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Honors Introduction to Analysis I Homework IV Solution February 23, 2009 Problem 1 A rational function is a ratio of two polynomials with real coefficients, R(x) = P (x)/Q(x), Q 6= 0. Equality between rational functions and the operations of addition and multiplication are defined similarly to the case of the usual fractions (i.e. the rational numbers). Prove that the set of rational functions with these operations is a field. You may assume standard facts about fractions and polynomials. Solution. Just check all the axioms, and take into account that the 0 polynomial is the polynomial with all the coefficients 0. Problem 2 Using only the axioms of an ordered field, prove that the field of complex numbers, with the usual operations, cannot be ordered (you need to show that it is impossible to select a subset P of “positive” complex numbers consistently with the properties listed in Theorem 2.2.2). Solution. The idea of the proof is that in an ordered field, x2 is positive for any non-zero element x. Indeed, according to the axioms of an ordered field, either x or −x is positive. Since x2 = (−x)2 , it is always a product of two positive elements, hence itself positive. Now apply this to the complex numbers x = i (imaginary unit) and x = 1. We obtain that i2 = −1 and 12 = 1 are both positive. This is a contradiction: for example, their sum is 0, but the sum of two positive elements must be positive. Hence the field of complex numbers cannot be ordered. Problem 3 Compute the sup, inf, limsup, liminf, and all the limit points of the following sequences: • xn = 1/n + (−1)n • xn = 1 + (−1)n /n • xn = (−1)n + 1/n + 2 sin(nπ/2) Solution. • sup = 3/2, inf = −1, lim inf = −1, lim sup = 1, E = {−1, 1} We consider the subsequences x2n = 1 + 1/2n (even terms - monotonically decreasing) and x2n−1 = −1 + 1/(2n − 1) (odd terms - monotonically increasing). sup xn = sup x2n = 3/2, inf xn = inf x2n−1 = −1. Now lim sup xn = lim x2n = 1 and lim inf xn = lim x2n−1 = −1. Since each term converges to one of these two values, then E = {−1, 1} and these are the only limit points of the sequence. • sup = 3/2, inf = 0, lim inf = 1, lim sup = 1, E = {1} (the sequence converges to 1) We consider the subsequences x2n = 1 + 1/2n (even terms - monotonically decreasing) and x2n−1 = 1 − 1/(2n − 1) (odd terms - monotonically decreasing). sup xn = sup x2n = 3/2, inf xn = inf x2n−1 = x1 = 0. Now lim sup xn = lim x2n = 1 and lim inf xn = lim x2n−1 = 1. Hence one can see that the sequence is convergent (equivalently, since it is monotonically decreasing and bounded, it is convergent), so E = {, 1} and this is the only limit point of the sequence. 1 • sup = 3/2, inf = −1, lim inf = −1, lim sup = 1, E = {−1, 1} We consider the subsequences x4n = 1 + 1/4n, x4n−1 = −3 + 1/(4n − 1), x4n−2 = 1 + 1/(4n − 2) and x4n−3 = 1 + 1/(4n − 3). {x4n−1 } is monotonically increasing, and the others are decreasing. inf xn = inf x4n−1 = −3, sup xn = x1 = 2, lim sup xn = lim x4n = lim x4n−2 = lim x4n−3 = 1 and lim inf xn = lim x4n−1 = −3. Since each term converges to one of these two values, then E = {−3, 1} and these are the only limit points of the sequence. Problem 4 If a bounded sequence is the sum of a monotone increasing and a monotone decreasing sequence, does it follow that the sequence converges? What if the two sequencese are bounded? ( −n, if n = 2k − 1 Solution. No. Consider xn = n and yn = (so it’s −1, −1, −3, −3, ...). One can see that −(n − 1), if n = 2k ( 0, if n = 2k − 1 zn = xn + yn = , so it is bounded, but as it alternates, it doesn’t converge. 1, if n = 2k If {xn } and {yn } are bounded, since they are monotonous, they will converge. Hence their sum converges (limit of sums is sum of limits). Problem 5 If E is a set and y a point that is the limit of two sequences, {xn } and {yn } such that xn is in E and yn is an upper bound for E, prove that y = sup E. Is the converse true? Solution. Upper bound Assume there exist x ∈ E such that x > y. As lim yn = y, for n such that x − y > 1/n, there is m such that |yk − y| ≤ 1/n for k ≥ m. So there is an yn0 such that n0 ≥ m and y < yn0 < x. But every yn is un upper bound for E, so yn0 ≥ x (contradiction). Hence y is un upper bound of E. Least upper bound Assume there is z another upper bound, such that z < y. Then xn ≤ z, for every n. Taking the limit of xn , we get that y ≤ z (contradiction). Hence y sup E. The converse is not true. There are sets E and sequences {xn } and {yn }, such that y = sup E is a limit of both, but {xn } is not in E and yn is not an upper bound for E. Problem 6 Prove that lim sup{xn +yn } ≤ lim sup{xn }+lim sup{yn } if both limsups are finite, and give an example where equality doesn’t hold. Solution. Since xj ≤ supj>k xj and yj ≤ supj>k yj , we have that xj + yj ≤ supj>k xj + supj>k yj , for any j ≥ k. Then if we take the sup over all these j’s, we get: supj>k (xj + yj ) ≤ supj>k (supj>k xj + supj>k yj ) = supj>k (supj>k xj ) + supj>k (supj>k yj ) = supj>k xj + supj>k yj (since the sup is taken on the same set). Now we have an inequality of convergent sequences, namely supj>k (xj + yj ), supj>k xj and supj>k yj , so we can take the limit k → ∞ and get: lim sup(xj + yj ) ≤ lim (sup xj + sup yj ) = lim sup xj + lim sup yj k→∞ j>k k→∞ j>k j>k k→∞ j>k k→∞ j>k Hence lim sup(xn + yn ) ≤ lim sup xn + lim sup yn To see that inequality can hold strictly, take xn = (−1)n and yn = (−1)n+1 . Then lim sup(xn + yn ) = 0 < 2 = lim sup xn + lim sup yn Problem 7 Construct a sequence whose set of limit points is exactly the set of integers. Solution. We can take 0, −1, 0, 1, −2, −1, 0, 1, 2, −3, −2, −1, 0, 1, 2, 3, ..... This sequence contains every integer infinitely often so it has has every integer as a limit point (or equivalently, one can take subsequences consisting of constant terms, so the subsequence consisting only of k, which converge to k). Since any two integers differ by distance at least one, any limit point must also be an integer. 2 Problem 8 Can there exist a sequence whose set of limit points is exactly 1, 1/2, 1/3, ... Solution. No, because 0 has to be a limit point also, but is not in the set. To see this, take a subsequence xlk converging to 1/n0 , for some n0 ≥ 1. Then for any n, there is an m, such that |xlk − 1/n0 | ≤ 1/n, for any k ≥ m. But then given any t, we can find an n such that 1/n + 1/n0 ≤ 1/t. Then for that n, there is an m such that |xlk | = |xlk − 1/n0 + 1/n0 | ≤ |xlk − 1/n0 | + |1/n0 | ≤ 1/n + 1/n0 ≤ 1/t, for k ≥ m. Hence the subsequence converges to 0, but we can do this for every n0 . So we can find arbitrarily many subsequences converging to 0, which means that 0 has to be a limit point. But it is not given in the set. Hence there isn’t such a sequence. 3