Set Theory - Springer

Appendix ASet Theory

A.1 Sets

The purpose of this appendix is to introduce some of the basic ideas and terminologiesfrom set theory which are essential to our present work. In this naive treatment, wedo not intend to give a complete and precise analysis of set theory, which belongsto the foundations of mathematics and to mathematical logic. Rather, we shall dealwith sets on an intuitive basis. We remark that this usage can be formally justified.

Intuitively, a set is a collection of objects called members of the set. The terms“collection” and “family” are used as synonyms for “set”. If an object x is a memberof a set X , we write x ∈ X to express this fact. Implicit in the idea of a set is thenotion that a given object either belongs or does not belong to the set. The statement“x is not a member of X” is indicated by x /∈ X . The objects which make up a setare usually called the elements or points of the set.

Given two sets A and X , we say that A is a subset of X if every element of A isalso an element of X . In this case, we write A ⊆ X (read A is contained in X ) orX ⊇ A (read X contains A). The sets A and X are said to be equal, written A = X ,if A ⊆ X and X ⊆ A. We call A a proper subset of X (written A ⊂ X ) if A ⊆ Xbut A �= X . The empty or null set which has no element is denoted by ∅. We havethe inclusion ∅ ⊆ X for every set X . The set whose only element is x is called asingleton, denoted by {x}. Note that ∅ �= {∅}. If P (x) is a statement about elementsin X, that is either true or false for a given element of X, then the subset of all thex ∈ X for which P (x) is true is denoted by {x ∈ X | P (x)} or {x ∈ X : P (x)}.

In a particular mathematical discussion, there is usually a set which consists ofall primary elements under consideration. This set is referred to as the “universe”.To avoid any logical difficulties, all the sets we consider in this section are assumedto be subsets of the universe X.

The difference of two sets A and B, denoted by A − B, is the set {x ∈ A | x /∈ B}. IfB ⊆ A, the complement of B in A is A − B. Notice that the complement operation isdefined only when one set is contained in the other, whereas the difference operationdoes not have such a restriction. The union of two sets A and B is the set A ∪ B =© Springer Nature Singapore Pte Ltd. 2019T. B. Singh, Introduction to Topology,https://doi.org/10.1007/978-981-13-6954-4

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412 Appendix A: Set Theory

{x | x belongs to at least one ofA, B}. The intersection of two sets A and B is the setA ∩ B = {x | xbelongs to both A and B}. When A ∩ B = ∅, the sets A and B arecalled disjoint; otherwise we say that they intersect.

Proposition A.1.1 (a) X − (X − A) = A.(b) B ⊆ A ⇔ X − A ⊆ X − B.(c) A = B ⇔ X − A = X − B.

Proposition A.1.2 A ∪ A = A ∩ A.

Proposition A.1.3 A ∪ B = B ∪ A, A ∩ B = B ∩ A.

Proposition A.1.4 A ⊆ B ⇔ A = A ∩ B ⇔ B = A ∪ B.

Proposition A.1.5 (a) A ∪ (B ∪ C) = (A ∪ B) ∪ C.(b) A ∩ (B ∩ C) = (A ∩ B) ∩ C.

Proposition A.1.6 (a) A ∩ (B ∪ C) = (A ∩ B) ∪ (∩C).(b) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

Proposition A.1.7 (a) A − B = A ∩ (X − B).

(b)X − (A ∪ B) = (X − A) ∩ (X − B)

X − (A ∩ B) = (X − A) ∪ (X − B)

}(De Morgan’s laws).

(c) (A − B) ∪ (B − A) = (A ∪ B)− (A ∩ B).(d) If X = A ∪ B and A ∩ B = ∅, then B = X − A.

Let J be a nonempty set, and suppose that a set A j is given for each j ∈ J . Thenthe collection of sets {A j | j ∈ J }, also written as {A j } j∈J , is called an indexedfamily of sets, and J is called an indexing set for the family. The union of an indexedfamily {A j | j ∈ J } of subsets of a set X is the set

⋃j∈J

A j = {x ∈ X | x ∈ A j for some j in J },

and the intersection is the set

⋂j∈J

A j = {x ∈ X | x ∈ A j for every j in J }.

The union of the sets A j is also denoted by⋃{A j | j ∈ J }, and their intersection by⋂{A j | j ∈ J }. If there is no ambiguity about the indexing set, we simply use

⋃A j

for the union and⋂

A j for the intersection. If J = {1, . . . , n}, n > 0 an integer, thenwe write

⋃nj=1 A j for the union of A1, . . . , An , and

⋂nj=1 A j for their intersection.

Observe that any nonempty collection C of sets can be considered an indexed familyof sets by “self-indexing”: The indexing set isC itself and one assigns to each S ∈ Cthe set S. Accordingly, the foregoing definitions become

Appendix A: Set Theory 413

⋃ {S : S ∈ C } = {x | x ∈ S for some S in C } and⋂ {S : S ∈ C } = {x | x ∈ S for every S in C }.

Ifwe allow the collectionC to be the empty set, then, by convention,⋃ {S : S ∈ C } =

∅ and⋂ {S : S ∈ C } = X , the specified universe of the discourse.

Proposition A.1.8 Let A j , j ∈ J , be a family of subsets of a set X. Then we have

(a) X −⋃j A j =⋂

j

(X − A j

), and X −⋂

j A j =⋃j

(X − A j

).

(b) If K ⊂ J , then⋃

k∈K Ak ⊆⋃j∈J A j and

⋂k∈K Ak ⊇⋂

j∈J A j .

A.2 Functions

With each two objects x, y, there corresponds a newobject (x, y), called their orderedpair. This is another primitive notion that we will use without a formal definition.Ordered pairs are subject to the condition: (x, y) = (

x ′, y′)⇔ x = x ′ and y = y′.

Accordingly, (x, y) = (y, x)⇔ x = y. The first (resp. second) element of an orderedpair is called the first (resp. second) coordinate. Given two sets X1 and X2, theirCartesian product X1 × X2 is defined to be the set of all ordered pairs (x1, x2), wherex j ∈ X j for j = 1, 2.Thus X1 × X2 = {(x1, x2) | x j ∈ X j for every j = 1, 2}.Notethat X1 × X2 = ∅ if and only if X1 = ∅ or X2 = ∅. When both X1 and X2 arenonempty, X1 × X2 = X2 × X1 ⇔ X1 = X2.

Let X and Y be two sets. A function f from the set X to Y (written f : X → Y )is a subset of X × Y with the following property: for each x ∈ X , there is one andonly one y ∈ Y such that (x, y) ∈ f . A function is also referred to as a mapping (orbriefly, a map). We write y = f (x) to denote (x, y) ∈ f , and say that y is the imageof x under f or the value of f at x . We also say that f maps (or carries) x into y orf sends (or takes) x to y. A function f from X to Y is usually defined by specifyingits value at each x ∈ X , and if the value at a typical point x ∈ X is f (x), we writex �→ f (x) to give f . We refer to X as the domain, and Y as the codomain of f . Theset f (X) = { f (x) | x ∈ X}, also denoted by im( f ), is referred to as the range of f .

The identity function on X , which sends every element of X to itself, is denotedby 1 or 1X . A map c : X → Y which sends every element of X to a single elementof Y is called a constant function. Notice that the range of a constant functionconsists of just one element. If A ⊂ X , the function i : A ↪→ X , a �→ a, is called theinclusion map of A into X . If f : X → Y and A ⊂ X , then the restriction of f toA is the function f |A : A → Y defined by ( f |A) (a) = f (a) ∀ a ∈ A. The otherway around, if A ⊂ X and g : A → Y is a function, then an extension of g over Xis a function G : X → Y such that G|A = g. The inclusion map i : A ↪→ X is therestriction of the identity map 1 on X to A.

Proposition A.2.1 Let X be a set and {A j | j ∈ J } a family of subsets of X withX =⋃

A j . If, for each j ∈ J , f j : A j → Y is a function such that f j |(

A j ∩ Ak) =


fk |(

A j ∩ Ak)

for all j, k ∈ J , then there exists a unique function F : X → Y whichextends each f j .

Proof Given x ∈ X , there exists an index j ∈ J such that x ∈ A j . We put F (x) =f j (x) if x ∈ A j . If x ∈ A j ∩ Ak , then f j (x) = fk (x), by our hypothesis. Thus F (x)

is uniquely determined by x , and we have a single-valued function F : X → Y . It isclearly an extension of each f j . The uniqueness of F follows from the fact that eachx ∈ X belongs to some A j ; consequently, any function X → Y, which agrees witheach f j on A j , will have to assume the value f j (x) at x . ♦

We say that a function f : X → Y is surjective (or a surjection or onto) ifY = f (X). If f (x) �= f

(x ′

)for every x �= x ′, then we say that f is injective (or

an injection or one-to-one). A function is bijective (or a bijection or a one-to-onecorrespondence) if it is both injective and surjective.

If f : X → Y and g : Y → Z are functions, then the function X → Z whichmapsx into g ( f (x)) is called their composition and denoted by g ◦ f or simply g f .

Proposition A.2.2 Suppose that f : X → Y and g : Y → X satisfy g f = 1X . Thenf is injective and g is surjective.

Given a function f : X → Y , a function g : Y → X such that g f = 1X and f g =1Y is called an inverse of f . If such a function g exists, then it is unique and we denoteit by f −1. It is clear from the preceding result that f has an inverse if and only if itis a bijection. Also, it is evident that

(f −1

)−1 = f .Let f : X → Y be a function. For a set A ⊆ X , the subset

f (A) = { f (x) | x ∈ A}

of Y is called the image of A under f and, for a set B ⊆ Y , the subset

f −1 (B) = {x ∈ X | f (x) ∈ B}

of X is called the inverse image of B in X under f .The power set of a set X is the family P (X) of all subsets of X . A function

f : X → Y induces a functionP (X)→P (Y ), A �→ f (A). It also induces a func-tion P (Y )→P (X), B �→ f −1 (B). The main properties of these functions aredescribed in the following.

Proposition A.2.3 Let f : X → Y be a function.

(a) A1 ⊆ A2 ⇒ f (A1) ⊆ f (A2),(b) f (

⋃j A j ) =⋃

j f (A j ),(c) f (

⋂j A j ) ⊆⋂

j f (A j ), and(d) Y − f (A) ⊆ f (X − A)⇔ f is surjective.

Proposition A.2.4 Let f : X → Y be a function.

(a) B1 ⊆ B2 ⇒ f −1(B1) ⊆ f −1(B2),


(b) f −1(⋃

j B j ) =⋃j f −1(B j ),

(c) f −1(⋂

j B j ) =⋂j f −1

(B j

), and

(d) f −1(Y − B) = X − f −1(B).

Proposition A.2.5 Let f : X → Y be a function. Then:

(a) For each A ⊆ X, f −1 ( f (A)) ⊇ A; in particular, f −1 ( f (A)) = A if f isinjective.

(b) For each B ⊆ Y , f(

f −1 (B)) = B ∩ f (X); in particular, f

(f −1 (B)

) = Bif f is surjective.

A.3 Cartesian Products

In the previous section, we have defined the Cartesian product of two sets. For thisoperation, the following statements are easily proved.

Proposition A.3.1 (a) X1 × (X2 ∪ X3) = (X1 × X2) ∪ (X1 × X3).(b) X1 × (X2 ∩ X3) = (X1 × X2) ∩ (X1 × X3).(c) X1 × (X2 − X3) = (X1 × X2)− (X1 × X3).(d) For Yi ⊆ Xi , i = 1, 2,

(X1 × X2)− (Y1 × Y2) = X1 × (X2 − Y2) ∪ (X1 − Y1)× Y2 =[(X1 − Y1)× (X2 − Y2)] ∪ [Y1 × (X2 − Y2) ∪ [(X1 − Y1)× Y2].

Proposition A.3.2 Let X j , j ∈ J , be a family of subsets of a set X, and let Yk,k ∈ K , be a family of subsets of a set Y. Then

(a) (⋃

j X j )× (⋃

k Yk) =⋃j,k

(X j × Yk

);

(b) (⋂

j X j )× (⋂

k Yk) =⋂j,k

(X j × Yk

);

With J = K,

(c) (⋂

j X j )× (⋂

j Y j ) =⋂j

(X j × Y j

)and

(d) (⋃

j X j )× (⋃

j Y j ) ⊃⋃j

(X j × Y j

).

It is obvious that (X1 × X2)× X3 �= X1 × (X2 × X3) when the sets X1, X2, andX3 are all nonempty. However, there is a canonical bijection

(X1 × X2)× X3 ←→ X1 × (X2 × X3)

given by (x1 × x2)× x3 ↔ x1 × (x2 × x3). Therefore, theCartesian product ofn setsX1, . . . , Xn (n > 2) may be defined by induction as X1 × · · · × Xn = (X1 × . . .×Xn−1)× Xn . A typical element x of X1 × · · · × Xn is written as x = (x1, . . . , xn),where xi ∈ Xi for very i = 1, . . . , n and referred to as the i th coordinate of x .Observethat two elements x = (x1, . . . , xn) and y = (y1, . . . , yn) in X1, . . . , Xn are equal if


and only if xi = yi for very i . Accordingly, an element of X1, . . . , Xn is called anordered n-tuple.

