1 Homework. Recommended Reading:

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1 Analysis MT43C Notes/Problems/Homework Recommended Reading: R. G. Bartle, D. R. Sherbert Introduction to real analysis, principal reference M. Spivak Calculus W. Rudin Principles of mathematical analysis S. Abbott Understanding analysis K. R. Stromberg Introduction to Classical Real Analysis T. W. Körner A Companion to Analysis Homework. This problem is about set operations. Recall the notation: N = {,, 3, 4, 5, 6,...} the set of natural numbers Z = N {0} N = {..., 3,,, 0,,, 3,...} the set of integers union of two sets. {,, 3} {, 3, 4, 5} = {,, 3, 4, 5} intersection. {,, 3} {, 3, 4, 5} = {, 3} \ difference. {,, 3} \ {, 3, 4, 5} = {} subset. If A and B are sets, we write A B if A is a subset of B, i.e. if each element of A is also an element of B. Thus {,, 3} {,, 3, 4, 5, 7} N Z but {, 0, } N Cartesian product. {,, 3} {, 3, 4, 5} = {(, ), (, 3), (, 4), (, 5), (, ), (, 3), (, 4), (, 5), (3, ), (3, 3), (3, 4), (3, 5)} # cardinality, number of elements, of a set. # {, 3, 8, 88} = 4 {x P (x)} set of all x satisfying the property P (x). {u u N, 5 n} = {5, 0, 5, 0,...} set of all natural numbers divisible by 5. {F (x) x M} where M is a set and F a function, set of all elements of the form F (x) with x M. {5k k N} = {5, 0, 5, 0,...} set of all natural numbers divisible by 5. {x x Z} = {..., ( ), ( ), 0,,,...} = {..., 4,, 0,, 4,...} = {0,, 4,...} set of perfect squares. Write the following sets by listing their elements. Thus, for example, ({,, 33} {0,,, 3}) \ ({3,, 5} {n N n > 0}) = {, 0,,, 3, 33} \ {3} = {, 0,,, 33} (a) ({,, 3, 4} {k k N}) {6, 7} Solution: = {, 4} {6, 7} = {(, 6), (, 7), (4, 6), (4, 7)} (b) {k k N} {3k k N} {n N n < 0} *** Solution: = {6,, 8} (c) {#(A B) A, B {n N n < 0}, #A = #B = } Solution: = {3, 4, 5, 6, 7, 8, 9, 0, } (d) {(x, y) x, y Z, x + y <, 0 < x < y} Solution: = {(, ), (, 3)} ***

2 . Which of the following statements a-d do not hold for all sets A, B, C? If the statement holds for all A, B, C, show the set in the Venn diagram. If the statement does not hold for all A, B, C find such sets as small as possible violating the statement (a small counterexample). Hint: Draw Venn diagramms. As an example, the formula (A B) C = A (B C) does not always hold. To see this, look at the Venn Diagram for three sets and then pick sets that the difference is not empty. In this example the sets differ by A \ C. Thus, if you take A = { } (or any other set with one element) and B = C =, we get (A B) C =, but A (B C) = { }. (a) (A \ B) (B \ A) = (A B) \ (A B) Solution: true (b) A (C \ (B \ A)) = (A C) \ B Solution: A (C \ (B \ A)) = A (C \ B) (A C) \ B, counterexample A = { }, B = { }, C = (c) A (B \ C) = (A B) \ (A C) Solution: A (B \ C) B \ C B \ (A C) = (A B) \ (A C), counterexample A = { }, B = C = 3. Which of the following statements 3a-3d do not hold for all sets A, B, C, D? If the statement holds for all A, B, C, show the set in the diagram below. If the statement does not hold for all A, B, C, D find such sets as small as possible violating the statement (a small counterexample). Hint: Look at a Venn-type diagram for products, D B A C Where is (A \ C) (B D) (for instance)? (a) (A B) (C D) = (A C) (B D) Solution: (A B) (C D) (A B) (C D) (A \ C D \ B) (C \ A B \ D) counterexample A = D = { }, B = C =. (b) (A B) (C D) = (A C) (B D) Solution: true = (A C) (B D) (c) (A B) \ (C D) = (A \ C) (B \ D) Solution: (A B) \ (C D) = ((A \ C) (B \ D)) ((A \ C) (B D)) ((A C) (B \ D)) counterexample A = B = D = { }, C = (d) A (B \ D) = (A B) \ (A D) Solution: true (A \ C) (B \ D) (e) (A \ C) (B \ D) = ((A B) \ (C B)) \ (A D) Solution: true Please hand up your solutions to -3 on Friday, /.

3 4. Recall that a function f is (a) injective if for a, b D (f), f(a) = f(b) implies a = b. (b) surjective, or onto B (a set) if for each b B there is a D (f) such that f(a) = b (c) a bijection of A with B (A, B sets) if for each a A there is a unique b and for each b B there is a unique a A such that f(a) = b. We denote by D (f) = {a u : (a, u f} the domain of f and by image (f) = {x u : (u, x) f} the image of f. For each of the following functions f i : A i B i, i = 4a... 4d, say whether f i is injective, onto B i, a bijection of A i with B i. If f i is a bijection of A i with B i find a formula for the inverse f i : B i A i. If f i is not injective, find x, x A i, x x so that f(x ) = f(x ) If f i is not onto B i, find an element in B i \ image (f i ). (a) f 4a : A 4a = Z B 4a = N, f 4a : x x + x 4. Solution: The map f 4a is not injective and not onto N: f 4a ( ) = f 4a (), 3 image (f 4a ). (b) f 4b : A 4b = Z N B 4b = Q, f 4b : (x, y) x y Solution: The map f 4b is onto Q but not injective. f 4b (4, ) = 4 = = f 4b(8, 4). (c) f 4c : A 4c = R B 4c = [, ) = {x R }, f 4c : x e e x Solution: This map is injective and not onto B 4c, there is no x R with f 4c (x) = B 4c The map is a bijection of R with (, ) = B 4c \ {}. The inverse is given by f 4c (y) = ln ((ln(y)) ) = ln ( ln(y)). (d) f 4d : A 4d = R B 4d = R, f 4d : x x 3 x Solution: This function f 4d is onto R but not injective: f 4d (0) = f 4d () = Let f = {(x, y) x, y R and x + y 4 = 6 and y 0} *** (a) What is f()? Solution: 4 (b) Draw f. Solution: y (0, ) ( 4, 0) (0, 0) domain of f image of f (4, 0) x (c) Domain and range of f are intervals. Find them. Solution: D (f) = [ 4, 4], image (f) = [0, ]. 6. Let A, B be sets and f : A B and g : B A be functions. Assume that the composition g f is the identity on A. Prove that f is injective. Hint: For A a set, the identity id A : A A is the function with id A (a) = a for all a A. Thus The composition v u of two functions u and v is id A = {(a, a) a A}. v u = {(x, y) z : (x, z) u and (z, y) v}.

4 Thus, if D (v) image (u), (v u)(x) = v(u(x)) for x D (u). Solution: In order to show that f is injective, we need to prove that x, y D (f), f(x) = f(y) = x = y. To this end, let x, y D (f) with f(x) = f(y). Then g(f(x)) = g(f(y)) since g is a function. Now, by assumption, g(f(x)) = id A (x) = x, g(f(y)) = id A (y) = y. Hence x = y. 7. Consider the functions f, g, Show that image (g f) = g(image (f)). Solution: A f B g C image (g f) = {g(f(a)) a A} = g({f(a) a A}) = g(image (f)). Please hand up your solutions to 4-7 on Friday, 6/.

