Topology MATH 4181 Fall Semester 1999webpages.iust.ac.ir/m_nadjafikhah/Courses/PoSeTo/roy.pdf · in...

TopologyMATH 4181

Fall Semester 1999

David C. RoysterUNC Charlotte

September 2, 1999

Chapter 1

Introduction to Topology

1.1 History1

Topology is thought of as a discipline that has emerged in the twentieth century.There are precursors of topology dating back into the 1600’s. Gottfried WilhelmLeibniz (1646–1716) was the first to foresee a geometry in which position, rather thanmagnitude was the most important factor. In 1676 Leibniz use the term geometriasitus(geometry of position) in predicting the development of a type of vector calculussomewhat similar to topology as we see it today.

The first practical application of topology was made in the year 1736 by the Swissmathematician Leonhard Euler (1707–1783) in the Konigsberg Bridge Problem.

Carl F. Gauss (1777–1855) predicted in 1833 that geometry of location wouldbecome a mathematical discipline of great importance. His studey of closed surfacessuch as the sphere and the torus and surfaces much like those encountered in multi-dimensional calculus may be considered as a harbinger of general topology. Gausswas also interested in knots, which are of current interest today in topology.

The word topology was first used by the German mathematics Joseph B. Listing(1808–1882) in the title of his book Vorstudien zur Topologie (Introductory Studiesin Topology), a textbook published in 1847. Listing book dealt with knots and sur-faces but failed to generate much interest in either the name or the subject matter.Throughout much of the nineteenth and early twentieth centuries, much of what nowfalls under the auspices of topology was studied under the name of analysis situs(analysis of position).

Bernard Riemann (1826–1866) was the first mathematician to foresee topologyin the generality it has achieved today. He initiated the study of connectivity of asurface, or the arrangement of holes in a surface. He used concepts in which thenumber of dimensions exceeded three, which at that time was generally conceded tobe the maximum number of dimensions involved with any geometric object.

Present-day topology can be traced to two primary sources: the development ofnon-euclidean geometry and the process of putting calculus on a firm mathematicalfoundation.

1The information here is taken from Principles of Topology by Fred H. Croom.

2

1.2. SETS AND SET OPERATIONS 3

1.2 Sets and Set Operations

We need some basic information about sets in order to study the logic and the ax-iomatic method. This is not a formal study of sets, but consists only of basic defini-tions and notation.

Braces { and } are used to name or enumerate sets. The roster method for namingsets is simply to list all of the elements of a set between a pair of braces. For examplethe set of integers 1, 2, 3, and 4 could be named

{1, 2, 3, 4}.

This does not work well for sets containing a large number of elements, though it canbe used. The more common method for this is known as the set builder notation.A property is specified which is held by all objects in a set. P (x), read P of x, willdenote a sentence referring to the variable x. For example,

x = 23

x is an odd integer.

1 ≤ x ≤ 4.

The set of all objects x such that x satisfies P (x) is denoted by

{x | P (x)}.

The set {1, 2, 3, 4} can be named

{x | 1 ≤ x ≤ 4, x ∈ Z} = {x ∈ Z | 1 ≤ x ≤ 4}.

From hence forth, the words object, element, and member mean the same thingwhen referring to sets. Sets will be denoted mainly by capital Roman letters andelements of the sets by small letters. The following have the same meaning:

a ∈ A

a is in set A

a is a member of set A

a is an element of set A

Likewise, a 6∈ A means that a is not an element of set A.A is a subset of B if every element of A is also an element of B. The following

have the same meaning:

A ⊂ B

Every element of A is an element of B

c©1999, David Royster Introduction to Topology For Classroom Use Only

4 CHAPTER 1. INTRODUCTION TO TOPOLOGY

If a ∈ A, then a ∈ B

A is included in B

B contains A

A is a subset of B

Note that a set is always a subset of itself.

If A and B are sets, then we say that A = B if A and B represent the same set:

A = B

A and B are the same set

A and B have the same members

A ⊂ B and B ⊂ A

The set which contains no elements is known as the empty set, and is denoted by∅. Note that for each set A, ∅ ⊂ A.

The intersection of two sets A and B is the set of all elements common to bothsets. The intersection is symbolized by A ∩ B or {x | x ∈ A and x ∈ B}. The unionof two sets A and B is the set of elements which are in A or B or both. The unionis symbolized by A ∪B or {x | x ∈ A or x ∈ B}.

1.2.1 Universal Sets and Compliments

When we are working in an area or on a certain problem, we always have a frame ofreference in which we are working called a universal set. In our geometry course, itwill be the set of points that lie on a plane. In calculus we consider the set of realnumbers, the set of real functions, the set of differentiable functions, and the set ofcontinuous functions as universal sets.

The complement of a set A is defined to be the set of all elements of the universalset which are not in A, and is symbolized by CA = A′ = Ac. Note that A ∪ Ac isalways the universal set, while A ∩ Ac = ∅.

The set difference of the sets A and B is defined to be all of those elements in Awhich are not in B. It is denoted by

A \B = {x ∈ A | x 3 B}.

Note that A \B and B \A will usually be different, and that even though A \B = ∅it need not follow that A = B.


1.3. PRODUCTS SETS 5

1.3 Products Sets

Let X and Y be sets. The set of all ordered pairs {(x, y) | x ∈ X and y ∈ Y } is theproduct set X×Y , or Cartesian product or direct product of X and Y . A slice of thisproduct set is {x}× Y or X ×{y} for a given x ∈ X or y ∈ Y . Examples of commonproduct spaces are the plane R2 = R × R, 3-space R3 = R × R2, a right circularcylinder, S1 × [0, 1], or the torus, S1 × S1.

Theorem 1.1 Let X and Y be sets and let A,C ⊂ X and B,D ⊂ Y .

a) A× (B ∩D) = (A ∩B)× (A ∩D).

b) A× (B ∪D) = (A ∪B)× (A ∪D).

c) A× (Y \D) = (A× Y ) \ (A×D).

d) (A×B) ∩ (C ×D) = (A ∩ C)× (B ∩D).

e) (A×B) ∪ (C ×D) ⊆ (A ∪ C)× (B ∪D).

f) (X × Y ) \ (A×B) = (X × (Y \B)) ∪ ((X \ A)× Y )

The concept of the product of two sets can be extended to more than two factors.If {Xi}ni=1 is a finite collection of sets, then their product is

X1 ×X2 × · · · ×Xn =n∏i=1

Xi = {(x1, x2, . . . , xn) | xi ∈ Xi for each i = 1, 2, . . . , n}.

For an infinite collection of sets, the product is defined by

∞∏i=1

Xi = {(x1, x2, x3, . . . ) | xi ∈ Xi for each i = 1, 2, . . . }.

1.4 Functions

A function f : X → Y is a rule which assigns to each x ∈ X a unique y ∈ Y and wesay y = f(x). If y = f(x) then y is called the image of x and x is called the preimageof y. The set X is the domain of f and Y is the range or codomain of f .

Let A ⊂ X. The set f(A) = {y ∈ Y | y = f(x) for some x ∈ A} is called theimage of A. The set f(X) is called the image of f . For B ⊂ Y , the set

f−1(B) = {x ∈ X | f(x) ∈ B}

is the inverse image of B under f . The set of points

Γ = {(x, f(x)) ∈ X × Y | x ∈ X}

is called the graph of the function f .



A function f : X → Y is injective if for distinct elements x1, x2 ∈ X, f(x1) 6= f(x2)in Y . Another way to think of this is to say that f is injective if f(x1) = f(x2) impliesthat x1 = x2.

If f(X) = Y , the function f is said to be surjective.A function that is surjective and injective is called a bijection. In this case we

have that f : X → Y is a bijection provided that each member of Y is the imageunder f of exactly one member of X. In this case the inverse function f−1 : Y → Xexists assigning to each element y ∈ Y its unique preimage x = f−1(y) in X.

The identity function iX : X → X from a set X to itself is the function definedby iX(x) = x for all x ∈ X. This function is often denoted by 1X .

If f : X → Y and g : Y → Z are functions, then the composite function g◦f : X →Z is defined by g ◦ f(x) = g(f(x)), for x ∈ X.

Definition 1.1 Let X be a set. A sequence in X is a function f : Z+ → X whosedomain is the set of all positive integers, Z+ or the set of positive integers less thanor equal to some given positive integer N . The sequence is called finite if its domainis {1, 2, . . . , N} and infinite if its domain is all positive integers.

1.5 Equivalence Relations

Let X be a set. A relation R on X is a subset of X × X. If (x, y) ∈ R we will saythat x is related to y by R and to write xRy.

A relation R on a set X is called reflexive, symmetric, or transitive if it satisfiesthe corresponding property below.

(a) The Reflexive Property : xRx for all x ∈ X.

(b) The Symmetric Property : If xRy, then yRx.

(c) The Transitive Property : If xRy and yRz, then xRz.

Definition 1.2 An equivalence relation on a set X is a relation on X which isreflexive, symmetric, and transitive.

Definition 1.3 Let ∼ denote an equivalence relation on X. For x ∈ X the set[x] = {y ∈ X | y ∼ x} is calle dth e equivalence class of x.

Proposition 1.1 Let X be a set and let ∼ denote an equivalence relation on X.

a) x ∈ [x] for each x ∈ X.

b) x ∼ y if and only if [x] = [y].

c) x 6∼ y if and only if [x] ∩ [y] = ∅.

d) For x, y ∈ X, [x] and [y] are either identical or disjoint.


1.6. CARDINALITY 7

1.6 Cardinality

We are often interested in how big sets are in relation to one another. Clearly, wecan tell the difference in sizes of two finite sets, but how do we differentiate betweentwo infinite sets? Are there different sizes of infinite sets? How do we compare setsto tell if one has a greater number of members?

Definition 1.4 a set A is finite if A is empty or if there is a bijection between Aand the set of integers from 1 to N for some positive integer N . In the latter case, Ais said to have N members. If a set is not finite, it is called infinite.

Definition 1.5 A set is denumerable or countably infinite if there is a bijectionbetween the set and the positive integers. A set which is either finite or denumerableis called countable. A set which is not countable is called uncountable.

Lemma 1.1 a) Each subset of a finite set is finite.

b) Each subset of a countable set is countable.

c) Each set which contains an infinite set is infinite.

d) Each set which contains an uncountable set is uncountable.

Example 1.6.1 1. The set Z+∪{0} of all non-negative integers is countable. Thebijection is f : Z+ ∪ {0} → Z

+ given by

f(n) = n+ 1, n ∈ Z+ ∪ {0}.

2. The set of all integers, Z is countable. The bijection g : Z→ Z+ is given by

g(n) =

{2n− 1 if n is positive

−2n if n is negative

3. The product set Z+ × Z+ is countable. One method is to use the CantorDiagonalization Method to count the ordered pairs (m,n). A second method isto define the function g : Z+ × Z+ → Z

+ by

g(m,n) = 2m3n, (m,n) ∈ Z+ × Z+.

Now, g is not surjective, but the Fundamental Theorem of Arithmetic on theunique factorization into primes guarantees that the function g is injective.Thus, there is a bijection from Z

+ × Z+ to a subset of Z+. Since every subsetof a countable set is countable, we have that Z+ × Z+ is countable.



Theorem 1.2 a) If {Ai}Ni=1 is a finite collection of finite sets, then both⋃Ni=1 Ai and

∏Ni=1 Ai are finite.

(Finite unions and finite products of finite sets are finite.)

b) If {Ai}∞i=1 is a countable collection of countable sets, then⋃∞i=1 Ai is

countable.(Countable unions of countable sets are countable.)

c) If {Ai}Ni=1 is a finite collection of countable sets, then both∏N

i=1 Ai iscountable.(Finite products of countable sets are countable.)

Why didn’t we claim that a countable product of countable sets is countable?Mainly because it is not true, as is seen in the following example.

Example 1.6.2 Let Ai = {0, 1} for i = 1, 2, . . . . Let

U =∞∏i=1

Ai = {(a1, a2, a3, . . . ) | ai = 0 or 1}.

Assume that U is countable. Then there is a bijection f : Z+ → U . For an elementa = (a1, a2, a3, . . . ) ∈ U we shall refer to a1 as the first coordinate, a2 as the secondcoordinate, and so forth.

We can list all of the elements in U using the bijection f . They are {f(1), f(2), F (3), . . . }.Consider the following element in U . Define x = (x1, x2, x3, . . . ) as follows:

xi =

{0 if the ith coordinate of f(i) is 1

1 if the ith coordinate of f(i) is 0

Then, we have that for each positive integer i, x 6= f(i) for they differ in the ith

coordinate. This means that x is not in the exhaustive list of elements we have in ourbijection. That is this bijection is not surjective. This contradiction show us that Ucannot be countable.

Theorem 1.3 The set of rational numbers is countable.

Proof: There are several ways of proving this. One method is to use the CantorDiagonalization Method to count the rationals.

(Method 1) List the positive rational numbers in rows where the first row consistsof all positive rational numbers with 1 as a denominator, the second row lists allpositive rational numbers with 2 as a denominator, and so on:

11

21

31

41

. . .12

22

32

42

. . .13

23

33

43

. . .14

24

34

44

. . .

......

......

. . .


1.6. CARDINALITY 9

Clearly, we have each positive rational number in here numerous times, but thediagonalization method will still show that there are a countable number of elementsin this array. The positive rationals form a subset of this array, thus there must bea countable number of positive rationals. This will yield that there are a countablenumber of rational numbers.

(Method 2) Every rational number can be expressed uniquely in lowest terms asm/n where m and n are integers with no common positive divisor other than 1, andn is positive. Consider the function m/n 7→ (m,n) from the set of rational numbersinto Z × Z. This function is injective since the ordered pair (m,n) determines onlyone rational number m/n. Thus, the set of rational numbers is equivalent to a subsetof the countable set Z× Z, and is hence countable.

Theorem 1.4 The set of real numbers is uncountable.

Proof: We will make use of the example of the countable product above. Eachelement in

∏∞i=1 0, 1 is a sequence consisting of 0’s and 1’s. Each of these sequences

represents a unique real number between 0 and 1, by the correspondence

(a1, a2, a3, . . . ) 7→ 0.a1a2a3 . . . .

This is a one-to=one correspondence. Thus, the set of real numbers between 0 and1 representable as a decimal using only 0’s and 1’s is an uncountable set. Thus, Rcontains an uncountable set and hence is uncountable.

Theorem 1.5 The set of irrational numbers is uncountable.

Proof: Since the set of real numbers is the union of the set of rational numbers andthe set of irrational numbers, if the set of irrationals were countable, then we wouldhave that the real numbers are countable. That failing to be true, implies that theirrationals must be uncountable.


Chapter 2

Metric Spaces

2.1 Definition and Some Examples

Definition 2.1 Let X be a set and d : X×X → R+ a function satisfying the following

properties. For all x, y, z ∈ X,

a) d(x, y) = 0 if and only if x = y.

b) d(x, y) = d(y, x).

c) d(x, z) ≤ d(x, y) + d(y, z).

Then d is called a metric or distance function on X and d(x, y) is called thedistance from x to y. The set X with a metric d is called a metric space and isdenoted by (X, d).

Note that these properties are modeled on the distance functions that we have onR and R2. Doing so we usually call property (c) the Triangle Inequality.

Example 2.1.1 The real line, R is a metric space using the standard distance func-tion, the absolute value: d(a, b) = |a− b|. The above properties are standard proofsabout the absolute value function.

Example 2.1.2 The plane, R2, with the usual Euclidean distance formula is a metricspace. If P = (x1, y1) and Q = (x2, y2), then

d(P,Q) =√

(x2 − x1)2 + (y2 − y1)2.

Example 2.1.3 These are special cases of the general Euclidean n-space,

Rn = {(a1, a2, . . . , an) | ai ∈ R}.

10

2.1. DEFINITION AND SOME EXAMPLES 11

The distance formula here is the usual distance formula for Euclidean n-space:

d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) =

(n∑i=1

(xi − yi)2

)1/2

.

d is called the usual metric on Rn.To show that d is a metric, we need two standard results about vectors in Rn.

First, let a ∈ Rn. The norm ‖a‖ is the distance from a to the origin O = (0, 0, . . . , 0):

‖a‖ = d(a,O) =

(n∑i=1

a2i

)1/2

.

Theorem 2.1 (Cauchy-Schwarz Inequality) For any points a, b ∈ Rn

|a · b| ≤ ‖a‖‖b‖.

Theorem 2.2 (The Minkowski Inequality) For any points a, b ∈ Rn

‖a+ b‖ ≤ ‖a‖+ ‖b‖.

The distance between two points is given by d(a, b) = ‖a− b‖.The first two conditions making d a metric are easily seen to be satisfied. We only

need check the Triangle Inequality. Let x, y, z ∈ Rn

d(x, z) = ‖x− z‖ = ‖x− y + y − z‖≤ ‖x− y‖+ ‖y − z‖= d(x, y) + d(y, z)

Example 2.1.4 [The Taxicab Metric] Define a function d′ : R2×R2 → R as follows.If x = (x1, x2) and y = (y1, y2), then

d′(x, y) = |x1 − y1|+ |x2 − y2|.

This is called the taxicab metric because the distance is measured along line segmentsparallel to the coordinate axes.

Clearly, d′(x, x) = 0 and if d′(x, y) = 0, then |x1 − y1| + |x2 − y2| = 0 whichmeans |x1 − y1| = 0 and |x2 − y2| = 0. This implies that x1 = y1 and x2 = y2,and x = y. Because of the basic properties of the absolute value, it is obvious thatd′(x, y) = d′(y, x). The Triangle Inequality follows because of the validity of theTriangle Inequality with the absolute value on the real line.

What is the following set?

U = {x = (x1, x2) ∈ R2 | d′(x,O) = 1}.

We can define an analogous metric, called the taxicab metric, on Rn.

d′(x, y) =n∑i=1

|xi − yi|.


12 CHAPTER 2. METRIC SPACES

Example 2.1.5 [The Max Metric on Rn] Another metric for Rn is given by takingthe largest of the differences of the coordinates of x and y.

d′′(x, y) = max{|xi − yi|}ni=1.

Example 2.1.6 [The Discrete Metric] For any set X, define

d(x, y) =

{0 if x = y

1 if x 6= y

This defines a metric on X, called the discrete metric. It is usually of little use, exceptfor counterexamples. It does show, though, that every set can be assigned a metric.

Example 2.1.7 Let C [a, b] denote the set of all continuous real-valued functionsdefined on the interval [a, b]. For f, g ∈ C [a, b] define

ρ(f, g) =

∫ b

a

|f(x)− g(x)| dx.

The fact that ρ is a metric follows from the usual properties of the Riemann integral.This metric measures the distance between two functions to be the area between thetwo graphs from x = a to x = b.

Example 2.1.8 For the set C [a, b] define ρ′ by

ρ′(f, g) = lub{|f(x)− g(x)| | x ∈ [a, b]}.

The metric is called the supremum metric or the uniform metric for C [a, b]. Itmeasures the distance between f and g to be the supremum of the vertical distancesfrom points (x, f(x)) to (x, g(x)) on the graphs of f and g on the closed interval [a, b].

Definition 2.2 A number u is an upper bound for a set A of real numbers providedthat a ≤ u for all a ∈ A. If there is a smallest upper bound u0 for A, that is an upperbound that is less than or equal to all other upper bounds for A, then u0 is calledthe least upper bound or supremum of A. The least upper bound for a set A isdenoted by lubA or supA.

Definition 2.3 A number ` is an lower bound for a set A of real numbers providedthat ` ≤ a for all a ∈ A. If there is a largest lower bound `0 for A, that is a lowerbound that is less than or equal to all other lower bounds for A, then `0 is called thegreatest lower bound or infimum of A. The greatest lower bound for a set A isdenoted by glbA or inf A.



A very basic property of the real numbers is included in the following two state-ments:The Least Upper Bound Property: Every non-empty set of real numbers whichhas an upper bound has a least upper bound.The Greatest Lower Bound Property: Every non-empty set of real numberswhich has a lower bound has a greatest lower bound.

We will accept the first property as an axiom of the real number system. Thesecond property follows from the first.

Definition 2.4 Let (X, d) be a metric space and let A be a non-empty subset ofX. If {d(x, y)‖x, y ∈ A} has an upper bound, then A is said to be bounded, andlub{d(x, y)‖x, y ∈ A} is called the diameter of A. For completeness, we define thediameter of the empty set to be 0. If the set X is bounded, then we call (X, d) abounded metric space.

If x ∈ X, then the distance from x to A is defined by

d(x,A) = glb{d(x, y) | y ∈ A}.


Chapter 2

Metric Spaces


Definition 2.1 Let X be a set and d : X×X → R+ a function satisfying the following



b) d(x, y) = d(y, x).

c) d(x, z) ≤ d(x, y) + d(y, z).



Example 2.1.1 The real line, R is a metric space using the standard distance func-tion, the absolute value: d(a, b) = |a− b|. The above properties are standard proofsabout the absolute value function.

Example 2.1.2 The plane, R2, with the usual Euclidean distance formula is a metricspace. If P = (x1, y1) and Q = (x2, y2), then

d(P,Q) =√

(x2 − x1)2 + (y2 − y1)2.

Example 2.1.3 These are special cases of the general Euclidean n-space,

Rn = {(a1, a2, . . . , an) | ai ∈ R}.

10



d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) =

(n∑i=1

(xi − yi)2

)1/2

.



‖a‖ = d(a,O) =

(n∑i=1

a2i

)1/2

.

Theorem 2.1 (Cauchy-Schwarz Inequality) For any points a, b ∈ Rn

|a · b| ≤ ‖a‖‖b‖.

Theorem 2.2 (The Minkowski Inequality) For any points a, b ∈ Rn

‖a+ b‖ ≤ ‖a‖+ ‖b‖.




Example 2.1.4 [The Taxicab Metric] Define a function d′ : R2×R2 → R as follows.If x = (x1, x2) and y = (y1, y2), then

d′(x, y) = |x1 − y1|+ |x2 − y2|.




U = {x = (x1, x2) ∈ R2 | d′(x,O) = 1}.


d′(x, y) =n∑i=1

|xi − yi|.



Example 2.1.5 [The Max Metric on Rn] Another metric for Rn is given by takingthe largest of the differences of the coordinates of x and y.

d′′(x, y) = max{|xi − yi}ni=1.

Example 2.1.6 [The Discrete Metric] For any set X, define

d(x, y) =

{0 if x = y

1 if x 6= y


Example 2.1.7 Let C [a, b] denote the set of all continuous real-valued functionsdefined on the interval [a, b]. For f, g ∈ C [a, b] define

ρ(f, g) =

∫ b

a

|f(x)− g(x)| dx.


Example 2.1.8 For the set C [a, b] define ρ′ by

ρ′(f, g) = lub{|f(x)− g(x)| | x ∈ [a, b]}.


Definition 2.2 A number u is an upper bound for a set A of real numbers providedthat a ≤ u for all a ∈ A. If there is a smallest upper bound u0 for A, that is an upperbound that is less than or equal to all other upper bounds for A, then u0 is calledthe least upper bound or supremum of A. The least upper bound for a set A isdenoted by lubA or supA.

Definition 2.3 A number ` is an lower bound for a set A of real numbers providedthat ` ≤ a for all a ∈ A. If there is a largest lower bound `0 for A, that is a lowerbound that is less than or equal to all other lower bounds for A, then `0 is called thegreatest lower bound or infimum of A. The greatest lower bound for a set A isdenoted by glbA or inf A.


2.2. CONTINUOUS FUNCTIONS 13



Definition 2.4 Let (X, d) be a metric space and let A be a non-empty subset ofX. If {d(x, y)‖x, y ∈ A} has an upper bound, then A is said to be bounded, andlub{d(x, y)‖x, y ∈ A} is called the diameter of A. For completeness, we define thediameter of the empty set to be 0. If the set X is bounded, then we call (X, d) abounded metric space.


d(x,A) = glb{d(x, y) | y ∈ A}.

Theorem 2.3 Let {(Xi, di)}ni=1 be a finite collection of metric spaces and let

X =n∏i=1

Xi.

For each pair of points x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn) in X, let d : X×X →R be defined by

d(x, y) =

(n∑i=1

(di(xi, yi))2

)1/2

.

Then (X, d) is a metric space. The metric d defined above is called the productmetric on X.

2.2 Continuous Functions

In topology we are concerned with how spaces are changed when stretched, bent,twisted and modified — but not torn. We do so by studying the maps that doso. Our friend here is the continuous map. In your study of calculus, you saw thatcontinuous functions did many things. At the time you were more interested in specialcontinuous functions — the differential functions. We here are more interested in themore general function.

In calculus, we saw that a continuous function was one that did not do too muchdamage to the domain in the range. By this, we mean that if two points were closein the domain, then their images were not too far apart in the image. We saw thisintuitively through looking at graphs and looking at limits. To insure specificity, weneed the definition of continuity due to Cauchy and Weierstrauss. It is one with whichyou are familiar.



Definition 2.5 Let f : (X, d)→ (Y, d′) be a function between two metric spaces. Leta ∈ X. We say that f is continuous at a if given any ε > 0 there is a δ > 0 sothat d′(f(x), f(a)) < ε whenever d(x, a) < δ. We say that f is continuous if it iscontinuous at a ∈ X for all a ∈ X.

This clearly depends on the metric in each of the two spaces. A change of met-ric might change the continuity of the given function. Will it? Is continuity thatdependent on the metric in the domain or the range?

Let’s check two well-known functions that we think should be continuous andmake certain that they are continuous under this definition.

Example 2.2.1 Let f : (X, d) → (Y, d′) be given by f(x) = b for all x ∈ X whereb ∈ Y is a constant. This is just the constant function.

To show that f is continuous, we need to show that if we are given any ε > 0, thenwe can find a δ > 0 so that whenever d(x1, x2) < δ then d′(f(x1), f(x2)) < ε. In thiscase, this is easy. This is because d′(f(x1), f)(x2)) = d′(b, b) = 0 < ε for any choiceof x1, x2 ∈ X. Thus, it does not matter what we may choose for δ. You could takeδ = ε or δ = 1. Regardless, whenever d(x1, x2) < δ then d′(f(x1), f(x2)) = 0 < ε, andwe are done.

Example 2.2.2 Let 1X : (X, d) → (X, d) denote the identity map from X to itselfgiven by 1X(x) = x. We claim that this function is continuous.

Again, to show this we are given an ε > 0. We then need to find a δ > 0 so thatwhenever d(x1, x2) < δ then d(1X(x1), 1X(x2)) < ε. However, since 1X(x1) = x1 and1X(x2) = x2, it is easy to see that if we take δ ≤ ε, then if d(x1, x2) < δ it followsthat d(1X(x1), 1X(x2)) = d(x1, x2) < δ ≤ ε. Thus, 1X is a continuous function.

Example 2.2.3 This time we will be working with the same underlying set, but wewill place a different metric on it. Will this make a difference?

Let X = Rn with the usual metric. Let Y = Rn with the maximum metric,

d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) = max1≤i≤n

{|xi − yi|}.

Define h : (X, d) → (Y, d′′) by h(x1, x2, . . . , xn) = (x1, x2, . . . , xn). It is the identitymap on the underlying set, but it does not carry the same metric information. Is hcontinuous? Is h−1 continuous?

