some notes for ma342 james cruickshankjc/docs/topologynotes.pdf · james cruickshank. notation...

$: Some Notes for MA342 James Cruickshankjc/docs/TopologyNotes.pdf · James Cruickshank. Notation \)" means \implies that". \," means \if and only if". N denotes the set of natural numbers$
Some Notes for MA342

James Cruickshank

Notation• “⇒” means “implies that”.• “⇔” means “if and only if”.• N denotes the set of natural numbers.• Z denotes the set of integers.• Q denotes the set of rational numbers.• R denotes the set of real numbers.• C denotes the set of complex numbers.• R+ denotes the set of strictly positive real numbers.

Contents

Chapter 0. Some Motivation for Topology 51. Knots 52. Networks and Graphs 53. The Fundamental Theorem of Algebra 5

Chapter 1. Set Theory 71. Sets 72. Operations on Sets 73. The Algebra of Sets 94. Functions 9

Chapter 2. Topological Spaces 131. Motivation 132. The Definition and Some First Examples 133. Bases 154. Closed Sets 165. Interior and Closure 17

Chapter 3. Continuity and Homeomorphisms 191. Continuous Functions 192. Homeomorphisms 21

Chapter 4. New Spaces From Old 231. Subspaces 232. Quotient Spaces 243. Product Spaces 264. Interesting Examples of Spaces 27

Chapter 5. Connectedness and Compactness 291. Connectedness 292. Compactness 30

3

CHAPTER 0

Some Motivation for Topology

I will mention a few problems that have ’topological connections’.The idea is to give an impression of the breadth of application of thesubject rather than to explain the detail of each application. Topol-ogy is the underlying mathematical theory that connects all of theseproblems. Our course will not necessarily explain how to solve theseproblems, but we will see how to put these problems into the appro-priate mathematical context.

1. Knots

Can we classify different types of knots. What does that mean?Recently biologists have become interested in study knots that arise inprotein structures.

2. Networks and Graphs

Suppose that Anne, Barry and Cathy work in three different officeand live in three different houses. Is it possible to find paths for eachperson from house to office so that no pair of the paths intersect.

3. The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that if n ≥ 1 anda0, · · · an−1 ∈ C then there is some z ∈ C such that

zn + an−1zn−1 + · · ·+ a1z + a0 = 0

Note that this is not true over the field of real numbers. The FTA isan existence result only - it says nothing about the location of z Manyquestions about the existence of solutions of equations ar topological.

5

CHAPTER 1

Set Theory

1. Sets

Topology is typically one of the first courses in which the student ismostly focused on logical and set theoretical concepts, rather than oncomputational issues. In this chapter we will review some of the basicnotions of set theory that will be used in the rest of the course.

1.1. Equality and containment. We will take the notion of aset as a primitive concept. We write x ∈ X if x is an element of theset X (or if x belongs to the set X).

If X1 and X2 are sets, then we write

X1 = X2

if, x ∈ X1 ⇔ x ∈ X2.We write A ⊂ X if every element of A is also an element of X.

1.2. The set builder notation. There are various common waysin which sets are specified or described. One very common way isto describe a set is to give properties that characterise its elements.For example, we might say that X is the set of all integers that areone greater than the square of an integer. Rather than writing outthe description as in the previous sentence, it is more common to usemathematical shorthand as follows.

X = {a ∈ Z : a = n2 + 1 for some n ∈ Z}.

The curly braces “{” and “}” enclose a description of the elements ofa set. In this context the colon “:” is shorthand for “such that”. Wecould also specify the same set in the following way.

X = {n2 + 1 : n ∈ Z}.

2. Operations on Sets

The union of sets A and B, denoted A ∪B, is defined by

A ∪B = {x : x ∈ A or x ∈ B}.7

8 1. SET THEORY

The intersection of A and B, denoted A ∩B, is defined by

A ∩B = {x : x ∈ A and x ∈ B}.Sometimes, we would like to discuss the union or intersection of

infinitely many sets. Thus, suppose that {Ai, i ∈ I} is a collection ofsets. Then x ∈

⋃i∈I Ai ⇔ x belongs to at least one Ai0 . Similarly

x ∈⋂i∈I Ai ⇔ x belongs to every Ai. In this context, the set I is

called an indexing set.In the special case where the indexing set is a set of consecutive

integers, we often use the notations⋃i=ni=mAi or

⋂i=ni=mAi.

The complement of A in B, denoted B − A, is defined by

B − A = {x ∈ B : x /∈ A}.Given sets A and B, the (cartesian) product of A and B, denoted

A×B, is defined by

A×B = {(a, b) : a ∈ A and b ∈ B}.In other wordsA×B is the set of all ordered pairs whose first componentis an element of A and whose second component is an element of B.Note that in this context (a, b) denotes an ordered pair and not aninterval of R.

2.1. Equivalence relations. An equivalence relation on a set Xis a subset R of X ×X that satisfies the following three conditions.

(1) For all x ∈ X, (x, x) ∈ R (i.e. R is reflexive).(2) If (x, y) ∈ R then (y, x) ∈ R (i.e. R is symmetric).(3) If (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R (i.e. R is transi-

tive).

If R is an equivalence relation on X, we will write x ∼R y if (x, y) ∈R, or sometimes just x ∼ y (if no possible confusion can arise fromsuppressing the R in the notation). If R is an equivalence relation onX, and x ∈ X, then the equivalence class of x, denoted by [x]R or [x],is defined by

[x] = {y ∈ X : x ∼ y}.The set of all equivalence classes, denoted by X/ ∼, is defined by

X/ ∼= {[x] : x ∈ X}.

2.2. Problems.

(1) Give a concise description of the set∞⋃i=1

[1

n, 1]

4. FUNCTIONS 9

where [a, b] denotes {x ∈ R : a ≤ x ≤ b}.(2) True or false: If A ∩B 6= ∅ and B ∩ C 6= ∅ then A ∩ C 6= ∅.(3) Suppose that {Ai : i ∈ I} is a collection of sets. What is

meant by the notation∏

i∈I Ai?(4) Suppose that R1 and R2 are both equivalence relations on a

set X. Prove that R1 ∩ R2 is also an equivalence relation onX.

(5) Let A be a subset of X × X. Prove that there is a uniqueequivalence relation RA with the property that RA is a subsetof any equivalence relation that contains A (note: In this sit-uation, we say that RA is the equivalence relation generatedby A)

3. The Algebra of Sets

Theorem 3.1. Suppose that A, B and C are sets. Then

(1) A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)(2) A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C)

Proof. INSERT PROOF HERE �

3.1. De Morgan’s Laws. The conclusions of the following theo-rem are often referred to as De Morgan’s Laws.

Theorem 3.2. Suppose that X and Ai, i ∈ I are sets. Then

(1) X −⋃i∈I Ai =

⋂i∈I(X − Ai)

(2) X −⋂i∈I Ai =

⋃i∈I(X − Ai)

Proof. We will prove the first conclusion and leave the proof ofthe second conclusion to the reader. We have x ∈ X −

⋃i∈I Ai ⇔

x ∈ X and x /∈⋃i∈I Ai. But x /∈

⋃i∈I Ai ⇔ x /∈ Ai for all i ∈ I.

