Due date: Monday, February 6, 2017.

Modern Analysis Homework 3 Solutions Due date: Monday, February 6, 2017. 1. If A R define A = {x R : x A}. Let A be a nonempty set of real numbers, assume A is bounded above. Prove that A is bounded below and inf( A) = sup A. Solution. Notice that x A if and only if x A. Since A is non empty and bounded above, sup A exists. Let σ = sup A. If x A, then x A, hence x σ, thus x = ( x) σ, proving σ is a lower bound of A. In particular, A is bounded below and inf( A) σ. Let τ = inf( A), so σ τ. Just as we proved that σ was a lower bound of A, we can now prove that τ is an upper bound of A, hence σ τ, implying σ τ. It follows that σ = τ. 2. Let A, B be two sets of real numbers; assume both are nonempty, assume R = A B, and a < b for all a A, b B. Prove: A is bounded above B is bounded below and sup A = inf B. Solution. Let s concentrate on one of the sets, say B. Since A is not empty and every element of A is less than every element of B, it follows that B is bounded below (by all the elements of A). As a non-empty set that is bounded below, inf B exists. Because every element of A is a lower bound of B we have also x inf B for all x A, thus inf B is an upper bound of A. It follows that sup A infb. Assume sup A < infb. There is then x, sup A < x < inf B; for example x = (sup A + inf B)/2. But x > sup A implies x / A; x < inf B implies x / B, contradicting that R = A B. This proves that we must have inf B = sup A. bounded above 3. Let D be a non=empty set and let f : D R. We say f is bounded below bounded bounded above if and only if the range f(d) = {f(x) : x D} is bounded below. bounded One defines (if they exist) sup f(x) = sup f(d), x D inf f(x) = inf f(d). x D Prove one of the two following statements (they are both essentially equivalent) and answer the posed question: Let f, g : D R be bounded above. Prove f + g is bounded above and sup(f(x) + g(x)) sup f(x) + sup g(x). (1) x D x D x D

2 Can one replace in (1) by =? Solution. For every x D we have f(x) sup x D f(x), g(x) sup x D g(x), hence f(x) + g(x) sup x D f(x) + sup x D g(x). It follows that sup x D f(x)+sup x D g(x) is an upper bound for the range of f + g, hence sup x D (f(x) + g(x)) sup x D f(x) + sup x D g(x). Equality does not necessarily hold. For example if D = {0, 1}, f(0) = 0, f(1) = 1, g(0) = 1, g(1) = 0. Then f(x) + g(x) = 1 for all x D, thus sup x D f(x) + g(x) = 1 < 2 = 1 + 1 = sup x D f(x) + sup x D g(x). Let f, g : D R be bounded below. Prove f + g is bounded below and inf (f(x) + g(x)) inf f(x) + sup g(x). (2) x D x D x D Can one replace in (2) by =? Solution. For every x D we have f(x) inf x D f(x), g(x) inf x D g(x), hence f(x)+g(x) inf x D f(x)+inf x D g(x). It follows that inf x D f(x)+inf x D g(x) is a lower bound for the range of f +g, hence inf x D (f(x) + g(x)) inf x D f(x) + inf x D g(x). Equality does not necessarily hold. For example if D = {0, 1}, f(0) = 0, f(1) = 1, g(0) = 1, g(1) = 0. Then f(x) + g(x) = 1 for all x D, thus inf x D f(x) + g(x) = 1 > 0 = 0 + 0 = inf x D f(x) + inf x D g(x). 4. Let G be a subset of R; assume 0 G and assume that whenever x, y G, then x y G. Let G + = {x G : x > 0}. Prove: If G + is not empty and inf G + = 0, then G is dense in R in the sense that if a, b R and a < b, there exists g G such that a < g < b. Hint: prove that G is closed under addition and that if x G, then mx G for all m Z. Then imitate the proof that the rational numbers are dense in R. Incidentally,examples of sets G satisfying the hypotheses are; R, Q, G = {m + nα : m, n Z} where α is an irrational number, and many more. Solution. Incidentally, in my notes on some rules on how to write proofs, I wrote out most of the proof of this exercise. We show first that G is closed under addition. In fact, since 0 G, if x G we also have x G. If x, y G, then x, y G, hence x+y = y ( x) G. We can now prove by induction that if x G, then nx G for all n N. This is clearly true if n = 1, so assume it holds for some n N. Then nx G, x G implies (n + 1)x = nx + x G. By induction nx G

