4130 HOMEWORK 4. , a 2

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1 4130 HOMEWORK 4 Due Tuesday March 2 (1) Let N N denote the set of all sequences of natural numbers. That is, N N = {(a 1, a 2, a 3,...) : a i N}. Show that N N = P(N). We use the Schröder-Bernstein Theorem. First, there is an injection from P(N) to N N, because we may regard a subset of N as a sequence of zeroes and ones, or equivalently as a sequence of 1 s and 2 s, and this gives the desired injection. The hard part is showing that there is an injection N N P(N). To see this, note that a sequence of natural numbers is the same thing as a function N N. But by definition, a function is a special kind of subset of N N. In this way, we get an injection N N P(N N). But it was shown in class that N N and N have the same cardinality, and hence so do their power sets. In this way, we get the desired injection N N P(N). Working through the above proof, we can explicitly write down an injection if we like. An example is: (a 1, a 2,...) {2 3 a 1, a 2,...} N. (2) Let {x n } be a Cauchy sequence of rational numbers. Regarding {x n } as a sequence of real numbers, show that {x n } converges to the real number x defined as the equivalence class of the sequence {x n }. The hardest part of this is working out what to prove in the first place. Let ε > 0 be a real number. Choose A N with 3/2A < ε. Since {x n } is a Cauchy sequence of rational numbers, there exists N N such that if m, n > N then x n x m < 1/A. In other words, if m, n > N then 1/A < x m x n < 1/A. Now fix some m > N and consider the sequence of rational numbers {y n } where y n = x m x n 1

2 for n 1. We have chosen N so that if n > N then 1/A < y n < 1/A. This implies that for n > N, we have and y n + 3/2A > 1/2A 3/2A y n > 1/2A. The first inequality is the statement that the real number defined as the equivalence class of the Cauchy sequence {y n + 3/2A} is positive. (Ex: why is this sequence Cauchy?) If we let y = [{y n }], then we have an inequality in the real numbers Similarly, the other inequality gives y + 3/2A > 0. 3/2A y > 0. Putting these together, we have, for n > N, 3/2A < y < 3/2A. But y = [{x m x n } n 1 ] = [{x m, x m,...}] [{x 1, x 2,...}] which is the real number x m x, by definition of how the rationals are embedded in the reals. Thus, we have and hence 3/2A < x m x < 3/2A x m x < 3/2A < ε. This holds for any given m > N. So we have shown that for all ε > 0 we can find N N such that if m > N then x m x < ε, as required. (3) Section # 4. Let x R be the equivalence class of the Cauchy sequence {x n } where x i Q. We construct a sequence q n of rational numbers which is increasing and converges to x. To start with, take q 1 Q with x > q 1 > x 1. Now suppose we have constructed q 1,..., q n 1 with q n 1 < x. Let q n be a rational number with x > q n > max{q n 1, x 1/n}. Then the sequence {q n } is increasing and q n < x for all n. Also, 2

3 q n > x 1/n, so x q n = x q n < 1/n, and thus {q n } x. In particular, {q n } is Cauchy. We now have an increasing Cauchy sequence of rational numbers {q n }, and it remains to show that this sequence is equivalent to {x n }. For this, let ε > 0. Choose N 1 N such that if n > N 1 then x q n < ε/2. By the previous problem, we can choose N 2 N such that if n > N 2 then x x n < ε/2. Now if n > max{n 1, N 2 } then x n q n x n x + x q n < ε, which shows that the sequences {x n } and {q n } are equivalent sequences of rationals. (4) Show that every subset S of R which is bounded below has a greatest lower bound. (Hint: see p. 75 of the textbook.) Let S be a subset of R which is bounded below. Let S = { x : x S}. Then S is bounded above, so it has a least upper bound s := sup( S). We claim that s is the greatest lower bound of S. First, s is a lower bound because if x S then x s and so x s. Also, if b is any other lower bound for S, then b is an upper bound for S, and so b s, whence b s and so s is the greatest lower bound, as required. (5) Find, if they exist, the supremum (least upper bound) and infimum (greatest lower bound) of the following subsets of R. (a) {1, 2, 3}. Sup = 3, inf = 1. (b) (0, 1) {2} [3, 4) = {x R : 0 < x < 1 or x = 2 or 3 x < 4}. Sup = 4, inf = 0. (c) {1 1 n : n N}. (d) Q. The set is {0, 1/2, 2/3,...}. The supremum is 1 and the infimum is the smallest element, which is 0. This set is not bounded above or below, and so it does not have a finite sup or inf. (6) Prove Theorem in the textbook. 3

4 Let {x n } and {y n } be sequences of real numbers. (a) Suppose {x k } x and {y k } y. We must show that {x k + y k } x + y. Let ε > 0. Then there exists N 1 N such that if k > N 1 then x k x < ε/2. There also exists N 2 such that if k > N 2 then y k y < ε/2. Now suppose k > max{n 1, N 2 }. Then (x k + y k ) (x + y) x k x + y k y < ε. Next, we need to show that {x k y k } xy. convergent sequence of real numbers is bounded. We have x k x k x + x. To do this, we first show that a Taking ε = 1, we know that there exists N such that if k > N then x k x < 1. Therefore, if k > N then x k < 1 + x. It follows that for all k, x k B where B = max{ x 1, x 2,..., x N, 1 + x }. Thus, every convergent sequence of real numbers is bounded. Now we show that {x k y k } xy. Let ε > 0 and choose B such that x k B for all k and y B. Choose N 1 such that if k > N 1 then x k x < ε/2b. Choose N 2 such that if n > N 2 then y k y < ε/2b. Then if N > max{n 1, N 2 } then x k y k xy = x k (y k y) + y(x k x) x k y k y + y x k x < ε. Next, suppose y 0. Then for all ε > 0 there exists N N such that if n > N then y n y < ε. So y n (y ε, y + ε). If y > 0 then y ε > 0 for some ε > 0. If y < 0 then y + ε < 0 for some ε > 0. In either case, we have y n 0 for sufficiently large n, as desired. We show that {1/y k } 1/y. As stated in the book, we may as well assume that y k 0 for all k; otherwise we can neglect the terms with y k = 0. By convergence of {y n }, there exists N N such that if n > N then Then if n > N, we have y n y < y /2. y n y y y n > y y /2 = y /2. 4

5 Therefore, 1 1 y n y = y n y 2 y n y y y 2 n y, for n > N. Let ε > 0. Choose N 2 such that if n > N 2 then y n y < ε y 2 /2. Then for n > max{n, N 2 }, we get as required. 1 y n 1 y < ε Now the statement that {x k /y k } x/y follows from writing x k /y k = x k (1/y k ). (b) Suppose there is m N such that x k y k for k > m. Suppose for a contradiction that x < y. Then let 0 < ε < (y x)/2. There exists N 1 N such that if n > N 1 then x n x < ε and there exists N 2 N such that if n > N 2 then y n y < ε. Therefore, if n > max{n 1, N 2, m} then x n < x + ε < x + (y x)/2 = (x + y)/2 = y (y x)/2 < y ε < y n which contradicts x n y n. 5