Therefore x ∈ X −⋃i∈I Ai ⇔ x ∈ X − Ai for all i ∈ I. Therefore

x ∈ X −⋃i∈I Ai ⇔ x ∈

⋂i∈I(X − Ai) as required. �

Informally, De Morgan’s Laws say that the complement of the unionis the intersection of the complements, and that the complement of theintersection is the union of the complements.

4. Functions

We will also take the notion of function as a primitive notion. Wewrite f : X → Y if f is a function whose domain is X and whosecodomain is Y . Throughout this section we will assume that f : X →Y .

A function f is injective if f(x) = f(y) ⇒ x = y. A functionf : X → Y is surjective if, for every y ∈ Y , there exists x ∈ X

10 1. SET THEORY

such that y = f(x). A function is bijective if it is both injective andsurjective.

Given that A ⊂ X, the image of A under f , denoted f(A) is definedby

f(A) = {y : y = f(a) for some a ∈ A}.Given that B ⊂ Y , the preimage of B under f , denoted f−1(B), is

defined by

f−1(B) = {x ∈ X : f(x) ∈ B}

4.1. Sequences. A sequence in X is a function whose domain isN and whose codomain is X (i.e. a function f : N→ X). The value ofthe function at n is called the nth term of the sequence and is usuallydenoted by fn (rather than f(n)).

We often write things like “. . . (xn)∞n=1 is a sequence in ...”, whichindicates that xn is the nth term of the given sequence. Sometimeswe will just write (xn) instead of (xn)∞n=1 when there is no possibleconfusion.

Suppose that (xn)∞n=1 is a sequence in X. A subsequence of (xn)is a sequence of the form (xnk

)∞k=1, where (nk) is a strictly increasingsequence in N. In other words, ni+1 > ni for all i ∈ N. We can think ofa subsequence of (xn) as the sequence that is left after we delete someof the terms of (xn) (we can delete infinitely many terms, but we mustleave infinitely terms undeleted).

For example, suppose that xn = 1n. So

(xn) = (1,1

2,1

3,1

4, . . . ).

Now suppose nk = k2 + 1. So

(nk) = (2, 5, 10, 17, . . . ).

The

(xnk) = (

1

2,1

5,

1

10,

1

17, . . . ).

4.2. Problems.

(1) Let f : N→ R be defined by f(a) =√a. What is f−1(Q)?

(2) Suppose that f : R→ R is quadratic polynomial function.(a) Prove that f is not surjective.(b) Prove that f is not injective.

(3) Let g : R → R be defined by g(x) = x3 + ax2 + bx + c wherea, b and c are real constants. Prove that g is bijective if andonly if a2 < 3b.

4. FUNCTIONS 11

(4) Let X be a set and let XN denote the set of all sequences inX. Show that there is a natural bijective function

XN → X ×X ×X × . . .

CHAPTER 2

Topological Spaces

1. Motivation

Suppose that (X, d) is a metric space (i.e. d is a metric on the setX). Recall that a subset A ⊂ X is said to be d-open in X if, givenany x ∈ A, there exists ε > 0 such that d(x, z) < ε ⇒ z ∈ A. Thisnotion of “openness” of a subset plays a crucial role in metric spacetheory. So it is natural to ask if it is possible to axiomatise this notionof “openness”. In other words, is it possible to write down a list ofproperties which somehow capture all the ‘essential’ features of d-opensubsets of X?

2. The Definition and Some First Examples

Definition 2.1. Let X be a set. We say that T is a topology onX if T is a set of subsets of X that satisfies the following properties.

(1) ∅ ∈ T and X ∈ T .(2) If Gi ∈ T for all i ∈ I then

⋃i∈I Gi ∈ T .

(3) If G1 ∈ T and G2 ∈ T , then G1 ∩G2 ∈ T .

The elements of T are called open subsets of X (with respect to T ). Atopological space is a pair (X, T ) where T is a topology on the set X.

Informally, we can summarise the properties of a topology by sayingthat any union of open sets is open and that the intersection of twoopen sets is open.

Example 2.2. LetX = {a, b, c} and let T = {∅, {a}{a, b}, {a, c}, X}.Then T is a topology on X.

Example 2.3. Let P (X) denote the set of all subsets of a set X.Then P (X) is a topology on X. This topology is called the discretetopology.

Example 2.4. Let X be a set and let

Xfc = {G ⊂ X : X −G is finte} ∪ {∅}We will show that Xfc is a topology on X. Clearly ∅ ∈ Xfc. AlsoX −X = ∅ which is finite, so X ∈ Xfc.

13

14 2. TOPOLOGICAL SPACES

Now suppose that Gi ∈ Xfc for i ∈ I. If⋃i∈I Gi 6= ∅ then Gi0 6= ∅

for some i0 ∈ I. So X −Gi0 is finite. We have

X −⋃i∈I Gi =

⋂i∈I(X −Gi) by de Morgan’s Laws

⊂ X −Gi0

Thus X −⋃i∈I Gi is finite (since it is a subset of a finite set), and⋃

i∈I Gi ∈ Xfc.Suppose that G1 ∈ Xfc and G2 ∈ Xfc. If either of G1 or G2 is the

empty set, then so is G1 ∩ G2. So suppose that G1 6= ∅ and G2 6= ∅.Then

X − (G1 ∩G2) = (X −G1) ∪ (X −G2)

by de Morgan’s Laws. Therefore X − (G1 ∩G2) is finite (since it is theunion of two finite sets), and G1 ∩G2 ∈ Xfc as required.

The topology Xfc is called the finite complement topology on X.

2.1. The topology induced by a metric. Let (X, d) be a metricspace. Recall that a subset G of X is said to be d-open if, for all x ∈ Gthere exists ε > 0 such that d(x, y) < ε⇒ y ∈ G. Let

Td = {G ⊂ X : G is d-open}

Lemma 2.5. Td is a topology on X.

Proof. It is clear that ∅ and X are both in Td.Now suppose that Gi is d-open for i ∈ I. Let x ∈

⋃i∈I Gi. So

x ∈ Gi0 for some i0 ∈ I. Therefore, there is some ε > 0 such thatd(x, y) < ε⇒ y ∈ Gi. Clearly d(x, y) < ε⇒ y ∈

⋃i∈I Gi, so

⋃i∈I Gi is

d-open.Suppose that G1 ∈ Td and G1 ∈ Td. Let x ∈ G1∩G2. There is some

ε1 > 0 such that d(x, y) < ε1 ⇒ y ∈ G1. Similarly there is some ε2 > 0such that d(x, y) < ε2 ⇒ y ∈ G2. Now, let ε = min{ε1, ε2}. Clearly,ε > 0 and d(x, y) < ε⇒ y ∈ G1 ∩G2. So G1 ∩G2 is d-open.

