Test 1 ITA- Homework Problems + Solution Skeleton
Give an example of each of the following, or state that such a request is impossible by referencing the proper theorem(s): (a) sequences (xn) and (yn), which both diverge, but whose sum (xn + yn) converges; (b) sequences (xn) and (yn), where (xn) converges, (yn) diverges, and (xn + yn) converges; (c) a convergent sequence (bn) with bn ̸= 0 for all n such that (1/bn) diverges; d) an unbounded sequence (an) and a convergent sequence (bn) with (an − bn) bounded; (e) two sequences (an) and (bn), where (anbn) and (an) converge but (bn) does not.
(a) (xn) = n and (yn) = −n diverge but xn + yn = 0 converges (b) Impossible, the algebraic limit theorem implies lim(xn+yn)−lim(xn) = lim yn therefore (yn) must converge if (xn) and (xn + yn) converge. (c) bn = 1/n has bn → 0 and 1/bn diverges. If bn → b ̸= 0 then 1/bn → 1/b, but since b = 0 ALT doesn't apply. (d) Impossible, |bn| is convergent and therefore bounded (Theorem 2.3.2) so |bn| ≤ M1, and |an − bn| ≤ M2 is bounded, therefore |an| ≤ |an − bn| + |bn| ≤ M1 + M2 must be bounded. (e) bn = n and an = 0 works. However if (an) → a, a ̸= 0 and (anbn) → p then the ALT would imply (bn) → p/a.
(a) Prove that the sequence defined by yn = sup {ak : k ≥ n} converges. (b) The limit superior of (an), or lim sup an, is defined by lim sup an = lim yn where yn is the sequence from part (a) of this exercise. Provide a reasonable definition for lim inf an and briefly explain why it always exists for any bounded sequence. (c) Prove that lim inf an ≤ lim sup an for every bounded sequence, and give an example of a sequence for which the inequality is strict. (d) Show that lim inf an = lim sup an if and only if lim an exists. In this case, all three share the same value.
(a) (yn) is decreasing and converges by the monotone convergence theorem. (b) Define lim inf an = lim zn for zn = inf{an : k ≥ n}. zn converges since it is increasing and bounded. (c) Obviously inf{ak : k ≥ n} ≤ sup{an : k ≥ n} so by the Order Limit Theorem lim inf an ≤ lim sup an. (d) If lim inf an = lim sup an then the squeeze theorem (Exercise 2.3.3) implies an converges to the same value, since inf{ak≥n} ≤ an ≤ sup{ak≥n}.
(a) Show how induction can be used to conclude that (A1 ∪ A2 ∪ · · · ∪ An) c = A c 1 ∩ A c 2 ∩ · · · ∩ A c n for any finite n ∈ N (b) It is tempting to appeal to induction to conclude [∞ i=1 Ai !c = \∞ i=1 A c i but induction does not apply here. Induction is used to prove that a particular statement holds for every value of n ∈ N, but this does not imply the validity of the infinite case. To illustrate this point, find an example of a collection of sets B1, B2, B3, . . . where Tn i=1 Bi ̸= ∅ is true for every n ∈ N, but T∞ i=1 Bi ̸= ∅ fails. (c) Nevertheless, the infinite version of De Morgan's Law stated in (b) is a valid statement. Provide a proof that does not use induction.
(a) 1.2.5 Is our base case, Assume (A1 ∪ · · · ∪ An) c = Ac 1 ∩ · · · ∩ Ac n . We want to show the n + 1 case. Using associativity we have ((A1 ∪ · · · ∪ An) ∪ An+1) c = (A1 ∪ · · · ∪ An) c ∩ A c n+1 = (A c 1 ∩ · · · ∩ A c n ) ∩ A c n+1 = A c 1 ∩ · · · ∩ A c n ∩ A c n+1 (b) B1 = {1, 2, . . . }, B2 = {2, 3, . . . }, . . . (c) First suppose x ∈ ( T∞ i=1 Ai) c , then x /∈ T∞ i=1 Ai meaning x /∈ Ai for some i, which is the same as x ∈ Ac i for some i, meaning x ∈ S∞ i=1 Ac i . This shows \∞ i=1 Ai ! ⊆ [∞ i=1 A c i Now suppose x ∈ S∞ i=1 Ac i meaning x /∈ Ai for some i, which is the same as x /∈ T∞ i=1 Ai implying x /∈ ( T∞ i=1 Ai) c . This shows inclusion the other way and completes the proof.
Give an example of each or state that the request is impossible. When a request is impossible, provide a compelling argument for why this is the case. (a) Two sets A and B with A ∩ B = ∅,sup A = sup B,sup A /∈ A and sup B /∈ B. (b) A sequence of nested open intervals J1 ⊇ J2 ⊇ J3 ⊇ · · · with T∞ n=1 Jn nonempty but containing only a finite number of elements. (c) A sequence of nested unbounded closed intervals L1 ⊇ L2 ⊇ L3 ⊇ · · · with T∞ n=1 Ln = ∅. (An unbounded closed interval has the form [a,∞) = {x ∈ R : x ≥ a}.) (d) A sequence of closed bounded (not necessarily nested) intervals I1, I2, I3, . . . with the property that TN n=1 In ̸= ∅ for all N ∈ N, but T∞ n=1 In = ∅.
