Analysis MT432C Notes/Problems/Homework

Recommended Reading:

R. G. Bartle, D. R. Sherbert Introduction to real analysis, principal referenceM. Spivak CalculusW. Rudin Principles of mathematical analysisS. Abbott Understanding analysisK. R. Stromberg Introduction to Classical Real AnalysisT. W. Korner A Companion to Analysis

1 Homework

1. This problem is about set operations. Recall the notation:

• N = {1, 2, 3, 4, 5, 6, . . .} the set of natural numbers

• Z = −N ∪ {0} ∪ N = {. . . ,−3,−2,−1, 0, 1, 2, 3, . . .} the set of integers

• ∪ union of two sets. {1, 2, 3} ∪ {2, 3, 4, 5} = {1, 2, 3, 4, 5}• ∩ intersection. {1, 2, 3} ∪ {2, 3, 4, 5} = {2, 3}• \ difference. {1, 2, 3} \ {2, 3, 4, 5} = {1}• ⊂ 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 {1, 2, 3} ⊂ {1, 2, 3, 4, 5, 7} ⊂ N ⊂ Z but {−1, 0, 1} 6⊂ N• × Cartesian product.{1, 2, 3}×{2, 3, 4, 5} = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 2), (3, 3), (3, 4), (3, 5)}• # cardinality, number of elements, of a set. # {2, 3, 8, 88} = 4

• {x P (x)} set of all x satisfying the property P (x).{u u ∈ N, 5|n} = {5, 10, 15, 20, . . .} 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) withx ∈M .{5k k ∈ N} = {5, 10, 15, 20, . . .} set of all natural numbers divisible by 5.{x2 x ∈ Z} = {. . . , (−2)2, (−1)2, 02, 12, 22, . . .} = {. . . , 4, 1, 0, 1, 4, . . .} = {0, 1, 4, . . .} set ofperfect squares.

Write the following sets by listing their elements. Thus, for example,

({11, 22, 33} ∪ {20, 21, 22, 23}) \ ({23, 11, 5} ∩ {n ∈ N n > 20})

= {11, 20, 21, 22, 23, 33} \ {23} = {11, 20, 21, 22, 33}

(a) ({1, 2, 3, 4} ∩ {2k k ∈ N})× {6, 7} ***

Solution: = {2, 4} × {6, 7} = {(2, 6), (2, 7), (4, 6), (4, 7)}(b) {2k k ∈ N} ∩ {3k k ∈ N} ∩ {n ∈ N n < 20} ***

Solution: = {6, 12, 18}(c) {#(A ∩B) A,B ⊂ {n ∈ N n < 20} ,#A = #B = 11}

Solution: = {3, 4, 5, 6, 7, 8, 9, 10, 11}(d) {(x, y) x, y ∈ Z, x2 + y2 < 11, 0 < x < y}

Solution: = {(1, 2), (1, 3)}

2. Which of the following statements 2a-2d do not hold for all sets A,B,C? If the statement holds forall A,B,C, show the set in the Venn diagram. If the statement does not hold for all A,B,C findsuch 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)” doesnot always hold. To see this, look at the Venn Diagram for three sets and then pick sets that thedifference is not empty. In this example the sets differ by A \ C. Thus, if you take A = {∅} (or anyother set with one element) and B = C = ∅, we get (A ∪B) ∩ C = ∅, but A ∪ (B ∩ C) = {∅} 6= ∅.

(a) (A \B) ∪ (B \ A) = (A ∪B) \ (A ∩B)

Solution: true

(b) A ∪ (C \ (B \ A)) = (A ∪ C) \BSolution: 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 holdsfor all A,B,C, show the set in the diagram below. If the statement does not hold for all A,B,C,Dfind such sets as small as possible violating the statement (a small counterexample).

Hint: Look at a Venn-type diagram for products,

A C

B

D

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)

= (A ∪ C)× (B ∪D)

counterexample A = D = {∅}, B = C = ∅.(b) (A×B) ∩ (C ×D) = (A ∩ C)× (B ∩D)

Solution: true

(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))

⊃ (A \ C)× (B \D)

counterexample A = B = D = {∅}, C = ∅(d) A× (B \D) = (A×B) \ (A×D)

Solution: true

(e) (A \ C)× (B \D) = ((A×B) \ (C ×B)) \ (A×D)

Solution: true

Please hand up your solutions to 1-3 on Friday, 12/2.

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 ∈ Bthere 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 fi : Ai → Bi, i = 4a . . . 4d, say whether fi is injective, onto Bi, abijection of Ai with Bi.

If fi is a bijection of Ai with Bi find a formula for the inverse f−1i : Bi → Ai.

If fi is not injective, find x1, x2 ∈ Ai, x1 6= x2 so that f(x1) = f(x2)

If fi is not onto Bi, find an element in Bi \ image (fi).

(a) f4a : A4a = Z→ B4a = N, f4a : x 7→ x2 + x4.

Solution: The map f4a is not injective and not onto N: f4a(−1) = f4a(1), 3 6∈ image (f4a).

(b) f4b : A4b = Z× N→ B4b = Q, f4b : (x, y) 7→ xy

Solution: The map f4b is onto Q but not injective. f4b(4, 2) = 42

= 2 = f4b(8, 4).

(c) f4c : A4c = R→ B4c = [1,∞) = {x ∈ R ≥ 1}, f4c : x 7→ e√ex

Solution: This map is injective and not onto B4c, there is no x ∈ R with f4c(x) = 1 ∈ B4c

The map is a bijection of R with (1,∞) = B4c \ {1}. The inverse is given by

f−14c (y) = ln

((ln(y))2

)= ln (2 ln(y)) .

(d) f4d : A4d = R→ B4d = R, f4d : x 7→ x3 − xSolution: This function f4d is onto R but not injective: f4d(0) = f4d(1) = 0.

5. Let f = {(x, y) x, y ∈ R and x2 + y4 = 16 and y ≥ 0} ***

(a) What is f(2)?

Solution: 4√

12

(b) Draw f .

Solution:

x

y

(−4, 0)

domain of f

(4, 0)(0, 0)

image of f

(0, 2)

(c) Domain and range of f are intervals. Find them.

Solution: D (f) = [−4, 4], image (f) = [0, 2].

6. Let A,B be sets and f : A→ B and g : B → A be functions. Assume that the composition g ◦ f isthe identity on A. Prove that f is injective.

Hint: For A a set, the identity idA : A→ A is the function with idA(a) = a for all a ∈ A. Thus

idA = {(a, a) a ∈ A} .

The composition v ◦ u of two functions u and v is

v ◦ u = {(x, y) ∃z : (x, z) ∈ u and (z, y) ∈ v} .

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)) = idA(x) = x, g(f(y)) = idA(y) = y. Hencex = y.

7. Consider the functions f, g,

Af−→ B

g−→ C

Show that image (g ◦ f) = g(image (f)).

Solution:

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, 26/2.

8. Prove from the axioms R1-R11, the definitions DFR1-5 and the lemmas L1-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 secondcase, again by L4, we have b = 0 or c = 0.

(b) If x ∈ P , then 1x∈ P .

Solution: Let x ∈ P . By R10, x 6= 0, hence 1x

exists. By R10, 1x∈ P or 1

x= 0 or 1

x∈ P .

Now if 1x

= 0 then 1 = 1x· x = 0 by bu DFR-2 and L1, contradicting 1 ∈ R \ 0 from R6.

If 1x∈ −P , then by L7, − 1

x∈ P hence, by R5 and R8, −1 = x ·

(− 1x

)∈ P , hence 1 ∈ −P . By

R10 this contradicts 1 ∈ P from L6.

The only remaining case by R10 is 1x∈ P .

9. If X is a real number we write

X+ =

{X if X ≥ 00 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 thatM ≥ A (M ≤ A), in which case we write M = maxA (M = minA).

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 Xi in x, y, z ∈ R in terms of thefunctions given with it (and arithmetic operations).

(a) (Example) X9a(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|

2

(b) (Example) X9b(x) = x− in terms of x 7→ x+:

x− = x+ − x

(c) X9c(x) = |x| in terms of x+, x− ***

Solution: |x| = x+ + x−

(d) X9d(x) = |x| in terms of x 7→ x+.

Solution: X9d(x) = |x| = 2x+ − x(e) X9e(x) = min {|x− k| k ∈ Z} in terms of [·], 〈·〉, |·|

Solution: X9e(x) =∣∣⟨x− 1

2

⟩− 1

2

∣∣(f) X9f (x) = sgn(sin(x)) in terms of sgn, [·], 〈·〉, |·|

Solution:

X9f (x) = sgn(sin(x)) = sgn

(∣∣∣∣⟨ x

2π− 1

4

⟩− 1

2

∣∣∣∣− 1

4

)10. 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 involvingpowers and binomial coefficients. Derive this.

Hint: Try decompositions in two sets, i.e. try to count

Y2 = {(A,B) A,B ⊂ X,X = A ∪B, A ∩B = ∅}

This is in bijection with the set of all subsets of X, because if (A,B) ∈ Y2, then B = X \ A.

Solution: There is a bijection

YΦ−→ {1, 2, 3}X = {f : X → {1, 2, 3}}

given by assigning to (A,B,C) ∈ Y the function

Φ(A,B,C) : X → {1, 2, 3} with

[Φ(A,B,C)] (x) =

1 if x ∈ A2 if x ∈ B3 if x ∈ C

To see that Φ is a bijection, look at the function

Ψ: {1, 2, 3}X → Y

given byΨ(f) = (f(1), f(2), f(3))

Clearly, Ψ ◦ Φ = idY and Φ ◦Ψ = id{1,2,3}X It follows that

#Y = # {1, 2, 3}X = 3#X .

We could also have counted Y using binomial coefficients. Recall that for n ∈ N0, “n choose k”,(nk

)= # {A ⊂ {1, . . . , n} #A = k} =

n!

k!(n− k)!

is the number of k-element subsets of an n-element set.

If X has n elements, then for each k, 0 ≤ k ≤ n we have

(nk

)choices for A so that #A = k. Fixing

A and l, 0 ≤ l ≤ n− k leaves

(n− kl

)choices for B so that #B = l. For C we then have no further

choices because C = X \ A \B. Summing up all these choices, we get

3n = #Y =n∑k=0

(nk

) n−k∑l=0

(n− kl

)

11. Sketch the following subsets of R2:

(a) {(x, y) ∈ R2 x > [y]}

Solution:

x

y

{([y] , y) y ∈ R} {(x, y) ∈ R2 x > [y]}

(b) {(x, y) ∈ R2 |x|p + |y|p < 1} for p = 1/2, 1, 2, 4.

