Set theory and the construction of numbers

Set Theory and the Construction of Numbers

Robert Andersen

Department of Mathematics

University of Wisconsin - Eau Claire

1

I

INTRODUCTION

Modern mathematics is couched in the language and steeped in the theory of sets. Sets are the heart and soul of mathematics. Sets should be well understood by any serious student of the subject. What we attempt to do here is to present the Zermelo Fraenkel Axioms for Set Theory and develop a model for those axioms. We call this model the number model, as our usual concept of numbers is a consequence of this model. There is an initial temptation when developing a theory to define all of the terms. This temptation is quickly extinguished when we realize its futility. To define a new term we must define it in terms of previous terms; thus we are faced with two possibilities, either having circular definitions or having an infinite regression of definitions. Either case is unsatisfactory. Therefore we begin with undefined terms, or as I prefer “dictionary” defined terms. That is, definitions taken directly from a standard dictionary. A set is a group of persons or things classed or belonging together. We may paraphrase this as: A set is a collection of objects. However, as we shall see later, not every collection can be regarded as a set. Collections, either sets or non-sets, are often referred to as spaces to avoid repetitive rhetoric. The objects, persons, or things that make up the set we shall call elements. A single one of these objects is of course an element. The aggregate of elements is the set. If a particular element is a member of the aggregate we say it is an element of the set. Let a represent a set and x one of its element. We express this fact with the notation, x ∈ a, and we read this as x is an element of a. In developing an axiomatic discussion of sets it may be tempting to posit

2 the existence of a set. This is troublesome for the following reason. In any well defined axiom system the axioms are chosen to be independent. An axiom is independent of the other axioms if a model can be constructed, using all other axioms, and replacing the axiom in question with another that is its denial or implies its denial. An axiom system is independent if all of its axioms are independent. If our axiom system includes an axiom that asserts the existence of a set, then to demonstrate its independence, we must construct a model that assumes the validity of all other axioms, which are statements about sets, and an axiom that denies the existence of a set. It is not necessary to posit the existence of a set since examples of concrete sets are ubiquitous. An examination of my right front pocket reveals a collection of keys, indeed a set exists.

3

II

THE AXIOMS

The classical collection of Axioms for set theory are the Zermelo Fraenkel Axioms (ZF Axioms).

The Axiom of Extension ZF1 Axiom of extension.

Two sets are equal if and only if they have

identically the same elements. ∀a∀b(a = b) ⇔ ∀x(x ∈ a ⇔ x ∈ b). This axiom not only defines the term equal and its associated symbol, =, but also tells us that it is the elements that uniquely define a set. The set may however have different descriptions. For example consider those planets that orbit the sun closer than the earth, and those planets of our solar system with no natural satellite. A simple check of an almanac reveals that exactly the same collection of planets satisfies both descriptions. Let a be any set. Let P (x) be a proposition about an arbitrary element x from a; that is, for every x ∈ a, P (x) is a statement that is either true or false. An example of a non-proposition in the variable x is x ∈ b. In this example b is also an indeterminate. Without specifying b we have no way of knowing whether any particular element of our initial set a is in b or not.

The Axiom Schema of Specification ZF2 Axiom schema of specification. For every set a and proposition P (x) there is a set b that consists of those elements of a where P (x) is true. ∀a∃b(x ∈ b ⇔ x ∈ a ∧ P (x)). ZF2 is not regarded as a single axiom but rather as a collection of axioms, hence the term axiom schema is used. Each possible proposition P (x)

4 produces a separate axiom. Several of the Zermelo Fraenkel Axioms are in fact axiom schema and will be identified as such. By convention, braces { and } are used when representing a set, with the elements listed between the braces. The pair of braces should be considered as a single symbol, a single brace without its complement is meaningless in set theory. The braces can be thought of as the purse that holds the coins. The purse is by no means a member of the collection of coins, and the braces are not elements of a set. The braces simple mean that the elements listed between them are to be considered a set. {x, y, z} is the set that consists of the letters x, y, z (the commas, or course, are punctuation). When we use an axiom of specification we express the set b (of ZF2) by {x ∈ a | P (x)}, which is read: The set of all x in a such that P (x). The set a is called the Universal Set and P (x) is the proposition that must be satisfied for an element to be included in the set b. Let a be an arbitrary set, for example let a be the collection of nations in the United Nations. Now let P (x) be the statement: x is not x, e.g., Canada is not Canada. At first glance this statement may seem absurd, but in reality it is not absurd but simply false for every element of our set. Using our notation we write the set specified as: {x ∈ a | x 6= x}. By virtue of the axiom schema of specification this set exists, but no element of a satisfies the proposition x 6= x. Hence we must conclude that the set is vacuous and we call it the Empty or Null Set. The empty set is not just an interesting or pathological footnote to the axiom of specification, but is absolutely crucial to the study of sets. In fact we may at this time choose

5 to disregard any other set (and I choose to do so) and focus our attention solely on the empty set and other sets that may be derived from it by the axioms. We reserve the symbols ∅ and { } for the empty set. We may state this result as a theorem. Theorem 2.1 There exists a set that has no elements. Here is the formal proof. Proof Let ∅ = {x ∈ a | x 6= x}. Since for any set a no element satisfies the proposition x 6= x, we conclude ∅ has no elements. An interesting result known as Russell’s Paradox can now be presented. Since sets themselves can be regarded as objects, it seems quite reasonable to consider a collection of sets as a set itself. Let s be the collection of all sets and consider the following set. r = {x ∈ s | x 6∈ x}. Now the set r is either an element of itself or it is not. Assume that it is, i.e. we assume r ∈ r. Then by the defining proposition of the set r we see that it must be the case that r 6∈ r, contrary to the assumption. Thus we may assume that the other case is true, that r is not an element of itself, i.e r 6∈ r, but we see that if that is true we must necessarily have r ∈ r. Again this presents a contradiction. Clearly the collection, s, is much too large or encompassing to be considered a set. This unfortunate result is handled by an appropriate axiom yet to be stated. (See ZF9) At this stage the only sets that we have are real, tangible objects and the empty set. We wish to present axioms that allow us to expand our collection of sets to more abstract objects. The next few axioms allow us to construct new sets from preexisting sets.

6

The Axiom of Pairing ZF3 Axiom of pairing.

If a and b are sets then there exists a set c such

that a ∈ c and b ∈ c. ∀a∀b∃c(a ∈ c ∧ b ∈ c). There is an important fact to note here. The axiom specifies two elements that are in c but says nothing about what else may be in c, in this sense c is not a well defined set. But by the use of an axiom of specification we may construct a well defined set. Let c be a set guaranteed to exist by the axiom of pairing and consider the following set: d = {x ∈ c | x = a, or x = b}. Clearly d = {a, b}. Another point to note here is that a and b may be the same set, that is a = b. In this case we have the following: d = {a, b} = {a, a} = {a} The last equation is true by virtue of the axiom of extension. Since both sets on each side of the equal sign have exactly the same elements the statement is true. Using only the empty set, we can now construct a new set. Let a = b = ∅, and we have d = {∅}. We may continue ad infinitum constructing new sets from previously constructed sets, for example let a = ∅ and b = {∅}, and we now have d0 = {∅, {∅}}.

7

The Axiom of Unions ZF4 Axiom of unions.

For every set c there is a set a where, if b ∈ c

and x ∈ b, then x ∈ a. ∀c∃a∀b(x ∈ b ∧ b ∈ c ⇒ x ∈ a). Here we regard the set c as a collection of sets. We form a new set by including every element of each of the member sets of c. If c is not a collection of sets then the statement is vacuously true and a is possibly empty. Again a is not a well defined set, the axiom does not exclude other possible elements that do not satisfy our stated condition. An application of the axiom of specification (ZF2) will form a set that contains only those elements that are contained in the sets of c. Let a0 represent that set whose elements consist of only those elements of a that are elements of members of c. We write a0 =

[

b or simply a0 =

[ c

b∈c

which is read a prime is the union of c. If c consists of only two elements, b1 , b2 , then we write b1 ∪ b2 . A closely related concept to the union of sets is the intersection of sets. Let c be a collection of sets, we define the intersection of this collection of sets by

\ b∈c

b = {x ∈

[

| x ∈ b for every b in c}.

c

The reader may question why we choose the Union of c for our Universal set when any element of c would suffice. By choosing the Union we eliminate the problem of devising some mechanism to choose some element of c to be our Universal set.

