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Homework 1. Solutions

1. Show that the discrete metric satisfies the properties of a metric.

The discrete metric is defined by the formula

d(x, y) =

{

1 if x 6= y0 if x = y

}

.

It is clearly symmetric and non-negative with d(x, y) = 0 if and only

if x = y. It remains to establish the triangle inequality

d(x, y) ≤ d(x, z) + d(z, y).

If x = y, then the left hand side is zero and the inequality certainly

holds. If x 6= y, then the left hand side is equal to 1. Since x 6= y,we must have either z 6= x or else z 6= y. Thus, the right hand side

is at least 1 and the triangle inequality holds in any case.


2. Compute the distances d1(f, g) and d∞(f, g) when f, g ∈ C[0, 1]are the functions defined by f(x) = x2 and g(x) = x3.

Since x2 ≥ x3 for all x ∈ [0, 1], the first distance is given by

d1(f, g) =

∫

1

0

(x2 − x3) dx =

[

x3

3−

x4

4

]1

0

=1

3−

1

4=

1

12.

To compute the second distance, we need to find the maximum of

h(x) = x2 − x3, 0 ≤ x ≤ 1.

Since h′(x) = 2x− 3x2 = x(2− 3x), it easily follows that

d∞(f, g) = h(2/3) = 4/9− 8/27 = 4/27.


3. Sketch the open ball B((0, 0), 1) in the metric space (R2, d∞).

The open ball B((0, 0), 1) contains the points (x, y) that satisfy

d∞((x, y), (0, 0)) = max{|x|, |y|} < 1.

Now, the maximum of two numbers is smaller than 1 if and only if

the two numbers are both smaller than 1. This gives the condition

|x| < 1 and |y| < 1.

Thus, the open ball B((0, 0), 1) is the interior of the square whose

vertices are located at the points (±1,±1).


4. Let A = {x ∈ R : x > 0}. Is this set bounded in (R, d) when d is

the usual metric? Is it bounded when d is the discrete metric?

This set is not bounded with respect to the usual metric. If it were

bounded, then we would have

A ⊂ (x− r, x+ r)

for some x ∈ R and some r > 0. This is not the case because

|x|+ r ∈ A, |x|+ r /∈ (x− r, x+ r).

To show that A is bounded with respect to the discrete metric, we

note that A is contained in B(x, 2) = R for any x ∈ R.


5. Consider a metric space (X, d) whose metric d is discrete. Show

that every subset A ⊂ X is open in X.

Let x ∈ A and consider the open ball B(x, 1). Since d is discrete,

this open ball is equal to {x}, so it is contained entirely within A.


1. Let (X, d) be a metric space. Given a point x ∈ X and a real

number r > 0, show that U = {y ∈ X : d(x, y) > r} is open in X.

Let y ∈ U . Then ε = d(x, y)− r is positive and we have

z ∈ B(y, ε) =⇒ d(y, z) < ε

=⇒ r + d(y, z) < d(x, y) ≤ d(x, z) + d(z, y)

=⇒ r < d(x, z)

=⇒ z ∈ U.

This shows that B(y, ε) ⊂ U and that the set U is open.


2. Show that each of the following sets is closed in R.

A = [0,∞), B = Z, C = {x ∈ R : sinx ≤ 0}.

The complements of the given sets can be expressed in the form

R−A = (−∞, 0) =⋃

n∈N

(−n, 0),

R−B =⋃

x∈Z

(x, x+ 1),

R− C = {x ∈ R : sinx > 0} =⋃

k∈Z

(2kπ, 2kπ + π).

These are all unions of open intervals, so they are all open in R.

Thus, the given sets A,B,C are all closed in R.


3. Find a collection of closed subsets of R whose union is not closed.

Needless to say, there are several examples. One typical example is

An = [1/n, 2] =⇒⋃

n∈N

An = (0, 2].

More generally, let {xn} be any strictly decreasing sequence of real

numbers whose limit x is finite. Then it is easy to see that

An = [xn, y] =⇒⋃

n∈N

An = (x, y].


4. Let (X, dX) and (Y, dY ) be metric spaces. Assuming that dX is

discrete, show that any function f : X → Y is continuous.