Solution: These sets are the interiors of the contours in the picture below

x

y

p = 1/2

p = 1p = 2

p = 4

(c) {(x, y) ∈ R2 max {|x− 1| , |2y − 6|} < 4}Solution: We can rewrite this as{

(x, y) ∈ R2 max {|x− 1| , |2y − 6|} < 4}

={

(x, y) ∈ R2 |x− 1| < 4 and |y − 3| < 2}

= (−3, 5)× (1, 5) ,

the interior of the rectangle below.

x

x

−3 1 5

1

3

5

Please hand up your solutions to 8-11 on Friday, 11/3.

12. Prove from the axioms R1-R11, the definitions DFR1-5, the lemmas L1-L8 from the Appendix andthe axioms for the natural numbers (in particular induction):

(a) If x, y are positive real numbers with x < y and n ∈ N, then xn < yn.

Hint: Here xn is defined recursively: x1 := 1, xn+1 = x · xn for n ∈ N.

Solution: Let Y (n) be the statement “ 0 < x < y =⇒ xn < yn ”. We prove Y (n) for alln by induction:

For Y (1) is trivial.

Assume n ∈ N is so that Y (n) holds. Let x, y ∈ R, 0 < x < y. By Y (n) we then have xn < yn.Since 0 < x we have from L8 and the recursive definition of xn+1 and yn+1 that

xn+1 = x · xn < x · yn and x · yn < y · yn = yn+1 .

Since < is transitive (L7), this implies

xn+1 < yn+1 .

(b) If n ∈ N and x, y ∈ R thenxn+1 − yn+1 = (x− y) · pn(x, y)

where pn(x, y) is a polynomial of degree n in x and y. Find a formula for pn.

Solution: pn(x, y) =∑

k+l=n xkyl

13. For each of the following sets Xi, i = 13c . . . 13f , find the supremum supXi and an expression xi(ε)such that for any ε > 0, xi(ε) ∈ Xi is as required in the definiton of the supremum, i.e.

supXi − ε < xi(ε) ≤ supXi .

(a) (Example) X13a ={− 1n2 n ∈ N

}. The supremum is

supX13a = 0 and x13a(ε) = − 1[1 + 1√

ε

]2

(b) (Example) X13b =

{1

3k1+ · · ·+ 1

3knn ∈ N, k1, . . . kn ∈ N, 0 < k1 < k2 < . . . < kn

}. The

supremum is

supX13b =1

2

Consider the elements xn of X13b given by

xn =n∑j=1

1

3j=

1

2

(1− 1

3n

)∈ X13b .

We have1

2> xn =

1

2− 1

2 · 3n>

1

2− ε

if

3n >1

2ε.

Since 3n > n for all n ∈ N, this last estimate holds if n > 1ε. We can therefore use n = 1 +

[1ε

]and get

x13b(ε) = x1+[ 1ε ]=

1

2

(1− 1

31+[ 1ε ]

)(c) X13c = (−4,−3) ***

Solution: supX13c = −3, x13c(ε) = max{−3− ε

2,−31

2

}(d) X13d = {x ∈ R x3 = x}. ***

Solution: X13d = {−1, 0, 1} hence supX13d = 1 = maxX13d and we can take x13d(ε) = 1.

(e) X13e =

{uv

u2 + v2u, v ∈ N, u < v

}.

Solution: Since 0 ≤ (u− v)2 = u2 + v2 − 2uv, we have uv < u2+v2

2for all u, v ∈ R. Hence 1

2is

an upper bound for X13e. We claim that this is sharp i.e. supX13e = 12. To see this, let ε > 0.

Then, for u ∈ N,

X13e 3 xu :=u(u+ 1)

u2 + (u+ 1)2=

u2 + u

2u2 + 2u+ 1=

1

2−

12

2u2 + 2u+ 1>

1

2− 1

4u2>

1

2− ε

if u > 12√ε. Hence we can take x13e(ε) = x

1+[

12√ε

]

(f) X13f =

{n∑i=1

n

n2 + i2n ∈ N

}Hint:

n

n2 + i2=

1n

1 + i2

n2

arctan(1) =

∫ 1

0

dx

1 + x2.

Solution:

Ln =n∑i=1

1

n× 1

1 + i2

n2

=n∑i=1

n

n2 + i2

is a lower sum for∫ 1

01

1+x2dx = arctan(1) = π/4, hence π/4 is an upper bound for X13f . This

is also sharp because an upper sum for the same integral is

Un =n−1∑i=0

n

n2 + i2

In particular,

Un ≥∫ 1

0

1

1 + x2dx = π/4 ≥ Ln

and since

Un − Ln =n−1∑i=0

n

n2 + i2−

n∑i=1

n

n2 + i2=

1

n− 1

2n=

1

2n< ε

if n > 12ε

, we can take

x13f (ε) = L1+[ 12ε ]

=

1+[ 12ε ]∑

i=1

1 +[

12ε

](1 +

[12ε

])2+ i2

.

Please hand up your solutions to 12-13 on Friday, 8/4

14. Which of the sequences below converge? Prove your answer.

(a)

(7k + 3 + 1

k

2k + 5k

)k∈N

Solution: We first rewrite the fraction,

7k + 3 + 1k

2k + 5k

=7 + 3

k+ 1

k2

2 + 5k2

and then apply the theorem about the limit of a quotient,

limk→∞

7 + 3k

+ 1k2

2 + 5k2

=limk→∞

(7 + 3

k+ 1

k2

)limk→∞

(2 + 5

k2

) =limk→∞ 7 + limk→∞

3k

+ limk→∞1k2

limk→∞ 2 + limk→∞5k2

=7 + 0 + 0

2 + 0=

7

2.

since the limit in the denominator is 2 6= 0.

(b)(m

5−[m

5

])m∈N

(c)

(m∏q=1

(1− 1

(q + 1)2

))m∈N

Hint: If n,m ∈ N, N ≤ m then

m∏q=n

aq = an × an+1 × · · · × am .

Thus the first terms of this sequence are

1− 1

4,

(1− 1

4

)(1− 1

9

),

(1− 1

4

)(1− 1

9

)(1− 1

25

), · · ·

Solution: Abbreviate

am =m∏q=1

(1− 1

(m+ 2)2

).

Thus

am+1 = am ×(

1− 1

(q + 1)2

)< am

Hence the sequence decreases. Since am > 0 for all m, 0 is a lower bound. By the monotonicconvergence theorem, the sequence converges.

15. For each of the following subsets of Ai ⊂ R, i = 15b . . . 15d determine the set A of cluster points ofAi:

(a) (example) The set of cluster points of A15a ={x x ∈ R, |x− [x]| < 1

4

}is A15a ={

x x ∈ R, |x− [x]| ≤ 14

}(b) (example) The set of cluster points of A15b =

{1m

+ 1n

n,m ∈ N}

is A15b = 1N ∪ {0}.

(c) A15c = Q ***

(d) A15d = Z ***

(e) A15e = R \(

1N ∪ {0}

)(f) A15f = R \ Z

16. Let f : R+ → R be a function so that x ≤ f(x) ≤ x+ x2 for all x ∈ R+. Prove that

limx→0

f(x)

x

exists.

17. Recall the definition of “cluster point”: A cluster point of a set A ⊂ R is a point x0 ∈ R such thatfor every ε > 0 there is aε ∈ A \ {x0} with |x− x0| < ε.

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 (an)n∈N ∈ (A \ {a})N with limn→∞

an = 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 supA are cluster points of A.

(d) A finite subset of R has no cluster points.

Please hand up your solutions to 14-17 on Friday, 22/4

2 Examples

1. 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 a ≤ L}” is true (We usecapitals 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) ({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 = 25) and ((K,M ∈ A,L ∈ S,K ≤ L ≤M)⇒ L ∈ A)}

Solution:= {S \ {z} , S \ {a}}

= {{a, . . . , y} , {b, . . . , z}}

2. Which of the following statements 2a-2d do not hold for all sets A,B,C? If the statement holds forall A,B,C, show the set in the Venn diagram. If the statement does not hold for all A,B,C findsuch 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)” doesnot always hold. To see this, look at the Venn Diagram for three sets and then pick sets that thedifference is not empty. In this example the sets differ by A \ C. Thus, if you take A = {∅} (or anyother set with one element) and B = C = ∅, we get (A ∪B) ∩ C = ∅, but A ∪ (B ∩ C) = {∅} 6= ∅.

(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 fi : Ai → Bi, i = 3a . . . 3d, say whether fi is injective, onto Bi, abijection of Ai with Bi.

If fi is a bijection of Ai with Bi find a formula for the inverse f−1i : Bi → Ai.

If fi is not injective, find x1, x2 ∈ Ai, x1 6= x2 so that f(x1) = f(x2)

If fi is not onto Bi, find an element in Bi \ image (fi).

(a) f3a : A3a = R+ → B3a = R, f3a : x 7→ e√x

Hint: R+ is the set of positive real numbers.

Solution: not onto R. The image of f3a is R(f3a) = (1,∞), hence 0 ∈ R \ image (f3a)

(b) f3b : A3b = N→ B3b = N, f3b : x 7→{

3x2

if x is evenx+5

2if x is odd

Solution: not onto B3b, 1 ∈ B3b \ image (f3b). The map is not injective, f3b(2) = 3 = f3b(1).

(c) f3c : A3c = Q→ B3c = N, f3c : x 7→ min{n n ∈ N and ∃z ∈ Z : x = z

n

}Hint: If ∅ 6= A ⊂ N then minA is the element a ∈ A such that a ≤ x for all x ∈ A. Thus, forinstance, min {34, 66, 3, 9} = 3.

Solution: onto B3c, not injective. f(

12

)= f

(32

)= 2.

(d) f3d : A3d = Z→ B3d = 2Z, f3d : x 7→ {x, x+ 1, x+ 2, x+ 3}.Hint: 2Z = {U U ⊂ Z} is the power set of Z, i.e. the set of all subsets of Z.

Solution: The map f3d is injective but not onto 2Z: ∅ ∈ B3d \ image (f3d).

4. For a real number x we write

[x] := max {k ∈ Z k ≤ x} and {x} = x− [x] (2.1)

for the integer part (or floor) and the fractional part of x. Also recall the definition of the absolutevalue. For a real number x,

|x| :={

x if x ≥ 0−x if x ≤ 0

.

Example:[2.3] = 2, {2.3} = 0.3, [−9/2] = −5, {−9/2} = 1/2. Consider the functions

sin : R→ [−1, 1] and arcsin : [−1, 1]→[−π

2,π

2

]We have seen in class that sin ◦ arcsin = id[−1,1] but arcsin ◦ sin 6= idR. Find a formula forarcsin(sin(x)) in terms of [·] and {·}, and the absolute value | · |.Solution: If x ∈

[−π

2, 3π

2

]we have two cases:

x ∈[−π

2,π

2

]: arcsin(sin(x)) = x

x ∈[π

2,3π

2

]: arcsin(sin(x)) =

π

2−(x− π

2

).

which we can put together:

x ∈[π

2,3π

2

]: arcsin(sin(x)) =

π

2−∣∣∣x− π

2

∣∣∣ .Since sin is 2π-periodic, i.e. sin(u) = sin(2π + u) for all u ∈ R, we can write any x ∈ R as

x = 2πk + x with k ∈ Z, x ∈[−π

2,3π

2

].