We now extend the definition of the Cartesian product to an arbitrary indexedfamily of sets {Xα | α ∈ A}. Of course, the new definition of the product of the setsXα forα ∈ {1, 2}must reduce to the earlier notionof theCartesianproduct of two sets.Observe that each ordered pair (x1, x2) in X1 × X2 may be considered as defininga function x : {1, 2} → X1 ∪ X2 with x (1) = x1 and x (2) = x2. Accordingly, theCartesian product of the Xα is defined to be the set

∏α Xα (or simply written

as∏

Xα) of all functions x : A →⋃α Xα such that x (α) ∈ Xα for each α ∈ A.

Occasionally, we denote the product∏

α Xα by∏

α∈A Xα or∏{Xα | α ∈ A} to avoid

any ambiguity about the indexing set. We call Xα the αth factor (or coordinate set)of

∏Xα. If x ∈∏

Xα, then x (α) is called the αth coordinate of x , and is oftendenoted by xα.

If the family {Xα} has n sets, n a positive integer, then it may be indexed by the set{1, . . . , n}. In this case, we have two definitions of its Cartesian product; the one inthe present sense will be denoted by

∏ni=1 Xi . Notice that the function

∏2i=1 Xi →

X1 × X2, x �→ (x1, x2), is a bijection. This shows that the notion of the product∏α Xα is a reasonable generalization of the notion of the Cartesian product of two

sets. Similarly, the mapping x �→ (x1, . . . , xn) gives a bijection between∏n

i=1 Xi →X1 × · · · × Xn , and allows us to identify the two products of X1, . . . , Xn .We usuallycall an element x ∈∏n

i=1 Xi an ordered n-tuple. In general, an element x ∈∏α∈A Xα

is referred to as an A-tuple and written as (xα). It is obvious that two elements x, yin

∏α∈A Xα are equal if and only if xα = yα for all α ∈ A. If one of the sets Xα is

empty, then so is∏

Xα. On the other hand, if {Xα} is a nonempty family of nonemptysets, it is not quite obvious that

∏α Xα �= ∅. In fact, a positive answer to the question

of the existence of an element in∏

α Xα is one of the set-theoretic axioms, knownas follows.

Theorem A.3.3 (The Axiom of Choice) If {Xα | α ∈ A} is a nonempty family ofnonempty sets, then there exists a function

c : A →⋃α

Xα

such that c (α) ∈ Xα for each α ∈ A (c is called a choice function for the family{Xα}).

This axiom is logically equivalent to a number of interesting propositions; onesuch proposition is the following.

Theorem A.3.4 (Zermelo’s Postulate) Let {Xα | α ∈ A} be a family of nonemptypairwise disjoint sets. Then there exists a set C consisting of exactly one elementfrom each Xα.

For, if c is a choice function for the family {Xα | α ∈ A} in Theorem A.3.4, thenthe setC = c(A) is a desired set. Conversely, if {Xα | α ∈ A} is a family of nonempty


sets, then Yα = {α} × Xα is nonempty for every α and Yα ∩ Yβ �= ∅ for all α �= βin A. By the above postulate, there exists a set C which consists of exactly oneelement from each Yα. Accordingly, for each α ∈ A, we have a unique xα ∈ Xα suchthat (α, xα) ∈ C. Obviously, C ⊆⋃

α Yα ⊆ A × (⋃α Xα

), and we have a function

c : A →⋃α Xα defined by c(α) = xα for all α ∈ A.

Later, we will discuss some more theorems equivalent to the axiom of choice.It should be noted that the sets Xα in the definition of the Cartesian product∏

α∈A Xα need not to be different from one another; indeed, it may happen that theyare all the same set X . In this case, the product

∏α∈A Xα may be called the Cartesian

product of A copies of X or the Cartesian Ath power of X , and denoted by X A. Noticethat X A is just the set of all functions A → X.

Proposition A.3.5 If Yα ⊆ Xα for every α ∈ A, then∏

α Yα ⊆∏α Xα. Conversely,

if each Xα �= ∅ and∏

α Yα ⊆∏α Xα, then Yα ⊆ Xα for every α.

Proposition A.3.6 Let {Xα | α ∈ A} be a family of nonempty sets, and let Uα, Vα ⊆Xα for every α. Then we have

(a)∏

Uα ∪∏Vα ⊆∏

(Uα ∪ Vα).(b)

∏Uα ∩∏

Vα =∏(Uα ∩ Vα).

If {Xα | α ∈ A} is a family of nonempty sets, then for each β ∈ A, we have themapping pβ :∏α∈A Xα → Xβ given by x �→ xβ . It is easy to see that each of thesemaps is surjective. The map pβ is referred to as the projection onto the βth factor. ForUβ ⊆ Xβ , p−1β

(Uβ

)is the product Uβ ×∏

α �=β Xα, referred to as a slab in∏

Xα.

Proposition A.3.7 If B ⊂ A, then⋂

β∈B p−1β

(Uβ

) =∏Yα, where Yα = Xα if α ∈

A − B while Yα = Uα for α ∈ B.

A.4 Equivalence Relations

Arelation on a set X is a subset R ⊆ X × X . If (x, y) ∈ R,wewrite x Ry. The relationΔ = {(x, x) | x ∈ X} is called the identity relation on X . This is also referred to asthe diagonal. If R is a relation on X , and Y ⊂ X , then R ∩ (Y × Y ) is called therelation induced by R on Y.

Given a set X , a relation R on X is

(a) reflexive if x Rx ∀ x ∈ X (equivalently, Δ ⊆ R),(b) symmetric if x Ry ⇒ y Rx ∀ x, y ∈ X , and(c) transitive if x Ry and y Rz ⇒ x Rz, ∀ x, y, z ∈ X .

An equivalence relation on the set X is a relation which is reflexive, symmetric,and transitive. An equivalence relation is usually denoted by the symbol ∼, read“tilde.” Suppose that ∼ is an equivalence relation on X . Given an element x ∈ X ,the set [x] = {y ∈ X | y ∼ x} is called the equivalence class of x . It is clear that


X equals the union of all the equivalence classes, and two equivalence classes areeither disjoint or identical. A partition of a set X is a collection of nonempty, disjointsubsets of X whose union is X . With this terminology, the family of equivalenceclasses of X determined by ∼ is a partition of X . Conversely, given a partition E ofX , there is an equivalence relation∼ on X such that the equivalence classes of∼ areprecisely the sets of E. This relation is obtained by declaring x ∼ y if both x and ybelong to the same partition set.

Given an equivalence relation∼ on X , the set of all equivalence classes [x], x ∈ X ,is called the quotient set of X by ∼, and is denoted by X/∼. Thus we have anotherimportantmethod of forming new sets fromold ones. Themapπ : X → X/∼ definedby π (x) = [x] is called the projection of X onto X/∼.

If R is a binary relation on a set X , then there is an equivalence relation ∼ on Xdefined by x ∼ y if and only if one of the following is true: x = y, x Ry, y Rx , or thereexist finitely many points z1, . . . , zn+1 in X such that z1 = x , zn+1 = y and eitherzi Rzi+1 or zi+1Rzi for all 1 ≤ i ≤ n. It is called the equivalence relation generatedby R.

A.5 Finite and Countable Sets

For counting objects in a set, we use the natural numbers (or positive integers)1, 2, 3, . . .. When it is feasible, the process of counting a set X requires putting it inone-to-one correspondence with a set {1, 2, . . . , n} consisting of the natural numbersfrom 1 to n. Also, counting of sets in essence serves to determine if one of two givensets has more elements than the other. For this purpose, it may be easier to pair offeach member of one set with a member of the other and see if any elements areleft over in one of the sets rather than count each. The following definition makesprecise the notion of “sets having the same size.” We say that two sets X and Y areequipotent (or have the same cardinality) if there exists a bijection between them.Clearly, the relation of equipotence between sets is an equivalence relation on anygiven collection of sets.

We shall assume familiaritywith the set of natural numbersN = {1, 2, 3, . . .} (alsodenoted by Z+), and the usual arithmetic operations of addition and multiplicationinN. We shall also assume the notion of the order relation “less than”< onN and the“well-ordering property”: Every nonempty subset of N has a smallest element. Forany n ∈ N the set {1, 2, . . . , n} consisting of the natural numbers from 1 to n will bedenoted by Nn . The set Nn is referred to as an initial segment of N. A set X is finiteif it is either empty or equipotent to a set Nn for some n ∈ N. X is called infinite if itis not finite. When X is equipotent to a set Nn we say that X contains n elements orthe cardinality of X is n. The cardinality of the empty set is 0. Of course, we mustjustify that the cardinality of a finite set X is uniquely determined by it. To see this,we first prove the following.


Proposition A.5.1 For any natural numbers m < n, there is no injection from Nn

to Nm.

Proof We use induction on n to establish the proposition. If n = 2, then m = 1, andthe proposition is obviously true. Now, let n > 2 and assume that the propositionis true for n − 1. We show that it is true for n. If possible, suppose that there isan injection f : Nn → Nm , where m < n. If m is not in the image of Nn−1, thenf |Nn−1 is an injection Nn−1 → Nm−1, contrary to our inductive assumption. So wemay further assume that f ( j) = m, j �= n. Then f (n) �= m. We define a mappingg : Nn−1 → Nm−1 by

g (i) ={

f (i) for i �= jf (n) for i = j.

It is clear that g is an injection which, again, contradicts our inductive hypothesis.Therefore, there is no injection Nn → Nm . ♦

The preceding proposition is sometimes called the “Pigeonhole Principle.” Thecontrapositive of the above proposition states that if there is an injection Nm →Nn , then m ≤ n. As an immediate consequence of this, we see that the cardinalityof a finite set X is uniquely determined. For, if f : X → Nn and g : X → Nm arebijections, then the composite f g−1 : Nm → Nn is a bijection. Hence, m ≤ n andn ≤ m which implies that m = n. It is now evident that two finite sets X and Y areequipotent if and only if they have the same number of elements. Notice that forinfinite sets, the idea of having the same “number of elements” becomes quite vague,whereas the notion of equipotence retains its clarity.

Next, we observe the following useful property of finite sets.

Proposition A.5.2 An injective mapping from a finite set to itself is also surjective.

Proof Let X be nonempty finite set, and f : X → X be injective. Let x ∈ X be anarbitrary point. Write x = x0, f (x0) = x1, f (x1) = x2, and so on. Since X is finite,we must have xm = xn for some positive integers m < n. Because f is injective, weobtain x0 = xn−m ⇒ x = f (xn−m−1). Thus f is surjective. ♦Proposition A.5.3 A proper subset of a finite set X is finite and has cardinality lessthan that of X.

Proof If X has n elements, then there is a bijection X → Nn . Consequently, eachsubset of X is equipotent to a subset of Nn , and thus it suffices to prove that everyproper subset ofNn is finite and has atmost n elements.We use induction on n to showthat any proper subset of Nn is finite and has at most n − 1 elements. If n = 1, thenN1 = {1} and its only proper subset is the empty set ∅. Obviously, the proposition istrue in this case. Suppose now that every proper subset of Nk (k > 1) has less thank elements. We show that each proper subset Y of Nk+1 has at most k elements. Ifk + 1 does not belong to Y , then either Y = Nk or it is a proper subset of Nk ; inthe latter case, Y has at most k elements, by the induction assumption. If k + 1 is inY , then Y − {k + 1} ⊂ Nk . Therefore, either Y = {k + 1} or there exists an integer


m < k and a bijection f : Y − {k + 1} → Nm , by the induction assumption again. Inthe latter case, we define g : Y → Nm+1 by setting g (y) = f (y) for all y �= k + 1,and g (k + 1) = m + 1. Clearly g is a bijection. So Y is finite and has m + 1 ≤ kelements. ♦

The contrapositive of this proposition states that if a subset Y ⊆ X is infinite, thenso is X.

Theorem A.5.4 For any nonempty set X, the following statements are equivalent:

(a) X is finite.(b) There is a surjection Nn → X for some n ∈ N.(c) There is an injection X → Nn for some n ∈ N.

Proof (a)⇒ (b): Obvious.(b)⇒ (c): Let f : Nn → X be a surjection. Then f −1 (x) �= ∅ for every x ∈ X.

So, for each x ∈ X , we can choose a number c (x) ∈ f −1 (x). Since { f −1 (x) | x ∈X} is a partition of Nn , the function x �→ c (x) is an injection from X to Nn .

(c)⇒ (a): If there is an injection f : X → Nn , then X is equipotent to f (X) ⊆Nn . By the preceding proposition, there exists a positive integerm ≤ n and a bijectionf (X)→ Nm . It follows that X is equipotent to Nm and hence finite. ♦Proposition A.5.5 If X is a finite set, then there is no one-to-one mapping of X ontoany of its proper subsets.