(a) A = Q ∩ (0, 1), B = I ∩ (0, 1). A ∩ B = ∅, sup A = sup B = 1 and 1 ∈/ A, 1 ∈/ B. (b) Defining Ji = (ai , bi), A = {an : n ∈ N}, B = {bn : n ∈ N}, T∞ n=1 Jn will at least contain (sup A, inf B). Thus, a necessary condition to meet the request is sup A = inf B. Ji = (−1/n, 1/n) satisfies this condition (sup A = inf B = 0) and by inspection, T∞ n=1 Jn = {0}, which meets the request. (c) Ln = [n, ∞) has T∞ n=1 Ln = ∅ (d) Impossible. Let Jn = Tn k=1 Ik and observe the following (i) Since TN n=1 In ̸= ∅ we have Jn ̸= ∅. (ii) Jn being the intersection of closed intervals makes it a closed interval. (iii) Jn+1 ⊆ Jn since In+1 ∩ Jn ⊆ Jn (iv) T∞ n=1 Jn = T∞ n=1 ( Tn k=1 Ik) = T∞ n=1 In By (i), (ii) and (iii) the Nested Interval Property tells us T∞ n=1 Jn ̸= ∅. Therefore by (iv) T∞ n=1 In ̸= ∅.
Let (an) → 0, and use the Algebraic Limit Theorem to compute each of the following limits (assuming the fractions are always defined): (a) lim 1+2an 1+3an−4a 2 n (b) lim (an+2)2−4 an (c) lim 2 an +3 1 an +5 .
(a) Apply the ALT lim 1 + 2an 1 + 3an − 4a 2 n = lim (1 + 2an) lim (1 + 3an − 4a 2 n ) = 1 + 2 lim(an) 1 + 3 lim an − 4 lim a 2 n = 1 Showing a 2 n → 0 is easy so I've omitted it (b) Apply the ALT lim (an + 2)2 − 4 an ! = lim a 2 n + 2an an = lim(an + 2) = 2 + lim an = 2 (c) Multiply the top and bottom by an then apply the ALT lim 2 an + 3 1 an + 5! = lim 2 + 3an 1 + 5an = 2 + 3 lim an 1 + 5 lim an = 2
Decide which of the following represent true statements about the nature of sets. For any that are false, provide a specific example where the statement in question does not hold. (a) If A1 ⊇ A2 ⊇ A3 ⊇ A4 ··· are all sets containing an infinite number of elements, then the intersection ∞n=1 An is infinite as well. (b) If A1 ⊇ A2 ⊇ A3 ⊇ A4 ··· are all finite, nonempty sets of real numbers, then the intersection ∞n=1 An is finite and nonempty. (c) A∩(B∪C)=(A∩B)∪C. (d) A∩(B∩C)=(A∩B)∩C. (e) A∩(B∪C)=(A∩B)∪(A∩C).
(a) False, consider A1 = {1, 2, . . . }, A2 = {2, 3, . . . }, ... has T∞ n=1 An = ∅. (b) True, because we eventually reach Aj = {x} and get stuck (c) False, A = ∅ gives ∅ = C. (d) True, intersection is associative. (e) True, draw a diagram
Supply rebuttals to the following complaints about the proof of Theorem 1.6.1. (a) Every rational number has a decimal expansion, so we could apply this same argument to show that the set of rational numbers between 0 and 1 is uncountable. However, because we know that any subset of Q must be countable, the proof of Theorem 1.6.1 must be flawed. (b) Some numbers have two different decimal representations. Specifically, any decimal expansion that terminates can also be written with repeating 9's. For instance, 1/2 can be written as .5 or as .4999 . . . Doesn't this cause some problems?
(a) False, since the constructed number has an infinite number of decimals it is irrational. (b) No, since if we have 9999 . . . and change the nth digit 9992999 = 9993 is still different.
(i) A sequence (an) is eventually in a set A ⊆ R if there exists an N ∈ N such that an ∈ A for all n ≥ N. (ii) A sequence (an) is frequently in a set A ⊆ R if, for every N ∈ N, there exists an n ≥ N such that an ∈ A. (a) Is the sequence (−1)n eventually or frequently in the set {1}? (b) Which definition is stronger? Does frequently imply eventually or does eventually imply frequently? (c) Give an alternate rephrasing of Definition 2.2.3B using either frequently or eventually. Which is the term we want? (d) Suppose an infinite number of terms of a sequence (xn) are equal to 2 . Is (xn) necessarily eventually in the interval (1.9, 2.1)? Is it frequently in (1.9, 2.1)?
(a) Frequently, but not eventually. (b) Eventually is stronger, it implies frequently. (c) (xn) → x if and only if xn is eventually in any ϵ-neighborhood around x. (d) (xn) is frequently in (1.9, 2.1) but not necessarily eventually (consider xn = 2(−1)n ).
(a) Give an example of a countable collection of disjoint open intervals. (b) Give an example of an uncountable collection of disjoint open intervals, or argue that no such collection exists.
(a) I1 = (0, 1), I2 = (1, 2) and in general In = (n − 1, n) (b) Let A denote this set. Intuitively no such collection should exist since each In has nonzero length. The key here is to try and show A ∼ Q instead of directly showing A ∼ N. For any nonempty interval In the density theorem tells us there exists an r ∈ Q such that r ∈ In. Assigning each I ∈ A a rational number r ∈ I proves I ⊆ Q and thus I is countable.