We have shown that unions of d-open subsets are d-open and thatthe intersection of any two d-open subsets is d-open. Therefore Td is atopology on X. �

The topology Td is called the topology induced by the metric d, orsometimes just the metric topology.

Example 2.6. The standard metric on Rn is defined by

d(x, y) =√

(x1 − y1)2 + · · ·+ (xn − yn)2

where x = (x1, . . . , xn) and y = (y1, . . . , yn) are points in Rn. Thetopology induced by this metric is called the standard topology on Rn,or the usual topology on Rn.

3. BASES 15

3. Bases

Definition 3.1. Let B be a collection of subsets of X. We saythat B is basis (for a topology on X) if the following two statementsare true.

(1)⋃B∈B B = X.

(2) Given B1 ∈ B and B2 ∈ B and x ∈ B1 ∩ B2, there is someB3 ∈ B such that x ∈ B3 and B3 ⊂ B1 ∩B2,

Elements of B are called basic (open) subsets of X.

Note that the second property in Definition 3.1 is a weaker condi-tion than being closed under finite intersection. In other words, anycollection that closed under finite intersections also has this property.

Example 3.2. Let

B = {(a, b) where a and b are real numbers with a < b}.

Then B is a basis for a topology on R.

Proposition 3.3. Let B be a basis as defined above. Then thecollection

T = {G ⊂ X : G is a union of basic subsets} ∪ {∅}

is a topology on X.

Proof. By construction ∅ ∈ T . Also, by the definition of a basisX is a union of basic subsets. Therefore X ∈ T .

Now suppose that, for each i ∈ I, Gi ∈ T . So each Gi is a unionof basic subsets of X. Then clearly

⋃i∈I Gi is also a union of basic

subsets of X. So⋃i∈I Gi ∈ T .

Finally, suppose that G1 ∈ T and G2 ∈ T . Let x ∈ G1 ∩G2. Sincex ∈ G1, there is some basic subset B1 such that x ∈ B1 and B1 ⊂ G1.Similarly, there is some basic subset B2 such that x ∈ B2 and B2 ⊂ G2.Now x ∈ B1 ∩ B2, so by the definition of a basis there is some basicset Bx such that x ∈ Bx and Bx ⊂ B1 ∩ B2. But B1 ∩ B2 ⊂ G1 ∩ G2.So for each x ∈ G1 ∩ G2 there is a basic set Bx such that x ∈ Bx andBx ⊂ G1 ∩G2. Clearly,

G1 ∩G2 =⋃

x∈G1∩G2

Bx,

so G1 ∩G2 ∈ T as required. �

We say that the topology described in Proposition 3.3 is the topol-ogy generated by the basis B.


Example 3.4. (The Sorgenfrey topology) Consider the followingcollection of subsets of R.

B = {[a, b) : a < b}.Recall that [a, b) = {x ∈ R : a ≤ x < b}. We will show that B isa basis. Clearly, R =

⋃∞n=1[−n, n). Now, suppose that [a1, b1) ∈ B

and [a2, b2) ∈ B. It is easy to check that [max a1, a2,min{b1, b2}) =[a1, b1) ∩ [a2, b2) (the reader should verify this). So in fact B is closedunder finite intersections.

The topology generated by this basis is called the Sorgenfrey topol-ogy on R.

3.1. Problems.

(1) Prove that (0, 1) is an open subset of R with respect to theSorgenfrey topology.

(2) Prove that every subset of R that is open with respect to theusual topology is also open with respect to the Sorgenfreytopology.

(3) Let (X, d) be a metric space. Recall that if x ∈ X and r ∈ R,then B(x, r) = {y ∈ X : d(x, y) < r}. Let

B = {B(x, r) : x ∈ X, r ∈ R}.Show that B is a basis. Show that the topology generated bythis basis is the same as the topology described in Section 2.1.

(4) Let X be a set and let B = {{x} : x ∈ X}. Show that B is abasis. What is the topology generated by B?

4. Closed Sets

Let (X, T ) be a topological space. A closed subset of X is a subsetof X whose complement in X is open, In other words A is a closedsubset of X if and only if X − A ∈ T .

For example, the interval [a, b] is a closed subset of R with respectto the usual topology, since [a, b] = R− ((−∞, a) ∪ (b,∞)).

Using de Morgan’s Laws, we can translate many of the propertiesof open sets into corresponding properties of open sets. For example

Proposition 4.1. Let (X, T ) be a topological space. Then

(1) ∅ and X are closed subsets of X.(2) If, for each i ∈ I, is a closed subset of X, then

⋂i∈I Ai is a

closed subset of X.(3) If A1 and A2 are closed subsets of X, then A1 ∪A2 is a closed

subset of X.

5. INTERIOR AND CLOSURE 17

Proof. �

Example 4.2. (The Cantor Set) For each positive integer n, let

An =

12(3n−1)⋃j=0

[2j

3n,2j + 1

3n

].

Now, for each n ∈ N, An is a closed subset of R with respect to theusual topology, since it is a finite union of closed intervals. We definethe Cantor ternary set by letting

C =∞⋂n=1

An

Since each An is closed, we see that C is also a closed subset of Rwith respect to the usual topology. The reader should try to provethat C consists of all real numbers in [0, 1] that cannot have the digit2 occurring in any base 3 expansion of that number (note that somenumbers may have non unique base 3 expansions).

A topological space is said to be T1 if every singleton subset is aclosed subset. Note that we have already seen some examples of spacesthat are not T1. However, Rn with its usual topology is certainly T1.Indeed, any metric space is T1.

5. Interior and Closure

Throughout this section (X, T ) is a topological space.

Definition 5.1. Let A be a subset of X. The interior of A, denotedby A or int(A) is the union of all of the open sets, contained in A. Theclosure of A, denoted A or cl(A) is the intersection of all of the closedsets that contain A.

Example 5.2. Consider R with the usual topology. Then Q = ∅and Q = R

Example 5.3. Consider R with the Sorgenfrey topology. Thenint([0, 1)) = [0, 1) and cl([0, 1)) = [0, 1).

We say that a subset A of X is dense in X if A = X.Now, we would like to characterise the points that lie in the interior

or closure of a set A.

Proposition 5.4. Let A be a subset of X. Then

(1) x ∈ A if and only if there is some open subset B such thatx ∈ B and B ⊂ A.


(2) x ∈ A if and only if every open subset that contains x intersectsA nontrivially. (i.e. B open and x ∈ B, ⇒ B ∩ A 6= ∅).

Proof. By definition A =⋃G⊂A,GopenG, so (1) follows immedi-

ately from the definition. For (2) observe first that, given subsets Aand C of X, we have A ⊂ C ⇔ X − C ⊂ X − A. Therefore we canargue that x ∈ A⇔ x belongs to every closed set that contains A⇔ xdoes not belong to any open set contained in X − A ⇔ x does notbelong to any open subset of X that is disjoint from A⇔ any open setthat contains x is not disjoint from A. (Recall that sets are disjoint ifand only if they have empty intersection.) �

Now, we consider the case where the topology is induced by a metric

Theorem 5.5. Let (X, d) be a metric space and suppose that A ⊂X. Then x ∈ A if and only if there is a sequence in A that convergesto x.