3 for all n N. If now m Z and m < 0, let n = m. so n G. Then mx = nx = 0 nx G. Finally, 0x = 0 G. Assume now G + and inf G + = 0. Let a, b R, a < b. Then b a > 0 = inf G + ; thus b a cannot be a lower bound of G +, hence there is x G + such that x < b a. Notice that by virtue of being in G +, x > 0. Claim: There is m Z such that mx a < (m + 1)x. There are many ways of establishing this claim. Here is one, not necessarily the shortest one. The Archimedean property states that N is not bounded above; an immediate consequence is that Z is not bounded below. There is thus some integer M such that M < a/x. There is also (again by the Archimedean property) an integer N such that a/x < N. The set of integers {k : M k N} is a finite set (of N M + 1 integers), thus so is its subset C = {k Z : M k a/x}. This set is not empty (M is in it) so being finite it contains its supremum; that is, there is an integer m such that m a/x and k m if M k a/x. Now m + 1 / C since it is larger than sup C. But m + 1 > M; so m + 1 / C implies m + 1 > a/x. We thus have mx a < (m = 1)x as claimed. Since x G, m Z, we have that g = (m + 1)x G. Also a < g by construction. Finally g = (m + 1)x = mx + x a + x < a + (b a) = b. 5. Let α R and let G + = {x : x = m + nα for some m, n Z and x > 0}. Prove: inf G + = 0 if and only if α is irrational. I won t grade this exercise. Notice that this set G satisfies the conditions of exercise 4. Let d = inf G +. Assume first d > 0. Claim: Then G = {md : m Z}. In fact, first of all we notice that in this case d G +. In fact, suppose d G +. Now 3d/2 > d, so there exists x G +, 3d/2 > x d; since d / G, it actually is 3d/2 > x > d. Since x > d, there is y G, x > y > d. Now 0 < x y < 3d 2 d d = 2 ; contradicting the definition of d since x y G + thus cannot be less than the inf of G +. With d G, we have already G {md : m Z}. For the converse inclusion let x G. Since d > 0 proceeding as in the proof of Exercise 4, there is m Z such that md x < (m + 1)d. Now 0 x md < d. If x md, then x md is an element of G + that is less than d; not possible. It follows that G = {md : m Z} as claimed. Since 1 = 1 + 0α G, there exists m Z such that 1 = md so d = 1/m. Since α = 0 + 1 α G, there is n such that α = nd = n/m Q. This concludes the proof that if inf G + > 0, then α Q.

4 Conversely, assume α = a/b is in Q; a, b Z; we may assume that gcd(a, b) = 1 and b > 0. If n Z, we can invoke the division algorithm to write na = bq + r, where b, q Z, 0 r < b. Thus m + nα = m + na b = m + q + r b. Notice that r b < 1, so the only way that m + nα > 0 is if m + q 0. From this it becomes obvious that the infimum of G is precisely 1/b > 0. 6. Find a real number M such that Solution. sup{ x 3 21x 2 + 86x + 264 : 2 x 10} M. x 3 21x 2 + 86x + 264 x 3 + 21 x 2 + 86 x + 264. It should be clear that if 2 x 10, then 0 x 10, so the largest value of the right hand side of the displayed inequality will be reached for x = 10. Thus x 3 21x 2 +86x+264 x 3 +21 x 2 +86 x +264 10 3 +21 10 2 +86 10+264 = 4224. Thus M = 4224 works. 7. The online textbook defines convergence of a sequence {x n } of real numbers to a number x R as follows: Prove the following statements are equivalent for a sequence {x n } of real numbers and x R. (a) The sequence {x n } converges to x in the sense of Definition 2.1.2. (b) For every ɛ > 0 there is M R such that x n x < ɛ for all n M. (c) For every ɛ > 0 there is M N such that x n x ɛ for all n > M. (d) For every r > 0 the set {n N : x n / (x r, x + r)} is finite. Proof. (a) (b) Assume (a). Let ɛ > 0 be given. By (a), there is M N such that x n x < ɛ if n M. Since M N implies M R, it should be clear that (b) will hold.

5 (b) (c) Assume (b). Let ɛ > 0 be given. By (b), there is M R such that x n x < ɛ if n M. Let N = max{1, M }. Then N N and N > M. If n N then n M, hence x n x < ɛ; in particular, x n x ɛ. This proves (c). (c) (d) Assume (c). let r > 0 be given. Let ɛ = r/2; then ɛ > 0 and by (c) there exists M N such that x n x ɛ < r if n M. Thus x n / (x r, x + r) implies n < M; there are at most M 1 terms of the sequence not in (x r, x + r). (d) follows. (d) (a) Assume (d). Let ɛ > 0 be given. Let r = ɛ so that by (d) the interval (x ɛ, x + ɛ)contains all but a finite number of terms of the sequence. Since there is only a finite number of terms of the sequence not in this interval, there is a largest index M for which x M (x ɛ, x + ɛ); implying that x n (x ɛ, x + ɛ); i.e., x n x < ɛ for n M + 1. 8. Let {a n }, {b n } be two sequences of real numbers, both converging to the same limit L. Form a new sequence {c n } by { ak, if n = 2k 1 is odd, c n = b k, if n = 2k is even. Prove {c n } converges to L. Solution. In the proof provided below, if you extract what is written in red, it is the definition of limit. The red sentences (or equivalent ones) will be present in any proof based on the definition of limit. Of course, they need to be connected by valid and legal statements. The things I wrote in blue are comments; they are not really part of the proof and if I were a student (instead of the instructor) and this a homework to be handed in, I would remove them. Let ɛ > 0 be given. Because {a n } converges to L, there is M 1 N such that a n L < ɛ if n M 1. Similarly, because {b n } converges to L, there is M 2 N such that b n L < ɛ if n M 2. Now here is a little subtlety, or thing to think about. Let s say M 1 = M 2 = 10. It could be; or it could be something else, but let s suppose it is. That means that after ten terms into the two sequences we are at ɛ from L. But now look at the composite sequence, that goes a 1, b 1, a 2, b 2, a 3, b 3, a 4, b 4, a 5, b 5, a 6, b 6,.... After 10 terms we are only 5 terms deep into either original sequence. We might still be at more than ɛ from L. This should explain the factor of 2 in the definition of M. Set M = 2 max{m 1, M 2 }. Assume n M, so n 2M 1 and n 2M 2. If n = 2k 1 is odd, then n 2M 1 implies 2k 1 2M 1, thus 2k 2M 1 +1;

6 in particular k M 1. Then c n L = a k L < ɛ. If n = 2k is even, then n 2M 2 implies k M 2 and hence c n L = b k L < ɛ. Since a natural number is either even or od, we see that in every case c n L < ɛ. We are done.