8 If we are constructing the intersection of only two sets, b1 , b2 , then we write b1 ∩ b2 . Definition We say the sets a and b are disjoint, if a ∩ b = ∅. Definition For two sets a and b we say a is a subset of b, written a ⊂ b, if and only if every element of a is an element of b; i.e. a ⊂ b ⇔ (x ∈ a ⇒ x ∈ b). Theorem 2.2 a = b if and only if a ⊂ b and b ⊂ a. This theorem follows immediately from the definition of subset and the axiom of extension (ZF1). Note that it is vacuously true that ∅ ⊂ a for every set a. If a ⊂ b and a 6= b, we say that a is a proper subset of b, and we write a ( b. Definition For sets a and b, the complement of b in a, written a − b is the set a − b = {x ∈ a | x 6∈ b}. Theorem 2.3 De Morgan laws If c is a set and b is a collection of subsets [ \ \ [ a = (c − a). a = (c − a) and c − of c, then c− a∈b

a∈b

a∈b

a∈b

Proof h [ [ a a ⇒ x ∈ c and x 6∈ x∈c− a∈b

a∈b

⇒ x ∈ c and ∀a ∈ b x 6∈ a ⇒

∀a ∈ b x ∈ (c − a) i \ ⇒ x ∈ (c − a) a∈b

⇒ c−

[

a∈b

a⊂

\

(c − a)

a∈b

9 On the other hand h x∈

\

(c − a) ⇒ ∀ a ∈ b x ∈ (c − a)

a∈b

⇒ x ∈ c and ∀ a ∈ b x 6∈ a [ ⇒ x ∈ c and x 6∈ a [ i a ⇒ x∈c− ⇒

\

a∈b

a∈b

(c − a) ⊂ c −

a∈b

[

a

a∈b

By Theorem 2.2 we have the result c−

[

a=

c−

\

(c − a).

a∈b

a∈b

The proof that

\

a=

a∈b

[

(c − a)

a∈b

is similar and left to the reader.

The Axiom of Power Sets ZF5 Axiom of power sets.

For any set a there is a set b such that if

x ⊂ a, then x ∈ b. ∀a∃b(x ⊂ a ⇒ x ∈ b). Again we must apply an axiom of specification (ZF2)to construct a well defined set whose elements are only the subsets of a. This set is called the power set of a and is written P(a). By virtue of the axiom of pairing (ZF3) for any set a we may construct the singleton set {a}. Now by virtue of the axiom of unions(ZF4) we may

10 construct the set a ∪ {a}. This set is known as the successor of a and is written a + 1. We formally define the successor by: Definition For any set a the successor of a is the set a + 1 given by: a + 1 = a ∪ {a}. Examples: ∅ + 1 = ∅ ∪ {∅} = {∅} ∅ + 1 + 1 = {∅} ∪ {{∅}} = {∅, {∅}} For convenience we name these sets, ∅ = 0, {∅} = 1, {∅, {∅}} = 2. Thus we see: {} = 0 {0} = 1 {0, 1} = 2 .. . {0, 1, . . . , n − 1} = n .. . Remark: The notation a + 1 + 1 + · · · + 1 is unambiguous as its only interpretation is (. . . ((a + 1) + 1) + · · · + 1). To associate differently, e.g. a + (1 + 1 + · · · + 1) has no meaning. The notation a + n refers to the nth successor of a. That is a + n = a + 1| + 1 +{z· · · + 1}. n

11

The Axiom of Infinity The following axiom is a powerful statement that allows for the construction of arbitrarily large sets and allows us to regard unbounded classes of numbers as sets. ZF6 Axiom of infinity.

There exists a set, a, that contains ∅, and the

successor of each of its elements. ∃a(∅ ∈ a ∧ (x ∈ a ⇒ x + 1 ∈ a). A set that satisfies the axiom of infinity is called a successor set. Theorem 2.4 If a and b are successor sets, then a ∩ b is a successor set. Proof (a, b successor sets) ⇒ (∅ ∈ a and ∅ ∈ b) ⇒ (∅ ∈ a ∩ b). (α ∈ a ∩ b) ⇒ (α ∈ a and α ∈ b) ⇒ α + 1 ∈ a and α + 1 ∈ b) ⇒ α + 1 ∈ a ∩ b).

We generalize this theorem to include arbitrary intersections and hence the above theorem becomes a corollary to the following theorem. Theorem 2.5 If A is an arbitrary collection of successor sets, then

\

is a

A

successor set.

\ Proof Let A be a collection of successor sets. ∅ ∈ a ∀a ∈ A ⇒ ∅ ∈ a. a∈A \ \ a. α∈ a ⇒ α ∈ a ∀a ∈ A ⇒ α + 1 ∈ a ∀a ∈ A ⇒ α + 1 ∈ a∈A

a∈b

Let Ω be a successor set, and let A = {a ∈ P(Ω)|a is a successor set }, \ that is A is the collection of all successor subsets of Ω. We let ω = . A

We notice ω is unique regardless of the initial choice of Ω since if we let Ω1 and Ω2 be two successor sets we have Ω1 ∩ Ω2 is a successor set and Ω1 ∩ Ω2 ⊂ Ω1 and Ω1 ∩ Ω2 ⊂ Ω2 .

12 We also see ω is the minimal successor set. Thus we see ω = {0, 1, 2, . . . }. We can now construct the successors of ω, ω + 1 = {0, 1, 2, . . . , ω} and ω + 1 + 1 = {0, 1, 2, . . . , ω, ω + 1} We name ω + 1 + 1 = ω + 2 and ω + 1| + ·{z · · + 1} = ω + n. n

It is important to observe here that the successor of a set is not a successor set.

The Ordering of Sets, Cartesian Products and Functions Definition We say that a is less than b, written a < b, iff a ∈ b. We should note here that sets and elements are somewhat synonymous, they differ only in their relationship to each other. Now we wish to deal with the situation where we have a set with two or more elements and we wish to designate one element as the first element, and another as the second and so forth. Let us begin with the easiest case, a set with two elements. Let x and y be elements, by virtue of the axiom of pairing (ZF3) we can construct the two sets {x, y}, and {x}. Again using the axiom of pairing we construct the set {{x}, {x, y}}. For simplicity of notation we use (x, y) to represent the set {{x}, {x, y}}. Definition The ordered pair, (x, y), is the set {{x}, {x, y}}. We regard the ordered triple (x, y, z) as the ordered pair ((x, y), z) = {{{{x}, {x, y}}}, {{{x}, {x, y}}, z}}.

13 Thus we may inductively define the ordered n-tuple by (x1 , x2 , · · · , xn ) = ((x1 , x2 , · · · , xn−1 ), xn ). We now want to consider certain subsets of the collection of all ordered pairs formed from two sets. Definition The Cartesian product of two sets a and b is a × b = {z ∈ P(P(a ∪ b)) | z = {{x}, {x, y}} where x ∈ a, y ∈ b}. For simplicity of notation we write a × b = {(x, y)|x ∈ a, y ∈ b}. Definition A function from a set a to a set b is a subset, f , of a × b that satisfies the following two condition: 1. ∀x ∈ a ∃(x, y) ∈ f , and 2. if (x, y) ∈ f and (x, z) ∈ f then y = z. To work with functions efficiently, it helps to name the related sets. The set a is called the domain of f , and b is called the codomain of f . The range of f is the set {y ∈ b|(x, y) ∈ f }. If c ⊂ a, then the image of c under f is the set {y ∈ b | x ∈ c and (x, y) ∈ f }. If d ⊂ b, then the preimage of d under f is the set {x ∈ a|(x, y) ∈ f and y ∈ d}. We write f (c) for the image of c under f , and f −1 (d) for the preimage of d under f . Also, if the range of a function f is equal to its codomain, we say the function is onto its codomain.