It turns out that both are continuous! To prove this, let’s first look at h−1 : (Y, d′′)→(X, d). We are given an ε > 0. We need to find a δ > 0 so that if

d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < δ

then d(h−1(x1, x2, . . . , xn), h−1(y1, y2, . . . , yn)) < ε.


2.3. OPEN SETS AND CLOSED SETS 15

To say that d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < δ means that |xi− yi| < δ for alli = 1, . . . , n. Thus,

d(h−1(x1, x2, . . . , xn), h−1(y1, y2, . . . , yn)) = d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) (2.1)

=

(n∑i=1

|xi − yi|2)1/2

(2.2)

<

(n∑i=1

δ2

)1/2

= δ√n (2.3)

Thus, we need δ√n < ε, or take δ <

ε√n

.

Now, to show that h is continuous we are given an ε > 0. We need to find δ sothat whenever d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < δ, we have that

d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < ε.

To say that d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < δ means that (∑n

i=1 |xi − yi|2)1/2

<δ. Thus, each of the differences |xi−yi|must be less than δ, and the largest of these dif-ferences is still less than δ. Thus, in order for d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < ε,we need only choose δ < ε.

2.3 Open Sets and Closed Sets

Definition 2.6 Let (X, d) be a metric space, a ∈ X, and r > 0 a positive real number.The open ball Bd(a; r) with center a and radius r is the set

Bd(a; r) = {x ∈ X | d(a, x) < r}.

When there is only one metric under consideration, we will simplify the notation toB(a; r).

Definition 2.7 Let (X, d) be a metric space and let a ∈ X. A subset N ⊂ X is aneighborhood of a if there is a δ > 0 so that B(a; δ) ⊂ N . The collection Na of allneighborhoods of a point a ∈ X is called a complete system of neighborhoods ofthe point a.

Definition 2.8 A subset U of a metric space (X, d) is an open set with respect tothe metric d provided that U is a union of open balls. The family of all open setsdefined in this way is called the topology for X generated by d. A subset C ⊂ Xis said to be closed (with respect to d) if its complement X \ C is an open set (withrespect to d).



Thus, a neighborhood of a and an open set containing a need not be the samething. However, if U is an open set containing a, then U is a neighborhood of a.

Theorem 2.4 The following statements are equivalent (TFAE) for a subset U of ametric space (X, d).

a) U is an open set;

b) for each x ∈ U there is an εx > 0 so that B(x; εx) ⊂ U .

c) for each x ∈ U , d(x,X \ U) > 0, if U 6= X.

Proof: What this means is that Statements (a) and (b) are equivalent, (b) and (c)are equivalent, and (a) and (c) are equivalent. We can show this by proving that (a)is equivalent to (b) and then that (b) is equivalent to (c). In condition (c) we willassume that U 6= X since the distance from the empty set is not defined.

Assume that U is an open set and let x ∈ U . Since U is the union of open balls,then x ∈ B(a; r) ⊂ U . Then d(x, a) < r. We want to center an open ball at x andhave it contained in U . Choose εx ≤ r − d(x, a). Then B(x; εx) ⊂ B(a; r) for thefollowing reason: If y ∈ B(x; εx),

d(y, a) ≤ d(y, x) + d(x, a) < εx + d(x, a) ≤ r − d(x, a) + d(x, a) = r.

Thus, B(x; εx) is an open ball of positive radius centered at x and contained in U .Thus (a) =⇒ (b).

To show that (b) =⇒ (a), since each x ∈ U lies in an open ball contained in U ,U is the union of these open balls.

To see that (b) =⇒ (c), let B(x; εx) ⊂ U . Then any point within distance εx ofx is in U , so the distance from x to X \U must be at least εx. Thus, d(x,X \U) > 0for each x ∈ U .

Assuming that (c) holds, d(x,X \ U) = αx > 0 depending on x. This means thatthe distance from x to a point outside U must be at least αx, so any point withindistance αx of x must be in U . This means B(x;αx) ⊂ U .

Note that we have just shown that for each a ∈ X and for each δ > 0, the openball B(a; δ) is a neighborhood of each of its points.

2.3.1 Neighborhoods and Continuous Functions

How do we plan to use this information? While our definition of continuity is precise,it requires some specificity and does not look generalizible. What I mean by this isthat the definition seems to rely specifically on the definition of the metric, and itwill be hard to realign our definition when we have to move away from metric spaces.

Theorem 2.5 Le f : (X, d) → (Y, d′). f is continuous at a ∈ X if and only if foreach neighborhood M of f(a) there is a corresponding neighborhood N of a, such that

f(N) ⊂M,


2.4. LIMITS 17

or equivalently

N ⊂ f−1(M).

Proof: First, let’s suppose that f is continuous at a ∈ X and let M be a neigh-borhood of f(a). This means that for some ε > 0 Bd′(f(a); ε) ⊂ M . Since f iscontinuous at a ∈ X we know that we can find δ > 0 so that if d(x, a) < δ thend′(f(a), f(x)) < ε. This says that Bd(a; δ) ⊂ f−1(M) and we already have seen thatBd(a; δ) is a neighborhood of a ∈ X. Thus, if f is continuous, we have found acorresponding neighborhood to M .

Now, suppose that for any neighborhood, M , of f(a) we can find a neighborhoodN of a so that f(N) ⊂ M . Let ε > 0 be given to you. You must find a δ > 0 sothat whenever d(a, x) < δ we have d′(f(a), f(x)), ε. Now, let M = Bd′(f(a); ε). Mis a neighborhood of f(a), so we know that there is a neighborhood N ⊂ X of a sothat f(N) ⊂M . Since N is a neighborhood of a, it must contain a δ-ball centered ata, Bd(a; δ) ⊂ N , by the definition of a neighborhood. Thus, if we have x ∈ Bd(a; δ),then f(x) ∈ M = Bd′(f(a); ε). The other way of writing this is: if d(a, x) < δ thend′(f(a), f(x)) < ε. Therefore, f is continuous at a ∈ X.

2.4 Limits

Recall that a sequence is just a function a : Z+ → (X, d). We want to discuss whathappens to the sequence as we let n go to infinity; in other words what happens tothe sequence as we look further and further into the range of a. Let us first recallthe definitions in the real numbers and then try to set them up so that we can easilygeneralize them to arbitrary metric spaces.

Let {ai} be a sequence of real numbers. A real number L is said to be the limit ofthe sequence {an} if, given any ε > 0, there is a positive integer N such that whenevern > N , |an − L| < ε. In this case we say that the sequence converges to L and write

limn→∞

an = L.

How can we generalize this to an arbitrary metric space? It should not be hard,because all we used in the definition was the distance function in the real numbers.We will use the distance function in our metric space similarly.

Definition 2.9 Let {xn} be a sequence in the metric space (X, d). We say that thissequence {xn} converges to x ∈ X if given any ε > 0 there is a positive N ∈ Z+ sothat whenever n > N , d(x, xn) < ε. In this case we will write limxn = x.

Lemma 2.1 Let (X, d) be a metric space and {xn} be a sequence in X. Thenlimxn = x ∈ X if and only if for each neighborhood V of x there is an integerN > 0 so that xn ∈ V whenever n > N .



This is nothing but applying the definitions of convergence and neighborhood, andits proof will be omitted.

If S is a set of infinite points and there is at most a finite number of elementsof S for which a certain statement is false, thent he statement is said to be true foralmost all of S. Thus, we may phrase the above lemma by saying that the sequence{xn} converges to x if each neighborhood of x contains almost all of the poins of thesequence.

One reason for looking a sequences is the concept of continuity. In calculus wedefine a function to be continuous at a ∈ R if the following conditions were met:

1. limx→a f(x) exists;

2. f(a) exists;

3. limx→a f(x) = f(a).

It suffices to check this for all sequences appproaching a (a fact to be proven later).Thus, we find that we can show that f is continous at a if for each sequence {xn} → a,we have that {f(xn)} → f(a).

We are able to extend this result to arbitrary metric spaces.

Theorem 2.6 Let f : (X, d) → (Y, d′). f is continous at a point a ∈ X if and onlyif whenever limxn = x we have lim f(xn) = f(x).

The proof is straightforward.

Proof: Assume that f is continuous and let {xn} → x in X. Let ε > 0 and letM = Bd′(f(x); ε) ∈ V . There is a neighborhood U of x in X, such that f(U) ⊂ M .Since U is a neighborhood there is a δ > 0 so that Bd(x

′δ) ⊂ U . Now, {xn} → x so forthis δ there is a positive integer N so that whenever n > N we have xn ∈ Bd(x

′δ) ⊂ U .Thus, f(xn) ∈ f(U) ⊂ M . Therefore, for any neighborhood M of f(x) there is apositive integer N so that whenever n > N we have d′(f(x), f(xn) < ε, which impliesthat the sequence {f(xn)} converges to f(x).

To prove the other direction,

2.5 Open Sets and Closed Sets Revisited

Remember that we defined an open set as a set that is the union of open balls. A setis closed if its complement is open.

Theorem 2.7 The open subsets of a metric space (X, d) have the following proper-ties:

1. X and ∅ are open sets.

2. The union of any family of open sets is open.


2.5. OPEN SETS AND CLOSED SETS REVISITED 19

3. The intersection of a finite family of open sets is open.

Proof: These are straightforward.

1. The whole space X is open since it is the union of all open balls with allpossible centers and radii. The empty set is open since it is the union of theempty collection of open balls.

2. If {Uα | α ∈ A} is a family of open sets in X, then each Uα is a union of openballs. Then

⋃α∈A Uα is the union of all of the open balls that comprise each Uα

and is hence open.

3. Let {Ui | i = 1, . . . , n} be a finite collection of open sets and let x ∈⋂ni=1 Ui.

Then, by our previous theorem there exist εi, i = 1, . . . , n so that Bd(x; εi) ⊂ Uiand

n⋂i=1

Bd(x; εi) ⊂n⋂i=1

Ui.

Let ε = min{εi | i = 1, . . . , n}. Then,⋂ni=1 Bd(x; εi) = Bd(x; ε). Thus, Bd(x; ε)

is an open ball centered at x and contained in⋂ni=1 Ui. Thus,

⋂ni=1 Ui is open.

Theorem 2.8 The closed subsets of a metric space (X, d) have the following proper-ties:

1. X and ∅ are closed sets.

2. The intersection of any family of closed sets is closed.

3. The union of a finite family of closed sets is closed.

This follows from our previous theorem and complements.

Definition 2.10 Let (X, d) be a metric space and A a subset of X. A point x ∈ X isa limit point or accumulation point of A provided that every open set containingx contains a point of A distinct from x. The set of limit points of A is called itsderived set, denoted by A′.

Lemma 2.2 Let (X, d) be a metric space and A a subset of X. A point x ∈ X is alimit point of A if and only if d(x,A \ {x}) = 0.

Lemma 2.3 A subset A of a metric space (X, d) is closed if and only if A containsall its limit points.



Proof: Let A be closed and let x be a limit point of A. If x 6∈ A then X \ A is anopen set containing x but containing no other point of A. Thus, x could not be alimit point of A. This means that if x is a limit point of A, then it must be a memberof A.

Now suppose that A contains all of its limit points. To show that A is closed, wemust show that X \ A is open. If x ∈ X \ A, then x is not a limit point of A. Thus,there is some open set Ux containing x but no other point of A. Then X \ A is theunion of all of these sets. Hence, X \ A is open and A is closed.

What is the connection between limit points and the limit of a sequence?

Theorem 2.9 Let (X, d) be a metric space and A a subset of X.

1. A point x ∈ X is a limit point of A if and only if there is a sequence of distinctpoints of A which converges to x.

2. The set A is closed if and only if each convergent sequence of points of A con-verges to a point of A.

Corollary 1 Let x be a limit point of a subset A of a metric space X. Then everyopen set containing x contains infinitely many members of A.

2.6 Interior, Closure, and Boundary

Definition 2.11 Let A be a subset of a metric space (X, d). A point x ∈ A is aninterior point of A if there is an open set U which contains x and is contained inA; x ∈ U ⊂ A. The interior of A, denoted intA, is the set of all interior points ofA.

Note that for the open set U in the definition, every point of U is an interior pointof A. Thus, the interior of A contains every open set contained in A and is the unionof this family of open sets. This means two things:

1. the interior of a set A is an open set, and

2. the interior of a set A is the largest open set contained in A.

Item (2) above means that if U is open and U ⊂ A, then U ⊂ intA.

Example 2.6.1

Let X = R with the usual metric.

1. For a, b ∈ R with a < b

int(a, b) = int[a, b) = int(a, b] = int[a, b] = (a, b).


2.6. INTERIOR, CLOSURE, AND BOUNDARY 21

2. The interior of a finite set is empty, since such a set cannot contain any openinterval.

3. The interior of the set of irrational numbers is empty, since each open intervalmust contain some rational number. Likewise, the interior of the set of rationalsis empty. If the rationals contained an open interval, then the set of rationalswould have to be uncountable, since an open interval is uncountable.

4. int∅ = ∅; intR = R.

Definition 2.12 The closure A of a subset of a metric space (X, d) is the union ofthe set A and the set of its limit points:

A = A ∪ A′

where A′ is the derived set of A.

Example 2.6.2



(a, b) = [a, b) = (a, b] = [a, b] = [a, b].

2. The closure of a finite set is itself, since the set of limit points of a finite set isempty.

3. The closure of the set of rational numbers is R. Likewise, the closure of theset of irrationals is R. Since every open interval contains both rational andirrational numbers.

4. ∅ = ∅; R = R.

While the interior of a set is the largest open set contained in the set, the closurehas a similar property described in the next theorem.

Theorem 2.10 If A ⊂ X, then A is a closed set and is a subset of every closed setcontaining A.

This says that the closure of a set is the smallest closed set containing the set.

Proof: To show that A is closed, we need to show that it contains all of its limitpoints. Suppose that x 6∈ A. Then there is an open set U containing x so thatU ∩A = ∅. Now, this means that U cannot contain a limit point of A either, since ifan open set contains a limit point of A it must contain some other point of A also.Thus, U contains no point of A, so x is not a limit point of A. This means that all ofthe limit points of A must be contained in A. Thus, A is closed.

Suppose now that F is a closed subset of X and A ⊂ F . Then we can show thatA ⊂ F and, since F contains all of its limit points, then F = F ∪ F ′ = F . Thus,A ⊂ F fore every closed set F containing A.



Since this shows that A is the smallest closed set containing A, we can easily showthat A is the intersection of all closed sets containing A.

Theorem 2.11 Let A be a subset of the metric space (X, d).

1. A is open if and only if A = int A.

2. A is closed if and only if A = A.

Definition 2.13 Let A be a subset of the metric space (X, d). A point x ∈ X is aboundary point of A provided that x ∈ A∩X \ A. The set of boundary points of Ais called the boundary of A and is denoted by ∂A.

The industrious reader will readily work to show that the following statementsare equivalent for a subset A of X and a points x in the metric space (X, d).

1. x ∈ ∂A,

2. x ∈ (A \ int A),

3. Every open set containing x contains a point of A and a point of X \ A.

4. Every neighborhood of x contains a point of A and a point of X \ A.

5. d(x,A) = d(x,X \ A) = 0.

6. x ∈ A ∩X \ A.

Example 2.6.3 1. Let X = R with the usual metric. For a, b ∈ R with a < b

∂(a, b) = ∂[a, b) = ∂(a, b] = ∂[a, b] = {a, b}.

2. In Rn

∂B(a; ε) = {x ∈ Rn | d(a, x) = ε}.

3. The boundary of the set of all points in Rn having only rational coordinates isRn.

4. For any metric space (X, d),

∂∅ = ∂X = ∅.


Chapter 3

Topological Spaces


We want to generalize the concepts that we developed in studying the metric spaces.We want to remove our reliance on a distance function. We were able to define mostof what we wanted to do in metric spaces by defining our concepts in terms of theopen sets. This was especially true of our study of continuous functions.

We will use the results that we proved about open sets as our basis for the gener-alization. We will define open sets as sets that satisfy certain conditions.

Definition 3.1 Let X be a set and T a family of subsets of X satisfying the followingproperties.

a) The set X and ∅ belong to T ,

b) The union of any family of members of T is a member of T .

c) The intersection of any finite family of members of T is a member ofT .

Then T is called a topology for X and the members of T are called open sets.The ordered pair (X,T ) is called a topological space, or simply a space.

If we use the terminology open sets instead of member of T , then the definition ofa topological space may be restated as follows: A family of subsets of X is a topologyfor X means that:

a) Both X and ∅ are open sets.

b) The union of any family of open sets is an open set.

c) The intersection of any finite family of open sets is open.

23

24 CHAPTER 3. TOPOLOGICAL SPACES

Example 3.1.1 The usual topology for the real line R is the topology generated byits usual metric. We shall refer to the real line with the usual topology as simply thereal line or R.

Example 3.1.2 The usual topology on Rn is the topology generated by the usualmetric on Rn. It is also the topology generated by the taxicab metric and the maxmetric. Thus, the usual topology does not distinquish the metric determining it fromthe other two. We shall refer to Rn with the usual topology as Euclidean n-space, orsimply Rn.

Example 3.1.3 For any set X we take T = 2X to be the set of all subsets of X.This clearly satisfies all of the properties of a topology, since we have included everypossible subset in the topology. This is called the discrete topology. Note that it is thetopology generated by the discrete metric. Also, note that this is the largest possiblecollection of open subsets of X.

Example 3.1.4 At the opposite extreme, we may take T = {∅, X}. This is calledthe trivial topology, or indiscrete topology, on X. This is the smallest collection ofopen sets on X.

Example 3.1.5 Let X be a set. We shall take T to consist of ∅, X, and all sets Uso that X \ U is a finite set. Then T is a topology on X called the cofinite topology,or finite complement topology. This is really of interest only when X is an infinite set.When X is a finite set, this is the same as the discrete topology.

Definition 3.2 A subset F of a topological space X is closed if X \ F is an openset.

Theorem 3.1 The closed sets of a topological space X have the following properties:

a) X and ∅ are closed.

b) The intersection of any family of closed sets is closed.

c) The union of any finite family of closed sets is closed.

Definition 3.3 Let (X,T ) be a topological space and let A ⊂ X. A point x in X isa limit point of A if every open set containing x contains a point of A distinct fromx. The set of limit points of A is called the derived set of A, denoted A′.


3.2. INTERIOR, CLOSURE AND BOUNDARY 25

Example 3.1.6 Let X = {a, b, c, d}. Let T0 be the indiscrete topology; T1, thediscrete topology; T2 = {∅, X, {a}, {b}, {a, b}}; andT3 = {∅, X, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}. The reader should verify thatT2 and T3 are topologies on X. Let A = {a, b}, B = {c}, and C = {d}. We want tofind the limit points of these sets in the different topologies.

T0 T1 T2 T3

A X ∅ {c, d} {d}B {a, b, d} ∅ {d} {d}C {a, b, c} ∅ {c} ∅

Theorem 3.2 A subset A of a topological space X is closed if and only if A containsall of its limit points.

This is no surprise, and is proven exactly the way in which we proved it earlierin a metric space. We were careful there not to use the distance function, but to usethe open sets.

Definition 3.4 Let X be a topological space and let {xn} be a sequence of pointsin X. We say that {xn} converges to the point x ∈ X, or x is the limit of thesequence, if for each open set U containing x there is a positive integer N so thatxn ∈ U for all n ≥ N .

Sequences are not as fundamental in general topological spaces as they are inmetric spaces. The following example may show why.

Example 3.1.7 Consider R with the cofinite topology. Let {xn} be any sequence ofreal numbers. Let a ∈ R be any real number. Then {xn} converges to a, because ifU is any open set containing a, then R \ U is a finite set. Since {xn} is an infiniteset, we must have that infinitely many members of {xn} lie in U . Thus, there is apositive integer N such that if n ≥ N xn ∈ U . Thus, {xn} converges to a. However,a was any arbitrary real number. This means that {xn} converges to every realnumber. What is more (and maybe worse) is that {xn} was an arbitrary sequence.This means that every sequence converges to every real number. There are no non-convergent sequences and sequences do not have unique limits.

3.2 Interior, Closure and Boundary

Just as before we will define the interior, closure, and the boundary.

Definition 3.5 Let A be a subset of the topological space X. A point x ∈ A is aninterior point of A if there is an open set Uso that x ∈ U ⊂ A. A is called aneighborhood of x. The interior of A, denoted by A◦, is the set of all interiorpoints of A.



The closure, A, of A is the union of A and it set of limit points:

A = A ∪ A′.

A point x ∈ X is a boundary point of A if x ∈ A∩X \ A. The set of boundarypoints of A is called the boundary of A and is denoted by ∂A.

Theorem 3.3 For any subsets A,B of a topological space X

a) The interior of A is the union of all open sets contained in A and is thelargest open set contained in A.

b) A is open if and only if A = A◦.

c) If A ⊂ B, then A◦ ⊂ B◦.

d) (A ∩B)◦ = A◦ ∩B◦.

Proof: We will offer only a proof for (d). The others follow from closely from whatwe did in the case of a metric space.

Since A ∩B is a subset of both A and B, then by (c) (A ∩B)◦ ⊂ A◦ ∩B◦. Now,A◦ ∩ B◦ is an open set and is a subset of A ∩ B. Thus by (a), A◦ ∩ B◦ ⊂ (A ∩ B)◦.This completes the proof.

Theorem 3.4 For any subsets A,B of a topological space X

a) The closure of A is the intersection of all closed sets containing in Aand is the smallest closed set containing in A.

b) A is closed if and only if A = A.

c) If A ⊂ B, then A ⊂ B.

d) (A ∪B) = A ∪B.

We leave this to the reader to prove.

Theorem 3.5 Let A be a subset of a topological space X.

a) ∂A = A ∩X \ A = ∂(X \ A).

b) ∂A, A◦, and (X \ A)◦ are pairwise disjoint sets whose union is X.

c) ∂A is a closed set.

d) A = A◦ ∪ ∂A.

e) A is open if and only if ∂A ⊂ (X \ A).


3.3. BASIS FOR A TOPOLOGY 27

f) A is closed if and only if ∂A ⊂ A.

g) A is open and closed if and only if ∂A = ∅.

Proof: Parts (a)–(d) follow immediately from the definitions.(e) If A is open then A = A◦. Now by (b) A◦ and ∂A are disjoint. Thus, A and

∂A are disjoint. This implies that ∂A ⊂ X \ A. Now, if ∂A ⊂ X \ A then no pointof A is a boundary point of A. Thus, every point of A is an interior point of A andA = A◦. Thus, A is open.

(f) This follows from our duality of open and closed sets.(g) If A is both open and closed, then ∂A ⊂ A ∩ (X \ A) = ∅. If ∂A = ∅ then

clearly ∂A ⊂ A — meaning A is closed — and ∂A ⊂ (X \A) — meaning A is open.

Definition 3.6 A set A in a topological space X is dense if A = X. If X has acountable dense set, then X is a separable space.

Example 3.2.1 1. The reals with the usual topology is separable, since the ra-tionals are dense.

2. Euclidean n-space is separable, since the set of points having only rationalcoordinates is dense and countable.

3. The reals with the cofinite topology is separable, since every countable infiniteset is dense.

Definition 3.7 A subset B of a space X is nowhere dense if (B)◦ = ∅.

Note that a finite subset of a metric space is nowhere dense. In other topologicalspaces, we will see more interesting examples.

3.3 Basis for a Topology

It appears that a topology can be relatively large. In fact, for an infinite set thediscrete topology consists of all subsets of the space, so it would be prohibitive tohave to check all subsets. We have seen though that we can get by with just checkingsome of the sets. For the discrete topology we have usually only checked the singletonsets. For a metric space we were able to do everything we wanted by working withthe open balls. In fact we defined all open sets in terms of the open balls. Can we dothis in general? Can we find a certain collection of subsets that will generate all of theelements of the topology, just like the open balls generate the metric topology? Theanswer is yes, because we can take T as this generating set. This begs the answer,because we are looking for a smaller collection than the whole topology.



Definition 3.8 Let (X,T ) be a topological space. A basis B for T is a subcollectionof T with the property that each member of T is a union of members of B. Themembers of B are called basic open sets and T is the topology generated by B.

Example 3.3.1 Most of what we have seen is based on metric spaces.

1. The collection of all open intervals is a basis for the usual topology on the reals.

2. The collection of all open balls is a basis for the metric topology on the metricspace (X, d).

3. For any set X the collection of all singleton sets {x} is a basis for the discretetopology.

Definition 3.9 Let (X,T ) be a topological space. A local basis at a ∈ X is asubcollection Ba of T such that

a) a belongs to each member of Ba, and

b) each open set containing a contains a member of Ba.

Definition 3.10 A space X is first countable if there is a countable local basis ateach point of X. The space X is second countable if the topology for X has acountable basis.

Note that every second countable space is first countable because if there is acountable basis B, then the number of these sets containing any given point a ∈ Xis at most countable.

Theorem 3.6 Every second countable space is separable.

Proof: Let X be a second countable space with a countable basis B. Let A be aset formed by choosing one element from each non-empty element of B. Each pointof X is a limit point of some point in A by the definition of a basis. Thus, A is densein X.

Theorem 3.7 a) Every metric space is first countable.

b) Every separable metric space is second countable.

The proof is left to the reader.We have been starting with a topology and asking if there is a basis for it. We

could be starting with a collection of open sets and asking if it forms a basis for atopology. Not every collection of open sets will work. When is a collection of opensets a basis for a topology on X?



Theorem 3.8 A family B of subsets of a set X is a basis for a topology on X if andonly if both of the following hold:

a) The union of members of B is X.

b) For each B1, B2 ∈ B and x ∈ B1 ∩B2, there is a member Bx of B suchthat x ∈ Bx ⊂ B1 ∩B2.

Example 3.3.2 [Sorgenfrey Line] Let B be the collection of all half-open intervalsof R of the form [a, b), a < b. Clearly, the union of all of these intervals is R. If wetake two of these sets and intersect them, we can find another of these sets in theintersection. Thus, these sets form a basis for a topology on the real line, called thehalf-open interval topology T ′′ for R. R with this topology is called the Sorgenfreyline. The Sorgenfrey line has the property that it is first countable and separable,but not second countable.


We were able to move away from the epsilon-delta definition of continuity of a functionin between two metric spaces by using open balls. We will use this as our startingpoint for general topological spaces.

Definition 3.11 A function f : (X,T )→ (Y,T ′) is continuous at a point a ∈ Xif for each open set V in Y containing f(a) there is an open set U ⊂ X containing aso that f(U) ⊂ V , or equivalently U ⊂ f−1(V ).

Now, we are more interested in the situation where the function is continous atevery point of X. This means that for every open set V in Y and every point a ∈ Xwith f(a) ∈ V , there is an open set Ua ⊂ X with a ∈ Ua and f(Ua) ⊂ V . Equivalently,we have a ∈ Ua ⊂ f−1(V ). This means that for each point in f−1(V ) we can find anopen set containing that point and contained in f−1(V ). Thus, f−1(V ) must be openin X for each open set V ⊂ Y . This leads us to a more general definition.

Definition 3.12 A function f : (X,T ) → (Y,T ′) is continuous if for each openset V in Y f−1(V ) is an open set in X.