We can express x as

x = 2π

{x+ π

2

2π

}− π

2.

Thus for all x ∈ R,

arcsin(sin(x)) =π

2−∣∣∣x− π

2

∣∣∣ =π

2−∣∣∣∣(2π

{x+ π

2

2π

}− π

2

)− π

2

∣∣∣∣ =π

2−∣∣∣∣2π{x+ π

2

2π

}− π

∣∣∣∣ .

x

sin

range ofarcsin

domain ofarcsin

arcsin ◦ sin

5. How many elements do the following sets have? ***

(a) A5a = {(a, b, c) a, b, c ∈ Z and a2 + b2 + c2 < 5}Solution: There are many ways of counting this set. For instance, if (a, b, c) ∈ A5a thena, b, c = −2,−1, 0, 1, 2. If one of the a, b, c is equal to ±2 then the others must be 0. This gives3 × 2 = 6 possibilities. If none of the a, b, c is equal to ±2, i.e. they are all −1, 0 or 1, thenthere is no further restriction:

{−1, 0, 1}3 ⊂ A5a

which gives 33 further possibilities. It follows that

#A5a = 6 + 33 = 33 .

(b)⋃

n∈N,n<4

⋃m∈N,m<4

{x2 + y2 x, y ∈ N, 1 ≤ x ≤ m, 1 ≤ y ≤ n

}Solution: Let

X(m,n) :={x2 + y2 x, y ∈ N, 1 ≤ x ≤ m, 1 ≤ y ≤ n

}thus we have to count ⋃

n∈N,n<4

⋃m∈N,m<4

X(m,n) .

The key observation to simplify this is that

X(m,n) ⊂ X(k, l) whenever m ≤ k and n ≤ l .

It follows that ⋃n∈N,n<4

⋃m∈N,m<4

X(m,n) =⋃

n∈N,n<4

X(n, 3) = X(3, 3) .

Now

X(3, 3) ={x2 + y2 x, y ∈ N, 1 ≤ x ≤ 3, 1 ≤ y ≤ 3

}={x2 + y2 x, y ∈ N, 1 ≤ x ≤ y ≤ 3

}={

12 + 12, 12 + 22, 12 + 32, 22 + 22, 22 + 32, 32 + 32}

= {2, 5, 10, 8, 13, 18} ,

hence the set has 6 elements.

6. Recall the definitions 4.6 of maximum, minimum, supremum, infimum. For each of the followingsets Yi ⊂ R, i = 6a . . . 6e determine maxYi,minYi, supYi, inf Yi if they exist.

(a) Y6a ={

1n2+n

n ∈ N}

Solution: maxY6a = supY6a = 12, minY6a does not exist, inf Y6a = 0.

(b) Y6b ={

(−1)z + 11+z2

z ∈ Z}

Solution: maxY6b = supY6b = 2, minY6b does not exist, inf Y6b = −1.

(c) Y6c = {x2 + x3 x ∈ R, 0 ≤ x < 2}Solution: maxY6c does not exist, supY6c = 12, minY6c = inf Y6c = 0.

(d) Y6d ={nm

n,m ∈ N}

Solution: maxY6d, supY6d do not exist, minY6d does not exist, inf Y6d = 0.

(e) Y6e ={

5 + k√

2−[k√

2]

k ∈ Z}

Solution: maxY6e does not exist, supY6e = 6, minY6e does not exist, inf Y6e = 5.

7. Prove that for n ∈ N,n∑k=1

2k − 1 = n2

Solution: For n ∈ N let P (n) be the statement

P (n) :n∑k=1

2k − 1 = n2 .

Clearly, P (1) holds. We now show that for all n ∈ N, P (n) implies P (n+ 1). Thus, assuming P (n),we compute

(n+ 1)2 = 1 + 2n+ n2 = 1 + 2n+n∑k=1

2k − 1 = 2(n+ 1)− 1 +n∑k=1

2k − 1 =n+1∑k=1

2k − 1

which proves P (n+ 1).

8. Let a, b ∈ R. Prove that

max {a, b} =a+ b

2+

∣∣∣∣a− b2

∣∣∣∣ .What is the analogous formula for min {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

2+a− b

2=a+ b

2+

∣∣∣∣a− b2

∣∣∣∣ .Similarly, if b ≥ a,

max {a, b} = b =a+ b

2− a− b

2=a+ b

2+

∣∣∣∣a− b2

∣∣∣∣ .The analogous formula for min is

min {a, b} =a+ b

2−∣∣∣∣a− b2

∣∣∣∣ .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) {x ∈ R x3 − x > 0}Solution: max,min, sup do not exist, inf = {x ∈ R x3 − x > 0} = −1.

We plot the function R→ R, x 7→ x3 − x = x(x2 − 1),

{(x, x3 − x) x ∈ R}

{x ∈ R x3 − x > 0} = (−1, 0) ∪ (1,∞)

(b){x ∈ R x− [x] > 1

2

}Solution: max,min, sup, inf do not exist.

We plot the function R→ R, x 7→ x− [x]− 12, {

(x, x− [x]− 12) x ∈ R

}

{x ∈ R x− [x]− 1

2> 0

}=⋃s∈Z

(1

2+ s, 1 + s

)

(c) {x ∈ R ∃z ∈ {0, 1, 2, 3} : |x− 3z| < 1}Solution: max,min do not exist.

inf {x ∈ R ∃z ∈ {0, 1, 2, 3} : |x− 3z| < 1} = −1 ,

sup {x ∈ R ∃z ∈ {0, 1, 2, 3} : |x− 3z| < 1} = 10 ,

We can write this set as

{x ∈ R ∃z ∈ {0, 1, 2, 3} : |x− 3z| < 1} = {x ∈ R ∃z ∈ {0, 3, 6, 9} : |x− z| < 1}

that is the set of all real numbers having distance less than 1 from one of the points 0, 3, 6, 9.We first draw these points and then the intervals of length 2 around them:

0

(−1, 1)

3

(2, 4)

6

(5, 7)

9

(8, 10)

Hint: Look at Example 33.

10. 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, thenMN 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 = {−1}(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 Brespectively. Let y ∈ A+ B. Then by definition of A+ B, y of the form y = a+ b with a ∈ Aand 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 4.1, a ≤M , b ≤ N means that

M − a,N − b ∈ P

But since P + P ⊂ P by R-11, we have that

(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 .

11. For each of the following sets Yi ⊂ R, i = 11a . . . 11d say whether maxYi,minYi, supYi, inf Yi exist, ***and if so, find the value. Also find the minimal interval I so that Yi ⊂ I. Thus in each case, youhave to find and interval I so that I ⊃ Yi and J 6⊃ Yi whenever J is an interval with J ⊂ I.

Hint: See example 34

(a) Y11a = [−2, 3) ∪ NSolution: inf Y11a = minY11a = −2 exist, maxY11a and supY11a do not exist, the set has noupper bound. The minimal interval containing Y11a is I11a = [−2,∞)

(b) Y11b = (−3,−1)

Solution: inf Y11b = −3 and supY11b = −1 exist, maxY11b and minY11b do not exist. Theminimal interval containing Y11b is I11b = Y11b = (−3,−1)

(c) Y11c =⋂

u∈R,u>5

(−∞, u)

Solution: Y11c = (−∞, 5] = I11c, hence inf Y11c and minY11c do not exist, the set has no lowerbound. supY11c = maxY11c = 5 exist.

(d) Y11d =⋂

x∈R+,x<3

(−∞, 1

x

]Solution: Y11d =

⋂x∈R+,x<3

(−∞, 1

x

]=

⋂y∈R+,y>1/3

(−∞, y] =

(−∞, 1

3

]= I11d hence, similar to

problem 11c, inf Y11d and minY11d do not exist and supY11d = maxY11d = 13

12. Prove the following statement or provide a counterexample: ***

“If A is a nonempty bounded set of real numbers and ∅ 6= B ⊂ A with supB = supA, thensupA 6= sup (A \B).”

Hint: See example 35.

Solution: The statement is wrong. A counterexample is A = R− = (−∞, 0), B = −1/N ={−1n

n ∈ N}

. ThensupA = supB = supA \B = 0 .

13. For all n ∈ N let An be a set. Prove or provide a counterexample for the following: ***

∀n ∈ N : An+1 ⊂ An =⇒⋂n∈N

An 6= ∅ ?

Does this change if, in addition, the An are required to be closed intervals, i.e. intervals of the form

An = [an, bn] with an, bn ∈ R, an ≤ bn ?

Solution: Without further assumptions on the sets An, the statement is wrong: For a counterex-ample, let An = (0, 1/n). Then⋂

n∈N

An =

{x ∈ R ∀n ∈ N : 0 < x <

1

n

}= ∅

by the Archimedean Property of R.

However, if the An = [an, bn] are closed intervals with An+1 ⊂ An for all n, then⋂n∈NAn 6= ∅. To

see this, first observe that since An+1 ⊂ An for all n, we have estimates

a1 ≤ a2 ≤ a3 . . . ≤ an ≤ an+1 ≤ . . . ≤ bn+1 ≤ bn ≤ . . . ≤ b3 ≤ b2 ≤ b1

In particular, the sets A = {an n ∈ N} and B = {bn n ∈ N} are bounded. By completeness, supAand inf B exist. We have supA ≤ inf B and⋂

n∈N

An = [supA, inf B] 6= ∅ .

14. Guess the limits of the sequences ***(1

k

)k∈N

,(2−n)n∈N ,

(9q7 + 3q2 + q + 1

5q7 − 34q + 2−q

)q∈N

Solution: The limits of these sequences are 0, 0, 9/5 respectively.

15. If ∅ 6= A ⊂ R, A bounded above, s ∈ R, then supA is the real number such that

(i) supA is an upper bound for A and

(ii) ∀ε > 0∃aε ∈ A : supA− ε < aε

For each of the following sets Ai ⊂ R, i = 15b . . . 15d find supAi and a formula expressing aε interms of ε such that 15ii holds.

(a) For example, if A15a =

{3− 1

nn ∈ N

}then supA15a = 3 and

aε = 3− 1

2 +[

1ε

]satisfies 15ii. To see this, first note that

2 +

[1

ε

]∈ N hence 3− 1

2 +[

1ε

] ∈ A15a

for all ε > 0. Then, as2 + [x] > x for all x ∈ R

we have

aε = 3− 1

2 +[

1ε

] > 3− 11ε

= 3− ε .