14 We express a function, f , with domain a and codomain b by f : a → b. Also for a function f , if (x, y) ∈ f , then we express y as f (x). We may write y = f (x), or (x, f (x)) to represent the element (x, y) of f . We wish to generalize the cartesian product to arbitrary collections of sets. To do so we introduce the concept of indexing one set by another. Definition A function I from a set Λ onto a set a is said to index the set a by Λ. The set Λ is called the index and a is the indexed set. If I(λ) = a, then we write aλ for I(λ). Definition Let a be a non-empty set (we remind the reader here, that the elements of sets are considered to be sets) indexed by a set Λ. The cartesian product of a is defined to be the collection, Π, of all functions with domain [ b, satisfying the condition f (λ) ∈ bλ . We write Λ and codomain b∈a

c=

Y

bλ .

λ∈Λ

For clarity we present some examples here. Let a = {{1, 2}, {2, 3}}. Then we have the four functions f1 = {({1, 2}, 1), ({2, 3}, 2)} f2 = {({1, 2}, 1), ({2, 3}, 3)} f3 = {({1, 2}, 2), ({2, 3}, 2)} f4 = {({1, 2}, 2), ({2, 3}, 3)} This notation is cumbersome but we can represent these functions by the following ordered pairs without loss of any information. f1 = (1, 2),

f2 = (1, 3),

f3 = (2, 2),

f4 = (2, 3).

15 Hence the cartesian product is Y = {(1, 2), (1, 3), (2, 2), (2, 3)}. In this example where there are only two elements in a we may write {1, 2} × {2, 3} = {(1, 2), (1, 3), (2, 2), (2, 3)}. We leave as an exercise for the reader to demonstrate that the cartesian product of the following collection a = {{1, 2}, {2, 3}, {3, 4}} can be represented by Y αλ = {(1, 2, 3), (1, 2, 4), (1, 3, 3), (1, 3, 4), (2, 2, 3), (2, 2, 4), (2, 3, 3), (2, 3, 4)}. λ∈3

Let a = {{0, 1}} be indexed by ω. The cartesian product of a with respect to this indexing can be represented by the set of infinite strings of zeros and ones. That is Y

{0, 1}i = {(b0 , b1 , · · · )|bi = 0 or 1 ∀i ∈ ω}.

i∈ω

Any function, f , that satisfies the conditions in the previous definition is called a choice function. The rationale for this name is clear, as the function chooses an element from each set. Y Let a be a set and c = bλ . The projection map pbλ : c → bλ is the λ∈Λ

function defined by pbλ (x) = x(λ). For our original example we may compute p{1,2} (f1 ) = f1 ({1, 2}) = 1

16 and p{1,2} (f3 ) = f3 ({1, 2}) = 2. We leave as an exercise to compute p{1,2} (f2 ) and p{1,2} (f4 ).

The Axiom of Choice The next axiom is known as the Axiom of Choice. We give three formulations of the statement of this axiom. ZF7 Axiom of Choice. I.

For every nonempty set whose elements are nonempty sets there exists a choice function.

II.

If {ai } is a family of nonempty sets, indexed by a nonempty set I, then there exists a family {xi } with i ∈ I such that xi ∈ ai for each i ∈ I.

III.

The cartesian product of a nonempty collection of nonempty sets is Y nonempty. (∀a 6= ∅ ∨ x ∈ a ⇒ x 6= ∅) ⇒ 6= ∅. x∈a

Theorem 2.6 The three previous statements are equivalent. Proof I. ⇒ II. Let A be a collection of disjoint nonempty sets. We have [ [ A⊂ P( ai ). By I. there exists a choice function f on P( ai ). Let b be i∈I

i∈I

the image of A. Pick an element a ∈A, f (a) ∈ a ∩ b since f (a) ∈ a. Let y ∈ b where y 6= f (a) thus we have y = f (a0 ) where a0 6= a, and thus y ∈ a0 . Since a and a0 are disjoint y 6∈ a. Thus the only element of b ∩ a is f (a). II. ⇒ I. Define the choice function to be {(ai , xi ) | i ∈ I}.

17 I ⇐⇒ III. Since the cartesian product is the collection of choice functions, if a choice function exists then the cartesian product is non-empty. Conversely if the cartesian product is non-empty its elements are choice functions, thus a choice function exists.

The Axiom Schema of Replacement ZF8 Axiom schema of replacement.

If P (x, y) is a proposition such that

for each x in a set a, P (x, y) and P (x, z) implies that y = z, then there exists a set b such that y ∈ b if and only if there exists an x in a such that P (x, y). ∀x ∈ a(P (x, y) ∧ P (x, z) ⇒ y = z) ⇒ (∃b ∧ (y ∈ b ⇔ ∃x ∈ a ∧ P (x, y))). ZF8 allows the construction of new sets in the following way. If a set a exists and a rule that assigns to elements of a other pre-existing elements or sets that may or may not be elements of other sets, then there exists a set b that contains only those elements. This axiom schema will be heavily relied upon in the next chapter.

The Axiom Schema of Restriction ZF9 Axiom schema of restriction.

Let S(x) be any proposition involv-

ing x that does not involve y or z. If there exists an x such that S(x) is true, then there exists a y such that S(y) is true and, for all z, if z ∈ y then S(z) is false. If we take S(x) to be the statement x ∈ a, then we have the following statement, which we call the axiom of regularity. Axiom of regularity.

Every nonempty set a contains an element b such

that a ∩ b = ∅. ∀a 6= ∅∃b ∈ a ∧ a ∩ b = ∅.

18 Two important lemmas follow from the axiom of regularity. Lemma 2.7 For each set a, a 6∈ a. Proof The proof is indirect. We assume that there exists a set a such that a ∈ a. We thus have a ∈ {a} ∩ a. However by the axiom of regularity {a} contains an element whose intersection with {a} is empty. Since a is the only element we have {a} ∩ a = ∅, which contradicts our original assumption. Corollary There does not exist a set of all sets. Proof If there existed a set of all sets it would have to be an element of itself which would contradict the previous lemma.

.

Thus we see the axiom of regularity is the response to Russell’s paradox. Lemma 2.8

No two sets can be elements of each other.

Proof Again the proof is indirect. Assume that a and b are sets such that a ∈ b and b ∈ a. We thus have a ∈ {a, b}∩b and b ∈ {a, b}∩a. By the Axiom of regularity we must have an element x ∈ {a, b} such that x ∩ {a, b} = ∅. Since our only two choices are a or b we must have either {a, b} ∩ a = ∅ or {a, b} ∩ b = ∅ which contradicts our assumption. These two lemmas can be replaced by a more general theorem (Theorem 3.3) that will be stated and proved in the next chapter. Meanwhile we can use Lemma 2.8 to prove the following “cancellation” law. Theorem 2.9 If x + 1 = y + 1, then x = y. Proof x + 1 = y + 1 ⇒ x ∪ {x} = y ∪ {y} ⇒ either x = y or x ∈ y and y ∈ x. The latter case is a contradiction to lemma 2.8. Since the collection of all sets is an object that can be contemplated the study of collections can be extended to include collections that are not sets.

19 Collections that may or may not be sets are called classes. It is not within the scope of this book to study classes but we give the following definition. Definition A collection that is not a set is called a proper class. Exercises Prove the following “distributive” laws. 1. a ∪

¡\

¢ \¡ ¢ bλ = a ∪ bλ

λ∈Λ

2. a ∩

¡[

λ∈Λ

¢ [¡ ¢ bλ = a ∩ bλ

λ∈Λ

λ∈Λ

Also show 3. (a, b) = (c, d) ⇒ a = c and b = d.

20

III

ORDINAL NUMBERS Order Relations

Definition For any set a, each subset of a × a is called a relation on a. If R is a relation on a set a and if (x, y) ∈ R then we write xRy. Definition A relation that satisfies the following conditions: R xRx ∀x ∈ a A xRy & yRx ⇒ x = y. T xRy & yRz ⇒ xRz is called an order relation, or simply an order. Condition R is called reflexive, A is called antisymmetric, and T is called transitive. Let a be a set. In our context the elements of a are themselves sets. We see that R = {(x, y) | x ⊂ y} is an order relation on a. We leave the verification as an exercise for the reader. With an order relation we use the symbol x ¹ y instead of xRy. If x ¹ y we say that x precedes y or x is less than or equal to y. Definition If x ¹ y and y 6¹ x, then we say x is strictly less than (or simply less than) y, and we write x ≺ y. A set together with an order is said to be an ordered set. We often refer to an ordered set as a partially ordered set to differentiate it from linearly ordered and well ordered sets which we now define.