Theorem 3.9 Let f : X → Y be a function on the topological spaces X and Y andlet a ∈ X. The following are equivalent.

a) f is continuous at a.

b) For each open set V ∈ Y containing f(a), there is an open set U in Xsuch that a ∈ U ⊂ f−1(V ).

c) For each neighborhood V of f(a), f−1(V ) is a neighborhood of a.



The proof is left to the reader.

Theorem 3.10 Let f : X → Y be a function of topological spaces. The following areequivalent.

(i) f is continuous.

(ii) For each closed subset C ⊂ Y , f−1(C) is closed in X.

(iii) For each subset A ⊂ X, f(A) ⊂ f(A).

(iv) There is a basis B for the topology of Y so that f−1(B) is open in X for eachbasic open set B ∈ B.

Proof: To show that (i) implies (ii) will require the duality between open and closedsets. If C ⊂ Y is closed, the Y \C is open in Y . Since f is continuous, f−1(Y \C) isopen in X. Hence X\(f−1(Y \ C)) is open. If x ∈ X\(f−1(Y \ C)) then f(x) 6∈ Y \C,or f(x) ∈ C. Thus, X\(f−1(Y \ C)) ⊂ f−1(C). The opposite inclusion is clear. Thus,f−1(C) = X \ (f−1(Y \ C)) is closed. A similar analysis shows that (ii) implies (i).

To show that (ii) implies (iii), let A ⊂ X. Then f(A) is a closed subset of Y .Hence, f−1(f(A)) is a closed subset of X. Now, A ⊂ f−1(f(A)) so A ⊂ f−1(f(A)).Thus, f(A) ⊂ f(A).

To show that (iii) implies (ii), let C be a closed subset of Y . Then,

f(f−1(C)) ⊂ ff−1(C) ⊂ C ⊂ C

so f−1(C) ⊂ f−1(C) making f−1(C) a closed set.For the last equivalence, (i) clearly implies (iv). We need to prove the opposite

implication. Let O be an open set in Y . Then by the definition of a basis, O = ∪α∈IBα

for some subcollection {Bα}α∈I of the basis B. Then

f−1(O) = f−1

(⋃α∈I

Bα

)=⋃α∈I

f−1(Bα).

Since each f−1(Bα) is open in X and the union of any family of open sets is open,then f−1(O) is open in X and f is continuous.

Theorem 3.11 If f : X → Y and g : Y → Z are continuous functions, then g ◦f : X → Z is continuous.

Definition 3.13 A function f : X → Y is a homeomorphism if

a) f is one-to-one, (injective)

b) f is onto, (surjective)


3.5. SUBSPACES 31

c) f is continuous,

d) f−1 is continuous.

Topological spaces are topologically equivalent or homeomorphic if there home-omorphism from f from X onto Y .

The continuity of the function and its inverse are extremely important! A prop-erty P of topological spaces is a topological property or topological invariantprovided that if space X has property P , then so does every space which is homeo-morphic to X.

Theorem 3.12 Separability is a topological property.

Theorem 3.13 First countability and second countability are topological properties.

Definition 3.14 A topological space is metrizable provided that the topology on Xis generated by a metric.

Theorem 3.14 Metrizability is a topological property.

Since R and (0, 1) are homeomorphic, the property of being a bounded metricspace is not a topological property. Likewise, distance is not a topological invariant.

3.5 Subspaces

Let (X,T ) be a topological space and let A be a subset of X. The relative topologyor subspace topology T ′ on A determined by T consists of all sets of the formO ∩ A for which O is an open set of T .

T ′ = {O ∩ A | O ∈ T }.

The members of T ′ are called relatively open sets in A, and (A,T ′) is called asubspace of (X,T ).

Note that this is actually a topology for A.

∅ = ∅ ∩ A A = X ∩ A,

so both ∅ and A are open in A. If {Uα} are open in A, then Uα = Oα ∩ A and⋃α

Uα =⋃α

(Oα ∩ A) =

(⋃a

lphaOα

)∩ A

is relatively open since the union of any family of open sets is open in X. For anyfinite family of open sets Ui = Oi ∩ A, we have

n⋂i=1

Ui =n⋂i=1

(Oi ∩ A) =

(n⋂i=1

Oi

)∩ A

is relatively open since the intersection of any finite family of open sets is open in X.



Lemma 3.1 Let (a,T ′) be a subspace of the topological space (X,T ). A subset Fof A is closed in the subspace topology on A if and only if F = C ∩A for some closedsubset C of X.

Example 3.5.1 1. The closed interval [a, b] with a < b is a subspace of R withthe usual topology. The open sets containing a are sets of the form [a, c) witha < c < b.

2. The subset of Rn+1 consisting of all (n + 1)-tuples (x1, x2, . . . , xn, xn+1) withxn+1 = 0 is homeomorphic to Rn

3. Let a < b < c < d. Let A = [a, b]∪ (c, d) be considered as a subspace of the realline. Then the subset [a, b] of A is both relatively open and relatively closed.It is clearly closed because [a, b] = [a, b] ∩ A and [a, b] is closed in the real line.It is open because for 0 < ε < c − b, [a, b] = (a − ε, b + ε) ∩ Y . Thus, we seethat since (c, d) is the complement of this set that is both relatively open andrelatively closed, then we see that (c, d) is both relatively open and relativelyclosed.

Definition 3.15 A property P to topological spaces is hereditary provided that ifX has property P , then every subspace of X has this property.

Example 3.5.2 1. First countability and second countability are hereditary prop-erties. If X has a countable basis, then intersecting these basis elements with Awill give a countable basis for the subspace topology. First countable is similar.

2. Separability is not hereditary. Let A ⊂ R2 consist of the x-axis and the pointa = (0, 1). Define a topology T on A by taking the empty set and all subsetsof A that contain the singleton set {a}. Then (X,T ) is separable because thesingleton set {a} is dense. Every point except a is a limit point of {a}. However,the subspace topology on R (the x-axis) is the discrete topology, so (R,T ′) isnot separable.

3.6 Hausdorff Spaces

A topological space X is a Hausdorff space if for each pair of distinct pointsa, b ∈ X there exist disjoint open sets U and V such that a ∈ U and b ∈ V .

Example 3.6.1 1. Every metric space is Hausdorff. We proved this as a home-work problem, but it is simple. Let r = d(a, b) and then take two open balls ofradius r/2 centered at a and b respectively.

2. The real line with the co-finite topology is not Hausdorff. Likewise, the real linewith the co-countable topology is not Hausdorff.


3.6. HAUSDORFF SPACES 33

3. Take any set with more than one point and give it the indiscrete (trivial) topol-ogy. It is not Hausdorff.

4. The space in Example 2 is not Hausdorff, because you can never separate afrom any other point.

5. [The Zariski Topology] Let n be a positive integer and consider the family Pof all polynomials in n real variables x1, x2, . . . , xn. For p ∈P let Z(p) denoteits solution set in Rn:

Z(p) = {(x1, x2, . . . , xn) ∈ Rn | p(x1, x2, . . . , xn) = 0}.

Let B be the collection of sets that are complements of some Z(p) for somep ∈P. This forms a basis for a topology on Rn called the Zariski topology.

For the real line, n = 1, this is just the co-finite topology. This is because eachfinite set of real numbers is the solution set for some polynomial in one realvariable. If A = {a1, a2, . . . , an}, then

p(x) = (x− a1)(x− a2) . . . (x− an)

is a polynomial with A as its solution set. Likewise, the set of solutions of apolynomial in one real variable of dimension n is at most n.

For n > 1 this is not the co-finite topology. For example, the line y = a in R2

is the solution set to the polynomial in two variables

p(x, y) = y − a.

Note that this is not a finite set. However, each finite set can serve as thesolution set of a polynomial.

Now, Rn with the Zariski topology is not Hausdorff. Assume that P = (a1, a2, . . . , an)and Q = (b1, b2, . . . , bn) are two distinct points in Rn.

Theorem 3.15 1. The property of being a Hausdorff space is topological and hered-itary.

2. In a Hausdorff space a sequence {xn}∞n=1 cannot converge to more than onepoint.

Proof: We will prove 1 only and leave the other to the reader.Suppose that X is Hausdorff and f : X → Y is a homeomorphism. For a 6= b ∈ Y ,

we have that f−1(a) and f−1(b) are distinct points in X. Thus, there are disjointopen sets U, V ⊂ X so that f−1(a) ∈ U and f−1(b) ∈ V . Hence, f(U) and f(V ) aredisjoint open sets of Y containing a and b respectively.

To show that Hausdorff is hereditary, assume that X is Hausdorff and that A ⊂ X.Let a 6= b ∈ A. Then, a 6= b ∈ X and there are disjoint open sets U, V ⊂ X so thata ∈ U and b ∈ V . Then, U ∩ A and V ∩ A are disjoint relatively open subsets of Acontaining a and b respectively.


TopologyMATH 4181

Fall Semester 1999

David C. RoysterUNC Charlotte

November 11, 1999

Chapter 1

Introduction to Topology

1.1 History1

Topology is thought of as a discipline that has emerged in the twentieth century.There are precursors of topology dating back into the 1600’s. Gottfried WilhelmLeibniz (1646–1716) was the first to foresee a geometry in which position, rather thanmagnitude was the most important factor. In 1676 Leibniz use the term geometriasitus(geometry of position) in predicting the development of a type of vector calculussomewhat similar to topology as we see it today.

The first practical application of topology was made in the year 1736 by the Swissmathematician Leonhard Euler (1707–1783) in the Konigsberg Bridge Problem.

Carl F. Gauss (1777–1855) predicted in 1833 that geometry of location wouldbecome a mathematical discipline of great importance. His studey of closed surfacessuch as the sphere and the torus and surfaces much like those encountered in multi-dimensional calculus may be considered as a harbinger of general topology. Gausswas also interested in knots, which are of current interest today in topology.

The word topology was first used by the German mathematics Joseph B. Listing(1808–1882) in the title of his book Vorstudien zur Topologie (Introductory Studiesin Topology), a textbook published in 1847. Listing book dealt with knots and sur-faces but failed to generate much interest in either the name or the subject matter.Throughout much of the nineteenth and early twentieth centuries, much of what nowfalls under the auspices of topology was studied under the name of analysis situs(analysis of position).

Bernard Riemann (1826–1866) was the first mathematician to foresee topologyin the generality it has achieved today. He initiated the study of connectivity of asurface, or the arrangement of holes in a surface. He used concepts in which thenumber of dimensions exceeded three, which at that time was generally conceded tobe the maximum number of dimensions involved with any geometric object.

Present-day topology can be traced to two primary sources: the development ofnon-euclidean geometry and the process of putting calculus on a firm mathematicalfoundation.

1The information here is taken from Principles of Topology by Fred H. Croom.

2

1.2. SETS AND SET OPERATIONS 3

1.2 Sets and Set Operations

We need some basic information about sets in order to study the logic and the ax-iomatic method. This is not a formal study of sets, but consists only of basic defini-tions and notation.

Braces { and } are used to name or enumerate sets. The roster method for namingsets is simply to list all of the elements of a set between a pair of braces. For examplethe set of integers 1, 2, 3, and 4 could be named

{1, 2, 3, 4}.

This does not work well for sets containing a large number of elements, though it canbe used. The more common method for this is known as the set builder notation.A property is specified which is held by all objects in a set. P (x), read P of x, willdenote a sentence referring to the variable x. For example,

x = 23

x is an odd integer.

1 ≤ x ≤ 4.

The set of all objects x such that x satisfies P (x) is denoted by

{x | P (x)}.

The set {1, 2, 3, 4} can be named

{x | 1 ≤ x ≤ 4, x ∈ Z} = {x ∈ Z | 1 ≤ x ≤ 4}.

From hence forth, the words object, element, and member mean the same thingwhen referring to sets. Sets will be denoted mainly by capital Roman letters andelements of the sets by small letters. The following have the same meaning:

a ∈ A

a is in set A

a is a member of set A

a is an element of set A

Likewise, a 6∈ A means that a is not an element of set A.A is a subset of B if every element of A is also an element of B. The following

have the same meaning:

A ⊂ B

Every element of A is an element of B



If a ∈ A, then a ∈ B

A is included in B

B contains A

A is a subset of B

Note that a set is always a subset of itself.

If A and B are sets, then we say that A = B if A and B represent the same set:

A = B

A and B are the same set

A and B have the same members

A ⊂ B and B ⊂ A

The set which contains no elements is known as the empty set, and is denoted by∅. Note that for each set A, ∅ ⊂ A.

The intersection of two sets A and B is the set of all elements common to bothsets. The intersection is symbolized by A ∩ B or {x | x ∈ A and x ∈ B}. The unionof two sets A and B is the set of elements which are in A or B or both. The unionis symbolized by A ∪B or {x | x ∈ A or x ∈ B}.

1.2.1 Universal Sets and Compliments

When we are working in an area or on a certain problem, we always have a frame ofreference in which we are working called a universal set. In our geometry course, itwill be the set of points that lie on a plane. In calculus we consider the set of realnumbers, the set of real functions, the set of differentiable functions, and the set ofcontinuous functions as universal sets.

The complement of a set A is defined to be the set of all elements of the universalset which are not in A, and is symbolized by CA = A′ = Ac. Note that A ∪ Ac isalways the universal set, while A ∩ Ac = ∅.

The set difference of the sets A and B is defined to be all of those elements in Awhich are not in B. It is denoted by

A \B = {x ∈ A | x 3 B}.

Note that A \B and B \A will usually be different, and that even though A \B = ∅it need not follow that A = B.


1.3. PRODUCTS SETS 5

1.3 Products Sets

Let X and Y be sets. The set of all ordered pairs {(x, y) | x ∈ X and y ∈ Y } isthe product set X × Y , or Cartesian product or direct product of X and Y . A sliceof this product set is {x} × Y or X × {y} for a given x ∈ X or y ∈ Y . Examplesof common product spaces are the plane R2 = R × R, 3-space R3 = R × R2, a rightcircular cylinder, S1 × [0, 1], or the torus, S1 × S1.

Theorem 1 Let X and Y be sets and let A,C ⊂ X and B,D ⊂ Y .

a) A× (B ∩D) = (A ∩B)× (A ∩D).

b) A× (B ∪D) = (A ∪B)× (A ∪D).

c) A× (Y \D) = (A× Y ) \ (A×D).

d) (A×B) ∩ (C ×D) = (A ∩ C)× (B ∩D).

e) (A×B) ∪ (C ×D) ⊆ (A ∪ C)× (B ∪D).

f) (X × Y ) \ (A×B) = (X × (Y \B)) ∪ ((X \ A)× Y )

The concept of the product of two sets can be extended to more than two factors.If {Xi}ni=1 is a finite collection of sets, then their product is

X1 ×X2 × · · · ×Xn =n∏i=1

Xi = {(x1, x2, . . . , xn) | xi ∈ Xi for each i = 1, 2, . . . , n}.

For an infinite collection of sets, the product is defined by

∞∏i=1


1.4 Functions

A function f : X → Y is a rule which assigns to each x ∈ X a unique y ∈ Y and wesay y = f(x). If y = f(x) then y is called the image of x and x is called the preimageof y. The set X is the domain of f and Y is the range or codomain of f .

Let A ⊂ X. The set f(A) = {y ∈ Y | y = f(x) for some x ∈ A} is called theimage of A. The set f(X) is called the image of f . For B ⊂ Y , the set

f−1(B) = {x ∈ X | f(x) ∈ B}

is the inverse image of B under f . The set of points

Γ = {(x, f(x)) ∈ X × Y | x ∈ X}

is called the graph of the function f .



A function f : X → Y is injective if for distinct elements x1, x2 ∈ X, f(x1) 6= f(x2)in Y . Another way to think of this is to say that f is injective if f(x1) = f(x2) impliesthat x1 = x2.

If f(X) = Y , the function f is said to be surjective.A function that is surjective and injective is called a bijection. In this case we

have that f : X → Y is a bijection provided that each member of Y is the imageunder f of exactly one member of X. In this case the inverse function f−1 : Y → Xexists assigning to each element y ∈ Y its unique preimage x = f−1(y) in X.

The identity function iX : X → X from a set X to itself is the function definedby iX(x) = x for all x ∈ X. This function is often denoted by 1X .

If f : X → Y and g : Y → Z are functions, then the composite function g◦f : X →Z is defined by g ◦ f(x) = g(f(x)), for x ∈ X.

Definition 1 Let X be a set. A sequence in X is a function f : Z+ → X whosedomain is the set of all positive integers, Z+ or the set of positive integers less thanor equal to some given positive integer N . The sequence is called finite if its domainis {1, 2, . . . , N} and infinite if its domain is all positive integers.

1.5 Equivalence Relations

Let X be a set. A relation R on X is a subset of X × X. If (x, y) ∈ R we will saythat x is related to y by R and to write xRy.

A relation R on a set X is called reflexive, symmetric, or transitive if it satisfiesthe corresponding property below.

(a) The Reflexive Property : xRx for all x ∈ X.

(b) The Symmetric Property : If xRy, then yRx.

(c) The Transitive Property : If xRy and yRz, then xRz.

Definition 2 An equivalence relation on a set X is a relation on X which isreflexive, symmetric, and transitive.

Definition 3 Let ∼ denote an equivalence relation on X. For x ∈ X the set [x] ={y ∈ X | y ∼ x} is calle dth e equivalence class of x.

Proposition 1 Let X be a set and let ∼ denote an equivalence relation on X.

a) x ∈ [x] for each x ∈ X.

b) x ∼ y if and only if [x] = [y].

c) x 6∼ y if and only if [x] ∩ [y] = ∅.

d) For x, y ∈ X, [x] and [y] are either identical or disjoint.


1.6. CARDINALITY 7

1.6 Cardinality

We are often interested in how big sets are in relation to one another. Clearly, wecan tell the difference in sizes of two finite sets, but how do we differentiate betweentwo infinite sets? Are there different sizes of infinite sets? How do we compare setsto tell if one has a greater number of members?

Definition 4 a set A is finite if A is empty or if there is a bijection between A andthe set of integers from 1 to N for some positive integer N . In the latter case, A issaid to have N members. If a set is not finite, it is called infinite.

Definition 5 A set is denumerable or countably infinite if there is a bijectionbetween the set and the positive integers. A set which is either finite or denumerableis called countable. A set which is not countable is called uncountable.

Lemma 1 a) Each subset of a finite set is finite.

b) Each subset of a countable set is countable.

c) Each set which contains an infinite set is infinite.

d) Each set which contains an uncountable set is uncountable.

Example 1 1. The set Z+ ∪ {0} of all non-negative integers is countable. Thebijection is f : Z+ ∪ {0} → Z

+ given by

f(n) = n+ 1, n ∈ Z+ ∪ {0}.

2. The set of all integers, Z is countable. The bijection g : Z→ Z+ is given by

g(n) =

{2n− 1 if n is positive

−2n if n is negative

3. The product set Z+ × Z+ is countable. One method is to use the CantorDiagonalization Method to count the ordered pairs (m,n). A second method isto define the function g : Z+ × Z+ → Z

+ by

g(m,n) = 2m3n, (m,n) ∈ Z+ × Z+.

Now, g is not surjective, but the Fundamental Theorem of Arithmetic on theunique factorization into primes guarantees that the function g is injective.Thus, there is a bijection from Z

+ × Z+ to a subset of Z+. Since every subsetof a countable set is countable, we have that Z+ × Z+ is countable.



Theorem 2 a) If {Ai}Ni=1 is a finite collection of finite sets, then both⋃Ni=1 Ai and

∏Ni=1 Ai are finite.

(Finite unions and finite products of finite sets are finite.)

b) If {Ai}∞i=1 is a countable collection of countable sets, then⋃∞i=1 Ai is

countable.(Countable unions of countable sets are countable.)

c) If {Ai}Ni=1 is a finite collection of countable sets, then both∏N

i=1 Ai iscountable.(Finite products of countable sets are countable.)

Why didn’t we claim that a countable product of countable sets is countable?Mainly because it is not true, as is seen in the following example.

Example 2 Let Ai = {0, 1} for i = 1, 2, . . . . Let

U =∞∏i=1

Ai = {(a1, a2, a3, . . . ) | ai = 0 or 1}.

Assume that U is countable. Then there is a bijection f : Z+ → U . For an elementa = (a1, a2, a3, . . . ) ∈ U we shall refer to a1 as the first coordinate, a2 as the secondcoordinate, and so forth.

We can list all of the elements in U using the bijection f . They are {f(1), f(2), F (3), . . . }.Consider the following element in U . Define x = (x1, x2, x3, . . . ) as follows:

xi =

{0 if the ith coordinate of f(i) is 1

1 if the ith coordinate of f(i) is 0

Then, we have that for each positive integer i, x 6= f(i) for they differ in the ith

coordinate. This means that x is not in the exhaustive list of elements we have in ourbijection. That is this bijection is not surjective. This contradiction show us that Ucannot be countable.

Theorem 3 The set of rational numbers is countable.

Proof: There are several ways of proving this. One method is to use the CantorDiagonalization Method to count the rationals.

(Method 1) List the positive rational numbers in rows where the first row consistsof all positive rational numbers with 1 as a denominator, the second row lists allpositive rational numbers with 2 as a denominator, and so on:

11

21

31

41

. . .12

22

32

42

. . .13

23

33

43

. . .14

24

34

44

. . .

......

......

. . .


1.6. CARDINALITY 9

Clearly, we have each positive rational number in here numerous times, but thediagonalization method will still show that there are a countable number of elementsin this array. The positive rationals form a subset of this array, thus there must bea countable number of positive rationals. This will yield that there are a countablenumber of rational numbers.

(Method 2) Every rational number can be expressed uniquely in lowest terms asm/n where m and n are integers with no common positive divisor other than 1, andn is positive. Consider the function m/n 7→ (m,n) from the set of rational numbersinto Z × Z. This function is injective since the ordered pair (m,n) determines onlyone rational number m/n. Thus, the set of rational numbers is equivalent to a subsetof the countable set Z× Z, and is hence countable.

Theorem 4 The set of real numbers is uncountable.

Proof: We will make use of the example of the countable product above. Eachelement in

∏∞i=1 0, 1 is a sequence consisting of 0’s and 1’s. Each of these sequences

represents a unique real number between 0 and 1, by the correspondence

(a1, a2, a3, . . . ) 7→ 0.a1a2a3 . . . .

This is a one-to=one correspondence. Thus, the set of real numbers between 0 and1 representable as a decimal using only 0’s and 1’s is an uncountable set. Thus, Rcontains an uncountable set and hence is uncountable.

Theorem 5 The set of irrational numbers is uncountable.

Proof: Since the set of real numbers is the union of the set of rational numbers andthe set of irrational numbers, if the set of irrationals were countable, then we wouldhave that the real numbers are countable. That failing to be true, implies that theirrationals must be uncountable.


Chapter 2

Metric Spaces


Definition 6 Let X be a set and d : X×X → R+ a function satisfying the following



b) d(x, y) = d(y, x).

c) d(x, z) ≤ d(x, y) + d(y, z).



Example 3 The real line, R is a metric space using the standard distance function,the absolute value: d(a, b) = |a− b|. The above properties are standard proofs aboutthe absolute value function.

Example 4 The plane, R2, with the usual Euclidean distance formula is a metricspace. If P = (x1, y1) and Q = (x2, y2), then

d(P,Q) =√

(x2 − x1)2 + (y2 − y1)2.

Example 5 These are special cases of the general Euclidean n-space,

Rn = {(a1, a2, . . . , an) | ai ∈ R}.

10



d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) =

(n∑i=1

(xi − yi)2

)1/2

.



‖a‖ = d(a,O) =

(n∑i=1

a2i

)1/2

.

Theorem 6 (Cauchy-Schwarz Inequality) For any points a, b ∈ Rn

|a · b| ≤ ‖a‖‖b‖.

Theorem 7 (The Minkowski Inequality) For any points a, b ∈ Rn

‖a+ b‖ ≤ ‖a‖+ ‖b‖.




Example 6 [The Taxicab Metric] Define a function d′ : R2 × R2 → R as follows. Ifx = (x1, x2) and y = (y1, y2), then

d′(x, y) = |x1 − y1|+ |x2 − y2|.




U = {x = (x1, x2) ∈ R2 | d′(x,O) = 1}.


d′(x, y) =n∑i=1

|xi − yi|.



Example 7 [The Max Metric on Rn] Another metric for Rn is given by taking thelargest of the differences of the coordinates of x and y.

d′′(x, y) = max{|xi − yi}ni=1.

Example 8 [The Discrete Metric] For any set X, define

d(x, y) =

{0 if x = y

1 if x 6= y


Example 9 Let C [a, b] denote the set of all continuous real-valued functions definedon the interval [a, b]. For f, g ∈ C [a, b] define

ρ(f, g) =

∫ b

a

|f(x)− g(x)| dx.


Example 10 For the set C [a, b] define ρ′ by

ρ′(f, g) = lub{|f(x)− g(x)| | x ∈ [a, b]}.


Definition 7 A number u is an upper bound for a set A of real numbers providedthat a ≤ u for all a ∈ A. If there is a smallest upper bound u0 for A, that is an upperbound that is less than or equal to all other upper bounds for A, then u0 is calledthe least upper bound or supremum of A. The least upper bound for a set A isdenoted by lubA or supA.

Definition 8 A number ` is an lower bound for a set A of real numbers providedthat ` ≤ a for all a ∈ A. If there is a largest lower bound `0 for A, that is a lowerbound that is less than or equal to all other lower bounds for A, then `0 is called thegreatest lower bound or infimum of A. The greatest lower bound for a set A isdenoted by glbA or inf A.





Definition 9 Let (X, d) be a metric space and let A be a non-empty subset of X.If {d(x, y)‖x, y ∈ A} has an upper bound, then A is said to be bounded, andlub{d(x, y)‖x, y ∈ A} is called the diameter of A. For completeness, we definethe diameter of the empty set to be 0. If the set X is bounded, then we call (X, d) abounded metric space.


d(x,A) = glb{d(x, y) | y ∈ A}.

Theorem 8 Let {(Xi, di)}ni=1 be a finite collection of metric spaces and let

X =n∏i=1

Xi.

For each pair of points x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn) in X, let d : X×X →R be defined by

d(x, y) =

(n∑i=1

(di(xi, yi))2

)1/2

.

Then (X, d) is a metric space. The metric d defined above is called the productmetric on X.


In topology we are concerned with how spaces are changed when stretched, bent,twisted and modified — but not torn. We do so by studying the maps that doso. Our friend here is the continuous map. In your study of calculus, you saw thatcontinuous functions did many things. At the time you were more interested in specialcontinuous functions — the differential functions. We here are more interested in themore general function.

In calculus, we saw that a continuous function was one that did not do too muchdamage to the domain in the range. By this, we mean that if two points were closein the domain, then their images were not too far apart in the image. We saw thisintuitively through looking at graphs and looking at limits. To insure specificity, weneed the definition of continuity due to Cauchy and Weierstrauss. It is one with whichyou are familiar.



Definition 10 Let f : (X, d)→ (Y, d′) be a function between two metric spaces. Leta ∈ X. We say that f is continuous at a if given any ε > 0 there is a δ > 0 sothat d′(f(x), f(a)) < ε whenever d(x, a) < δ. We say that f is continuous if it iscontinuous at a ∈ X for all a ∈ X.