Let x ∈ X and ε > 0 be given. Then we have

y ∈ B(x, 1) =⇒ y = x

=⇒ f(y) = f(x)

=⇒ f(y) ∈ B(f(x), ε).

This shows that f is continuous at the point x.


1. Consider the sequence of functions defined by fn(x) =x

1+n2x2 for

all x ≥ 0. Show that this sequence converges uniformly on [0,∞).

It is clear that fn(x) converges pointwise to the zero function. To

show that it converges uniformly, we compute the supremum of

gn(x) = |fn(x)− 0| =x

1 + n2x2

on the interval [0,∞). Using the quotient rule, we get

g′n(x) =1 + n2x2 − 2n2x2

(1 + n2x2)2=

1− n2x2

(1 + n2x2)2

and so gn attains its maximum at the point x = 1/n. In particular,

the supremum is gn(1/n) = 1/(2n) and it goes to zero as n → ∞.


2. Let (X, d) be a metric space, let fn : X → R be a sequence of

continuous functions such that fn → f uniformly on X and let xn be

a sequence of points of X with xn → x. Show that fn(xn) → f(x).

Let ε > 0 be given. Then there exists an integer N1 such that

|fn(x)− f(x)| < ε/2 for all n ≥ N1 and all x ∈ X.

Since fn → f uniformly, the limit f is continuous. Since xn → x,we must also have f(xn) → f(x). Pick an integer N2 such that

|f(xn)− f(x)| < ε/2 for all n ≥ N2.

Given any integer n ≥ max{N1, N2}, we must then have

|fn(xn)− f(x)| ≤ |fn(xn)− f(xn)|+ |f(xn)− f(x)| < ε.


3. Let f : X → Y be a function between metric spaces and let xn be

a Cauchy sequence in X. Show that f(xn) must also be Cauchy, if fis Lipschitz continuous. Is the same true, if f is merely continuous?

Suppose that f is Lipschitz continuous with constant L > 0. Given

any ε > 0, we can then find an integer N such that

d(xm, xn) < ε/L for all m,n ≥ N .

Since f is Lipschitz continuous, this also implies that

d(f(xm), f(xn)) ≤ L · d(xm, xn) < ε for all m,n ≥ N

and so f(xn) is Cauchy. When f is merely continuous, the result is

not true. For instance, xn = 1/n is Cauchy in R and f(x) = 1/x is

continuous, but f(xn) = n is certainly not Cauchy.


4. Show that (X, d) is complete, if the metric d is discrete.

Suppose that xn is a Cauchy sequence in X. Then there exists an

integer N such that

d(xm, xn) < 1 for all m,n ≥ N .

Since the metric d is discrete, this actually means that

xm = xn for all m,n ≥ N .

Given any ε > 0, we must then have

d(xn, xN ) = d(xN , xN ) < ε for all n ≥ N .

In other words, we must have xn → xN as n → ∞.


1. Let f : R → R be continuous and let A = {x ∈ R : f(x) ≥ 0}.Show that A is closed in R and conclude that A is complete.

The set U = (−∞, 0) is open in R because it can be written as

U = (−∞, 0) =⋃

n∈N

(−n, 0)

and this is a union of open intervals. Since f is continuous,

f−1(U) = {x ∈ R : f(x) ∈ U} = {x ∈ R : f(x) < 0}

is then open in R. Thus, the complement of f−1(U) is closed and

this means that A is closed. Since R is complete and A is a closed

subset of R, we conclude that A is complete.


2. Suppose f : [a, b] → [a, b] is a differentiable function such that

L = supa≤x≤b

|f ′(x)|

satisfies L < 1. Show that f has a unique fixed point in [a, b].

Let x, y ∈ [a, b]. Using the mean value theorem, one finds that

|f(x)− f(y)| = |f ′(c)| · |x− y| ≤ L · |x− y|

for some point c between x and y. Since L < 1 by assumption, this

shows that f is a contraction on [a, b]. On the other hand, [a, b] is aclosed subset of R and thus complete. It follows by Banach’s fixed

point theorem that f has a unique fixed point in [a, b].