5 8. Prove from the axioms R-R, the definitions DFR-5 and the lemmas L-L8 from the Appendix: (a) If a, b, c R with a (b c) = 0, then a = 0 or b = 0 or c = 0. *** Solution: Let a, b, c R with a (b c) = 0. By L4 we have a = 0 or b c = 0. In the second case, again by L4, we have b = 0 or c = 0. (b) If x P, then P. x Solution: Let x P. By R0, x 0, hence exists. By R0, P or = 0 or P. x x x x Now if = 0 then = x = 0 by bu DFR- and L, contradicting R \ 0 from R6. x x If P, then by L7, P hence, by R5 and R8, = x ( x x x) P, hence P. By R0 this contradicts P from L6. The only remaining case by R0 is P. x 9. If X is a real number we write X + = { X if X 0 0 if X 0 and X = { 0 if X 0 X if X 0. Also recall some notation/conventions: If A R and M R we write M A (M A) if M a (M a) holds for all a A. A set A R has a maximum (minimum) if there is M A so that M A (M A), in which case we write M = max A (M = min A). For x R we also write [x] = max {k Z k x} x = x [x] For each of the following i = 9c... 9f, write the given expression X i in x, y, z R in terms of the functions given with it (and arithmetic operations). (a) (Example) X 9a (x, y, z) = max {x, y, z} min {x, y, z} in terms of : max {x, y, z} min {x, y, z} = x y + y z + x z (b) (Example) X 9b (x) = x in terms of x x + : x = x + x (c) X 9c (x) = x in terms of x +, x Solution: x = x + + x (d) X 9d (x) = x in terms of x x +. Solution: X 9d (x) = x = x + x (e) X 9e (x) = min { x k k Z} in terms of [ ],, Solution: X 9e (x) = x (f) X 9f (x) = sgn(sin(x)) in terms of sgn, [ ],, Solution: ( x X 9f (x) = sgn(sin(x)) = sgn π 4 ) 4 0. Let X be a finite set and *** Y := {(A, B, C) A, B, C X, X = A B C, A B = A C = B C = } Express #Y in terms of #X. Counting the set Y in two different ways gives formula involving powers and binomial coefficients. Derive this. Hint: Try decompositions in two sets, i.e. try to count Y = {(A, B) A, B X, X = A B, A B = }

6 This is in bijection with the set of all subsets of X, because if (A, B) Y, then B = X \ A. Solution: There is a bijection given by assigning to (A, B, C) Y the function To see that Φ is a bijection, look at the function given by Y Φ {,, 3} X = {f : X {,, 3}} Φ(A, B, C): X {,, 3} with if x A [Φ(A, B, C)] (x) = if x B 3 if x C Ψ: {,, 3} X Y Ψ(f) = (f(), f(), f(3)) Clearly, Ψ Φ = id Y and Φ Ψ = id {,,3} X It follows that #Y = # {,, 3} X = 3 #X. We could also have counted Y using binomial coefficients. Recall that for n N 0, n choose k, ( ) n n! = # {A {,..., n} #A = k} = k k!(n k)! is the number of k-element subsets of an n-element set. ( ) n If X has n elements, then for each k, 0 k n we have choices for A so that #A = k. Fixing ( ) k n k A and l, 0 l n k leaves choices for B so that #B = l. For C we then have no further l choices because C = X \ A \ B. Summing up all these choices, we get 3 n = #Y = n ( ) n n k ( ) n k k l k=0 l=0. Sketch the following subsets of R : (a) {(x, y) R x > [y]} y {([y], y) y R} {(x, y) R x > [y]} x Solution:

7 (b) {(x, y) R x p + y p < } for p = /,,, 4. Solution: These sets are the interiors of the contours in the picture below y p = 4 p = / x p = p = (c) {(x, y) R max { x, y 6 } < 4} Solution: We can rewrite this as { (x, y) R max { x, y 6 } < 4 } = { (x, y) R x < 4 and y 3 < } the interior of the rectangle below. x = ( 3, 5) (, 5), x Please hand up your solutions to 8- on Friday, /3.

8 . Prove from the axioms R-R, the definitions DFR-5, the lemmas L-L8 from the Appendix and the axioms for the natural numbers (in particular induction): (a) If x, y are positive real numbers with x < y and n N, then x n < y n. Hint: Here x n is defined recursively: x :=, x n+ = x x n for n N. Solution: Let Y (n) be the statement 0 < x < y = x n < y n. We prove Y (n) for all n by induction: For Y () is trivial. Assume n N is so that Y (n) holds. Let x, y R, 0 < x < y. By Y (n) we then have x n < y n. Since 0 < x we have from L8 and the recursive definition of x n+ and y n+ that Since < is transitive (L7), this implies (b) If n N and x, y R then x n+ = x x n < x y n and x y n < y y n = y n+. x n+ < y n+. x n+ y n+ = (x y) p n (x, y) where p n (x, y) is a polynomial of degree n in x and y. Find a formula for p n. Solution: p n (x, y) = k+l=n xk y l 3. For each of the following sets X i, i = 3c... 3f, find the supremum sup X i and an expression x i (ɛ) such that for any ɛ > 0, x i (ɛ) X i is as required in the definiton of the supremum, i.e. sup X i ɛ < x i (ɛ) sup X i. (a) (Example) X 3a = { n n N }. The supremum is sup X 3a = 0 and x 3a (ɛ) = ] [ + ɛ (b) (Example) X 3b = supremum is { } n N, k k,... k n N, 0 < k < k <... < k n. The 3 kn Consider the elements x n of X 3b given by x n = n j= sup X 3b = 3 j = ( 3 ) X n 3b. We have > x n = 3 > n ɛ if 3 n > ɛ. Since 3 n > n for all n N, this last estimate holds if n >. We can therefore use n = + [ ] ɛ ɛ and get x 3b (ɛ) = x +[ ɛ] = ( ) 3 +[ ɛ] (c) X 3c = ( 4, 3) Solution: sup X 3c = 3, x 3c (ɛ) = max { 3 ɛ, 3 } (d) X 3d = {x R x 3 = x}. Solution: X 3d = {, 0, } hence sup X 3d = = max X 3d and we can take x 3d (ɛ) =. *** ***

9 { uv (e) X 3e = u + v } u, v N, u < v. Solution: Since 0 (u v) = u + v uv, we have uv < u +v for all u, v R. Hence is an upper bound for X 3e. We claim that this is sharp i.e. sup X 3e =. To see this, let ɛ > 0. Then, for u N, X 3e x u := u(u + ) u + (u + ) = if u >. Hence we can take x ɛ 3e(ɛ) = x [ ] + ɛ { n } n (f) X 3f = n N n + i Hint: Solution: i= L n = u + u u + u + = u + u + > 4u > ɛ n n + i = n + i arctan() = n i= 0 n = + i n n dx + x. n i= n n + i is a lower sum for dx = arctan() = π/4, hence π/4 is an upper bound for X 0 +x 3f. This is also sharp because an upper sum for the same integral is In particular, and since if n >, we can take ɛ U n L n = U n n i=0 0 U n = n i=0 n n + i + x dx = π/4 L n n n n + i x 3f (ɛ) = L +[ ɛ] = i= n n + i = n n = n < ɛ +[ ɛ] Please hand up your solutions to -3 on Friday, 8/4 i= + [ ] ɛ ( [ + ɛ]). + i