Proof Let X be a finite set and Y a proper subset of X. Suppose that X has n elements.Then there is a bijection f : X → Nn for some n ∈ N. Also, there is a positiveinteger m < n and a bijection g : Y → Nm , by Proposition A.5.3. If h : X → Y isa bijection, then the composite gh f −1 : Nn → Nm would be an injection, whichcontradicts Proposition A.5.1. ♦

By the preceding proposition, a finite set cannot be equipotent to one of its propersubsets. As n �→ n + 1 is a bijection between N and N− {1}, the set N is not finite.In fact, this is the characteristic property of infinite sets. To establish this, we needthe following.

Theorem A.5.6 (Principle of Recursive Definition) Let X be a set and f : X → Xbe a function. Given a point x0 ∈ X, there is a unique function g : N→ X such thatg (1) = x0 and g (n + 1) = f (g (n)) for all n ∈ N.

This is one of the most useful ways of defining a function on N, which we willtake for granted.

Theorem A.5.7 Let X be a set. The following statements are equivalent:

(a) X is infinite.(b) There exists an injection N→ X.(c) X is equipotent to one of its proper subsets.


Proof (a)⇒ (b): Let F be the family of all finite subsets of X. By the axiomof choice, there exists a function c from the collection of all nonempty subsetsY ⊆ X to itself such that c (Y ) ∈ Y. Since X is not finite, X − F �= ∅ for all F ∈ F.So c (X − F) ∈ X − F. It is obvious that F ∪ {c (X − F)} is a member of F forevery F ∈ F. Accordingly, F �→ F ∪ {c (X − F)} is a function of F into itself.Set G (1) = {c (X)} and G (n + 1) = G (n) ∪ {c (X − G (n))} for every n ∈ N. Bythe principle of recursive definition, G is a function N→ F. We show that thefunction φ : N→ X defined by φ (n) = c (X − G (n)) is an injection. It is easilyseen, by induction, that G (m) ⊆ G (n) for m ≤ n. Therefore, for m < n, we haveφ (m) = c (X − G (m)) ∈ G (m + 1) ⊆ G (n), while φ (n) ∈ X − G (n).

(b)⇒ (c): Let φ : N→ X be an injection, and put Y = φ (N). Consider the map-ping ψ : X → X defined by

ψ (x) ={

φ(1+ φ−1 (x)

)if x ∈ Y, and

x if x /∈ Y.

We assert that ψ is a bijection between X and X − {φ (1)}. The surjectivity ofψ is clear. To show that it is injective, assume that ψ (x) = ψ

(x ′

). Then both x

and x ′ belong to either Y or X − Y . If x, x ′ ∈ Y, then we have φ(1+ φ−1 (x)

) =φ

(1+ φ−1

(x ′

)), and it follows from the injectivity of φ that x = x ′. And, if

x, x ′ ∈ X − Y, then x = x ′, by the definition of ψ.(c)⇒ (a): This is contrapositive of Proposition A.5.5. ♦A set X is called countably infinite (or denumerable, or enumerable) if X is

equipotent to the set N of natural numbers. X is called countable if it is either finiteor denumerable; otherwise, it is called uncountable.

For example, the set Z of all integers is countably infinite, since f : N→ Z,defined by

f (n) ={

n/2 if n is even, and− (n − 1) /2 if n is odd,

is a bijection.We conclude from Theorem A.5.7 that every infinite set contains a countably

infinite subset. The following proposition shows that no uncountable set can be asubset of a countable set.

Proposition A.5.8 Every subset of N is countable.

Proof Let M ⊆ N. If M is finite, then it is countable, by definition. If M = N, thenM is a countably infinite set, since the identity function from N to M is a bijection.So assume that M is an infinite proper subset of N. Then we define a function fromN to M by recursion as follows: Let f (1) be the smallest integer in M , whichexists by the well-ordering property of N. Suppose that we have chosen integersf (1), f (2), . . . , f (k). Since M is infinite, M − f {1, 2, . . . , k} is nonempty forevery k ∈ N.Wedefine f (k + 1) to be the smallest integer of this set. By the principleof recursive definition, f (k) is defined for all k ∈ N. Clearly, f (1) < f (2) < · · ·


so that f is injective. Now, we see that f is surjective, too. For each m ∈ M , theset {1, 2, . . . , m} is finite. Since the set { f (k) | k ∈ N} is infinite, there is a k ∈ N

such that f (k) > m. So we can find the smallest integer l ∈ N such that f (l) ≥ m.Then f ( j) < m for each j < l; accordingly, m /∈ f {1, . . . , l − 1}. By the definitionof f (l), we have f (l) ≤ m, and the equality m = f (l) follows. Thus f : N→ Mis a bijection. ♦

It follows that a subset of N is either finite or countably infinite. Note that anyset which is equipotent to a countable set is countable. Accordingly, any subset of acountable set is countable.

Theorem A.5.9 For any nonempty set X, the following statements are equivalent:

(a) X is countable.(b) There is a surjection N→ X.(c) There is an injection X → N.

Proof (a)⇒ (b): If X is countably infinite, then there is a bijection between N

and X , and we are done. If X is finite, then there exists an n ∈ N and a bijectionf : {1, . . . , n} → X. We define g : N→ X by setting

g (m) ={

f (m) for 1 ≤ m ≤ n, andf (n) for m > n.

Clearly, g is a surjection.(b)⇒ (c): Let g : N→ X be a surjection. Then g−1 (x) is nonempty for every x ∈

X . So it contains a unique smallest integer h (x), say. Then x �→ h (x) is a mappingh : X → N, which is injective, since g−1 (x) ∩ g−1 (y) �= ∅ whenever x �= y.

(c)⇒ (a): Suppose that h : X → N is an injection. Then h : X → h (X) is abijection. By Proposition A.5.8, h (X) is countable, and therefore X is countable. ♦

By definition, a countable set is the range of a (finite or infinite) sequence, andthe converse follows from the preceding theorem. Thus the elements of a countableset X can be listed as x1, x2, . . ., and such a listing is called an enumeration of X.Observe that the range of a function of a countable set is countable, and the domainof an injective function into a countable set is countable.

Lemma A.5.10 For any finite number factors, N× · · · × N is countably infinite.

Proof Let 2, 3, . . . , pk be the first k prime numbers, where k is the number of factorsin N× · · · × N. Define a mapping f : N× · · · × N→ N by f (n1, n2, . . . , nk) =2n13n2 · · · pnk

k . By the fundamental theorem of arithmetic, f is injective, and henceN× · · · × N is countable. It is obvious that n �→ (n, 1, . . . , 1) is an injection N→N× · · · × N so that N× · · · × N is infinite. ♦

It is immediate from the preceding lemma that a finite product of countable sets Xi

is countable, for if fi : N→ Xi are surjections for 1 ≤ i ≤ k, then so is the mapping


(n1, . . . , nk) �→ ( f1 (n1) , . . . , fk (nk)) of N× · · · × N into X1 × · · · × Xk . Noticethat if each Xi is nonempty and some X j is countably infinite, then

∏k1 Xi is also

countably infinite, since x j �→(

x01 , . . . , x0

j−1, x j , x0j+1, . . . , x0

k

), where x0

i ∈ Xi are

fixed elements for i �= j , is an injection X j →∏k1 Xi . We also see that if each Xi

is finite, then so is∏k

1 Xi . In fact, if any X j = ∅, then∏k

1 Xi = ∅, and, if Xi hasmi > 0 elements, then

∏k1 Xi contains m1 · · ·mk elements. By induction on k, it

suffices to establish this proposition in the case k = 2. Let Nmi = {1, 2, . . . , mi }and fi : Nmi → Xi be bijections, i = 1, 2. Then f1 × f2 : Nm1 × Nm2 → X1 × X2,(a, b) �→ ( f1 (a) , f2 (b)) is a bijection. Define a mapping g : Nm1 × Nm2 → Nm1m2

by g (a, b) = (a − 1) m2 + b. It is easily verified that g is a bijection, so X1 × X2

has m1m2 elements.However, an infinite product of even finite sets is not countable. For example, let

X = {0, 1}, and consider the countable product XN =∏∞1 Xn , where each Xn = X .

Note that an element in XN is a sequence s : N→ X. Let f : N→ XN be any function.Write tn = 1− f (n)n . Then t = 〈tn〉 ∈ XN and t �= f (n) for all n ∈ N. Thus f isnot surjective. Since f is an arbitrary function N→ XN, XN cannot be countable,by Theorem A.5.9.

This method of proof was first used by G. Cantor, and is known as Cantor’sdiagonal process. We use this technique to prove the following

Example A.5.1 The setRof all real numbers is uncountable. Since anuncountable setcannot be a subset of a countable set, it is enough to show that the unit interval I ⊂ R isuncountable. ByTheoremA.5.9, it suffices to show that there is no surjectivemappingf : N→ I . Let f : N→ I be any function. We use the decimal representation ofreal numbers to write f (n) = 0.an1an2 · · · . Here each ani is a digit between 0 and9. This representation of a number is not necessarily unique, but if a number has twodifferent decimal representations, then one of these representations repeats 9s fromsome place onward and the other repeats 0s from some place onward. We definea new real number r whose decimal representation is 0.b1b2 · · · , where bn = 3 ifann �= 3, and bn = 5 otherwise. It is clear that r has a unique decimal representationand differs from f (n) in the nth decimal place for every n ∈ N. So r �= f (n) for alln ∈ N. Obviously, r belongs to I , and thus f is not surjective.

Proposition A.5.11 The union of a countable family of countable sets is countable.

Proof Let A be a countable set, and suppose that for every α ∈ A, Eα is a countableset. Put X =⋃

Eα. We show that X is countable. If X = ∅, there is nothing toprove. So we assume that X �= ∅. Then A �= ∅, and we may further assume thatEα �= ∅ for every α ∈ A, since the empty set contributes nothing to the union ofthe Eα. Since A is countable, there is a surjection N→ A, n �→ αn . Since Eαn is anonempty countable set, there is a surjection fn : N→ Eαn . Write fn (m) = xnm forevery m ∈ N. Then we have a mapping φ : N× N→ X defined by φ (n, m) = xnm .Clearly, φ is surjective, for X =⋃

n Eαn . By Lemma A.5.10, N× N is countable,and hence X is countable. ♦


If the indexing set A and the sets Eα in the preceding proposition are all finite,then it is easily verified that the union

⋃α Eα is finite.

Example A.5.2 The setQ of all rational numbers is countably infinite. LetQ+ denotethe set of all positive rationals, and Q− denote the set of all negative rationals. ThenQ = Q+ ∪Q− ∪ {0}. Obviously,Q+ andQ− are equipotent, and therefore it sufficesto prove thatQ+ is countably infinite. AsN ⊂ Q+,Q+ is infinite. There is a surjectivemapping g : N× N→ Q+ defined by g (m, n) = m/n. By Lemma A.5.10, thereis a bijection f : N→ N× N, so the composition g f : N→ Q+ is surjective. ByTheorem A.5.9, Q+ is countable.

Since R is uncountable, we see that the set R−Q of all irrational numbers isuncountable.

Theorem A.5.12 The family of all finite subsets of a countable set is countable.

Proof Let X be a countable set, andF (X) be the family of all finite subsets of X. IfX is a finite set having n elements, then every subset of X is finite, andF (X) has 2n

members. If X is countably infinite, then there is a bijection betweenF (X) and thefamilyF (N) of all finite subsets ofN. So it suffices to prove thatF (N) is countablyinfinite. It is obvious that n �→ {n} is an injection N→ F (N), soF (N) is infinite.To see that it is countable, consider the sequence 〈2, 3, . . . , pk, . . .〉of primenumbers.If F = {n1, n2, . . . , nk} ⊂ N, then the ni can be indexed so that n1 < n2 < · · · < nk .Put ν (F) = 2n13n2 · · · pnk

k . Then F �→ ν (F) defines an injection F (N)− {∅} →N, by the fundamental theorem of arithmetic. By Theorem A.5.9, F (N)− {∅} iscountable, and soF (N) is countable. ♦

It must be noted that the family of all subsets of a countably infinite set is notcountable. This follows from the following.

Theorem A.5.13 (Cantor (1883))For any set X, there is no surjection X →P (X).

Proof Assume that there is a surjection f : X →P (X) and consider the set S ={x ∈ X | x /∈ f (x)}. Then S ⊆ X , so there exists an x ∈ X such that S = f (x). Now,if x ∈ S, then x /∈ f (x), and if x /∈ S, then x ∈ f (x). Thus, in either case, we obtaina contradiction, and hence the theorem. ♦

We end this section with the following theorem, which will be proved later inSect. A.8.

Theorem A.5.14 (Bernstein–Schröeder) Let X and Y be sets. If there exists injec-tions X → Y and Y → X, then there exists a bijection X → Y .

A.6 Orderings

Definition A.6.1 Let X be a set. An order (or a simple order) on X is a binaryrelation, denoted by ≺, such that


(a) if x, y ∈ X , then one and only one of the statements x = y, x ≺ y, y ≺ x istrue, and

(b) x ≺ y and y ≺ z ⇒ x ≺ z.

X together with a definite order relation defined in it is called an ordered set.