3. Show that there is a unique real number x such that cos x = x.Hint: Such a number must lie in [−1, 1]. Use the previous problem.

Since f(x) = cos x is between −1 and 1 for all x, every fixed point

of f must lie in the interval [−1, 1]. Note that

L = sup−1≤x≤1

|f ′(x)| = sup−1≤x≤1

| sinx| = sup0≤x≤1

sinx = sin 1

is strictly less than 1. In view of the previous problem, f must have

a unique fixed point in [−1, 1], so it has a unique fixed point in R.

Thus, there is a unique real number x such that f(x) = x.


4. Show that the set Q of all rational numbers is not complete.

Consider a sequence of rational numbers that converges to√2, say

x1 = 1.4

x2 = 1.41

x3 = 1.414

and so on. This sequence is convergent in R, so it is Cauchy, but it

is not convergent in Q because its limit is irrational.


1. Let (X,T ) be a topological space and let A,B be subsets of X.

Show that the closure of their union is given by A ∪B = A ∪B.

Since A ∪ B is a closed set that contains A ∪ B and A ∪B is the

smallest closed set that contains A ∪B, we must certainly have

A ∪B ⊂ A ∪B.

To prove the opposite inclusion, we note that Theorem 2.5 gives

x ∈ A ∪B =⇒ every neighbourhood of x intersects A or B

=⇒ every neighbourhood of x intersects A ∪B

=⇒ x ∈ A ∪B.


2. Find two open intervals A,B ⊂ R such that A ∩B 6= A ∩B.

Pick any real numbers a < b < c and consider the open intervals

A = (a, b), B = (b, c).

Since A ∩B = ∅, one also has A ∩B = ∅. On the other hand,

A ∩B = [a, b] ∩ [b, c] = {b}.


3. Let (X,T ) be a topological space and let A ⊂ X. Show that

∂A = ∅ ⇐⇒ A is both open and closed in X.

If A is both open and closed in X, then the boundary of A is

∂A = A ∩X −A = A ∩ (X −A) = ∅.

Conversely, suppose that ∂A = ∅. Then Theorem 2.6 gives

A◦ = A.

Since A◦ ⊂ A ⊂ A by definition, these three sets are equal, so

A◦ = A = A =⇒ A is both open and closed in X.


4. Let (X,T ) be a topological space and let A ⊂ X. Show that

X −A = X −A◦.

Using Theorem 2.5, one finds that

x ∈ X −A ⇐⇒ every neighbourhood of x intersects X −A

⇐⇒ no neighbourhood of x is contained in A

⇐⇒ x /∈ A◦

⇐⇒ x ∈ X −A◦.


1. Let (Y, T ) be a topological space and let A ⊂ Y . Show that A is

open in Y if and only if every point of A has a neighbourhood which

lies within A. Hint: We know that A is open if and only if A = A◦.

As we already know, a set A is open if and only if A = A◦. On the

other hand, one always has A◦ ⊂ A by definition, so a set A is open

if and only if A ⊂ A◦. This is true precisely when every point of Ahas a neighbourhood which lies within A.


2. Let (X,T ) be a Hausdorff space. Show that the set

A = {(x, y) ∈ X ×X : x 6= y}

is open in X ×X. Hint: Use the previous problem with Y = X ×X.

Let (x, y) be an arbitrary point of A. Then x 6= y and there exist

sets U, V which are open in X with x ∈ U , y ∈ V and U ∩ V = ∅.

Now, the product U × V is a neighbourhood of (x, y) such that

(a, b) ∈ U × V =⇒ a ∈ U and b ∈ V

=⇒ a 6= b

=⇒ (a, b) ∈ A.

It is thus a neighbourhood of (x, y) which lies within A. Using the

previous problem, we conclude that A is open in X ×X.


3. Suppose X is a Hausdorff space that has finitely many elements.

Show that every subset of X is open in X. Hint: Use Theorem 2.14.

If A is a subset of X, then its complement is a finite subset of a

Hausdorff space X, so it is closed in X. Since the complement of Ais closed in X, we conclude that A is open in X.


