Mathematics 242 – Principles of Analysis Problem Set 1 Solutions September 9, 2005

1.2.1. a) As in the proof of Theorem 1.1.1, we argue by contradiction. Suppose \sqrt{3} is rational, so \sqrt{3} = \frac{m}{n}, where the fraction is in lowest terms. This says m^{2} = 3n^{2}. So 3 divides m^{2}, which implies 3 divides m: m = 3k for some integer k. Hence 9k^{2} = 3n^{2}, so 3k^{2} = n^{2}. Hence 3 divides n by the same reasoning. This is a contradiction because we assumed that \frac{m}{n} was in lowest terms, so m and n have no common (integer) factors. The same reasoning does work to show that \sqrt{6} is irrational since if 6 divides n^{2} for some integer, then 6 must divide n.

b) If 4 divides n^{2}, then it is not necessarily true that 4 divides n. For instance 4 divides 6^{2} = 36, but 4 does not divide 6.

1.2.2. a) False: Consider the sets A_{n} = [0, 1/n). Then each A_{n} contains infinitely many real numbers, but ∩_{n=1}^{∞}A_{n} = {0} is not infinite.

b) True: The intuitive reason is: Since each A_{n}, including A_{1}, is finite, there can be only a finite number of strict inclusions in the chain. In other words, after a finite number of steps, the chain “stabilizes,” and we have all the A_{n}’s the same for n ≥ N: A_{N} = A_{N+1} = · · ·. The intersection is the set A_{N}.

c) False: Let A = {1, 3}, B = {1}, C = {2}. Then A∩(B∪C) = {1}, but (A∩B)∪C = {1, 2}.

d) True: This can be seen with a Venn diagram, for instance.

e) True: This can be seen with a Venn diagram, for instance.

1.2.3c. Let x ∈ (A ∪ B)^{c}, then x is not an element of either B or C (since the elements of A ∪ B are the x that are elements of A, or of B, or of both sets). Saying this another way, x is not an element of A and not an element of B. Hence x ∈ A^{c} ∩ B^{c}. This shows (A ∪ B)^{c} ⊆ A^{c} ∩ B^{c}. Now we show the other inclusion. Let x ∈ A^{c} ∩ B^{c}. This says x is not an element of A and also not an element of B. Hence x is not an element of A ∪ B (see above). This shows x ∈ (A ∪ B)^{c}, so A^{c} ∩ B^{c} ⊆ (A ∪ B)^{c}. Since both inclusions are true, the sets are equal.

1.2.6. a) f(A) = [0, 4] and f(B) = [1, 16]. f(A) ∩ f(B) = [1, 4] = f(A ∩ B) since A ∩ B = [1, 2]. f(A) ∪ f(B) = [0, 16] = f(A ∪ B).

b) The key observation here is that f(x) = f(−x), so f(A) and f(B) can have equal elements coming from unequal (opposite sign) x’s. An “extreme” example: Let A = [1, 2] and B = [−2, −1]. Then f(A) = f(B) = [1, 4], so f(A) ∩ f(B) = [1, 4], but A ∩ B = ∅, so f(A ∩ B) = ∅.

c) If y ∈ g(A∩B), then y = g(x) for some x ∈ A∩B. This x is in both A and B. This means y = g(x) for some x ∈ A and also y = g(x) for some x ∈ B. Hence y ∈ g(A) ∩ g(B) and g(A ∩ B) ⊆ g(A) ∩ g(B).

d) The general statement we can make here is that g(A ∪ B) = g(A) ∪ g(B). Proof: Let y ∈ g(A ∪ B). Then y = g(x) for some x ∈ A ∪ B. This means x ∈ A or x ∈ B. Hence y = g(x) for some x ∈ A or some x ∈ B, which implies y ∈ g(A) ∪ g(B). Hence g(A∪B) ⊆ g(A)∪g(B). For the other inclusion, let y ∈ g(A)∪g(B). Then either y = g(x) for some x ∈ A or some x ∈ B (or both – but note that it might be g(x) = g(x′) for some x ≠ x′ with x ∈ A and x′ ∈ B). Either way, y = g(x) for some x ∈ A ∪ B. So the other inclusion holds as well.

1.2.7. a) f^{−1}(A) = [−2, 2] and f^{−1}(B) = [−1, 1] (note that only the subinterval [0, 1] ⊂ [−1, 1] is “hit” by f though). f^{−1}(A ∪ B) = f^{−1}([−1, 4]) = [−2, 2] = f^{−1}(A) ∪ f^{−1}(B). f^{−1}(A ∩ B) = f^{−1}([0, 1]) = [−1, 1] = f^{−1}(A) ∩ f^{−1}(B).

b) To show f^{−1}(A∪B) = f^{−1}(A)∪f^{−1}(B), first let x ∈ f^{−1}(A∪B), then f(x) ∈ A∪B, so f(x) ∈ A or f(x) ∈ B. This shows x ∈ f^{−1}(A) or x ∈ f^{−1}(B), so x ∈ f^{−1}(A) ∪ f^{−1}(B), showing that f^{−1}(A ∪ B) ⊆ f^{−1}(A) ∪ f^{−1}(B). For the opposite inclusion, we can just reverse the reasoning here.

To show f^{−1}(A∩B) = f^{−1}(A)∩f^{−1}(B), first let x ∈ f^{−1}(A∩B), then f(x) ∈ A∩B, so f(x) ∈ A and f(x) ∈ B. This shows x ∈ f^{−1}(A) and x ∈ f^{−1}(B), so x ∈ f^{−1}(A)∩f^{−1}(B), showing that f^{−1}(A∩B) ⊆ f^{−1}(A)∩f^{−1}(B). For the opposite inclusion, we can just reverse the reasoning here.

1.2.10. a) The base case is y_{1} < 4, which is true since y_{1} = 1. Now assume that y_{k} < 4 and consider y_{k+1} = (3y_{k} + 4)/4. Since y_{k} < 4, y_{k+1} < (3 · 4 + 4)/4 = 16/4 = 4. Hence y_{n} < 4 for all n.

b) We must show y_{n} ≤ y_{n+1} for all n. The base case is the statement y_{1} ≤ y_{2}, which is true since y_{1} = 1, and y_{2} = (3 + 4)/4 = 7/4. Now assume that y_{k} ≤ y_{k+1}. We will show y_{k+1} ≤ y_{k+2}. From the inequality y_{k} ≤ y_{k+1}, it follows that (3y_{k} + 4)/4 ≤ (3y_{k+1} + 4)/4, since adding the same quantity to both sides and multiplying by positive numbers preserve the inequality. But now we see (3y_{k} + 4)/4 = y_{k+1} and (3y_{k+1} + 4)/4 = y_{k+2}. So we are done by induction.

1.2.11. We argue by induction on n. The base case here is n = 0 (that is, A = ∅). There is exactly 1 subset of A, namely ∅. So A has 2^{0} = 1 subsets. Now assume that every set with k elements has exactly 2^{k} different subsets. Consider A with k + 1 elements, say A = {x_{1}, …, x_{k+1}}. Every subset of A either contains the “last element” x_{k+1}, or not. The ones that don’t are the same as the subsets of A′ = {x_{1}, …, x_{k}}. So by induction, there are 2^{k} of these subsets. Next consider the subsets of A that do contain x_{k+1}. If we remove x_{k+1} from one of these subsets, what is left is a subset of A′. Putting x_{k+1} “back in” never gives the same subset of A twice if the subsets of A′ were different. So by induction again there are also 2^{k} subsets of A containing x_{k+1}. Adding we find 2^{k} + 2^{k} = 2 · 2^{k} = 2^{k+1} different subsets of A.

1.2.8. a) For some real numbers satisfying a < b and for all n ∈ N, a + 1/n ≥ b.

b) Most natural form of negation: There exist two real numbers such that there is no rational number between them.

c) For some natural number n ∈ N, \sqrt{n} is not a natural number and \sqrt{n} is rational.

d) For some real number x ∈ R, x ≥ n for all n ∈ N. (Or: There is some real number greater than every natural number.)