The statement “x ≺ y” is read as “x precedes y” or “y follows x .” We also saythat “x is less than y” or “y is greater than x .” It is sometimes convenient to writey � x to mean x ≺ y. The notation x � y is used to indicate x ≺ y or x = y, that is,the negation of y ≺ x. It should be noticed that an order relation is irreflexive, thatis, for no x x ≺ x . However, if ≺ is an order on X, then the relation � satisfies thefollowing conditions:

(a) (Reflexivity) x � x ∀ x ∈ X ;(b) (Antisymmetry) x � y and y � x ⇒ x = y;(c) (Transitivity) x � y and y � z ⇒ x � z; and(d) (Comparability) x � y or y � x ∀ x, y ∈ X.

A relation having the properties (a)–(c) is called a partial ordering. A set togetherwith a definite partial ordering is called a partially ordered set. Two elements x, y ofa partially ordered set (X,�) are called comparable if x � y or y � x, and a subsetY ⊆ X in which any two elements are comparable is called a chain in X. A partiallyordered set that is also a chain is called a linearly or totally ordered set. Thus wehave associated with each simple order a total (or linear) order relation.

Conversely, there is a simple order relation associated with each total ordering.In fact, to any partial order relation � on X, there is associated a unique relation ≺given by

x ≺ y ⇔ x � y and x �= y.

It is transitive and has the property that (a) for no x ∈ X , the relation x ≺ x holds, and(b) if x ≺ y, then y ⊀ x for every x, y ∈ X . A transitive relation with this propertyis called a strict partial ordering. If � totally orders the set X , then ≺ is clearly asimple order (or a strict total ordering) on X. It follows that the notions of a total(resp. partial) order relation and a strict total (resp. strict partial) order relation areinterchangeable. If� is a partial order, we use the notation≺ to denote the associatedstrict partial order, and conversely. The relation ≤ is a total ordering on the set R

of real numbers, while the inclusion relation ⊆ on P (X) is not if X has more thanone element. Note that the set inclusion is always a partial ordering on any family ofsubsets of a set X.

It is obvious that the induced ordering on a subset of a partially ordered set is apartial ordering on that subset. If (X1,�1) and (X2,�2) are partially (totally) orderedsets, then the dictionary (or lexicographic) order relation on X1 × X2 defined by

(x1, x2) � (y1, y2) if x1 ≺1 y1 or if x1 = y1 and x2 �2 y2

is a partial (total) ordering.


Let (X,�) be a partially ordered set. If Y ⊆ X , then an element b ∈ X is an upperbound of Y if for each y ∈ Y, y � b. If there exists an upper bound of Y, then wesay that Y is bounded above. A least upper bound or supremum of Y is an elementb0 ∈ X such that b0 is an upper bound of Y and if c ≺ b0, then c is not an upperbound of Y . Observe that there is at most one such b0, by antisymmetry, and it may ormay not exist. We write b0 = sup Y when it exists. Notice that sup Y may or may notbelong to Y. If sup Y ∈ Y, it is referred to as the last (or largest or greatest) elementof Y. Analogously, an element a ∈ X is a lower bound of Y if a � y for all y ∈ Y. Ifthere exists a lower bound of Y, we say that Y is bounded below. A greatest lowerbound or infimum of Y is an element a0 ∈ X which is a lower bound of Y and ifa0 ≺ c, then c is not a lower bound of Y . It is clear that there is at most one such a0,and it may or may not exist. When it exists, we write a0 = inf Y. If inf Y ∈ Y, wecall it the first (or smallest or least) element of Y. It is obvious that sup Y is the firstelement of the set {x ∈ X | y � x for all y ∈ Y }, and inf Y is the last element of theset {x ∈ X | x � y for all y ∈ Y }.

A partially ordered set X is said to have the least upper bound property if eachnonempty subset Y ⊆ X with an upper bound has a least upper bound. Analogously,X is said to have the greatest lower bound property if each nonempty subset Y ⊆ Xwhich is bounded below has a greatest lower bound. If X is a set, thenP (X) orderedby inclusion has the least upper bound property.

Proposition A.6.2 If an ordered set X has the least upper bound property, then italso has the greatest lower bound property.

An ordered set X which has the least upper bound (equivalently, the greatest lowerbound) property is called order complete.

Let (X,�) be a partially ordered set. An element a ∈ X is called a minimalelement of X if for every x ∈ X , x � a ⇒ x = a, that is, no x ∈ X which is distinctfrom a precedes a. Similarly, an element b ∈ X is called a maximal element of X iffor every x ∈ X , b � x ⇒ x = b, that is, no x ∈ X which is distinct from b followsb. It should be noted that if X has a last (first) element, then that element is the uniquemaximal (minimal) element of X . If � is a total order, a maximal element is the lastelement, but there are partially ordered sets with unique maximal elements whichare not last elements.

If x, y ∈ X and x ≺ y, we say that x is a predecessor of y (or y is a successorof x). If x ≺ y and there is no z ∈ X such that x ≺ z ≺ y, then we say that x is animmediate predecessor of y (or y an immediate successor of x).

Ordinals

A partially ordered set X is called well ordered (or an ordinal) if each nonemptysubset A ⊆ X has the first element.

The set N of the positive integers is well ordered by its natural ordering ≤. Onthe other hand, the set R of real numbers is not well ordered by the usual ordering≤. And, if X has more than one element, then the partial ordering⊆ inP (X) is nota well ordering.


Clearly, any subset of a well-ordered set is well ordered in the induced ordering.The empty set ∅ is considered to be a well-ordered set. If (X1,�1) and (X2,�2) arewell-ordered sets, then the dictionary order relation on X1 × X2 is a well ordering.

It was E. Zermelo who first formulated the axiom of choice (though it was beingused by mathematicians without explicit formulation) and established that every setcan be well ordered (1904). In fact, we have the following.

Theorem A.6.3 The following statements are equivalent:

(a) The Axiom of Choice.(b) Well-Ordering Principle: Every set can be well ordered.(c) Zorn’s Lemma: Let X be a partially ordered set. If each chain in X has an

upper bound, then X has a maximal element.(d) Hausdorff Maximal Principle: If X is a partially ordered set, then each

chain in X is contained in a maximal chain, that is, for each chain C in X,there exists a chain M in X such that C ⊆ M and M is not properly containedin any other chain which contains C.

We do not wish to prove this theorem and refer the interested reader to the textsby Dugundji [3] and Kelley [6]. We remark that there are several other statementsequivalent to the axiom of choice individually.

We note that a well-ordered set (W,�) is totally ordered. For, if x, y ∈ W, then thesubset {x, y} ⊆ W has a first element; accordingly, either x � y or y � x . Moreover,W is order complete. In fact, given a nonempty set X ⊆ W with an upper bound, thefirst element of the set of all upper bounds of X in W is sup X.

Each element x of W that is not the last element of W has an immediate successor.This is first element of the set {y ∈ W | x ≺ y}, usually denoted by x + 1. However,x need not have an immediate predecessor. For example, consider the setN of positiveintegers with the usual ordering ≤. Choose an element ∞ /∈ N, and put W = N ∪{∞}. Define a relation � on W by x � y if x, y ∈ N and x ≤ y, x � ∞ for allx ∈ N, and∞ �∞. Then (W,�) is a well-ordered set, and∞ has no immediatepredecessor in W . It is obvious that∞ is the last element of W. We say that N ∪ {∞}has been obtained from N by adjoining∞ as the last element.

For each x ∈ W, the set S (x) = {w ∈ W | w ≺ x} is called the section (or initialsegment) determined by x . The first element of W is usually denoted by 0; it isobvious that S (0) = ∅. Each section S (x) in W obviously satisfies the condition:y ∈ S (x) and z � y ⇒ z ∈ S (x). Conversely, if S is a proper subset of W satisfyingthis condition, then S is a section in W . For, if x is the least element of W − S, theny ≺ x ⇔ y ∈ S. So S = S (x). It is clear that a union of sections in W is either asection in W or equals W. Also, it is immediate that an intersection of sections in Wis a section in W.

A.6.4 (Principle of Transfinite Induction) Let (W,�) be a well-ordered set. If X isa subset of W such that S (y) ⊆ X implies y ∈ X for every y ∈ W , then X = W .

Proof The first element 0 of W is in X, since ∅ = S (0) ⊂ X . If W − X �= ∅, thenit has a first element y0, say. So S (y0) ⊆ X , which forces y0 ∈ X, by our hypothesis.This contradiction establishes the theorem. ♦


The preceding theorem is generally used in the following form: For each x ∈ W,let P (x) be a proposition. Suppose that (a) P (0) is true, and (b) for each x ∈ W,P (x) is also true whenever P (y) is true for all y ∈ S (x). Then P (x) is true forevery x ∈ W.

Since each element of N other than 1 has an immediate predecessor, the inductionprinciple on N is equivalent to the following statement.

Let P (n) be a proposition defined for each n ∈ N. If P (1) is true, and for eachn > 1, the hypothesis “P (n − 1) is true” implies that P (n) is true, then P (n) is truefor every n ∈ N.

Let (W,�) and(W ′,�′) be well-ordered sets. A mapping f : W → W ′ is said

to be order-preserving if f (x) �′ f (y) whenever x � y in W. We call W and W ′isomorphic (or of the same order type) if there is a bijective order-preserving mapf : W → W ′; such a map f is called an isomorphism. An order-preserving injectionf : W → W ′ is called a monomorphism. It should be noted that an isomorphism ofa partially ordered set X onto a partially ordered set X ′ is a bijection f : X → X ′such that x � y ⇔ f (x) �′ f (y). However, if X and X ′ are totally ordered, oneneeds only the implication x � y ⇒ f (x) �′ f (y). For, this also implies the reverseimplication f (x) �′ f (y)⇒ x � y: If x � y, then y ≺ x ⇒ f (y) ≺′ f (x), con-tradicting the antisymmetry of �′.

The ordinal N is isomorphic to the well-ordered set ω = {0, 1, 2, . . .} of nonneg-ative integers in its natural order. We note that a well-ordered set may be isomorphicto one of its proper subsets. For example, the well-ordered set N is isomorphic to theset E of even integers; n �→ 2n is a desired isomorphism. However, no well-orderedset W can be isomorphic to a section in it; this is immediate from the following fact.

Proposition A.6.5 If f is a monomorphism of a well-ordered set W into itself, thenf (w) � w for every w ∈ W.

Proof Let X = {x ∈ W | f (x) ≺ x}. If X �= ∅, then it has a first element, say x0. Asx0 ∈ X , we have f (x0) ≺ x0 whence f ( f (x0)) ≺ f (x0). This implies that f (x0) ∈X , which contradicts the definition of x0. Therefore X = ∅. ♦Corollary A.6.6 Two sections in a well-ordered set W are isomorphic if and onlyif they are identical.

Proof Suppose that S (x) and S(x ′

)are isomorphic sections inW and let f : S (x)→

S(x ′

)be an isomorphism. If S (x) �= S

(x ′

), then we have either x ≺ x ′ or x ′ ≺ x .

By interchanging S (x) and S(x ′

), if necessary, we may assume that x ′ ≺ x . Then

x ′ ∈ S (x) which implies that f(x ′

) ≺ x ′. This contradicts Proposition A.6.5. ♦Corollary A.6.7 The only one-to-one order-preserving mapping of a well-orderedset onto itself is the identity mapping.

Proof Suppose that W is a well-ordered set and f : W → W is an isomorphism.If the set S = {x ∈ W | f (x) �= x} is nonempty, then S has a least element a, say.By PropositionA.6.5, we have a ≺ f (a). Since f is surjective, a = f (x) for somex ∈ W . So f (x) ≺ f (a)⇒ x ≺ a. Then, by the definition of a, f (x) = x and weobtain a = x ≺ a, a contradiction. Therefore S = ∅, and f is the identity mapping.


Theorem A.6.8 If X and Y are well-ordered sets, then exactly one of the followingstatements holds.

(a) X is isomorphic to Y.(b) X is isomorphic to a section in Y.(c) Y is isomorphic to a section in X.

Proof We first show that the three possibilities are mutually exclusive. If (a) and(b) occur together, then Y is isomorphic to a section in it, a contradiction. For thesame reason, (a) and (c) cannot occur together. If f : X → S (y) and g : Y → S (x)

are isomorphisms, then g f : X → S (x) is a monomorphism with g f (x) ≺ x . Thiscontradicts PropositionA.6.5, so (b) and (c) also cannot occur together.

Now, we prove that one of the above three possibilities does occur. Suppose thatneither of (a) and (b) holds, and let Σ be the family of all sections S (xα) in Xsuch that there exists an isomorphism fα of S (xα) onto a section in Y or onto Yitself. For every pair of indices α and β, S (xα) ∩ S

(xβ

)is a section in X , say S(xγ),

and fα(S

(xγ

))and fβ

(S

(xγ

))are clearly isomorphic. If α �= β, then one of these

is certainly a section in Y, and hence the other is also a section in Y. Therefore, byCorollary A.6.6, we have fα

(S

(xγ

)) = fβ(S

(xγ

)), and then Corollary A.6.7 shows

that fα (x) = fβ (x) for all x ∈ S(xγ

). Hence, there is a function

f :⋃α S (xα)→⋃α fα (S (xα))

defined by f (x) = fα (x) if x ∈ S (xα). It is easily checked that f is an isomorphism.Moreover,

⋃α S (xα) is either X or a section in it, and

⋃α fα (S (xα)) is either Y or

a section in Y. If⋃

α S (xα) = X, then we have alternative (a) or (b), contrary to ourassumption. So

⋃α S (xα) = S (x0) for some x0 ∈ X , and S (x0) ∈ Σ . We assert that

f : S (x0)→ Y is surjective. Assume otherwise. Then we have f (S (x0)) = S (y0)for some y0 ∈ Y. Obviously, we can extend f to an isomorphism S (x0) ∪ {x0} →S (y0) ∪ {y0} by defining f (x0) = y0. Note that S (y0) ∪ {y0} is either a section inY or coincides with Y (if y0 is the last element of Y ). If x0 were the last element ofX , then we would have alternative (a) or (b). Therefore S (x0) ∪ {x0} is a memberof the family Σ. Then, from the definition of S(x0), it follows that x0 ∈ S(x0), acontradiction. Hence, our assertion and the alternative (c) holds. ♦Corollary A.6.9 Any subset of a well-ordered set W is isomorphic to either a sectionin W or W itself.

Proof Let X be a subset of W. Then X is a well-ordered set under the inducedorder. If there is an isomorphism f of W onto a section S (x0) in X, then thereexists a monomorphism g : X → X defined by f , which satisfies g (x0) ≺ x0. Thiscontradicts Proposition A.6.5. Therefore, one of the two possibilities (a) and (b) inTheorem A.6.8 must hold. ♦


A.7 Ordinal Numbers

It is evident that the relation of isomorphism between ordinals is an equivalencerelation.Ordinal numbers are objects uniquely associatedwith the isomorphic classesof well-ordered sets. A natural way to define them is to consider these isomorphicclasses themselves as the ordinal numbers. The main drawback with this definition isthat isomorphic classes of well-ordered sets are unfortunately not sets; accordingly,logical contradictions arise when such large classes are collected into sets. To avoidsuch difficulties, ordinal numbers are defined to be well-ordered sets such that eachisomorphic class of well-ordered sets contains exactly one ordinal number.

Definition A.7.1 An ordinal number is a well-ordered set (W,�) such that

(a) if X ∈W and x ∈ X , then x ∈W,(b) for every X, Y ∈W, one of the possibilities: X = Y , X ∈ Y or Y ∈ X holds,

and(c) X � Y ⇔ X ∈ Y or X = Y.

The fact that ordering� in the above definition is actually a well ordering followsfrom one of the axioms in set theory:

Every nonempty set X contains an element y such that x /∈ y for all x ∈ X. (∗)As a consequence of this axiom, we see that no nonempty set can be a member of

itself, and both the relations X ∈ Y and Y ∈ X cannot hold simultaneously for anytwo sets X , Y. So only one of the three alternatives described in the condition (b) canoccur. It is now obvious that� is reflexive and antisymmetric. To see the transitivityof this relation, suppose that X � Y and Y � Z in W. Then we must have one ofthe relations X = Z , X ∈ Z or Z ∈ X . If Z ∈ X , then we cannot have X = Y orY = Z , for Y /∈ X and Z /∈ Y . So X ∈ Y and Y ∈ Z . But, this leads to the violationof Axiom (∗) by the set {X, Y, Z}. Hence X � Z . Using the above axiom, we alsosee that every nonempty subset of W has a least member.

The empty set ∅ is obviously an ordinal number. To construct a few more ordinalnumbers, observe that if W is an ordinal number, then so is W ∪ {W}. Thus {∅},{∅, {∅}}, {∅, {∅}, {∅, {∅}}}, etc. are first few ordinal numbers. Henceforth, we willgenerally denote ordinal numbers by small Greek letters.

Proposition A.7.2 Every element of an ordinal number is an ordinal number.

Proof Let β be an ordinal number and α ∈ β. We show that α is also an ordinalnumber. Suppose that x ∈ y and y ∈ α. Then we have either x = α or x ∈ α orα ∈ x . If x = α, then we would have x ∈ y and y ∈ x , a contradiction. Now, ifα ∈ x , then the subset {x, y,α} ⊂ β fails to have a least element. So x ∈ α, and thecondition DefinitionA.7.1(a) holds. Next, suppose that x, y ∈ α. Then both x, y aremembers of β. So one of the possibilities x = y, x ∈ y or y ∈ x must hold. Thus thecondition (b) in DefinitionA.7.1 is satisfied by α. ♦Lemma A.7.3 If α and β are ordinal numbers, then α ∈ β if and only if α ⊂ β.


Proof Ifα ∈ β, then x ∈ α⇒ x ∈ β, by DefinitionA.7.1. Moreover, α = β ⇒ α ∈α, a contradiction. So we have α ⊂ β. Conversely, assume that α ⊂ β. Then β − αis nonempty and has a least element λ, say. We observe that α = λ. Suppose thatx ∈ λ. Then x ∈ β, for λ ∈ β. If x /∈ α, then we have λ � x . This implies that eitherλ = x or λ ∈ x , and contradicts the fact that λ is an ordinal number, by A.7.2. Onthe other hand, if x ∈ α, then x �= λ and λ /∈ x , for λ /∈ α. Since both x and λ aremembers of β, we have x ∈ λ. Thus α = λ ∈ β. ♦Proposition A.7.4 If α and β are any two ordinal numbers, then α ⊆ β or β ⊆ α.

Proof Suppose that α and β are two ordinal numbers. Then γ = α ∩ β = {x | x ∈α and x ∈ β} is an ordinal number. If γ �= α,β, then γ ⊂ α and γ ⊂ β. By thepreceding lemma, we have γ ∈ α and γ ∈ β. Thus γ ∈ α ∩ β = γ, a contradiction.So γ = α or γ = β; accordingly, we have α ⊆ β or β ⊆ α. ♦Theorem A.7.5 (a) Every nonempty set of ordinal numbers has a least element

with respect to inclusion.(b) The union of a set of ordinal numbers is an ordinal number.

Proof (a): Let Σ be a nonempty set of ordinal numbers. Choose α ∈ Σ . If T =α ∩Σ is empty, then σ ∈ Σ implies that σ �⊂ α, by Lemma A.7.3. Hence α ⊆ σ,and α is the least element of Σ . If T �= ∅, then T contains an element β such thatβ ∈ τ or β = τ for every τ ∈ T , sinceα is well ordered. Thus β ⊆ τ for every τ ∈ T .And, if σ ∈ Σ − T , then we clearly have β ⊂ α ⊆ σ. Thus, in this case, β is theleast element of Σ .

(b): Let Σ be a set of ordinal numbers. Then X =⋃α∈Σ α is a set. It is obvious

that the condition (a) of DefinitionA.7.1 holds in X . Furthermore, if x, y ∈ X , thenthere exist ordinal numbers α,β ∈ Σ with x ∈ α and y ∈ β. By Proposition A.7.4,we have α ⊆ β or β ⊆ α. To be specific, suppose that α ⊆ β. Then x, y ∈ β and,therefore, they satisfy the condition (b) of DefinitionA.7.1. Lastly, we show that X iswell ordered by the relation x � y ⇔ x ∈ y or x = y. It is obvious that� is a linearordering on X and, by Proposition A.7.2, every element of X is an ordinal number.Then, for each nonempty subset Y of X , the part (a) of the theorem shows that thereis an element z ∈ Y such that z ⊆ y for all y ∈ Y . By Lemma A.7.3, we deduce thatz is the least element of Y . Thus (b) holds. ♦

Given two ordinal numbers α,β, we write α ≤ β if and only if α ⊆ β. Then,by Proposition A.7.4, we have either α ≤ β or β ≤ α. Moreover, it is immediatefrom PropositionA.7.5 that any nonempty set of ordinal numbers is well ordered bythe relation ≤. We also see, by Proposition A.7.2 and Lemma A.7.3, that an ordinalnumber α consists of all those ordinal numbers which precede it. It follows that anytwo distinct ordinal numbers cannot be isomorphic. For, ifα < β, thenα is the section{γ ∈ β | γ < α} in β. Therefore, by Proposition A.6.5, α cannot be isomorphicto β.

Notice that the classO of all ordinal numbers is not a set. For, ifO were a set, thenX =⋃

α∈O αwould be an ordinal number, by the preceding theorem. Consequently,T = X ∪ {X} is an ordinal number, and we obtain the contradiction X < T ≤ X .


Theorem A.7.6 Every well-ordered set W is isomorphic to a unique ordinal number.

Proof Clearly, if W is isomorphic to two ordinal numbers α and β, then they them-selves are isomorphic, and we must have α = β.

To find the desired ordinal number, let X be the collection of all elements x ∈ Wfor which there are ordinal numbers αx (depending upon x) such that the sectionS (x) in W is isomorphic to αx . Put Ψ (x) = αx . If x, y ∈ X and x = y, then theordinal numbers αx and αy are isomorphic. Hence αx = αy , and it follows thatΨ is single-valued on X . By an axiom in set theory, we see that X is a set, andso is im(Ψ ). Observe that Ψ is injective, too. For, if Ψ (x) = Ψ (y), then S (x)

and S (y) are isomorphic. By Corollary A.6.6, we deduce that S (x) = S (y), whichforces x = y. Thus Ψ : X → im(Ψ ) is a bijection. We show that it is, in fact, anisomorphism. Denote the ordering in W by�, and suppose that x, y ∈ X and y ≺ x .Let f : S (x)→ Ψ (x) be an isomorphism. Then f (y) < Ψ (x), and f induces anisomorphism of S (y) onto {α ∈ Ψ (x) | α < f (y)} = f (y). By the definition ofΨ , Ψ (y) = f (y) < Ψ (x). Conversely, suppose that β = Ψ (y) < Ψ (x) and letg : Ψ (x)→ S (x) be an isomorphism. Then β ∈ Ψ (x) and g|β is an isomorphism ofβ onto S (g (β)). SoΨ (g (β)) = Ψ (y). SinceΨ is injective, we have y = g (β) ≺ x .This proves our claim.

We next show that im(Ψ ) is an ordinal number. Since the class O of all ordinalnumbers is not a set, we have O �= im(Ψ ). So there exists an ordinal number ν /∈im(Ψ ). If ν contains elements which do not belong to im(Ψ ), then we have sucha least element λ, by TheoremA.7.5(a); otherwise, we put λ = ν. We observe thatim(Ψ ) = {α ∈ O | α < λ} = λ. Clearly, α < λ⇒ α ∈ im(Ψ ). On the other hand,if α ∈ im(Ψ ), then there exists an x ∈ W and an isomorphism g : Ψ (x) = α→S (x). Consequently, for every β < α, y = g (β) ≺ x and g induces an isomorphismbetween β and S(y). So β = Ψ (y) ∈ im(Ψ ), and this implies that α < λ. Thus haveim(Ψ ) = λ.

To finish the proof, we need to show that X = W . If X �= W , then W − X hasa least element, say w0. We observe that X = S (w0). Suppose x ∈ X , w ∈ W andw ≺ x . Then there exists an isomorphism f : S(x)→ Ψ (x), which clearly inducesan isomorphism between S(w) and the ordinal number f (w). Therefore, w ∈ X andwe conclude that X is the section S (w0) in W . Since Ψ : X → λ is an isomorphism,we have w0 ∈ X , a contradiction. ♦

By the preceding theorem, it is clear that each well-ordered set determines aunique ordinal number.

If an ordinal number α is nonempty, then the least element a of α must be ∅,for x ∈ a would imply that x ∈ α and x < a. It follows that ∅ is the least ordi-nal number, denoted by 0. We denote the ordinal number {∅} by 1, {∅, {∅}} by2, {∅, {∅}, {∅, {∅}}} by 3, and so on. Notice that the ordinal number denoted bya positive integer n is determined by the well-ordered set {0, 1, . . . , n − 1}. More-over, we see that 1 is the immediate successor of 0, 2 is the immediate successorof 1, etc. In general, for each ordinal number α, α ∪ {α} is an ordinal number,


which is the least ordinal number > α. Thus α ∪ {α} is the immediate succes-sor of α, usually denoted by α+ 1. However, there are ordinal numbers, such asω =⋃∞

n=1 n, which do not have immediate predecessors, and they are called limitordinal numbers. Observe that ω is isomorphic to the well-ordered set of all non-negative integers with the usual ordering, and each ordinal number α < ω containsa finite number of elements. We refer to an ordinal number containing a finite num-ber of elements as a finite ordinal number. It is obvious that a finite ordinal numbercoincides with some ordinal number n, and thus ω is the smallest ordinal numbergreater than every finite ordinal number. Accordingly, ω is called the first infiniteordinal number. Using the construction of successors, we obtain the sequence ofordinal numbers ω + 1,ω + 2, . . .. By TheoremA.7.5(b),

⋃n(ω + n) is an ordi-

nal number, denoted by 2ω = ω + ω. Thus, there is a sequence of ordinal num-bers 0, 1, . . . ,ω,ω + 1, . . . , 2ω, 2ω + 1, . . . , 2ω + ω = 3ω, . . .. The ordinal num-ber

⋃n nω is denoted by ω2,

⋃n

(ω2 + nω

)is denoted by 2ω2, and so on. Notice

that these are all countable ordinal numbers.We conclude this section by showing the existence of an uncountable ordinal

number. To this end, it suffices to construct an uncountable well-ordered set, sinceevery well-ordered set determines a unique ordinal number.