4. Suppose f : X → Y is both continuous and injective. Suppose also

that Y is Hausdorff. Show that X must be Hausdorff as well.

Suppose x, y are distinct points in X. Then f(x), f(y) are distinct

points in Y by injectivity. Since Y is Hausdorff, we can always find

sets U, V which are open in Y with

f(x) ∈ U, f(y) ∈ V, U ∩ V = ∅.

It follows by continuity that f−1(U), f−1(V ) are open in X with

x ∈ f−1(U), y ∈ f−1(V ), f−1(U) ∩ f−1(V ) = ∅.

This shows that the space X is Hausdorff as well.


1. Show that the unit circle C = {(x, y) ∈ R2 : x2 + y2 = 1} is

connected. Hint: Use polar coordinates and Theorem 2.15.

In terms of polar coordinates, the points on the unit circle are the

points that have the form (cos θ, sin θ). Consider the function

f : [0, 2π) → R2, f(θ) = (cos θ, sin θ).

Each component of f is continuous, so f itself is continuous. Since

the domain of f is an interval, it is connected, so the image of f is

connected as well. In other words, the unit circle C is connected.


2. Let U = {(x, y) ∈ R2 : xy 6= 0}. How many connected components

does this set have? Hint: One of the components is (0,∞) × (0,∞).

To say that xy 6= 0 is to say that each of x, y is either positive or

negative. Thus, U can be expressed as the union of the sets

U1 = (0,∞)× (0,∞), U2 = (0,∞) × (−∞, 0),

U3 = (−∞, 0)× (0,∞), U4 = (−∞, 0)× (−∞, 0).

Each of these sets is a product of intervals and thus connected. It

remains to show that they are actually connected components.

Let A be a connected subset of U and let pi : A → R denote

the projection on the ith variable. By continuity, each pi(A) is an

interval that does not contain 0, so it is contained in either (−∞, 0)or (0,∞). Thus, A itself is contained in one of the sets Uj.


3. Show that there is no surjective continuous function f : A → B, if

A = (0, 3) ∪ (3, 6), B = (0, 1) ∪ (1, 2) ∪ (2, 3).

Hint: Look at the restriction of f on each subinterval. You will need

to use Theorem 2.11 and also some parts of Theorem 2.15.

Note that A = A1 ∪ A2 is the union of two disjoint open intervals

and B = B1 ∪B2 ∪B3 is the union of three disjoint open intervals.

Since f : A → B is continuous, each restriction f : Ai → B must be

continuous, so each f(Ai) must be connected.

Since f(Ai) is a connected subset of B1 ∪B2 ∪ B3, it lies within

either B1 or B2∪B3. If it actually lies in B2∪B3, then it lies within

either B2 or B3. Thus, each f(Ai) is contained in a single Bj and

the function f is not surjective, contrary to assumption.


4. Let (X,T ) be a topological space and suppose A1, A2, . . . , An are

connected subsets of X such that Ak ∩ Ak+1 is nonempty for each k.Show that the union of these sets is connected. Hint: Use induction.

When n = 1, the union is equal to A1 and this set is connected by

assumption. Suppose that the result holds for n sets and consider

the union of n+ 1 sets. This union has the form

U = A1 ∪ · · · ∪An ∪An+1 = B ∪An+1,

where B is connected by the induction hypothesis. Since An+1 has

a point in common with An, it has a point in common with B. In

particular, the union B ∪An+1 is connected and the result follows.


1. Show that the set A = {(x, y) ∈ R2 : x2 + 4y4 ≤ 4} is compact.

First of all, the set A is bounded because its points satisfy

x2 ≤ x2 + 4y4 ≤ 4 =⇒ |x| ≤ 2,

4y4 ≤ x2 + 4y4 ≤ 4 =⇒ |y| ≤ 1.

To show that A is also closed, we consider the function

f : R2 → R, f(x, y) = x2 + 4y4.

Since f is continuous and (−∞, 4] is closed in R, its inverse image

is closed in R2. This means that A is closed in R

2. Since A is both

bounded and closed in R2, we conclude that A is compact.


2. Show that the set B is a compact subset of R2 when

B = {(x, y) ∈ R2 : x ≥ 0, y ≥ 0, x+ y ≤ 1}.

First of all, the set B is bounded because its points satisfy

0 ≤ x ≤ x+ y ≤ 1, 0 ≤ y ≤ x+ y ≤ 1.

To show that B is also closed, we consider the functions

f1(x, y) = x, f2(x, y) = y, f3(x, y) = x+ y.

These are all continuous functions and it is easy to see that

B = f−1

1([0,∞)) ∩ f−1

2([0,∞)) ∩ f−1

3((−∞, 1]).

In particular, B is closed in R2 and also bounded, so it is compact.


3. Suppose A,B are compact subsets of a Hausdorff space X. Show

that A∩B is compact. Hint: Use the first two parts of Theorem 2.19.

Since A is a compact subset of a Hausdorff space X, it is actually

closed in X. This implies that A ∩B is closed in B. Being a closed

subset of a compact space, A ∩B must also be compact.


4. Let Cn be a sequence of nonempty, closed subsets of a compact

space X such that Cn ⊃ Cn+1 for each n. Show that the intersection

of these sets is nonempty. Hint: One has⋃

(X − Ci) = X −⋂

Ci.

Suppose the intersection is empty. Then we actually have

X = X −

∞⋂

i=1

Ci =

∞⋃

i=1

(X − Ci),

so the sets X − Ci form an open cover of X. Since X is compact,

it is covered by finitely many sets, say the first k. This gives

X =

k⋃

i=1

(X −Ci) = X −

k⋂

i=1

Ci = X − Ck,

so the set Ck must be empty, contrary to assumption.


1. Are any of the following sets homeomorphic? Explain.

A = (0, 1), B = [0, 1), C = [0, 1], D = [0,∞).

The set C is not homeomorphic to any of the other sets because Cis compact and the other sets are not. To show that B and D are

homeomorphic, we note that either of the functions

f(x) = x/(1− x), g(x) = tan(πx/2)

gives rise to a homeomorphism between B = [0, 1) and D = [0,∞).

Finally, we turn to A and B. Were these sets homeomorphic, we

would have a homeomorphism h : [0, 1) → (0, 1) and its restriction

on (0, 1) would also be a homeomorphism. This is not possible, as

the image of the restriction is (0, 1)−{h(0)} which is not connected.


2. Let (X, d) be a metric space and fix some y ∈ X. Show that the

function f : X → R defined by f(x) = d(x, y) is Lipschitz continuous.

Letting x, z ∈ X be arbitrary, we use the triangle inequality to get

f(x) = d(x, y) ≤ d(x, z) + d(z, y) = d(x, z) + f(z)

f(z) = d(z, y) ≤ d(z, x) + d(x, y) = d(x, z) + f(x).

Once we now combine these equations, we may conclude that

|f(x)− f(z)| ≤ d(x, z).

This shows that the function f : X → R is Lipschitz continuous.


3. Let xn ∈ ℓp denote the sequence whose first n2 entries are equal

to 1/n and all other entries are zero. For which values of 1 ≤ p ≤ ∞does this sequence converge to the zero sequence in ℓp?

Using the definition of the norm in ℓp, we find that

||xn − 0||pp =∞∑

i=1

|xni|p =

n2

∑

i=1

1

np= n2−p.

This expression converges to zero if and only if the exponent 2 − pis negative, hence if and only if p > 2.


4. Let en ∈ ℓ∞ denote the sequence whose nth entry is equal to 1and all other entries are zero. Show that {en}

∞

n=1is bounded but not

Cauchy and that the unit ball B = {x ∈ ℓ∞ : ||x||∞ ≤ 1} is closed

and bounded, but not compact.

First of all, {en}∞

n=1is bounded but not Cauchy because

||en||∞ = ||em − en||∞ = 1 (∗)

whenever m 6= n. It is clear that B is bounded. To show that B is

also closed, we note that the norm f : X → R is continuous in any

normed vector space and that B is the inverse image of (−∞, 1].

Finally, suppose that B is compact. Then the sequence {en} has

a convergent subsequence by Theorem 2.18. Such a subsequence is

actually Cauchy and this contradicts equation (∗).

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