(a) If x ∈ (A∩B)c, explain why x ∈ Ac ∪Bc. This shows that (A∩B)c ⊆ Ac ∪Bc. Prove the reverse inclusion (A ∩ B)c ⊇ Ac ∪ Bc, and conclude that (A ∩ B)c = Ac ∪ Bc. Show (A ∪ B)c = Ac ∩ Bc by demonstrating inclusion both ways
(a) If x ∈ (A ∩ B) c then x /∈ A ∩ B so x /∈ A or x /∈ B implying x ∈ Ac or x ∈ Bc which is the same as x ∈ Ac ∪ Bc . (b) Let x ∈ Ac ∪ Bc implying x ∈ Ac or x ∈ Bc meaning x /∈ A or x /∈ B implying x /∈ A ∩ B which is the same as x ∈ (A ∩ B) c (c) First let x ∈ (A ∪ B) c implying x /∈ A ∪ B meaning x /∈ A and x /∈ B which is the same as x ∈ Ac and x ∈ Bc which is just x ∈ Ac ∩Bc . Second let x ∈ Ac ∩Bc implying x ∈ Ac and x ∈ Bc implying x /∈ A and x /∈ B meaning x /∈ A ∪ B which is just x ∈ (A ∪ B)
Give an example of each of the following, or argue that such a request is impossible. (a) A sequence that has a subsequence that is bounded but contains no subsequence that converges. (b) A sequence that does not contain 0 or 1 as a term but contains subsequences converging to each of these values. (c) A sequence that contains subsequences converging to every point in the infinite set {1, 1/2, 1/3, 1/4, 1/5, . . .}. (d) A sequence that contains subsequences converging to every point in the infinite set {1, 1/2, 1/3, 1/4, 1/5, . . .}, and no subsequences converging to points outside of this set.
(a) Impossible, the Bolzano-Weierstrass theorem tells us a convergent subsequence of that subsequence exists, and that sub-sub sequence is also a subsequence of the original sequence. (b) (1 + 1/n) → 1 and (1/n) → 0 so (1/2, 1 + 1/2, 1/3, 1 + 1/3, . . .) has subsequences converging to 0 and 1. (c) Copy the finitely many previous terms before proceeding to a new term (1, 1/2, 1, 1/3, 1, 1/2, 1/4, 1, 1/2, 1/3, . . .) The sequence contains infinitely many terms in {1, 1/2, 1/3, . . . } hence subsequences exist converging to each of these values. (d) Impossible, the sequence must converge to zero which is not in the set. Proof: Let ϵ > 0 be arbitrary, pick N large enough that 1/n < ϵ/2 for n > N. We can find a subsequence (bm) → 1/n meaning |bm − 1/n| < ϵ/2 for some m. using the triangle inequality we get |bm − 0| ≤ |bn − 1/n| + |1/n − 0| < ϵ/2 + ϵ/2 = ϵ therefore we have found a number bm in the sequence am with |bm| < ϵ. This process can be repeated for any ϵ therefore a sequence which converges to zero can be constructed.
Give an example of each of the following, or state that the request is impossible. (a) A set B with inf B ≥ sup B. (b) A finite set that contains its infimum but not its supremum. (c) A bounded subset of Q that contains its supremum but not its infimum.
(a) Let B = {0} we have inf B = 0 and sup B = 0 thus inf B ≥ sup B. (b) Impossible, finite sets must contain their infimum and supremum. (c) Let B = {r ∈ Q | 1 < r ≤ 2} we have inf B = 1 ∈/ B and sup B = 2 ∈ B
(a) Prove if A1, . . . , Am are countable sets then A1 ∪ · · · ∪ Am is countable. (b) Explain why induction cannot be used to prove that if each An is countable, then S∞ n=1 An is countable. (c) Show how arranging N into the two-dimensional array 1 3 6 10 15 · · · 2 5 9 14 · · · 4 8 13 · · · 7 12 · · · 11 · · · . . . leads to a proof for the infinite case
(a) Let B, C be disjoint countable sets. We use the same trick as with the integers and list them as B ∪ C = {b1, c1, b2, c2, . . . } Meaning B ∪ C is countable, and A1 ∪ A2 is also countable since we can let B = A1 and C = A2 \ A1. Now induction: suppose A1 ∪ · · · ∪ An is countable, (A1 ∪ · · · ∪ An)∪An+1 is the union of two countable sets which by above is countable. (b) Induction shows something for each n ∈ N, it does not apply in the infinite case. (c) Rearranging N as in (c) gives us disjoint sets Cn such that S∞ n=1 Cn = N. Let Bn be disjoint, constructed as B1 = A1, B2 = A1 \ B1, . . . we want to do something like f(N) = f [∞ n=1 Cn ! = [∞ n=1 fn(Cn) = [∞ n=1 Bn = [∞ n=1 An Let fn : Cn → Bn be bijective since Bn is countable, define f : N → S∞ n=1 Bn as f(n) = f1(n) if n ∈ C1 f2(n) if n ∈ C2 . . . (i) Since each Cn is disjoint and each fn is 1-1, f(n1) = f(n2) implies n1 = n2 meaning f is 1-1. (ii) Since any b ∈ S∞ n=1 Bn has b ∈ Bn for some n, we know b = fn(x) has a solution since fn is onto. Letting x = f −1 n (b) we have f(x) = fn(x) = b since f −1 n (b) ∈ Cn meaning f is onto. By (i) and (ii) f is bijective and so S∞ n=1 Bn is countable. And since [∞ n=1 Bn = [∞ n=1 An We have that S∞ n=1 An is countable, completing the proof.
Show that if (xn) is a convergent sequence, then the sequence given by the averages yn = x1 + x2 + · · · + xn n also converges to the same limit. (b) Give an example to show that it is possible for the sequence (yn) of averages to converge even if (xn) does not.