(b) A15b = {x ∈ R |3x+ 1| < 1}.Solution: A15b =

(−23, 0), supA15b = 0, aε = max {−ε/2,−1/3}, for instance.

(c) A15c = {x ∈ R x3 < 3}.Solution: A15c =

(−∞, 3

√3), supA15c = 3

√3, aε = 3

√3− ε/2.

(d) A15d =

{−1

n2 + n+ 23n ∈ N

}.

Solution: supA15d = 0. If ε > 0, for supA15d − ε < aε we need to find nε ∈ N so that

supA15d − ε = −ε < aε =−1

n2ε + nε + 23

(2.2)

Since n2 ≤ n2 + n+ 23 for all n ∈ N, we have an estimate

−1

n2ε

<−1

n2ε + nε + 23

hence (2.2) holds if

−ε < −1

nε

which is true if nε > 1/ε, say nε = 1 + d1/εe which gives

aε =−1

n2ε + nε + 23

=−1

(1 + d1/εe)2 + (1 + d1/εe) + 23.

16. Let A ⊂ R be a bounded set. Prove that there is n ∈ N such that A ⊂ [−n, n].

Hint: Archimedean Axiom R-12

Solution: Since A is bounded, it is bounded above by u, say and below by l. Thus l ≤ a ≤ u forall a ∈ A. By the Archimedian Axiom, there are m,n ∈ N, such that u ≤ m and −l ≤ n. Thenmax {m,n} ∈ N and for all a ∈ A we have

a ≤ u ≤ m ≤ max {m,n}

anda ≥ l ≥ −n ≥ −max {m,n} .

ThereforeA ⊂ [−max {m,n} ,max {m,n}] .

17. For each of the following sequences (xn)n∈N, x = b . . . e find L := limn→∞ xn and a formula expressingnε in terms of ε such that 5.2 holds.

Hint: Recall the definition in (5.2). The floor and ceiling functions might be useful for this:

[x] := max {z ∈ Z z ≤ x} and dxe = min {z ∈ Z z ≥ x}

(a) For example, if an =1

2nwe try L = 0 and look at all n ∈ N such that∣∣∣∣ 1

2n− L

∣∣∣∣ =

∣∣∣∣ 1

2n

∣∣∣∣ < ε .

Since2n > n for all n ∈ N ,

we can estimate ∣∣∣∣ 1

2n

∣∣∣∣ < 1

n< ε

if n > 1ε. Thus we can take

nε :=

⌈1

ε

⌉.

Of course you could also use

nε :=

⌈log2

(1

ε

)⌉,

but we have not really defined log yet.

(b) bn =6n2 + 1

2n2 + n+ 1.

Solution: With L = 3 we estimate

|bn − L| =∣∣∣∣ 6n2 + 1

2n2 + n+ 1− 3

∣∣∣∣ =

∣∣∣∣ −3n− 2

2n2 + n+ 1

∣∣∣∣ =3n+ 2

2n2 + n+ 1<

5n

2n2<

3

n< ε

whenever

n >3

ε.

We can therefore take

nε := 1 +

[3

ε

].

(c) cn =1√∑nj=1 j

=1√

1 + 2 + · · ·+ n

Solution: We have 0 ≤ cn ≤ 1√n< ε if n > 1

ε2. Thus we can take nε = 1 +

⌈1

ε2

⌉(d) dn =

9n7 + 3n2 + n+ 1

5n7 − 34n+ 2−n

Solution: We estimate dn by

9

5≤ dn =

9 + 3n−5 + n−6 + n−7

5− 34n−6 + 2−nn−7≤ 9 + 5n−5

5− 34n−6

if 5n7 > 34n, for instance if n > 34. For such n, it then follows that∣∣∣∣dn − 9

5

∣∣∣∣ ≤ 9 + 5n−5

5− 34n−6− 9

5=

25n−5 + 9× 35n−6

5(5− 34n−6)≤ 500n−5

5=

100

n5< ε

if n > 5

√100ε

. We can therefore take

nε = max

{34,

⌈5

√100

ε

⌉}.

(e) en =(−2)n

n!.

Hint: Recall that n! = 1× 2× 3× · · · × (n− 1)× n.

Solution: We use a rough estimate for n! if n ≥ 4:

n! = 1× 2× 3× 4× 5× 6× · · · × n ≥ 1× 2× 3× 4× 4× 4× · · · × 4 = 4!× 4n−4 .

With L = 0, we get the estimate

|en − L| =∣∣∣∣(−2)n

n!

∣∣∣∣ =2n

n!≤ 2n

4!× 4n−4=

24

4!× 2n−4

4n−4=

2

3× 1

2n<

2

3× 1

n<

1

n< ε

if n ≥ 4 and n > 1ε. Thus we can take

nε = max

{4,

⌈1

ε

⌉}18. Prove that the sequence (

(−1)k +1

k2

)k∈N

is bounded but does not converge. Find a convergent subsequence.

Hint: Recall Definition 5.3

Solution: By the triangle inequality,∣∣∣∣(−1)k +1

k2

∣∣∣∣ ≤ ∣∣(−1)k∣∣+

∣∣∣∣ 1

k2

∣∣∣∣ = 1 +1

k2≤ 2

hence the sequence is bounded.

If L were the limit of the sequence, then for any ε > 0 there would be kε ∈ N such that for all k > kε,∣∣∣∣(−1)k +1

k2− L

∣∣∣∣ < ε .

In order to prove this wrong, let ε = 160

and let m,n > kε, m even, n odd. We would then have∣∣∣∣(−1)m +1

m2− L

∣∣∣∣ =

∣∣∣∣1 +1

m2− L

∣∣∣∣ < 1

60and

∣∣∣∣(−1)n +1

n2− L

∣∣∣∣ =

∣∣∣∣−1 +1

n2− L

∣∣∣∣ < 1

60

hence, by the triangle inequality,∣∣∣∣(1 +1

m2− L

)−(−1 +

1

n2− L

)∣∣∣∣ ≤ ∣∣∣∣1 +1

m2− L

∣∣∣∣+

∣∣∣∣−1 +1

n2− L

∣∣∣∣ < 2

60

But the left hand side of this is∣∣∣∣(1 +1

m2− L

)−(−1 +

1

n2− L

)∣∣∣∣ =

∣∣∣∣2 +1

m2− 1

n2

∣∣∣∣ > 1 ,

a contradiction. This proves that the sequence does not converge.

For a convergent subsequence extract the even entries: Then

b :=

(1

(2n)2

)2n∈N

=

((−1)2n +

1

(2n)2

)2n∈N

=

(1 +

1

(2n)2

)2n∈N

= (a2n)n∈N

is a subsequence of the given sequence a =((−1)k + 1

k2

)k∈N. Clearly, b converges to 1.

19. Prove that the sequence e−n2

cos(3n+ 16) + n+ 1∑i∈N,i2≤n

i

n∈N

(2.3)

converges.

Solution: First observe that

∑i∈N,i2≤n

i =

[√n]∑

i=1

=[√n] ([√n] + 1)

2≤ (√n+ 1)

√n

2=n+√n

2and ≥

√n(√n− 1)

2=n−√n

2.

Let a denote the sequence (2.3). Then

an =e−n

2cos(3n+ 16) + n+ 1∑

i∈N,i2<n

i≤ 2

n+ 2

n+√n≤ 2

n+ 2

n= 2 +

4

n

andan ≥ 2

n

n−√n≥ 2

n

n= 2

hence

2 ≤ an ≤ 2 +4

n

It follows that for n > nε :=⌈

4ε

⌉we have

|an − 2| ≤ 4

n< ε

which proves thatlimn→∞

an = 2 .

20. For each of the following sequences, say whether they inrease, decrease, converge. Prove your answer.(n+ 2

n+ 1

)n∈N

,

(3− 1

2k

)k∈N

,

(1

v− v)v∈N

Solution: The sequence(n+2n+1

)n∈N

decreases:

(n+ 1) + 2

(n+ 1) + 1− n+ 2

n+ 1=n+ 3

n+ 2− n+ 2

n+ 1=

−1

(n+ 2)(n+ 1)< 0

and converges to 1: ∣∣∣∣n+ 2

n+ 1− 1

∣∣∣∣ =1

n+ 1< ε

if n > 1ε.

The sequence(3− 1

2k

)k∈N increases:

3− 1

2k+1−(

3− 1

2k

)=

1

2k− 1

2k+1=

1

2k+1> 0

and converges to 3: ∣∣∣∣3− 1

2k+1− 3

∣∣∣∣ =1

2k<

1

k< ε

if k > 1ε.

The sequence(

1v− v)v∈N decreases:

1

v + 1− (v + 1)−

(1

v− v)

=1

v + 1− 1

v− 1 < 0

and diverges: To see this, assume the contrary, L = limv→∞1v− v, i.e. for all ε > 0 there is mε ∈ N

such that for all v > mε, ∣∣∣∣1v − v − L∣∣∣∣ < ε

hence

L− ε < 1

v− v < L+ ε ,

hence, since 0 < 1v≤ 1,

L− ε− 1 < −v < L+ ε ,

equivalently− L− ε < v < −L+ ε+ 1 . (2.4)

However, by the Archimedean Property, there are no L, ε ∈ R such that(2.4) holds for all v ∈ N.

21. Let (an)n∈N and (bn)n∈N be convergent sequences of real numbers. Prove that the sequence

(max {an, bn})n∈N

converges.

Hint: Look at Theorem 5.10.

Solution: Let ε > 0 be given. Since a and b converge to A respectively B, say, there are na, nb ∈ Nso that we have

|an − A| < ε for n > na ,

|bn −B| < ε for n > nb .

Hence, for n > max{na, nb

},

A− ε < an < A+ ε , B − ε < bn < B + ε

and thereforemax {A,B} − ε < max {an, bn} < max {A,B}+ ε

22. Find a convergent subsequence in (n3−[n

3

])n∈N

.

Hint: Recall definition 5.6 and look at example 39.

Solution: Let rn = 3n. Then (rn3−[rn

3

])n∈N

= (0)n∈N

converges to 0.

23. Prove that the function f : R → R, f(x) =√x2 + |x| is continuous at x0 = 0. Thus, for arbitrary

ε > 0 find δε satisfying (6.2), i.e. so that |x− x0| < δε implies that |f(x)− f(x0)| < ε.

Hint: See example 36.

Solution: For a given ε > 0 we want

|f(x)− f(x0)| =∣∣∣√x2 + |x| −

√02 + |0|

∣∣∣ =√x2 + |x| < ε . (2.5)

This is equivalent tox2 + |x| < ε2 . (2.6)

If |x| ≤ 1 then x2 ≤ |x|. Hence (2.6) and thus (2.5) are satisfied if

|x| = |x− x0| < δε := min

{1,ε2

2

}.