10 4. Which of the sequences below converge? Prove your answer. ( 7k ) k (a) k + 5 k k N Solution: We first rewrite the fraction, 7k k k + 5 k = k + k + 5 k and then apply the theorem about the limit of a quotient, lim + k k k + 5 = lim ( k ) ( k k = lim k 7 + lim k k lim k + 5 = = since the limit in the denominator is 0. ( m [ m ]) (b) 5 5 m N ( m ( ) ) (c) (q + ) q= m N Hint: If n, m N, N m then m a q = a n a n+ a m. q=n Thus the first terms of this sequence are (, ) ( ) k ), 3 + lim k k k lim k + lim k 5 k ( ) ( ) ( ) 4 9 5, Solution: Abbreviate Thus a m = m q= a m+ = a m ( ) (m + ) ( ) < a (q + ) m Hence the sequence decreases. Since a m > 0 for all m, 0 is a lower bound. By the monotonic convergence theorem, the sequence converges.. 5. For each of the following subsets of A i R, i = 5b... 5d determine the set A i : Â of cluster points of (a) (example) The set of cluster points of A 5a = { { } x x R, x [x] < 4} is Â5a = x x R, x [x] 4 (b) (example) The set of cluster points of A 5b = { m + n n, m N} is Â5b = N {0}. (c) A 5c = Q (d) A 5d = Z (e) A 5e = R \ ( N {0}) (f) A 5f = R \ Z *** *** 6. Let f : R + R be a function so that x f(x) x + x for all x R +. Prove that exists. f(x) lim x 0 x

11 7. Recall the definition of cluster point : A cluster point of a set A R is a point x 0 R such that for every ɛ > 0 there is a ɛ A \ {x 0 } with x x 0 < ɛ. Which of the following statements are true? Either prove or provide a counterexample. (a) x R is a cluster point of A R if there is a sequence (a n ) n N (A \ {a}) N with lim n a n = x. (b) Let A R and f : A R be continuous. Then the composition sgn f is continuous. (c) If A R is a non empty bounded set, then inf A and sup A are cluster points of A. (d) A finite subset of R has no cluster points. Please hand up your solutions to 4-7 on Friday, /4

12 Examples. We assume the existence of letters a, b, c,... z in the usual way. Let S denote the set of small letters, i.e. S = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z} with the usual order, for instance a a, a b, x z. Thus S = {L S capitals for variables, since the letters a z are the elements of S). Write the following sets by listing their elements. Thus, for example, ({a, b, c} {p, q, r}) \ ({b, q} {r}) = {a, b, c, p, q, r} \ {b, q, r} = {a, c, p} a L} is true (We use (a) ({a, b, c, d} {b, c, d, e, f}) {b, x} Solution: = {b, c, d} {b, x} = {(b, b), (b, x), (c, b), (c, x), (d, b), (d, x)} (b) (S \ {a, b, c, d, e, f}) {e, f, g, h, i, j} Solution: = {g, h, i, j} (c) {L S m L} {L S L p} Solution: = {m, n, o, p} (d) L w {K S L K} *** *** Solution: = {w, x, y, z} (e) {A S (#A = 5) and ((K, M A, L S, K L M) L A)} Solution: = {S \ {z}, S \ {a}} = {{a,..., y}, {b,..., z}}. Which of the following statements a-d do not hold for all sets A, B, C? If the statement holds for all A, B, C, show the set in the Venn diagram. If the statement does not hold for all A, B, C find such sets as small as possible violating the statement (a small counterexample). Hint: Draw Venn diagramms. As an example, the formula (A B) C = A (B C) does not always hold. To see this, look at the Venn Diagram for three sets and then pick sets that the difference is not empty. In this example the sets differ by A \ C. Thus, if you take A = { } (or any other set with one element) and B = C =, we get (A B) C =, but A (B C) = { }. (a) A \ (A \ B) = B Solution: A \ (A \ B) = A B B, counterexample A =, B = { } (b) A \ (B \ A) = A \ B Solution: A \ (B \ A) = A A \ B, counterexample A = { }, B = { } (c) A (B \ C) = (A B) \ (A C) Solution: true (d) (A B) (A \ B) = A Solution: true 3. For each of the following functions f i : A i B i, i = 3a... 3d, say whether f i is injective, onto B i, a bijection of A i with B i. If f i is a bijection of A i with B i find a formula for the inverse f i : B i A i. If f i is not injective, find x, x A i, x x so that f(x ) = f(x ) If f i is not onto B i, find an element in B i \ image (f i ). (a) f 3a : A 3a = R + B 3a = R, f 3a : x e x Hint: R + is the set of positive real numbers. Solution: not onto R. The image of f 3a is R(f 3a ) = (, ), hence 0 R \ image (f 3a )

13 { 3x if x is even (b) f 3b : A 3b = N B 3b = N, f 3b : x x+5 if x is odd Solution: not onto B 3b, B 3b \ image (f 3b ). The map is not injective, f 3b () = 3 = f 3b (). (c) f 3c : A 3c = Q B 3c = N, f 3c : x min { } n n N and z Z : x = z n Hint: If = A N then min A is the element a A such that a x for all x A. Thus, for instance, min {34, 66, 3, 9} = 3. Solution: onto B 3c, not injective. f ( ( ) = f 3 ) =. (d) f 3d : A 3d = Z B 3d = Z, f 3d : x {x, x +, x +, x + 3}. Hint: Z = {U U Z} is the power set of Z, i.e. the set of all subsets of Z. Solution: The map f 3d is injective but not onto Z : B 3d \ image (f 3d ). 4. For a real number x we write [x] := max {k Z k x} and {x} = x [x] (.) for the integer part (or floor) and the fractional part of x. Also recall the definition of the absolute value. For a real number x, { x if x 0 x :=. x if x 0 Example:[.3] =, {.3} = 0.3, [ 9/] = 5, { 9/} = /. Consider the functions [ sin: R [, ] and arcsin: [, ] π, π ] We have seen in class that sin arcsin = id [,] but arcsin sin id R. Find a formula for arcsin(sin(x)) in terms of [ ] and { }, and the absolute value. Solution: If x [ π, ] 3π we have two cases: [ x π, π ] : arcsin(sin(x)) = x [ π x, 3π ] : arcsin(sin(x)) = π ( x π ). which we can put together: x [ π, 3π ] : arcsin(sin(x)) = π x π. Since sin is π-periodic, i.e. sin(u) = sin(π + u) for all u R, we can write any x R as [ x = πk + x with k Z, x π, 3π ]. We can express x as Thus for all x R, arcsin(sin(x)) = π x π = π (π { x + π } x = π π π. { x + π } π ) π π = π π { x + π } π π. domain of arcsin range of arcsin sin arcsin sin x

14 5. How many elements do the following sets have? *** (a) A 5a = {(a, b, c) a, b, c Z and a + b + c < 5} Solution: There are many ways of counting this set. For instance, if (a, b, c) A 5a then a, b, c =,, 0,,. If one of the a, b, c is equal to ± then the others must be 0. This gives 3 = 6 possibilities. If none of the a, b, c is equal to ±, i.e. they are all, 0 or, then there is no further restriction: {, 0, } 3 A 5a which gives 3 3 further possibilities. It follows that (b) #A 5a = = 33. { x + y x, y N, x m, y n } n N,n<4 m N,m<4 Solution: Let X(m, n) := { x + y x, y N, x m, y n } thus we have to count X(m, n). n N,n<4 m N,m<4 The key observation to simplify this is that It follows that Now X(m, n) X(k, l) whenever m k and n l. n N,n<4 m N,m<4 X(m, n) = n N,n<4 X(n, 3) = X(3, 3). X(3, 3) = { x + y x, y N, x 3, y 3 } = { x + y x, y N, x y 3 } hence the set has 6 elements. = { +, +, + 3, +, + 3, } = {, 5, 0, 8, 3, 8}, 6. Recall the definitions 4.6 of maximum, minimum, supremum, infimum. For each of the following sets Y i R, i = 6a... 6e determine max Y i, min Y i, sup Y i, inf Y i if they exist. (a) Y 6a = { n N} n +n Solution: max Y 6a = sup Y 6a =, min Y 6a does not exist, inf Y 6a = 0. (b) Y 6b = { ( ) z + z Z } +z Solution: max Y 6b = sup Y 6b =, min Y 6b does not exist, inf Y 6b =. (c) Y 6c = {x + x 3 x R, 0 x < } Solution: max Y 6c does not exist, sup Y 6c =, min Y 6c = inf Y 6c = 0. (d) Y 6d = { n n, m N} m Solution: max Y 6d, sup Y 6d do not exist, min Y 6d does not exist, inf Y 6d = 0. (e) Y 6e = { 5 + k [ k ] k Z } Solution: max Y 6e does not exist, sup Y 6e = 6, min Y 6e does not exist, inf Y 6e = Prove that for n N, n k = n k=