(a) Let D = sup{|xn − x| : n ∈ N} and let 0 < ϵ < D, we have |yn − x| = x1 + · · · + xn n − x ≤ |x1 − x| + · · · + |xn − x| n ≤ D Let |xn − x| < ϵ/2 for n > N1 giving |yn − x| ≤ |x1 − x| + · · · + |xN1 − x| + · · · + |xn − x| n ≤ N1D + (n − N1)ϵ/2 n Let N2 be large enough that for all n > N2 (remember 0 < ϵ < D so (D − ϵ/2) > 0.) 0 < N1(D − ϵ/2) n < ϵ/2 Therefor |yn − x| ≤ N1(D − ϵ/2) n + ϵ/2 < ϵ Letting N = max{N1, N2} completes the proof as |yn − x| < ϵ for all n > N. (Note: I could have used any ϵ ′ < ϵ instead of ϵ/2, I just needed some room.) (b) xn = (−1)n diverges but (yn) → 0.
Let xn ≥ 0 for all n ∈ N. (a) If (xn) → 0, show that √ xn → 0. (b) If (xn) → x, show that √ xn → √xn
(a) Setting xn < ϵ2 implies √ xn < ϵ (for all n ≥ N of course) (b) We want | √ xn − √ x| < ϵ multiplying by (√ xn + √ x) gives |xn − x| < ( √ xn + √ x)ϵ since xn is convergent, it is bounded |xn| ≤ M implying p |xn| ≤ √ M, multiplying gives |xn − x| < √ xn + √ x ϵ ≤ √ M + √ x ϵ Since |xn − x| can be made arbitrarily small we can make this true for some n ≥ N. Now dividing by √ M + √ x gives us | √ xn − √ x| ≤ |xn − x| √ M + √ x < ϵ therefore | √ xn − √ x| < ϵ completing the pro
Let y1 = 6, and for each n ∈ N define yn+1 = (2yn − 6) /3 (a) Use induction to prove that the sequence satisfies yn > −6 for all n ∈ N. (b) Use another induction argument to show the sequence (y1, y2, y3, . . .) is decreasing.
(a) Suppose for induction that yn > −6, our base case clearly satisfies y1 > −6. then yn+1 = (2yn − 6)/3 =⇒ yn = (3yn+1 + 6)/2 > −6 =⇒ yn+1 > (2 · (−6) − 6)/3 = −6 Thus yn+1 > −6 (b) Suppose yn+1 < yn, the base case 2 < 6 works. Now yn+1 < yn =⇒ 2yn+1 < 2yn =⇒ 2yn+1 − 6 < 2yn − 6 =⇒ (2yn+1 − 6)/3 < (2yn − 6)/3 =⇒ yn+2 < yn+1 Thus (yn) is decreasing
(a) Consider the recursively defined sequence y1 = 1 yn+1 = 3 − yn and set y = lim yn. Because (yn) and (yn+1) have the same limit, taking the limit across the recursive equation gives y = 3 − y. Solving for y, we conclude lim yn = 3/2 What is wrong with this argument? (b) This time set y1 = 1 and yn+1 = 3− 1 yn . Can the strategy in (a) be applied to compute the limit of this sequence?
(a) The sequence yn = (1, 2, 1, 2, . . .) does not converge. (b) Yes, yn converges by the monotone convergence theorem since 0 < yn < 3 and yn is increasing.
For each series, find an explicit formula for the sequence of partial sums and determine if the series converges. (a) P∞ n=1 1 2n (b) P∞ n=1 1 n(n+1) (c) P∞ n=1 log n+1 n (In (c), log(x) refers to the natural logarithm function from calculus.)
(a) This is a geometric series, we can use the usual trick to derive sn. Let r = 1/2 for convenience sn = 1 + r + r 2 + · · · + r n rsn = r + r 2 + · · · + r n+1 rsn − sn = r n+1 − 1 =⇒ sn = r n+1 − 1 r − 1 This is the formula when n starts at zero, but the sum in question starts at one so we subtract the first term to correct this X∞ n=1 1 2 n = −1 +X∞ n=0 1 2 n = −1 + limn→∞ (1/2)n+1 − 1 1/2 − 1 = −1 + −1 −1/2 = 1 (b) We can use partial fractions to get 1 n(n + 1) = 1 n − 1 n + 1 Which gives us a telescoping series, most of the terms cancel and we get sn = 1 − 1 n + 1 Therefor X∞ n=1 1 n(n + 1) = limn→∞ 1 − 1 n + 1 = 44 (c) Another telescoping series, since log n + 1 n = log(n + 1) − log(n) therefore most of the terms cancel and we get sn = log(n + 1) Which doesn't converge
Recall that I stands for the set of irrational numbers. (a) Show that if a, b ∈ Q, then ab and a + b are elements of Q as well. (b) Show that if a ∈ Q and t ∈ I, then a + t ∈ I and at ∈ I as long as a ̸= 0. (c) Part (a) can be summarized by saying that Q is closed under addition and multiplication. Is I closed under addition and multiplication? Given two irrational numbers s and t, what can we say about s + t and st?
(a) Trivial. (b) Suppose a + t ∈ Q, then by (a) (a + t) − a = t ∈ Q contradicting t ∈ I. (c) I is not closed under addition or multiplication. consider (1 − √ √ 2) ∈ I by (b), and 2 ∈ I. the sum (1 − √ 2) + √ 2 = 1 ∈ Q ∈/ I. Also √ 2 · √ 2 = 2 ∈ Q ∈/ I.
Decide whether the following propositions are true or false, providing a short justification for each conclusion. (a) If every proper subsequence of (xn) converges, then (xn) converges as well. (b) If (xn) contains a divergent subsequence, then (xn) diverges. (c) If (xn) is bounded and diverges, then there exist two subsequences of (xn) that converge to different limits. (d) If (xn) is monotone and contains a convergent subsequence, then (xn) converges.