21 Definition An ordered set a that for every x, y ∈ a either x ¹ y or y ¹ x is said to be linearly ordered or totally ordered. Theorem 3.1 Trichotomy If a is a linearly ordered set and x, y ∈ a Exactly one of the following statements is true: i) x ≺ y. ii) y ≺ x. iii) x = y. Proof If a is linearly ordered then we have    i) (x ¹ y and y ¹ x) ⇒ x = y or    x ¹ y or y ¹ x ⇒ ii) (x ¹ y and y ¹ 6 x) ⇒ x ≺ y or     iii) (x 6¹ y and y ¹ x) ⇒ y ≺ x. We observe that neither x ¹ y and x 6¹ y, nor y ¹ x and y 6¹ x can be true at the same time, thus the statements are pairwise inconsistent. If ∃x ∈ a such that x ¹ y ∀y ∈ a, then we say x is the least element in a. Definition An ordered set a is said to be well ordered if and only if whenever b is any nonempty subset of a, then b has a least element. Theorem 3.2 Every well ordered set is linearly ordered. Proof Let a be a well ordered set. Let x, y ∈ a. since a is well ordered the subset {x, y} has a least element thus either x ¹ y or y ¹ x. Hence a is linearly ordered.

Ordinal Numbers Definition Let a be a set and let x ∈ a, the section of x with respect to the set a is S(x) = {y ∈ a | y ≺ x}.

22 The weak section of x is ¯ S(x) = {y ∈ a | y ¹ x}. Definition An ordinal number is a well ordered set a where for all x ∈ a, S(x) = x. In the last section we constructed the sets {} = 0 {0} =

1

{0, 1} =

2 .. .

{0, 1, . . . , n − 1} =

n .. .

By virtue of the axiom of infinity we may construct the sets ω= ω+1=

{0, 1, · · · } {0, 1, · · · , ω}

ω + 2 = ω + 1 + 1 = {0, 1, · · · , ω, ω + 1} .. . We can easily verify that these sets satisfy the definition of ordinal numbers where the order relation is x ⊂ y. We now want to construct ‘higher order’ ordinal numbers. The axiom of infinity will not work for us, since it only guarantees the existence of ω. We now must appeal to the axiom schema of replacement (ZF8) to continue our constructions.

23 Let P (x, y) be the proposition: For x ∈ ω, y is the xth successor of ω, i.e. y = ω + x. Since ordinal successors are unique if z = ω + x we must have y = z. Thus by the axiom schema of replacement (ZF8) there exists a set b, such that ω + n ∈ b for every n ∈ ω, and conversely b contains only those elements. We see that b = {ω, ω + 1, ω + 2, · · · }. We now construct the union of ω and b. ω ∪ b = {0, 1, 2, · · · , ω, ω + 1, ω + 2 · · · } We name this set ω2. We may repeat this process and construct the set ω3 where P (x, y) is the proposition: y is the xth successor of ω2. We continue constructing sets ω4, ω5, · · · . For clarity of discussion we shall refer to ωn as the nth multiple of ω. We let ω0 = 0 and ω1 = ω. We now let P (x, y) be the proposition: y is the xth multiple of ω. Since successors are unique and multiples are unique collections of successors we have unique multiples. Thus we apply the axiom of replacement (ZF8) and construct the set b = {0, ω, ω2, · · · } We now apply the axiom of unions (ZF4) to construct the union of all sets of b. We designate this set by ω 2 . [

= ω2

b

We may visualize ω 2 by the following array.     0, 1, 2, · · ·         ω, ω + 1, ω + 2, · · ·  2 ω =  ω2, ω2 + 1 ω2 + 2, · · ·         ..   .

24 We note here, that an element of ω 2 is of the form ωn + m where n, m ∈ ω. We may now construct successors of ω 2 , ω 2 + n, and by the axiom of [ replacement form the set b = {ω 2 + n|n ∈ ω}. The set we call ω 2 + ω. We b [ 2 continue as above to form the set b = {ω +ωn|n ∈ ω}, and the set we call b

ω 2 2. We continue and construct the multiples of ω 2 , ω 2 n. Again by virtue of the axiom of replacement we construct a set b = {0, ω 2 , ω 2 2, ω 2 3, · · · }. And [ by virtue of the axiom of unions we construct the set ω 3 = . b

We may continue in this fashion constructing the sets ω 4 , ω 5 , · · · . We refer to the set ω n as the nth power of ω. We again apply the axiom of replacement and construct the set b = {0, ω, ω 2 , ω 3 , · · · } and the union of the sets of b form the set [ . ωω = b

We observe here that an element of ω ω can be expressed in the form n X n n−1 ω k Ak . Where n ∈ ω and An ∈ ω. ω An + ω An−1 + · · · + A0 ≡ k=0

This process of course does not stop here but continues. We may form ω

sets ω ω , · · · . The set ω ω

· ··

we call ²0 . We of course have no reason to believe

that we have exhausted all ordinal numbers, and may continue in this fashion ad infinitum.

Transfinite Induction Theorem 3.3 The principle of transfinite induction. If a is an ordinal number and b ⊂ a such that, for x ∈ a, S(x) ⊂ b ⇒ x ∈ b, then b = a.

25 Proof Suppose to the contrary that a is an ordinal number and b ⊂ a such that for x ∈ a, S(x) ⊂ b ⇒ x ∈ b but, there exists c ∈ a such that c 6∈ b. Then there exists a nonempty set y = {x ∈ a | x 6∈ b}. Since y is a nonempty subset of a well ordered set it must have a least element, a0 , and S(a0 ) ⊂ b, thus by our hypothesis a0 ∈ b which contradicts our assumption. Thus we must conclude b = a.

Properties of Ordinal Numbers Lemma 3.4 The elements of ordinal numbers are ordinal numbers. Proof If a is an ordinal number and b ∈ a, then b = S(b) ⊂ a. We notice that any subset of b is a subset of a, thus b is well ordered, furthermore for any c ∈ b we have c ∈ a and thus c = S(c). Therefore b is an ordinal number.

Theorem 3.5 Let a, b be ordinal numbers, then either a ( b, or b ( a, or a = b. Proof Either a = b, or a 6= b. If a 6= b, then either ∃x ∈ a where x 6∈ b or ∃x ∈ b where x 6∈ a. If ∃x ∈ a where x 6∈ b, let t = {x ∈ b|x ∈ a} ⊂ b ∩ a ⊂ a. We show t = b by transfinite induction, and hence b ⊂ a. Let x ∈ b such that S(x) ⊂ t, thus S(x) ( a. Since S(x) is a proper subset of a we have {y ∈ a|y 6∈ S(x)} 6= ∅. Let r be the least element of {y ∈ a|y 6∈ S(x)}, then r = S(x) = x. Since r ∈ a we have x ∈ a and thus x ∈ t. Hence by Transfinite induction t = b. By the symmetric argument if ∃x ∈ b where x 6∈ a we have a ⊂ b. Definition An upper bound for an ordered set C is an element β such that x ¹ β ∀x ∈ C.

26 Definition A supremum or least upper bound for an ordered set C is an element α such that α is an upper bound, and if γ is an upper bound, then α ¹ γ. We indicate the supremum of an ordered set C by sup C. When the elements of an ordered sets are regarded as numbers we will use the symbols ≤ and < for ¹ and ≺. Thus for ordinal numbers the symbols ≤, ¹ and ⊂ are equivalent, as are 1, i 6= 1

38 b : n → (aj+1 , b1 ), where b : n − 1 → (a1 , bj ), where n > 1. To verify that the map defined by this recursion is a bijection consider the following diagram. The arrows indicate the “order” in which the elements are to be “counted”.

(a1 , b1 ) (a1 , b2 ) (a1 , b3 ) (a1 , b4 ) · · ·   % % % y (a2 , b1 )

(a2 , b2 ) %

(a3 , b1 )

% (a3 , b2 )

% (a4 , b1 ) .. .

(a2 , b3 ) % (a3 , b3 ) %

(a4 , b2 ) .. .