Alternative: In order to satisfy (2.6) we could also solve a quadratic equation:

x2 + |x| < ε2 ⇐⇒ |x| < −1

2+

√1

4+ ε2 =: δε

which gives the optimal δε

24. Prove that the function f : R → R, f(x) = |17x+ 3| is continuous. Thus, for arbitrary x0 ∈ R andε > 0 find δx0,ε satisfying (6.4), i.e. so that |x− x0| < δx0,ε implies that |f(x)− f(x0)| < ε.

Hint: See example 36. By the triangle inequality,

||a| − |b|| ≤ |a− b|

for a, b ∈ R.

Solution: For x, x0 ∈ R we want

|f(x)− f(x0)| = ||17x+ 3| − |17x0 + 3|| < ε (2.7)

By the triangle inequality,

||17x+ 3| − |17x0 + 3|| ≤ |(17x+ 3)− (17x0 + 3)| = 17 |x− x0| .

Thus (2.7) holds if |x− x0| < ε17

=: δx0,ε.

25. Use the intermediate value theorem 6.5 to show that there are at least 4 real roots of the polynomial

p(x) = x8 − 3x6 + 1 .

You may assume that the function p : R→ R, x 7→ x8 − 3x6 + 1 is continuous.

Solution: We compute

p(0) = 1 > 0 , p(±1) = −1 < 0 , p(±2) = 256− 192 + 1 > 0 .

By the Intermediate Value Theorem, the function has zeros in

(−2,−1) , (−1, 0) , (0, 1) , (1, 2)

26. Prove that(n(

1− n√

2))

n∈Nconverges.

Solution:

limn→∞

n(

1− n√

2)

= limh→0

20 − 2h

h=

d

dt

∣∣∣∣∣t=0

2t =d

dt

∣∣∣∣∣t=0

et ln(2) = ln(2) .

27. Show that the square root function√

: R+0 → R is continuous.

Solution: For continuity at 0, note that for x ≥ 0,

x = |x− 0| < ε2 =⇒√x =

∣∣∣√x−√0∣∣∣ < ε .

For continuity at a > 0, let 0 < ε <√a. If x ∈ R+

0 satisfies

|x− a| < ε√a

i.e.0 ≤ a− ε

√a < x < a+ ε

√a

then

0 ≤√a− ε ≤

√a− ε

√a <√x <

√a+ ε

√a ≤√a+ ε

because of the monotonicity of the square root function and since(√a− ε

)2= a− 2

√aε+ ε2 ≤ a− ε

√a because ε <

√a and

a+ ε√a ≤

(√a+ ε

)2= a+ 2

√aε+ ε2 .

In particular ∣∣√x−√a∣∣ < ε .

We thus have shown that for every a ∈ R+0 and every ε > 0 there is δε (ε2 in case a = 0 and ε

√a in

case a > 0) such that for x ∈ R+0 , |x− a| < δε implies that |

√x−√a| < ε.

3 Some Problems for the Study week

1. 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) 6= f(a′) whenevera 6= a′.

2. 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

f(z) :=

{4z + 1 if z ≥ 0−4z + 3 if z < 0

is injective.

4. Define a surjective function g : R→ [0, 3).

Solution: The function g : R→ [0, 3),

x 7→ 3− 3

1 + x2

is onto [0, 3).

x

g3

5. Let f : R→ R be the function with f(x) = 2x5 − x3 − x for all x ∈ R. Determine f−1((−∞, 0)), i.e.the inverse image of R− = (−∞, 0) under f .

Solution: By definition,

f−1((−∞, 0)) = {x f(x) < 0} ={x 2x5 − x3 − x < 0

}={x 2x5 < x3 + x

}.

If x ≥ 1 or −1 ≤ x ≤ 0, then x5 ≥ x3 ≥ x, hence f(x) ≥ 0 and x is not in the set.

If 0 < x < 1 or x < −1 then x5 < x3 < x, hence f(x) < 0, x is in the set.

It follows thatf−1((−∞, 0)) = (−∞,−1) ∪ (0, 1) .

6. Let f : R→ R be given by f(x) = 2 |x| − |x+ 1|. Determine f([1, 2]) and f−1((−∞, 1).

Solution: We eliminate the absolute value:

f(x) = 2 |x| − |x+ 1| =

2x− (x+ 1) if x ≥ 0−2x− (x+ 1) if −1 ≤ x ≤ 0−2x+ (x+ 1) if x ≤ −1

=

x− 1 if x ≥ 0−3x− 1 if −1 ≤ x ≤ 0−x+ 1 if x ≤ −1

Thus, for x ∈ [1, 2], we have f(x) = x− 1, hence

f([1, 2]) = [0, 1]

In order to determinef−1((−∞, 1) = {x f(x) < 1}

we need to solve ‘f(x) < 1’ in the three cases x ≥ 0, −1 ≤ x ≤ 0 and x ≤ −1.

If x ≥ 0 then f(x) = x− 1 < 1 if x < 2.

If −1 ≤ x ≤ 0 then f(x)− 3x− 1 < 1 if −3x < 2, i.e. x > −2/3

If x ≤ −1 then f(x) = −x+ 1 < 1 if x > 0.

The union of these gives

f−1((−∞, 1) = [0, 2) ∪ (−2/3, 0] ∪ ∅ = (−2/3, 2) .

x

1

2 |x| − |x+ 1|

-2 -1 -2/3 1 2 3

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 6= ∅, m ∈ R. State what “m = maxA” means by definition.

Solution: For A ⊂ R, by definition, m = maxA, 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 realnumbers. Then supA ∪B = max {supA, supB}.”Solution: This is true. To prove this, let M := max {supA, supB}. Then M ≥ supA andM ≥ supB, 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 = supA or M = supB. By definition of the supremum, for any N < M there is x ∈ Aor x ∈ B, i.e. x ∈ A ∪B with N < x ≤M which shows that M is the least upper bound for A ∪B.

10. Say whether minimum and infimum of the set {x ∈ R x3 − x > 0} exist and if so, find their values.Prove your result.

Solution: If x ≤ −1 then x3 ≤ x, hence x 6∈ {x ∈ R x3 − x > 0} =: X. Therefore −1 is a lowerbound for X. If −1 < x < 0 then x3 > x hence x ∈ X. This also shows that −1 = inf X. Since−1 6∈ X, the minimum of X does not exist.

11. Prove or privide a counterexample to the following: If f : R → R is continuous and ∅ 6= 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) =1

1 + x2and A = [1,∞) .

Then f(A) = (0, 1/2] and inf f(A) = 0 but f(inf A) = f(1) = 1/2.

12. Let f : R → R be the function given by f(x) = sgn(x − [x] − 12). Determine the set of all x0 such

that f is not continuous at x0.

Solution: The function is not continuous at x0 if and only if x0 or x0 − 12

is an integer.

x

-2 -1 0 1 2 3

sgn(x− [x]− 12)

x− [x]− 12

13. What is a subsequence of a sequence?

Solution: A subsequence of the sequence (an)n∈N is a sequence (ark)k∈N where (rk)k∈N is a strictlyincreasing sequence of natural numbers.

14. Let a be a sequence of real numbers and L ∈ R. By definition, what does “limn→∞ an = L” mean?

Solution: “limn→∞ an = L” means that for every positive real number ε there is a natural numberm such that for all n > m we have |an − L| < ε.

15. State the Monotone Convergence Theorem.

Solution: A bounded monotone sequence converges.

16. Let (ak)k∈N be an increasing sequence of positive real numbers. Prove that

(1

an

)n∈N

converges.

Solution: Since (ak)k∈N increases, we have

1

an≤ 1

amwhenever n > m

hence

(1

an

)n∈N

decreases. Since an > 0 for all n, we also have 1an> 0, in particular,

(1

an

)n∈N

is

bounded. By the Monotone Convergence Theorem,

(1

an

)n∈N

converges

17. Prove that the sequence

n2

5−[n2

5

]+ 4n7

n4 + 7n5 + 9n7

n∈N

converges.

Solution: Abbreviate y = n2

5−[n2

5

]. Since 0 ≤ y − [y] < 1 for all real numbers y, we estimate∣∣∣∣ y + 4n7

n4 + 7n5 + 9n7− 4

9

∣∣∣∣ =|9y + 36n7 − 4(n4 + 7n5 + 9n7)|

9(n4 + 7n5 + 9n7)=|9y − 4n4 − 28n5|9(n4 + 7n5 + 9n7)

=4n4 + 28n5 − 9

9(n4 + 7n5 + 9n7)

≤ 32n5

81n7≤ n5

n7=

1

n2< ε

if n > 1√ε. Thus for any ε > 0 there is m := 1√

εsuch that for all n > m we have

∣∣∣∣ n25 −[n25 ]+4n7

n4+7n5+9n7 − 4/9

∣∣∣∣ <ε.

18. Prove that the sequence a =

(√k2k + k

3kk

)k∈N

converges.

Solution: We show that limk→∞

√k2k+k

3kk= 1

3. To this end, let ε > 0 be given. Then∣∣∣∣ak − 1

3

∣∣∣∣ =

∣∣∣∣∣√k2k + k

3kk− 1

3

∣∣∣∣∣ =

√k2k + k − kk

3kk=

1

3

(√1 +

k

kk− 1

)< ε

if √1 +

k

kk< 3ε+ 1

that is ifk

kk< (3ε+ 1)2 − 1

equivalentlykk

k>

1

(3ε+ 1)2 − 1.

Since kk/k > k − 1 for all k ∈ N, this holds whenever

k > 1 +1

(3ε+ 1)2 − 1.

Thus we have shown that for any ε > 0, there is kε ∈ R, namely kε := 1 + 1(3ε+1)2−1

, such that for all

k > kε we have∣∣ak − 1

3

∣∣ < ε.

19. Show that there is a sequence (bn)n∈N so that

∀n ∈ N : bn ∈ N and bn+1 > bn ,

and so that (sin(2bm) +

b3m

b6m + 5

)m∈N

converges.

Solution: Since sin(x) ∈ [−1, 1] for all x ∈ R, we have

−1 ≤ xn := sin(2n) +n3

n6 + 5≤ 2,

the sequence (xn)n∈N is bounded. By the Bolzano-Weierstrass Theorem, (xn)n∈N has a convergentsubsequence, i.e. there is an increasing sequence (bn)n∈N of natural numbers so that

(xbj)j∈N con-

verges as required.

20. 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 ε > 0 there isδ > 0 so that for all x ∈ A with |x− p| < δ we have |f(a)− f(p)| < ε.

21. State the Intermediate Value Theorem.

Solution: Let f : [a, b] → R be a continuous function, a, b ∈ R, a ≤ b. Also assume that y ∈[f(a), f(b)] or y ∈ [f(b), f(a)]. Then there is x ∈ [a, b] with f(x) = y.

22. Prove or provide a counterexample for the following: Let f : R→ R be continuous and assume thatf(0) > 0. Then there is u > 0 with f(u) > 0.

Solution: This is true: To prove this, let ε := f(0)/2. By continuity of f at 0, there is δ > 0 sothat for all u with |u| < δ, in particular for u = δ/2, we have

|f(u)− f(0)| < ε = f(0)/2

hence, by the triangle inequality,f(u) > f(0)/2 > 0 .

23. Let f : R → R be the function with f(x) = 4 |x3 − x| − 1. Find six pairwise disjoint intervals eachcontaining 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 somevalues of f and look for sign changes. Since f is even, we do so for positive arguments first. We have

f(0) = −1 < 0 , f(1/2) = 1/2 > 0 , f(1) = −1 , f(999) > 0 .

Therefore f changes sign in the intervals

(0, 1/2) , (1/2, 1) , (1, 999)

and, since f is even, f also changes sign on the intervals

(−1/2, 0) , (−1,−1/2) , (−999,−1) .

By the Intermediate Value Theorem, f has a zero in each of these six intervals.

x

f

24. Let f : R→ R be the function with f(x) = 2(x2)−4x2 + 12. Find four pairwise disjoint intervals each

of which contains at least one zero of f . You may assume without proof that x 7→ 2(x2) is continuous.

Solution: Since f is continuous, we can apply the Intermediate Value Theorem, looking for signchanges. Since f is even (i.e. f(x) = f(−x)) we need to find only 2 sign changes on the positive realaxis. Then there will be another two on the negative. We compute some values of f :

f(0) = 1 + 1/2 > 0 , f(1) = 2− 4 + 1./2 < 0 , f(2) = 24 − 16 + 1/2 = 1/2 > 0 .

Thus f changes sign in the intervals(0, 1) , (1, 2)

and by symmetry also in the intervals

(−1, 0) , (−2,−1) .

By the Intermediate Value Theorem, f has at least one zero in each of these four intervals.

25. Prove that the function f : R→ R with f(x) = 3√x is continuous at 8.

Solution: For given ε > 0|f(x)− f(2)| =

∣∣ 3√x− 2

∣∣ < ε

is equivalent to

(2− ε)3 = 8− 4ε+ 2ε2 − ε3 < x < (2 + ε)3 = 8 + 4ε+ 2ε2 + ε3

i.e.− 4ε+ 2ε2 − ε3 < x− 8 < 4ε+ 2ε2 + ε3 (3.1)

If ε < 1, then ε2, ε3 < ε and

−4ε+ 2ε2 − ε3 < −ε and ε < 4ε+ 2ε2 + ε3 .

In particular (3.1) holds if

−ε < x− 8 < ε equivalently |x− 8| < ε

Thus, withδε := min {ε, 1}

we have shown that |x− 8| < δε implies that |f(x)− f(8)| < ε. f is therefore continuous at 8.

26. Let g : R→ R be the function such that g(1/n) = 1 for all n ∈ N and g(x) = 0 if there is no n ∈ Nwith x = 1/n. Prove that g is not continuous at 0.

Solution: We show that for ε = 1/2 there is no δ > 0 such “|x− 0| = |x| < δ” implies that“|g(x)− g(0)| = |g(x)| < ε = 1/2”.

Assume otherwise, i.e. there were such a δ > 0. Then by the Archimedean Property of R, we canfind n ∈ N with n > 1

δ, hence 0 < 1/n < δ. Clearly, g(1/n) = 1, hence

|g(1/n)− g(0)| 6< ε =1

2

despite |1/n− 0| = 1/n < δ.

27. What is a function?

Solution: A function is a set f of pairs such that if (a, b) ∈ f and (a, c) ∈ f , then a = c.

28. Let f : R → R be the function with f(x) = x2 + cos(x) for all x ∈ R. Is f injective? Prove youranswer.

Solution: f is not injective, f(1) = f(−1)

29. For which α ∈ R is the function fα : R→ R with fα(x) = α |x|+ 5x injective? Prove your answer.

Solution: For x ∈ R,

fα(x) =

{(5 + α)x if x ≥ 0(5− α)x if x ≤ 0

If α > 5, then 15−α < 0 < 1

5+αbut fα

(1

5−α

)= 1 = fα

(1

5+α

).

If α = 5, then fα(0) = fα(−1) = 0.

If −5 < α < 5 then fα(x) > fα(y) whenever x > y.

If α = −5, then fα(0) = fα(1) = 0.

If α < −5, then −15−α < 0 < −1

5+αbut fα

( −15−α

)= −1 = fα

( −15+α

).

Thus fα is injective if and only if α < 5.

30. Prove that for all x ∈ R, [x] := sup {z ∈ Z z ≤ x} exists.

31. Prove that for all x ∈ R we have x ∈ [[x] , [x] + 1).

32. Prove that the sequence

(12k

k +√k2 +

√k

)k∈N

converges.

Solution: We will show that limk→∞12k

k+√k2+√k

= 6. To this end, let ε > 0 be given. Then

∣∣∣∣∣ 12k

k +√k2 +

√k− 6

∣∣∣∣∣ =

∣∣∣∣∣6k − 6√k2 +

√k

k +√k2 +

√k

∣∣∣∣∣ = 6

∣∣∣∣∣k −√k2 +

√k

k +√k2 +

√k

∣∣∣∣∣ < ε .

holds if ∣∣∣∣∣k −√k2 +

√k

k +√k2 +

√k

∣∣∣∣∣ < ε

6. (3.2)

Since

k +

√k2 +

√k ≥ 2k

the estimate (3.2) holds if∣∣∣∣∣k −√k2 +

√k

2k

∣∣∣∣∣ =

∣∣∣∣∣∣12 −√

1 +√k

k2

2

∣∣∣∣∣∣ =

√1 +

√k

k2

2− 1

2<ε

6

This, in turn, holds if

1 +

√k

k2<( ε

3+ 1)2

equivalently (1(

ε3

+ 1)2 − 1

) 23

< k

33. Prove thatsup

{x x4 < 16

}= 2 .

Solution: Start with the definition, in this case of sup, say as ”least upper bound”. Thus you needto show

(a) 2 is an upper bound for {x x4 < 16}, i.e. u > 2⇒ u 6∈ {x x4 < 16}, and

(b) if b is an upper bound for {x x4 < 16} then b ≥ 2, equivalently, if s < 2 then s is not an upperbound for {x x4 < 16}.

To see the first claim, let u ∈ R. We have shown in class that if a, b, c ∈ R, c ∈ P , a > b thenca > cb. Thus, if u > 2, we have u, 2 > 0, and applying this rule 4 times we get

u4 = u · u3 > 2 · u3 > 2 · 2 · u2 > 2 · 2 · 2 · u > 2 · 2 · 2 · 2 = 16 (3.3)

which shows u 6∈ {x x4 < 16}.In order to show the second claim, let s ∈ R, s < 2. If s ≤ 0 then s is not an upper bound because14 = 1 < 16, hence s < 1 ∈ {x x4 < 16}. Let d = 2+s

2. We have

s < d < 2 because s =s

2+s

2<s

2+

2

2= d <

2

2+

2

2= 2

ands4 < s3 · d < · · · d4 < · · · < 24 = 16

by the same argument as for (3.3), hence s < d but d ∈ {x x4 < 16}, hence s is not an upper boundfor {x x4 < 16}.

34. Consider the set

Y =

{x ∈ R x4 − 2x2 +

3

4< 0

}Find maxY , minY , inf Y , supY if they exist. Also determine the minimal interval I with Y ⊂ I.

Solution: For x ∈ R,

x4 − x2 +3

4= (x2 − 1)2 − 1

4< 0

if and only if

−1

2< x2 − 1 <

1

2equivalently

−1

2+ 1 =

1

2< x2 <

1

2+ 1 =

3

2that is

−√

3√2< x < − 1√

2or

1√2< x <

√3√2.

Therefore {x ∈ R x4 − 2x2 +

3

4< 0

}=

(−√

3√2,− 1√

2

)∪

(1√2,

√3√2

)⊂

(−√

3√2,

√3√2

)︸ ︷︷ ︸

I

and I is minimal with Y ⊂ I. It also follows that minY and maxY do not exist, and that

inf Y = −√

3√2

and supY =

√3√2.

35. Prove that if ∅ 6= B ⊂ A ⊂ R, A bounded, supB 6= supA, then supA = supA \B.

Solution: Since A \ B ⊂ A, any upper bound for A is an upper bound for A \ B. In particular,supA is an upper bound for A \B.

We now show that supA is the least upper bound for A \ B. Thus we need to show that for everyc ∈ R, c < supA there is x ∈ A \B such that c < x. To see this, let c ∈ R, c < supA be arbitrary.

Since B ⊂ A, we must have supB ≤ supA. By assumption the two suprema are different, hencesupB < supA.

Thus we havec′ = max {c, supB} < supA ,

c′ is not an upper bound for A and there is x ∈ A with c′ < x. But then supB < x and since supBis an upper bound for B, x 6∈ B. It follows that x ∈ A \B and c ≤ c′ < x.

36. Prove that the square function f : R→ R, x 7→ x2, is continuous.

Solution: For x, x0 ∈ R and given ε > 0 we want

|f(x)− f(x0)| =∣∣x2 − x2

0

∣∣ = |(x+ x0)(x− x0)| < ε (3.4)

Now if |x− x0| < 1, then by the triangle inequality

|x+ x0| = |2x0 + (x− x0)| ≤ 2 |x0|+ 1

and|(x+ x0)(x− x0)| < (2 |x0|+ 1) |x− x0| < ε

if additionally |x− x0| < ε2|x0|+1

. We define

δx0,ε := min

{1,

ε

2 |x0|+ 1

}Then if |x− x0| < δx0,ε we have (3.4).

37. Prove that the square root function R+0 → R, x 7→

√x, is continuous.

Solution: For ε > 0 and x0 > 0, we have ∣∣√x−√x0

∣∣ < ε (3.5)

if|x− x0| < δ1

ε,x0:= ε√x0 < ε(

√x+√x0) (3.6)

since 0 <√x+√x0, because (√

x+√x0

) (√x−√x0

)= x− x0 .

For continuity at x0 = 0, simply observe that in this case∣∣√x−√x0

∣∣ =∣∣√x∣∣ < ε

if|x| < δε,0 := ε2 . (3.7)

In all cases, we have shown that for any ε > 0 and x0 ≥ 0, we have∣∣√x−√x0

∣∣ < ε, (3.5), if|x− x0| < δε,x0 , where δε,x0 is given in (3.6) for x0 > 0 and (3.7) for x0 = 0.

38. 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

∀x, x′ ∈ (a, b), x ≤ x′ : f(x) ≤ f(x′)

and f is onto (u, v) means that

∀w ∈ (u, v)∃x ∈ (a, b) : f(x) = y .