This clearly depends on the metric in each of the two spaces. A change of met-ric might change the continuity of the given function. Will it? Is continuity thatdependent on the metric in the domain or the range?

Let’s check two well-known functions that we think should be continuous andmake certain that they are continuous under this definition.

Example 11 Let f : (X, d)→ (Y, d′) be given by f(x) = b for all x ∈ X where b ∈ Yis a constant. This is just the constant function.

To show that f is continuous, we need to show that if we are given any ε > 0, thenwe can find a δ > 0 so that whenever d(x1, x2) < δ then d′(f(x1), f(x2)) < ε. In thiscase, this is easy. This is because d′(f(x1), f)(x2)) = d′(b, b) = 0 < ε for any choiceof x1, x2 ∈ X. Thus, it does not matter what we may choose for δ. You could takeδ = ε or δ = 1. Regardless, whenever d(x1, x2) < δ then d′(f(x1), f(x2)) = 0 < ε, andwe are done.

Example 12 Let 1X : (X, d)→ (X, d) denote the identity map from X to itself givenby 1X(x) = x. We claim that this function is continuous.

Again, to show this we are given an ε > 0. We then need to find a δ > 0 so thatwhenever d(x1, x2) < δ then d(1X(x1), 1X(x2)) < ε. However, since 1X(x1) = x1 and1X(x2) = x2, it is easy to see that if we take δ ≤ ε, then if d(x1, x2) < δ it followsthat d(1X(x1), 1X(x2)) = d(x1, x2) < δ ≤ ε. Thus, 1X is a continuous function.

Example 13 This time we will be working with the same underlying set, but we willplace a different metric on it. Will this make a difference?

Let X = Rn with the usual metric. Let Y = Rn with the maximum metric,

d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) = max1≤i≤n

{|xi − yi|}.

Define h : (X, d) → (Y, d′′) by h(x1, x2, . . . , xn) = (x1, x2, . . . , xn). It is the identitymap on the underlying set, but it does not carry the same metric information. Is hcontinuous? Is h−1 continuous?

It turns out that both are continuous! To prove this, let’s first look at h−1 : (Y, d′′)→(X, d). We are given an ε > 0. We need to find a δ > 0 so that if

d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < δ

then d(h−1(x1, x2, . . . , xn), h−1(y1, y2, . . . , yn)) < ε.


2.3. OPEN SETS AND CLOSED SETS 15

To say that d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < δ means that |xi− yi| < δ for alli = 1, . . . , n. Thus,

d(h−1(x1, x2, . . . , xn), h−1(y1, y2, . . . , yn)) = d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) (2.1)

=

(n∑i=1

|xi − yi|2)1/2

(2.2)

<

(n∑i=1

δ2

)1/2

= δ√n (2.3)

Thus, we need δ√n < ε, or take δ <

ε√n

.

Now, to show that h is continuous we are given an ε > 0. We need to find δ sothat whenever d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < δ, we have that

d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < ε.

To say that d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < δ means that (∑n

i=1 |xi − yi|2)1/2

<δ. Thus, each of the differences |xi−yi|must be less than δ, and the largest of these dif-ferences is still less than δ. Thus, in order for d′′((x1, x2, . . . , xn), (y1, y2, . . . , yn)) < ε,we need only choose δ < ε.

2.3 Open Sets and Closed Sets

Definition 11 Let (X, d) be a metric space, a ∈ X, and r > 0 a positive real number.The open ball Bd(a; r) with center a and radius r is the set

Bd(a; r) = {x ∈ X | d(a, x) < r}.

When there is only one metric under consideration, we will simplify the notation toB(a; r).

Definition 12 Let (X, d) be a metric space and let a ∈ X. A subset N ⊂ X is aneighborhood of a if there is a δ > 0 so that B(a; δ) ⊂ N . The collection Na of allneighborhoods of a point a ∈ X is called a complete system of neighborhoods ofthe point a.

Definition 13 A subset U of a metric space (X, d) is an open set with respect tothe metric d provided that U is a union of open balls. The family of all open setsdefined in this way is called the topology for X generated by d. A subset C ⊂ Xis said to be closed (with respect to d) if its complement X \ C is an open set (withrespect to d).



Thus, a neighborhood of a and an open set containing a need not be the samething. However, if U is an open set containing a, then U is a neighborhood of a.

Theorem 9 The following statements are equivalent (TFAE) for a subset U of ametric space (X, d).

a) U is an open set;

b) for each x ∈ U there is an εx > 0 so that B(x; εx) ⊂ U .

c) for each x ∈ U , d(x,X \ U) > 0, if U 6= X.

Proof: What this means is that Statements (a) and (b) are equivalent, (b) and (c)are equivalent, and (a) and (c) are equivalent. We can show this by proving that (a)is equivalent to (b) and then that (b) is equivalent to (c). In condition (c) we willassume that U 6= X since the distance from the empty set is not defined.

Assume that U is an open set and let x ∈ U . Since U is the union of open balls,then x ∈ B(a; r) ⊂ U . Then d(x, a) < r. We want to center an open ball at x andhave it contained in U . Choose εx ≤ r − d(x, a). Then B(x; εx) ⊂ B(a; r) for thefollowing reason: If y ∈ B(x; εx),

d(y, a) ≤ d(y, x) + d(x, a) < εx + d(x, a) ≤ r − d(x, a) + d(x, a) = r.

Thus, B(x; εx) is an open ball of positive radius centered at x and contained in U .Thus (a) =⇒ (b).

To show that (b) =⇒ (a), since each x ∈ U lies in an open ball contained in U ,U is the union of these open balls.

To see that (b) =⇒ (c), let B(x; εx) ⊂ U . Then any point within distance εx ofx is in U , so the distance from x to X \U must be at least εx. Thus, d(x,X \U) > 0for each x ∈ U .

Assuming that (c) holds, d(x,X \ U) = αx > 0 depending on x. This means thatthe distance from x to a point outside U must be at least αx, so any point withindistance αx of x must be in U . This means B(x;αx) ⊂ U .

Note that we have just shown that for each a ∈ X and for each δ > 0, the openball B(a; δ) is a neighborhood of each of its points.

2.3.1 Neighborhoods and Continuous Functions

How do we plan to use this information? While our definition of continuity is precise,it requires some specificity and does not look generalizible. What I mean by this isthat the definition seems to rely specifically on the definition of the metric, and itwill be hard to realign our definition when we have to move away from metric spaces.

Theorem 10 Le f : (X, d) → (Y, d′). f is continuous at a ∈ X if and only if foreach neighborhood M of f(a) there is a corresponding neighborhood N of a, such that

f(N) ⊂M,


2.4. LIMITS 17

or equivalently

N ⊂ f−1(M).

Proof: First, let’s suppose that f is continuous at a ∈ X and let M be a neigh-borhood of f(a). This means that for some ε > 0 Bd′(f(a); ε) ⊂ M . Since f iscontinuous at a ∈ X we know that we can find δ > 0 so that if d(x, a) < δ thend′(f(a), f(x)) < ε. This says that Bd(a; δ) ⊂ f−1(M) and we already have seen thatBd(a; δ) is a neighborhood of a ∈ X. Thus, if f is continuous, we have found acorresponding neighborhood to M .

Now, suppose that for any neighborhood, M , of f(a) we can find a neighborhoodN of a so that f(N) ⊂ M . Let ε > 0 be given to you. You must find a δ > 0 sothat whenever d(a, x) < δ we have d′(f(a), f(x)), ε. Now, let M = Bd′(f(a); ε). Mis a neighborhood of f(a), so we know that there is a neighborhood N ⊂ X of a sothat f(N) ⊂M . Since N is a neighborhood of a, it must contain a δ-ball centered ata, Bd(a; δ) ⊂ N , by the definition of a neighborhood. Thus, if we have x ∈ Bd(a; δ),then f(x) ∈ M = Bd′(f(a); ε). The other way of writing this is: if d(a, x) < δ thend′(f(a), f(x)) < ε. Therefore, f is continuous at a ∈ X.

2.4 Limits

Recall that a sequence is just a function a : Z+ → (X, d). We want to discuss whathappens to the sequence as we let n go to infinity; in other words what happens tothe sequence as we look further and further into the range of a. Let us first recallthe definitions in the real numbers and then try to set them up so that we can easilygeneralize them to arbitrary metric spaces.

Let {ai} be a sequence of real numbers. A real number L is said to be the limit ofthe sequence {an} if, given any ε > 0, there is a positive integer N such that whenevern > N , |an − L| < ε. In this case we say that the sequence converges to L and write

limn→∞

an = L.

How can we generalize this to an arbitrary metric space? It should not be hard,because all we used in the definition was the distance function in the real numbers.We will use the distance function in our metric space similarly.

Definition 14 Let {xn} be a sequence in the metric space (X, d). We say that thissequence {xn} converges to x ∈ X if given any ε > 0 there is a positive N ∈ Z+ sothat whenever n > N , d(x, xn) < ε. In this case we will write limxn = x.

Lemma 2 Let (X, d) be a metric space and {xn} be a sequence in X. Then limxn =x ∈ X if and only if for each neighborhood V of x there is an integer N > 0 so thatxn ∈ V whenever n > N .



This is nothing but applying the definitions of convergence and neighborhood, andits proof will be omitted.

If S is a set of infinite points and there is at most a finite number of elementsof S for which a certain statement is false, thent he statement is said to be true foralmost all of S. Thus, we may phrase the above lemma by saying that the sequence{xn} converges to x if each neighborhood of x contains almost all of the poins of thesequence.

One reason for looking a sequences is the concept of continuity. In calculus wedefine a function to be continuous at a ∈ R if the following conditions were met:

1. limx→a f(x) exists;

2. f(a) exists;

3. limx→a f(x) = f(a).

It suffices to check this for all sequences appproaching a (a fact to be proven later).Thus, we find that we can show that f is continous at a if for each sequence {xn} → a,we have that {f(xn)} → f(a).

We are able to extend this result to arbitrary metric spaces.

Theorem 11 Let f : (X, d)→ (Y, d′). f is continous at a point a ∈ X if and only ifwhenever limxn = x we have lim f(xn) = f(x).

The proof is straightforward.

Proof: Assume that f is continuous and let {xn} → x in X. Let ε > 0 and letM = Bd′(f(x); ε) ∈ V . There is a neighborhood U of x in X, such that f(U) ⊂ M .Since U is a neighborhood there is a δ > 0 so that Bd(x

′δ) ⊂ U . Now, {xn} → x so forthis δ there is a positive integer N so that whenever n > N we have xn ∈ Bd(x

′δ) ⊂ U .Thus, f(xn) ∈ f(U) ⊂ M . Therefore, for any neighborhood M of f(x) there is apositive integer N so that whenever n > N we have d′(f(x), f(xn) < ε, which impliesthat the sequence {f(xn)} converges to f(x).

To prove the other direction,

2.5 Open Sets and Closed Sets Revisited

Remember that we defined an open set as a set that is the union of open balls. A setis closed if its complement is open.

Theorem 12 The open subsets of a metric space (X, d) have the following properties:

1. X and ∅ are open sets.

2. The union of any family of open sets is open.

3. The intersection of a finite family of open sets is open.


2.5. OPEN SETS AND CLOSED SETS REVISITED 19

Proof: These are straightforward.

1. The whole space X is open since it is the union of all open balls with allpossible centers and radii. The empty set is open since it is the union of theempty collection of open balls.

2. If {Uα | α ∈ A} is a family of open sets in X, then each Uα is a union of openballs. Then

⋃α∈A Uα is the union of all of the open balls that comprise each Uα

and is hence open.

3. Let {Ui | i = 1, . . . , n} be a finite collection of open sets and let x ∈⋂ni=1 Ui.

Then, by our previous theorem there exist εi, i = 1, . . . , n so that Bd(x; εi) ⊂ Uiand

n⋂i=1

Bd(x; εi) ⊂n⋂i=1

Ui.

Let ε = min{εi | i = 1, . . . , n}. Then,⋂ni=1 Bd(x; εi) = Bd(x; ε). Thus, Bd(x; ε)

is an open ball centered at x and contained in⋂ni=1 Ui. Thus,

⋂ni=1 Ui is open.

Theorem 13 The closed subsets of a metric space (X, d) have the following proper-ties:

1. X and ∅ are closed sets.

2. The intersection of any family of closed sets is closed.

3. The union of a finite family of closed sets is closed.

This follows from our previous theorem and complements.

Definition 15 Let (X, d) be a metric space and A a subset of X. A point x ∈ X isa limit point or accumulation point of A provided that every open set containingx contains a point of A distinct from x. The set of limit points of A is called itsderived set, denoted by A′.

Lemma 3 Let (X, d) be a metric space and A a subset of X. A point x ∈ X is alimit point of A if and only if d(x,A \ {x}) = 0.

Lemma 4 A subset A of a metric space (X, d) is closed if and only if A contains allits limit points.



Proof: Let A be closed and let x be a limit point of A. If x 6∈ A then X \ A is anopen set containing x but containing no other point of A. Thus, x could not be alimit point of A. This means that if x is a limit point of A, then it must be a memberof A.

Now suppose that A contains all of its limit points. To show that A is closed, wemust show that X \ A is open. If x ∈ X \ A, then x is not a limit point of A. Thus,there is some open set Ux containing x but no other point of A. Then X \ A is theunion of all of these sets. Hence, X \ A is open and A is closed.

What is the connection between limit points and the limit of a sequence?

Theorem 14 Let (X, d) be a metric space and A a subset of X.

1. A point x ∈ X is a limit point of A if and only if there is a sequence of distinctpoints of A which converges to x.

2. The set A is closed if and only if each convergent sequence of points of A con-verges to a point of A.

Corollary 1 Let x be a limit point of a subset A of a metric space X. Then everyopen set containing x contains infinitely many members of A.

2.6 Interior, Closure, and Boundary

Definition 16 Let A be a subset of a metric space (X, d). A point x ∈ A is aninterior point of A if there is an open set U which contains x and is contained inA; x ∈ U ⊂ A. The interior of A, denoted intA, is the set of all interior points ofA.

Note that for the open set U in the definition, every point of U is an interior pointof A. Thus, the interior of A contains every open set contained in A and is the unionof this family of open sets. This means two things:

1. the interior of a set A is an open set, and

2. the interior of a set A is the largest open set contained in A.

Item (2) above means that if U is open and U ⊂ A, then U ⊂ intA.

Example 14



int(a, b) = int[a, b) = int(a, b] = int[a, b] = (a, b).


2.6. INTERIOR, CLOSURE, AND BOUNDARY 21

2. The interior of a finite set is empty, since such a set cannot contain any openinterval.

3. The interior of the set of irrational numbers is empty, since each open intervalmust contain some rational number. Likewise, the interior of the set of rationalsis empty. If the rationals contained an open interval, then the set of rationalswould have to be uncountable, since an open interval is uncountable.

4. int∅ = ∅; intR = R.

Definition 17 The closure A of a subset of a metric space (X, d) is the union ofthe set A and the set of its limit points:

A = A ∪ A′

where A′ is the derived set of A.

Example 15



(a, b) = [a, b) = (a, b] = [a, b] = [a, b].

2. The closure of a finite set is itself, since the set of limit points of a finite set isempty.

3. The closure of the set of rational numbers is R. Likewise, the closure of theset of irrationals is R. Since every open interval contains both rational andirrational numbers.

4. ∅ = ∅; R = R.

While the interior of a set is the largest open set contained in the set, the closurehas a similar property described in the next theorem.

Theorem 15 If A ⊂ X, then A is a closed set and is a subset of every closed setcontaining A.

This says that the closure of a set is the smallest closed set containing the set.

Proof: To show that A is closed, we need to show that it contains all of its limitpoints. Suppose that x 6∈ A. Then there is an open set U containing x so thatU ∩A = ∅. Now, this means that U cannot contain a limit point of A either, since ifan open set contains a limit point of A it must contain some other point of A also.Thus, U contains no point of A, so x is not a limit point of A. This means that all ofthe limit points of A must be contained in A. Thus, A is closed.

Suppose now that F is a closed subset of X and A ⊂ F . Then we can show thatA ⊂ F and, since F contains all of its limit points, then F = F ∪ F ′ = F . Thus,A ⊂ F fore every closed set F containing A.



Since this shows that A is the smallest closed set containing A, we can easily showthat A is the intersection of all closed sets containing A.

Theorem 16 Let A be a subset of the metric space (X, d).

1. A is open if and only if A = int A.

2. A is closed if and only if A = A.

Definition 18 Let A be a subset of the metric space (X, d). A point x ∈ X is aboundary point of A provided that x ∈ A∩X \ A. The set of boundary points of Ais called the boundary of A and is denoted by ∂A.

The industrious reader will readily work to show that the following statementsare equivalent for a subset A of X and a points x in the metric space (X, d).

1. x ∈ ∂A,

2. x ∈ (A \ int A),

3. Every open set containing x contains a point of A and a point of X \ A.

4. Every neighborhood of x contains a point of A and a point of X \ A.

5. d(x,A) = d(x,X \ A) = 0.

6. x ∈ A ∩X \ A.

Example 16 1. Let X = R with the usual metric. For a, b ∈ R with a < b

∂(a, b) = ∂[a, b) = ∂(a, b] = ∂[a, b] = {a, b}.

2. In Rn

∂B(a; ε) = {x ∈ Rn | d(a, x) = ε}.

3. The boundary of the set of all points in Rn having only rational coordinates isRn.

4. For any metric space (X, d),

∂∅ = ∂X = ∅.


Chapter 3

Topological Spaces


We want to generalize the concepts that we developed in studying the metric spaces.We want to remove our reliance on a distance function. We were able to define mostof what we wanted to do in metric spaces by defining our concepts in terms of theopen sets. This was especially true of our study of continuous functions.

We will use the results that we proved about open sets as our basis for the gener-alization. We will define open sets as sets that satisfy certain conditions.

Definition 19 Let X be a set and T a family of subsets of X satisfying the followingproperties.

a) The set X and ∅ belong to T ,

b) The union of any family of members of T is a member of T .

c) The intersection of any finite family of members of T is a member ofT .

Then T is called a topology for X and the members of T are called open sets.The ordered pair (X,T ) is called a topological space, or simply a space.

If we use the terminology open sets instead of member of T , then the definition ofa topological space may be restated as follows: A family of subsets of X is a topologyfor X means that:

a) Both X and ∅ are open sets.

b) The union of any family of open sets is an open set.

c) The intersection of any finite family of open sets is open.

23


Example 17 The usual topology for the real line R is the topology generated by itsusual metric. We shall refer to the real line with the usual topology as simply thereal line or R.

Example 18 The usual topology on Rn is the topology generated by the usual metricon Rn. It is also the topology generated by the taxicab metric and the max metric.Thus, the usual topology does not distinquish the metric determining it from theother two. We shall refer to Rn with the usual topology as Euclidean n-space, orsimply Rn.

Example 19 For any set X we take T = 2X to be the set of all subsets of X.This clearly satisfies all of the properties of a topology, since we have included everypossible subset in the topology. This is called the discrete topology. Note that it is thetopology generated by the discrete metric. Also, note that this is the largest possiblecollection of open subsets of X.

Example 20 At the opposite extreme, we may take T = {∅, X}. This is called thetrivial topology, or indiscrete topology, on X. This is the smallest collection of opensets on X.

Example 21 Let X be a set. We shall take T to consist of ∅, X, and all sets U sothat X \ U is a finite set. Then T is a topology on X called the cofinite topology, orfinite complement topology. This is really of interest only when X is an infinite set.When X is a finite set, this is the same as the discrete topology.

Definition 20 A subset F of a topological space X is closed if X \F is an open set.

Theorem 17 The closed sets of a topological space X have the following properties:

a) X and ∅ are closed.

b) The intersection of any family of closed sets is closed.

c) The union of any finite family of closed sets is closed.

Definition 21 Let (X,T ) be a topological space and let A ⊂ X. A point x in X isa limit point of A if every open set containing x contains a point of A distinct fromx. The set of limit points of A is called the derived set of A, denoted A′.


3.2. INTERIOR, CLOSURE AND BOUNDARY 25

Example 22 Let X = {a, b, c, d}. Let T0 be the indiscrete topology; T1, the discretetopology; T2 = {∅, X, {a}, {b}, {a, b}}; andT3 = {∅, X, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}. The reader should verify thatT2 and T3 are topologies on X. Let A = {a, b}, B = {c}, and C = {d}. We want tofind the limit points of these sets in the different topologies.

T0 T1 T2 T3

A X ∅ {c, d} {d}B {a, b, d} ∅ {d} {d}C {a, b, c} ∅ {c} ∅

Theorem 18 A subset A of a topological space X is closed if and only if A containsall of its limit points.

This is no surprise, and is proven exactly the way in which we proved it earlierin a metric space. We were careful there not to use the distance function, but to usethe open sets.

Definition 22 Let X be a topological space and let {xn} be a sequence of points in X.We say that {xn} converges to the point x ∈ X, or x is the limit of the sequence,if for each open set U containing x there is a positive integer N so that xn ∈ U forall n ≥ N .

Sequences are not as fundamental in general topological spaces as they are inmetric spaces. The following example may show why.

Example 23 Consider R with the cofinite topology. Let {xn} be any sequence ofreal numbers. Let a ∈ R be any real number. Then {xn} converges to a, because ifU is any open set containing a, then R \ U is a finite set. Since {xn} is an infiniteset, we must have that infinitely many members of {xn} lie in U . Thus, there is apositive integer N such that if n ≥ N xn ∈ U . Thus, {xn} converges to a. However,a was any arbitrary real number. This means that {xn} converges to every realnumber. What is more (and maybe worse) is that {xn} was an arbitrary sequence.This means that every sequence converges to every real number. There are no non-convergent sequences and sequences do not have unique limits.

3.2 Interior, Closure and Boundary

Just as before we will define the interior, closure, and the boundary.

Definition 23 Let A be a subset of the topological space X. A point x ∈ A is aninterior point of A if there is an open set Uso that x ∈ U ⊂ A. A is called aneighborhood of x. The interior of A, denoted by A◦, is the set of all interiorpoints of A.



The closure, A, of A is the union of A and it set of limit points:

A = A ∪ A′.

A point x ∈ X is a boundary point of A if x ∈ A∩X \ A. The set of boundarypoints of A is called the boundary of A and is denoted by ∂A.

Theorem 19 For any subsets A,B of a topological space X

a) The interior of A is the union of all open sets contained in A and is thelargest open set contained in A.

b) A is open if and only if A = A◦.

c) If A ⊂ B, then A◦ ⊂ B◦.

d) (A ∩B)◦ = A◦ ∩B◦.

Proof: We will offer only a proof for (d). The others follow from closely from whatwe did in the case of a metric space.

Since A ∩B is a subset of both A and B, then by (c) (A ∩B)◦ ⊂ A◦ ∩B◦. Now,A◦ ∩ B◦ is an open set and is a subset of A ∩ B. Thus by (a), A◦ ∩ B◦ ⊂ (A ∩ B)◦.This completes the proof.

Theorem 20 For any subsets A,B of a topological space X

a) The closure of A is the intersection of all closed sets containing in Aand is the smallest closed set containing in A.

b) A is closed if and only if A = A.

c) If A ⊂ B, then A ⊂ B.

d) (A ∪B) = A ∪B.

We leave this to the reader to prove.

Theorem 21 Let A be a subset of a topological space X.

a) ∂A = A ∩X \ A = ∂(X \ A).

b) ∂A, A◦, and (X \ A)◦ are pairwise disjoint sets whose union is X.

c) ∂A is a closed set.

d) A = A◦ ∪ ∂A.

e) A is open if and only if ∂A ⊂ (X \ A).


3.3. BASIS FOR A TOPOLOGY 27

f) A is closed if and only if ∂A ⊂ A.

g) A is open and closed if and only if ∂A = ∅.

Proof: Parts (a)–(d) follow immediately from the definitions.(e) If A is open then A = A◦. Now by (b) A◦ and ∂A are disjoint. Thus, A and

∂A are disjoint. This implies that ∂A ⊂ X \ A. Now, if ∂A ⊂ X \ A then no pointof A is a boundary point of A. Thus, every point of A is an interior point of A andA = A◦. Thus, A is open.

(f) This follows from our duality of open and closed sets.(g) If A is both open and closed, then ∂A ⊂ A ∩ (X \ A) = ∅. If ∂A = ∅ then

clearly ∂A ⊂ A — meaning A is closed — and ∂A ⊂ (X \A) — meaning A is open.

Definition 24 A set A in a topological space X is dense if A = X. If X has acountable dense set, then X is a separable space.

Example 24 1. The reals with the usual topology is separable, since the rationalsare dense.

2. Euclidean n-space is separable, since the set of points having only rationalcoordinates is dense and countable.

3. The reals with the cofinite topology is separable, since every countable infiniteset is dense.

Definition 25 A subset B of a space X is nowhere dense if (B)◦ = ∅.

Note that a finite subset of a metric space is nowhere dense. In other topologicalspaces, we will see more interesting examples.

3.3 Basis for a Topology

It appears that a topology can be relatively large. In fact, for an infinite set thediscrete topology consists of all subsets of the space, so it would be prohibitive tohave to check all subsets. We have seen though that we can get by with just checkingsome of the sets. For the discrete topology we have usually only checked the singletonsets. For a metric space we were able to do everything we wanted by working withthe open balls. In fact we defined all open sets in terms of the open balls. Can we dothis in general? Can we find a certain collection of subsets that will generate all of theelements of the topology, just like the open balls generate the metric topology? Theanswer is yes, because we can take T as this generating set. This begs the answer,because we are looking for a smaller collection than the whole topology.



Definition 26 Let (X,T ) be a topological space. A basis B for T is a subcollectionof T with the property that each member of T is a union of members of B. Themembers of B are called basic open sets and T is the topology generated by B.

Example 25 Most of what we have seen is based on metric spaces.

1. The collection of all open intervals is a basis for the usual topology on the reals.

2. The collection of all open balls is a basis for the metric topology on the metricspace (X, d).

3. For any set X the collection of all singleton sets {x} is a basis for the discretetopology.

Definition 27 Let (X,T ) be a topological space. A local basis at a ∈ X is asubcollection Ba of T such that

a) a belongs to each member of Ba, and

b) each open set containing a contains a member of Ba.

Definition 28 A space X is first countable if there is a countable local basis ateach point of X. The space X is second countable if the topology for X has acountable basis.

Note that every second countable space is first countable because if there is acountable basis B, then the number of these sets containing any given point a ∈ Xis at most countable.

Theorem 22 Every second countable space is separable.

Proof: Let X be a second countable space with a countable basis B. Let A be aset formed by choosing one element from each non-empty element of B. Each pointof X is a limit point of some point in A by the definition of a basis. Thus, A is densein X.

Theorem 23 a) Every metric space is first countable.

b) Every separable metric space is second countable.

The proof is left to the reader.We have been starting with a topology and asking if there is a basis for it. We

could be starting with a collection of open sets and asking if it forms a basis for atopology. Not every collection of open sets will work. When is a collection of opensets a basis for a topology on X?



Theorem 24 A family B of subsets of a set X is a basis for a topology on X if andonly if both of the following hold:

a) The union of members of B is X.

b) For each B1, B2 ∈ B and x ∈ B1 ∩B2, there is a member Bx of B suchthat x ∈ Bx ⊂ B1 ∩B2.

Example 26 [Sorgenfrey Line] Let B be the collection of all half-open intervals ofR of the form [a, b), a < b. Clearly, the union of all of these intervals is R. If wetake two of these sets and intersect them, we can find another of these sets in theintersection. Thus, these sets form a basis for a topology on the real line, called thehalf-open interval topology T ′′ for R. R with this topology is called the Sorgenfreyline. The Sorgenfrey line has the property that it is first countable and separable,but not second countable.


We were able to move away from the epsilon-delta definition of continuity of a functionin between two metric spaces by using open balls. We will use this as our startingpoint for general topological spaces.

Definition 29 A function f : (X,T )→ (Y,T ′) is continuous at a point a ∈ X iffor each open set V in Y containing f(a) there is an open set U ⊂ X containing aso that f(U) ⊂ V , or equivalently U ⊂ f−1(V ).

Now, we are more interested in the situation where the function is continous atevery point of X. This means that for every open set V in Y and every point a ∈ Xwith f(a) ∈ V , there is an open set Ua ⊂ X with a ∈ Ua and f(Ua) ⊂ V . Equivalently,we have a ∈ Ua ⊂ f−1(V ). This means that for each point in f−1(V ) we can find anopen set containing that point and contained in f−1(V ). Thus, f−1(V ) must be openin X for each open set V ⊂ Y . This leads us to a more general definition.

Definition 30 A function f : (X,T )→ (Y,T ′) is continuous if for each open setV in Y f−1(V ) is an open set in X.

Theorem 25 Let f : X → Y be a function on the topological spaces X and Y andlet a ∈ X. The following are equivalent.

a) f is continuous at a.

b) For each open set V ∈ Y containing f(a), there is an open set U in Xsuch that a ∈ U ⊂ f−1(V ).

c) For each neighborhood V of f(a), f−1(V ) is a neighborhood of a.



The proof is left to the reader.

Theorem 26 Let f : X → Y be a function of topological spaces. The following areequivalent.

(i) f is continuous.

(ii) For each closed subset C ⊂ Y , f−1(C) is closed in X.

(iii) For each subset A ⊂ X, f(A) ⊂ f(A).

(iv) There is a basis B for the topology of Y so that f−1(B) is open in X for eachbasic open set B ∈ B.

Proof: To show that (i) implies (ii) will require the duality between open and closedsets. If C ⊂ Y is closed, the Y \C is open in Y . Since f is continuous, f−1(Y \C) isopen in X. Hence X\(f−1(Y \ C)) is open. If x ∈ X\(f−1(Y \ C)) then f(x) 6∈ Y \C,or f(x) ∈ C. Thus, X\(f−1(Y \ C)) ⊂ f−1(C). The opposite inclusion is clear. Thus,f−1(C) = X \ (f−1(Y \ C)) is closed. A similar analysis shows that (ii) implies (i).

To show that (ii) implies (iii), let A ⊂ X. Then f(A) is a closed subset of Y .Hence, f−1(f(A)) is a closed subset of X. Now, A ⊂ f−1(f(A)) so A ⊂ f−1(f(A)).Thus, f(A) ⊂ f(A).

To show that (iii) implies (ii), let C be a closed subset of Y . Then,

f(f−1(C)) ⊂ ff−1(C) ⊂ C ⊂ C

so f−1(C) ⊂ f−1(C) making f−1(C) a closed set.For the last equivalence, (i) clearly implies (iv). We need to prove the opposite

implication. Let O be an open set in Y . Then by the definition of a basis, O = ∪α∈IBα

for some subcollection {Bα}α∈I of the basis B. Then

f−1(O) = f−1

(⋃α∈I

Bα

)=⋃α∈I

f−1(Bα).

Since each f−1(Bα) is open in X and the union of any family of open sets is open,then f−1(O) is open in X and f is continuous.

Theorem 27 If f : X → Y and g : Y → Z are continuous functions, then g◦f : X →Z is continuous.

Definition 31 A function f : X → Y is a homeomorphism if

a) f is one-to-one, (injective)

b) f is onto, (surjective)


3.5. SUBSPACES 31

c) f is continuous,

d) f−1 is continuous.

Topological spaces are topologically equivalent or homeomorphic if there home-omorphism from f from X onto Y .

The continuity of the function and its inverse are extremely important! A prop-erty P of topological spaces is a topological property or topological invariantprovided that if space X has property P , then so does every space which is homeo-morphic to X.

Theorem 28 Separability is a topological property.

Theorem 29 First countability and second countability are topological properties.

Definition 32 A topological space is metrizable provided that the topology on X isgenerated by a metric.

Theorem 30 Metrizability is a topological property.

Since R and (0, 1) are homeomorphic, the property of being a bounded metricspace is not a topological property. Likewise, distance is not a topological invariant.

3.5 Subspaces

Let (X,T ) be a topological space and let A be a subset of X. The relative topologyor subspace topology T ′ on A determined by T consists of all sets of the formO ∩ A for which O is an open set of T .

T ′ = {O ∩ A | O ∈ T }.

The members of T ′ are called relatively open sets in A, and (A,T ′) is called asubspace of (X,T ).

Note that this is actually a topology for A.

∅ = ∅ ∩ A A = X ∩ A,

so both ∅ and A are open in A. If {Uα} are open in A, then Uα = Oα ∩ A and⋃α

Uα =⋃α

(Oα ∩ A) =

(⋃a

lphaOα

)∩ A

is relatively open since the union of any family of open sets is open in X. For anyfinite family of open sets Ui = Oi ∩ A, we have

n⋂i=1

Ui =n⋂i=1

(Oi ∩ A) =

(n⋂i=1

Oi

)∩ A

is relatively open since the intersection of any finite family of open sets is open in X.



Lemma 5 Let (a,T ′) be a subspace of the topological space (X,T ). A subset F ofA is closed in the subspace topology on A if and only if F = C ∩ A for some closedsubset C of X.

Example 27 1. The closed interval [a, b] with a < b is a subspace of R with theusual topology. The open sets containing a are sets of the form [a, c) witha < c < b.

2. The subset of Rn+1 consisting of all (n + 1)-tuples (x1, x2, . . . , xn, xn+1) withxn+1 = 0 is homeomorphic to Rn

3. Let a < b < c < d. Let A = [a, b]∪ (c, d) be considered as a subspace of the realline. Then the subset [a, b] of A is both relatively open and relatively closed.It is clearly closed because [a, b] = [a, b] ∩ A and [a, b] is closed in the real line.It is open because for 0 < ε < c − b, [a, b] = (a − ε, b + ε) ∩ Y . Thus, we seethat since (c, d) is the complement of this set that is both relatively open andrelatively closed, then we see that (c, d) is both relatively open and relativelyclosed.

Definition 33 A property P to topological spaces is hereditary provided that if Xhas property P , then every subspace of X has this property.

Example 28 1. First countability and second countability are hereditary proper-ties. If X has a countable basis, then intersecting these basis elements with Awill give a countable basis for the subspace topology. First countable is similar.

2. Separability is not hereditary. Let A ⊂ R2 consist of the x-axis and the pointa = (0, 1). Define a topology T on A by taking the empty set and all subsetsof A that contain the singleton set {a}. Then (X,T ) is separable because thesingleton set {a} is dense. Every point except a is a limit point of {a}. However,the subspace topology on R (the x-axis) is the discrete topology, so (R,T ′) isnot separable.

3.6 Hausdorff Spaces

A topological space X is a Hausdorff space if for each pair of distinct pointsa, b ∈ X there exist disjoint open sets U and V such that a ∈ U and b ∈ V .

Example 29 1. Every metric space is Hausdorff. We proved this as a homeworkproblem, but it is simple. Let r = d(a, b) and then take two open balls of radiusr/2 centered at a and b respectively.

2. The real line with the co-finite topology is not Hausdorff. Likewise, the real linewith the co-countable topology is not Hausdorff.


3.6. HAUSDORFF SPACES 33

3. Take any set with more than one point and give it the indiscrete (trivial) topol-ogy. It is not Hausdorff.

4. The space in Example 2 is not Hausdorff, because you can never separate afrom any other point.

5. [The Zariski Topology] Let n be a positive integer and consider the family Pof all polynomials in n real variables x1, x2, . . . , xn. For p ∈P let Z(p) denoteits solution set in Rn:

Z(p) = {(x1, x2, . . . , xn) ∈ Rn | p(x1, x2, . . . , xn) = 0}.

Let B be the collection of sets that are complements of some Z(p) for somep ∈P. This forms a basis for a topology on Rn called the Zariski topology.

For the real line, n = 1, this is just the co-finite topology. This is because eachfinite set of real numbers is the solution set for some polynomial in one realvariable. If A = {a1, a2, . . . , an}, then

p(x) = (x− a1)(x− a2) . . . (x− an)

is a polynomial with A as its solution set. Likewise, the set of solutions of apolynomial in one real variable of dimension n is at most n.

For n > 1 this is not the co-finite topology. For example, the line y = a in R2

is the solution set to the polynomial in two variables

p(x, y) = y − a.

Note that this is not a finite set. However, each finite set can serve as thesolution set of a polynomial.

Now, Rn with the Zariski topology is not Hausdorff. Assume that P = (a1, a2, . . . , an)and Q = (b1, b2, . . . , bn) are two distinct points in Rn.

Theorem 31 1. The property of being a Hausdorff space is topological and hered-itary.

2. In a Hausdorff space a sequence {xn}∞n=1 cannot converge to more than onepoint.

Proof: We will prove 1 only and leave the other to the reader.Suppose that X is Hausdorff and f : X → Y is a homeomorphism. For a 6= b ∈ Y ,

we have that f−1(a) and f−1(b) are distinct points in X. Thus, there are disjointopen sets U, V ⊂ X so that f−1(a) ∈ U and f−1(b) ∈ V . Hence, f(U) and f(V ) aredisjoint open sets of Y containing a and b respectively.

To show that Hausdorff is hereditary, assume that X is Hausdorff and that A ⊂ X.Let a 6= b ∈ A. Then, a 6= b ∈ X and there are disjoint open sets U, V ⊂ X so thata ∈ U and b ∈ V . Then, U ∩ A and V ∩ A are disjoint relatively open subsets of Acontaining a and b respectively.


Chapter 4

Connectedness

4.1 Connected and Disconnected Spaces

It is easier to define disconnected than connected, and we will do so.

Definition 34 A topological space X is disconnected or separated if it is theunion of two disjoint, non-empty open sets. Such a pair A,B of subsets of X is calleda separation of X. A space is connected if it is not disconnected.

A subspace Y of X is connected provided that it is a connected space whenassigned the subspace topology.

Example 30 1. A discrete space with more than one point is disconnected.

2. Any set with the indiscrete topology is connected, since there do not exist twonon-empty open sets.

3. Let A be the set of non-zero real numbers with the subspace topology. Then Ais disconnected since (−∞, 0) and (0,∞) form a separation.

4. Let B = R2 \ R be the plane minus the real axis. B is disconnected.

5. X = [0, 1] ∪ [2, 3] is disconnected.

6. Let X be the set of real numbers with the addition of a point (0, 1), and let thetopology T for X consist of ∅, X, and all subsets of X which contain a. Now,X is connected because every open set contains the point a. On the other hand,as a subspace of (X,T ) the real line is assigned the discrete topology and isdisconnected.

These last few examples show that the property of being connected is not heredi-tary.

34

4.2. HOW TO TELL IF A SPACE IS CONNECTED? 35

Example 31 The real line R with the usual topology is connected. Suppose other-wise that

R = A ∪B

where A ∩B = ∅ are both open and non-empty. Since

A = R \B and B = R \ A

we see that A and B are also both closed. Consider two points a, b ∈ R with a ∈ Aand b ∈ B. We may assume that a < b.

Put A+ = A ∩ [a, b]. Then A+ is a closed and bounded subset of R. Thus itmust contain its least upper bound c. Now, c 6= b since A and B have no points incommon. Thus, c < b. Thus, A contains no points of (c, b], placing (c, b] ⊂ B. Thus,c ∈ B. Now, B is closed, so c ∈ B. This implies that c ∈ A ∩ B, contradicting theassumption that A and B are disjoint. Thus, R is connected.

4.2 How to Tell If a Space is Connected?

Definition 35 Non-empty subsets A and B are separated sets if A∩B = A∩B = ∅.

Theorem 32 The following are equivalent for a topological space X.

1. X is disconnected.

2. X is the union of two disjoint, non-empty closed sets.

3. X is the union of two separated sets.

4. There is a continuous function from X onto the discrete two-point space {0, 1}.

5. X has a proper subset A which is both open and closed.

6. X has a proper subset A such that

A ∩ (X \ A) = ∅.

Proof: First, we will show that (1) ⇔ (2). Assume that X is disconnected. Thenthere are two nonempty, disjoint open sets A and B, so that X = A∪B. This meansthat X \A = B and X \B = A. Since both A and B are open, we now have that bothA and B are closed, and the result follows. The proof of the opposite implication isanalogous.(1) ⇔ (3): Assume that X is disconnected. We have then that there are twononempty, disjoint open sets A and B, so that X = A ∪ B. We just proved thatA and B are closed. Thus, A = A and B = B. Hence,

A ∩B = A ∩B = ∅ = A ∩B = A ∩B.


36 CHAPTER 4. CONNECTEDNESS

Again, the opposite implication is analogous.(1)⇔ (4): Assume that X is disconnected. Define a function f : X → {0, 1} by

f(x) =

{0 if x ∈ A1 if x ∈ B.

f is clearly onto. Since we have placed the discrete topology on {0, 1}, we only needto check f−1({0}) and f−1({1}) to check that f is continuous. Since f−1({0}) = Ais open and f−1({1}) = B is open, we see that f is continous.

Assume that there is a continous, onto function f : X → {0, 1}. Since {0, 1} hasthe discrete topology, the sets {0} and {1} are open and disjoint. Let A = f−1({0})and B = f−1({1}). These then form a separation for X.(1) ⇔ (5): Assume that X is disconnected and that A and B form a separation ofX. Then A is a proper subset which is both open and closed. If U ⊂ X is a propersubset which is both open and closed, then V = X \ U is open and U and V form aseparation of X.(1) ⇔ (6): Assume that X is disconnected and A and B form a separation of X.Then X \ A = B, A = A, and (X \ A) = B = B = X \ A. Hence

A ∩ (X \ A) = ∅.

If such a set A exists, then A and X \ A form a separation of X.

Corollary 2 The following statements are equivalent.

1. X is connnected.

2. X is not the union of two disjoint, non-empty closed sets.

3. X is not the union of two separated sets.

4. There is not continuous function from X onto a discrete two point space {0, 1}.

5. The only subsets of X that are both open and closed are X and ∅.

6. X has no proper subset A so that

A ∩ (X \ A) = ∅.

Theorem 33 Let X be a connected topological space and let f : X → Y be a contin-uous surjective function. Then Y is a connected space.

Proof: We will prove that if Y is not connected, thenX is not connected, completingthe proof by proving the contrapositive. If Y is disconnected, let A and B form theseparation of Y . Then Y = A ∪B and the sets f−1(A) and f−1(B)



(a) are open sets since f is continous;

(b) are disjoint since f is a function;

(c) are non-empty since f is surjective; and

(d) have union X because

X = f−1(Y ) = f−1(A ∪B) = f−1(A) ∪ f−1(B).

Thus, if Y is disconnected, then X is disconnected.

Corollary 3 If f : X → Y is a continuous function, then f(X), the image of f , isconnected.

Lemma 6 A subspace Y of a space X is disconnected if and only if there are opensets U and V in X such that

U ∩ Y 6= ∅, V ∩ Y 6= ∅, U ∩ V ∩ Y = ∅, Y ⊂ U ∪ V.

Theorem 34 If Y is a connected subspace of X, then Y is connected.

Proof: Suppose that Y is connected. Consider a continuous function f : Y → {0, 1}.We must show that f is not surjective. We know that the restriction f |Y is notsurjective, so that f maps Y either to {0} or {1}. Assume that f(Y ) = {0}. Since fis continuous, we know that

f(Y ) ⊂ (f(Y )) = {0} = {0},

so f is not surjective. Thus, Y is connected.

Corollary 4 Let Y be a connected subspace of X and Z a subspace so that Y ⊂ Z ⊂Y . Then Z is connected.

Example 32 Each interval of the real line is connected. We know that each openinterval of the real line is connected, being homeomorphic to the real line. A non-degenerate closed interval is the closure of an open interval, and is hence connected.A degenerate closed interval is a single point, which is connected. Any other intervalis trapped between an open interval and its closure, so it is connected. The emptyset, which is an interval, is connected because there are no non-empty subsets.



Example 33 [The Topologist’s Sine Curve] Let A = {(0, y) | −1 ≤ y ≤ 1} andB = {(x, y) | 0 < x ≤ 1, y = sin(π/x)}. Put T = A ∪ B. T is called the topologist’ssine curve. Note that B is connected, since it is the continuous image of (0, 1]. Notealso that T = B, so that T is connected.

–1

0

1

We should never believe that the union of two connected sets is connected, sinceone of our first examples of a disconnected space was two disjoint intervals. Whatwould it take for a union of connected sets to be connected?

Theorem 35 Let X be a space and let {Aα | α ∈ I} be a family of connected subsetsof X for which ∩α∈IAα is not empty. Then ∪α∈IAα is connected.

Proof: We shall use Lemma 6 to prove that Y = ∪α∈IAα is connected. Supposethat U and V are open sets of X so that:

U ∩ Y 6= ∅, U ∩ V ∩ Y = ∅, Y ⊂ U ∪ V.

We need to then show that Y ∩ V = ∅. This will show that Y is connected. SinceU ∩ Y 6= ∅, U must contain some point of some Aβ, for some β ∈ I. Since Aβ isconnected, then Aβ ⊂ U . If b ∈ ∩α∈IAα, then b must be in Aβ, so b ∈ U . Thus, Ucontains a point b in each Aα, α ∈ I. Since Aα is connected, the Aα ⊂ U for eachα ∈ I. Thus,

Y =⋃α∈I

Aα ⊂ U

so V ∩ Y = ∅.

Various variants of this theorem are as follows:



Lemma 7 Let X be a space, {Aα | α ∈ I} a family of connected subsets of X, andB a connected subset of X such that for each α ∈ I, Aα∩B 6= ∅. Then B∪ (∪α∈IAα)is connected.

Lemma 8 Let {An} be a sequence of connected subsets of a space X such that foreach integer n ≥ 1, An has at least one point in common with one of the precedingsets A1, . . . , An−1. Then ∪∞n=1An is connected.

Definition 36 A connected component, or component, of a topological space isa connected subset C of X which is not a proper subset of any connected subset of X.

Theorem 36 For a topological space X

1. Each point x ∈ X belongs to exactly one component. The component Cx con-taining x is the union of all the connected subsets of X which contain x, and isthus the largest connected subset of X containing the point x.

2. For points x, y ∈ X the components Cx and Cy are either disjoint or identical.

3. Every connected subset of X is contained in a component.

4. Each component is a closed set.

5. X is connected if and only if it has one component.

6. If C is a component of X and A and B form a separation for X, then C is asubset of A or a subset of B.

Example 34 1. For the subspace X = (0, 1) ∪ (2, 3) of R, there are two compo-nents, (0, 1) and (2, 3). Both components are closed sets with respect to thesubspace topology.

2. In a discrete space, each component consists of only one point.

3. For the set Q of rational numbers with the subspace topology, each componentconsists of only one point. Note, however, that the subspace topology is not thediscrete topology.

Example 35 Let X be the subspace of R2 consisting of the following sequence:

X = {(x, y) | 0 ≤ x ≤ 1, y =1

n, n = 1, . . . ,∞}

⋃([0,

1

2

)∪(

1

2, 1

])Now, [0, 1

2) is the component of X containing 0 and (1

2, 1] is the component of X

containing 1. Thus, 0 and 1 belong to different components. However, for anyseparation of X into disjoint non-empty open sets A and B whose union is X, both0 and 1 belong to A or to B.



Definition 37 A space X is totally disconnected if each component of X consistsof a single point.

Thus, a discrete space is totally disconnected, as is the subspace of rational num-bers in the real line.

4.3 Applications of Connectedness

You have been using the properties of connectedness in Calculus without knowing it.The Intermediate Value Theorem as well as several fixed point theorems all dependon connectedness.

Theorem 37 (Intermediate Value Theorem) Let f : [a, b] → R be a continuousreal-valued function on a closed interval, [a, b]. Let y0 be a real number between f(a)and f(b). Then there is number c ∈ [a, b] for which f(c) = y0.

Proof: The interval [a, b] is connected, so f([a, b]) is connected, since f is continuous.Thus, f([a, b]) is an interval in R. Therefore, any number, y0, between f(a) and f(b)must be in the image f([a, b]). This means that y0 = f(c) for some c ∈ [a, b].

You have used the following corollary numerous times.

Corollary 5 Let f : [a, b] → R be continuous for which one of f(a) and f(b) ispositive and the other is negative. Then the equation f(x) = 0 has at least one rootbetween a and b.

Theorem 38 (Fixed Point Theorem) Let f : [a, b]→ [a, b] be a continuous func-tion on the closed interval [a, b]. Then there is a c ∈ [a, b] so that f(c) = c.

Proof: If f(a) = a or if f(b) = b then we are done, so assume that f(a) 6= a andf(b) 6= b. Thus, a < f(a) and f(b) < b since the range of f is [a, b]. Define a newfunction g : [a, b]→ R by

g(x) = f(x)− x, x ∈ [a, b].

Then g(a) = f(a) − a > 0 and f(b) = f(b) − b < 0. and g is continuous. Thus, bythe corollary, there is a c ∈ [a, b] so that g(c) = 0 or f(c) − c = 0. Thus, there is ac ∈ [a, b] so that f(c0 = c.

If f : X → X is a function, then we say that x is a fixed point of f if f(x) = x. Atopological space X has the fixed point property if every continuous function f : X →X has at least one fixed point. We can restate the above theorem as follows:

Theorem 39 Every closed and bounded interval has the fixed-point property.


4.4. PATH-CONNECTED SPACES 41

Theorem 40 The fixed-point property is a topological invariant.

Example 36 The real line does not have the fixed-point property. An example isf(x) = x = 1. Thus, no open interval has the fixed-point property. The intervals ofthe form [a, b), (a, b], (−∞, b], and [a,∞) also do not have the fixed point property.

The n-sphere, Sn, n ≥ 1, does not have the fixed-point property since the functiong(x) = −x has no fixed point.

4.4 Path-Connected Spaces

Our initial idea of connectedness is really best summed up in the definition of pathconnected. We usually think of a space as being connected if we can get from anyone point to another without leaving the space. This is a stronger property thanconnectedness, though.

We use I to denote the unit interval, [0, 1].

Definition 38 A path in a space X is a continuous function f : I → X. The pointsf(0) and f(1) are the endpoints of the path. The path f is called a path from f(0)to f(1). If f is a path in X, the reverse path f is the path defined by

f(t) = f(1− t), t ∈ I.

Definition 39 A space X is path connected if for any pair of points x, y ∈ X thereis a path in X with initial point x and terminal point y. A subspace A of X is pathconnected provided that A is path connected with its subspace topology.

Example 37 Every interval on the real line is path connected. For a, b ∈ K definethe path by

p(t) = a(1− t) + bt, t ∈ I.

Example 38 The generalizations of the interval to subsets of Rn are called convexsets. A set C of Rn is convex if for any two points, a, b ∈ C, the line segment joininga and b lies entirely in C. Since a line segment defines a path, each convex set is pathconnected.

Theorem 41 Every path connected set is connected.

Proof: Suppose that X is path connected and let a ∈ X. For each x ∈ X let Cxdenote the image of the path from a to x. Since each Cx is connected and a ∈ Cx forall x ∈ X, then by our previous theorem

X =⋃x∈X

Cx

is connected.



Example 39 The Topologist’s Sine Curve, T , is connected but it is not path connnected.

Thus, while in R connected and path connected are identical, in Rn there are setsthat are connected, but not path connected. All is not lost though.

Theorem 42 Every open, connected subset of Rn is path connected.

There are analogous results for path connectedness as for connectedness. Likewise,one defines the path component of a space X as a path connected subset of X whichis not a proper subset of any path connected subset of X.

4.5 Locally Connected and Locally Path Connected

Spaces

The terms connected and path connected are global properties, i.e., properties thatapply to the whole space. Local topological properties are characteristics of a space”near” a particular point.

Definition 40 A topological space X is locally connected at a point p ∈ X if everyopen set containing p also contains a connected set which contains p. The space X islocally connected if it is locally connected at each point.

Theorem 43 Let X be a topological space.

a) X is locally connected at a point p ∈ X if and only if there is a localbasis at p consisting of connected open sets.

b) X is locally connected if and only if it has a basis of connected open sets.

Example 40 1. Any interval in R is connected and locally connected.

2. Rn is connected and locally connected for each integer n ≥ 0.

3. The subspace [0, 1] ∪ [2, 3] is locally connected, but not connected.

4. Let X = {(x, 0) | 0 ≤ x ≤ 1}, Y = {(0, y) | 0 ≤ y ≤ 1}, andZ =

{(1n, y)∣∣n ∈ Z+, 0 ≤ y ≤ 1

}. Let C = X ∪ Y ∪ Z. C is called the topolo-

gist’s comb.

The topologist’s comb is obviously connected, being path connected. A pathcan run up and down the vertical tines and across the base. However, C is notlocally connected at any point (0, t), 0 < t ≤ 1, since small open sets containingsuch points consist of collections of open vertical intervals.

5. The set Q of rational numbers is neither connected nor locally connected.


4.5. LOCALLY CONNECTED AND LOCALLY PATH CONNECTED SPACES43

Thus, from these examples, we see that one property does not imply the other.This is commonly true with global and local properties, but even this statement isnot infallible. Recall that second countability does imply first countability.

Theorem 44 A space X is locally connected if and only if for each open subsetO ⊂ X, each component of O is an open set.

Definition 41 A space X is locally path connected at a point p ∈ X if every openset containing p contains a path connected set containing p. The space X is locallypath connected if it is locally path connected at each point.

Theorem 45 Let X be a topological space.

a) X is locally path connected at a point p ∈ X if and only if there is alocal basis at p consisting of path connected open sets.

b) X is locally path connected if and only if it has a basis of path connectedopen sets.

Theorem 46 A space X is locally path connected if and only if for each open subsetO ⊂ X, each path component of O is an open set.

Theorem 47 If X is a connected, locally path connected space, then X is path con-nected.

Proof: For each point x ∈ X, let Px denote the path component of X to which xbelongs. Since X is an open set, Theorem 46 shows that each Px is open. Recall thatpath components are either disjoint or identical.

For a particular point a ∈ X, suppose that Pa 6= X. Then Pa and the union ofall Px for which x 6∈ Pa are disjoint, non-empty open subsets of X whose union isX. This would imply that X is disconnected. Since X is connected, we have thatPa = X and X must be path connected as well.


Chapter 5

Compactness

Compactness is the generalization to topological spaces of the property of closed andbounded subsets of the real line: the Heine-Borel Property. While compact may infer”small” size, this is not true in general. We will show that [0, 1] is compact while(0, 1) is not compact.

Compactness was introduced into topology with the intention of generalizing theproperties of the closed and bounded subsets of Rn.

5.1 Compact Spaces and Subspaces

Definition 42 Let A be a subset of the topological space X. An open cover for Ais a collection O of open sets whose union contains A. A subcover derived from theopen cover O is a subcollection O ′ of O whose union contains A.

Example 41 Let A = [0, 5] and consider the open cover

O = {(n− 1, n+ 1) | n = −∞, . . . ,∞}.

Consider the subcover P = {(−1, 1), (0, 2), (1, 3), (2, 4), (3, 5), (4, 6)} is a subcover ofA, and happens to be the smallest subcover of O that covers A.

Definition 43 A topological space X is compact provided that every open cover ofX has a finite subcover.

This says that however we write X as a union of open sets, there is always a finitesubcollection {Oi}ni=1 of these sets whose union is X. A subspace A of X is compactif A is a compact space in its subspace topology. Since relatively open sets in thesubspace topology are the intersections of open sets in X with the subspace A, thedefinition of compactness for subspaces can be restated as follows.Alternate Definition: A subspace A of X is compact if and only if every opencover of A by open sets in X has a finite subcover.

44

5.1. COMPACT SPACES AND SUBSPACES 45

Example 42 1. Any space consisting of a finite number of points is compact.

2. The real line R with the finite complement topology is compact.

3. An infinite set X with the discrete topology is not compact.

4. The open interval (0, 1) is not compact. O = {(1/n, 1) | n = 2, . . . ,∞} is anopen cover of (0, 1). However, no finite subcollection of these sets will cover(0, 1).

5. Rn is not compact for any positive integer n, since O = {B(0, n) | n =1, . . . ,∞} is an open cover with no finite subcover.

A sequence of sets {Sn}∞n−1 is nested if Sn+1 ⊂ Sn for each positive integer n.

Theorem 48 (Cantor’s Nested Intervals Theorem) If {[an, bn]}∞n=1 is a nestedsequence of closed and bounded intervals, then ∩∞n=1[an, bn] 6= ∅. If, in addition, thediameters of the intervals converge to zero, then the intersection consists of preciselyone point.

Proof: Since [an+1, bn+1] ⊂ [an, bn] for each n ∈ Z+, the sequences {an} and {bn} ofleft and right endpoints have the following properties:

(i) a1 ≤ a2 ≤ · · · ≤ an ≤ . . . and {an} is an increasing sequence;

(ii) b1 ≥ b2 ≥ · · · ≥ bn ≥ . . . and {bn} is a decreasing sequence;

(iii) each left endpoint is less than or equal to each right endpoint.

Let c denote the least upper bound of the left endpoints and d the greatest lowerbound of the right endpoints. The existence of c and d are guaranteed by the LeastUpper Bound Property. Now, by property (iii), c ≤ bn for all n, so c ≤ d. Sincean ≤ c ≤ d ≤ bn, then [c, d] ⊂ [an, bn] for all n. Thus, ∩∞n=1[an, bn] contains the closedinterval [c, d] and is thus non-empty.

If the diameters of [an, bn] go to zero, then we must have that c = d and c is theone point of the intersection.

Theorem 49 The interval [0, 1] is compact.

Proof: Let O be an open cover. Assume that [0, 1] is not compact. Then either[0, 1

2] or [1

2, 1] is not covered by a finite number of members of O. Let [a1, b1] be the

half that is not covered by a finite number of members of O.Apply the same reasoning to the interval [a1, b1]. One of the halves, which we

will call [a2, b2], is not finitely coverable by O and has length 14. We can continue

this reasoning inductively to create a nested sequence of closed intervals {[an, bn]}∞n=1,none of which is finitely coverable by O. Also, by construction, we have that

bn − an =1

2n,


46 CHAPTER 5. COMPACTNESS

so the diameters of these intervals goes to zero.By the Cantor Nested Intervals Theorem, we know that there is precisely one

point in the intersection of all of these intervals; p ∈ [an, bn], for all n. Since p ∈ [0, 1]there is an open interval O ∈ O with p ∈ O. Thus, there is a positive number, ε > 0so that (p− ε, p+ ε) ⊂ O. Let N be a positive integer so that 1/2N < ε. Then sincep ∈ [aN , bn] it follows that

[an, bn] ⊂ (p− ε, p+ ε) ⊂ O.

This contradicts the fact that [aN , bN ] is not finitely coverable by O since we justcovered it with one set from O. This contradiction shows that [0, 1] is finitely coverableby O and is compact.

Compactness is defined in terms of open sets. The duality between open andclosed sets and if Cα = X \Oα,

X \

(⋂α∈I

Cα

)=⋃α∈I

Oα

leads us to believe that there is a characterization of compactness with closed sets.

Definition 44 A family A of subsets of a space X has the finite intersectionproperty provided that every finite subcollection of A has non-empty intersection.

Theorem 50 A space X is compact if and only if every family of closed sets in Xwith the finite intersection property has non-empty intersection.

This says that if F is a family of closed sets with the finite intersection property,

then we must have that⋂F

Cα 6= ∅.

Proof: Assume that X is compact and let F = {Cα | α ∈ I} be a family of closedsets with the finite intersection property. We want to show that the intersectionof all members of F is non-empty. Assume that the intersection is empty. LetO = {Oα = X \ Cα | α ∈ I}. O is a collection of open sets in X. Then,⋃

α∈I

Oα =⋃α∈I

X \ Cα = X \⋂α∈I

Cα = X \ ∅ = X.

Thus, O is an open cover for X. Since X is compact, it must have a finite subcover;i.e.,

X =n⋃i=1

Oαi =n⋃i=1

(X \ Cαi) = X \n⋂i=1

Cαi .

This means that ∩ni=1Cαi must be empty, contradicting the fact that F has thefinite intersection property. Thus, if F has the finite intersection property, then theintersection of all members of F must be non-empty.

The opposite implication is left as an exercise.



Is compactness hereditary? No, because (0, 1) is not a compact subset of [0, 1]. Itis closed hereditary.

Theorem 51 Each closed subset of a compact space is compact.

Proof: Let A be a closed subset of the compact space X and let O be an open coverof A by open sets in X. Since A is closed, then X \ A is open and

O∗ = O ∪ {X \ A}

is an open cover of X. Since X is compact, it has a finite subcover, containing onlyfinitely many members O1, . . . , On of O and may contain X \ A. Since

X = (X \ A) ∪n⋃i=1

Oi,

it follows that

A ⊂n⋃i=1

Oi

and A has a finite subcover.

Is the opposite implication true? Is every compact subset of a space closed? Notnecessarily. The following though is true.

Theorem 52 Each compact subset of a Hausdorff space is closed.

Proof: Let A be a compact subset of the Hausdorff space X. To show that A isclosed, we will show that its complement is open. Let x ∈ X \A. Then for each y ∈ Athere are disjoint sets Uy and Vy with x ∈ Vy and y ∈ Uy. The collection of open sets{Uy | y ∈ A} forms an open cover of A. Since A is compact, this open cover has afinite subcover, {Uyi | i = 1, . . . , n}. Let

U =n⋃i=1

Uyi V =n⋂i=1

Vyi .

Since each Uyi and Vyi are disjoint, we have U and V are disjoint. Also, A ⊂ U andx ∈ V . Thus, for each point x ∈ X \ A we have found an open set, V , containing xwhich is disjoint from A. Thus, X \ A is open, and A is closed.

Corollary 6 Let X be a compact Hausdorff space. A subset A of X is compact ifand only if it is closed.

The following results are left to the reader to prove.

Theorem 53 If A and B are disjoint compact subsets of a Hausdorff space X, thenthere exist disjoint open sets U and V in X such that A ⊂ U and B ⊂ V .

Corollary 7 If A and B are disjoint closed subsets of a compact Hausdorff space X,then there exist disjoint open sets U and V in X such that A ⊂ U and B ⊂ V .



5.2 Compactness and Continuity

Theorem 54 Let X be a compact space and f : X → Y a continuous function fromX onto Y . Then Y is compact.

Proof: We will outline this proof. Start with an open cover for Y . Use the continuityof f to pull it back to an open cover of X. Use compactness to extract a finite subcoverfor X, and then use the fact that f is onto to reconstruct a finite subcover for Y .

Corollary 8 Let X be a compact space and f : X → Y a continuous function. Theimage f(X) of X in Y is a compact subspace of Y .

Corollary 9 Compactness is a topological invariant.

Theorem 55 Let X be a compact space, Y a Hausdorff space, and f : X → Y acontinuous one-to-one function. Then f is a homeomorphism.

5.3 Locally Compact and One-Point Compactifi-

cations

Is it always possible to consider a topological space as a subspace of a compact topolog-ical space? We can consider the real line as an open interval (they are homeomorphic).Can we always do something of this sort?

Definition 45 A space X is locally compact at a point x ∈ X provided that thereis an open set U containing x for which U is compact. A space is locally compactif it is locally compact at each point.

Note that every compact space is locally compact, since the whole space X satisfiesthe necessary condition. Also, note that locally compact is a topological property.However, locally compact does not imply compact, because the real line is locallycompact, but not compact.

Definition 46 Let X be a topological space and let ∞ denote an ideal point, calledthe point at infinity, not included in X. Let X∞ = X ∪∞ and define a topologyT∞ on X∞ by specifying the following open sets:

(a) the open sets of X, considered as subsets of X∞;

(b) the subsets of X∞ whose complements are closed, compact subsets of X; and

(c) the set X∞.

The space (X∞,T∞ is called the one point compactification of X.


5.3. LOCALLY COMPACT AND ONE-POINT COMPACTIFICATIONS 49

Theorem 56 Let X be a topological space and X∞ its one-point compactification.Then

a) X∞ is compact.

b) (X,T ) is a subspace of (X∞,T∞).

c) X∞ is Hausdorff if and only if X is Hausdorff and locally compact.

d) X is a dense subset of X∞ if and only if X is not compact.

Proof:

a) Any open cover O of X∞ must have a member U containing∞. Since thecomplement X∞ \ U is compact, it has a finite subcover {Oi}ni=1 derivedfrom O. Thus, U,O1, . . . , On is a finite subcover of X∞.

b) The fact that (X,T ) is the subspace topology in (X∞,T∞) basicallyfollows from the definition of the extended topology. It also requires thatwe look at what open sets containing the point at infinity look like. Onesuch set is U = X∞ itself and U ∩X = X is open in X. The second typeis a subset of X∞ so that X \ U is closed and compact in X. In this caseU ∩X is open since its complement is closed.

c) Suppose that X∞ is Hausdorff. Then X is Hausdorff since the propertyis hereditary. Now, let p ∈ X. Since X∞ is Hausdorff, there are open,disjoint sets U and V in X∞ so that∞ ∈ U and p ∈ V . Thus, V ⊂ X∞\Uand this latter set is closed and compact in X. Hence V ⊂ X∞ \ U , so Vis compact, since it is a closed subset of a compact set. Thus, X is locallycompact at p.

Now, suppose that X is Hausdorff and locally compact. To show that X∞is Hausdorff, we only need to be able to separate ∞ from any point inp ∈ X. Since X is locally compact, there is an open set O so that p ∈ Oand O is compact. Then O and X∞ \O are two disjoint open sets in X∞containing p and ∞ respectively.

d) If X is compact, then {∞} is an open set in X∞, since {∞} = X∞ \X.Thus, ∞ is not a limit point of X, and X 6= X∞. Hence, X is not dense.

If X is not dense in X∞, then X = X, since ∞ 6∈ X. Hence, {∞} is openin X∞. Thus, X is compact.

Example 43 What is the one-point compactification of the open interval (0, 1)? Youcan define a function f : (0, 1)∞ → S1 by

f(t) =

{(cos(2πt), sin(2πt)) if 0 < t < 1

(1, 0) if t =∞



This f is a one-to-one continuous function from (0, 1)∞ onto the unit circle. ByTheorem 55, this is a homeomorphism.


Chapter 5

Compactness

Compactness is the generalization to topological spaces of the property of closed andbounded subsets of the real line: the Heine-Borel Property. While compact may infer”small” size, this is not true in general. We will show that [0, 1] is compact while(0, 1) is not compact.

Compactness was introduced into topology with the intention of generalizing theproperties of the closed and bounded subsets of Rn.

5.1 Compact Spaces and Subspaces

Definition 5.1 Let A be a subset of the topological space X. An open cover for Ais a collection O of open sets whose union contains A. A subcover derived from theopen cover O is a subcollection O ′ of O whose union contains A.

Example 5.1.1 Let A = [0, 5] and consider the open cover

O = {(n− 1, n+ 1) | n = −∞, . . . ,∞}.

Consider the subcover P = {(−1, 1), (0, 2), (1, 3), (2, 4), (3, 5), (4, 6)} is a subcover ofA, and happens to be the smallest subcover of O that covers A.

Definition 5.2 A topological space X is compact provided that every open cover ofX has a finite subcover.

This says that however we write X as a union of open sets, there is always a finitesubcollection {Oi}ni=1 of these sets whose union is X. A subspace A of X is compactif A is a compact space in its subspace topology. Since relatively open sets in thesubspace topology are the intersections of open sets in X with the subspace A, thedefinition of compactness for subspaces can be restated as follows.Alternate Definition: A subspace A of X is compact if and only if every opencover of A by open sets in X has a finite subcover.

43


Example 5.1.2 1. Any space consisting of a finite number of points is compact.

2. The real line R with the finite complement topology is compact.

3. An infinite set X with the discrete topology is not compact.

4. The open interval (0, 1) is not compact. O = {(1/n, 1) | n = 2, . . . ,∞} is anopen cover of (0, 1). However, no finite subcollection of these sets will cover(0, 1).

5. Rn is not compact for any positive integer n, since O = {B(0, n) | n =1, . . . ,∞} is an open cover with no finite subcover.

A sequence of sets {Sn}∞n−1 is nested if Sn+1 ⊂ Sn for each positive integer n.

Theorem 5.1 (Cantor’s Nested Intervals Theorem) If {[an, bn]}∞n=1 is a nestedsequence of closed and bounded intervals, then ∩∞n=1[an, bn] 6= ∅. If, in addition, thediameters of the intervals converge to zero, then the intersection consists of preciselyone point.

Proof: Since [an+1, bn+1] ⊂ [an, bn] for each n ∈ Z+, the sequences {an} and {bn} ofleft and right endpoints have the following properties:

(i) a1 ≤ a2 ≤ · · · ≤ an ≤ . . . and {an} is an increasing sequence;

(ii) b1 ≥ b2 ≥ · · · ≥ bn ≥ . . . and {bn} is a decreasing sequence;

(iii) each left endpoint is less than or equal to each right endpoint.

Let c denote the least upper bound of the left endpoints and d the greatest lowerbound of the right endpoints. The existence of c and d are guaranteed by the LeastUpper Bound Property. Now, by property (iii), c ≤ bn for all n, so c ≤ d. Sincean ≤ c ≤ d ≤ bn, then [c, d] ⊂ [an, bn] for all n. Thus, ∩∞n=1[an, bn] contains the closedinterval [c, d] and is thus non-empty.

If the diameters of [an, bn] go to zero, then we must have that c = d and c is theone point of the intersection.

Theorem 5.2 The interval [0, 1] is compact.

Proof: Let O be an open cover. Assume that [0, 1] is not compact. Then either[0, 1

2] or [1

2, 1] is not covered by a finite number of members of O. Let [a1, b1] be the

half that is not covered by a finite number of members of O.Apply the same reasoning to the interval [a1, b1]. One of the halves, which we

will call [a2, b2], is not finitely coverable by O and has length 14. We can continue

this reasoning inductively to create a nested sequence of closed intervals {[an, bn]}∞n=1,none of which is finitely coverable by O. Also, by construction, we have that

bn − an =1

2n,



so the diameters of these intervals goes to zero.By the Cantor Nested Intervals Theorem, we know that there is precisely one

point in the intersection of all of these intervals; p ∈ [an, bn], for all n. Since p ∈ [0, 1]there is an open interval O ∈ O with p ∈ O. Thus, there is a positive number, ε > 0so that (p− ε, p+ ε) ⊂ O. Let N be a positive integer so that 1/2N < ε. Then sincep ∈ [aN , bn] it follows that

[an, bn] ⊂ (p− ε, p+ ε) ⊂ O.

This contradicts the fact that [aN , bN ] is not finitely coverable by O since we justcovered it with one set from O. This contradiction shows that [0, 1] is finitely coverableby O and is compact.

Compactness is defined in terms of open sets. The duality between open andclosed sets and if Cα = X \Oα,

X \

(⋂α∈I

Cα

)=⋃α∈I

Oα

leads us to believe that there is a characterization of compactness with closed sets.

Definition 5.3 A family A of subsets of a space X has the finite intersectionproperty provided that every finite subcollection of A has non-empty intersection.

Theorem 5.3 A space X is compact if and only if every family of closed sets in Xwith the finite intersection property has non-empty intersection.

This says that if F is a family of closed sets with the finite intersection property,

then we must have that⋂F

Cα 6= ∅.

Proof: Assume that X is compact and let F = {Cα | α ∈ I} be a family of closedsets with the finite intersection property. We want to show that the intersectionof all members of F is non-empty. Assume that the intersection is empty. LetO = {Oα = X \ Cα | α ∈ I}. O is a collection of open sets in X. Then,⋃

α∈I

Oα =⋃α∈I

X \ Cα = X \⋂α∈I

Cα = X \ ∅ = X.

Thus, O is an open cover for X. Since X is compact, it must have a finite subcover;i.e.,

X =n⋃i=1

Oαi =n⋃i=1

(X \ Cαi) = X \n⋂i=1

Cαi .

This means that ∩ni=1Cαi must be empty, contradicting the fact that F has thefinite intersection property. Thus, if F has the finite intersection property, then theintersection of all members of F must be non-empty.

The opposite implication is left as an exercise.



Is compactness hereditary? No, because (0, 1) is not a compact subset of [0, 1]. Itis closed hereditary.

Theorem 5.4 Each closed subset of a compact space is compact.

Proof: Let A be a closed subset of the compact space X and let O be an open coverof A by open sets in X. Since A is closed, then X \ A is open and

O∗ = O ∪ {X \ A}

is an open cover of X. Since X is compact, it has a finite subcover, containing onlyfinitely many members O1, . . . , On of O and may contain X \ A. Since

X = (X \ A) ∪n⋃i=1

Oi,

it follows that

A ⊂n⋃i=1

Oi

and A has a finite subcover.

Is the opposite implication true? Is every compact subset of a space closed? Notnecessarily. The following though is true.

Theorem 5.5 Each compact subset of a Hausdorff space is closed.

Proof: Let A be a compact subset of the Hausdorff space X. To show that A isclosed, we will show that its complement is open. Let x ∈ X \A. Then for each y ∈ Athere are disjoint sets Uy and Vy with x ∈ Vy and y ∈ Uy. The collection of open sets{Uy | y ∈ A} forms an open cover of A. Since A is compact, this open cover has afinite subcover, {Uyi | i = 1, . . . , n}. Let

U =n⋃i=1

Uyi V =n⋂i=1

Vyi .

Since each Uyi and Vyi are disjoint, we have U and V are disjoint. Also, A ⊂ U andx ∈ V . Thus, for each point x ∈ X \ A we have found an open set, V , containing xwhich is disjoint from A. Thus, X \ A is open, and A is closed.

Corollary 6 Let X be a compact Hausdorff space. A subset A of X is compact ifand only if it is closed.

The following results are left to the reader to prove.

Theorem 5.6 If A and B are disjoint compact subsets of a Hausdorff space X, thenthere exist disjoint open sets U and V in X such that A ⊂ U and B ⊂ V .

Corollary 7 If A and B are disjoint closed subsets of a compact Hausdorff space X,then there exist disjoint open sets U and V in X such that A ⊂ U and B ⊂ V .


5.2. COMPACTNESS AND CONTINUITY 47

5.2 Compactness and Continuity

Theorem 5.7 Let X be a compact space and f : X → Y a continuous function fromX onto Y . Then Y is compact.

Proof: We will outline this proof. Start with an open cover for Y . Use the continuityof f to pull it back to an open cover of X. Use compactness to extract a finite subcoverfor X, and then use the fact that f is onto to reconstruct a finite subcover for Y .

Corollary 8 Let X be a compact space and f : X → Y a continuous function. Theimage f(X) of X in Y is a compact subspace of Y .

Corollary 9 Compactness is a topological invariant.

Theorem 5.8 Let X be a compact space, Y a Hausdorff space, and f : X → Y acontinuous one-to-one function. Then f is a homeomorphism.

5.3 Locally Compact and One-Point Compactifi-

cations

Is it always possible to consider a topological space as a subspace of a compact topolog-ical space? We can consider the real line as an open interval (they are homeomorphic).Can we always do something of this sort?

Definition 5.4 A space X is locally compact at a point x ∈ X provided thatthere is an open set U containing x for which U is compact. A space is locallycompact if it is locally compact at each point.

Note that every compact space is locally compact, since the whole space X satisfiesthe necessary condition. Also, note that locally compact is a topological property.However, locally compact does not imply compact, because the real line is locallycompact, but not compact.

Definition 5.5 Let X be a topological space and let ∞ denote an ideal point, calledthe point at infinity, not included in X. Let X∞ = X ∪∞ and define a topologyT∞ on X∞ by specifying the following open sets:

(a) the open sets of X, considered as subsets of X∞;

(b) the subsets of X∞ whose complements are closed, compact subsets of X; and

(c) the set X∞.

The space (X∞,T∞ is called the one point compactification of X.



Theorem 5.9 Let X be a topological space and X∞ its one-point compactification.Then

a) X∞ is compact.

b) (X,T ) is a subspace of (X∞,T∞).

c) X∞ is Hausdorff if and only if X is Hausdorff and locally compact.

d) X is a dense subset of X∞ if and only if X is not compact.

Proof:

a) Any open cover O of X∞ must have a member U containing∞. Since thecomplement X∞ \ U is compact, it has a finite subcover {Oi}ni=1 derivedfrom O. Thus, U,O1, . . . , On is a finite subcover of X∞.

b) The fact that (X,T ) is the subspace topology in (X∞,T∞) basicallyfollows from the definition of the extended topology. It also requires thatwe look at what open sets containing the point at infinity look like. Onesuch set is U = X∞ itself and U ∩X = X is open in X. The second typeis a subset of X∞ so that X \ U is closed and compact in X. In this caseU ∩X is open since its complement is closed.

c) Suppose that X∞ is Hausdorff. Then X is Hausdorff since the propertyis hereditary. Now, let p ∈ X. Since X∞ is Hausdorff, there are open,disjoint sets U and V in X∞ so that∞ ∈ U and p ∈ V . Thus, V ⊂ X∞\Uand this latter set is closed and compact in X. Hence V ⊂ X∞ \ U , so Vis compact, since it is a closed subset of a compact set. Thus, X is locallycompact at p.

Now, suppose that X is Hausdorff and locally compact. To show that X∞is Hausdorff, we only need to be able to separate ∞ from any point inp ∈ X. Since X is locally compact, there is an open set O so that p ∈ Oand O is compact. Then O and X∞ \O are two disjoint open sets in X∞containing p and ∞ respectively.

d) If X is compact, then {∞} is an open set in X∞, since {∞} = X∞ \X.Thus, ∞ is not a limit point of X, and X 6= X∞. Hence, X is not dense.

If X is not dense in X∞, then X = X, since ∞ 6∈ X. Hence, {∞} is openin X∞. Thus, X is compact.

Example 5.3.1 What is the one-point compactification of the open interval (0, 1)?You can define a function f : (0, 1)∞ → S1 by

f(t) =

{(cos(2πt), sin(2πt)) if 0 < t < 1

(1, 0) if t =∞


5.3. LOCALLY COMPACT AND ONE-POINT COMPACTIFICATIONS 49

This f is a one-to-one continuous function from (0, 1)∞ onto the unit circle. ByTheorem 5.8, this is a homeomorphism.


Chapter 6

Product and Quotient Spaces

When we study topology, we do not differ from other areas of mathematics that much.We look at the different mathematical operations that are available to us and howthey affect the mathematical structure we are studying. We also want to see how ourmathematical structure affects these operations.

The standard operations are addition, subtraction, multiplication and division. Intopology addition corresponds to the union of spaces, A ∪ B, and subtraction corre-sponds to the set difference A \B. Note that addition is commutative, but differenceis not. How do multiplication and quotients carry to the area of topology. Actually,multiplication is quite simple to see, but quotients are not as easily introduced.

6.1 Finite Products

Let X and Y be sets. The Cartesian product of X and Y is denoted by X ×Y and isdefined to be the set of all ordered pairs of points whose first coordinate comes fromX and whose second coordinate comes from Y :

X × Y = {(a, b) | a ∈ X and b ∈ Y }.

Now, if X and Y are topological spaces, then we would like to see what naturaltopology we should have on the product space, X × Y . The topology on X × Yshould be built from the topologies on X and Y , but how? If U is open in X andV is open in Y , then we would want U × V to be open in X × Y . However, thecollection of all sets of this form is not closed under unions, and hence cannot bea topology. It is not difficult to see, in the plane, that the union of two of theserectangles (0, 1)× (0, 1)∪ (2, 3)× (2, 3) is not a set of the form U ×V , because U ×Vwould have four pieces, not two.

Let B = {Uα × Vβ | Uα is open in X and Vβ is open in Y }. We have just seenthat B is not a topology. It is, however, a basis for a topology on X × Y . We needto show that it satisfies the conditions of Theorem 4.8. We must show that the unionof the members of B is X × Y and that for each B1, B2 ∈ B and x ∈ B1 ∩B2, thereis a member Bx of B such that x ∈ Bx ⊂ B1 ∩B2.

51

52 CHAPTER 6. PRODUCT AND QUOTIENT SPACES

It should be clear that the union of the members of B is all of X × Y . For thesecond condition, let B1 = U1 × V1 and B2 = U2 × V2 where Ui is open in X and V1

is open in Y . Now,

B1 ∩B2 = (U1 × V1) ∩ (U2 × V2) = (U1 ∩ U2)× (V1 ∩ V2),

so that B1 × B2 ∈ B. This is stronger than what is required in Theorem ??. Thus,B is a basis for a topology on X × Y . The topology determined by B is called theproduct topology on X × Y .

We can generalize this definition:

Definition 47 Let (X1,T1), (X2,T2), dots, (Xn,Tn) be a finite collection of nonemptytopological spaces, and let X denote the Cartesian product

X =n∏i=1

Xi = X1 ×X2 ×X3 × · · · ×Xn.

Let B be the family of all subsets of X of the form

O =n∏i=1

Oi = O1 ×O2 ×O3 × · · · ×On

where each Oi is open in Xi. Then B is a basis for the product topology for X.The set X with the product topology is a product space. The spaces X1, X2, . . . ,Xn are called the coordinate spaces or factor spaces of X.

Each point x ∈ X is of the form

x = (x1, x2, . . . , xn), xi ∈ Xi.