(a2 , b4 ) · · · (a3 , b4 ) · · · %

(a4 , b3 ) .. .

(a4 , b4 ) · · · .. .

If b(n) = b(m), then the set of predecessors of b(n) is equal to the set of predecessors of b(m). Thus the section of n is equal to the section of m, and thus n = m. So b is 1 to 1. If (a, b) ∈ A × B, then (a, b) = (ak , bl ) for some k, and l in ω. Then by “counting” backwards through the recursion we may determine an n where b(n) = (a, b) Corollary

Y

ω for n ∈ ω is countable.

i∈n

Theorem 4.5

The countable union of countable sets is countable.

Proof Without loss of generality we may assume the sets are disjoint. This is in fact the most extreme case. We may now let aij represent the ith element of the j th set. The argument is now identical to the proof of theorem 4.4.

39 ω ω is countable. [ Proof ω ω = Where b = {0, ω, ω 2 , · · · }. Corollary

b

From the above arguments it is easily seen that all the ordinal numbers that have explicitly been constructed by the method outlined in chapter III must be countable. Not all ordinals are countable as we shall see in the next chapter.

Some Uncountable Sets Theorem 4.6

The countably infinite product of countable sets each of

which has cardinality of at least 2 is not countable. Proof Assume the conclusion is false. That is assume there exists a bijecY tion b from ω to Ai , where each Ai is countable. i∈ω

Let pn be the projection map onto the nth co-ordinate space. For all Y n ∈ ω pick an ∈ An − {(pn ◦ b)(n)}. Then the sequence (a0 , a1 , · · · ) ∈ Ai . i∈ω

We have for all n ∈ ω pn ((a0 , a1 · · · )) = an , but (pn ◦ b)(n) 6= an . Thus b(n) 6= (a0 , a1 , · · · ) ∀n ∈ ω. Which is a contradiction. Thus no bijection exists.

Exercises The following set of exercises leads to an alternate proof of the Schr¨oderBernstein Theorem. Let X be a partially ordered set and A a subset of X. An element x ∈ X is said to be an upper bound for A if x ≥ a ∀a ∈ A. Equivalently, an

40 element y ∈ X is said to be a lower bound for A if y ≤ a ∀a ∈ A. An upper bound u of A is the Supremum or least upper bound of A if u ≤ x for all upper bounds x of A. A lower bound l of A is the infimum or greatest lower bound of A if l ≥ y for all lower bounds y of A. We say that a partially ordered set X is complete if there exists a supremum and infimum for every subset of X. Let X and Y be partially ordered sets. A function f : X → Y is order preserving if x ≤ y ⇒ f (x) ≤ f (y). 1. If L is a complete partially ordered set and f : L → L is an order preserving function, show that there exists a ∈ L such that f (a) = a. 2. Let A be any set. Show that P(A) is a complete partially ordered set, where X ≤ Y if X ⊆ Y (we say that P(A) is ordered by inclusion). For simplicity of notation define X − A = A0 , that is A0 is the complement of A in X. 3. Let A and B be sets and f and g be functions such that f : A → B and g : B → A. Let h : P(A) → P(A) be defined by h(S) = [g(f (S)0 )]0 . Show that T ⊂ S ⇒ h(T ) ⊂ h(S). 4. Observe that h is an order preserving map, thus there exists S ⊂ L such that h(S) = S. that is g(f (s)0 )0 = X, and g(f (S)0 ) = S 0 . Now assume f and g are one to one and demonstrate the bijection from A to B.

41

V

ZORN’S LEMMA AND WELL ORDERING

We have shown that there are sets with cardinality greater than ℵ0 but we have yet to demonstrate that there exists ordinal numbers with cardinality greater than ℵ0 . We will accomplish this task by showing that every set can be well ordered, and that every well ordered set is order isomorphic to an ordinal number. By order isomorphic we shall mean the following. Definition Two partially ordered sets a and b are said to be order isomorphic if there exists a bijection between them that preserves order. That is if β : a−→b is the order preserving bijection then x ≤ y if and only if β(x) ≤ β(y). We write a ' b.

Zorn’s Lemma Theorem 5.1 Zorn’s Lemma: If X is a partially ordered set such that every chain in x has an upper bound in X, then X contains a maximal element. By chain we mean a totally or linearly ordered subset. In the hypothesis of Zorn’s Lemma the upper bound need not be in the chain. In the conclusion the maximal element is simply an element with no superior, that is if x ≤ y ⇒ x = y then x is maximal. There may in fact be elements in X that are not comparable to x, yet x may still be maximal. ¯ Proof Let X be a partially ordered set. For each x ∈ X, S(x) = {y ∈ X | ¯ is a function from X to P(X) since the y ≤ x} is the weak section of x. S ¯ is a collection of subsets section of any element is unique. The range R of S ¯ is one to one, that are partially ordered by inclusion, i.e. A ≤ B if A ⊆ B. S

42 ¯ ¯ since S(x) ⊆ S(y) if and only if x ≤ y. Thus if we find a maximal element ¯ ¯ S(z) in R then z is maximal in X. Also C is a chain in X if and only if S(C) is a chain in R. Let X be the set of all chains in X. Let Γ ∈ X . Since Γ is a chain it ¯ has an upper bound x by hypothesis thus Γ ⊆ S(x) for some x ∈ X. X is a non-empty collection of sets partially ordered by inclusion. Now if C is a [ chain in X then G = Γ ∈ X . G is an upper bound of C in X as each Γ∈C

element of C is dominated by G. Now let X be an arbitrary non-empty collection of subsets of a non-empty set X subject to 1. if A ∈ X and B ⊆ A then B ∈ X, and [ 2. if C is a chain in X then Γ ∈ X. Γ∈C

Notice that these are the exact conditions that our set X had in the previous discussion. Also notice that the first condition implies that ∅ ∈ X. Our task is to show that X has a maximal element. Now let φ be a choice function from the non-empty subsets of X to X, i.e φ : (P(X) − ∅)−→X. We note that φ is a function such that φ(A) ∈ A for all non-empty subsets A of X. For each set A ∈ X let Aˆ = {x ∈ X | A ∪ {x} ∈ X}. Define a function γ :X−→ X by the following:   A ∪ {φ(Aˆ − A)} if Aˆ − A 6= ∅ γ(A) =  A if Aˆ − A = ∅. We observe that Aˆ − A = ∅ if and only if A is maximal. Our task is now to show there exists a set A in X such that γ(A) = A. Since φ(Aˆ − A) is a

43 single element we notice that γ(A) contains at most one more element than A. We now define a tower as a subcollection T of X that satisfies the following conditions: 1. ∅ ∈ T , 2. if A ∈ T , then γ(A) ∈ T , and 3. if C is a chain in T , then

[

A ∈ T.

A∈C

We notice here that X satisfies the conditions for a tower, and thus towers exist. We can also easily verify that the intersection of a collection of towers is \ Tλ . a tower. Let {Tλ } be a collection of towers, since ∅ ∈ Tλ for all λ, ∅ ∈ λ \ Tλ , then A ∈ Tλ for all λ, thus γ(A) ∈ Tλ for all λ and thus If A ∈ λ \ \ Tλ , then C is a chain Tλ for all λ. And finally if C is a chain in γ(A) ∈ λ λ [ [ \ Tλ . It follows in Tλ for all λ, thus A ∈ Tλ for all λ, and thus A∈ A∈C

A∈C

λ

that the intersection of all towers T0 is the smallest tower. We now wish to show that T0 is a chain. We say that a set B in T0 is comparable if it is comparable with every set in T0 , that is, for all A ∈ T0 either A ⊆ B or B ⊆ A. To show that T0 is a chain we show that every set in T0 is comparable. We now let B be an arbitrary comparable set in T0 . Comparable sets do exist since ∅ is clearly comparable. Suppose A ∈ T0 and A is a proper subset of B. Since B is comparable we have γ(A) ⊆ B or B is a proper subset of γ(A). If B ⊂ γ(A) we have A as a proper subset of B and B a proper subset

44 of γ(A), but γ(A)−A is a singleton thus there cannot be a set between them. We conclude γ(A) ⊆ B. Now consider the collection U of all sets A in T0 where either A ⊆ B or γ(B) ⊆ A. The collection U is no larger than the collection of sets in T0 that are comparable with γ(B) since, if A ∈ U and since B ⊆ γ(B) we have either A ⊆ γ(B) or γ(B) ⊆ A. We now claim that U is a tower. We verify the three conditions: 1) ∅ ∈ U . 2) To show A ∈ U ⇒ γ(A) ∈ U Consider three cases i. A ⊂ B. ii. A = B. iii. γ(B) ⊆ A. For i. γ(A) ⊆ B by the preceding argument thus γ(A) ∈ U . For ii. γ(A) = γ(B) thus γ(B) ⊆ γ(A), therefore γ(A) ∈ U . For iii. γ(B) ⊆ A ⇒ γ(B) ⊆ γ(A), therefore γ(A) ∈ U . [ 3) Let C be a chain in U , if γ(B) ⊆ D for some D ∈ C, then γ(B) ⊆ D D∈C [ [ hence D ∈ U . If however D ⊆ B for all D ∈ C, then D ⊆ B. Thus D∈C D∈C [ we may conclude D ∈ U. D∈C

We thus conclude that U is a tower and is a subset of T0 which is the smallest tower hence we have U = T0 .