1.3.2. a) A real number t is the greatest lower bound of a set A, written t = inf(A) if

(i) t is a lower bound for A (that is, t ≤ x for all x ∈ A), and (ii) If c is any lower bound for A, then c ≤ t.

b) Statement: Let t be a lower bound for A ⊆ R. Then t = inf(A) if and only if for every ε > 0, there exists a ∈ A with t ≤ a < t+ε. (Note: in intuitive terms, this is another way of saying that no number greater than t is a lower bound for A.)

Proof: ⇒: Let t = inf(A). The contrapositive form of (ii) in the definition of part a says: if c > t, then c is not a lower bound for A. In other words, for any ε > 0, taking c = t + ε > t, there exists some a ∈ A with t ≤ a < t + ε.

⇐: Assume that t is a lower bound for A and that for every ε > 0, there exists a ∈ A with t ≤ a < t + ε. This says that no number greater than t is a lower bound for A. So if c is any lower bound for A, then c ≤ t. Hence by part a, t = inf(A).

1.3.3. a) Assuming A is nonempty (not stated explicitly!), the set B is bounded above. Hence the Axiom of Completeness implies that s = sup(B) exists in R. To show s = inf(A), we use the definition of inf(A) given in 1.3.2a above.

First we will show s ≤ a for all a ∈ A. We can argue by contradiction. If not, there would be some a ∈ A with a < s. But then by Lemma 1.3.7, there would be some b ∈ B with a < b ≤ s. This is impossible since b ∈ B is supposed to be a lower bound for A. Therefore, s ≤ a for all a ∈ A. Hence (i) of the definition of inf(A) must hold.

Next, let c be any lower bound of A. By definition this says c ∈ B. Hence by the definition of sup(B), c ≤ s. Hence (ii) of the definition of inf(A) holds for s as well.

b) We do not need to assert that greatest lower bounds exist separately because the proof in part a here shows that their existence follows from the existence of least upper bounds.

c) An alternate way to use the Axiom of Completeness to prove that sets bounded below have greatest lower bounds is to note that taking negatives reverses inequalities, turns upper bounds to lower bounds, etc. In fact if A is bounded below, then −A = {−x : x ∈ A} is bounded above, and sup(−A) = − inf(A).

1.3.4. Since B ⊂ A, any upper bound for A is also an upper bound for B. By definition, sup(A) is an upper bound for A and also for B. Therefore, sup(A) ≥ sup(B), since if t is any upper bound for B, then g ≥ sup(B).

1.3.6. a) A = {n ∈ N : n^{2} < 10} = {1, 2, 3}. inf(A) = 1 and sup(B) = 3.

b) B = {n/(m + n) : m, n ∈ N}. Note that 0 < n/(m + n) < 1 since m, n ≥ 1 are natural numbers). inf(B) = 0 (think of fixing n and letting m → ∞). sup(B) = 1 (think of fixing m = 1 and letting n → ∞.

c) C = {n/(2n + 1) : n ∈ N}. Note that n/(2n + 1) increases with n. So inf(C) = 1/3 (n = 1) and sup(C) = 1/2 (think of what happens as n → ∞).

d) D = {n/m : m, n ∈ N, m + n ≤ 10}. This is a finite set: inf(D) = 1/9, sup(D) = 9/1 = 9.

1.3.7. The first condition in the definition of sup(A) is true by hypothesis. So we only need to show that if b is any upper bound for A, then b ≥ a. But this follows immediately because a ∈ A.

1.3.8. Assume sup(A) < sup(B). Then by Lemma 1.3.7, there exists some b ∈ B such that sup(A) < b ≤ sup(B). This b is an upper bound for A since b > sup(A) ≥ a for all a ∈ A.

1.3.9. a) True (finite means the set has finitely many elements; the situation is like that in 1.3.6 above).

b) False: Take B from 1.3.6 b, C from 1.3.6 c, or any similar example.

c) False: An example: A = {1 − 1/n : n ∈ N} and B = {1 + 1/n : n ∈ N}. Note sup(A) = 1 = inf(B).

d) True. The intuitive idea is that s + t is an upper bound for the set of sums. Moreover, A contains elements arbitrarily close to s = sup(A) and B contains elements arbitrarily close to t = sup(B). Hence the set of sums A + B contains elements arbitrarily close to s + t.

e) False: Take A = {1} and B = {1 − 1/n : n ∈ N}. sup(A) = sup(B) = 1, but no element of B is an upper bound for A.

1.4.4. The first statement of the Archimedean Property is that if x is any real number then there exists a natural number n with n > x. So let ε > 0 be any positive number (as close to zero as we like). Applying the Archimedean Property, there exists some n > 1/ε, so 1/n < ε. This shows that no positive number is an lower bound for the set {1/n : n ∈ N}. But the set is obviously bounded below (0 is a lower bound since 1/n > 0 for all n), so the inf exists by the Axiom of Completeness. It follows that 0 = inf({1/n : n ∈ N}).

1.4.5. If there were some x ∈ ∩_{n=1}^{∞}(0, 1/n), then 0 < x < 1/n for all n ∈ N. This would say the positive number x would be a lower bound for the set A from the previous problem 1.4.4. But by that problem, 0 = inf({1/n : n ∈ N}). So there can be no such x.

2.2.1. a) Let ε > 0 and take N > 1/\sqrt{6ε} in N. If n ≥ N, then

|\frac{1}{6n^{2} + 1} − 0| = \frac{1}{6n^{2} + 1}

≤ \frac{1}{6n^{2}}

≤ \frac{1}{6N^{2}}

< ε

This shows that lim \frac{1}{6n^{2}+1} = 0.

b) Let ε > 0 and take N > 13/(4ε) in N. If n ≥ N, then

|\frac{3n + 1}{2n + 5} − \frac{3}{2}| = \frac{13}{4n + 10}

≤ \frac{13}{4n}

≤ \frac{13}{4N}

< ε

This shows that lim \frac{3n+1}{2n+5} = \frac{3}{2}.

c) Let ε > 0 and take N > 4/ε^{2} in N. If n ≥ N, then

|\frac{2}{\sqrt{n + 3}} − 0| = \frac{2}{\sqrt{n + 3}}

≤ \frac{2}{\sqrt{n}}

≤ \frac{2}{\sqrt{N}}

< ε This shows that lim \frac{2}{\sqrt{n+3}} = 0.

2.2.2. An example: Let x_{n} = (−1)^{n} and x = 0. (x_{n}) “verconges” to x since for ε = 2 (for instance), for all N, n ≥ N implies |x_{n} − x| = 1 < 2. That inequality is true for all n ∈ N, in fact.

This definition says that all the terms in the sequence (x_{n}) lie in some interval (x − ε, x + ε). In other words, saying (x_{n}) “verconges” to x just says that all the terms in the sequence lie in a bounded interval containing x.

2.2.3. a) Must show that at some US college, no student is at least seven feet tall (equivalently, all students are less than seven feet tall).

b) Must show that at some US college, every professor gives grades other than A or B to some students.

c) Must show that at all US colleges, some students are less than six feet tall.

2.2.4. Let ε < 1. Then for all N ∈ N, there exist n ≥ N such that |x_{n} − 0| = 1 > ε. Hence the sequence does not converge to 0. The rest of the question is asking about the set of ε for which it would be true that there exists some N ∈ N such that |x_{n} − 0| < ε for all n ≥ N. As long as ε ≥ 1, there is no problem finding such an N. You can take N = 1, in fact! But as above, for ε < 1, there is no such N.

2.2.6. a) larger. Reason: If N “works” for ε in the definition of convergence, and N′ > N, then n ≥ N′ implies n ≥ N also, so |x_{n} − a| < ε for all n ≥ N′. b) larger. If |x_{n} − a| < ε for all n ≥ N, and ε′ > ε, then we have |x_{n} − a| < ε′ too.

2.2.7. a) Definition: lim x_{n} = +∞ if for every positive number B (no matter how big), there exists an N such that whenever n ≥ N, x_{n} > B.