(a) True, removing the first term gives us the proper subsequence (x2, x3, . . .) which converges. This implies (x1, x2, . . .) also converges to the same value, since discarding the first term doesn't change the limit behavior of a sequence. (b) True, the divergent subsequence is unbounded, hence (xn) is also unbounded and divergent. (c) True, since xn is bounded lim sup xn and lim inf xn both converge. And since xn diverges Exercise 2.4.7 tells us lim sup xn ̸= lim inf xn. (d) True, The subsequence (xnk ) converges meaning it is bounded |xnk | ≤ M. Suppose (xn) is increasing, then xn is bounded since picking k so that nk > n we have xn ≤ xnk ≤ M. A similar argument app
Decide if the following statements about suprema and infima are true or false. Give a short proof for those that are true. For any that are false, supply an example where the claim in question does not appear to hold. (a) If A and B are nonempty, bounded, and satisfy A ⊆ B, then sup A ≤ sup B. (b) If sup A < inf B for sets A and B, then there exists a c ∈ R satisfying a < c < b for all a ∈ A and b ∈ B. (c) If there exists a c ∈ R satisfying a < c < b for all a ∈ A and b ∈ B, then sup A < inf B.
(a) True. We know a ≤ sup A and a ≤ sup B since A ⊆ B. since sup A is the least upper bound on A we have sup A ≤ sup B. (b) True. Let c = (sup A + inf B)/2, c > sup A implies a < c and c < inf B implies c < b giving a < c < b as desired. (c) False. consider A = {x | x < 1}, B = {x | x > 1}, a < 1 < b but sup A ̸< inf B since 1 ̸< 1.
Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit. (a) lim 2n+1/5n+4 = 2/5 . (b) lim 2n^2/n3+3 = 0. (c) lim sin(n 2 )/ n^1/3 = 0.
(a) We have 2n + 1 5n + 4 − 2 5 = 5(2n + 1) − 2(5n + 4) 5(5n + 4) = −3 5(5n + 4) = 3 5(5n + 4) < ϵ We now find n such that the distance is less then ϵ 3 5(5n + 4) < 1 n < ϵ =⇒ n > 1 ϵ You could also solve for the smallest n, which would give you 3 5(5n + 4) < ϵ =⇒ 5n + 4 > 3 5ϵ =⇒ n > 3 25ϵ − 4 5 I prefer the first approach, the second is better if you were doing numerical analysis and wanted a precise convergence rate. (b) We have 2n 2 n3 + 3 − 0 = 2n 2 n3 + 3 < 2n 2 n3 = 2 n < ϵ =⇒ n > 2 ϵ (c) We have sin(n 2 ) n1/3 ≤ 1 n1/3 < ϵ =⇒ n > 1 ϵ 3 Really slow convergence! if ϵ = 10−2 we would require n > 106
Given sets A and B, define A + B = {a + b : a ∈ A and b ∈ B}. Follow these steps to prove that if A and B are nonempty and bounded above then sup(A + B) = sup A + sup B (a) Let s = sup A and t = sup B. Show s + t is an upper bound for A + B. (b) Now let u be an arbitrary upper bound for A + B, and temporarily fix a ∈ A. Show t ≤ u − a. (c) Finally, show sup(A + B) = s + t. (d) Construct another proof of this same fact using Lemma 1.3.8.
(a) We have a ≤ s and b ≤ t, adding the equations gives a + b ≤ s + t. (b) t ≤ u − a should be true since u − a is an upper bound on b, meaning it is greater then or equal to the least upper bonud t. Formally a + b ≤ u implies b ≤ u − a and since t is the least upper bound on b we have t ≤ u − a. (c) From (a) we know s + t is an upper bound, so we must only show it is the least upper bound. Let u = sup(A + B), from (a) we have t ≤ u − a and s ≤ u − b adding and rearranging gives a + b ≤ 2u − s − t. since 2u − s − t is an upper bound on A + B it is less then the least upper bound, so u ≤ 2u − s − t implying s + t ≤ u. and since u is the least upper bound s + t must equal u. Stepping back, the key to this proof is that a + b ≤ s, ∀a, b implying sup(A + B) ≤ s can be used to transition from all a + b to a single value sup(A + B), avoiding the ϵ-hackery I would otherwise use. (d) Showing s + t − ϵ is not an upper bound for any ϵ > 0 proves it is the least upper bound by Lemma 1.3.8. Rearranging gives (s − ϵ/2) + (t − ϵ/2) we know there exists a > (s − ϵ/2) and b > (t − ϵ/2) therefore a + b > s + t − ϵ meaning s + t cannot be made smaller, and thus is the least upper bound.
(a) Explain why √xy ≤ (x + y)/2 for any two positive real numbers x and y. (The geometric mean is always less than the arithmetic mean.) (b) Now let 0 ≤ x1 ≤ y1 and define xn+1 = √ xnyn and yn+1 = xn + yn 2 Show lim xn and lim yn both exist and are equal
(a) We have √ xy ≤ (x + y)/2 ⇐⇒ 4xy ≤ x 2 + 2xy + y 2 ⇐⇒ 0 ≤ (x − y) 2 (b) The only fixed point is xn = yn so we only need to show both sequences converge. The inequality x1 ≤ y1 is always true since √ xnyn ≤ xn + yn 2 =⇒ xn+1 ≤ yn+1 Also xn ≤ yn implies (xn + yn)/2 = yn+1 ≤ yn, similarly √xnyn = xn+1 ≥ xn meaning both sequences converge by the monotone convergence theorem.
(a) Show (a, b) ∼ R for any interval (a, b). (b) Show that an unbounded interval like (a,∞) = {x : x > a} has the same cardinality as R as well. (c) Using open intervals makes it more convenient to produce the required 1-1, onto functions, but it is not really necessary. Show that [0, 1) ∼ (0, 1) by exhibiting a 1-1 onto function between the two sets.