Proof. Suppose first that x ∈ A. Let n ∈ N and recall thatB(x, 1

n) = {y ∈ X : d(x, y) < 1

n} is an open subset of X that contains x.

By Proposition 5.4 A∩B(x, 1n) 6= ∅, so we can choose an ∈ A∩B(x, 1

n).

It is clear that (an) is a sequence in A that converges to x.On the other hand, suppose that (an) is a sequence in A that con-

verges to x and let G be any open set containing x. There is someε > 0 such that d(x, y) < ε ⇒ y ∈ G. However, there is some N suchthat n > N ⇒ d(an, x) < ε. Therefore an ∈ G ∩ A for all n > N . Inparticular G ∩ A 6= ∅. So, by Proposition 5.4, x ∈ A. �

5.1. Problems.

(1) Prove that X − int(A) = cl(X − A).(2) Prove that X − cl(A) = int(X − A).

(3) Prove that A = A⇔ A is an open subset of X.(4) True or false: A ∩B = A ∩B.

(5) True or false: int(A ∩B) = A ∩ B.(6) True or false: A ∪B = A ∪B.

(7) True or false: int(A ∪B) = A ∪ B.

CHAPTER 3

Continuity and Homeomorphisms

1. Continuous Functions

So far we have been concerned with topological spaces ’in isolation’.Now we want to consider functions between such spaces. The topologyallows us to talk about continuity of such functions. Recall that if f :X → Y is a function and B ⊂ Y , then f−1(B) = {x ∈ X : f(x) ∈ B}(f−1(B) is called the preimage of B under f .)

Definition 1.1. Let (X, TX) and (Y, TY ) be topological spaces andlet f : X → Y be a function. We say that f is a continuous functionif, for every open subset B of Y , its preimage f−1(B) is an open subsetof X.

Notice, the relative simplicity of Definition 1.1 as compared to thetypical ε-δ definition of continuity that one sees in a first course inanalysis or in metric spaces. It can be distilled down to the simplemnemonic “continuous means preimage of open is open”. By contrast,the ε-δ definition of continuity in metric spaces is awkward to state andto use.

Of course, we must do some work to prove that this definition ofcontinuity is really an extension of the notion of continuity of functionsof metric spaces. That is the content of Theorem 1.4 below. Beforeproving that, we give some examples to illustrate Definition 1.1.

Example 1.2. Let f : R→ R be the function defined by f(x) = x.In this case, if B ⊂ R then f−1(B) = B. So f is continuous if and onlyif every set that is open with respect to the topology of the codomain isalso open with respect to the topology of the domain. For example, ifthe domain has the usual topology and the codomain has the Sorgenfreytopology, then f is not continuous. However, if the domain has theSorgenfrey topology and the codomain has the usual topology then fis continuous (why?)

Example 1.3. Let F be a field and let p : F→ F be a polynomialfunction. Suppose that the domain and codomain are equipped withthe finite complement topology, then p is continuous. Note that con-tinuity of polynomial functions in this context is a consequence of the

19

20 3. CONTINUITY AND HOMEOMORPHISMS

purely algebraic fact that nontrivial polynomials have finitely manyroots. Indeed this example has a far reaching generalisation in thesubject of algebraic geometry, where one considers the so called Zariskitopology and continuous functions with respect to that topology.

Now, we return to the comparison of continuity in the metric andtopological contexts.

Theorem 1.4. Let (X, dX) and (Y, dY ) be metric spaces and letf : X → Y . Then f is continuous is the sense of metric space theory ifand only if f is continuous in the sense of Definition 1.1 with respectto topologies induced by the metrics.

Proof. First suppose that f is continuous in the metric sense.Let B be an open subset of Y and let x ∈ f−1(B). Since B is openthere is some ε > 0 such that dY (f(x), y) < ε ⇒ y ∈ B. Now, sincef is continuous in the metric sense, there is some δ > 0 such thatdX(x, u) < δ ⇒ dY (f(x), f(u)) < ε. Therefore, if dX(x, u) < δ thenu ∈ f−1(B). Thus f−1(B) is an open subset of X.

Conversely, suppose that f is continuous in the sense of Definition1.1. Let x ∈ X and let ε > 0. The set B = {y ∈ Y : dY (f(x), y) < ε} isan open subset of Y , so, by assumption, f−1(B) is an open subset of X.Now, clearly x ∈ f−1(B), so there is some δ > 0 such that dX(x, u) <δ ⇒ u ∈ f−1(B). Therefore, dX(x, u) < ε⇒ dY (f(x), f(u)) < ε and fis continuous in the sense of metric spaces, as required. �

We observe that we can characterise also continuity using closedsets

Proposition 1.5. A function f : X → Y is continuous if and onlyif the preimage of every closed subset of Y is a closed subset of X.

1.1. Problems.

(1) Let f : R → R be defined by f(x) = x2. In each of thefollowing cases, decide whether or not f is continuous.(a) The domain has the usual topology and the codomain has

the discrete topology.(b) The domain has the discrete topology and the codomain

has the Sorgenfrey topology.(c) The domain has the Sogenfrey topology and the codomain

has the usual topology.(d) The domain has the usual topology and the codomain has

the finite complement topology.(e) The domain has the finite complement topology and the

codomain has the usual topology.

2. HOMEOMORPHISMS 21

(2) Prove that a function f : X → Y is continuous if and only if

f(A) ⊂ f(A) for all subsets A of X.

2. Homeomorphisms

In popular literature, topology is sometimes referred to as “rubbersheet geometry”. The following definition is the rigourous idea thatunderlies that intuitive notion.

Definition 2.1. Let (X, TX) and (Y, TY ) be topological spaces. Afunction f : X → Y is a homeomorphism if f is continuous, bijectiveand has a continuous inverse. If a homeomorphism f : X → Y existsthen we say that (X, TX) is homeomorphic to (Y, TY ) and we write(X, TX) ≡ (Y, TY ) (or sometimes just X ≡ Y ).

If X ≡ Y then X and Y are indistinguishable as topological spaces.One of the fundamental goals of topology is to develop methods fordeciding when two spaces are related by a homeomorphism. In general,it can be very difficult to decide.

Example 2.2. Consider the metric spaces (0, 1) and (0,∞) wherethe metric in each case is the standard metric (i.e. d(x, y) = |x −y| The function f : (0, 1) → (0,∞) defined by f(x) = 1

x− 1 is a

homeomorphism. Thus (0, 1) ≡ (0,∞). Observe that as metric spaces,these spaces are fundamentally different. Indeed the standard metric on(0, 1) is bounded whereas the standard metric on (0,∞) is unbounded.However as topological spaces, these two spaces are indistinguishable.

Proposition 2.3. The relation of homeomorphism is an equiva-lence relation.

Proof. We must show that “≡” is reflexive, symmetric and tran-sitive.