15 Solution: For n N let P (n) be the statement n P (n) : k = n. k= Clearly, P () holds. We now show that for all n N, P (n) implies P (n + ). Thus, assuming P (n), we compute n n n+ (n + ) = + n + n = + n + k = (n + ) + k = k which proves P (n + ). k= 8. Let a, b R. Prove that max {a, b} = a + b What is the analogous formula for min {a, b}? + a b Solution: We check the identity in the two cases a b and a b. If a b, then a b 0. Hence, by the definition of max and, max {a, b} = a = a + b + a b = a + b + a b. Similarly, if b a, max {a, b} = b = a + b The analogous formula for min is a b min {a, b} = a + b = a + b 9. For each of the following subsets of R decide whether Maximum, Minimum, Supremum, Infimum exist and if so, find these values. Prove your statements. Also sketch these sets. a b.. + k= a b (a) {x R x 3 x > 0} Solution: max, min, sup do not exist, inf = {x R x 3 x > 0} =. We plot the function R R, x x 3 x = x(x ), {(x, x 3 x) x R}. k= {x R x 3 x > 0} = (, 0) (, ) (b) { } x R x [x] > Solution: max, min, sup, inf do not exist. We plot the function R R, x x [x], { (x, x [x] ) x R} {x R x [x] } > 0 = ( ) + s, + s s Z

16 (c) {x R z {0,,, 3} : x 3z < } Solution: max, min do not exist. We can write this set as inf {x R z {0,,, 3} : x 3z < } =, sup {x R z {0,,, 3} : x 3z < } = 0, {x R z {0,,, 3} : x 3z < } = {x R z {0, 3, 6, 9} : x z < } that is the set of all real numbers having distance less than from one of the points 0, 3, 6, 9. We first draw these points and then the intervals of length around them: (, ) (, 4) (5, 7) (8, 0) Hint: Look at Example Recall Definition 4.5. Prove or find a counterexample to the following statements: (a) If A, B R are nonempty, M is an upper bound for A and N is an upper bound for B, then MN is an upper bound for AB := {ab a A, b B}. Solution: The statement is false: For a counterexample let M = N = 0, A = B = { } (b) If A, B R are nonempty, M is an upper bound for A and N is an upper bound for B, then M + N is an upper bound for A + B := {a + b a A, b B}. Solution: The statement is correct. Proof: Assume M and N are upper bounds for A and B respectively. Let y A + B. Then by definition of A + B, y of the form y = a + b with a A and b B. It follows that a M and b N, hence y = a + b M + N. To justify the last conclusion, recall that by the definition of in 4., a M, b N means that But since P + P P by R-, we have that M a, N b P (M a) + (N b) P. Using associativity and commutativity of the addition in R, we get that (M a) + (N b) = (M + N) (a + b) = M + N y P hence M + N y.. For each of the following sets Y i R, i = a... d say whether max Y i, min Y i, sup Y i, inf Y i exist, *** and if so, find the value. Also find the minimal interval I so that Y i I. Thus in each case, you have to find and interval I so that I Y i and J Y i whenever J is an interval with J I. Hint: See example 34 (a) Y a = [, 3) N Solution: inf Y a = min Y a = exist, max Y a and sup Y a do not exist, the set has no upper bound. The minimal interval containing Y a is I a = [, ) (b) Y b = ( 3, ) Solution: inf Y b = 3 and sup Y b = exist, max Y b and min Y b do not exist. The minimal interval containing Y b is I b = Y b = ( 3, )

17 (c) Y c = u R,u>5 (, u) Solution: Y c = (, 5] = I c, hence inf Y c and min Y c do not exist, the set has no lower bound. sup Y c = max Y c = 5 exist. (d) Y d = (, ] x x R +,x<3 Solution: Y d = (, ] ( = (, y] =, ] = I d hence, similar to x 3 x R +,x<3 y R +,y>/3 problem c, inf Y d and min Y d do not exist and sup Y d = max Y d = 3. Prove the following statement or provide a counterexample: *** If A is a nonempty bounded set of real numbers and = B A with sup B = sup A, then sup A sup (A \ B). Hint: See example 35. { Solution: The statement is wrong. A counterexample is A = R = (, 0), B = /N = n N }. Then n sup A = sup B = sup A \ B = For all n N let A n be a set. Prove or provide a counterexample for the following: *** n N : A n+ A n = n N A n? Does this change if, in addition, the A n are required to be closed intervals, i.e. intervals of the form A n = [a n, b n ] with a n, b n R, a n b n? Solution: Without further assumptions on the sets A n, the statement is wrong: For a counterexample, let A n = (0, /n). Then { A n = x R n N : 0 < x < } = n n N by the Archimedean Property of R. However, if the A n = [a n, b n ] are closed intervals with A n+ A n for all n, then n N A n. To see this, first observe that since A n+ A n for all n, we have estimates a a a 3... a n a n+... b n+ b n... b 3 b b In particular, the sets A = {a n n N} and B = {b n n N} are bounded. By completeness, sup A and inf B exist. We have sup A inf B and A n = [sup A, inf B]. n N 4. Guess the limits of the sequences ( ) k k N, ( n ) n N, ( ) 9q 7 + 3q + q + 5q 7 34q + q q N *** Solution: The limits of these sequences are 0, 0, 9/5 respectively. 5. If A R, A bounded above, s R, then sup A is the real number such that (i) sup A is an upper bound for A and

18 (ii) ɛ > 0 a ɛ A : sup A ɛ < a ɛ For each of the following sets A i R, i = 5b... 5d find sup A i and a formula expressing a ɛ in terms of ɛ such that 5ii holds. (a) For example, if A 5a = {3 n } n N then sup A 5a = 3 and a ɛ = 3 + [ ] ɛ satisfies 5ii. To see this, first note that [ ] + N ɛ hence 3 for all ɛ > 0. Then, as we have + [ ɛ + [x] > x for all x R a ɛ = 3 + [ ] > 3 ɛ ɛ ] A 5a = 3 ɛ. (b) A 5b = {x R 3x + < }. Solution: A 5b = (, 0), sup A 3 5b = 0, a ɛ = max { ɛ/, /3}, for instance. (c) A 5c = {x R x 3 < 3}. Solution: A 5c = (, 3 3 ), sup A 5c = 3 3, a ɛ = 3 3 ɛ/. { } (d) A 5d = n + n + 3 n N. Solution: sup A 5d = 0. If ɛ > 0, for sup A 5d ɛ < a ɛ we need to find n ɛ N so that sup A 5d ɛ = ɛ < a ɛ = n ɛ + n ɛ + 3 (.) Since n n + n + 3 for all n N, we have an estimate n ɛ < n ɛ + n ɛ + 3 hence (.) holds if ɛ < n ɛ which is true if n ɛ > /ɛ, say n ɛ = + /ɛ which gives. a ɛ = n ɛ + n ɛ + 3 = ( + /ɛ ) + ( + /ɛ ) Let A R be a bounded set. Prove that there is n N such that A [ n, n]. Hint: Archimedean Axiom R- Solution: Since A is bounded, it is bounded above by u, say and below by l. Thus l a u for all a A. By the Archimedian Axiom, there are m, n N, such that u m and l n. Then max {m, n} N and for all a A we have a u m max {m, n} and Therefore a l n max {m, n}. A [ max {m, n}, max {m, n}].