Define a function πi : X → Xi by πi(x) = xi. This is called the projection map onthe ith coordinate space or the ith projection map.

Theorem 57 The projection maps πi : X → Xi from the product space to the coor-dinate spaces are continuous.

Proof: Let πi : X → Xi and let Ui ⊂ Xi be open. Then π−1i (Ui) = X1 × . . . Xi−1 ×

Ui×Xi+1× . . . Xn. This is a product of open sets, and is open. Thus, πi is continuous.

Theorem 58 Let f : Y → X be a function from a topological space to a productspace X =

∏ni=1 Xi. Then f is continuous if and only if πi ◦ f is continuous for each

projection map.

What properties are preserved by products?


6.1. FINITE PRODUCTS 53

Theorem 59 The product of a finite number of Hausdorff spaces is Hausdorff.

Proof: Let Xi, i = 1, . . . , n, be a Hausdorff space for each n. Let a = (a1, . . . , an)and b = (b1, . . . , bn) be distinct points in

∏ni=1 Xi. Since they are distinct, there is at

least one coordinate space, Xj, for which aj 6= bj. Since Xj is Hausdorff, there aredisjoint open sets, Uj and Vj in Xj so that aj ∈ Uj and bj ∈ Vj. Then

U = π−1j (Uj) and V = π−1

j (Vj)

are disjoint open sets of X containing a and b, respectively.

Theorem 60 The product of a finite number of connected spaces is connected.

Proof: The proof is most easily done by induction. We will prove it for the case ofn = 2 and indicate the manner of the rest of the proof.

Assume that X1 and X2 are connected and let x1 ∈ X1. Then, {x1} × X2 is asubspace of X1 ×X2 which is homeomorphic to X2 using the function

(x1, t) 7→ t, t ∈ X2.

Since X2 is connected, {x1} ×X2 is connected. Now a similar argument shows thatX1 × {t} is a connected set for each t ∈ X2. Now, {x1} ×X2 and X1 × {t} have thepoint (x− 1, t) in common. Thus, the set

({x1} ×X2) ∪⋃t∈X2

(X1 × {t}) = X1 ×X2

is connected.Now, by induction, suppose that

∏n−1i=1 Xi is connected and consider

∏ni=1 Xi,

where each Xi is connected. By this previous argument,∏n

i=1 Xi is connected sinceit is homeomorphic to the product of two connected spaces:

∏n−1i=1 Xi and Xn. Hence

the finite product of connected spaces is connected.

Theorem 61 The finite product of separable spaces is separable.

Theorem 62 The finite product of first countable spaces is first countable.

Theorem 63 The finite product of second countable spaces is second countable.

Theorem 64 If (X1, d1), . . . , (Xn, dn) are metric spaces, then the product topologyon the product space is generated by the product metric:

d ((x1, . . . , xn), (y1, . . . , yn)) =

(n∑i=1

di(xi, yi)2

)1/2

.



In order to prove an analogous result for compactness, we need the followinglemma.

Lemma 9 In order that a space X be compact it is sufficient that there exists a basisB for X such that every open cover of X by members of B has a finite subcover.

Theorem 65 The finite product of compact spaces is compact.

Proof: Using an inductive argument as in our last theorem, it suffices to show thatX1 ×X2 is compact if X1 and X2 are compact.

Let B = {U × V | U is open in X1 and V is open in X2} be the subbasis for theproduct topology. Let O be an open cover of X1 ×X2 composed of members of B.By the above lemma, the compactness of X1 ×X2 will be proved if it can be shownthat there is a finite subcover for X1 ×X2 derived from O.

For x ∈ X1, the subset {x} × X2 is compact and is therefore contained in theunion of a finite number of members, say U1 × V1, U2 × V2, . . . , Um × Vm of O, eachof which meets {x} ×X2. Then

Ux =m⋂i=1

Ui

is an open set in X1 containing x. Note that

Ux ×X2 = Ux ×

(m⋃i=1

Vi

)=

m⋃i=1

(Ux × Vi) ⊂m⋃i=1

(Ui × Vi).

This last inclusion is due to the fact that Ux ⊂ Ui for all i. Thus, for each x ∈ X1

there is an open set Ux containing x for which the set Ux × X2 is contained in theunion of a finite number of members of O.

This family of open sets {Ux | x ∈ X1} is an open cover for X1. Since X1 iscompact, this cover admits a finite subcover {Uxi}ri=1. Now,

X1 ×X2 =

(r⋃i=1

Uxi

)×X2 =

r⋃i=1

(Uxi ×X2)

is the union of a finite number of the sets of the form Uxi × X2. Each of these setsis contained in the union of a finite number of members of O. This forms a finitesubcover for X1 ×X2 of the members of O, making X1 ×X2 compact.

6.2 Arbitrary Products

We have previously defined a countable product of spaces as follows:

∞∏i=1



6.2. ARBITRARY PRODUCTS 55

We identify these with sequences of elements, each element coming from the appro-priate factor space. We had a different way to define sequences when we studiedcalculus. If you will recall a sequence is a function

f : Z+ → X =n⋃i=1

Xi, f(i) ∈ Xi.

We will use this idea to extend the definition of products to arbitrary indexing sets.

Definition 48 Let I be an index set and {Xα | α ∈ I} be a family of sets. TheCartesian product X =

∏α∈I Xα is the collection of all functions x with domain I

having the property that the value xα of x at α belongs to the set Xα:

X =∏α∈I

Xα =

{x : I →

⋃α∈I

Xα | x(α) = xα ∈ Xα for each α ∈ I

}

For α ∈ I the function πα : X → Xα defined by πα(x) = xα is called the projectionmap of X on the αth coordinate set.

Note that this does not differ from our previous definition if the index set I isfinite. We get exactly the same space.

We define the product topology on the space X =∏

αXα using the projectionmaps. Let S consist of all sets of the form π−1

α (Uα) where Uα is open in Xα. This setforms a subbasis for the product topology. This means that the basis for the producttopology for X consists of all finite intersections:

n⋂i=1

π−1αi

(Uαi), αi ∈ I, 1 ≤ i ≤ n.

Such a basic open set may be expressed in the product form as

n⋂i=1

π−1αi

(Uαi) =∏α∈I

Vα

where Vαi = Uαi for i = 1, . . . , n and Vα = Xα otherwise.Proofs of the following theorems are similar to there finite analogues.

Theorem 66 Let f : Y → X be a function from a space Y into a product spaceX =

∏αXα. Then f is continuous if and only if the composition πα ◦ f of f with

each projection map is continuous.

Theorem 67 The product of any family of Hausdorff spaces is a Hausdorff space.

Theorem 68 The product of an arbitrary collection of connected spaces is connected.



This proof is based on the previous proof, but is not exactly the same.To prove the analogous theorem for compact spaces, we need some further results.

The proofs are not trivial, but are not included here. The first is needed for thesecond.

Lemma 10 (The Alexander Subbasis Theorem) In order that a space X be com-pact, it is necessary and sufficient that there exist a subbasis S for X such that everyopen cover of X by members of S has a finite subcover.

Theorem 69 (The Tychonoff Theorem) The product of an arbitrary family ofcompact spaces is compact.

6.3 Quotient Spaces

Quotient spaces are sometimes called identification spaces, because they result froma gluing, or identification, process.

Definition 49 Let X be a space and let ∼ be an equivalence relation on X. Let X/ ∼denote the set of all equivalence classes [x] = {y ∈ X | x ∼ y}. X/ ∼ is called thequotient space of X by the equivalence relation ∼. Define a function q : X → X/ ∼by q(x) = [x]. This map q is called the quotient map. Define a set A ⊂ X/ ∼ to beopen if q−1(A) is open in X. This collection of open sets defines a topology on X/ ∼called the quotient topology.

A particularly simple example of an equivalence relation giving rise to interestingquotient spaces is the following. Let A ⊂ X and, for the sake of interest, let A havemore than one member. Define a relation, ∼A, on X by: x ∼A y if x, y ∈ A or ifx = y 6∈ A. Thus, every point of A is related to every other point of A, and if a pointis not in A it is related only to itself. The quotient space is denoted by X/A and iscalled the identification space obtained by identifying the members of A toa point.

Example 44 The quotient space obtained by identifying the two endpoints 0 and 1to a point is the circle, S1.

Example 45 The quotient space of D2 obtained by identifying the boundary circle,S1, to a point is homeomorphic to a sphere in R3.

Example 46 Let I2 = [0, 1]× [0, 1]. The quotient space of I2 obtained by identifyingthe pairs of points (0, x2) and (1, 1−x2), 0 ≤ x2 ≤ 1, is homeomorphic to the Mobiusstrip.


6.3. QUOTIENT SPACES 57

Example 47 Let X = I2 with equivalence relation ∼ defined as follows: (1) (x1, 0) ∼(x1, 1) for each x1 ∈ [0, 1], and (2) (0, x2) ∼ (1, x2) for each x2 ∈ [0, 1]. Then thequotient space X/ ∼ is homeomorphic to a torus in R3.

Example 48 Let ∼ be an equivalence relation on R defined by x ∼ y if and only ifx − y ∈ Z. In other words, x ∼ x + n where n ∈ Z. We can actually think of theseequivalence classes as

[x] = {x+ n | x ∈ [0, 1] and n ∈ Z}

We have reduced our consideration from R to [0, 1]. It is clear that [0] = [1], so thatR/ ∼ is homeomorphic to our first example, and R/ ∼ is topologically equivalent tothe circle. The quotient map here is equivalent to t 7→ (cos(t), sin(t)).

Example 49 Let X = I2 and define ∼ on X by:

1. (x1, 0) ∼ (x1, 1) for x1 ∈ [0, 1], and

2. (0, x2) ∼ (1, 1− x2) for x2 ∈ [0, 1].

This is slightly different from 46 in that in the last stage the circles are identified withreversed orientations. The quotient space is the Klein bottle.

Example 50 Define an equivalence relation on the circle, S1, by z ∼ −z, wherewe consider the circle as the set of all complex numbers of modulus 1 (or the set ofvectors in the plane with norm 1). Then the quotient space S1/ ∼ is topologicallyequivalent to S1.

Example 51 Let X = I2 and define ∼ on X by:

1. (x1, 0) ∼ (1− x1, 1) for x1 ∈ [0, 1], and

2. (0, x2) ∼ (1, 1− x2) for x2 ∈ [0, 1].

This is equivalent to taking the unit disk and applying the previous equivalencerelation to the bounding circle. The resulting space is called the projective plane, andcannot be reasonably drawn in three dimensions.

Example 52 Let X be a topological space and let A be the subset of X × I givenby A = X × {1}. The quotient space T (X) = (X × I)/A is called the cone over X.

Note that our definition of the quotient topology guarantees that the quotientmap is continuous. This is extremely useful in later results.



Theorem 70 Let X/ ∼ be a quotient space with quotient map q : X → X/ ∼. Then afunction f : X/ ∼→ Y is continuous if and only if the composite function f ◦ q : X →Y is continuous.

Quotient topologies are not as uncommon as you might expect. A lot of topologiesmay be considered as quotient topologies, especially considering the characterizationin the above theorem, Theorem 70

Let X be a space and let f : X → Y be a function onto the set Y . Define a subsetO in Y to be open if its inverse image f−1(O) is open in X. The family of open setsso defined on Y is called the quotient topology determined by f .

Note that this actually is a topology. Clearly, f−1(∅) = ∅ and f−1(Y ) = X, sothat those conditions are satisfied. If {Uα | α ∈ I} are open in Y , then

f−1

(⋃α∈I

Uα

)=⋃α∈I

f−1(Uα)

is open in X, so arbitrary unions of open sets are open in Y . If {Ui} are open in Y ,then

f−1

(n⋂i=1

Ui

)=

n⋂i=1

f−1(Ui)

is open in X, so the finite intersection of open sets is open.

Theorem 71 Let X and Y be spaces and f : X → Y a continuous, onto function.If f is either open or closed, then Y has the quotient topology determined by f .

Proof: Let T denote the topology on Y and let Tf denote the quotient topologyinduced by the function f . If U ∈ T , then, since f is continuous, f−1(U) is open.This puts U ∈ Tf . Thus, T ⊂ Tf .

Suppose that f is an open function and let U ∈ Tf . Then f−1(U) is open in X.Since f is an open function with respect to T , f(f−1(U) = U is an open set in thetopology T . Thus, Tf ⊂ T , and we are done.

Example 53 [Adjunction Spaces] Let X and Y be topological spaces and let A ⊂ Xbe non-empty. Assume that X and Y are disjoint and that there is a continuousfunction f : A→ Y . Define a function φ : X ∪ Y → (X \ A) ∪ Y by

φ(z) =

f(z) if z ∈ Az if z ∈ X \ Az if z ∈ Y

The topology on X ∪ Y is that a set is open if and only if its intersection with bothX and Y are open. Clearly, φ is onto. When we put the quotient topology inducedby φ on the space (X \ A) ∪ Y , we will denote this space by X ∪f Y and call it theadjunction space via f .


6.3. QUOTIENT SPACES 59

a) If Y is a single point, then attaching X to a single point by the function f : A→{a} is the same as shrinking A to a point.

b) Let I2 = I × I be the unit square in the plane. Let A = {(x, y) | x = 0 or x =1 and 0 ≤ y ≤ 1}. Let Y = I = [0, 1] and define f : I2 → Y by f(x, y) = y.Then I2 ∪f Y is a cylinder formed by identifying the two vertical edges of I2.

c) Let X be a topological space and let Y = {p0, p1} be a two point discretespace not in X × [−1, 1]. Let A = X × ({−1} ∪ {1}). Define f : A → Y bedefined by f(x,−1) = p0 and f(x, 1) = p1. Since Y has the discrete topology,f is continuous. The adjunction space Σ(X) = X × [−1, 1] ∪f Y is called thesuspension of X. The equator is the image of X × {0} in Σ(X). The image ofX × [0, 1] in Σ(X) is homeomorphic to the cone over X. Thus, the suspensionof X is two cones over X identified along the equator. As an example, Σ(S1) ishomeomorphic to the 2-sphere, S2.


MATH 4181 001 Fall 1999Problem Set 1

1. (Exercise 2, page 11) Prove that if A has precisely n distinct elements and B hasprecisely m distinct elements, where m and n are positive integers, then A × B hasprecisely mn distinct elements.

2. (Exercise 3, page 11) Let A and B be sets, both of which have at least two distinctmembers. Prove that there is a subset W ⊂ A×B that is not the product of a subsetof A with a subset of B.

3. (Exercise 1, page 14) Let f : A→ B be given. Prove the following:

(a) For each subset X ⊂ A, X ⊂ f−1(f(X)).

(b) For each subset Y ⊂ B, f(f−1(Y )) ⊂ Y .

(c) If f is injective, then for each subset X ⊂ A,

f−1(f(X)) = X.

(d) If f is surjective, then for each subset Y ⊂ B,

f(f−1(Y )) = Y.

4. (Exercise 4, page 14) Let f : A→ B be given.

(c) If Y is a subset of B then f−1(B \ Y ) = A \ (f−1(Y )).

(d) If X ⊂ A and Y ⊂ B, then

f(X ∩ f−1(Y )) = f(X) ∩ Y.

5. (Exercise 7, page 15) Let X be a set and A,B,C ⊂ X. The function χA : X → {0, 1}defined by

χA(x) =

{1 if x ∈ A0 if x 6∈ A

is called the characteristic function of A. Show

(a) χA∩B = χA · χB, where χA · χB(x) = χA(x)χB(x).

(b) χA∪B = χA + χB − χA∩B and find a similar expression for χA∪B∪C .

6. Prove that ∅ ×B = ∅ for each set B.

7. Give an example to show that f(A∩B) may not equal f(A)∩f(B). Show that equalityholds if f is injective.

1

MATH 4181 001 Fall 1999Problem Set 1 Solutions

1. (Exercise 2, page 11) Prove that if A has precisely n distinct elements and B hasprecisely m distinct elements, where m and n are positive integers, then A × B hasprecisely mn distinct elements.

We can prove this by appealing to a counting argument and showing that the productset is in one-to-one correspondence with an m × n matrix, which we know has mnelements.

2. (Exercise 3, page 11) Let A and B be sets, both of which have at least two distinctmembers. Prove that there is a subset W ⊂ A× B that is not the product of a subsetof A with a subset of B.

Let A = B = {a, b}. Then A × B consists of 4 elements: {(a, a), (a, b), (b, a), (b, b)}.The subset {(a, a), (b, b)} is not the product of a subset of A and a subset of B, becausethe subset of A would have to contain both a and b, as would the subset of B. Then,the product of those two subsets would be all of A×B.

3. (Exercise 1, page 14) Let f : A→ B be given. Prove the following:

(a) For each subset X ⊂ A, X ⊂ f−1(f(X)).

Let a ∈ X. Then f(a) ∈ f(X) by definition. Thus, a ∈ {x ∈ X | f(x) ∈ f(X)}.This is just the set f−1(f(X)). Thus, X ⊂ f−1(f(X)).

(b) For each subset Y ⊂ B, f(f−1(Y )) ⊂ Y .

Let y ∈ f(f−1(Y )). Thus, there is an element x ∈ X so that y = f(x). What ismore, x ∈ f−1(Y ), which means that f(x) ∈ Y . Thus, y ∈ Y and f(f−1(Y )) ⊂ Y .

(c) If f is injective, then for each subset X ⊂ A,

f−1(f(X)) = X.

We have shown thatX ⊂ f−1(f(X)), so it only remains to show that f−1(f(X)) ⊂X. Let y ∈ f(X). Then there is an x ∈ X so that f(x) = y. Let a ∈ f−1(f(x)),i.e., f(a) = f(x) = y. Since f is one-to-one a = x ∈ X. Thus, any element off−1(f(X)) is in X and hence f−1(f(X)) ⊂ X. It then follows that f−1(f(X)) =X.

(d) If f is surjective, then for each subset Y ⊂ B,

f(f−1(Y )) = Y.

Again, we only need to show that Y ⊂ f(f−1(Y )). Let y ∈ Y . We need to findan element x ∈ A so that y = f(x). What is more, we would like to show thatthis x actually belongs to f−1(Y ). Since f is surjective, there is an element x ∈ Aso that y = f(x), by the definition of surjectivity. Also, by definition of the set,x ∈ f−1(Y ) since f(x) = y ∈ Y . Thus, y ∈ f(f−1(Y )) and we are done.

MATH 4181 Problem Set 1 Solutions 2

4. (Exercise 4, page 14) Let f : A→ B be given.

(c) If Y is a subset of B then f−1(B \ Y ) = A \ (f−1(Y )).

We have to show the two set containments: f−1(B \ Y ) ⊂ A \ (f−1(Y )) andA \ (f−1(Y )) ⊂ f−1(B \ Y ).

Let x ∈ f−1(B\Y ). Then f(x) ∈ B\Y or f(x) 6∈ Y . This means that x 6∈ f−1(Y )or x ∈ A \ (f−1(Y )). Thus, f−1(B \ Y ) ⊂ A \ (f−1(Y )).

Let x ∈ A \ (f−1(Y )). Thus, x 6∈ f−1(Y ) and f(x) 6∈ Y . Thus, f(x) ∈ B \ f(Y ),making x ∈ f−1(B \ Y ). Thus, f−1(B \ Y ) ⊂ A \ (f−1(Y )) and we are done.

(d) If X ⊂ A and Y ⊂ B, then

f(X ∩ f−1(Y )) = f(X) ∩ Y.

Again, we have two set containments to show.

Let v ∈ f(X ∩ f−1(Y )). Then we know that v = f(u) where u ∈ X ∩ f−1(Y ).Thus, u ∈ X and u ∈ f−1(Y ). It then easily follows that f(u) ∈ f(X) andf(u) ∈ Y . Thus, v = f(u) ∈ f(X) ∩ Y . Therefore f(X ∩ f−1(Y )) ⊂ f(X) ∩ Y .

Now, let v ∈ f(X) ∩ Y . Then v ∈ Y and v ∈ f(X). Thus, there is a u ∈ X sothat f(u) = v. Now, f(u) ∈ Y , which places u ∈ f−1(Y ). Thus, u ∈ X ∩ f−1(Y )and this places v ∈ f(X ∩ f−1(Y )) and f(X) ∩ Y ⊂ f(X ∩ f−1(Y )). Thus thetwo sets are equal.

5. (Exercise 7, page 15) Let X be a set and A,B,C ⊂ X. The function χA : X → {0, 1}defined by

χA(x) =

{1 if x ∈ A0 if x 6∈ A

is called the characteristic function of A. Show

(a) χA∩B = χA · χB, where χA · χB(x) = χA(x)χB(x).

If x ∈ A ∩ B then x ∈ A and x ∈ B. Thus, χA∩B(x) = 1, χA(x) = 1 andχB(x) = 1. In this case, χA∩B = χA · χB.

If x 6∈ A ∩ B, then there are three cases: (1) x ∈ A and x 6∈ B; (2) x 6∈ A andx ∈ B; and (3) x 6∈ A and x 6∈ B. The proof for cases (1) and (2) are similar.

(1) If x ∈ A and x 6∈ B, then χA(x) = 1 and χB(x) = 0 and χA∩B(x) = 0. Thus,in this case χA∩B = χA · χB.

(3) If x 6∈ A and x 6∈ B, then χA(x) = 0 and χB(x) = 0 and χA∩B(x) = 0. Thus,in this case χA∩B = χA · χB.

Therefore, χA∩B = χA · χB.

(b) χA∪B = χA + χB − χA∩B and find a similar expression for χA∪B∪C .

If x ∈ A ∪ B, then x ∈ A, x ∈ B, or x lies in both. Assume that x ∈ A andx 6∈ B. Then χA∪B(x) = 1 and χA(x) = 1, χB(x) = 0, and χA∩B(x) = 0. Thus,

c© David Royster Introduction to Topology For Classroom Use Only


in this case χA∪B = χA + χB − χA∩B. If x ∈ B and x 6∈ A, then we have asimilar computation. If x ∈ A and x ∈ B, then χA(x) = χB(x) = χA∩B(x) =χA∪B(x) = 1. Then, χA∪B(x) = χA(x) + χB(x)− χA∩B(x) = 1 + 1− 1. ThereforeχA∪B = χA + χB − χA∩B.

What we do in this case is we see how many times each point is counted. Thepoints in A∩B are counted twice, once in χA and once in χB. Thus, for χA∪B∪Cthe points in A ∩ B, A ∩ C and B ∩ C are counted twice. When we pull themout though, we pull out the points in A∩B ∩C three times – one for each of theintersections. Thus,

χA∪B∪C = χA + χB + χC − χA∩B − χA∩C − χB∩C + χA∩B∩C .

6. Prove that ∅ ×B = ∅ for each set B.

Assume not, that is assume that there is a point in ∅×B. Then this point is an orderedpair (x, y) where x ∈ ∅ and y ∈ B. However, the empty set contains no elements, thusx 6∈ ∅. Thus, there is no point in ∅ ×B. Thus, ∅ ×B = ∅.

7. Give an example to show that f(A∩B) may not equal f(A)∩f(B). Show that equalityholds if f is injective.

Let f : R → R by f(x) = x2. Let A = [−2, 1] and B = [−1, 2]. Then, f(A) = [0, 4]and f(B) = [0, 4] and f(A)∩f(B) = [0, 4]. Now, A∩B = [−1, 1] and f(A∩B) = [0, 1].Thus, in this case, f(A ∩B) 6= f(A) ∩ f(B).

Clearly, we have that f(A ∩ B) ⊂ f(A) ∩ f(B) since it is contained in each. If f isinjective, we need to show the opposite inclusion. Let x ∈ f(A) ∩ f(B). Then thereis an a ∈ A and b ∈ B so that f(a) = x and f(b) = x. Thus, f(a) = f(b). Sincef is injective, this implies that a = b. Hence, a ∈ A ∩ B and x ∈ f(A ∩ B). Thus,f(A) ∩ f(B) ⊂ f(A ∩B) and we have that f(A) ∩ f(B) = f(A ∩B).

c© David Royster Introduction to Topology For Classroom Use Only


1. (Exercise 1, page 34) Let (X, d) be a metric space and let k > 0 be any positive realnumber. Let dk(x, y) = k · d(x, y) for any two points x, y ∈ X. Show that (X, dk) is ametric space.

2. (Exercise 8, page 35) Let Z be the set of integers. Let p bea positive prime integer.Given distinct integers m,n ∈ Z there is a unique integer t = t(m,n) such thatm−n = pt ·k, where k is an integer not divisible by p. Define a function d : Z×Z→ R

by

d(m,n) =

{0 if m = n1pt

if m 6= n

Prove that (Z, d) is a metric space. Let p = 3. What is the set of elements n ∈ Z suchthat d(0, n) < 1? What is the set of elemnets m ∈ Z such that d(0,m) < 1

3?

3. (Exercise 1, page 39) Let X be the set of continuous functions f : [a, b] → R. Let d∗

be the metric on X given by

d∗(f, g) =

∫ b

a

|f(t)− g(t)| dt,

for f, g ∈ X. For each element f ∈ X set

I(f) =

∫ b

a

f(t) dt.

Prove that this function I : (X, d∗)→ (R, d) is continuous.

4. For P = (−2, 1) and Q = (3, 4) in R2, compute the distance from P to Q in each ofthe following metrics:

(a) usual;

(b) taxicab;

(c) maximum;

(d) discrete.

5. Let B = {P = (x1, x2) ∈ R2 | x21 + x2

2 ≤ 1}. Compute the diameter of B in each of thefollowing metrics:

(a) usual;

(b) taxicab;

(c) maximum;

(d) discrete.

1


1. (Exercise 1, page 34) Let (X, d) be a metric space and let k > 0 be any positive realnumber. Let dk(x, y) = k · d(x, y) for any two points x, y ∈ X. Show that (X, dk) is ametric space.

We have to check the three conditions for a metric space.

(1) dk(x, y) = k · d(x, y) ≥ k · 0 = 0 since k > 0, and if dk(x, y) = 0, then k · d(x, y) = 0which implies that d(x.y) = 0. This means that x = y since d is a metric.

(2) dk(x, y) = k · d(x.y) = k · d(y, x) = dk(y.x).

(3) Let x, y, z ∈ X,

dk(x, z) = k · d(x, z) ≤ k · (d(x, y) + d(y, z))

= kd(x, y) + kd(y, z)

= dk(x, y) + dk(y, z)

Thus, dk is a metric.

2. (Exercise 8, page 35) Let Z be the set of integers. Let p be a positive prime integer.Given distinct integers m,n ∈ Z there is a unique integer t = t(m,n) such that m−n =pt · k, where k is an integer not divisible by p. Define a function d : Z× Z→ R by

d(m,n) =

{0 if m = n1pt

if m 6= n

Prove that (Z, d) is a metric space. Let p = 3. What is the set of elements n ∈ Z suchthat d(0, n) < 1? What is the set of elements m ∈ Z such that d(0,m) < 1

3?

First, we need to prove that it is a metric space.

(1) From its definition d(m,n) = 0 or d(m,n) = 1/pt > 0. Thus, d(m,n) ≥ 0. Again,according to the definition, the only time that d(m,n) = 0 is if m = n. Thus, d satisfiesthe first condition on being a metric.