Let IX : X → X denote the identity function (i.e. IX(x) = x for allx ∈ X). Clearly, IX is a homeomorphism. So X ≡ X.

Suppose that X ≡ Y . So there is a homeomorphism f : X → Y .It is easy to check that f−1 : Y → X is also a homeomorphism. SoY ≡ X.

Finally, suppose that X ≡ Y and Y ≡ Z. So there are homeomor-phisms f : X → Y and g : Y → Z. Now, it is clear that g ◦ f : X → Zis a homeomorphism, so X ≡ Z. �

Example 2.4. Consider (0,∞) with its usual topology and (0, 1)with its usual topology. The function f : (0, 1) → (0,∞) given byf(t) = 1

t− 1 is a homeomorphism.

22 3. CONTINUITY AND HOMEOMORPHISMS

Example 2.5. On the other hand (R,U) is not homeomorphic to(R,P) where U is the usual topology and P is the discrete topology.This follows from the fact that if f : R→ R is continuous with respectto the usual topology on the domain and the discrete topology on thecodomain, then f is a constant function and therefore is not a bijection.

2.1. Problems.

(1) Let U be the usual topology on [0, 2] and define

B = U ∪ {[1, a) : a > 1}.(a) Show that B is a basis for a topology on [0, 2]. Denote

this topology by T .(b) Prove that ([0, 2], T ) ≡ [0, 1) ∪ [2, 3] where [0, 1) ∪ [2, 3]

has the usual topology.

CHAPTER 4

New Spaces From Old

1. Subspaces

Let (X, T ) be a topological space and let A be a subset of X. Wedefine

TA = {G ∩ A : G ∈ T }.

Proposition 1.1. The collection TA is a topology on A.

Proof. We verify that TA has all the properties required of a topol-ogy as follows. First ∅ = ∅ ∩A and A = X ∩A. Therefore ∅ ∈ TA andA ∈ TA.

Now suppose that Ui ∈ TA for i ∈ I. So for each i ∈ I there is someopen subset Gi of X such that Ui = Gi ∩ A. Now⋃

i∈I

Ui =⋃i∈I

(Gi ∩ A) = (⋃i∈I

Gi) ∩ A ∈ TA.

Finally, suppose that U1 and U2 are in TA. So U1 = G1 ∩ A andU2 = G2 ∩ A for some open subsets G1 and G2 of X. Then

U1 ∩ U2 = (G1 ∩ A) ∩ (G2 ∩ A) = (G1 ∩G2) ∩ A ∈ TA.�

The topology TA is called the induced topology, or the subspacetopology, on A.

Example 1.2. Consider R with its usual topology. The inducedtopology on the subset Z is the discrete topology. We can prove thisby observing that for any n ∈ Z, we have

{n} = (n− 1

2, n+

1

2) ∩ Z.

Therefore {n} is an open subset of Z with respect to the subspacetopology. However, if every singleton subset is open, then the topologymust be the discrete topology.

Example 1.3. Again consider R with the usual topology and con-sider the subset X = {1

2, 13, 14, . . . } and Y = X ∪ {0}. the induced

topology on X is the discrete topology. The induced topology on Y

23

24 4. NEW SPACES FROM OLD

is not the discrete topology, since any open subset of Y containing 0,must have finite complement in Y .

2. Quotient Spaces

It is recommended that the reader review the section on equivalencerelations (Section 2.1) before proceeding.

We begin by observing that given a set X and a subset A ⊂ X,there is an injective function i : A→ X defined by i(x) = x called theinclusion function. In a sense, every injective function can be thoughtof as an inclusion function.

In a similar (but dual) way, surjective functions correspond to equiv-alence relations. Given a function p : Y → B, we can define an equiv-alence relation on Y by x ∼p y ⇔ p(x) = p(y). We leave it as aneasy exercise to check that ∼p is indeed an equivalence relation. Onthe other hand, given an equivalence relation on a set Y , we have thecanonical function

p : Y → Y/ ∼defined by p(y) = [y]. This function is sometimes called the quotientmap.

2.1. The quotient topology. Throughout this section, let (X, T )be a topological space. Suppose that p : X → A is a function (of sets).Define

Tp = {G ⊂ A : p−1(G) ∈ T }.

Lemma 2.1. The collection Tp forms a topology on A.

Proof. First, observe that ∅ = p−1(∅) and X = p−1(A), so ∅ ∈ Tpand A ∈ Tp.

Now suppose that Gi ∈ Tp for i ∈ I. So p−1(Gi) ∈ T for all i ∈ I.Then p−1(

⋃i∈I Gi) =

⋃i∈I p

−1(Gi) ∈ T , since T is closed under unions.Therefore

⋃i∈I Gi ∈ Tp.

Similarly, if G1 ∈ Tp and G2 ∈ Tp. Then p−1(G1 ∩G2) = p−1(G1)∩p−1(G2) ∈ T , since p−1(G1) ∈ T and p−1(G2) ∈ T .. Therefore G1 ∩G2 ∈ Tp. �

The topology Tp is called the quotient topology on X.

Lemma 2.2. With notation as above, the function p : X → A isa continuous function, with respect to the topologies T and Tp on thedomain and codomain respectively.

Proof. �

2. QUOTIENT SPACES 25

It is useful to be able to think about the quotient topology in thecontext of equivalence relations as well. Suppose that ∼ is an equiv-alence relation on X and let p : X → X/ ∼ be the quotient map (asdescribed above). In this case, the inverse image under p of a subsetG of X/ ∼ is just the union of all the equivalence classes of ∼ thatare members of G. Thus, we see that the quotient topology can bedescribed as follows.

Tp = {G ⊂ X/ ∼:⋃

[x]∈G

[x] is an open subset of X}.

In this context, we often refer to the topological space (X/ ∼, Tp) as thequotient space. The quotient space construction is a very importantway of constructing examples of topological spaces with interestingproperties.

Example 2.3. Define an equivalence relation on R by x ∼ y ⇔x−y ∈ Z. The quotient space R/ ∼ is homeomorphic to S1 = {(x, y) ∈R2 : x2 + y2 = 1}. The proof of this fact will be given below.

Example 2.4. Consider Rn+1−{0}. We can define an equivalencerelation on this space by (x0, . . . , xn) ∼ (y0, . . . , yn) if and only if thereis some λ 6= 0 such that (y0, . . . , yn) = λ(x0, . . . , xn). We leave it to thereader to check that this is indeed an equivalence relation. We define

RPn =Rn+1 − {0}

∼.

If Rn+1 − {0} is equipped with its usual topology, then RPn can betopologised as a quotient space. This space is called the n-dimensional(real) projective space.

Now suppose that ∼ is an equivalence relation on X. Let (B,S)be a topological space and suppose that f : X → B has the propertythat f(x) = f(y) whenever x ∼ y. In this situation, we can define afunction f : X/ ∼→ B, by f([x]) = f(x). The function f is said to beinduced by f .

Theorem 2.5. With the notation as above, if f is continuous, thenf is continuous with respect to the quotient topology on X/ ∼.