19 7. For each of the following sequences (x n ) n N, x = b... e find L := lim n x n and a formula expressing n ɛ in terms of ɛ such that 5. holds. Hint: Recall the definition in (5.). The floor and ceiling functions might be useful for this: [x] := max {z Z z x} and x = min {z Z z x} (a) For example, if a n = we try L = 0 and look at all n N such that n L n = n < ɛ. Since n > n for all n N, we can estimate < n < ɛ if n > ɛ. Thus we can take nɛ := Of course you could also use n ɛ := n ɛ. ( ) log ɛ but we have not really defined log yet. (b) b n = 6n + n + n +. Solution: With L = 3 we estimate b n L = 6n + n + n + 3 = 3n n + n + = 3n + n + n + < 5n n < 3 n < ɛ whenever We can therefore take (c) c n = n j= j = n n > 3 ɛ. n ɛ := + [ ] 3 ɛ Solution: We have 0 c n n < ɛ if n > ɛ. Thus we can take n ɛ = + (d) d n = 9n7 + 3n + n + 5n 7 34n + n Solution: We estimate d n by 9 5 d n = 9 + 3n 5 + n 6 + n n 6 + n n n n 6 if 5n 7 > 34n, for instance if n > 34. For such n, it then follows that d n n n = 5n n 6 500n 5 5(5 34n 6 ) 5 if n > We can therefore take ɛ { } 5 00 n ɛ = max 34,. ɛ., ɛ = 00 n 5 < ɛ

20 (e) e n = ( )n. n! Hint: Recall that n! = 3 (n ) n. Solution: We use a rough estimate for n! if n 4: n! = n = 4! 4 n 4. With L = 0, we get the estimate e n L = ( ) n n! = n n! n 4! 4 if n 4 and n >. Thus we can take ɛ n 4 = 4 { n ɛ = max 4, 8. Prove that the sequence (( ) k + k ) 4! n 4 4 = n 4 3 < n 3 n < n < ɛ } ɛ k N is bounded but does not converge. Find a convergent subsequence. Hint: Recall Definition 5.3 Solution: By the triangle inequality, ( )k + ( ) k + k = + k hence the sequence is bounded. If L were the limit of the sequence, then for any ɛ > 0 there would be k ɛ N such that for all k > k ɛ, ( )k + k L < ɛ. In order to prove this wrong, let ɛ = and let m, n > k 60 ɛ, m even, n odd. We would then have ( )m + m L = + m L < and 60 ( )n + n L = + n L < 60 hence, by the triangle inequality, ( + m ) ( L + ) n L + m L + + n L < 60 But the left hand side of this is ( + m ) L k ( + n L ) = a contradiction. This proves that the sequence does not converge. + m >, n For a convergent subsequence extract the even entries: Then ( ) ( b := = ( ) n + ) ( = + ) (n) (n) (n) n N n N n N = (a n ) n N is a subsequence of the given sequence a = ( ( ) k +. Clearly, b converges to. k )k N

21 9. Prove that the sequence e n cos(3n + 6) + n + i converges. Solution: First observe that i N,i n n N (.3) i N,i n i = [ n] i= = [ n] ([ n] + ) ( n + ) n = n + n and n( n ) = n n. Let a denote the sequence (.3). Then a n = e n cos(3n + 6) + n + i i N,i <n n + n + n n + n = + 4 n and hence n a n n n n n = a n + 4 n It follows that for n > n ɛ := 4 ɛ we have a n 4 n < ɛ which proves that lim a n =. n 0. For each of the following sequences, say whether they inrease, decrease, converge. Prove your answer. ( ) n +, (3 ) ( ), n + k v v n N Solution: The sequence ( ) n+ decreases: n+ n N and converges to : if n > ɛ. The sequence ( 3 k )k N increases: k N (n + ) + (n + ) + n + n + = n + 3 n + n + n + = 3 k+ n + n + = n + < ɛ (3 k ) v N (n + )(n + ) < 0 = k k+ = k+ > 0 and converges to 3: if k > ɛ. 3 3 k+ = < k k < ɛ

22 The sequence ( v v) v N decreases: (v + ) v + ( ) v v = v + v < 0 and diverges: To see this, assume the contrary, L = lim v v, i.e. for all ɛ > 0 there is m v ɛ N such that for all v > m ɛ, v v L < ɛ hence L ɛ < v v < L + ɛ, hence, since 0 < v, L ɛ < v < L + ɛ, equivalently L ɛ < v < L + ɛ +. (.4) However, by the Archimedean Property, there are no L, ɛ R such that(.4) holds for all v N.. Let (a n ) n N and (b n ) n N be convergent sequences of real numbers. Prove that the sequence converges. Hint: Look at Theorem 5.0. (max {a n, b n }) n N Solution: Let ɛ > 0 be given. Since a and b converge to A respectively B, say, there are n a, n b N so that we have a n A < ɛ for n > n a, Hence, for n > max { n a, n b}, b n B < ɛ for n > n b. A ɛ < a n < A + ɛ, B ɛ < b n < B + ɛ and therefore max {A, B} ɛ < max {a n, b n } < max {A, B} + ɛ. Find a convergent subsequence in ( n [ n ]) 3 3 n N. Hint: Recall definition 5.6 and look at example 39. Solution: Let r n = 3n. Then converges to 0. ( rn [ 3 rn ]) = (0) 3 n N n N 3. Prove that the function f : R R, f(x) = x + x is continuous at x 0 = 0. Thus, for arbitrary ɛ > 0 find δ ɛ satisfying (6.), i.e. so that x x 0 < δ ɛ implies that f(x) f(x 0 ) < ɛ. Hint: See example 36. Solution: For a given ɛ > 0 we want f(x) f(x 0 ) = x + x = x + x < ɛ. (.5) This is equivalent to x + x < ɛ. (.6)

23 If x then x x. Hence (.6) and thus (.5) are satisfied if { } x = x x 0 < δ ɛ := min, ɛ. Alternative: In order to satisfy (.6) we could also solve a quadratic equation: x + x < ɛ x < ɛ =: δ ɛ which gives the optimal δ ɛ 4. Prove that the function f : R R, f(x) = 7x + 3 is continuous. Thus, for arbitrary x 0 R and ɛ > 0 find δ x0,ɛ satisfying (6.4), i.e. so that x x 0 < δ x0,ɛ implies that f(x) f(x 0 ) < ɛ. Hint: See example 36. By the triangle inequality, for a, b R. Solution: For x, x 0 R we want By the triangle inequality, a b a b f(x) f(x 0 ) = 7x + 3 7x < ɛ (.7) 7x + 3 7x (7x + 3) (7x 0 + 3) = 7 x x 0. Thus (.7) holds if x x 0 < ɛ 7 =: δ x 0,ɛ. 5. Use the intermediate value theorem 6.5 to show that there are at least 4 real roots of the polynomial p(x) = x 8 3x 6 +. You may assume that the function p: R R, x x 8 3x 6 + is continuous. Solution: We compute p(0) = > 0, p(±) = < 0, p(±) = > 0. By the Intermediate Value Theorem, the function has zeros in 6. Prove that Solution: ( ( n n )) converges. n N (, ), (, 0), (0, ), (, ) ( lim n n ) 0 h = lim n h 0 h 7. Show that the square root function : R + 0 R is continuous. = d t = d e t ln() = ln(). dt t=0 dt t=0 Solution: For continuity at 0, note that for x 0, x = x 0 < ɛ = x = x 0 < ɛ. For continuity at a > 0, let 0 < ɛ < a. If x R + 0 satisfies x a < ɛ a i.e. 0 a ɛ a < x < a + ɛ a

