Advanced Calculus I Chapter 3 Homework Solutions November 20, 2009

First Assignment

27. Prove that every set of the form { x | a < x < b } is open and every set of the form{ x | a ≤ x ≤ b } is closed.

Pick x ∈ (a, b). Set δ = 1

2min{x−a, b−x}, and note that δ > 0 (since

a < x < b). Then a < x − δ < x + δ < b, so that (x − δ, x + δ) ⊂ (a, b),so that (a, b) is open. [Compare #3 from Chapter 1.]

Note that if a = −∞, then the proof above (with δ = 1

2min{1, b− x})

shows that (−∞, b) is also open. Similarly, we can show that (a,∞)and (−∞,∞) are also open. To show that [a, b] is closed, we will show

that the complement (−∞, a)∪(b,∞) is open. From the previous comments

we know that each component set is open. It therefore suffices to show

that the union of two open sets is open.

To this end, let A and B be two open sets, and pick x ∈ A∪B. Then

either x ∈ A or x ∈ B (or both). Without loss, we will assume that

x ∈ A. Then since A is open, there is some δ > 0 so that (x−δ, x+δ) ⊂ A ⊂ A ∪ B. Thus A ∪ B is open.

30. Suppose f : R → R is continuous and fix a point r0 ∈ R . Prove that { x | f(x) 6= r0 }is an open set.

We will show that the complement E = { x | f(x) = r0 } is closed, by

showing that it contains all its accumulation points. Suppose x0 is

an accumulation point of E. Then there is a sequence of points {xn}∞

n=1

converging to x0 with xn ∈ E for all n ≥ 1. But then f(xn) = r0

for all n ≥ 1. Since f is continuous, this means that f(x0) = f(limxn) =lim f(xn) = lim r0 = r0. Thus x0 ∈ E, so E is closed, so its complement

is open.

31. Suppose f : [a, b] → R and g : [a, b] → R are both continuous, and consider the setT = { x | f(x) = g(0) } . Prove that T is closed.

Consider the function h(x) = (f − g)(x) = f(x)− g(x). Note that h is

continuous. The set T is exactly the set where h(x) = 0. We showed

in the previous problem that such sets are closed.

34. Find an open cover of (1, 2) with no finite subcover.

Set Gn =(

1 + 1

n, 2

)

. Then (1, 2) ⊂⋃

Gn, so the Gn form an open cover

of (1, 2). But any finite subcover has a maximum index N , and so covers

only(

1 + 1

N, 2

)

.

35. Let E be compact and nonempty. Prove that E is bounded and that sup E and inf Eboth belong to E .

We will show that sup E is contained in E. The proof for inf E is similar.

We showed in class that an unbounded set cannot be compact (the cover

Gn = (−n, n) suffices to show this). Thus E has an upper bound, and

hence has a finite supremum, which we shall call s = sup E. Suppose

for contradiction that s /∈ E, and consider the cover Gn =(

−∞, s − 1

n

)

.

The union of all these sets covers (−∞, s), which, since s is an upper

bound for E, covers E. But no finite subcover suffices to cover E.

This is because any finite subcover has a maximum index N , and so only

covers(

−∞, s − 1

N

)

. If this sufficed to cover E, then s− 1

Nwould

be an upper bound for E, contradicting that s is the least upper bound.

Thus s /∈ E implies E not compact. The result follows.

Second Assignment

19. Let f, g : D → R be uniformly continuous. Prove that f + g : D → R is uniformlycontinuous. What can be said about fg? Justify.

Let ǫ > 0 be given. Then there is some δ1 > 0 so that |f(x)−f(y)| <ǫ/2 whenever |x − y| < δ1. Similarly, there is some δ2 > 0 so that

|g(x) − g(y)| < ǫ/2 whenever |x − y| < δ2. Let δ = min{δ1, δ2}. Then

|x − y| < δ implies

|(f + g)(x) − (f + g)(y)| = |f(x) + g(x) − f(y) − g(y)|

≤ |f(x) − f(y)| + |g(x) − g(y)| < ǫ/2 + ǫ/2 = ǫ.

The functions f(x) = x and g(x) = sin x are both uniformly continuous

on all of R, but their product fg(x) = x sin x is not.

20. Let f : A → B and g : B → C be uniformly continuous. What can be said aboutg ◦ f : A → C ? Justify.

The composition is also uniformly continuous. Let ǫ > 0 be given.

Then there is some δ′ > 0 so that |g(x)−g(y)| < ǫ whenever |x−y| < δ′.Because f is also uniformly continuous, there is some δ > 0 so that

|f(x) − f(y)| < δ′ whenever |x − y| < δ. Thus for x, y ∈ A we have

|x − y| < δ ⇒ |f(x) − f(y)| < δ′ ⇒ |g(f(x)) − g(f(y))| < ǫ.

23. A function f : R → R is periodic if there is a real number h 6= 0 such that f(x + h) =f(x) for all x ∈ R . Prove that if f : R → R is periodic and continuous, then f is uniformlycontinuous.

24. Suppose A is bounded and not compact. Prove that there is a function that is continuouson A but not uniformly continuous. Give an example of a set that is not compact, but everyfunction continuous on that set is uniformly continuous.

Since A is bounded and not compact, it must not be closed. Hence there

is some point a that is an accumulation point of A but not in A. Define

f(x) = 1

x−a. This function is continuous wherever it is defined. Since

it is only undefined at a, and a /∈ A, that means f is continuous on

A. But f is not uniformly continuous on A because the image of the

bounded set A∩[a−1, a] (which is nonempty since a is an accumulation

point) is unbounded.

The set N is not compact because it is not bounded. On the other hand,

every function fN → R is both continuous and uniformly continuous

because every point of N is isolated (take δ < 1/2; then x, y ∈ N

with |x − y| < δ implies x = y, so f(x) = f(y), so |f(x) − f(y)| = 0 <ǫ).

40. Give an example of a function fR → R that is continuous and bounded but notuniformly continuous.

The function f(x) = sin(x2) has this property. Intuitively, it looks

like sin(x), except that as we move away from the origin, the numbers

actually being plugged in to the sine function is the square of the

actual x at which we lie. Thus the sine waves are getting closer and

closer together as we move away form the origin. In particular, for

any δ > 0, there is some interval of the form (x−δ, x+δ) that contains

an entire period, and hence has points that take on all values between

1 and −1. It follows that there is no single δ > 0 that works everywhere

for any ǫ that is less than 2.

Third Assignment

41. Find an interval of length 1 that contains a root of the equation xex = 1

44. Suppose f : [a, b] → [a, b] is continuous. Prove that there is at least one fixed point in[a, b] (a fixed point is a point x such that f(x) = x).