Proof. Suppose that U is an open subset of B and consider its

preimage f−1

(U). We must show that this an open subset of X/ ∼.

Now, by construction f = f ◦p. Therefore f−1(U) = p−1(f−1

(U)). Butf is continuous. Therefore f−1(U) is an open subset of X. Therefore

p−1(f−1

(U)) is an open subset of X. So by the definition of the quotient


topology f−1

(U) is an open subset of X/ ∼. Therefore f is a continuousfunction. �

Now, we can prove the assertion of Example 2.3. Define f : R→ S1

by f(x) = (cos(2πx), sin(2πx)). It is clear that f(x) = f(y)⇔ x− y ∈Z ⇔ x ∼ y where ∼ is the equivalence relation defined in Example2.3. Thus f induces a function f : R/ ∼→ S1. By Theorem 2.5 fis continuous. Moreover f is surjective since f is surjective and f =f ◦p. Also f is injective since f([x]) = f([y])⇒ (cos(2πx), sin(2πx)) =(cos(2πy), sin(2πy)) ⇒ x − y ∈ Z ⇒ [x] = [y]. So f is a continuous

bijection. It remains to show that f−1

is continuous. That is, we mustshow that if G is an open subset of R/ ∼ then f(G) is an open subsetof S1. We can do this as follows. Since G is open in R/ ∼, it mustbe that U = p−1(G) is an open subset of R. But f(U) = f(G). Now,we claim that f(U) is an open subset of S1. This follows from the factthat the image under f of any open interval of R is an open subset ofS1 - since f maps basic open subsets to open subsets - it maps anysubset to an open subset (see problems below).

2.2. Problems.

(1) Let X and Y be topological spaces and let BX be a basis forthe topology on X and let BY be a basis for the topology onY . Prove the following assertions(a) A function f : X → Y is continuous if and only if, for

every basic open subset G of Y , f−1(G) is an open subsetof X.

(b) Suppose that for every basic open subset U of X, f(U) isan open subset of Y . Then, for any open subset V of X,f(V ) is an open subset of Y .

3. Product Spaces

Throughout this section, let (X, TX) and (Y, TY ) be topologicalspaces. Recall that the cartesian product of the sets X and Y is theset

X × Y = {(x, y) : x ∈ X, y ∈ Y }.In other words, X×Y is the set of all possible ordered pairs where thefirst element of the pair comes from X and the second element of thepair comes from Y .

Given that X and Y are equipped with topologies, it seems naturalto wonder if there is an obvious way to topologise X × Y . Let B ={U × V : U ∈ TX , V ∈ TY } (in other words B consist of all subsets

4. INTERESTING EXAMPLES OF SPACES 27

X × Y that are products of an open subset of X and an open subsetof Y . One might hope that this forms a topology on X × Y . However,we leave it to the reader to check that B is not generally closed underunions and is therefore not a topology. However,

Proposition 3.1. The collection B forms a basis for a topology onX × Y .

Proof. We show that B satisfies all the conditions required byDefinition 3.1.

First X ∈ TX and Y ∈ TY . Therefore X×Y ∈ B. Also ∅ = ∅×∅ ∈B.

Now suppose that U1 × V1 ∈ B and U2 × V2 ∈ B. Then U1 × V1 ∩U2 × V2 = (U1 ∩ U2)× (V1 ∩ V2). Therefore U1 × V1 ∩ U2 × V2 ∈ B. Inparticular, condition (2) of Definition 3.1 is satisfied. �

The topology generated by the basis B is called the product topol-ogy on X × Y .

Example 3.2. Let U be the usual topology on R. Then the producttopology on R× R is the same as the usual topology on R× R.

3.1. Problems.

(1) Let PX and PY be the discrete topologies on X and Y respec-tively. Show that the product topology on X × Y is also thediscrete topology on X × Y .

4. Interesting Examples of Spaces

In this section, we consider various examples of topological spacesthat arise as special cases of the constructions we have described above(subspaces, quotient spaces and product spaces) and we see that thesame topological space can often arise naturally in more than one way.We will not prove the existence of many of the homeomorphisms whoseexistence are asserted in this section. It is left to the reader to verifytheir existence.

4.1. Surfaces. Recall that S1 = {(x, y) ∈ R2 : x2 + y2 = 1}. Weassume that S1 is equipped with the topology it inherits as a subspaceof R2. We can define the torus, Σ, to be the product space S1 × S1.The normal way to visualise the torus is as a subspace of R3. Thus, let

X = {((3 + cos t) cos s, (3 + cos t) sin s, sin t) ∈ R3 : s, t ∈ R}

and suppose that X is topologised as a subspace of R3. Then Σ ≡ X.


Another common way to construct the torus is to start with theunit square

I2 = {(x, y) ∈ R2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}and define an equivalence relation on I2 as follows: Let ∼ be the equiv-alence relation generated by the following relations. (0, y) ∼ (1, y) and(x, 0) ∼ (x, 1). Now Σ ≡ I2/ ∼.

Motivated by the last construction of Σ, we can construct otherexamples of surfaces by gluing the edges of polygons as indicated inthe diagram below.

4.2. Topological Groups. Recall that a group is a set G togetherwith a group multiplication that satisfies certain properties. The groupmultiplication gives a function µ : G×G→ G defined by µ(g, h) = ghand the existence of unique inverses gives a function ι : G→ G definedby ι(g) = g−1. If G comes equipped with a topology, then G × Ghas the product topology. A topological group is a group for whichthe functions µ and ι are continuous. For example, consider additivegroup of real numbers. This is a topological group. Another classicalexample is the group GL(R, n) consisting of n× n invertible matrices

with real entries. This set is a subspace of Rn2and so inherits the

subspace topology. The group operation is matrix multiplication andwe leave it to the reader to prove that multiplication and inversion inthis case are both continuous functions.

There is a vast and important theory of topological groups. Theseobjects arise naturally is many areas of applied and pure mathematics.

CHAPTER 5

Connectedness and Compactness

Throughout this chapter (X, T ) is a topological space.

1. Connectedness

Definition 1.1. We say that X is connected if no proper subsetof X is both closed and open. We say that X is disconnected if thereis a proper subset of X that is both closed and open.

Equivalently, X is disconnected if there are disjoint open nonemptysubsets A and B such that X = A ∪ B. We call such a pair {A,B} adisconnection of X. So X is connected if there is no such disconnection.

Example 1.2. Consider the subspace [0, 1] ∪ [2, 3] of R (with itsusual topology). Then {[0, 1], [2, 3]} is a disconnection of this space.

Example 1.3. The space of rational numbers Q (a subspace of R)is disconnected, since (−∞,

√2)∩Q is a subset of Q that is both closed

and open (why is it closed?).

Example 1.4. The space R (with its usual topology is connected.To see this, suppose that {A,B} is a disconnection of R. Define f :R→ R by

f(x) =

{−1 if x ∈ A1 if x ∈ B

For C ⊂ R, f−1(C) is either ∅, A, B or R, all of which are opensubsets of R. Therefore f is continuous. But clearly f violates theIntermediate Value Theorem.