(a) We will start by finding f : (−1, 1) → R and then transform it to (a, b). Example 1.5.4 gives a suitable f f(x) = x x 2 − 1 The book says to use calculus to show f is bijective, first we will examine the derivative f ′ (x) = x 2 − 1 − 2x 2 (x 2 − 1)2 = − x 2 + 1 (x 2 − 1)2 The denominator and numerator are positive, so f ′ (x) < 0 for all x ∈ (0, 1). This means no two inputs will be mapped to the same output, meaning f is one to one (a rigorous proof is beyond our current ability) To show that f is onto, we examine the limits lim x→1− x x 2 − 1 = −∞ lim x→−1+ x x 2 − 1 = +∞ Then use the intermediate value theorem to conclude f is onto. Now we shift f to the interval (a, b) g(x) = f 2x − 1 b − a − a Proving g(x) is also bijective is a straightforward application of the chain rule. (b) We want a bijective h(x) such that h(x) : (a,∞) → (−1, 1) because then we could compose them to get a new bijective function f(h(x)) : (a,∞) → R. Let h(x) = 2 x − a + 1 − 1 We have h : (a,∞) → (1, −1) since h(a) = 1 and limx→∞ h(x) = 1. Meaning that f(h(x)) : (a,∞) → R is our bijective map. (c) With countable sets adding a single element doesn't change cardinality since we can just shift by one to get a bijective map. we'll use a similar technique here to essentially outrun our problems. Define f : [0, 1) → (0, 1) as f(x) = ----- 1/2 if x = 0 1/4 if x = 1/2 1/8 if x = 1/4 . . . x otherwise Now we prove f is bijective by showing y = f(x) has exactly one solution for all y ∈ (0, 1). If y = 1/2 n then the only solution is y = f(1/2 n−1 ) (or x = 0 in the special case n = 1), If y ̸= 1/2 n then the only solution is y = f(y).
For some additional practice with nested quantifiers, consider the following invented definition: Let's call a sequence (xn) zero-heavy if there exists M ∈ N such that for all N ∈ N there exists n satisfying N ≤ n ≤ N + M where xn = 0 (a) Is the sequence (0, 1, 0, 1, 0, 1, . . .) zero heavy? (b) If a sequence is zero-heavy does it necessarily contain an infinite number of zeros? If not, provide a counterexample. (c) If a sequence contains an infinite number of zeros, is it necessarily zeroheavy? If not, provide a counterexample. (d) Form the logical negation of the above definition. That is, complete the sentence: A sequence is not zero-heavy if ....
(a) Yes. Choose M = 1; since the sequence has a 0 in every two spaces, for all N either xn = 0 or xn+1 = 0. (b) Yes. If there were a finite number of zeros, with the last zero at position K, then choosing N > K would lead to a contradiction. (c) No, consider (0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, . . .) where the gap between 0's grows indefinitely. For any value of M, for large enough N the gap between zeros will be greater than M. Then we simply choose N so that xN is the first 1 in a streak of at least M + 1 1's. (d) A sequence is not zero-heavy if for all M ∈ N, there exists some N ∈ N such that for all n ∈ N, N ≤ n ≤ N + M, xn ̸= 0.
Give an example of each or state that the request is impossible. For any that are impossible, give a compelling argument for why that is the case. (a) A sequence with an infinite number of ones that does not converge to one. (b) A sequence with an infinite number of ones that converges to a limit not equal to one. (c) A divergent sequence such that for every n ∈ N it is possible to find n consecutive ones somewhere in the sequence
(a) an = (−1)n (b) Impossible, if lim an = a ̸= 1 then for any n ≥ N we can find a n with an = 1 meaning ϵ < |1 − a| is impossible. (c) an = (1, 2, 1, 1, 3, 1, 1, 1, . . .)
Given a function f : D → R and a subset B ⊆ R, let f −1 (B) be the set of all points from the domain D that get mapped into B; that is, f −1 (B) = {x ∈ D : f(x) ∈ B}. This set is called the preimage of B. (a) Let f(x) = x 2 . If A is the closed interval [0, 4] and B is the closed interval [−1, 1], find f −1 (A) and f −1 (B). Does f −1 (A ∩ B) = f −1 (A) ∩ f −1 (B) in this case? Does f −1 (A ∪ B) = f −1 (A) ∪ f −1 (B)? (b) The good behavior of preimages demonstrated in (a) is completely general. Show that for an arbitrary function g : R → R, it is always true that g −1 (A∩B) = g −1 (A)∩g −1 (B) and g −1 (A ∪ B) = g −1 (A) ∪ g −1 (B) for all sets A, B ⊆ R
(a) f −1 (A) = [−2, 2], f −1 (B) = [−1, 1], f −1 (A ∩ B) = [−1, 1] = f −1 (A)∩f −1 (B), f −1 (A ∪ B) = [−2, 2] = f −1 (A) ∪ f −1 (B) (b) First let x ∈ g −1 (A∩B) meaning g(x) ∈ A∩B implying g(x) ∈ A and g(x) ∈ B which is the same as x ∈ g −1 (A) and x ∈ g −1 (B) meaning x ∈ g −1 (A) ∩ g −1 (B). Second let x ∈ g −1 (A) ∩ g −1 (B), this is the same as x ∈ g −1 (A) and x ∈ g −1 (B) which is the same as g(x) ∈ A and g(x) ∈ B implying g(x) ∈ A∩B and thus x ∈ g −1 (A∩B). Thus g −1 (A ∩ B) = g −1 (A) ∩ g −1 (B). Seeing g −1 (A ∪ B) = g −1 (A) ∪ g −1 (B) is obvious. see 1.2.7 (d).