45 Now let B be a comparable set, we form U as above and since U = T0 for any A ∈ T0 we have either A ⊆ B ⊆ γ(B) or γ(B) ⊆ A. We thus conclude that if B is a comparable set, then γ(B) is comparable also. We have that ∅ is comparable and γ maps comparable sets to comparable sets. Now since the union of a chain of comparable sets is comparable we may conclude that the comparable sets constitutes a tower, and hence they exhaust all of T0 . [

T0 is a chain, thus if A =

B ∈ T0 we have γ(A) ⊆ A, since the union

B∈T0

A includes all the sets in T0 . We always have A ⊆ γ(A), thus we conclude that A = γ(A). This is the condition that we noted earlier needed to be shown to complete the proof. Definition A well ordered set A is a continuation of a well ordered set B if i) B ⊂ A ii) B = S(a) for some a ∈ A iii) For a, b ∈ B, a ≤B b iff a ≤A b. The reason for the third condition is that a set may have more than one ordering, and for continuation we want B to have the same ordering as A when restricted to B.

The Well Ordering and Counting Theorems Theorem 5.2 The Well Ordering Theorem Every set can be well ordered.

46 An important point to be noted here is that the set may be presented with an ordering that is not a well ordering. The Well Ordering Theorem says we may disregard any previously assigned ordering that the set may have and endow it with a new ordering that is a well ordering. Before we prove this theorem we make this note. We shall regard an ordered set X as the pair (X, 0. Let d be the least element of {y|(x, y) ∈ [a, b] and b > 0}. We then let the unique element (c, d) ∈ [a, b] be the canonical representative of the rational number [a, b]. We again have the natural order defined on the rational numbers given by [x, y] < [z, w] iff xw < yz.

69 Theorem 7.7 The ordering of rational numbers is well defined. Proof Let [x, y] = [a, b], [z, w] = [c, d], and [x, y] < [z, w] with b, d, y, w > 0. We thus have ay = bx, dz = cw and xw < yz ⇒ bdxyw2 < bdy 2 zw ⇒ ady 2 w2 < bcy 2 w2 ⇒ ad < bc ⇒ [a, b] < [c, d]. Hence the ordering is well defined. We define addition and multiplication as follows, [x, y] + [z, w] = [xw + zy, yw] [x, y] · [z, w]

= [xz, yw].

Theorem 7.8 Addition and multiplication of rational numbers are well defined. Proof Let [x, y] = [a, b] and [z, w] = [c, d]. For addition we have ay = bx and dz = cy ⇒ aydw = bxdw and bydz = bycw ⇒

bdxw + bdyz = adyw + bcyw

⇒

[xw + yz, yw] = [ad + bc, bd].

Hence addition is well defined. For multiplication we have ay = bx and dz = cy ⇒

acyw = bdxz

⇒

[ac, bd] = [xz, yw].

70 Hence multiplication is well defined. Just as there is a natural embedding of the natural numbers into the integers, there is the natural embedding of the integers into the rational numbers given by the injection, J : x ,→ [x, 1]. It is easy to verify that this injection is an embedding.

A Field We leave it the reader to verify the following properties of Q. 1. a + b ∈ Q ∀a, b ∈ Q. 2. ab ∈ Q ∀a, b ∈ Q. 3. (a + b) + c = a + (b + c) ∀a, b, c ∈ Q. 4. (ab)c = a(bc) ∀a, b, c ∈ Q. 5. a + b = b + a ∀a, b ∈ Q. 6. ab = ba ∀a, b ∈ Q. 7. a(b + c) = ab + ac ∀a, b, c ∈ Q. 8. ∃e ∈ Q such that a + e = e + a = a ∀a ∈ Z 9. ∃u ∈ Q such that au = ua = a ∀a ∈ Q 10. ∀a ∈ Q ∃(−a) ∈ Q such that a + (−a) = (−a) + a = 0 11. ∀a ∈ Q, a 6= 0, ∃a−1 such that aa−1 = a−1 a = 1

71 The first 10 properties are identical to the properties of an Integral Domain. Property 11∗ is replaced by property 11 where a−1 is called the multiplicative inverse, or reciprocal. Any set with two binary operations satisfying these 11 properties is called a Field. Exercise Show that every field is an integral domain. That is to say, every field has no zero divisors.

Differences and Quotients Definition The difference between integers or rational numbers a and b is a + (−b), which is written a − b. Definition The quotient of two rational numbers a and b is a · b−1 , which a is written . b We can see that difference and quotient can be regarded as a binary operations, we also notice that neither operation is commutative nor associative. Exercise For any two rational numbers p < q, show that p
0, then pr ≤ qr. 2. If p, q, r are rational numbers where p ≤ q and r < 0, then pr ≥ qr. 3. If p and q are positive rational numbers, then

p q ≥ 1 ⇒ ≤ 1. q p

72

Mathematical Induction The next theorem is a special case of transfinite induction, that is widely used in many situations. Before we state and prove the theorem we need two small lemmas that we present as exercises. Exercise 1. Define the map φ : N → Z+ by φ(a) = [a + 1, 0]. Show that φ(a) + 1 = φ(a + 1). Exercise 2. Show that Z+ is order isomorphic to ω. Theorem 7.9 The Principle of Mathematical Induction If T ⊆ Z+ such that the following conditions are true: i. 1 ∈ T ii. if k ∈ T , then k + 1 ∈ T , then T = Z+ . Proof Consider the order preserving bijection φ : ω → Z+ defined by φ(a) = [a + 1, 0]. Let A = φ−1 (T ). Let x ∈ ω such that S(x) ⊂ A. If x = 0, then φ(x) = 1 ∈ T ⇒ x ∈ A. If x 6= 0, then S(x) ⊂ A

⇒x−1∈A ⇒ φ(x − 1) ∈ T ⇒ φ(x − 1) + 1 ∈ T ⇒ φ(x) ∈ T ⇒ x ∈ A.

Thus by Transfinite Induction A = ω ⇒ T = φ(A) = φ(ω) = Z+ .

73

The Cardinality of Integers and Rational Numbers

Theorem 7.10 Both the integers and the rational numbers are countable. Proof By Theorem 4.4 we know that N × N is countable. there exists the natural embedding of Z into N × N by identifying [a, b] with its canonical representative (a − b, 0) or (0, b − a). Thus there exists a bijection from Z to a subset of a countable set, hence Z is countable. Again by Theorem 4.4 we know that Z × Z is countable, and there exists the natural embedding of Q into Z × Z by identifying [a, b] with its canonical representative, (c, d) where d is positive and minimal. Thus there exists a bijection from Q to a subset of a countable set, hence Q is countable. Let Z∗ represent the image of the embedding of Z into Q. Also let Z+ ∗ and Z− ∗ represent respectively the images of the embeddings of the positive and negative integers into the rationals.

The Archimedian Property Theorem 7.11 Archimedian Property ∀r ∈ Q ∃n ∈ Z+ ∗ such that r < n. Proof Let r = [a, b], if r ≤ 1, then r < 2 and we are done. If r > 1, then without loss of generality, both a, and b are positive integers, and b(a + 1) ≥ a + 1 > a ⇒ [a, b] < [a + 1, 1] ∈ Z+ ∗. Lemma For positive rational numbers r and s, if r > s, then r−1 < s−1 .