Consider x_{n} = \sqrt{n}. Given B > 0, let N > B^{2} in N. Then for all n ≥ N, \sqrt{n} ≥ \sqrt{N} > B. Hence lim \sqrt{n} = +∞.

b) The given sequence does not converge to +∞, because if B > 0, there is no N such that x_{n} > B for all n ≥ N (the zero terms always fail this inequality).

2.2.8. a) frequently, not eventually.

b) eventually implies frequently, so eventually is stronger.

c) We can say: (x_{n}) converges to x if for every ε > 0, (x_{n}) is eventually in the interval (x − ε, x + ε).

d) We can say the sequence is frequently in the interval, but not necessarily eventually in the interval. For instance, consider x_{n} = \{ \begin{matrix}2 & if n even \\ 3 & if n odd \end{matrix}. This sequence is not eventually in the interval (1.9, 2.1) because of the infinitely many terms equal to 3, but it is frequently in that interval.

Mathematics 242 – Principles of Analysis Problem Set 4 Solutions October 17, 2005

2.3.8. a) Let x_{n} = (−1)^{n} and y_{n} = (−1)^{n+1}. Then (x_{n}) and (y_{n}) diverge. But x_{n} + y_{n} = 0 for all n, so the sum series converges.

b) Impossible by Theorem 2.3.3 (ii) and (i): Note that y_{n} = −x_{n} + (x_{n} + y_{n}). If (x_{n}) and (x_{n} + y_{n}) converge, so does (−x_{n}), and hence so does y_{n}.

c) Example: b_{n} = 1/n.

d) Impossible by Theorem 2.3.2. If (b_{n}) is convergent, then (b_{n}) is bounded: |b_{n}| ≤ B for all n. If (a_{n} − b_{n}) is also bounded, |a_{n} − b_{n}| ≤ A, then

|a_{n}| = |a_{n} − b_{n} + b_{n}| ≤ |a_{n} − b_{n}| + |b_{n}| ≤ A + B

for all n. So |a_{n}| would also have to be bounded.

e) Example: a_{n} = 1/n, b_{n} = n. Note lim a_{n} = 0, lim a_{n}b_{n} = 1, but (b_{n}) is not convergent.

2.4.2. a) First note that if x < 4, then 1/(4 − x) > 0. We have x_{1} = 3 and x_{2} = 1. Moreover if x_{k} > x_{k+1}, then 4 − x_{k} < 4 − x_{k+1}, and 1/(4 − x_{k}) > 1/(4 − x_{k+1}). This shows x_{k+1} > x_{k+2}. By induction, therefore, (x_{n}) is a decreasing sequence bounded below by zero. By Theorem 2.4.2 (the Monotone Convergence Theorem), (x_{n}) is convergent.

b) This should be intuitively clear, since (x_{n+1}) is just the same sequence, “shifted by one” in the index. Here is a formal proof: If (x_{n}) converges to a, then for all ε > 0, there exist N ∈ N such that |x_{n} − a| < ε for all n ≥ N. If n ≥ N, then n + 1 > n ≥ N too, so |x_{n+1} − a| < ε whenever n ≥ N. Hence x_{n+1} also converges to a.

c) Say L = lim x_{n} = lim x_{n+1}. Then we have L = 1/(4 − L), so L^{2} − 4L + 1 = 0. The smaller positive root is L = \frac{4−\sqrt{12}}{2} = 2 − \sqrt{3}. (This is the limit of the sequence since the other root 2 + \sqrt{3} > 3. Since x_{1} = 3 and the sequence is decreasing 2 + \sqrt{3} cannot be the limit.)

2.4.5. a) By the method of Lagrange Multipliers from MATH 241, the function (x + y)/2 is minimized at the points on the constraint curve xy = 2 with x, y > 0 where x = y = \sqrt{2} and the minimum value is (\sqrt{2} + 2/\sqrt{2})/2 = \sqrt{2}. It follows that if x > \sqrt{2}, then (x + 2/x)/2 > \sqrt{2}. Now we argue by induction. The base case is n = 1 and x_{1} = 2 > \sqrt{2}. By the above, if x_{k} > \sqrt{2}, then x_{k+1} = (x_{k} + 2/x_{k})/2 > \sqrt{2} also. Therefore x_{n} > \sqrt{2} for all n, so x_{n}^{2} > 2.

Then

x_{n} − x_{n+1} = \frac{1}{2} (x_{n} − \frac{2}{x_{n}} ) = \frac{1}{2} \frac{x_{n}^{2} − 2}{x_{n}} > 0

since x_{n}^{2} > 2 for all n. It follows that (x_{n}) is a monotone decreasing sequence. By the Monotone Convergence Theorem (2.4.2), it converges to some L in R. As in 2.4.2 and 2.4.4 above, L satisfies

L = \frac{1}{2} (L + \frac{2}{L}) ⇒ 2L^{2} = L^{2} + 2,

so L^{2} = 2. It follows that L = +\sqrt{2}, since all the terms of the sequence are positive.

b) The sequence defined by x_{n+1} = \frac{1}{2} (x_{n} + \frac{c}{x_{n}} ) converges to \sqrt{c} by the same kind of argument given for part a.

2.5.3. a) An example: x_{n} = \{ \begin{matrix}1/n & if n is odd \\ 1 + 1/n & if n is even. \end{matrix}

b) Impossible: Let x_{n_{k}} be the convergent subsequence, say lim_{k→∞} x_{n_{k}} = L. Let ε > 0. Then there exists a K such that whenever k ≥ K, |x_{n_{k}} − L| < ε. Since the whole sequence is monotone this implies that for all n ≥ N = n_{K}, |x_{n} − L| < ε also. Thus, (x_{n}) converges to L also.

c) One way: Let (x_{m}) the sequence produced as follows:

1, 1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4, 1, 1/2, 1/3, 1/4, 1/5, 1, …

(repeating the same pattern of starting over from 1 and adding one additional 1/n on each iteration). Then (x_{m}) has subsequences converging to 1/n for each n ∈ N because each 1/n appears infinitely many times in the x_{m} sequence.

Alternate Method: Take a sequence consisting of all rational numbers in the interval (0, 1), listed in some order. Then by the Density Theorem for Q, there are subsequences converging to every real number in (0, 1), including the ones given(!)

d) An example: x_{n} = \{ \begin{matrix}n & if n is even \\ 1/n & if n is odd, \end{matrix} (x_{n}) is unbounded because of the odd-

numbered terms, but the even-numbered subsequence converges to zero.

e) Impossible: The bounded subsequence is itself a bounded sequence, so the Bolzano-Weierstrass Theorem says it has a convergent subsequence. That subsequence is also a subsequence of the original sequence.

2.5.4. Aiming for a contradiction, suppose that (a_{n}) is a bounded sequence such that every convergent subsequence converges to a, but the whole sequence does not converge to a. Then by the definition, there exists ε > 0 such that for all N ∈ N, |a_{n} − a| ≥ ε for some n ≥ N. For the rest of this solution we will pick a particular ε > 0 as above. For N = 1, let n_{1} ∈ N be such that a_{n_{1}} satisfies |a_{n_{1}} − a| ≥ ε. Then with N = n_{1} + 1, let n_{2} ≥ n_{1} + 1 satisfy |a_{n_{1}} − a| ≥ ε. In the same way, we produce a strictly increasing sequence n_{k} of natural numbers, and a subsequence a_{n_{k}} such that |a_{n_{k}} − a| ≥ ε for all k ≥ 1. The (a_{n_{k}}) is a subsequence of the original (a_{n}) sequence. Hence it is also bounded. Then the Bolzano-Weierstrass theorem implies that (a_{n_{k}}) has a convergent subsequence. This is also a subsequence of the original sequence, so by hypothesis it must be converging to a. But that is a contradiction because we have |a_{n_{k}} − a| ≥ ε as above for all k.

2.5.6. Since (a_{n}) is a bounded sequence, the set S is bounded above. So sup(S) exists in R. By Lemma 1.3.7, given any ε > 0, there is some x ∈ S with s − ε < x ≤ s. By the definition of S, x ∈ S means there are infinitely many n such that s − ε < x < a_{n} ≤ s. So we argue as follows: For each k ∈ N in turn, taking ε = \frac{1}{k} in the above, we let n_{k} be the smallest integer > n_{k−1} with s − \frac{1}{k} < a_{n_{k}}. This gives a subsequence a_{n_{k}} converging to s, since we have s − \frac{1}{k} < a_{n_{k}} ≤ s for all k. We have lim_{k→∞} s − \frac{1}{k} = s, so by the Limit Squeeze Theorem, lim_{k→∞} a_{n_{k}} = s.

2.6.1. a) We just need to exhibit a non-monotone convergent sequence. An example: x_{n} = 1 + \frac{(−1)^{n}}{n}.

b) An example x_{n} = n.

c) Impossible: Theorem 2.6.4 (the Cauchy Criterion) says a Cauchy sequence coverges. By Theorem 2.5.2, every subsequence of a convergent sequence converges to the same limit as the whole sequence.

d) An example x_{n} = \{ \begin{matrix}n & if n is odd \\ 0 & if n is even \end{matrix}. The whole sequence is unbounded because of the odd-numbered terms, but the even-numbered subsequence is Cauchy (constant, in fact!)

2.6.3. a) The difference is the condition |s_{n+1} − s_{n}| < ε rather than |s_{m} − s_{n}| < ε. This condition only deals with pairs of consecutive terms in the sequence, not with arbitrary pairs “far enough out”.

b) The sequence s_{n} = ∑_{k=1}^{n} \frac{1}{n} (partial sums of the harmonic series) is pseudo-Cauchy, since s_{n+1} − s_{n} = \frac{1}{n+1}. Given any ε > 0, we can make \frac{1}{n+1} < ε by taking n ≥ N for suitably large N. However, it is not convergent, because the harmonic series diverges (see Example 2.4.5 and the class notes).

I. A) We know ∑_{k=1}^{∞} \frac{1}{\sqrt{k}} diverges by an example we did in class. Hence ∑_{k=1}^{∞} \frac{1}{\sqrt{3k}} diverges too, since the terms are just a constant times the terms in the first series. For all k ≥ 1, \sqrt{2k + 1} ≤ \sqrt{2k + k} = \sqrt{3k}. Hence \frac{1}{\sqrt{2k+1}} ≥ \frac{1}{\sqrt{3k}}. By the Comparison Test, ∑_{k=1}^{∞} \frac{1}{\sqrt{2k+1}} diverges, since the terms are greater than the terms in a divergent series of positive terms.

B) We know ∑_{k=1}^{∞} \frac{1}{k^{2}} converges by an example we did in class. Hence ∑_{k=1}^{∞} \frac{1}{4k^{2}} converges too, since the terms are just a constant times the terms in the first series. For all k ≥ 1, (2k + 1)^{2} > 4k^{2}. Hence \frac{1}{(2k+1)^{2}} ≤ \frac{1}{4k^{2}} . By the Comparison Test, ∑_{k=1}^{∞} \frac{1}{(2k+1)^{2}} converges, since the terms are less than the terms in a convergent series of positive terms.

C) This is a geometric series with ratio −2/9. Since | − 2/9| < 1, it converges.

D) Diverges, since the terms do not go to zero:

lim_{n→∞} \frac{n^{3} + n}{2n^{3} + n + 1} = \frac{1}{2}.

(Recall, if ∑ a_{n} converges, then lim_{n→∞} a_{n} = 0 by 2.7.3 in text.)

II. A) This is a geometric series with ratio sin(x). Since | sin(x)| ≤ 1 for all real x, it converges for all x such that sin(x) ≠ −1, 1. For x ∈ [0, 2π], this is all x except x = π/2, 3π/2. The sum is \frac{sin x}{1−sin x}.

B) This is also a geometric series with ratio e^{x}. It Converges for all x such that 0 < e^{x} < 1, so for x < 0. The sum is \frac{e^{x}}{1−e^{x}}.

2.7.3. a) (By contraposition) If ∑ p_{n} and ∑ q_{n} both converge, then ∑ p_{n} + q_{n} also converges by Theorem 2.7.1 (the Algebraic Limit Theorem for series). But p_{n} + q_{n} = a_{n} by definition, so ∑ a_{n} converges too.

b) ∑ a_{n} conditionally convergent means ∑ a_{n} converges but ∑ |a_{n}| diverges. Now by definition, a_{n} = p_{n} + q_{n} and |a_{n}| = p_{n} − q_{n}. We will use the Algebraic Limit Theorem for series to derive a contradiction if either ∑ p_{n} or ∑ q_{n} is convergent. Suppose ∑ p_{n} converges. Then from ∑ a_{n} convergent, we get ∑ q_{n} = ∑(a_{n} − p_{n}) is also convergent. Hence ∑ |a_{n}| = ∑(p_{n} − q_{n}) is convergent too. But that contradicts the hypothesis, so ∑ p_{n} cannot be convergent. The same argument shows ∑ q_{n} cannot be convergent either.

2.7.4. Take x_{n} = y_{n} = 1/n. Then ∑ x_{n} and ∑ y_{n} are harmonic series (divergent). But ∑ x_{n}y_{n} is the series ∑ \frac{1}{n^{2}} , which we know converges.

2.7.9. a) Given any r < 1, there exist r′ with r < r′ < 1, since for instance we can take r′ = (1 + r)/2. The hypothesis lim |\frac{a_{n+1}}{a_{n}} | = r < 1 says that if we take ε = r′ − r > 0 in the definition of convergence of a sequence, then there exists some N such that

||\frac{a_{n+1}}{a_{n}} | − r| < r′ − r

whenever n ≥ N. But this last inequality implies

|\frac{a_{n+1}}{a_{n}} | < r′

so |a_{n+1}| < |a_{n}|r′ for all n ≥ N.

For future reference, we note that starting from n = N, it follows by induction that |a_{N+k}| ≤ |a_{N}|(r′)^{k} for all k ≥ 0. The base case is k = 0 which is clear. If the statement |a_{N+ℓ}| ≤ |a_{N}|(r′)^{ℓ} is true for some ℓ, then

|a_{N+ℓ+1}| ≤ |a_{N+ℓ}|r′ ≤ |a_{N}|(r′)^{ℓ} · r′ = |a_{N}|(r′)^{ℓ+1}.

Hence |a_{N+k}| ≤ |a_{N}|(r′)^{k} for all k ≥ 0.

b) Because ∑_{k=0}^{∞}(r′)^{k} is a geometric series with ratio 0 < r′ < 1, this series converges.

c) The series ∑ |a_{n}| converges for the following reason: The finitely many terms with n < N are irrelevant for convergence. Once n ≥ N, say n = N + k with k ≥ 0, we have from the last part of part a: |a_{n}| ≤ |a_{N}|(r′)^{k}. Hence the “tail” of the series:

∑_{n=N}^{∞} |a_{n}| = ∑_{k=0}^{∞} |a_{N+k}|

converges by comparison with the convergent geometric series |a_{N}| ∑_{k=0}^{∞}(r′)^{k} from part b.

III. Take a_{n} = (x − 1)^{n}/3^{n} in the Ratio Test. We have lim |\frac{a_{n+1}}{a_{n}} | = \frac{1}{3}|x − 1|. The Ratio Test says the series converges absolutely for all x with |x − 1| < 3 or −2 < x < 4. At x = −2, the series is ∑(−1)^{n} which diverges (terms do not go to zero). The series also diverges at x = 4. The interval of convergence is (−2, 4) in this example.

3.2.1. a) The hypothesis is used when dealing with ε = min{ε_{i}}. Since we are taking the minimum of a finite set of positive numbers, we get one of the elements of the set, which is still strictly positive. If there were an infinite collection of open sets the intersection could be empty (for instance if inf({ε_{i}}) = 0), or the intersection could be non-open (see the next part).

b) An example: Let O_{n} = (0 − 1/n, 1 + 1/n), so O_{1} ⊃ O_{2} ⊃ O_{3} ⊃ · · ·. As n → ∞, lim −1/n = 0 and lim 1 + 1/n = 1 so the intersection ∩_{n=1}^{∞} is [0, 1].