Proposition A.7.7 There is an uncountable ordinal (W,�) with the last element ω1

such that, for each w ∈ W other than ω1, the section S (w) = {x ∈ W | x ≺ w} iscountable.

Proof Let X be any uncountable set. By the well-ordering principle, there is a well-ordering � for X . If X does not have a last element, we choose an element∞ /∈ Xand construct a well-ordered set X ∪ {∞} by adjoining∞ to X as the last element.This is done by extending the relation on X to X ∪ {∞}: Define x � ∞ for all x ∈ X .So we can assume that the well-ordered set (X,�) has a last element. Let Y be theset of elements y ∈ X such that the set {x ∈ X | x � y} is uncountable. Then Y isnonempty, and therefore has a least element. If ω1 denotes the least element of Y,then W = {x ∈ X | x � ω1} is the desired set. ♦

The ordinal W in the preceding proposition is unique in the sense that, if W ′ is anyordinal with the same properties, then there is an isomorphism of W onto W ′. For, byTheorem A.6.8, we can assume that there is a monomorphism f : W → W ′. Thenf (W ) is clearly uncountable, and f (ω1) is its last element. If f (ω1) is distinctfrom the last element ω′1 of W ′, then f (W ) would be countable, being a sectionS

(w′

)for some w′ < ω′1. This contradiction shows that f (ω1) = ω′1, and so f is an

isomorphism of W onto W ′.Also, observe that the section S (ω1) in W is uncountable, and any countable

ordinal is isomorphic to a section S (x) in W for some x ≺ ω1. Another notable factabout W is that if A ⊂ W is countable and does not contain ω1, then sup A ≺ ω1. Tosee this, note that the set Xa = {x ∈ W | x � a} is countable for each a ∈ A. SinceA is countable, so is Y =⋃

a Xa . Clearly, W �= Y , for W is uncountable. Let w0 bethe least element of W − Y . Then we have y ∈ Y ⇔ y ≺ w0. It follows that w0 has


only a countable number of predecessors, and therefore w0 ≺ ω1. Clearly, w0 is anupper bound for A, so sup A ≺ ω1.

The ordinal number determined by the ordinal W in the preceding propositionis called the first (or least) uncountable ordinal number, and is denoted by Ω . Thesection {α ∈ Ω | α < Ω}, usually denoted by [0,Ω), is the set of all countableordinal numbers and has no largest element. For, if α ∈ [0,Ω), then its immediatesuccessor α+ 1 is countable. So α+ 1 < Ω , since [0,Ω) is uncountable. Accord-ingly, Ω is also a limit ordinal; in fact, there are uncountably many limit ordinals in[0,Ω].

A.8 Cardinal Numbers

As seen in Sect.A.5, two finite sets X and Y are equipotent if and only if they havethe same number of elements. We associate with each set X an object

∣∣X∣∣ such that

two sets X and Y are equipotent (that is, X and Y have the same cardinality) if andonly if

∣∣X∣∣ = ∣∣Y ∣∣.

Isomorphic ordinals are obviously equipotent, and we observe that the converse isalso true for finite ordinals. We first show that an ordinal X consisting of n elements,n any nonnegative integer, is isomorphic to a section S (n) in ω. For, if X = ∅, thenit is S (0). So assume that X �= ∅, and define a mapping f : X → ω as follows. Mapthe first element x1 of X into the integer 0. If X − {x1} �= ∅, denote the first elementof X − {x1} by x2 and put f (x2) = 1. If X �= {x1, x2}, denote the first element ofX − {x1, x2} by x3 and put f (x3) = 2. Since X consists of finitely many elementsonly, this process terminates after finitely many steps. Thus, we obtain a positiveinteger n such that X = {x1, . . . , xn}, and there is a mapping f : X → ω given byf (xi ) = i − 1, i = 1, 2, . . . , n. Clearly, x1 < · · · < xn so that f is an isomorphismof X onto the section S (n) inω. It follows that any two finite ordinals both having thesame number of elements are isomorphic. However, this may not be true for ordinalshaving infinitely many elements. For example, ifω ∪ {q} is the ordinal obtained fromthe ordinal ω by adjoining q as the last element, then ω ∪ {q} is not isomorphic toω, by Proposition A.6.5. But φ : ω ∪ {q} → ω defined by φ (n) = n + 1, φ (q) = 0is a bijection.

By the well-ordering principle, every set X can be well-ordered and, by TheoremA.7.6, it has the cardinality of an ordinal number. We define

∣∣X∣∣ to be the least

ordinal number such that X is equipotent to∣∣X

∣∣. Clearly, the object ∣∣X∣∣ is uniquely

determined by X , and is called the cardinal number of X .If X is finite and has n elements, then

∣∣X∣∣ is the ordinal number n. The cardinal

number∣∣N∣∣ is denoted by ℵ0, and the cardinal number

∣∣R∣∣ is denoted by ℵ1 andalso by c, called the cardinality of the continuum. The mapping x �→ x/(1− |x |) isa bijection between the open interval (−1, 1) and R, and the mapping x �→ (2x −a − b)/(b − a) is a bijection between an open interval (a, b) and (−1, 1). Thereforethe cardinality of any open interval (a, b) is c. By Theorem A.5.14, we see that thecardinality of a closed interval or a half open interval is also c.


Proof of Theorem A.5.14: Let f : X → Y be an injection. By the definition of car-dinal number

∣∣Y ∣∣, there is a bijection ψ : Y → ∣∣Y ∣∣. Then the composition ψ f is aninjection of X into

∣∣Y ∣∣; consequently, there is a well ordering, say, � on X such thatψ f is an isomorphism of (X,�) into

∣∣Y ∣∣. If o(X,�) denotes the ordinal numberdetermined by the well-ordered set (X,�), then, by Corollary A.6.9, we concludethat o(X,�) ≤ ∣∣Y ∣∣. On the other hand, ∣∣X

∣∣ ≤ o(X,�), by the definition of∣∣X

∣∣. Sincethe relation ≤ between ordinal numbers is transitive, we have

∣∣X∣∣ ≤ ∣∣Y ∣∣. Similarly,

we also have∣∣Y ∣∣ ≤ ∣∣X

∣∣ if there is an injection Y → X . It follows that∣∣X

∣∣ = ∣∣Y ∣∣,and hence X and Y are equipotent.

Given two sets X and Y , we have∣∣X

∣∣ <∣∣Y ∣∣ if and only if there exists an injection

X → Y but there is no bijection between X and Y. Accordingly, it is logical to saythat X has fewer elements than Y when

∣∣X∣∣ <

∣∣Y ∣∣. As seen in §A.5, there is nosurjection from N to I while there is an injection N→ I , n → 1/n. So ℵ0 < c.

Cardinal Arithmetic

The sum of two cardinal numbers α and β, denoted by α+ β, is the cardinality ofthe disjoint union of two sets A and B, where

∣∣A∣∣ = α and

∣∣B∣∣ = β.

Obviously, n + ℵ0 = ℵ0, for N = {1, 2, . . . , n} ∪ {n + 1, n + 2, . . .}. Since N isthe union of disjoint sets {1, 3, 5, . . .} and {2, 4, 6, . . .}, we obtain ℵ0 + ℵ0 = ℵ0.Similarly, we derive ℵ0 + · · · + ℵ0 = ℵ0.

Considering the interval [0, 2) as the union of the intervals [0, 1) and [1, 2), wesee that c + c = c. Therefore, for any integer n ≥ 0,

c ≤ n + c ≤ ℵ0 + c ≤ c + c = c

which implies that n + c = ℵ0 + c = c + c = c.In fact, for any infinite cardinal numberα, we haveα+ ℵ0 = α. To prove this, we

first observe that every infinite set X contains a countably infinite subset. Choose anelement x1 ∈ X . Since X �= {x1}, we can choose an element x2 ∈ X such that x2 �=x1.We still have X − {x1, x2} �= ∅. Consequently, we can choose an element x3 ∈ Xsuch that x3 �= x1, x2. Assume that we have chosen n distinct elements x1, x2, . . . , xn

of X . Since X is infinite, it is possible to choose an xn+1 ∈ X that is distinct fromeach of the xi . Repeating this process indefinitely, we obtain a sequence 〈x1, x2, . . .〉of distinct points of X . The set {x1, x2, . . .} constructed in this way is obviouslycountably infinite. Now, given an infinite cardinal number α, find a set X with α =∣∣X

∣∣. Then there exists a countably infinite set Y ⊆ X . Writing β = ∣∣X − Y∣∣, we have

α = ∣∣Y ∣∣+ ∣∣X − Y∣∣ = ℵ0 + β, so α+ ℵ0 = β + ℵ0 + ℵ0 = β + ℵ0 = α.

Let M be a set and suppose that, for each m ∈ M , αm is a cardinal number. Then∑m∈M αm denotes the cardinal number of the union of pairwise disjoint sets Am ,

where∣∣Am

∣∣ = αm .Clearly, the setN can be written as the disjoint union of countably many countable

sets:


1 3 5 7 · · ·2 6 10 14 · · ·4 12 20 28 · · ·8 24 40 56 · · ·...

......

... · · ·

So ℵ0 + ℵ0 + · · · = ℵ0.If α1,α2,α3, . . ., are cardinal numbers such that 1 ≤ αn ≤ ℵ0 for every n =

1, 2, 3, . . ., then, by Proposition A.5.11, we derive

α1 + α2 + · · · = ℵ0.

The decomposition of the interval [0,∞) into the intervals [n − 1, n), wheren = 1, 2, . . ., shows that c + c + · · · = c.

The product of two cardinal numbers α and β, denoted by αβ, is the cardinalityof the Cartesian product of two sets A and B, where

∣∣A∣∣ = α and

∣∣B∣∣ = β. If μ is a

cardinal number, then the μth power of α, denoted by αμ, is the cardinal number ofthe set AM , where M is a set with

∣∣M∣∣ = μ.Suppose that for every m ∈ M , there is the same cardinal number αm = α. Then∑m∈M αm = αμ and

∏m∈M αm = αμ, where μ = ∣∣M∣∣.

For the first formula, we note that A × M =⋃m∈M Am , where Am = A × {m}.

Obviously, Am is equivalent to A, and the family {Am | m ∈ M} is pairwise disjoint.So

∣∣A × M∣∣ =∑

m∈M

∣∣Am

∣∣ and we have∑

m∈M αm = αμ.With Am = A for all m ∈ M , we have

∏m∈M Am = AM , and the second formula

follows.Using the above results, we obtain

nℵ0 = ℵ0 + · · · + ℵ0 = ℵ0,ℵ0ℵ0 = ℵ0 + ℵ0 + · · · = ℵ0,nc = c + · · · + c = c,ℵ0c = c + c + · · · = c,

where n is any positive integer.

Proposition A.8.1 c = 2ℵ0 .

Proof By definition, 2ℵ0 is the cardinality of 2N, the set of all sequences whose termsare the digits 0 and 1 only (that is, the dyadic sequences). Interestingly, only these twodigits are required to write the binary (or dyadic) representation of a real number x in[0, 1]; specifically, as the binary expansion, the representation x = 0.a1a2 · · · an · · ·means the number x =∑

an/2n , where each an is either 0 or 1. Note that, as in thecase of decimal expansion, certain numbers in [0, 1] have two binary expansions(e.g., 1/2 = 0.100 · · · = 0.011 · · · ). Hence we deduce that c ≤ ∣∣2N∣∣ ≤ 2c = c. ♦


Note that ℵ20 = ℵ0ℵ0 = ℵ0 and ℵn0 = ℵ0 · · · ℵ0 = ℵ0, by induction on n. Similar

results can be proved for any transfinite cardinal number α. With this end in view,we first establish the following.

Proposition A.8.2 For any infinite cardinal number α, 2α = α.

Proof Let X be an infinite set with∣∣X

∣∣ = α. Denote the two-point set {0, 1} by 2.Then, for any set A, 2× A is the union of disjoint sets {0} × A and {1} × A and sowehave 2

∣∣A∣∣ = ∣∣A

∣∣+ ∣∣A∣∣ = ∣∣2× A

∣∣. Now, consider the family F of all pairs (A, f )

such that A ⊆ X and f : A → 2× A is a bijection. By Theorem A.5.7, X contains acountably infinite set A and, by Proposition A.5.11, the set 2× A is also countable.Hence, there is a bijection f : A → 2× A, and (A, f ) belongs to F . Considerthe binary relation ≤ on F defined by (A, f ) ≤ (B, g) if A ⊆ B and f = g|B. Itis easily verified that ≤ is a partial ordering and every chain in F has an upperbound inF . So the Zorn’s lemma applies, and we get a maximal member (M, h) inF . Then

∣∣M∣∣+ ∣∣M∣∣ = ∣∣M∣∣. We show that∣∣M∣∣ = ∣∣X

∣∣. If X − M is infinite, then itcontains a countably infinite set B. As above, there exists a bijection g between B and2× B. Thenwe have a bijection k : B ∪ M → 2× (B ∪ M) defined by k|B = g andk|M = h. Thus (B ∪ M, k) belongs toF , and this contradicts the maximality of M .Therefore, X − M must be finite; accordingly, M is infinite. If Y ⊆ M is a countablyinfinite set (which does exist), then

∣∣Y ∪ (X − M)∣∣ = ∣∣Y ∣∣+ ∣∣X − M

∣∣ = ∣∣Y ∣∣ and weobtain ∣∣X

∣∣ = ∣∣Y ∪ (X − M) ∪ (M − Y )∣∣

= ∣∣Y ∪ (X − M)∣∣+ ∣∣M − Y

∣∣= ∣∣Y ∣∣+ ∣∣M − Y

∣∣ = ∣∣M∣∣.♦

As an immediate consequence of this result, we see that if α is an infinite cardinalnumber, then, by induction, nα = α for every integer n > 0. And, for a cardinalnumber β ≤ α, α+ β = α, since α ≤ α+ β ≤ 2α = α.

Proposition A.8.3 For any infinite cardinal number α, α2 = αα = α.

Proof Let X be an infinite setwith∣∣X

∣∣ = α and letF be the family of all pairs (A, f )

such that A ⊆ X and f : A → A × A is a bijection. Clearly, X contains a countablyinfinite set A and, by Lemma A.5.10, the set A × A is also countable. So there is abijection f between A and A × A, and thus the pair (A, f )belongs toF .WepartiallyorderF by (A, f ) ≤ (B, g) if A ⊆ B and f = g|B. Given a chain C = {Ai , fi } inF , put B =⋃

Ai and define f : B → B × B by setting f (x) = fi (x) if x ∈ Ai .Then f is clearly a bijection so that the pair (B, f ) belongs toF . It is obvious that(B, f ) is an upper bound for the chain C . By the Zorn’s lemma, we have a maximalmember (M, h) in F . Since h : M → M × M is a bijection,

∣∣M∣∣ = ∣∣M∣∣∣∣M∣∣. Weobserve that

∣∣X∣∣ = ∣∣M∣∣. As M ⊆ X ,

∣∣M∣∣ ≤ ∣∣X∣∣. If ∣∣M∣∣ <

∣∣X∣∣, then we find that∣∣M∣∣ <

∣∣X − M∣∣. For, ∣∣X − M

∣∣ ≤ ∣∣M∣∣ implies that

∣∣X∣∣ = ∣∣M∣∣+ ∣∣X − M

∣∣ ≤ ∣∣M∣∣+ ∣∣M∣∣ = ∣∣M∣∣,


by the preceding proposition. This contradicts our assumption that∣∣M∣∣ <

∣∣X∣∣. Let

j : M → X − M be an injection and put Y = j (M). Then∣∣Y ∣∣ = ∣∣M∣∣. Since ∣∣Y ∣∣ is

infinite, we have 3∣∣Y ∣∣ = ∣∣Y ∣∣, and it follows that there are three disjoint sets A, B, and

C contained in Y such that Y = A ∪ B ∪ C and∣∣Y ∣∣ = ∣∣A

∣∣ = ∣∣B∣∣ = ∣∣C∣∣. So ∣∣A

∣∣ =∣∣M × Y∣∣, ∣∣B

∣∣ = ∣∣Y × M∣∣ and ∣∣C∣∣ = ∣∣Y × Y

∣∣; accordingly, there is a bijection k :Y → (M × Y ) ∪ (Y × M) ∪ (Y × Y ). Since M ∩ Y = ∅, we have a bijection M ∪Y → (M ∪ Y )× (M ∪ Y )which extends h (and k). This contradicts the maximalityof M , and hence

∣∣M∣∣ = ∣∣X∣∣. ♦

If α is an infinite cardinal number, then, by induction on n, we see that αn = αfor every integer n > 0. Moreover, for any cardinal number β ≤ α, we have α ≤αβ ≤ αα = α, and therefore αβ = α. In particular, notice that ℵ0α = α. We usethese results to prove a generalization of Theorem A.5.12.

Theorem A.8.4 Let X be an infinite set and F be the family of all finite subsets ofX. Then

∣∣F ∣∣ = ∣∣X∣∣.

Proof Since the mapping X → F , x �→ {x}, is an injection, we have∣∣X

∣∣ ≤ ∣∣F ∣∣.To see the opposite inequality, for each integer n > 0, let Fn denote the family ofall those subsets of X , which contain exactly n elements. Now, for each set F ∈Fn , choose an element φ(F) in X × · · · × X = Xn (n copies), which has all itscoordinates in F . Of course, there are several choices for φ(F), we pick any oneof these. Then φ : Fn → Xn is an injection, and so

∣∣Fn

∣∣ ≤ ∣∣Xn∣∣ = ∣∣X

∣∣n = ∣∣X∣∣.

Obviously, we have F =⋃n Fn and therefore

∣∣F ∣∣ ≤∑n

∣∣Fn

∣∣ ≤∑n

∣∣X∣∣ = ℵ0∣∣X

∣∣ = ∣∣X∣∣.

♦Let X be a set and A ⊆ X . The function f A : X → {0, 1} defined by

f A (x) ={0 for x /∈ A, and1 for x ∈ A

is called the characteristic function of A. The function which is zero everywhere isthe characteristic function of the empty set, and the function which is identically 1 onX is the characteristic function of X . The set of all functions X → {0, 1} is denotedby 2X. Obviously, every element of 2X is a characteristic function on X .

Proposition A.8.5 For any set X, there is a bijection between the set P (X) of allits subsets and 2X.

Proof For a set A ⊆ X , let f A denote the characteristic function of A. Then themapping φ : 2X →P (X), f �→ f −1(1), is clearly surjective. It is also injective.For, if f �= g are in 2X , then we have f (x) �= g (x) for some x ∈ X . It follows that xbelongs to one of the sets f −1 (1) and g−1 (1), but not to the other. So φ ( f ) �= φ (g),and φ is an injection. ♦


From the above proposition, it is clear that∣∣P (X)

∣∣ = 2|X |. The function x �→ {x}is obviously an injection from X intoP (X), but there is no bijection between thesesets, by Theorem A.5.13. Therefore

∣∣X∣∣ <

∣∣P (X)∣∣, and thus we have established

the following.

Proposition A.8.6 For any set X,∣∣X

∣∣ < 2|X |.

Appendix BFields R, C, and H

B.1 The Real Numbers

We shall not concern ourselves here with the construction of the real number systemon the basis of a more primitive concept such as the positive integers or the rationalnumbers. Instead, we assume familiarity with the system R of real numbers as anordered field which is complete (that is, it has the least upper bound property). Wereview and list the essential properties of R, however, in aid of the reader.

With the usual addition and multiplication, the set R has the following properties.

Theorem B.1.1 (a) R is an abelian group under addition, the number 0 acts as theneutral element.(b) R− {0} is an abelian group under multiplication, the number 1 acts as the

multiplicative identity (unit element).(c) For all a, b, c ∈ R, a (b + c) = ab + ac.

A field is a set F containing at least two elements 1 �= 0 together with two binaryoperations called addition and multiplication, denoted by+ and · (or juxtaposition),respectively, which satisfy TheoremB.1.1. The element 0 is the identity element foraddition, and 1 acts as the multiplicative identity.

With this terminology, the set R is a field under the usual addition and multipli-cation. This field has an order relation < which satisfies (a) a + b < a + c if b < c,and (b) ab > 0 if a > 0 and b > 0. A field which also has an order relation satisfyingthese two conditions is called an ordered field. The set Q of all rational numbers isanother example of an ordered field.

We call an element a of an ordered field positive if a > 0, and negative if a < 0.

Theorem B.1.2 The following statements are true in every ordered field:

(a) a > 0⇒ −a < 0, and vice versa.(b) a > 0 and b < c ⇒ ab < ac,

a < 0 and b < c ⇒ ab > ac.

© Springer Nature Singapore Pte Ltd. 2019T. B. Singh, Introduction to Topology,https://doi.org/10.1007/978-981-13-6954-4

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442 Appendix B: Fields R, C, and H

(c) a �= 0⇒ a2 > 0. In particular 1 > 0.(d) a > 0⇒ a−1 > 0,

a < 0⇒ a−1 < 0.(e) ab > 0⇒ either both a > 0 and b > 0 or both a < 0 and b < 0.(f) ab < 0⇒ either both a < 0 and b > 0 or both a > 0 and b < 0.

The ordered field R has the least upper bound property: If S ⊆ R is nonemptyand bounded above, then sup S exists in R.

This is also called the completeness property of R. It follows that R is a completeordered field. Moreover, R has no gaps: If a < b, then c = (a + b)/2 satisfies a <

c < b. Thus we see that R is a linear continuum (refer to Exercise3.1.11).Using the completeness property, it can be shown that every nonempty set of real

numbers with a lower bound has a infimum. Another important consequence of thecompleteness property of R is described by the following.

Theorem B.1.3 (Archimedean Property) If a real number x > 0, then given anyreal number y, there exists a positive integer n such that nx > y.

Using this property of R, one can prove that, for any two real numbers x < y,there is a rational number r ∈ Q such that x < r < y. This fact is usually stated bysaying that Q is dense in R.

The absolute value of a real number x , denoted by |x |, is defined by |x | = x forx ≥ 0, and |x | = −x for x < 0.

Proposition B.1.4 For any x, y ∈ R, we have

(a) |x | ≥ 0 (b) |x | = 0⇔ x = 0.(c) | − x | = |x |. (d) |xy| = |x ||y|.(e) |x | ≤ y ⇔ −y ≤ x ≤ y. (f) − |x | ≤ x ≤ |x |.(g) |x + y| ≤ |x | + |y|. (h)

∣∣|x | − |y|∣∣ ≤ |x − y|.(i) |x − y| ≤ |x | + |y|.

By (d), |x | = √x2.

B.2 The Complex Numbers

The set R2 of all ordered pairs (x, y) of real numbers turns into a field under the

following addition and multiplication:

(x, y)+ (x ′, y′

) = (x + x ′, y + y′

),

(x, y)(x ′, y′

) = (xx ′ − yy′, xy′ + yx ′

).

The element (0, 0) acts as the neutral element for addition, and the element (1, 0)plays the role of multiplicative identity. It is routine to check that R

2 is a field

Appendix B: Fields R, C, and H 443

under these definitions. It is usually denoted by C, and its elements are referred toas the complex numbers. It is readily verified that (x, 0)+ (y, 0) = (x + y, 0), and(x, 0) (y, 0) = (xy, 0). This shows that the complex numbers of the form (x, 0) forma subfield of C, which is isomorphic to R under the correspondence x �→ (x, 0). Wecan therefore identify this subfield of C with the real field and regard R ⊂ C.

Writing ı = (0, 1), we have ı2 = −1 and (x, y) = (x, 0)+ (y, 0) (0, 1) = x +yı , using the identification x ↔ (x, 0). Thus C = {x + yı |x, y ∈ R}, where ı2 =−1. If z = x + yı is a complex number, then we call x the real part of z (denoted byRe(z)), and y the imaginary part of z (denoted by Im(z)).

If z = x + yı ∈ C, its conjugate is defined to be the complex number z = x − yı .For any complex numbers z and w, we have

z + w = z + w, zw = zw.

Observe that zz is a positive real number unless z = 0.The absolute value |z| of a complex number z is defined to be

√zz (the nonnegative

square root). Clearly, |z| > 0 except when z = 0, and |0| = 0. It is also obvious that|Re (z) | ≤ |z| and |z| = |z|. If z andw are any two complex numbers, then it is easilychecked that |zw| = |z||w| and |z + w| ≤ |z| + |w|.B.2.1 (Schwarz Inequality) If z1, . . . , zn andw1, . . . , wn are complex numbers, thenwe have ∣∣ ∑n

1 z jw j

∣∣2 ≤ (∑n1 |z j |2

) (∑n1 |w j |2

).

Proof Put α =∑n1 |z j |2, β =∑n

1 |w j |2 and γ =∑n1 z jw j . We need to show that

αβ − |γ|2 ≥ 0. If α = 0 or β = 0, this is trivial. We therefore assume that α �= 0 �=β. Then α,β > 0 and we have

∑n1 |βz j − γw j |2 = ∑n

1

(βz j − γw j

) (β z j − γw j

)= β2α− β|γ|2 = β

(αβ − |γ|2).

The left-hand side of the first equality is obviously nonnegative, and therefore wemust have αβ − |γ|2 ≥ 0. ♦

B.3 The Quaternions

Byusing the usual scalar and (nonassociative) vector product inR3, Hamilton defined

(in 1843) a multiplication inR4, which together with componentwise addition makes

it into a skew field. This field has proven to be fundamental in several areas ofmathematics and physics. We intend to discuss here some basic facts about it. Recallthat a skew field (also called a division ring) satisfies all field axioms except thecommutativity of multiplication.