24 then 0 a ɛ a ɛ a < x < a + ɛ a a + ɛ because of the monotonicity of the square root function and since ( a ɛ ) = a aɛ + ɛ a ɛ a because ɛ < a and a + ɛ a ( a + ɛ ) = a + aɛ + ɛ. In particular x a < ɛ. We thus have shown that for every a R + 0 and every ɛ > 0 there is δ ɛ (ɛ in case a = 0 and ɛ a in case a > 0) such that for x R + 0, x a < δ ɛ implies that x a < ɛ. 3 Some Problems for the Study week. By definition, what is an injective function? Solution: A function f : A B is injective if for all a, a A we have f(a) f(a ) whenever a a.. For a function f and a set B, state when a f is onto B. Solution: A function f is onto B if for each b B there is x in the domain of f such that f(x) = b. 3. Define an injective function f : Z N. Solution: For instance, the function f : Z N with { 4z + if z 0 f(z) := 4z + 3 if z < 0 is injective. 4. Define a surjective function g : R [0, 3). Solution: The function g : R [0, 3), is onto [0, 3). x x 3 g x 5. Let f : R R be the function with f(x) = x 5 x 3 x for all x R. Determine f ((, 0)), i.e. the inverse image of R = (, 0) under f. Solution: By definition, f ((, 0)) = {x f(x) < 0} = { x x 5 x 3 x < 0 } = { x x 5 < x 3 + x }. If x or x 0, then x 5 x 3 x, hence f(x) 0 and x is not in the set. If 0 < x < or x < then x 5 < x 3 < x, hence f(x) < 0, x is in the set. It follows that f ((, 0)) = (, ) (0, ).

25 6. Let f : R R be given by f(x) = x x +. Determine f([, ]) and f ((, ). Solution: We eliminate the absolute value: x (x + ) if x 0 f(x) = x x + = x (x + ) if x 0 x + (x + ) if x x if x 0 = 3x if x 0 x + if x Thus, for x [, ], we have f(x) = x, hence In order to determine f([, ]) = [0, ] f ((, ) = {x f(x) < } we need to solve f(x) < in the three cases x 0, x 0 and x. If x 0 then f(x) = x < if x <. If x 0 then f(x) 3x < if 3x <, i.e. x > /3 If x then f(x) = x + < if x > 0. The union of these gives f ((, ) = [0, ) ( /3, 0] = ( /3, ). x x /3 3 x 7. State when a subset A of R is bounded. Solution: A R is bounded if there is M R such that a < M for all a A. 8. Let A R, A, m R. State what m = max A means by definition. Solution: For A R, by definition, m = max A, if m A and m x for all x A. 9. Prove or provide a counterexample to the following: Let A, B be non empty bounded sets of real numbers. Then sup A B = max {sup A, sup B}. Solution: This is true. To prove this, let M := max {sup A, sup B}. Then M sup A and M sup B, hence M a for all a A and M b for all b B, hence M x for all x A B. This shows that M is an upper bound for the union A B. We have M = sup A or M = sup B. By definition of the supremum, for any N < M there is x A or x B, i.e. x A B with N < x M which shows that M is the least upper bound for A B. 0. Say whether minimum and infimum of the set {x R x 3 x > 0} exist and if so, find their values. Prove your result. Solution: If x then x 3 x, hence x {x R x 3 x > 0} =: X. Therefore is a lower bound for X. If < x < 0 then x 3 > x hence x X. This also shows that = inf X. Since X, the minimum of X does not exist.

26 . Prove or privide a counterexample to the following: If f : R R is continuous and = A R +, then f(inf A) = inf f(a). Solution: This is wrong. For a counterexample, let f : R R be given by f(x) = + x and A = [, ). Then f(a) = (0, /] and inf f(a) = 0 but f(inf A) = f() = /.. Let f : R R be the function given by f(x) = sgn(x [x] ). Determine the set of all x 0 such that f is not continuous at x 0. Solution: The function is not continuous at x 0 if and only if x 0 or x 0 is an integer. sgn(x [x] ) x [x] x 3. What is a subsequence of a sequence? Solution: A subsequence of the sequence (a n ) n N is a sequence (a rk ) k N where (r k ) k N is a strictly increasing sequence of natural numbers. 4. Let a be a sequence of real numbers and L R. By definition, what does lim n a n = L mean? Solution: lim n a n = L means that for every positive real number ɛ there is a natural number m such that for all n > m we have a n L < ɛ. 5. State the Monotone Convergence Theorem. Solution: A bounded monotone sequence converges. ( ) 6. Let (a k ) k N be an increasing sequence of positive real numbers. Prove that Solution: Since (a k ) k N increases, we have a n n N converges. whenever n > m a n a m ( ) hence decreases. Since a n > 0 for all n, we also have a a n > 0, in particular, n n N ( ) bounded. By the Monotone Convergence Theorem, converges [ ] n 5 7. Prove that the sequence n + 4n 7 5 n 4 + 7n 5 + 9n 7 n N a n converges. n N ( ) [ ] Solution: Abbreviate y = n n. Since 0 y [y] < for all real numbers y, we estimate 5 5 y + 4n 7 n 4 + 7n 5 + 9n = 9y + 36n7 4(n 4 + 7n 5 + 9n 7 ) 9(n 4 + 7n 5 + 9n 7 ) a n n N = 9y 4n4 8n 5 9(n 4 + 7n 5 + 9n 7 ) = 4n4 + 8n 5 9 9(n 4 + 7n 5 + 9n 7 ) is

27 3n5 8n n5 7 n = 7 n < ɛ if n > ɛ. Thus for any ɛ > 0 there is m := n ɛ such that for all n > m we have [ ] 5 n +4n 5 7 4/9 n 4 +7n 5 +9n 7 < ɛ. ( ) k k + k 8. Prove that the sequence a = converges. 3k k k N k Solution: We show that lim k +k k =. To this end, let ɛ > 0 be given. Then 3k k 3 a k 3 = k k + k ) k 3k k 3 = k + k k k = ( + kkk 3k k 3 < ɛ if + k k k < 3ɛ + that is if equivalently k k k < (3ɛ + ) k k k > (3ɛ + ). Since k k /k > k for all k N, this holds whenever k > + (3ɛ + ). Thus we have shown that for any ɛ > 0, there is k ɛ R, namely k ɛ := + k > k ɛ we have ak 3 < ɛ. 9. Show that there is a sequence (b n ) n N so that n N : b n N and b n+ > b n, and so that ( ) sin( bm ) + b3 m b 6 m + 5 m N converges. Solution: Since sin(x) [, ] for all x R, we have x n := sin( n ) + n3 n 6 + 5, (3ɛ+), such that for all the sequence (x n ) n N is bounded. By the Bolzano-Weierstrass Theorem, (x n ) n N has a convergent subsequence, i.e. there is an increasing sequence (b n ) n N of natural numbers so that ( x bj )j N converges as required. 0. By definition, when is a function f : A R, A R, continuous? Solution: For A R a function f : A R is continuous if for every p A and every ɛ > 0 there is δ > 0 so that for all x A with x p < δ we have f(a) f(p) < ɛ.. State the Intermediate Value Theorem. Solution: Let f : [a, b] R be a continuous function, a, b R, a b. [f(a), f(b)] or y [f(b), f(a)]. Then there is x [a, b] with f(x) = y. Also assume that y