3.2.2. This is similar to Example 3.2.9 (i)

a) Limit points {−1, 1}.

b) No, since for instance 1 ≠ B.

c) No, since for instance 4/5 ∈ B (take n = 4) but there is no open interval V_{ε}(4/5) contained in B. (The closest points to 4/5 in B are 2/3 and 6/7 and none of the points in (2/3, 4/5) or (4/5, 6/7) are in B.

d) Yes. By what we said in part c, 4/5 is an isolated point in B. All the other points in B are isolated as well.

e) \overline{B} = B ∪ \{−1, 1\}.

3.2.3. a) Neither. Let a be any rational number and ε > 0 be any positive real number. The set V_{ε}(a) = (a − ε, a + ε) contains both rational and irrational numbers. Every real number is a limit of a sequence of rational numbers so any irrational number is a limit point of Q not contained in Q.

b) Closed, not open. If ε > 0, then V_{ε}(1) never consists only of natural numbers.

c) Open, not closed. 0 is a limit point of {x ∈ R : x > 0} not contained in the set, since lim 1/n = 0.

d) Neither. There is no V_{ε}(1) contained in (0, 1], since any such set contains numbers > 1 also. 0 is a limit point of (0, 1] not contained in the set.

e) Neither. This set is the set of partial sums of the convergent p-series with p = 2. The sum of the series, π^{2}/6, is a limit point not contained in the set, so it is not closed. Every point of the set is isolated point, so the set is not open.

3.3.1. By the Heine-Borel Theorem (3.3.4), K is closed and bounded. Since K is bounded, inf(K) and sup(K) exist in R. Moreover, as we know, there are sequences of elements of K converging to inf(K) and to sup(K). Since K is closed, this implies inf(K) and sup(K) are elements of K.

3.3.5. a) Not compact – take any sequence of rational numbers converging to an irrational limit. b) Not compact – as in a, but take the irrational number in the interval [0, 1]. (Note this set is bounded, but not closed). c) Not compact – take any unbounded sequence like x_{n} = n. (This set is closed, but not bounded.) d) Compact (any finite set is compact). e) Not compact – 0 is a limit point not contained in the set, since lim 1/n = 0. f) Compact – lim \frac{n}{n+1} = 1, which is in the set.

Mathematics 242 – Principles of Analysis Solution for Discussion 2 October 31, 2005

C) 1) We use the following notation and facts. For sets A, B ⊂ R and c ∈ R, write A + B = {x + y : x ∈ A, y ∈ B} and similarly cA = {cx : x ∈ A}. If B = {b} is a single point, then we write A + b, which is just the translation (shift) of A by b along the number line. If A = [r, s] and B = [u, v] are closed intervals, then note A + B = [r + u, s + v] and cA = [cr, cs]. The ordinary distributive law in R implies that r(A + B) = rA + rB and (r + s)A = rA + sA.

We show that C_{n} + C_{n} = [0, 2] for all n ≥ 1 by induction. The base case is n = 1, where C_{1} = [0, 1/3] ∪ [2/3, 1]. Hence

C_{1} + C_{1} = ([0, 1/3] + [0, 1/3]) ∪ ([0, 1/3] + [2/3, 1]) ∪ ([2/3, 1] + [2/3, 1])
= [0, 2/3] ∪ [2/3, 4/3] ∪ [4/3, 2]
= [0, 2]

Now, assume as induction hypothesis that C_{k} + C_{k} = [0, 2], and consider C_{k+1}. The key observation we need here is that C_{k+1} is the union of two scaled and translated copies of C_{k}:

C_{k+1} = (1/3)C_{k} ∪ ((1/3)C_{k} + 2/3).

Hence, using the induction hypothesis:

C_{k+1} + C_{k+1} = ((1/3)C_{k} + (1/3)C_{k}) ∪ ((1/3)C_{k} + (1/3)C_{k} + 2/3)
∪ ((1/3)C_{k} + 2/3 + (1/3)C_{k} + 2/3)
= (1/3)[0, 2] ∪ ((1/3)[0, 2] + 2/3) ∪ ((1/3)[0, 2] + 4/3)
= [0, 2/3] ∪ [2/3, 4/3] ∪ [4/3, 2]
= [0, 2].

Hence C_{n} + C_{n} = [0, 2] for all n ≥ 1.

2) By part 1, we know that given s ∈ [0, 2], for each n there are x_{n}, y_{n} ∈ C_{n} such that s = x_{n} + y_{n}. We want to show that s = x + y for some x, y ∈ C = ∩_{n=1}^{∞}C_{n}. The sequences (x_{n}) and (y_{n}) live in the bounded set [0, 1]. Hence by the Bolzano-Weierstrass Theorem, there exists a convergent subsequence (x_{n_{k}}). By Bolzano-Weierstrass again, from the sequence (y_{n_{k}}) we can select a convergent subsequence (y_{n_{k_{ℓ}}}) and (x_{n_{k_{ℓ}}}) is a subsequence of a convergent sequence so it converges too. Let x = lim_{ℓ→∞} x_{n_{k_{ℓ}}} and y = lim_{ℓ→∞} y_{n_{k_{ℓ}}}. Since s = x_{n} + y_{n} for all n, we still have s = x + y by the Algebraic Limit Theorem.

Then to finish, we must show that x, y ∈ C = ∩_{n=1}^{∞}C_{n}. To do this in a “nice” way, we will use some of the ideas from class on Friday, 10/28. First for each ℓ, we have the set C_{n_{k_{ℓ}}}, which is a finite union of closed intervals, hence closed. We claim that x, y ∈ C_{n_{k_{ℓ}}}. This follows because of the fact that the C_{n} are nested: C_{1} ⊃ C_{2} ⊃ C_{3} ⊃ · · ·. For all ℓ′ > ℓ, x_{n_{k_{ℓ′}}} ∈ C_{n_{k_{ℓ}}} too. This means x, y are limit points of convergent sequences in the closed set C_{n_{k_{ℓ}}}. Hence x, y are in C_{n_{k_{ℓ}}}.

Next, we claim that this implies that x, y are in C_{n} for all n. The reason is that because n_{k_{ℓ}} is an increasing sequence in N, given any n there exists an ℓ such that n_{k_{ℓ}} > n and hence x, y ∈ C_{n_{k_{ℓ}}} ⊂ C_{n}). It follows that x, y ∈ C, since they are in C_{n} for all n.

Mathematics 242 – Principles of Analysis Problem Set 6 Solutions November 11, 2005

4.2.1. a) Given ε > 0, let δ = ε/2. Then for all x with 0 < |x − 2| < δ, we have

|(2x + 4) − 8| = 2|x − 2|
< 2(ε/2)
= ε.

This shows lim_{x→2} 2x + 4 = 8.

b) Given ε > 0, let δ = \sqrt[3]{ε}. Then for all x with 0 < |x| < δ, we have

|x^{3} − 0| = |x|^{3}
< (\sqrt[3]{ε})^{3}
= ε.

This shows lim_{x→0} x^{3} = 0.

c) Given ε > 0, let δ = min(1, ε/19). Then for all x with 0 < |x − 2| < δ, we have 1 < x < 3 so |x^{2} + 2x + 4| < 3^{2} + 2 · 3 + 4 = 19 (because x^{2} + 2x + 4 is increasing for all x > −1). Therefore

|x^{3} − 8| = |x − 2||x^{2} + 2x + 4|
< (ε/19) · 19
= ε.

This shows lim_{x→2} x^{3} = 8.

d) Given ε > 0, let δ = π − 3 ≐ .14 · · ·. Then for all x with 0 < |x − π| < δ, we have 3 < x < 2π − 3 ≐ 3.28 ⋯, so [[x]] = 3 for all such x and

|[[x]] − 3| = 0 < ε.

