Iowa State University. Instructor: Alex Roitershtein Summer Homework #1. Solutions

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1 Math 501 Iowa State University Introduction to Real Analysis Department of Mathematics Instructor: Alex Roitershtein Summer 015 EXERCISES FROM CHAPTER 1 Homework #1 Solutions The following version of the definition of the complement set is convenient: for each x X and A X we have either x A or x A c but not both Usually, to verify identity of two sets A = B we will first show that A B and then that B A (a) On one hand, if x A, then x A c x (A c ) c Thus A (A c ) c On the other hand, if x (A c ) c then x A c, x A (b) If x (A B) c then x A B either x A c or X B c Thus x A c B c On the other hand, if x A c B c then either x A c or X B c, which means either x A or X B In any case, x A B x (A B) c Using the first De Morgan s with A c in place of A and B c in place of B yields (A c B c ) c = (A c ) c (B c ) c = A B Taking complement on the both sides, we obtain A c B c = (A B) c, as desired (c) Here is a very nice picture I found in the Internet: Figure 1: Illustration to the first de Morgan s law 1

2 (d) The proof in (a) can be carried over nearly verbatim to prove the following Let I be any, possibly uncountable, indexing set Then ( ) c ( ) i I A i = i I A c c i and i I A i = i I A c i 1 (a) Since S is bounded and non-empty, sup S exists and is a real number Thus, if b is an upper bound for S and S [b ε, b] =, then b ε is also an upper bound for S and hence sup S b ε The contradiction show that S [b ε, b] (b) No For instance, consider S = {1, 3}, b = 3 and ε = 1 (c) By the definition of the cut, any element in B is an upper bound for A Thus, by the definition of the supremum, sup A y for any y B It follows that sup A x (1) On the other hand, clearly z sup A for any z A Since sup A A by the definition of the cut, it follows that sup A x Thus, by virtue of (1), sup A = x 15 As the hint in the textbook suggests, let A = {s > 0 : s n x} and y = sup A Clearly, A is bounded from above (for instance, max{1, x} is an upper bound for A) and therefore y (0, + ) Suppose that y n > x Then, by the result in Exercise 14 there exists u < y such that Since u < y, in fact y n u n < yn x 0 < y n u n < yn x u n = y n (y n u n ) y n yn x = yn + x Thus u is an upper bound bound for A, which is impossible since u < y = sup A This contradiction shows that in fact y n x Now suppose that y n < x Then, by the result in Exercise 14 there exists u > y such that y n u n < x yn > x

3 Since u > y, in fact 0 < u n y n < x yn u n = y n + (u n y n ) y n + x yn = yn + x Thus u A, which is impossible since u > y = sup A This contradiction shows that in fact y n = x 8(a) First, assume that S = {(x, y) R : f(x) y} is a convex set Choose any x 1, x R and let y 1 = f(x 1 ), y = f(x ) Then, (x 1, y 1 ) S and (x, y ) S, for any λ [0, 1] we have ( λx1 + (1 λ)x, λy 1 + (1 λ)y ) S, () < x which implies f(λx 1 + (1 λ)x ) λy 1 + (1 λ)y = λf(x 1 ) + (1 λ)f(x ) Thus f is a convex function Conversely, if f is convex then for any x 1, x R, f(λx 1 + (1 λ)x ) λf(x 1 ) + (1 λ)f(x ) Therefore, if y 1 f(x 1 ) and y f(x ) then f(λx 1 + (1 λ)x ) λf(x 1 ) + (1 λ)f(x ) λy 1 + (1 λ)y, () holds Thus S is a convex set 36(a) Let P n denote the set of all polynomials of degree n N with integer coefficients and E n denote the set of all roots of these polynomials The function f : P n Z n+1 mapping a polynomial to the vector of its coefficients ( n ) f a k t k = (a 0, a 1,, a n ) Z n+1 k=0 is a bijection Hence P n Z n+1 N Since each polynomial in P n has at most n roots, E n P n N Thus n N E n N by Corollary 18 in Chapter 1 of the textbook 4 Let A n {a n, a n+1, } and s n = sup A n, so that lim sup a n = lim s n 3

4 (a) Clearly, A n+1 A n s n is not increasing Furthermore, since A n A for any n N we have It follows that < inf A x sup A < n N, x A n < inf A s n sup A <, n N Thus s n is a bounded non-increasing sequence converges (b) If sup A = +, then for any M > 0 there is an index n N such that a n M This implies that there are actually infinitely many such indexes n Thus s n M for all n N This implies lim sup a n = lim s n M, and, since M > 0 is arbitrary, lim sup a n = + (c) If lim a n =, then for any M > 0 there is N N such that a n M for all n > N This implies s n M for all n > N It follows that lim sup a n = lim s n M, and, since M > 0 is arbitrary, lim sup a n = (d) (i) Let B = {b 1, b, } and, for n N, B = {b n, b n+1, } Since sup (a k + b k ) sup a k + sup b k, taking the limit in both sides of this inequality, we obtain lim sup (a n + b n ) lim sup a n + lim sup b n as long as the right-hand side is well-defined, namely unless one of the lim sup s on the right-hand side equals plus infinity while the other one is minus infinity In other words, the inequality holds true unless one of the sequences a n and b n is unbounded from above while the other one is converging to (ii) If c 0, then lim sup(ca n ) = lim sup(ca k ) = lim c sup a k = c lim = c lim sup a n Furthermore, since sup( A) = inf A, sup a k lim sup(ca n ) = lim sup( c a k ) = lim c sup( a k ) = c lim = c lim inf a k 4 inf a k

5 Thus the identity lim sup (ca k ) = c lim sup a k doen t hold for c < 0 unless lim sup a k = lim inf a k (3) Remark It can be verified that (3) is equivalent to the existence of lim a n To this end, recall the notation s n = sup a k and let t n = inf a k Suppose that a = lim s n = lim t n If a = + then, by definition of the limit applied to t n, for all M > 0 there exists N N such that n N implies t n > M a n t n > M Thus lim = + Similarly, if a = then, by definition of the limit applied to s n, for all M > 0 there exists N N such that n N implies s n < M a n s n < M Thus lim = Finally, if a R then by definition of the limit applied separately to both s n and t n, for all ε > 0 there exists N N such that n N implies Thus lim = a (e) Define lim inf a n = lim inf a k a ε t n s n < a + ε, a ε t n a n s n < a + ε Remark This definition of lim inf is natural since it is mimicking the definition of the lim sup It is all also useful for at least the following two reasons Firstly, the result we proved in the class identifies lim inf a n as the infimum of the set of all possible limits of converging subsequences of a n (we called the latter sequential liminf) Secondly, combining this theorem with the result in the above remark one obtains that lim a n = a [, + ] if and only if (3) holds and, furthermore, both sides of the identity in (3) are equal to a 44(a) By the definition of the norm, x + y + x y = ( x x y + y ) + ( x + x y + y ) = x + y 5