28 . Prove or provide a counterexample for the following: Let f : R R be continuous and assume that f(0) > 0. Then there is u > 0 with f(u) > 0. Solution: This is true: To prove this, let ɛ := f(0)/. By continuity of f at 0, there is δ > 0 so that for all u with u < δ, in particular for u = δ/, we have f(u) f(0) < ɛ = f(0)/ hence, by the triangle inequality, f(u) > f(0)/ > Let f : R R be the function with f(x) = 4 x 3 x. Find six pairwise disjoint intervals each containing at least one zero of f. Hint: The function is even. Solution: f is continuous. In order to apply the Intermediate Value Theorem, we compute some values of f and look for sign changes. Since f is even, we do so for positive arguments first. We have f(0) = < 0, f(/) = / > 0, f() =, f(999) > 0. Therefore f changes sign in the intervals (0, /), (/, ), (, 999) and, since f is even, f also changes sign on the intervals ( /, 0), (, /), ( 999, ). By the Intermediate Value Theorem, f has a zero in each of these six intervals. f x 4. Let f : R R be the function with f(x) = (x ) 4x +. Find four pairwise disjoint intervals each of which contains at least one zero of f. You may assume without proof that x (x) is continuous. Solution: Since f is continuous, we can apply the Intermediate Value Theorem, looking for sign changes. Since f is even (i.e. f(x) = f( x)) we need to find only sign changes on the positive real axis. Then there will be another two on the negative. We compute some values of f: f(0) = + / > 0, f() = 4 +./ < 0, f() = / = / > 0. Thus f changes sign in the intervals and by symmetry also in the intervals (0, ), (, ) (, 0), (, ). By the Intermediate Value Theorem, f has at least one zero in each of these four intervals.

29 5. Prove that the function f : R R with f(x) = 3 x is continuous at 8. Solution: For given ɛ > 0 f(x) f() = 3 x < ɛ is equivalent to ( ɛ) 3 = 8 4ɛ + ɛ ɛ 3 < x < ( + ɛ) 3 = 8 + 4ɛ + ɛ + ɛ 3 i.e. If ɛ <, then ɛ, ɛ 3 < ɛ and 4ɛ + ɛ ɛ 3 < x 8 < 4ɛ + ɛ + ɛ 3 (3.) 4ɛ + ɛ ɛ 3 < ɛ and ɛ < 4ɛ + ɛ + ɛ 3. In particular (3.) holds if Thus, with ɛ < x 8 < ɛ equivalently x 8 < ɛ δ ɛ := min {ɛ, } we have shown that x 8 < δ ɛ implies that f(x) f(8) < ɛ. f is therefore continuous at Let g : R R be the function such that g(/n) = for all n N and g(x) = 0 if there is no n N with x = /n. Prove that g is not continuous at 0. Solution: We show that for ɛ = / there is no δ > 0 such x 0 = x < δ implies that g(x) g(0) = g(x) < ɛ = /. Assume otherwise, i.e. there were such a δ > 0. Then by the Archimedean Property of R, we can find n N with n >, hence 0 < /n < δ. Clearly, g(/n) =, hence δ despite /n 0 = /n < δ. 7. What is a function? g(/n) g(0) ɛ = Solution: A function is a set f of pairs such that if (a, b) f and (a, c) f, then a = c. 8. Let f : R R be the function with f(x) = x + cos(x) for all x R. Is f injective? Prove your answer. Solution: f is not injective, f() = f( ) 9. For which α R is the function f α : R R with f α (x) = α x + 5x injective? Prove your answer. Solution: For x R, { (5 + α)x if x 0 f α (x) = (5 α)x if x 0 If α > 5, then < 0 < but f ( ) ( 5 α 5+α α 5 α = = fα 5+α). If α = 5, then f α (0) = f α ( ) = 0. If 5 < α < 5 then f α (x) > f α (y) whenever x > y. If α = 5, then f α (0) = f α () = 0. If α < 5, then < 0 < but f ( ) ( 5 α 5+α α 5 α = = fα 5+α). Thus f α is injective if and only if α < Prove that for all x R, [x] := sup {z Z z x} exists. 3. Prove that for all x R we have x [[x], [x] + ).

30 ( ) k 3. Prove that the sequence k + k + k Solution: We will show that lim k holds if Since the estimate (3.) holds if k+ k k + k + k 6 = k k N k + k 6k 6 k + converges. = 6. To this end, let ɛ > 0 be given. Then k + k = 6 k k + k k + k + k < ɛ. k + k k k + k k + k + k < ɛ 6. (3.) k + k k + k k = k + k k + k k = + k k < ɛ 6 This, in turn, holds if equivalently + k k < ( ɛ 3 + ) ( ) 3 ( ɛ + ) < k Prove that sup { x x 4 < 6 } =. Solution: to show Start with the definition, in this case of sup, say as least upper bound. Thus you need (a) is an upper bound for {x x 4 < 6}, i.e. u > u {x x 4 < 6}, and (b) if b is an upper bound for {x bound for {x x 4 < 6}. x 4 < 6} then b, equivalently, if s < then s is not an upper To see the first claim, let u R. We have shown in class that if a, b, c R, c P, a > b then ca > cb. Thus, if u >, we have u, > 0, and applying this rule 4 times we get which shows u {x x 4 < 6}. u 4 = u u 3 > u 3 > u > u > = 6 (3.3) In order to show the second claim, let s R, s <. If s 0 then s is not an upper bound because 4 = < 6, hence s < {x x 4 < 6}. Let d = +s. We have and s < d < because s = s + s < s + = d < + = s 4 < s 3 d < d 4 < < 4 = 6 by the same argument as for (3.3), hence s < d but d {x for {x x 4 < 6}. x 4 < 6}, hence s is not an upper bound

31 34. Consider the set Y = {x R x 4 x + 34 } < 0 Find max Y, min Y, inf Y, sup Y if they exist. Also determine the minimal interval I with Y I. Solution: For x R, x 4 x = (x ) 4 < 0 if and only if < x < equivalently + = < x < + = 3 that is 3 < x < or < x < 3. Therefore {x R x 4 x + 34 } ( ) ( 3 < 0 =,, ) ( 3 3, ) 3 } {{ } I and I is minimal with Y I. It also follows that min Y and max Y do not exist, and that 3 3 inf Y = and sup Y =. 35. Prove that if B A R, A bounded, sup B sup A, then sup A = sup A \ B. Solution: Since A \ B A, any upper bound for A is an upper bound for A \ B. In particular, sup A is an upper bound for A \ B. We now show that sup A is the least upper bound for A \ B. Thus we need to show that for every c R, c < sup A there is x A \ B such that c < x. To see this, let c R, c < sup A be arbitrary. Since B A, we must have sup B sup A. By assumption the two suprema are different, hence sup B < sup A. Thus we have c = max {c, sup B} < sup A, c is not an upper bound for A and there is x A with c < x. But then sup B < x and since sup B is an upper bound for B, x B. It follows that x A \ B and c c < x. 36. Prove that the square function f : R R, x x, is continuous. Solution: For x, x 0 R and given ɛ > 0 we want f(x) f(x 0 ) = x x 0 = (x + x0 )(x x 0 ) < ɛ (3.4) Now if x x 0 <, then by the triangle inequality and if additionally x x 0 < x + x 0 = x 0 + (x x 0 ) x 0 + (x + x 0 )(x x 0 ) < ( x 0 + ) x x 0 < ɛ ɛ x 0. We define + Then if x x 0 < δ x0,ɛ we have (3.4). { δ x0,ɛ := min, } ɛ x 0 +

32 37. Prove that the square root function R + 0 R, x x, is continuous. Solution: For ɛ > 0 and x 0 > 0, we have x x 0 < ɛ (3.5) if since 0 < x + x 0, because x x 0 < δ ɛ,x 0 := ɛ x 0 < ɛ( x + x 0 ) (3.6) ( x + x0 ) ( x x0 ) = x x0. For continuity at x 0 = 0, simply observe that in this case x x 0 = x < ɛ if x < δ ɛ,0 := ɛ. (3.7) In all cases, we have shown that for any ɛ > 0 and x 0 0, we have x x 0 < ɛ, (3.5), if x x 0 < δ ɛ,x0, where δ ɛ,x0 is given in (3.6) for x 0 > 0 and (3.7) for x 0 = Let a < b, u < v and f : (a, b) (u, v) be increasing and onto (u, v). Prove that f is continuous. Hint: By definition, f is increasing if and only if and f is onto (u, v) means that x, x (a, b), x x : f(x) f(x ) w (u, v) x (a, b) : f(x) = y. Proof: We will show continuity at an arbitrary x 0 (a, b). To this end, let ɛ > 0 be given. Without loss of generality, we can assume u + ɛ < f(x 0 ) < v ɛ. Then, by surjectivity, i.e. there are α, β (a, b), so that and since f increases, we must have u < f(x 0 ) ɛ }{{} =f(α) f(x 0 ) ± ɛ f((a, b)) < f(x 0 ) < f(x 0 ) + ɛ < v }{{} =f(β) a < α < x 0 < β < b. In particular, 0 < min {β x 0, x 0 α} and, again since f increases, we must have 39. Consider the sequence a defined recursively by f(x) f(x 0 ) < ɛ if x x 0 < min {β x 0, x 0 α} a := 0, a := and for n, a n+ := For any L N 0 find a subsequence that converges to L. Solution: The first elements of the sequence are 0,, 0,,, 0,,, 3, 0,,, 3, 4, 0,,,... { an + if a n max {a j j < n} 0 if a n > max {a j j < n}

33 We have a n = 0 if and only if n is the sum of the first k natural numbers for some k, i.e. n k = k j = j= k(k + ) for some k. We have a n = L N if and only if n = L + n k for some k L. Thus a subsequence of a convergent to L (because it is constant) is (a rk ) k N where r k = n k+l + L = 40. Let f : R R be the function with (k + L )(k + L) + L = (k + L) k + L f(x) = { x + if x Q 0 if x Q. Prove that f is nowhere continuous, i.e. there is no x 0 R such that f is continuous at x 0. Solution: Let x 0 R be arbitrary. We show that f is not continuous at x 0. To this end, let ɛ = f(x 0) = ( + x 0). Let δ > 0 be given. We show that there is x R with x x 0 < δ but f(x) f(x 0 ) > ɛ. To see this, let n N be so that n > δ and q = [nx 0] Z. Then If x 0 Q then x 0 δ < q n x 0 q + / n f(q/n) f(x 0 ) = q + n < x 0 + δ. 4. Prove that the function f : R R with f(x) = x + if x 0 x if x = 0 is continuous at 0. Solution: Let ɛ > 0 be given. Then ɛ > f(x) f(0) = x if x 0 x + 0 if x = 0 (3.8) if x = 0. For x 0, note that < x + < x + because < x + < ( x + ) = + 4 x + 4x. For x 0, (3.8) holds if x + ɛ > x = ( x + ) ( x + ) x + = + x +

34 This in turn holds if or, if which follows from = + x + + x + > ɛ ɛ > + x + x < ( ɛ ) 4. Prove the following characterisation of the supremum of a nonempty set of real numbers: A real number a is the supremum of a nonempty set A of real numbers if and only if (a) a is an upper bound for A and (b) there is a sequence of elements of A converging to a. Solution: Let = A R and a R. We will write x A if x y for all y A. Then Definition 4.6 says that a = sup A if and only if x a is equivalent to x A. First assume that a = sup A. From Definition 4.6 we thus have that a is an upper bound for A and that for each n N, there is a n A with a < a n n. Thus we have a < a n n a, hence n (a n ) n N a. Now assume a A and let a n A, n N, be so that lim n a n = a. Now if x a then x A. If x < a then ɛ := a x > 0 and since lim n a n = a, there is n ɛ so that a n a = a a n < a x for all n > n ɛ. By the triangle inequality, we have in particular that x < a n for n > n ɛ and therefore x A. 43. Let a 0 A R and f : A R. Prove that the following are equivalent: (a) f is continuous at a 0. (b) Whenever (a n ) n N is a sequence with a n A for all n N and lim n a n = a 0, then lim n f(a n ) = f(a 0 ). Solution: Let a 0 A R and f : A R. We first assume 43a and show 43b. To this end, let (a n ) n N A N and lim n a n = a 0. Let ɛ > 0 be given. By continuity of f at a 0, there is δ ɛ such that for x A, the implication x a 0 < δ ɛ = f(x) f(a 0 ) < ɛ (3.9) holds. By the defintion of convergence, there is n δɛ N so that the implication holds. Putting (3.9) and (3.0) together gives n > n δɛ = a n a 0 < δ ɛ (3.0) n > n δɛ = f(a n ) f(a 0 ) < ɛ. Thus we have proved that (f(a n )) n N converges to f(a 0 ). We prove the converse, 43b 43a, by contradiction. Thus we assume that f is not continuous at a 0, i.e. ɛ > 0 δ > 0 x δ A : x δ a 0 and f(x δ ) f(a 0 ) ɛ. In particular, for each n N, we can take δ = /n and let a n = x /n A be so that a n a 0 < n and f(a n) f(a 0 ) ɛ. Thus we have lim n a n = a 0 but the sequence (f(a n )) n N does not converge to f(a 0 ), contradicting 43a.

35 44. Let g : R R be a continuous function with g(0) = 0. Let M > 0 and f : R [ M, M] be any function (not necessarily continuous). Prove that the product function gf : R R, x g(x)f(x), is continuous at 0. Solution: Let ɛ > 0 be given. Now for all x, since f(x) < M, the estimate holds if f(x)g(x) < ɛ (3.) g(x) < ɛ/m. (3.) By continuity of g, there is δ = δ g ɛ/m > 0 such that for all x R with x < δg ɛ/m equation (3.) and therefore (3.) hold. This proves continuity of fg at 0. 4 Real Numbers 4. Order properties Addition and Multiplication of real numbers satsify the same rules as that of rational numbers. For a complete list of axioms, see the appendix 7. Here we only take a closer look at the axioms of order and completenes. The order on R is definded through a subset ( positive numbers ) such that P R \ {0} R = P {0} P, P P = Also the set P is required to be closed under addition and multiplication, P + P P P P Definition 4. We write x y, equivalently y x, if and only if x y P {0}. Similarly, x > y, equivalently y < x are defined to mean x y P. Theorem 4. defines a total order on R, i.e. for all x, y, z R we have x y or y x x y and y x = x = y x y and y z = x z We define the absolute value of x R by x if x > 0 x := 0 if x = 0 x if x < 0.