This shows lim_{x→π}[[x]] = 3.

4.2.3. a) Let x_{n} = 1/n. Then |x_{n}|/x_{n} = 1 for all n ≥ 1, and lim |x_{n}|/x_{n} = 1. On the other hand, let y_{n} = −1/n. Then |y_{n}|/y_{n} = −1 for all n ≥ 1, and lim |y_{n}|/y_{n} = −1. Since we have two sequences converging to the same limit lim x_{n} = 0 = lim y_{n}, but lim f(x_{n}) ≠ lim f(y_{n}), Corollary 4.2.5 implies that lim_{x→0} |x|/x does not exist.

b) Let x_{n} be any sequence of rational numbers converging to 1 (for instance x_{n} = 1 + 1/n). Then lim g(x_{n}) = 1, since g(x_{n}) = 1 for all n ≥ 1. On the other hand, let y_{n} be any sequence of irrational numbers converging to 1 (for instance y_{n} = 1 + \sqrt{2}/n. Then lim g(y_{n}) = 0, since g(y_{n}) = 0 for all n ≥ 1. Corollary 4.2.5 implies that lim_{x→1} g(x) does not exist.

4.3.1. a) Given ε > 0, let δ = ε^{3}. Then for all x with |x| < δ, we have

|\sqrt[3]{x} − 0| = \sqrt[3]{|x|}
< \sqrt[3]{ε^{3}}
= ε.

This shows lim_{x→0} \sqrt[3]{x} = 0 = \sqrt[3]{0}. So g(x) = \sqrt[3]{x} is continuous at x = 0.

b) Given ε > 0, let δ = min(|c|, c^{2/3}ε). Then for all x with |x − c| < δ, first since δ < |c|, c − |c| < x < c + |c|. So, it follows that x and c have the same sign (both + or both −). Then by the identity a^{3} − b^{3} = (a − b)(a^{2} + ab + b^{2}) and the fact that all terms in the denominator on the right of the first line are positive:

|\sqrt[3]{x} − \sqrt[3]{c}| = |\frac{(x − c)}{x^{2/3} + \sqrt[3]{c}\sqrt[3]{x} + c^{2/3}} |

≤ \frac{|x − c|}{c^{2/3}}

< \frac{c^{2/3}ε}{c^{2/3}}

= ε

This shows lim_{x→c} \sqrt[3]{x} = \sqrt[3]{c}. So g(x) = \sqrt[3]{x} is continuous at x = c.

4.3.2. a) Let ε > 0. Since g is continuous at f(c), there exists a δ > 0 such that if |y − f(c)| < δ, then |g(y) − g(f(c))| < ε. But then since f is continuous at c, for this δ, there exists a δ_{1} > 0 such that if |x−c| < δ_{1}, then |f(x)−f(c)| < δ. For any such x, taking y = f(x) in the first set of inequalities, |f(x) − f(c)| < δ implies |g(f(x)) − g(f(c))| < ε. This shows that g(f(x)) = (g ∘ f)(x) is continuous at c.

b) Let x_{n} be any sequence converging to c in the domain of g ∘ f. Then since f is continuous at c, part (iv) of Theorem 4.3.2 implies that f(x_{n}) is a sequence converging to f(c) in the range of f. Using the same part of the theorem, since g is continuous at f(c), g(f(x_{n})) is a sequence converging to g(f(c)). This shows that for any sequence x_{n} converging to c, (g ∘ f)(x_{n}) is a sequence converging to (g ∘ f)(c). By the equivalence of parts (i) and (iv) in the theorem, this implies that g ∘ f is continuous at c.

4.3.9. a) Let ε > 0 and take δ = ε/c. Then for any a ∈ R, for all x with |x − a| < δ,

|f(x) − f(a)| ≤ c|x − a| < c(ε/c) = ε.

Hence f is continuous at all a ∈ R. (Note: this would be true even if c > 1!)

b) (Note: for notational convenience, I’m starting the sequence from n = 0 rather than n = 1 as in the text. Also, instead of writing |y_{m} − y_{n}| as in the usual definition of Cauchy sequences, it will be easier to take m > n and write m = n + k.) Define the sequence (y_{n}) by picking y_{0} arbitrarily, then iterating: y_{n+1} = f(y_{n}) for all n ≥ 0. Let n be arbitrary and k ≥ 1. Then by an easy induction argument |y_{n+k} − y_{n}| =

|f(y_{n+k−1}) − f(y_{n−1})| ≤ c|y_{n+k−1} − y_{n−1}| ≤ c^{2}|y_{n+k−2} − y_{n−2}| ≤ ⋯ ≤ c^{n}|y_{k} − y_{0}| We might be tempted to say that lim c^{n} = 0, so this tends to zero, but we must still deal with the fact that |y_{k} − y_{0}| depends on k. We can deal with |y_{k} − y_{0}| as follows. By the triangle inequality, and similar reasoning to the above:

|y_{k} − y_{0}| ≤ |y_{k} − y_{k−1}| + |y_{k−1} − y_{k−2}| + ⋯ + |y_{1} − y_{0}|

≤ (c^{k−1} + c^{k−2} + ⋯ + 1)|y_{1} − y_{0}|

≤ \frac{1}{1 − c}|y_{1} − y_{0}|

(where in the last line we have used the sum formula for the convergent geometric series ∑ c^{n} and eliminated the dependence on k). Putting everything together we have

|y_{n+k} − y_{n}| ≤ \frac{c^{n}}{1 − c}|y_{1} − y_{0}|.

Now (since the factors |y_{1} − y_{0}| · \frac{1}{1−c} are fixed) we can argue that for any ε > 0, there exists some N such that n ≥ N (hence n + k ≥ N too) implies \frac{c^{n}}{1−c}|y_{1} − y_{0}| < ε, so the sequence y_{n} is Cauchy.

c) Let y be the limit of the sequence y_{n} from part b. Since f is continuous (part a), we have by part (iv) of Theorem 4.3.2

f(y) = f(lim y_{n}) = lim f(y_{n}) = lim y_{n+1} = y.

In other words, y is a fixed point of f.

For the last part of this question, we must show that there is only one fixed point of f. So suppose there are two: y, y′ with f(y) = y and f(y′) = y′. By hypothesis on f:

|y − y′| = |f(y) − f(y′)| ≤ c|y − y′|

with 0 < c < 1. This is only possible if |y − y′| = 0. So y = y′ and the fixed point is unique.

d) This follows from parts b and c. No matter where we start x = y_{0}, part b says we get a Cauchy, hence convergent sequence by iterating f. Part c says the limit is a fixed point of f, but f has only one fixed point.

4.3.12. a) Let c ∈ C. We claim that there exist sequences x_{n}, y_{n} with lim x_{n} = c = lim y_{n} with x_{n} ∈ C and y_{n} ∉ C. The easiest way to see this is to recall one of the facts we proved in Discussion 2: c is an element of the Cantor set if and only it has a ternary expansion

c = (.d_{1}d_{2} ⋯)_{3} = \frac{d_{1}}{3} + \frac{d_{2}}{3^{2}} + ⋯

containing ternary digits d_{i} that are all either 0 or 2.. We can get a sequence x_{n} ∈ C converging to c by taking

x_{n} = \frac{d_{1}}{3} + \frac{d_{2}}{3^{2}} + ⋯ + \frac{d_{n}}{3^{n}}

with 0’s the rest of the way. Similarly we can get a sequence y_{n} ∉ C by replacing the nth digit in the ternary expansion of c by a 1. Then y_{n} → c also since the difference between y_{n} and c is ±\frac{1}{3^{n}} and goes to zero as n → ∞. This construction gives lim g(x_{n}) = 1 ≠ 0 = lim g(y_{n}). Hence g is not continuous at c by Corollary 4.2.5.

b) The complement of C is open, since C is closed. Hence for every c in the complement of C, there is an open interval (c−δ, c+δ) centered at c contained in the complement. The value of g at all points in that interval is 0 so g is continuous at c (reason: |g(x) − g(c)| = |0 − 0| < ε whenever |x − c| < δ.)