Theorem 1.5. Suppose that X is a connected space and that f :X → Y is a continuous surjective function. Then Y is also a connectedspace.

Proof. If {A,B} is a disconnection of Y , then {f−1(A), f−1(B)}is a disconnection of X. �

Corollary 1.6. Suppose that X ≡ Y . Then X is connected if andonly if Y is connected.

Proof. Homeomorphisms are continuous and surjective. �

29

30 5. CONNECTEDNESS AND COMPACTNESS

1.1. Path Connectedness. Sometimes connectedness can be aslightly awkward property to deal with. It is quite common in certainareas of mathematics to use a slightly stronger property. This propertyalso has the added feature that it corresponds more closely to ourintuitive notion of what connectedness should mean.

Definition 1.7. A space X is said to be path connected if, for allx ∈ X and y ∈ X, there is a continuous function g : [0, 1] → X suchthat g(0) = x and g(1) = y.

Proposition 1.8. If X is path connected, then X is connected.

Proof. �

However, there are spaces that are connected but not path con-nected.

Example 1.9. Let S be the following subspace of R2 (with theusual topology).

S = {t, sin(1

t) : t > 0} ∪ {(0, y) : −1 ≤ y ≤ 1}

This space is known as the (closed) topologist’s sine curve. It is con-nected but not path connected.

2. Compactness

In set theory, an important property that may be possessed by aset is that of being finite. Many combinatorial arguments rest on thefiniteness of a certain set and the theory of finite sets is much simplerthan that of infinite sets. The topological analogue of finiteness iscompactness.

Let (X, T ) be a topological space. A collection of subset of subsetsof X is said cover X if its union is X - in that case the collection iscalled a cover of the set X. A subcover of a cover of X is a subcollectionthat also covers X. An open cover of X is a collection of open subsetsthat covers X.

Definition 2.1. A topological space (X, T ) is compact if everyopen cover of X has a finite subcover.

Example 2.2. Let R be the space of real numbers with the usualtopology. The collection {(−n, n) : n ∈ N} is an example of an opencover of R. Clearly this open cover has no finite subcover. Thus R isnot compact.

Example 2.3. Let X be a set and let Tfc be the finite complementtopology on X. Then (X, Tfc) is a compact topological space.

2. COMPACTNESS 31

Proposition 2.4. If f : X → Y is a continuous surjective functionand X is compact, then Y is also compact.

Proof. Let O be an open cover of Y and define

Q{f−1(U) : U ∈ O}.

It is clear that Q is an open cover of X. So by assumption, it musthave a finite subcover. That is, there are open set Ui ∈ O, i = 1, . . . , ksuch that

X = f−1(U1) ∪ · · · ∪ f−1(Uk).Now since f is surjective f(f−1(A)) = A, for any A ⊂ Y . Therefore

Y = f(X)= f(f−1(U1) ∪ · · · ∪ f−1(Uk))= f(f−1(U1)) ∪ · · · ∪ f(f−1(Uk))= U1 ∪ · · · ∪ Uk

So O has a finite subcover as required. �

Corollary 2.5. If X ≡ Y then X is compact if and only if Y iscompact.

Proof. Homeomorphisms are continuous and surjective. �

2.1. Review of sequences. Suppose that (xn) is a sequence inX.

Definition 2.6. We say that (xn) converges to x ∈ X if, for everyopen subset U containing x, there is a positive integer NU such thatn > NU ⇒ xn ∈ U .

If a sequence (xn) converges to x, we write (xn)→ x or limn→∞ xn =x.

Note that in spaces that are not metrically induced, sequences canbehave in pathological ways.

Example 2.7. Consider the topological space (N, Tfc) where Tfcis the finite complement topology let xn = n. Then for all m ∈ N,(xn) → m. In other words, the sequence converges to every point inthe space.

In the case of metric spaces, it is easy to see that

Lemma 2.8. Suppose that (Y, d) is a metric space and that (yn) isa sequence in Y . Then (yn)→ y if and only if for every ε > 0, there issome Nε such that n > Nε ⇒ d(y, yn) < ε.


Proposition 2.9. Suppose that f : Y → Z is a continuous func-tion and that (yn)→ y for some sequence (yn) in Y . Then (f(yn))→f(y).

Proof. Suppose that U is an open subset containing f(y). Then,by continuity, f−1(U) is an open subset ofX that contains y. Therefore,there is some NU such that n > NU ⇒ yn ∈ f−1(U). Thereforen > NU ⇒ f(yn) ∈ U . �

For metric spaces we have the following converse to Proposition 2.9.

Proposition 2.10. Suppose that Y and Z are metric spaces andthat f : Y → Z has the property that for every convergent sequence(yn) in Y , (f(yn))→ f(limn→∞ yn). Then f is a continuous function.

Proof. Suppose that f is not continuous. So there is some closedsubset B of Z, such that f−1(B) is not a closed subset of Y . Therefore,there is some y ∈ Y − f−1(B) such that every open set containing ymeets f−1(B). Now, for each positive integer n, choose yn ∈ B(y, 1

n)∩

f−1(B). Clearly (yn) → y. But (f(yn)) is a sequence in the closedsubset B and f(y) is not in B. Therefore (f(yn)) cannot converge tof(y) contradicting our hypothesis. �

Note that the conclusion of Proposition 2.10 is not necessarily trueif we only assume that Y and Z are topological spaces.

2.2. Sequential Compactness. Recall also that a subsequenceof a sequence (xn) is a sequence (xnk

) where nk is a strictly increas-ing sequence of integers. So, in particular, a sequence that does notconverge may have a convergent subsequence.

Definition 2.11. A topological space X is said to be sequentiallycompact if every sequence in X has a convergent subsequence.

Example 2.12. The open interval (0, 1) (with the usual topology)is not compact since the sequence ( 1

n) has no convergent subsequence.

Theorem 2.13. (Bolzan-Weierstrass) Any closed bounded intervalof real numbers is sequentially compact.

Proof. Let [a, b] be a closed bounded interval and let (xn) be asequence in [a, b].

We define a sequence (nk) of natural numbers and sequences (ak)and (bk) of real numbers inductively as follows. Let n1 = 1, a1 = aand b1 = b. Now suppose that ni, ai and bi have been chosen for somei ≥ 1. Moreover, suppose inductively that ai ≤ bi and that the interval[ai, bi] contains infinitely many terms of the sequence (xn). Now, at

2. COMPACTNESS 33

least one of the intervals [ai,ai+bi

2] or [ai+bi

2, bi] must contain infinitely

many terms of the sequence (xn). Choose ai+1 and bi+1 so that [ai, bi]is one of the aforementioned intervals that does contain infinitely manyterms of (xn). Clearly, we can choose a positive integer ni+1 so thatni+1 > ni and xni+1

∈ [ai+1, bi+1]. Thus, we have chosen ni+1, ai+1

and bi+1. Moreover, [ai+1, bi+1] contains infinitely many terms of thesequence (xn).

Now, we claim that the sequence (xnk) is convergent. Notice that

by construction

ai ≤ ai+1 ≤ bi+1 ≤ bi

for all i. So (ak) is a bounded increasing sequence of real numbers and istherefore convergent. Similarly, (bk) is a bounded decreasing sequenceof real numbers and is therefore also convergent. Moreover, for eachi, bi+1 − ai+1 = bi−ai

2. Therefore bi − ai = b−a

2i→ 0 as i → ∞. Thus

lim(ak) = lim(bk). Moreover, for each i, ai ≤ xni< bi. So clearly, (xnk

)must be convergent as a sequence in R. Let x = lim(xnk

). Since [a, b]is a closed subset of R, x ∈ [a, b]. Therefore (xnk

) must be convergentas a sequence in [a, b]. �

We can easily generalise this to higher dimensions. A subset A ofRm is said to be bounded if there is some M such that d(0, a) ≤M forall a ∈ A.

Corollary 2.14. A subspace of Rm is sequentially compact if andonly if it is a closed and bounded subset of Rn.

Proof. Let A be a closed bounded subset of Rn and let (xn) be

a sequence in A. Suppose that for each n, xn = (x(1)n , x

(2)n , . . . , x

(m)n ),

where x(i)n ∈ R. Then (x

(1)n ) is a bounded sequence of real numbers

and thus, by Theorem 2.13, it has a convergent subsequence (x(1)nk ).

Now, (x(2)nk ) is a bounded sequence of real numbers and so, by Theorem

2.13, it has a convergent subsequence (x(1)nkl

). Continuing in this way

we eventually construct a sequence Nk such that (x(i)Nk

) is convergentfor all i. Therefore (xNk

) is a convergent subsequence of the originalsequence. Since A is closed, it contains all its limit points, so (xNk

)converges as a sequence in A.

Conversely suppose that A is not bounded, then, for every positiveinteger n, we can find xn ∈ A such that d(0, xn) > n. Clearly, (xn)has no convergent subsequence. If A is not closed then we can find asequence (xn) in A that converges (in Rn) to a point outside A. As asequence in A, (xn) has no convergent subsequence. �


2.3. Compactness versus Sequential Compactness. In gen-eral there is no relationship between compactness and sequential com-pactness. There are spaces that are compact but not sequentially com-pact and there are spaces that are sequentially compact but not com-pact. However, we will focus especially on the case of metric spaces andin this case we can show that in fact the two properties are equivalent.

Proposition 2.15. If Y is a compact metric space then it is se-quentially compact.

Proof. Suppose that (yn) be a sequence in Y that has no conver-gent subsequence. For k = 1, 2, . . . , let

Gn = Y − {yn, yn+1, yn+2, . . . }.First, we observe that each Gk is in fact an open subset of Y . To seethis, suppose that {yn, yn+1, yn+2, . . . } is not closed. Then by Propo-sition 5.4, we can find a sequence (zm) in {yn, yn+1, yn+2, . . . } thatconverges to a point outside {yn, yn+1, yn+2, . . . }. Clearly this impliesthe existence of a convergent subsequence of (yn) contradicting our as-sumption. So {yn, yn+1, yn+2, . . . } is closed and Gn is thus an opensubset of Y . Now observe that Gi ⊂ Gi+1 for all i. So

G1 ⊂ G2 ⊂ G3 ⊂ . . . .

Now, we claim that⋃∞i=1Gi = Y . Suppose not, then there is some

y ∈ Y such that y /∈ Gn for all n. That means that y ∈ {yn, yn+1, . . . }for all n. Therefore, y occurs infinitely often in the sequence (yn) whichcontradicts the assumption that (yn) has no convergent subsequence.Therefore {Gn : n = 1, 2, . . . } is an open cover of the compact space Yand therefore has a finite subcover. But since

G1 ⊂ G2 ⊂ G3 ⊂ . . . ,

that means that Gl = Y for some l which is clearly false. Thereforeour assumption must be false and we have shown that Y is indeedsequentially compact. �

The proof of the converse of Proposition 2.15 is a little bit trickier.

Lemma 2.16. Suppose that O is an open cover of a sequentiallycompact metric space, then there is some strictly positive real numberr such that for every x ∈ X, B(x, r) is contained in some open setbelonging to O.

Proof. Suppose the conclusion is false. Then for each positiveinteger n, we can choose xn ∈ X such that B(xn,

1n) is not contained in

any open set in O. By assumption (xn) has a subsequence (xnk) that

2. COMPACTNESS 35

converges to some x. Now x ∈ G where G ∈ O. So there is some ε > 0such that B(x, ε) ⊂ G. Now choose k so that d(xnk

, x) < ε2

and so that1nk< ε

2. If z ∈ B(xnk

, 1nk

), then

d(x, z) ≤ d(x, xnk) + d(xnk

, z)< ε

2+ ε

2= ε

Therefore B(xnk, 1nk

) ⊂ B(x, ε) ⊂ G which contradicts our choice ofxnk

.�

Lemma 2.17. Let X be a sequentially compact metric space and letε be any positive real number. There are finitely many points x1, . . . , xksuch that X = B(x1, ε) ∪ · · · ∪B(xk, ε).

Proof. Suppose that the conclusion is false. We will construct asequence with no convergent subsequence, which contradicts our hy-pothesis. Choose y1 ∈ X. By assumption X 6= B(y1, ε) so choosey2 ∈ X − B(y1, ε). Now suppose that we have chosen y1, . . . , yn.By assumption X − B(y1, ε) ∪ · · · ∪ B(yn, ε) is not empty, so chooseyn+1 ∈ X −B(y1, ε) ∪ · · · ∪B(yn, ε).

Now, we claim that the sequence (yn) constructed above has noconvergent subsequence. Observe that if m > n, then ym /∈ B(yn, ε).Thus for all m and n, d(ym, yn) > ε. Clearly, no subsequence of asequence with this property can converge. �

Proposition 2.18. If X is a sequentially compact metric spacethen X is compact.

Proof. Let O be an open cover of X. By Lemma 2.16, we canchoose r > 0 such that for all x ∈ X, B(x, r) is contained in someelement of O. Now, by Lemma 2.17, we can find x1, . . . , xk such thatX = B(x1, r) ∪ · · · ∪ B(xk, r). Now, for each xi, let Gi be an elementof O that contains B(xi, r). So

X = B(x1, r) ∪ · · · ∪B(xk, r) ⊂ G1 ∪ · · · ∪Gk.

Therefore {G1, . . . , Gk} is a finite subcover of O. �

2.4. Problems.

(1) Let O be an open cover of a space X and define a functionL : X → R as follows.

L(x) = min{1, sup{r ∈ R : B(x, r) ⊂ G,G ∈ O}}.(a) Show that L(x) > 0 for all x ∈ X.(b) Show that L is a continuous function.


(c) Use these facts, together with the fact that all continuousreal valued functions on sequentially compact spaces arebounded, to provide another proof of Lemma 2.16.

some notes for ma342 james cruickshankjc/docs/topologynotes.pdf · james cruickshank. notation...

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