(2) If d(m,n) = 0 then m = n and d(n,m). Now, if d(m,n) 6= 0, then m − n = ptk,k 6= 0. Thus, n−m = pt · (−k) and hence d(m,n) = 1/pt = d(n,m).

(3) If m,n, r ∈ Z, then let m− n = ptk, n− r = ps`. Hence,

m− r = (m− n) + (n− r)= ptk + ps`

Without loss of generality we may assume that s < t

= ps(pt−sk + `) = psα


and p does not divide α.

Therefore,

d(m, r) =1

ps

≤ 1

pt+

1

ps

= d(m,n) + d(n, r)

The argument is similar if t < s.

Thus, d is a metric on Z.

Now, let p = 3. Now, if m,n ∈ Z, then write m − n = 3tk and if m 6= n thend(m,n) = 1/3t. Let A = {n ∈ Z | d(n, 0) < 1}. Now, this means that we write n = 3tkand we want 1/3t < 1. Thus, we want t > 0. Thus, A = {n ∈ Z | 3|n} which are themultiples of 3.

Let B = {m ∈ Z | d(m, 0) < 1/3}. With our analysis above that means that we needto find all numbers m = 3tk so that t > 1. Thus, we must have that t ≥ 2 and Bconsists of all of the multiples of 9.

3. (Exercise 1, page 39) Let X be the set of continuous functions f : [a, b]→ R. Let d∗ bethe metric on X given by

d∗(f, g) =

∫ b

a

|f(t)− g(t)| dt,

for f, g ∈ X. For each element f ∈ X set

I(f) =

∫ b

a

f(t) dt.

Prove that this function I : (X, d∗)→ (R, d) is continuous.

Let ε > 0 be given. We need to find δ > 0 so that if d∗(f, g) < δ then d(I(f), I(g)) < ε.From a theorem in calculus, we know that

∣∣∫ f(x) dx∣∣ ≤ ∫ |f(x)| dx. Thus, we have

d(I(f), I(g)) =

∣∣∣∣∫ b

a

f(x)− g(x) dx

∣∣∣∣≤∫|f(x)− g(x)| dx

= d∗(f, g)

Thus, if we take δ = ε, then d(I(f), I(g)) ≤ d∗(f, g) < δ = ε. Thus, I is continuous.

c© David C. Royster Introduction to Topology For Classroom Use Only


4. For P = (−2, 1) and Q = (3, 4) in R2, compute the distance from P to Q in each ofthe following metrics:

(a) usual;

d(P,Q) =√

(3− (−2))2 + (4− 1)2 =√

34.

(b) taxicab;

d′(P,Q) = |3− (−2)|+ |4− 1| = 8.

(c) maximum;

d′′(P,Q) = max {|3− (−2)|, |4− 1|} = 5.

(d) discrete.

Since P 6= Q, d(P,Q) = 1.

5. Let B = {P = (x1, x2) ∈ R2 | x21 + x2

2 ≤ 1}. Compute the diameter of B in each of thefollowing metrics:

(a) usual;

The diameter is the least upper bound of the distances apart of two points in thisset. In the usual metric, we know that the greatest distance will be diametricallyopposed points. This gives a diameter of 2.

(b) taxicab;

Let P = (x, y), Q = (a, b) ∈ B. Then |x|+ |y| ≤ 1 and |a|+ |b| ≤ 1. Now,

d′(P,Q) = |x− a|+ |y − b|≤ |x|+ |a|+ |y|+ |b|≤ (|x|+ |y|) + (|a|+ |b|) = 2

Thus, the least upper bound of these numbers is 2. The diameter of the set is 2.

(c) maximum;

Let P = (x, y), Q = (a, b) ∈ B. Then max{|x|, |y|} ≤ 1 and max{|a|, |b|} ≤ 1.Now,

d′′(P,Q) = max{|x− a|, |y − b|}≤ max{|x|+ |a|, |y|+ |b|}≤ 2

Thus, the least upper bound of these numbers is 2. The diameter of the set is 2.

(d) discrete.

In the discrete metric the greatest distance apart two points can be is 1. Thusthe diameter must be 1.



1. (Exercise 6, page 46) Let a and b be distinct points in a metric space X. Prove thatthere are neighborhoods Na and Nb of a and b respectively such that Na ∩Nb = ∅.

2. Let {xn} be a sequence in the metric space X and assume that {xn} converges tox ∈ X. Prove that this limit is unique, i.e., if limxn = x and limxn = L then x = L.

3. Show that a finite subset of a metric space has no limit points and is therefore a closedset.

4. Let (X, d) be a metric space with the discrete metric. Prove

(a) Every subset of X is open.

(b) Every subset of X is closed.

(c) No subset of X has a limit point.

1



1. (Exercise 6, page 46) Let a and b be distinct points in a metric space X. Prove thatthere are neighborhoods Na and Nb of a and b respectively such that Na ∩Nb = ∅.Since a and b are distinct points, d(a, b) = r > 0. Let Na = Bd(a; r/3) and Nb =Bd(b; r/3). Now, clearly, Na ∩Nb = ∅.

2. Let {xn} be a sequence in the metric space X and assume that {xn} converges to x ∈ X.Prove that this limit is unique, i.e., if limxn = x and limxn = L then x = L.

We will use the result from above to prove this. Assume that x 6= L. Then thereare disjoint neighborhoods Nx and NL of x and L, respectively. In fact, if we letr = d(x, L) 6= 0 (by assumption), then we may take each neighborhood to be the ballof radius r/3 centered at each respective point. Now, since the sequence converges tox we know that for any ε > 0 there is a positive integer Nε so that for any n > Nε

xn ∈ Bd(x; ε). This must be true for ε = r/3. Thus, there is a positive integer N sothat for all n > N , xn ∈ B(x; r/3). Applying the same analysis to the limit L of thesequence {xn} gives us a positive integer M so that for all n > M xn ∈ B(L; r/3).

Let P = max{M,N}. Now, if n > P we must have that xn ∈ B(x; r/3) and xn ∈B(L; r/3), or xn ∈ B(x; r/3) ∩ B(L; r/3) = ∅. This is impossible, so we must havex = L.

3. Show that a finite subset of a metric space has no limit points and is therefore a closedset.

Let A = {x1, . . . , xn} be a finite set in the metric space (X, d). Let y 6∈ A, then letdk = d(y, xk). Since y 6∈ A dk 6= 0 for k = 1, . . . , n. Let r = min{d1, . . . , dn}. Then forany ε < r, B(y; ε) does not contain any elements of A, because they are all further awayfrom y than ε. Thus, y is not a limit point of A. If y ∈ A we use the same techniqueto construct a ball about y which contains no other points of A. Thus showing that yis not a limit point of A. Therefore, A has no limit points, A′ = ∅. This then meansthat A contains all of its limit points, since ∅ ⊂ A. Therefore by our theorem, A mustbe closed.

4. Let (X, d) be a metric space with the discrete metric. Prove

(a) Every subset of X is open.

Note that B(x; 1/2) = {x} is an open set. Thus, each singleton set is an openset. Every set is a union of its elements and therefore is a union of open sets.Therefore every set is open.

(b) Every subset of X is closed.

For any set F , we know that X \F is open from above. Thus, by definition, F isclosed.



(c) No subset of X has a limit point.

Using the same open set as the first part shows that each point has an openneighborhood that contains no other point of any other set. Thus, each pointcannot be a limit point of any set. Thus, no subset of X has a limit point.



1. (Exercise 4, page 74) Let (X,T ) be a topological space. Prove that ∅ and X are closedsets, that a finite union of closed sets is a closed set, and that an arbitrary intersectionof closed sets is a closed set.

2. (Exercise 6, page 75) Prove that in a discrete topological space, each subset is simul-taneously open and closed.

3. Show that a topological space (X,T ) is discrete if and only if each set consisting ofonly one point is open.

4. Let X be a set and let T ′ the finite complement topology for X.

(a) Show that (X,T ′) is discrete if and only if X is a finite set.

(b) Show that if A is an infinite subset of X, then every point of X is a limit pointof A.

5. Let X be a set. The countable complement topology, or co-countable topology, T ′′ forX consists of X, ∅ and all subsets O of X for which X \O is a countable set.

(a) Show that T ′′ is a topology on X.

(b) For the space (X,T ′′), show that a countable set A of X has a derived set A′ = ∅and that an uncountable set B has B′ = X.

(c) Show that the intersection of any countable family of members of T ′′ is a memberof T ′′.

6. Let X = {a, b} be a two-element set and let T = {∅, {a}, {a, b}}. Show that T is atopology on X and identify the limit points of each subset of X. (This space is calledSierpenski space.)

7. How many different topologies are there for a set with three members?

1



1. (Exercise 4, page 74) Let (X,T ) be a topological space. Prove that ∅ and X are closedsets, that a finite union of closed sets is a closed set, and that an arbitrary intersectionof closed sets is a closed set.

Let (X,T ) be a topological space. To show that a set is closed we must show thatits compliment is open. Since X \ X = ∅ is open, X must be closed. Likewise, sinceX \ ∅ = X is open, the empty set is closed.

Let {Fi}ni=1 be a finite collection of closed sets. To show that their union is closed, wemust show that the complement of the union is open.

X \

(n⋃i=1

Fi

)=

n⋂i=1

(X \ Fi)

and each of the sets X \ Fi is open since Fi is closed. Since the finite intersectionof open sets is open, we have then that X \ (

⋃ni=1 Fi) is open, hence making

⋃ni=1 Fi

closed.

Let {Fα | α ∈ I} be a collection of closed sets. To show that the arbitrary intersectionis closed, we must show that its complement is open.

X \

(⋂α∈I

Fα

)=⋂α∈I

(X \ Fα)

and each of the sets X \ Fα is open since Fα is closed. Since the arbitrary union ofopen sets is open, we have then that X \

(⋂nα∈I Fα

)is open, hence making

⋂nα∈I Fα

closed.

2. (Exercise 6, page 75) Prove that in a discrete topological space, each subset is simulta-neously open and closed.

We already know that each subset is open, by the definition of the discrete topology.Let A ⊂ X. Then X \A is a subset of X and hence open. Thus, A is closed. Therefore,every subset of a discrete topological space is simultaneously open and closed.

3. Show that a topological space (X,T ) is discrete if and only if each set consisting ofonly one point is open.

If (X,T ) is discrete then every subset is open. Hence the singleton sets are open.

If (X,T ) has the property that the singleton sets are open, then each subset of Xbeing a union of its elements is a union of open sets and is thus open. Thus, everysubset is open, making (X,T ) a discrete topological space.



4. Let X be a set and let T ′ the finite complement topology for X.

(a) Show that (X,T ′) is discrete if and only if X is a finite set.

If X is a finite set, then the complement of a singleton set is finite. Thus, the sin-gleton sets are open. By the previous problem that gives X the discrete topology.

If the finite complement topology and the discrete topology coincide on the spaceX, then the singleton set {a} must be open (by the discrete topology). Since itis open, its complement must be finite (by the finite complement topology). ButX = {x} ∪ (X \ {x}. Each of the two sets on the right are finite, making X afinite set.

(b) Show that if A is an infinite subset of X, then every point of X is a limit pointof A.

Let A be an infinite set and let x ∈ X. If U is any open set containing x, thenX \U is a finite set. If A ⊂ X \U , then we would have a finite set with an infinitesubset — which is impossible. Thus, A ∩ U 6= ∅. This means that x is a limitpoint of A. Since x ∈ X was arbitrary, each point of X is a limit point of A.

5. Let X be a set. The countable complement topology, or co-countable topology, T ′′ forX consists of X, ∅ and all subsets O of X for which X \O is a countable set.

(a) Show that T ′′ is a topology on X.

Clearly, ∅ and X are in T ′′, by definition.

Let {Oα | α ∈ I} be a collection of open sets. Then, X \ Oα is countable in X.To see that

⋃α∈I Oα is open, we need look at X \

⋃α∈I Oα.

X \⋃α∈I

Oα =⋂α∈I

(X \Oα) ⊂ X \Oα.

Since each complement is countable, and the subset of a countable set is countable,we have that X \

⋃α∈I Oα is countable and

⋃α∈I Oα is open.

Let {Oi}ni=1 be a finite collection of open sets. Then, X \ Oi is countable in X.To see that

⋂ni=1 Oi is open, we need look at X \

⋂ni=1 Oi.

X \n⋂i=1

Oi =n⋂i=1

(X \Oi) ⊂ X \On.

Since each complement is countable, and the union of a finite number of countablesets is countable, we have that X \

⋂ni=1 Oi is countable and

⋂ni=1 Oi is open.

(b) For the space (X,T ′′), show that a countable set A of X has a derived set A′ = ∅and that an uncountable set B has B′ = X.

Let A be countable and let x ∈ X. If x 6∈ A, then X \A is an open set containingx and A ∩X \ A = ∅.If x ∈ A, then (X\A)∪{x} is an open set containing a and (A\{x})∩(X\A∪{x} =∅.



In neither case is x a limit point of A. Hence, no points are limit points of A, orA′ = ∅.If B is uncountable, let x ∈ X be any point and U ⊂ X be any open set withx ∈ U . Since X \ U is countable, we cannot have B ⊂ X \ U . This means thatU ∩B 6= ∅. Thus, each point is a limit point for B and B′ = X.

(c) Show that the intersection of any countable family of members of T ′′ is a memberof T ′′.

Let {Oi}∞i=1 be a countable collection of open sets. Then, X \ Oi is countable inX. To see that

⋂∞i=1 Oi is open, we need look at X \

⋂∞i=1 Oi.

X \∞⋂i=1

Oi =∞⋂i=1

(X \Oi) ⊂ X \O1.

Since each complement is countable, and the union of a countable number ofcountable sets is countable, we have that X \

⋂∞i=1 Oi is countable and

⋂∞i=1 Oi is

open.

6. Let X = {a, b} be a two-element set and let T = {∅, {a}, {a, b}}. Show that T is atopology on X and identify the limit points of each subset of X. (This space is calledSierpenski space.)

Clearly, T contains the empty set and the whole space. It is easy to see that it isclosed under unions and intersections. Thus, it is a topology on X.

For the limit points we have

LimitPoints

∅ ∅{a} ∅{b} ∅{a, b} {b}



7. How many different topologies are there for a set with three members?

Since there are 8 subsets of this space, there are at most 28 possibilities for topologies.Fortunately, there are not that many.

It is probably best just to list the collections that form a topology on this set X ={a, b, c}.

T1 = {∅, X} T2 = {∅, {a}, X}T3 = {∅, {b}, X} T4 = {∅, {c}, X}T5 = {∅, {a, b}, X} T6 = {∅, {a, c}, X}T7 = {∅, {b, c}, X} T8 = {∅, {a}, {a, b}, X}T9 = {∅, {a}, {a, c}, X} T10 = {∅, {a}, {b, c}, X}T11 = {∅, {b}, {a, b}, X} T12 = {∅, {b}, {a, c}, X}T13 = {∅, {b}, {b, c}, X} T14 = {∅, {c}, {a, b}, X}T15 = {∅, {c}, {a, c}, X} T16 = {∅, {c}, {b, c}, X}T17 = {∅, {a}, {b}, {a, b}, X} T18 = {∅, {a}, {c}, {a, c}, X}T19 = {∅, {b}, {c}, {b, c}, X} T20 = {∅, {a}, {a, b}, {a, c}, X}T21 = {∅, {b}, {a, b}, {b, c}, X} T22 = {∅, {c}, {a, c}, {b, c}, X}T23 = {∅, {a}, {b}, {a, b}, {b, c}, X} T24 = {∅, {a}, {b}, {a, b}, {a, c}, X}T25 = {∅, {a}, {c}, {a, b}, {a, c}, X} T26 = {∅, {a}, {c}, {a, c}, {b, c}, X}T27 = {∅, {b}, {c}, {a, b}, {b, c}, X} T28 = {∅, {b}, {c}, {a, c}, {b, c}, X}T29 = {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, X}



1. Let f : X → Y be a function. Then f is an open mapping if for each open setO ⊂ X, f(O) is open in Y .

(a) Give an example of a mapping that is continous, but not open.

(b) Give an example of a mapping that is open, but not continous.

(c) Prove that a one-to-one, onto mapping f : X → Y is a homeomorphism if andonly if f and f−1 are open mappings.

2. Prove that a finite subset A of a Hausdorff space X has no limit points. Conclude thatA must be closed.

3. If X is a space which is homeomorphic to a subspace A of a space Y , then X is said tobe embedded in Y . Give an example of spaces A and B for which A can be embeddedin B and B can be embedded in A, but A and B are not homeomorphic. (Simpleexamples can be found in R.)

4. Prove that every countable subset of R is totally disconnected.

5. Give examples of subsets A and B in R2 to illustrate each of the following. A drawingis sufficient.

(a) A and B are connected, but A ∩B is disconnected.

(b) A and B are connected, but A \B is disconnected.

(c) A and B are disconnected, but A ∪B is connected.

(d) A and B are connected and A ∩B 6= ∅, but A ∪B is disconnected.

6. Definition: A Hausdorff space X is 0-dimensional if X has a basis B of sets whichare simultaneously open and closed.

Prove that every 0-dimensional space is totally disconnected.

7. Prove:

(a) The property of being totally disconnected is a topological invariant but not acontinuous invariant.

(b) The property of being totally disconnected is hereditary.

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1. Let f : X → Y be a function. Then f is an open mapping if for each open setO ⊂ X, f(O) is open in Y .

(a) Give an example of a mapping that is continuous, but not open.

Let X = Y = R. Let TX denote the discrete topology on X and TY denote theusual, metric topology on Y . Define f : (X,TX)→ (Y,TY ) by f(x) = x. Then, fis a continuous mapping, because the inverse image of any set is open in (X,TX).However, f is not open since f({0}) = {0} which is not open in (Y,TY ).

(b) Give an example of a mapping that is open, but not continuous.

Use the same notation as in (a) and define g : (Y,TY ) → (X,TX) by g(x) = x.Then, g is open, since the image of any set is open in (X,TX), but g is notcontinuous since, for example, the inverse image of the open set {0} is not openin (Y,TY ).

(c) Prove that a one-to-one, onto mapping f : X → Y is a homeomorphism if andonly if f and f−1 are open mappings.

Let f be a homeomorphism. Then we know that f and f−1 are both continuous.We also know that f = (f−1)−1, so that the statement that f is open is equivalentto saying that f−1 is continuous. Likewise, the statement that f−1 is open isequivalent to saying that f is continuous. Thus, f and f−1 are open mappings.

If we know that f is one-to-one and onto, then assuming that f and f−1 are openmappings makes f and f−1 both continuous. Thus, f is a homeomorphism.

2. Prove that a finite subset A of a Hausdorff space X has no limit points. Conclude thatA must be closed.

Let A = {x1, . . . , xn} and let p be a limit point of A. Then, for every open set, U ,containing p we must have that U ∩A \ {p} 6= ∅. Since X is Hausdorff, there are opensets Ui containing p and Vi containing xi so that Ui ∩ Vi = ∅. Let U = ∩ni=1Ui and letV = ∪ni=1Vi. Then,

(i) p ∈ U ;

(ii) A ⊂ V ;

(iii) U ∩ V = ∅.

Thus, we found a neighborhood of p which does not intersect A. Hence, p cannot be alimit point of A and A has no limit points.

Since A′ = ∅, we have that A = A ∪ A′ = A, making A closed.



3. If X is a space which is homeomorphic to a subspace A of a space Y , then X is said tobe embedded in Y . Give an example of spaces A and B for which A can be embeddedin B and B can be embedded in A, but A and B are not homeomorphic. (Simpleexamples can be found in R.)

Let A = (0, 1) and let B = [0, 1]. Clearly, A can be embedded in B using the identityfunction, f(x) = x. Define a map g : B → A by g(x) = 1

2x+ 1

4. This maps B onto the

interval [14, 3

4] embedding it into A.

Thus, we can embed A into B and B into A. However, we know that the two intervals(0, 1) and [0, 1] are not homeomorphic.

4. Prove that every countable subset of R is totally disconnected.

We shall prove in Problem 7a that the property of being totally disconnected is atopological invariant. Since we have shown that Q is totally disconnected. Since Q iscountable it is homeomorphic to every countable subset of R. Hence, every countablesubset of R is totally disconnected.

5. Give examples of subsets A and B in R2 to illustrate each of the following. A drawingis sufficient.

(a) A and B are connected, but A ∩B is disconnected.

Let A be the segment on the x-axis, A = {(x, 0) | −1 ≤ x ≤ 1}. Let B denotethe upper hemisphere of the unit circle: B = {(x, y) | x2 + y2 = 1 and y ≥ 1}.Then each of A and B is connected but the intersection is the two disjoint points(−1, 0) and (1, 0) which are disconnected.

(b) A and B are connected, but A \B is disconnected.

Let A be the rectangle in the plane with vertices at (2, 1), (−2, 1), (−2,−1) and(2,−1). Let B be the square in the plane with vertices (1, 1), (−1, 1), (−1,−1)and (1,−1). Then A \B is two disconnected squares.

(c) A and B are disconnected, but A ∪B is connected.

Take A to be the two vertical sides of the unit square: A = {(1, t) | −1 ≤ t ≤1} ∪ {−1, t) | −1 ≤ t ≤ 1}. Take B to be the two horizontal sides of the unitsquare: B = {t, 1) | −1 ≤ t ≤ 1} ∪ {t,−1) | −1 ≤ t ≤ 1}. Then, each of A and Bis disconnected, but A ∪B is connected.

(d) A and B are connected and A ∩B 6= ∅, but A ∪B is disconnected.

Let A = (0, 1) and B = (1, 2). Then A and B are connected,

A ∩B = [0, 1] ∩ [1, 2] = {1} 6= ∅,

but A ∪B = (0, 1) ∪ (1, 2) is disconnected.



6. Definition: A Hausdorff space X is 0-dimensional if X has a basis B of sets whichare simultaneously open and closed.

Prove that every 0-dimensional space is totally disconnected.

Let X be a 0-dimensional space. Let C be a component of X containing the pointx ∈ X. Let y ∈ C. Since X is Hausdorff, there are disjoint open sets, U and V ,containing x and y, respectively. Now, each of these open sets consists of a union ofbasic open sets, thus, we can separate x and y by disjoint basic open sets. Thus, thereare sets Ux and Vy, disjoint and open and closed, since X is 0-dimensional. Then, Ccontains sets which are both open and closed, making C not connected. The only wayin which this can be prevented is for C = {x}. Thus, each component consists of asingle point and X is totally disconnected.

7. Prove:

(a) The property of being totally disconnected is a topological invariant but not acontinuous invariant.

Let f : X → Y be a homeomorphism and assume that X is totally disconnected.Let C be a connected component of Y . Then f−1(C) is a connected subset of X,being the continuous image of a connected set. If x ∈ f−1(C), then f−1(C) lies inthe connected component of X containing x. Since X is totally disconnected, thiscomponent is x. Thus, f−1(C) is a single point. Since f is a homeomorphism, Cconsists of a single point. Thus, Y is totally disconnected.

Let (X,T ) denote the reals with the discrete topology. Then (X,T ) is totallydisconnected. Let (Y,S ) denote the reals with the usual topology. Define f : X →Y by f(x) = x. This function is continuous, since the domain has the discretetopology, but the image space is connected, not totally disconnected.

(b) The property of being totally disconnected is hereditary.

Let X be totally disconnected and let A ⊂ X. Let C ⊂ A be the component ofA containing x ∈ A. Let y ∈ C. Now, y is disconnected from x in X, since X istotally disconnected. The same disconnection in X will disconnect x and y in A.Hence, C cannot contain more than one point, and A is totally disconnected.



1. Identify the two path components of the topologist’s sine curve. Show that one of thecomponents is closed and the other is not.

2. Prove that path connectedness is a topological invariant. Can you prove that it is acontinuous invariant?

3. Prove that local connectedness is a topological invariant, but that the continuous imageof a locally connected set need not be locally connected.

4. Prove that every locally path connected space is locally connected.

5. Prove that every finite subset of a topological space is compact.

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1. Identify the two path components of the topologist’s sine curve. Show that one of thecomponents is closed and the other is not.

The two path components of the topologist’s sine curve are the component alongthe y-axis, A = {(0, y) | −1 ≤ y ≤ 1}, and the sin(π/x) curve component: B ={(x, sin(π/x)) | 0 < x ≤ 1}. We proved in class that there is no path joining any pointof A to any point of B. The component A is closed, but B does not contain its limitpoints, which consist of A, and hence B is not closed.

2. Prove that path connectedness is a topological invariant. Can you prove that it is acontinuous invariant?

Let f : X → Y be a homeomorphism and assume that X is path connected. Leta, b ∈ Y be any two distinct points. Since f is onto, there are points a′, b′ ∈ X so thatf(a′) = a and f(b′) = b. Since X is path connected, there is a path, p : I → X, so thatp is continuous, p(0) = a′ and p(1) = b′. Let q = f ◦ p. Then q is continuous, being thecomposition to two continuous functions; q(0) = f(a′) = a and q(1) = f(b′) = b. Thus,any two points of Y are connected by a path, meaning that Y is path connected.

All we required in the above proof is that f was onto and continuous. Thus, thecontinuous image of a path connected set is path connected, using the same proof.

3. Prove that local connectedness is a topological invariant, but that the continuous imageof a locally connected set need not be locally connected.

Let f : X → Y be a homeomorphism and assume that X is locally connected. Letp ∈ Y contained in U ⊂ Y an open set. We need to show that there is an open,connected set containing p and contained in U . Let x = f−1(p). Then x ∈ f−1(U)which is open in X. Since X is locally connected, there is an open connected set,C, containing x and contained in f−1(U). We then know that f(C) is connected andp = f(x) ∈ f(C). Since f is a homeomorphism, f is an open map and f(C) is open.Hence, Y is locally connected.

Note that this time, we did need the fact that f is a homeomorphism to guaranteethat f(C) was open. Thus, we should not expect the continuous image of a locallyconnected set to be locally connected, because not every continuous map is open. As anexample, take the identity function that maps the rationals with the discrete topologyto the rationals with the usual topology. We showed in class that the rationals withthe usual topology is not locally connected. However, any discrete space is locallyconnected, specifically because each point is an open set.

4. Prove that every locally path connected space is locally connected.

Let X be a locally path connected space and let x ∈ X be contained in an open setU . Now, U contains an open path connected set containing x, and any path connected



set is connected. Thus, U contains an open connected set containing X. Hence, X islocally connected.

5. Prove that every finite subset of a topological space is compact.

Let A = {a1, a2, . . . , an} be a finite set in X. Let O be an open cover of A. Each aimust be contained in at least one open set Oi from the open cover O. Hence, we onlyneed the sets {O1, O2, . . . , On} to cover A. This makes A compact.



1. In the product space X×X, the set ∆ = {(x, x) | x ∈ X} is called the diagonal. Provethat a space X is Hausdorff if and only if the diagonal of X ×X is a closed set.

2. Prove that if X is connected, then every quotient space of X is connected.

3. Prove that if X is compact, then every quotient space of X is compact.

4. Give an example of a Hausdorff space which has a quotient space that is not Hausdorff.

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Topology MATH 4181 Fall Semester 1999webpages.iust.ac.ir/m_nadjafikhah/Courses/PoSeTo/roy.pdf · in...

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