Proof: We will show continuity at an arbitrary x0 ∈ (a, b). To this end, let ε > 0 be given. Withoutloss of generality, we can assume u+ ε < f(x0) < v − ε. Then, by surjectivity,

f(x0)± ε ∈ f((a, b))

i.e. there are α, β ∈ (a, b), so that

u < f(x0)− ε︸ ︷︷ ︸=f(α)

< f(x0) < f(x0) + ε︸ ︷︷ ︸=f(β)

< v

and since f increases, we must have

a < α < x0 < β < b .

In particular, 0 < min {β − x0, x0 − α} and, again since f increases, we must have

|f(x)− f(x0)| < ε if |x− x0| < min {β − x0, x0 − α}

•

39. Consider the sequence a defined recursively by

a1 := 0, a2 := 1 and for n ≥ 2, an+1 :=

{an + 1 if an ≤ max {aj j < n}

0 if an > max {aj j < n}

For any L ∈ N0 find a subsequence that converges to L.

Solution: The first elements of the sequence are

0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, . . .

We have an = 0 if and only if n is the sum of the first k natural numbers for some k, i.e.

nk =k∑j=1

j =k(k + 1)

2

for some k. We have

an = L ∈ N if and only if n = L+ nk for some k ≥ L .

Thus a subsequence of a convergent to L (because it is constant) is (ark)k∈N where

rk = nk+L−1 + L =(k + L− 1)(k + L)

2+ L =

(k + L)2 − k + L

2

•

40. Let f : R→ R be the function with

f(x) =

{1

x2+1if x ∈ Q

0 if x 6∈ Q .

Prove that f is nowhere continuous, i.e. there is no x0 ∈ R such that f is continuous at x0.

Solution: Let x0 ∈ R be arbitrary. We show that f is not continuous at x0. To this end, let

ε =f(x0)

2=

1

2(1 + x20).

Let δ > 0 be given. We show that there is x ∈ R with

|x− x0| < δ but |f(x)− f(x0)| > ε .

To see this, let n ∈ N be so that n > 12δ

and q = [nx0] ∈ Z. Then

x0 − δ <q

n≤ x0 ≤

q +√

2/2

n≤ q + 1

n< x0 + δ .

If x0 6∈ Q then|f(q/n)− f(x0)| =

41. Prove that the function f : R→ R with

f(x) =

√

2 |x|+ 1− 1

|x|if x 6= 0

1 if x = 0

is continuous at 0.

Solution: Let ε > 0 be given. Then

ε > |f(x)− f(0)| =

∣∣∣∣√2|x|+1−1

|x| − 1

∣∣∣∣ if x 6= 0

0 if x = 0(3.8)

if x = 0. For x 6= 0, note that

1 <√

2 |x|+ 1 < 2 |x|+ 1 because 1 < 2 |x|+ 1 < (2 |x|+ 1)2 = 1 + 4 |x|+ 4x2 .

For x 6= 0, (3.8) holds if

ε >

∣∣∣∣∣√

2 |x|+ 1− 1

|x|− 1

∣∣∣∣∣ =

∣∣∣∣∣∣ (2 |x|+ 1)− 1

|x|(

1 +√

2 |x|+ 1) − 1

∣∣∣∣∣∣ =

∣∣∣∣∣ 2

1 +√

2 |x|+ 1− 1

∣∣∣∣∣

= 1− 2

1 +√

2 |x|+ 1

This in turn holds if2

1 +√

2 |x|+ 1> 1− ε

or, if2

1− ε> 1 +

√2 |x|+ 1

which follows from

|x| <(

21−ε − 1

)2 − 1

2

42. 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 ∅ 6= A ⊂ R and a ∈ R. We will write x ≥ A if x ≥ y for all y ∈ A. Then Definition4.6 says that a = supA if and only if x ≥ a is equivalent to x ≥ A.

First assume that a = supA. From Definition 4.6 we thus have that a is an upper bound for Aand that for each n ∈ N, there is an ∈ A with a − 1

n< an. Thus we have a − 1

n< an ≤ a, hence

(an)n∈Nn→∞−→ a.

Now assume a ≥ A and let an ∈ A, n ∈ N, be so that limn→∞ an = a. Now if x ≥ a then x ≥ A. Ifx < a then ε := a − x > 0 and since limn→∞ an = a, there is nε so that |an − a| = a − an < a − xfor all n > nε. By the triangle inequality, we have in particular that x < an for n > nε and thereforex 6≥ A.

43. Let a0 ∈ A ⊂ R and f : A→ R. Prove that the following are equivalent:

(a) f is continuous at a0.

(b) Whenever (an)n∈N is a sequence with an ∈ A for all n ∈ N and limn→∞ an = a0, thenlimn→∞ f(an) = f(a0).

Solution: Let a0 ∈ A ⊂ R and f : A → R. We first assume 43a and show 43b. To this end, let(an)n∈N ∈ AN and limn→∞ an = a0. Let ε > 0 be given. By continuity of f at a0, there is δε suchthat for x ∈ A, the implication

|x− a0| < δε =⇒ |f(x)− f(a0)| < ε (3.9)

holds. By the defintion of convergence, there is nδε ∈ N so that the implication

n > nδε =⇒ |an − a0| < δε (3.10)

holds. Putting (3.9) and (3.10) together gives

n > nδε =⇒ |f(an)− f(a0)| < ε .

Thus we have proved that (f(an))n∈N converges to f(a0).

We prove the converse, 43b ⇒ 43a, by contradiction. Thus we assume that f is not continuous ata0, i.e.

∃ε > 0∀δ > 0∃xδ ∈ A : |xδ − a0| and |f(xδ)− f(a0)| ≥ ε .

In particular, for each n ∈ N, we can take δ = 1/n and let an = x1/n ∈ A be so that

|an − a0| <1

nand |f(an)− f(a0)| ≥ ε .

Thus we have limn→∞ an = a0 but the sequence (f(an))n∈N does not converge to f(a0), contradicting43a.

44. Let g : R → R be a continuous function with g(0) = 0. Let M > 0 and f : R → [−M,M ] be anyfunction (not necessarily continuous). Prove that the product function gf : R → R, x 7→ g(x)f(x),is continuous at 0.

Solution: Let ε > 0 be given. Now for all x, since |f(x)| < M , the estimate

|f(x)g(x)| < ε (3.11)

holds if|g(x)| < ε/M . (3.12)

By continuity of g, there is δ = δgε/M > 0 such that for all x ∈ R with |x| < δgε/M equation (3.12) and

therefore (3.11) hold. This proves continuity of fg at 0.

4 Real Numbers

4.1 Order properties

Addition and Multiplication of real numbers satsify the same rules as that of rational numbers. For acomplete list of axioms, see the appendix 7. Here we only take a closer look at the axioms of order andcompletenes. The order on R is definded through a subset

P ⊂ R \ {0}

(“positive numbers”) such that

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 4.1 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 4.2 ≤ 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| :=

x if x > 00 if x = 0−x if x < 0

.

The sign of x ∈ R is

sgn(x) :=

1 if x > 00 if x = 0−1 if x < 0

.

Thus for all x ∈ R,x = sgn(x) |x| .

Theorem 4.3 For all x, y ∈ R we have the triangle inequality

|x+ y| ≤ |x|+ |y|

4.2 Archimedean Property

The set N of natural numbers (in R) is defined as

N :=⋂

1∈A⊂R, 1+A⊂A

A

i.e. the intersection of all subsets A ⊂ R such that 1 ∈ A and 1 + x ∈ A whenever x ∈ A. Thus

N = {1, 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1 + 1, . . .}

The Archimedean Property of the real numbers is the fact that this set has no upper bound. Thus

∀x ∈ R∃n ∈ N : n ≥ x

An important consequence of this is

Theorem 4.4 For all ε ∈ R, ε > 0 there is n ∈ N such that

1

n< ε .

Proof: Let ε > 0 be given. By the Archimedean Property, there is n ∈ N such that n > 1ε, hence ε < 1

n. •

4.3 Completeness

Definition 4.5 Let A ⊂ R, A 6= ∅. Then

1. b ∈ R is called an upper bound for A if ∀a ∈ A : a ≤ b.

2. l ∈ R is called an lower bound for A if ∀a ∈ A : a ≥ l.

3. A subset A ⊂ R is bounded above/bounded below/bounded if A has an upper bound/lower bound/both.

Definition 4.6 Let A ⊂ R, A 6= ∅. The maximum, the minimum, the supremum and the infimum of Aare the elements maxA,minA, supA, inf A ∈ R characterized, if they exist, by the properties

1. maxA ∈ A and ∀a ∈ A : a ≤ maxA

2. minA ∈ A and ∀a ∈ A : a ≥ minA

3. supA ∈ R and ∀x ∈ R : (x ≥ supA⇔ ∀a ∈ A : x ≥ a)

4. inf A ∈ R and ∀x ∈ R : (x ≤ inf A⇔ ∀a ∈ A : x ≤ a)

The completeness axiom for the real numbers states that every bounded nonempty set of real numbers hasa supremum. Thus, if µ ∈ R, ∅ 6= A ⊂ R are so that ∀a ∈ A : a ≤ µ, then there is s ∈ R (“the supremumof A”) such that, for x ∈ R,

x ≥ s ⇐⇒ ∀a ∈ A : x ≥ a

(This in turn is equivalent to “x < s ⇐⇒ ∃a ∈ A : x < a”.)

Theorem 4.7 The Archimedian Property R-12 is a consequence of Completeness R-13 (and R-1-R-11)

Proof: For a set A ⊂ R and z ∈ R we write z ≥ A if z is an upper bound for A, i.e. ∀a ∈ A : z ≥ a. Fory ∈ R we have

y ≥ N ⇐⇒ y + 1 ≥ N . (4.8)

Assume there is an upper bound for N, i.e. x ∈ R with x ≥ N, i.e. x ≥ n for all n ∈ N. Then theassumption of the completeness axiom R-13 is satisfied with µ = x and A = N. By completenes R-13 and(4.8) we have the equivalences

y ≥ s ⇐⇒ y ≥ N ⇐⇒ y + 1 ≥ N ⇐⇒ y + 1 ≥ s

i.e. y ≥ s if and only y + 1 ≥ s, which is false for y = s− 1 for instance. •

5 Sequences and Limits

Definition 5.1 A sequence of real numbers, or a sequence in R is a function N→ R. For

f ∈ RN := {f f : N→ R a function}

we writef = (f(n))n∈N = (fn)n∈N

A sequence (an)n∈N of real numbers converges to L ∈ R if

lim a = limn→∞

andef⇐⇒ ∀ε > 0∃nε ∈ N∀n > nε : |an − L| < ε . (5.2)

Definition 5.3 A sequence a ∈ RN is bounded if

R(a) = {an n ∈ N}

is a bounded subset of R.