I.

4.5.2. a) False: Let f(x) = x^{2} (continous for all x). f((−1, 1)) = [0, 1), which is not open (but is bounded of course!)

b) False: See part a.

c) True: Follows from the Extreme and Intermediate Value Theorems: Assume f is continuous on [a, b]. By the Extreme Value Theorem, f attains a maximum s and a minimum t on [a, b]. Say f(x_{1}) = s and f(x_{2}) = t. Then the Intermediate Value Theorem implies that for all real r with t < r < s, there exist x in the interval with endpoints x_{1} and x_{2} with f(x) = r. Hence f([a, b]) is the closed interval [t, s].

4.5.3. There are no such functions. Assume that there was a continuous f : R → R with f(R) = Q. Let r_{1} < r_{2} be any two rational numbers. Then there are x_{1} and x_{2} with f(x_{1}) = r_{1} and f(x_{2}) = r_{2}. But in the interval (r_{1}, r_{2}), by the Density Theorems there are irrationals too: there always exist irrational s with r_{1} < s < r_{2}. By the Intermediate Value Theorem, applied to f on the interval with endpoints x_{1}, x_{2}, there exists some x with f(x) = s. This is a contradiction because we assumed f took only rational values.

4.5.7. If f(0) = 0 or f(1) = 1 then there is nothing to prove. So we will assume 0 < f(0) ≤ 1 and 0 ≤ f(1) < 1. Consider the function g(x) = f(x) − x. We know g is continuous on [0, 1] by the Algebraic Continuity Theorem (Theorem 4.3.4). We have g(0) = f(0)−0 > 0, but g(1) = f(1) − 1 < 0. Hence by the Intermediate Value Theorem, there must be some c in (0, 1) where g(c) = f(c) − c = 0. This means f(c) = c, so c is a fixed point of f.

5.2.3. Let

f(x) = \{ \begin{matrix}x^{2} & if x is rational \\ 0 & if x is irrational \end{matrix}

We claim f is differentiable at 0 (f′(0) = 0), but f is not differentiable at any other point. To see f′(0) = 0, note that

\frac{f(x) − f(0)}{x − 0} = \{ \begin{matrix}x & if x is rational \\ 0 & if x is irrational \end{matrix}

Hence lim_{x→0} \frac{f(x)−f(0)}{x−0} = 0, and f is differentiable at 0 with f′(0) = 0.

On the other hand, f is not differentiable at any other x ∈ R because it is not even continuous there. Note that if c ≠ 0, then there is a sequence x_{n} → c with x_{n} ∈ Q for all n, and another sequence y_{n} → c with y_{n} ∉ Q for all n. Then lim_{n→∞} f(x_{n}) = lim_{n→∞} x_{n}^{2} = c^{2}, but lim_{n→∞} f(y_{n}) = 0. Since c ≠ 0, this says lim_{x→c} g(x) does not exist.

5.2.4. Note that the two “halves” of f_{a} are continuous separately for all a (but f_{a}(0) is undefined as stated if a < 0). a) lim_{x→0^{+}} x^{a} = 0 for all a > 0, so f_{a}(x) is continuous at x = 0 for all a > 0.

b) f_{a} is differentiable at x = 0 as long as a > 1, because

lim_{x→0} \frac{f_{a}(x) − f_{a}(0)}{x − 0} = lim_{x→0} \{ \begin{matrix}x^{a−1} & if x rational \\ 0 & if x irrational \end{matrix} = 0.

The derivative function is f′_{a}(x) = \{ \begin{matrix}ax^{a−1} & if x ≥ 0 \\ 0 & if x < 0 \end{matrix}, so the derivative function is contin-

uous for all a > 1 by the same reasoning as in part a (it’s a constant times the function f_{a−1}(x)).

c) By the last observation in part b, f′_{a} is differentiable at x = 0 as long as a − 1 > 1, or a > 2.

II. A) We will use the fact that if f is a polynomial of odd degree, then as x → ±∞, one

of the limits is +∞ and the other is −∞ (which is which depends on the sign of the leading coefficient). Hence if B is sufficiently large f(B)f(−B) < 0. So by the Intermediate Value Theorem f has a root in the interval [−B, B].

B) If f has even degree, then as x → ±∞, both of the limits are either +∞ or −∞ (which it is again depends on the sign of the leading coefficient). Hence if f(a) < 0 < f(b), the Intermediate Value Theorem implies there is a root of f in [a, b]. But then f must change sign again on (−∞, a) if lim_{x→−∞} f(x) = +∞ (leading coefficient positive), or on (b, +∞) if lim_{x→+∞} f(x) = −∞ (leading coefficient negative). It follows that in any case, f has at least two distinct real roots.

C) False: There could be a single double root, as in the case of f(x) = x^{2}.

III. Consider f(x) = x^{2} − 3^{x}. f is continuous on all of R since it is a sum of continuous functions. Note that f(−1) = 1 − 1/3 = 2/3 > 0 while f(0) = 0 − 1 = −1 < 0. Hence by the Intermediate Value Theorem, there is some c ∈ (−1, 0) such that f(c) = 0, and hence c^{2} = 3^{c}.

IV. A) This is like the proof of 4.5.7 above. Consider g(x) = f(x) − x^{2}. We have g(0) =

f(0) − 0 < 0 and g(1) = f(1) − 1 > 0. Hence by the Intermediate Value Theorem, there is some c ∈ (0, 1) where g(c) = 0, so f(c) = c^{2}.

B) Consider h(x) = f(x)−g(x). We have h(0) = f(0)−g(0) < 0 and h(1) = f(1)−g(1) > 0. Hence by the Intermediate Value Theorem, there is some c ∈ (0, 1) where h(c) = 0, so f(c) = g(c).

5.3.2. Call the closed interval A. By the Mean Value Theorem, we have for all x, y ∈ A f(x) − f(y) = f′(c)(x − y) for some c ∈ (x, y). Taking absolute values |f(x) − f(y)| = |f′(c)||x − y|. By hypothesis, f′ is continuous on A, hence by the Extreme Value Theorem, it attains a maximum s = max |f′| < 1. Therefore, |f(x) − f(y)| ≤ s|x − y| for all x, y ∈ A. It follows that f is contractive on A.

Mathematics 242 – Principles of Analysis Solutions – Problem Set 8 December 7, 2005

5.3.3. a) h is differentiable, therefore continuous on [0, 3]. Hence the function g(x) = h(x) − x is also continuous on [0, 3]. We have g(0) = h(0) = 1, and g(3) = h(3) − 3 = −1. So h changes sign on [0, 3]. By the Intermediate Value Theorem, there is some d ∈ (0, 3) where g(d) = 0, so h(d) = d.

b) By the Mean Value Theorem, there is some c ∈ (0, 3) where 1 = h(3) − h(0) = h′(c)(3 − 0), so h′(c) = 1/3.

c) By an argument similar to that for part b, there is some b ∈ (1, 3) where 0 = h(3) − h(1) = h′(b)(3 − 1), so h′(b) = 0. Now by the intermediate value property for derivatives (the Darboux theorem 5.2.7), since 0 < 1/4 < 1/3, there is some x in the interval where h′(x) = 1/4.

5.3.4. a) We are given that f, g are continuous on [a, b] and differentiable on (a, b). Consider h(x) = (f(b)−f(a))g(x)−(g(b)−g(a))f(x). h is also continuous on [a, b] and differentiable on (a, b) by the usual algebraic properties of limits and derivatives. By the Mean Value Theorem, there exists some c ∈ (a, b) where h(b) − h(a) = h′(c)(b − a). But h(b) = g(a)f(b) − f(a)g(b) and h(a) = f(b)g(a) − f(a)g(b) = h(b). Hence h′(c) = 0. But h′(c) = (f(b) − f(a))g′(c) − (g(b) − g(a))f′(c), which implies

(1)  \frac{f(b) − f(a)}{g(b) − g(a)} = \frac{f′(c)}{g′(c)},

which is what we wanted to show.

b) Consider the parametric curve defined by x = f(t), y = g(t) for t ∈ [a, b]. Then part a says that there is some c where the slope of the tangent line:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g′(c)}{f′(c)}

equals the slope of the secant line through the endpoints (f(a), g(a)) and (f(b), g(b)): \frac{g(b)−g(a)}{f(b)−f(a)} (of course, the equation in (1) is obtained by taking reciprocals on both sides).