74 Proof First we note that if r = [a, b], then r−1 = [b, a], this is immediate by computing [a, b] · [b, a] = [ab, ab] = [1, 1]. Now let r = [a, b] and s = [c, d]. [a, b] > [c, d]

⇔ ad > bc ⇔ [d, c] > [b, a] ⇔ s−1 > r−1 .

For any integer, a, the product of a with itself b times, where b is a positive integer is denoted ab . Exercise Prove that for any positive integer, n, there exists an integer of the form 2m for some positive integer m, such that 2m > n. (Hint: use induction). Solution For n = 1, 1 < 2 = 21 . Now assume n < 2m for some m, then n + 1 < 2m + 1 < 2m + 2m = 2m+1 . Exercise Prove that for any positive rational number, q, there exists a rational number of the form 2n , where 2n > q, and where n is the embedded image of a positive integer. Solution Let q = (a, b) where a and b are positive integers. We then have (a, b) ≤ (a, 1) < (2n , 1) for some n.

The Division Algorithm Theorem 7.12 The division algorithm If a and d are integers with d > 0, then there exist unique integers q and r such that a = dq + r and 0 ≤ r < d. Proof This result is a consequence of the well ordering property. Let S = {x ∈ Z|x = a − dn ∀n ∈ Z}, and let S 0 = {x ∈ S|x ≥ 0}, S 0 is thus

75 the embedded image of some subset of N, and thus if it is non-empty it must have a least element. If a ≥ 0, then let n = 0, and thus x = a − 0 = a ≥ 0. Thus a ∈ S 0 . If a < 0, then let n = a, thus x = a−ad = a(1−d) ≥ 0. Thus a−ad ∈ S 0 . Thus S 0 6= ∅. Since S 0 6= ∅, and is embedded image of a subset of N, S 0 has a least element. Let the least element be r. Thus we have r = a − dq ≤ s ∀s ∈ S 0 and a = dq + r where r ≥ 0. We now must show r < d. We have a − d(q + 1) = a − dq − d = r − d, thus r − d ∈ S. Since r is the least element in S 0 and r − d < r we have r − d < 0 ⇒ r < d. We thus have a = dq + r with 0 ≤ r < d. Now we have to show that q and r are unique. Suppose a = dq1 + r1 and a = dq2 + r2 where 0 ≤ r1 < d and 0 ≤ r2 < d. Without loss of generality we may assume r1 ≤ r2 . We thus have 0 ≤ r2 − r1 < r2 < d. We note that 0 ≤ r2 − r1 = a − dq2 − a + dq1 = d(q1 − q2 ). Thus r2 − r1 is a multiple of d and non-negative. We thus have 0 ≤ r2 − r1 < d and r2 − r1 = d(q1 − q2 ) ⇒ 0 ≤ d(q1 − q2 ) < d ⇒ 0 ≤ q1 − q2 < 1 ⇒ q1 − q2 = 0. Thus q1 = q2 , and thus r1 − r2 = 0 ⇒ r1 = r2 .

76

Exercises 1. Verify that the relation (a, b) ≡ (c, d) ⇔ a+d = b+c is an equivalence relation on N × N, but not on α × α where α > ω. 2. Verify that the relation (x, y) ≡ (z, w) ⇔ xw = yz is an equivalence relation on Z × (Z − {0}) For any Integral Domain D show: 3. a · e = e ∀a ∈ D and where e is the additive identity. 4. u · (−u) = −u, where u is the multiplicative identity. 5. (−u) · (−u) = u, where u is the multiplicative identity. 6.

If a, z ∈ D and a + z = a, then show z = e. Solution to 3: a = a · u = a · (u + e) = a + a · e ⇒ a · e = e.

7. If v, a ∈ F , where F is a field and a · v = a, then show v = u. 8. Show that every Field is an Integral Domain. Mathematical Induction is often used to prove certain identities. Exercise 9 and its solution exemplifies how this is done. Exercise 10 is left as practice. n X

n(n + 1) ∀n ∈ Z+ . 2 k=1 ( ¯ n ) ¯ X n(n + 1) ¯ k= Solution to 9: Let A = n ¯ ¯ 2 k=1

9. Show that

i.

1 X k=1

k=

k=1=

1(2) thus 1 ∈ A. 2

77 ii. Assume m ∈ A, then m+1 X

k =m+1+

k=1

m X

k =m+1+

k=1

m(m + 1) (m + 1)(m + 2) = . 2 2

Thus m + 1 ∈ A. Therefore by Mathematical Induction A = N, and n X

k=

k=1

10. Show that

n(n + 1) ∀n ∈ Z+ . 2

n X k=1

2k − 1 = n2 ∀n ∈ Z+ .

78

VIII

REAL NUMBERS

It is a well known fact, and a standard exercise, that

√

2 is not rational.

This means the equation x2 − 2 = 0 has no solutions in the rational numbers, hence the need to extend the rational numbers to a larger set that would include the solutions to such equations. The next set we develop is the set the real numbers which we will indicate by R.

Dedekind Cuts and Real Numbers Definition A Cut or Dedekind cut of the rational numbers is a subset A of the rational numbers such that 1. A 6= ∅, and A 6= Q. 2. If a ∈ A, and b < a, then b ∈ A. 3. if a ∈ A ∃ a0 ∈ A, such that a < a0 . From this definition we see that a cut forms a partition of the rational numbers into two non-empty subsets, the cut and its complement, where every element of the cut is less than every element of its complement. The cut is called the lower set and its complement is called the upper set. Definition The set of Real Numbers, R, is the collection of all cuts of the rational numbers. We define a natural ordering of the real numbers by the following.

79 Definition For two real numbers, A and A0 , we say A < A0 if A ⊂ A0 . We emphasize the inclusion is proper, i.e., A 6= A0 . Theorem 8.1 For any rational number r the set z = {p|p < r} is a cut, and hence a real number. Proof We demonstrate that z satisfies the three conditions of the definition of a cut. 1) r − 1 < r ⇒ r − 1 ∈ z Thus z 6= ∅. Also r 6< r ⇒ r 6∈ z ⇒ z 6= Q. 2) If q ∈ z and p < q, then p < r, thus p ∈ z. 3) If q ∈ z, then q < r ⇒ q
r, then r ∈ xˆ.

80

Trichotomy Lemma 8.2 The trichotomy property for real numbers For any real number xˆ exactly one of the following is true: xˆ > ˆ0, xˆ = ˆ0, or xˆ < ˆ0. Proof Let xˆ be a real number, then exactly one of the following is true: ˆ0 ⊂ xˆ, xˆ = ˆ0, or xˆ ⊂ ˆ0. Definition If xˆ > 0 then we say xˆ is a positive real number. If xˆ < 0 then we say xˆ is a negative real number.

Addition We define addition of real numbers by the following: Definition xˆ + yˆ = {r|r < p + q, p ∈ xˆ, and q ∈ yˆ}. Theorem 8.3 The sum xˆ + yˆ is a real number. Proof We show that xˆ + yˆ satisfies the definition of a cut. 1) Since xˆ and yˆ are not empty there exists p ∈ xˆ and q ∈ yˆ, and p + q − 1 < p + q ⇒ xˆ + yˆ 6= ∅. Also since ∃c 6∈ xˆ and ∃d 6∈ yˆ we have c > t ∀t ∈ xˆ and d > w ∀w ∈ yˆ. We have c + d > t + w ∀t + w ∈ xˆ + yˆ. Thus c + d 6∈ xˆ + yˆ, and so xˆ + yˆ 6= Q. 2) Let a ∈ xˆ + yˆ and b < a. We have b < a < p + q where p ∈ xˆ and q ∈ yˆ. Thus b ∈ xˆ + yˆ. 3) Let a ∈ xˆ + yˆ, thus a < p + q where p ∈ xˆ and q ∈ yˆ. There exists r > p such that r ∈ xˆ, thus p + q < r + q. Thus a < p + q ∈ xˆ + yˆ.