444 Appendix B: Fields R, C, and H

The mapping R4 → R× R

3, (x0, x1, x2, x3) �→ (x0, (x1, x2, x3)), is a bijection.If we define the vector space structure on R× R

3 over R componentwise:

(a, x)+ (b, y) = (a + b, x + y) and c (a, x) = (ca, cx) ,

then this mapping becomes an isomorphism. Consequently, we can identify R×R

3 = H with R4, and call its elements quaternions. If q = (a, x), we refer to a as

the real part of q and x as the vector part of q. There are canonical monomorphismsR→ H, a �→ (a, 0), andR

3 → H, x �→ (0, x), of vector spaces. Hence, we identifythe real number a with the quaternion (a, 0), and the vector x with the quaternion(0, x). Then a quaternion (a, x) can be written as a + x . Accordingly, for any twoquaternions q = a + x and r = b + y, we have q + r = a + b + x + y, and if c isa real, then cq = ca + cx .

If x, y ∈ R3 ⊂ H, we first define xy = −x · y + x × y, where · is the usual scalar

product, and × is the usual vector product in R3. Notice that xy is in general an

element ofH. It is easily checked that thismultiplication of vectors inH is associative.As the multiplication in H ought to be distributive, we set

qr = ab + ay + bx + xy

for q = a + x and r = b + y. We leave it to the reader to verify the following con-ditions for all quaternions q, r, s, and real c:

q (cr) = c (qr) = (cq) r, q (rs) = (qr) s,q (r + s) = qr + qs, (r + s) q = rq + sq.

The quaternion 1 = 1+ 0 acts as the multiplicative identity: 1q = q = q1 for everyq ∈ H.

To complete the proof that H is a skew field with the above addition and multi-plication, it remains to verify that every nonzero quaternion q has a multiplicativeinverse. With this end in view, we define the conjugate of q = a + x as q = a − x .Observe that q + r = q + r , cq = cq , qr = r q for any quaternions q, r, s, and realc. Also, it is straightforward to see that qq = qq = a2 + x · x (a real). We define themodulus of q to be |q| = √qq . Notice that |q| is the Euclidean norm of q when it isconsidered as an element of R

4. For q, r ∈ H, we have

|qr |2 = (qr) (qr) = qrr q = q|r |2q = |r |2qq = |r |2|q|2,

which implies that |qr | = |q||r |. Clearly, |q| = 0⇔ q = 0 and, if q �= 0, thenq

(q/|q|2) = 1 = (

q/|q|2) q. Thus q/|q|2 is the inverse of q in H, usually denotedby q−1.

Observe that x ∈ R3 is a unit vector⇔ x · x = 1⇔ x2 = −1, and two vectors

x, y ∈ R3 are orthogonal⇔ x · y = 0⇔ xy = −yx . A right-handed orthonormal

system in R3 is an ordered triple ı, j, k of vectors in R

3 such that ı, j, k are ofunit length, mutually orthogonal and ı × j = k. So, if ı, j, k form a right-handed

Appendix B: Fields R, C, and H 445

orthonormal system, then ı2 = j 2 = k2 = −1 and ıjk = −1. Conversely, theseconditions imply that ı, j, k are of unit length, and ıj = k whence ı · j = 0 andı × j = k. Thus ı, j, k form a right-handed orthonormal system.

Suppose now that ı, j, k is a right-handed orthonormal system in R3. Then any

vector x ∈ R3 can be written uniquely as x = x1ı + x2j + x3k, xi ∈ R; accordingly,

any quaternion q can be expressed uniquely as

q = q0 + q1ı + q2j + q3k, qi ∈ R.

Clearly, we have ıj = k = −j ı , jk = ı = −kj , kı = j = −ık. Using these rules,we obtain the following formula for the product of two elements q = q0 + q1ı +q2j + q3k, q ′ = q ′0 + q ′1ı + q ′2j + q ′3k in H:

qq ′ = (q0q

′0 − q1q

′1 − q2q

′2 − q3q

′3

)+ (q0q

′1 + q1q

′0 + q2q

′3 − q3q

′2

)ı +(

q0q′2 + q2q

′0 + q3q

′1 − q1q

′3

)j + (

q0q′3 + q3q

′0 + q1q

′2 − q2q

′1

)k.

We also note that q = q0 − q1ı − q2j − q3k and |q| =√(

q20 + q2

1 + q22 + q2

3

).

For any unit vector x ∈ R3, the set of quaternions a + bx , a, b ∈ R, is a subfield of

H isomorphic to C under the mapping a + bx �→ a + bı . In particular, the subfieldof quaternions with no j and k components is identified withC, and we regardC as asubfield of H. Thus, we have field inclusions R ⊂ C ⊂ H. It is obvious that any realnumber commutes with every element of H. Conversely, if a quaternion q commuteswith every element of H, then q ∈ R. We emphasize, however, that the elements ofC do not commute with the elements of H.

References

1. G.E. Bredon, Topology and Geometry (Springer, New York, 1993)2. R. Brown, Elements of Modern Topology (McGraw-Hill, London, 1968)3. J. Dugundji, Topology (Allyn and Bacon, Boston, 1965)4. D.B. Fuks, V.A. Rokhlin, Beginner’s Course in Topology (Springer, Heidelberg, 1984)5. I.M. James, Topological and Uniform Spaces (Springer, New York, 1987)6. J.L. Kelley, General Topology (van Nostrand, New York, 1955)7. W.S. Massey, Algebraic Topology: An Introduction (Springer, New York, 1967)8. G.McCarty, Topology: An Introduction with Application to Topological Groups (McGraw-Hill,

New York, 1967)9. D. Montgomery, L. Zippin, Topological Transformation Groups (Interscience Publishers, New

York, 1955)10. J.R. Munkres, Topology: A First Course (Prentice-Hall, NJ, 1974)11. L. Pontriajagin, Topological Groups (Princeton University Press, New York, 1939)12. I.M. Singer, J.A. Thorpe, Lecture Notes on Elementary Topology and Geometry (Scott, Fores-

man and Company, IL, 1967)13. E.H. Spanier, Algebraic Topology (McGraw-Hill, New York, 1967)14. L.A. Steen, J.A. Seebach, Counterexamples in Topology (Springer, New York, 1978)15. R.C. Walker, The Stone-Cech Compactification (Springer, New York, 1974)16. S. Willard, General Topology (Addison-Wesley, MA, 1970)


447

https://doi.org/10.1007/978-981-13-6954-4

Index

SymbolsFσ set, 11Gδ set, 11T0-space, 91T1-space, 91T2-space, 89T3-space, 181T4-space, 186T3 1

2-space, 196

σ-discrete family, 212σ-locally finite family, 206

AAccumulation or cluster point

of a set, 12Action

effective, 299free, 299of a group, 297transitive, 299trivial, 299

Adherent point, 11Adjunction space, 156Admissible nbd, 374Affine map, 276Attaching map, 156

BBaire category theorem, 231Baire space, 230Basis, 17Bing metrization theorem, 218Boundary, 13Boundary point, 13Bounded metric, 5

Bounded set, 5Box topology, 43

CCantor set, 13Cardinal number, 434Cauchy sequence, 219Chain, 425Clopen set, 8Closed ball, 10Closed function, 34Closed set, 8Closure, 11Cluster or accumulation point

of a filter, 87of a net, 82of a sequence, 77

Coarser topology, 8Coarse topology, 145Cocountable topology, 7Cofinal, 83Cofinite topology, 7Coherent topology, 164Coherent union, 164Coinduced topology, 163Comb space, 64Compactification, 199Compact space, 96Compact subset, 96Completely normal space, 195Completely regular space, 196Complete metric space, 219Cone, 142Connected component, 59Connected set, 52Connected space, 51


449

https://doi.org/10.1007/978-981-13-6954-4

450 Index

Continuity at a point, 3, 31Continuous function, 4, 29Continuum, 104Contractible space, 318Contraction, 318Convergence

continuous, 235pointwise, 235uniform, 235

Convergent filter, 87Convergent net, 81Convergent sequence, 77Convex set, 64Coset space, 282Countably compact space, 106Covering map or

projection, 374Covering of a set, 95Cross section, 309Curve, 66Cylinder, 40

DDeck transformation, 394Decomposition space, 138Deformation retract, 321Degree of a loop, 340Dense set, 13Derived set, 12Diagonal, 42Diameter of a set, 5Dictionary order, 425Disconnected space, 51Discrete family, 212Discrete set, 27Discrete space, 7Discrete topology, 7Dunce cap, 325

EEmbedding, 36Equicontinuous family, 261Equivalence relation, 417Equivalent metric, 21Euclidean group, 295Evaluation map, 216, 243

FFamily separating

points, 216points and closed sets, 162

Filter, 86Filter base, 87Finer topology, 8Fine topology, 145Finite intersection property, 97First countable space, 107First or smallest or least

element, 426Fixed point, 300Fort space, 10Fréchet space, 49Free abelian group, 351Free group, 353Free product, 346

with amalgamation, 359Function, 413

associate, 251bijective, 414characteristic, 438injective, 414surjective, 414

Fundamental domain, 309Fundamental group, 331

GGeneralized Heine–Borel

theorem, 100Gluing lemma, 32Graph of a function, 42Group

general linear, 269orthogonal, 276special linear, 278special orthogonal, 278special unitary, 278symplectic, 277topological, 267unitary, 277

HHausdorff space, 81Hawaiian earring, 152Heine–Borel theorem, 96Hereditary property, 90Hilbert space, 3Homeomorphism, 33Homogeneous space, 270Homotopic maps, 316Homotopy, 316Homotopy class, 317Homotopy equivalence, 317Homotopy inverse, 317

Index 451

IIdentification map, 137Identification space, 137Identification topology, 137Indiscrete space, 7Indiscrete topology, 7Induced topology, 161Inductive topology, 164Inessential map, 318Inner production C

n and Hn , 2

Interior, 11Interior point, 11Invariant subset, 302Isolated point, 13Isometry, 38Isotropy subgroup, 299

JJoin, 143

KKlein bottle, 128k-space (or compactly

generated space), 119

LLast or largest or greatest

element, 426Lebesgue number, 112Left translation, 270Lens space, 305Lifting of a map, 381Limit point of a set, 12Lindelöf space, 177Linear continuum, 57Linear or total

ordering, 425Local homeomorphism, 377Locally closed, 116Locally compact space, 114Locally connected space, 69Locally finite family, 32Locally metrizable space, 218Locally path-connected space, 72Loop, 320Lower limit topology, 20

MManifold, 213Mapping cone, 159

Mapping cylinder, 159Maximal element, 426Metric space, 2Metric topology, 7Metrizable space, 47Minimal element, 426Möbius band, 127Monodromy action, 389

NNagata–Smirnov metrization

theorem, 215Neighborhood, 9Neighborhood basis, 22Net, 80Normal space, 186Nowhere dense, 230Null homotopic map, 318

OOne-point compactification, 117Open ball, 4Open function, 34Open set, 4, 7Orbit, 299Orbit map, 305Orbit space, 304Order relation, 424Order topology, 16Ordinal, 426Ordinal number, 430

first infinite, 433first uncountable, 434

PParacompact space, 205Partial ordering, 425Partition of unity, 212Path, 63Path component, 66Path-connected space, 64Perfectly normal space, 195Perfect map, 121Perfect set, 13Pointed space, 151Precompact space, 109Principal filter, 87Product space, 39Product topology, 39, 43Proper map, 121Pseudocompact space, 114

452 Index

Pseudo-metrizable space, 215

QQuasi-component, 61Quotient space, 126Quotient topology, 126

RRank

of a free abelian group, 353of a free group, 355

Reduced suspension, 154Refinement, 205Reflection, 313Regular covering, 398Regular space, 181Relative homotopy, 321Relative topology, 25Retract, 321Right translation, 270Rotation, 311

SSaturated set, 126, 137Second countable space, 169Semilocally simply connected, 405Separable space, 175Separated sets, 53Sequence, 77Sequentially compact space, 107Sequential space, 174Set

countable, 421of first category, 231of second category, 231order complete, 426orthonormal, 277well ordered, 426

Sheet, 374Sierpinski space, 7Simple ordering, 424Simply connected space, 334Skew field, 443Smash product, 153Stereographic projection, 34Stone–Cech compactification, 199Strong deformation retract, 321Subbasis, 16Subcovering, 95Subnet, 83Subsequence, 78

Subspace, 5, 25Subspace topology, 25Sup metric, 6Suspension, 142Symmetric nbd, 271Symplectic space, 3

TTelevision topology, 24Tietze extension theorem, 191Topological invariant, 35Topologically complete space, 220Topological space, 7Topological sum, 149Topologist’s sine curve, 57Topology, 7

admissible, 243compact-open, 244of compact convergence, 258of pointwise convergence, 236of uniform convergence, 238of uniform convergence on compacta,258

Torus, 41, 128Totally bounded space, 109Totally disconnected space, 60Tychonoff theorem, 99Tychonoff topology, 43

UUltrafilter, 87Uniform homeomorphism, 228Uniformly continuous map, 113Uniformly convergent sequence, 80Uniformly equivalent metric, 228Uniform metric, 50Unitary space, 3Unit n-cube, 5Unit n-disc, 5Unit n-sphere, 5Universal covering space, 388Universal (ultra) net, 84Upper limit topology, 20Urysohn embedding theorem, 191Urysohn lemma, 188Urysohn metrization theorem, 190

WWeak topology, 164Wedge, 151Winding number, 341

Set Theory - Springer

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