7.2.2. a) L(f, P) = (3·1/2)+(4·1/2)+(5·1) = 17/2. U(f, P) = (4·1/2)+(5·1/2)+(7·1) = 23/2, and U(f, P) − L(f, P) = 3.

b) We get U(f, P) − L(f, P) = 22/2 − 18/2 = 2.

c) The regular partition with any n ≥ 5 is one.

7.2.3. If f(x) = c is constant, then for each subinterval in the partition m_{k} = M_{k} = c. Hence L(f, P) = ∑_{k} c(x_{k} − x_{k−1}). This sum telescopes to L(f, P) = c(b − a). Similarly U(f, P) = c(b − a). Since U(f, P) − L(f, P) = 0 for all P, the criterion from Theorem 7.2.8 is certainly met, and f is integrable on [a, b]. The integral is ∫_{a}^{b} f = c(b − a).

7.2.4. a) This follows directly from Theorem 7.2.8. If lim_{n→∞} U(f, P_{n}) − L(f, P_{n}) = 0, then for any ε > 0, there exists N such that whenever n ≥ N, U(f, P_{n}) − L(f, P_{n}) < ε.

b) P_{n} = {i/n : 0 ≤ i ≤ n}, so

U(f, P_{n}) = ∑_{i=1}^{n} \frac{i}{n} · \frac{1}{n} = \frac{1}{n^{2}} ∑_{i=1}^{n} i = \frac{n^{2} + n}{2n^{2}} = \frac{1}{2} + \frac{1}{2n}

Similarly: b) P_{n} = {i/n : 0 ≤ i ≤ n}, so

L(f, P_{n}) = ∑_{i=0}^{n−1} \frac{i}{n} · \frac{1}{n} = \frac{1}{n^{2}} ∑_{i=0}^{n−1} i = \frac{n^{2} − n}{2n^{2}} = \frac{1}{2} − \frac{1}{2n}

c) From part b, we have U(f, P_{n})−L(f, P_{n}) = \frac{1}{n}. Hence lim_{n→∞} U(f, P_{n})−L(f, P_{n}) = 0 and part a implies that f(x) = x is integrable on [0, 1]. The value of ∫_{0}^{1} x is

lim_{n→∞} U(f, P_{n}) = lim_{n→∞} L(f, P_{n}) = 1/2.

7.3.1. a) No matter which partition we take, every subinterval will contain points where f(x) = 1 so, m_{i} = inf{f(x) : x_{i−1} ≤ x ≤ x_{i}} = 1 for all i. Therefore L(f, P) = ∑_{i=1}^{n}(x_{i} − x_{i−1}) = x_{n} − x_{0} = 1.

b) We can take any partition where the subinterval containing x = 1 has width < 1/10. A “lazy” example is the partition P = {0, 10/11, 1} Then U(f, P) = 1 · 10/11 + 2 · 1/11 = 12/11 = 1 + 1/11 < 1 + 1/10.

c) Along the lines of part b, given ε > 0, let δ < ε and consider P = {0, 1 − δ, 1}. Then U(f, P) = 1 · (1 − δ) + 2 · δ = 1 + δ < 1 + ε.

7.3.4. a) This can either be proved by an argument following the ideas of the proof of Theorem 7.3.2, or by using that theorem and some of the results in the next section. Here is the second approach: Say g has the same value as f except at c ∈ [a, b]. Then g and f are equal on [d, b] for all d with b > d > c. This says g is integrable on [c, b] by Theorem 7.3.2. Similarly g is integrable on [a, c]. Finally Theorem 7.4.1 implies g is integrable on [a, b].

Next, we will show ∫_{a}^{b} f = ∫_{a}^{b} g. Say f(c) − g(c) > 0 first for simplicity. The difference f(x) − g(x) is zero except at x = c and integrable by Theorem 7.4.2. If we take a partition P = \{a, c − δ, c + δ, b\} with δ < \frac{ε}{2(f(c)−g(c))} then U(f − g, P) = (f(c) − g(c)) · 2δ < ε. Since

ε can be any positive number, this shows ∫_{a}^{b} f − g = 0, so ∫_{a}^{b} f = ∫_{a}^{b} g. The other case f(c)−g(c) < 0 follows by an argument parallel to this but using L(f −g, P), or U(g−f, P).

b) Suppose f(x) equals g(x) except at x = c_{1}, …, c_{n}. Then ∫_{a}^{b} f = ∫_{a}^{b} g follows by an induction argument in which we alter one point at a time. The base case is part a. The induction step uses exactly the same reasoning as the base case.

c) The Dirichlet function f(x) = \{ \begin{matrix}1 & if x rational \\ 0 & if x irrational \end{matrix} is an example of this. The function g(x) = 0 is integrable on [0, 1] since it is constant, hence continuous. The set of rational numbers in [0, 1] is countable, hence we can get f from g by changing a countable number of values. But we know f is not integrable according to our definition.

7.3.5. Let 0 < c < 1. Then in the interval [c, 1] there are only finitely many numbers of the form 1/n for n a natural number. This implies that f is integrable on [c, 1] by part b of Exercise 7.3.4. But then f is also integrable on [0, 1] by Theorem 7.3.2. (This can also be proved directly of course, using a proof like the proof of 7.3.2.) The value is ∫_{0}^{1} f = 0. For any 0 < ε < 1, note that there are at most 2/ε integers n with 1/n > ε/2. Let P be the partition containing 0, ε/2 and points 1/n − ε^{2}/8, 1/n + ε^{2}/8 for each of the n with 1/n > ε/2. Then

U(f, P) = 1 · ε/2 + ∑_{{n:1/n>ε/2}} 1 · ε^{2}/4 < ε/2 + (2/ε) · ε^{2}/4 = ε.

This shows that ∫_{0}^{1} f = 0 since there are partitions P with U(f, P) < ε for all ε > 0.

7.4.2. By the “usual case” of this additivity result, if c ≤ a ≤ b, then ∫_{c}^{b} f = ∫_{c}^{a} f + ∫_{a}^{b} f. But then Definition 7.4.3 says ∫_{a}^{c} f = − ∫_{c}^{a} f, so ∫_{a}^{b} = ∫_{a}^{c} f + ∫_{c}^{b} f works here too.

7.4.4. a) False: Consider the function f(x) = \{ \begin{matrix}1 & if x rational \\ −1 & if x irrational \end{matrix}. |f| is the constant

function with value 1 for all x and hence is integrable on [0, 1]. But f is not integrable on [0, 1] for the same reason the Dirichlet function is not: U(f, P) = 1 for all partitions P since every subinterval contains rational numbers, while L(f, P) = −1 for all partitions since every subinterval contains irrational numbers.

b) False: The function from Exercise 7.3.5 is a counterexample.

c) True: If g is continuous on [a, b], g(x) ≥ 0, and g(x_{0}) > 0, then take ε = g(x_{0})/2 in the definition of continuity. There exists δ > 0 such that for all x with |x − x_{0}| < δ, |g(x) − g(x_{0})| < ε/2. This implies g(x) > ε/2 > 0 for all x with |x − x_{0}| < δ. By an “area under the curve” estimate, we get

∫_{a}^{b} g ≥ (ε/2)(2δ) = εδ > 0.

d) True: Since we are given ∫_{a}^{b} f > 0, we know f is integrable, and sup{L(f, P) : P } is positive. This means that for some partitition P, L(f, P) > 0. Hence for some subinterval [c, d] in that partition we have m_{i} = inf{f(x) : x ∈ [c, d]} > 0. Take δ = m_{i}. Then f(x) ≥ δ for all x ∈ [c, d].