Given a function f and a subset A of its domain, let f(A) represent the range of f over the set A; that is, f(A) = {f(x) : x ∈ A}. (a) Let f(x) = x 2 . If A = [0, 2] (the closed interval {x ∈ R : 0 ≤ x ≤ 2}) and B = [1, 4], find f(A) and f(B). Does f(A ∩ B) = f(A) ∩ f(B) in this case? Does f(A ∪ B) = f(A) ∪ f(B)? (b) Find two sets A and B for which f(A ∩ B) ̸= f(A) ∩ f(B). (c) Show that, for an arbitrary function g : R → R, it is always true that g(A ∩ B) ⊆ g(A) ∩ g(B) for all sets A, B ⊆ R (d) Form and prove a conjecture about the relationship between g(A∪B) and g(A)∪g(B) for an arbitrary function g
(a) f(A) = [0, 4], f(B) = [1, 16], f(A∩B) = [1, 4] = f(A)∩f(B) and f(A∪B) = [0, 16] = f(A) ∪ f(B) (b) A = {−1}, B = {1} thus f(A ∩ B) = ∅ but f(A) ∩ f(B) = {1} (c) Suppose y ∈ g(A ∩ B), then ∃x ∈ A ∩ B such that g(x) = y. But if x ∈ A ∩ B then x ∈ A and x ∈ B, meaning y ∈ g(A) and y ∈ g(B) implying y ∈ g(A) ∩ g(B) and thus g(A ∩ B) ⊆ g(A) ∩ g(B). Notice why it is possible to have x ∈ g(A)∩g(B) but x /∈ g(A∩B), this happens when something in A \ B and something in B \ A map to the same thing. If g is 1-1 this does not happen. (d) I conjecture that g(A ∪ B) = g(A) ∪ g(B). To prove this we show inclusion both ways, First suppose y ∈ g(A∪B). then either y ∈ g(A) or y ∈ g(B), implying y ∈ g(A)∪g(B). Now suppose y ∈ g(A) ∪ g(
Let A1, A2, A3, . . . be a collection of nonempty sets, each of which is bounded above. (a) Find a formula for sup (A1 ∪ A2). Extend this to sup (Sn k=1 Ak). (b) Consider sup (S∞ k=1 Ak). Does the formula in (a) extend to the infinite case?
(a) sup (Sn k=1 Ak) = sup {sup Ak | k = 1, . . . , n} (b) In general no, since S∞ k=1 Ak may be unbounded, for example with An = [n, n + 1].
Assume (an) is a bounded sequence with the property that every convergent subsequence of (an) converges to the same limit a ∈ R. Show that (an) must converge to a.
(a2, a3, . . .) Is a convergent subsequence, so obviously if (a2, a3, . . .) → a then (an) → a also
(b) Find an efficient proof for all the cases at once by first demonstrating (a + b) 2 ≤ (|a| + |b|) 2 (c) Prove |a − b| ≤ |a − c| + |c − d| + |d − b| for all a, b, c, and d
(b) (a + b) 2 ≤ (|a| + |b|) 2 reduces to 2ab ≤ 2|a||b| which is true as the left side can be negative but the right side can't. and since squaring preserves inequality this implies |a + b| ≤ |a| + |b|. (c) I would like to do this using the triangle inequality, I notice that (a−c)+(c−d)+(d−b) = a − b. Meaning I can use the triangle inequality for multiple terms |a − b| = |(a − c) + (c − d) + (d − b)| ≤ |a − c| + |c − d| + |d − b| The general triangle inequality is proved by repeated application of the two variable inequality |(a + b) + c| ≤ |a + b| + |c| ≤ |a| + |b| + |c|
What happens if we reverse the order of the quantifiers in Definition 2.2.3? Definition: A sequence (xn) verconges to x if there exists an ϵ > 0 such that for all N ∈ N it is true that n ≥ N implies |xn − x| < ϵ Give an example of a vercongent sequence. Is there an example of a vercongent sequence that is divergent? Can a sequence verconge to two different values? What exactly is being described in this strange definition?
Firstly, since we have for all N ∈ N we can remove N entirely and just say n ∈ N. Our new definition is Definition: A sequence (xn) verconges to x if there exists an ϵ > 0 such that for all n ∈ N we have |xn − x| < ϵ. In other words, a series (xn) verconges to x if |xn−x| is bounded. This is a silly definition though since if |xn − x| is bounded, then |xn − x ′ | is bounded for all x ′ ∈ R, meaning if a sequence is vercongent it verconges to every x ′ ∈ R. Put another way, a sequence is vercongent if and only if it is bounded.
The limit of a sequence, when it exists, must be unique. Prove Theorem 2.2.7. To get started, assume (an) → a and also that (an) → b. Now argue a = b
If a ̸= b then we can set ϵ small enough that having both |an − a| < ϵ and |an − b| < ϵ is impossible. Therefore a = b. (Making this rigorous is trivial and left as an exercise to the reader)
Consider the sequence given by bn = n − √ n2 + 2n. Taking (1/n) → 0 as given, and using both the Algebraic Limit Theorem and the result in Exercise 2.3.1, show lim bn exists and find the value of the limit.