Definition 5.4 Monotonicity: a ∈ RN increases if

∀n ∈ N : an+1 ≥ an (5.5)

In case of the reverse estimate one says the sequence decreases. A monotone sequences is a sequence thatincreases or decreases. If the estimate in (5.5) is sharp (< resp. > instead of ≤ and ≥) the sequencein/decreases strictly.

Definition 5.6 A subsequence of a is a sequence of the form

a ◦ r = (arn)n∈N

where r ∈ NN is a strictly increasing sequence of natural numbers,

1 ≤ r1 < r2 < r3 < r4 < · · · rn < rn+1 < · · ·

Theorem 5.7 (Monotone Convergence) A bounded monotone sequence converges. Thus if b ∈ R anda ∈ RN is a sequence of real numbers such that an+1 ≥ an ≤ b for all n ∈ N, then

limn→∞

an = sup {an n ∈ N} ≤ b

exists. Similarly, if l ∈ R and a ∈ RN is a sequence of real numbers such that an+1 ≤ an ≥ b for all n ∈ N,then

limn→∞

an = inf {an n ∈ N} ≥ l

exists.

Proof: (for increasing sequences.) The supremum exists by the completeness of R. By the definition ofthe supremum, for each ε > 0, there is nε ∈ N such that

sup {an n ∈ N} − ε ≤ anε ≤ sup {an n ∈ N}

But since the sequence increases, and sup {an n ∈ N} is an upper bound, we have

∀n ≥ nε : sup {an n ∈ N} − ε ≤ anε ≤ an ≤ sup {an n ∈ N} .

In particular∀n ≥ nε : |an − sup {an n ∈ N}| < ε

that is convergence as claimed. •

Theorem 5.8 (Bolzano-Weierstrass) Every bounded sequence contains a convergent subsequence.Thus if a = (an)n∈N ∈ RN and M ∈ R are so that for all n ∈ N, |an| < M then there is a strictlyincreasing sequence r ∈ NN such that

(arn)n∈N

converges.

Proof: Let (an)n∈N be bounded. By completeness of R the suprema

sn := sup {ak k ≥ n} , n ∈ N

exist. We now define (rn)n∈N recursively, starting with r1 := 1. Next, assuming rn already definded, wechoose rn+1 > rn so that

srn+1 −1

n+ 1< arn+1 ≤ srn+1 (5.9)

Since (sn)n∈N decreases and is bounded, it converges. Taking limits over (5.9) shows that (arn)n∈N alsoconverges to the same limit. •

Theorem 5.10 Let a, b be convergent sequences of real numbers so that

limn→∞

bn 6= 0 6= bn for all n ∈ N .

Then

limn→∞

anbn

=limn→∞ anlimn→∞ bn

.

Proof: Since a and b both converge,

A := limk→∞

ak , B := limk→∞

bk 6= 0 ,

say, for any α, β > 0 there are naα, nbβ ∈ N so that

∀n > naα : |an − A| < α

∀n > nbβ : |bn −B| < min

{β,|B|2

}Consequently, for n > max

{naα, n

bβ

}we have∣∣∣∣anbn − A

B

∣∣∣∣ =

∣∣∣∣anB − AbnbnB

∣∣∣∣ =

∣∣∣∣(an − A)B − A(bn −B)

bnB

∣∣∣∣ < α |B|+ |A| β|B|2 /2

=2

|B|α +

2 |A||B|2

β

Since B 6= 0, for any ε > 0 we can choose αε, βε > 0 so that

2

|B|α <

ε

2and

2 |A||B|2

β <ε

2

be given. Then for any n > nε = max{naαε , n

bβε

}we have∣∣∣∣anbn − A

B

∣∣∣∣ < ε

•

6 Continuity

Definition 6.1 Let A ⊂ R be a subset and f : A→ R a function. Then f is continuous at x0 ∈ A if

∀ε > 0∃δε > 0∀x ∈ (x0 − δε, x0 + δε) ∩ A : f(x) ∈ (f(x0)− ε, f(x0) + ε) . (6.2)

Thus f : A→ R is continuous at x0 ∈ A if for each ε > 0 there is δε > 0 such that

|x− x0| < δε implies that |f(x)− f(x0)| < ε .

Definition 6.3 A function f : A → R is continuous if f is continuous at x0 for all x0 ∈ A. Thus f iscontinuous if

∀x0 ∈ A, ε > 0∃δx0,ε > 0∀x ∈ (x0 − δx0,ε, x0 + δx0,ε) ∩ A : f(x) ∈ (f(x0)− ε, f(x0) + ε) . (6.4)

Theorem 6.5 (Intermediate Value Theorem) Let a, b ∈ R, a ≤ b, and f : [a, b] → R be continuous.If sgn(f(a)) 6= sgn(f(b)) then there is x ∈ [a, b] such that f(x) = 0.

An immediate consequence is that a continuous function f : [a, b] → R assumes all values between f(a)and f(b),

[min {f(a), f(b)} ,max {f(a), f(b)}] ⊂ f([a, b]) .

Proof: We may assume a < b and also f(a) < 0 < f(b). Otherwise the statement is trival or we mayreplace f by −f .

Assume 0 ∈ [f(a), f(b)] \ f([a, b]). Then, by completeness of R,

A := sup {x ∈ [a, b] f(x) < 0}

exists since the set contains a and is bounded above by b. Since a ≤ A ≤ b, f(A) is defined and byassumption, f(A) 6= 0. Let ε < |f(A)|. Since f is continuous at A, there is δ > 0 such that for all x with

A− δ < x < A+ δ (6.6)

we havef(A)− ε < f(x) < f(A) + ε .

If f(A) < 0 then by our choice of ε, f(A) + ε < 0 and all x with (6.6) lie in {x ∈ [a, b] f(x) < 0}.

If f(A) > 0 then f(A)−ε > 0 and therefore all x satisfying (6.6) are upper bounds for {x ∈ [a, b] f(x) < 0}

In both cases A can not be the supremum of this set. •

Theorem 6.7 Let A ⊂ R and f : A→ R be continuous. Assume [a, b] ⊂ A. Then there are u, v ∈ R suchthat

f([a, b]) = [u, v]

7 Appendix: Axioms for The Real Numbers

The real numbers are a quadruple (R,+, ·, P ) consising of a set R, a subset P ⊂ R and two functions+, · : R× R→ R such that

R-1.∀x, y, z ∈ R : x+ (y + z) = (x+ y) + z

R-2.∃0 ∈ R∀x ∈ R : 0 + x = x+ 0 = x (7.1)

R-3. If 0 is as in (7.1) then∀x ∈ R∃y ∈ R : x+ y = y + x = 0

R-4.∀x, y ∈ R : x+ y = y + x

R-5.∀x, y, z ∈ R : x · (y · z) = (x · y) · z

R-6.∃1 ∈ R \ {0} ∀x ∈ R : 1 · x = x · 1 = x (7.2)

R-7. If 1 is as in (7.2) then∀x ∈ R \ {0} ∃z ∈ R : x · z = z · x = 1

R-8.∀x, y ∈ R : x · y = y · x

R-9. [Distributive Law]

∀x, y, z ∈ R : x · (y + z) = (x · y) + (x · z) = (y + z) · x

R-10. [Trichotomy]R = −P ∪ {0} ∪ P , P ⊂ R \ {0} , P ∩ −P = ∅

We write x ≥ y, equivalently y ≤ x, if and only if x− y ∈ P ∪ {0}.

R-11.P + P ⊂ P ⊃ P · P

R-12. [Archimedean Property] The set N of natural numbers (in R) is defined as

N :=⋂

1∈A⊂R, 1+A⊂A

A

i.e. the intersection of all subsets A ⊂ R such that 1 ∈ A and 1 + x ∈ A whenever x ∈ A. Then

∀x ∈ R∃n ∈ N : n ≥ x

R-13. [Completeness] If µ ∈ R, ∅ 6= A ⊂ R are so that ∀a ∈ A : a ≤ µ, then there is s ∈ R (“the supremumof A”) such that, for x ∈ R,

x ≥ s ⇐⇒ ∀a ∈ A : x ≥ a

(This in turn is equivalent to “x < s ⇐⇒ ∃a ∈ A : x < a”.)

Definition 7.3

DFR-1. Let x ∈ R. We write −x for the additive inverse of x, i.e. the element y ∈ R is so that x+ y = 0 =y + x.

DFR-2. Let x ∈ R× := R \ {0}. We write 1x

= 1/x for the multiplicative inverse of x, i.e. the element z ∈ Ris so that x · z = 0 = z · x.

DFR-3. We write xy for x · y.

DFR-4. (Subtraction) For x, y ∈ R we define the difference

x− y := x+ (−y)

DFR-5. (Division) For x, y ∈ R, y 6= 0, we define the quotient

x

y:= x · 1

y

Lemma 7.4 Let a, b, c ∈ R. Then

L-1. 0 · a = 0Proof: We compute

0 · a+ aR−7= 0 · a+ 1 · a R−9

= (0 + 1) · a R−3= 1 · a R−7

= a (7.5)

hence

0 · a R−2= 0 · a+ (a+ (−a))

R−1= (0 · a+ a) + (−a)

(7.5)= a+ (−a)

R−2= 0

L-2. (−1) · a = (−a)Proof:

L-3. a · b = 1 =⇒ b = 1a

Proof: Assume a · b = 1 (XXX). If a = 0 then by L1, a · b = 0 but by R-7, 0 6= 1. Hence we musthave a 6= 0. By R-7, 1

aexists and

1

aXXX

=1

a· (a · b) R−5

=

(a · 1

a

)· b R−6

= 1 · b R−7= b

L-4. a · b = 0 =⇒ a = 0 or b = 0

L-5. ≥ is an order relation.

L-6. a2 ∈ P or a = 0, in particular 1 ∈ P.Proof: If a ∈ P then by R-11, a2 ∈ P.

The case a = 0 is trivial

If a ∈ −P then by (L-7) −a ∈ P, hence

a2 = −(−a2)(L-2)= (−1) · (−a2)

(L-2)= (−1) · ((−1) · a2)

R−5,R−8= ((−1) · a) · ((−1) · a)

(L-2)= (−a)2 ∈ P

by R-11.

L-7. −(−a) = a, if a 6= 0 then 1

( 1a)

= a

Proof: By definition, −(−a) has the property that

(−(−a)) + (−a) = 0

hence

a = a+ ((−(−a)) + (−a))R−1,R−4

= ((a+ (−a)) + (−(−a))R−2= 0 + (−(−a))

R−3= −(−a)

L-8. If a ≥ b then a+ c ≥ b+ c. If a ≥ b and c ≥ 0, then a · c ≥ b · c. Similarly, if a > b then a+ c ≥ b+ c.If a > b and c > 0, then a · c > b · c.