81 Corollary xˆ + yˆ = {p + q|p ∈ xˆ, and q ∈ yˆ}. Proof From part 2) we have {p + q|p ∈ xˆ, and q ∈ yˆ} ⊆ {r|r < p + q, p ∈ xˆ, and q ∈ yˆ}. Now if r < p + q where p ∈ xˆ and q ∈ yˆ, then r − p < q, thus r − p ∈ yˆ and so r = p + (r − p) where p ∈ xˆ and (r − p) ∈ yˆ, thus {r|r < p + q, p ∈ xˆ, and q ∈ yˆ} ⊆ {p + q|p ∈ xˆ, and q ∈ yˆ}. Thus xˆ + yˆ = {p + q|p ∈ xˆ, and q ∈ yˆ}. We leave to the reader as an exercise to show addition of real numbers is commutative and associative; i.e., xˆ + yˆ = yˆ + xˆ, and xˆ + (ˆ y + zˆ) = (ˆ x + yˆ) + zˆ for all real numbers xˆ, yˆ, zˆ. Exercises For p, q ∈ Q, and A ∈ N show i) pˆ + qˆ = p[ + q. X X \ ii) pˆi = pi . i∈A

i∈A

Theorem 8.4 For any real number xˆ and the real number ˆ0, xˆ + ˆ0 = xˆ. Proof: If r ∈ xˆ + ˆ0, then r = p + q where p ∈ xˆ and q < 0, thus r = p + q < p ⇒ r ∈ xˆ, thus xˆ + ˆ0 ⊆ xˆ. If r ∈ xˆ, then there is a rational number s ∈ xˆ, such that s > r, thus r − s < 0, thus r − s ∈ ˆ0, and (r − s) + s = r, thus xˆ ⊆ xˆ + ˆ0. Hence xˆ + ˆ0 = xˆ. We say ˆ0 is the additive identity. Definition If for two real numbers xˆ, and yˆ we have xˆ + yˆ = ˆ0, then we say yˆ is the additive inverse or opposite of xˆ. We use the notation −ˆ x to indicate the additive opposite of xˆ.

82 Theorem 8.5 Every real number has an additive inverse. Proof Let xˆ be a real number and let y = {q ∈ Q|q + p < 0 ∀ p ∈ xˆ}. We first show that y is a real number, i.e., y is a cut. 1) Since xˆ 6= Q ∃ d ∈ Q such that d 6∈ xˆ. Thus d > p ∀ p ∈ xˆ. Thus −d < −p ∀ p ∈ xˆ. Thus p + (−d) < 0 ∀ p ∈ xˆ. Thus −d ∈ y. Thus y 6= ∅. Also we notice for any p ∈ xˆ, p + (−p) = 0, thus −p 6∈ y, thus y 6= Q. 2) Let q ∈ y and a < q. We thus have a + p < q + p < 0 ∀ p ∈ xˆ. Thus a ∈ y. 3) Let q ∈ y, thus q + p < 0 ∀p ∈ xˆ. Now assume p + q + r ≥ 0 for some p ∈ xˆ and ∀r ∈ Q+ . Thus p + r 6∈ xˆ ∀r ∈ Q+ . But ∀p ∈ xˆ ∃s > p such that s ∈ xˆ. Let r = s − p, and we thus have p + r = s ∈ xˆ, which contradicts our assumption. Thus ∀p ∈ xˆ ∃r ∈ Q+ such that q + p + r < 0, thus q + r ∈ y, and q < q + r. Thus y is a cut and we write y = yˆ. We now show that xˆ + yˆ = ˆ0. If r ∈ xˆ + yˆ, then r = p + q < 0, thus ˆ xˆ + yˆ ⊆ 0. If s ∈ ˆ0, and p ∈ xˆ is arbitrary, then s − p ∈ Q and p + s − p = s < 0, thus s − p ∈ yˆ. Thus s = p + s − p ∈ xˆ + yˆ, and thus ˆ0 ⊆ xˆ + yˆ, and thus xˆ + yˆ = ˆ0. Lemma 8.6 −(−ˆ x) = xˆ. Proof We have xˆ + (−ˆ x) = ˆ0 = −(−ˆ x) + (−ˆ x) ⇒ ⇒

xˆ + (−ˆ x) + xˆ = −(−ˆ x) + (−ˆ x) + xˆ xˆ = −(−ˆ x)

83 Lemma 8.7 xˆ > ˆ0 if and only if −ˆ x < ˆ0. Proof Assume xˆ > ˆ0 then 0 ∈ xˆ, so ∃p ∈ xˆ such that p > 0. Now assume −ˆ x ≥ ˆ0, then ∃q ∈ −ˆ x such that q > −p. Thus q + p > 0, and thus xˆ + −ˆ x > ˆ0, which is a contradiction. Now assume −ˆ x < ˆ0, then ∃q < 0 such that q > p ∀p ∈ −ˆ x. So −q < −p ∀p ∈ −ˆ x. Thus p + −q < 0 ⇒ −q ∈ xˆ. Since we have 0 < −q, we ˆ have 0 ∈ xˆ, and thus xˆ > 0. Corollary −ˆ0 = ˆ0. Lemma 8.8 −(ˆ x + yˆ) = −ˆ x + (−ˆ y ). Proof We have −ˆ x + (−ˆ y ) = {r + s|r ∈ −ˆ x and s ∈ −ˆ y }. Thus for r ∈ −ˆ x and s ∈ −ˆ y , r + p < 0 ∀p ∈ xˆ and s + q < 0 ∀q ∈ yˆ. Thus r + s + p + q < 0 ∀(p + q) ∈ xˆ + yˆ. Thus −ˆ x + (−ˆ y ) = {r + s|(r + s) + (p + q) < 0, r ∈ −ˆ x, s ∈ −ˆ y ∀(p + q) ∈ (ˆ x + yˆ)}. Now we also have −(ˆ x + yˆ) = {t|t + (p + q) < 0 ∀((p + q) ∈ xˆ + yˆ}. Since (r + s) ∈ Q ∀r ∈ −ˆ x and s ∈ −ˆ y we have (r + s) ∈ −(ˆ x + yˆ). Thus we have −ˆ x + (−ˆ y ) ⊆ −(ˆ x + yˆ). Now let t ∈ −(ˆ x + yˆ), then t + (p + q) < 0 ∀p ∈ xˆ and q ∈ yˆ. Also t + (p + q) t + (p + q) t + (p + q) < < 0. So − p + p < 0 ∀p ∈ xˆ and q ∈ yˆ. 2 2 t + (p + q) t + (p + q) Thus − p ∈ −ˆ x. Also − q + q < 0 ∀p ∈ xˆ and q ∈ yˆ. 2 2 t + (p + q) t + (p + q) t + (p + q) − q ∈ −ˆ y . And we have t = −p+ − q. Thus 2 2 2 Thus t ∈ −ˆ x + (−ˆ y ). Thus −(ˆ x + yˆ) ⊂ −ˆ x + (−ˆ y ). Thus we have −(ˆ x + yˆ) = −ˆ x + (−ˆ y ).

84

Multiplication We now define multiplication of real numbers. First we define the product of two positive real numbers by: Definition For xˆ, yˆ > ˆ0, xˆ · yˆ = {r|r < pq, p ∈ xˆ, q ∈ yˆ, and p > 0, q > 0}. For ease of notation we will often use juxtaposition to indicate the operation of multiplication. That is xˆyˆ ≡ xˆ · yˆ. Theorem 8.9 If xˆ and yˆ are positive real numbers, then xˆyˆ is a positive real number. Proof We show xˆyˆ is a cut. 1) xˆ, yˆ > ˆ0 ⇒ ∃p > 0, q > 0, p ∈ xˆ and q ∈ yˆ, thus 0 < pq, thus 0 ∈ xˆyˆ, thus xˆyˆ 6= ∅. Now ∃a 6∈ xˆ and ∃b 6∈ yˆ, thus a > p ∀p ∈ xˆ and b > q ∀q ∈ yˆ, thus ab > pq ∀p ∈ xˆ and q ∈ yˆ. Hence ab 6∈ xˆyˆ. Hence xˆyˆ 6= Q. 2) Let r ∈ xˆyˆ, r < pq for some p ∈ xˆ and q ∈ yˆ. If s < r, then s < pq and thus s ