α 6= ∅ and α 6= Q. Chapter 1. The Real and Complex Number Systems 10 (II) If p ∈ α and q ∈ Q such that q < p, then q ∈ α. Suppose that “(III) If p ∈ α, then p < r for some r ∈ α” is deleted in the definition of a cut, see p. 17. Then it is easy to see that Step 2 and Step 3 are still satisfied. For Step 4, we still have the definition of addition of cuts: if α, β ∈ R, then α + β = {r + s | r ∈ α, s ∈ β}. We define 0 ∗ = {p ∈ Q | p ≤ 0}. It is clear that 0∗ satisfies (I) and (II), but it has the maximum element 0. It is also obvious that the addition so defined satisfies axioms (A1) to (A3). Let α ∈ R. If r ∈ α and s ∈ 0 ∗ , then either r + s < r or r + s = r which imply that r + s ∈ α, i.e., α + 0∗ ⊆ α. For any p ∈ α, we always have p + 0 = p so that p ∈ α + 0∗ , i.e., α ⊆ α + 0∗ . Hence we have α = α + 0∗ and the addition satisfies (A4). Let α = {p ∈ Q | p < 0} ∈ R. Assume that there was β ∈ R such that α + β = 0∗ . Since α 6= ϕ, we have p ∈ α and then β contains an element q ∈ Q such that p + q = 0. Since p is a negative rational, q must be a positive rational. By Theorem 1.20(b), we have 0 < q1 < q for some q1 ∈ Q. By (II), we have q1 ∈ β. However, we have p + q1 > 0 so that p + q1 ∈/ 0 ∗ by definition. This contradicts the assumption and hence we have the fact that the addition does not satisfy (A5). This completes the proof of the problem.  CHAPTER 2 Basic Topology 2.1 The empty set and properties of algebraic numbers Rudin Chapter 2 Exercise 1. Problem 2.1 Proof. Assume that there was a set A such that ∅ is not a subset of it. Then ∅ contains an element which is not in A. However, ∅ has no element by definition. Hence no such element exists and ∅ must be a subset of every set. This completes the proof of the problem.  Rudin Chapter 2 Exercise 2. Problem 2.2 Proof. We don’t use the hint to prove the result. For each positive integer n, we let Pn be the set of all polynomials of degree less than or equal to n with integer coefficients. Then it is easy to check that the mapping f : Pn → Z n+1 defined by f(a0z n + a1z n−1 + · · · + an−1z + an) = (a0, a1, . . . , an) is bijective. By Example 2.5, it is clear that Z is countable. Thus it follows from Theorem 2.13 that Z n+1 is countable and thus Pn is also countable. For each p(z) ∈ Pn, we let Bp(z) be the set of all roots of p(z). Since a polynomial p(z) ∈ Pn has at most n (distinct) roots, Bp(z) is a finite set. By the corollary to Theorem 2.12, the set S = [ p(z)∈Pn Bp(z) is at most countable. Since every positive integer is an algebraic number (consider the polynomial z −n), the set of all algebraic numbers A is infinite. Since A is a subset of S, we have A is countable, completing the proof of the problem.