I'm going to find the value of the limit before proving it. We have n − √ n2 + 2n = n − p (n + 1)2 − 1 For large n, p (n + 1)2 − 1 ≈ n + 1 so lim bn = −1. Factoring out n we get n 1 − p 1 + 2/n . Tempting as it is to apply the ALT here to say (bn) → 0 it doesn't work since n diverges. How about if I get rid of the radical, then use the ALT to go back to what we had before? (n − √ n2 + 2n)(n + √ n2 + 2n) = n 2 − (n 2 + 2n) = −2n Then we have bn = n − √ n2 + 2n = −2n n + √ n2 + 2n = −2 1 + p 1 + 2/n Now we can finally use the algebraic limit theorem! lim −2 1 + p 1 + 2/n! = −2 1 + p 1 + lim (2/n) = −2 1 + √ 1 + 0 = −1 Stepping back the key to this technique is removing the radicals via a difference of squares, then dividing both sides by the growth rate n and applying the ALT.
Finish the following proof for Theorem 1.5.7. Assume B is a countable set. Thus, there exists f : N → B, which is 1 − 1 and onto. Let A ⊆ B be an infinite subset of B. We must show that A is countable. Let n1 = min{n ∈ N : f(n) ∈ A}. As a start to a definition of g : N → A set g(1) = f (n1). Show how to inductively continue this process to produce a 1-1 function g from N onto A.
Let nk = min{n ∈ N | f(n) ∈ A, n /∈ {n1, n2, . . . , nk−1}} and g(k) = f(nk). since g : N → A is 1-1 and onto, A is countable
(a) Show that √ 2, q 2 + √ 2, r 2 + q 2 + √ 2, . . . converges and find the limit.
Let x1 = √ 2 and xn+1 = √ 2 + xn clearly x2 > x1. assuming xn+1 > xn gives 2 + xn+1 > 2 + xn ⇐⇒ p 2 + xn+1 > √ 2 + xn ⇐⇒ xn+2 > xn+1 Since xn is monotonically increasing and bounded the monotone convergence theorem tells us (xn) → x. Equating both sides like in 2.4.1 gives x = √ 2 + x ⇐⇒ x 2 − x − 2 = 0 ⇐⇒ x = 1 2 ± 3 2 Since x > 0 we must have x = 2.
Show that if xn ≤ yn ≤ zn for all n ∈ N, and if lim xn = lim zn = l, then lim yn = l as w
Let ϵ > 0, set N so that |xn − l| < ϵ/4 and |zn − l| < ϵ/4. Use the triangle inequality to see |xn − zn| < |xn − l| + |l − zn| < ϵ/2. Note that since xn ≤ yn ≤ zn, |yn − xn| = yn − xn ≤ zn − xn = |zn − xn|. Apply the triangle inequality again to get |yn − l| ≤ |yn − xn| + |xn − l| ≤ |zn − xn| + |xn − l| < ϵ/2 + ϵ/4 < ϵ
Let B be a set of positive real numbers with the property that adding together any finite subset of elements from B always gives a sum of 2 or less. Show B must be finite or countable.
Notice B ∩ (a, 2) is finite for all a > 0, since if it was infinite we could make a set with sum greater then two. And since B is the countable union of finite sets S∞ n=1 B ∩ (1/n, 2), B must be countable or finite.
Finish the proof of Theorem 1.4.5 by showing that the assumption α 2 > 2 leads to a contradiction of the fact that α = sup T
Recall T = {t ∈ R | t 2 < 2} and α = sup T. suppose α 2 > 2, we will show there exists an n ∈ N such that (α − 1/n) 2 > 2 contradicting the assumption that α is the least upper bound. We expand (α − 1/n) 2 to find n such that (α 2 − 1/n) > 2 2 < (α − 1/n) 2 = α 2 − 2α n + 1 n2 < α2 + 1 − 2α n Then 2 < α2 + 1 − 2α n =⇒ n(2 − α 2 ) < 1 − 2α. Since 2 − α 2 < 0 dividing reverses the inequality gives us n > 1 − 2α 2 − α2 This contradicts α 2 > 2 since we have shown n can be picked such that (α 2 − 1/n) > 2 meaning α is not the least upper bound.
(a) Prove that sqrt(3) is irrational.
Suppose for contradiction that p/q is a fraction in lowest terms, and that (p/q) ^2 = 3. Then p^ 2 = 3q ^2 implying p is a multiple of 3 . Write p as 3r for some r, substitute, implying q is also a multiple of 3 contradicting the assumption that p/q is in lowest terms.
Using Exercise 1.4.1, supply a proof that I is dense in R by considering the real numbers a − √ 2 and b − √ 2. In other words show for every two real numbers a < b there exists an irrational number t with a < t < b.
The density theorem lets us find a rational number r with a − √ 2 < r < b − √ 2, adding √ 2 to both sides gives a < r + √ 2 < b. From 1.4.1 we know r + √ 2 is irrational, so setting t = r + √ 2 gives a < t < b as desired.
Use a similar strategy to the one in Example 2.5.3 to show lim b 1/n exists for all b ≥ 0 and find the value of the limit. (The results in Exercise 2.3.1 may be assumed.)
To show b 1/n is monotone bounded consider two cases (I won't prove each rigorously to avoid clutter, you can if you want) (i) If b > 1 then b 1/n is decreasing, and bounded b 1/n > 1 ⇐⇒ b > 1 n (raise both to n) (ii) If b < 1 then b 1/n is increasing, and bounded b 1/n < 1 ⇐⇒ b < 1 n (raise both to n) therefore b 1/n converges for each b ≥ 0 by the monotone convergence theorem. to find the limit equate terms b 1/(n+1) = b 1/n =⇒ b = b n+1 n = b 1+1/n Raise both to (n + 1)th power =⇒ 1 = b 1/n Divide by b, this assumes b ̸= 0 Thus lim b 1/n = 1 if b ̸= 0, if b = 0 then lim b 1/n = 0.
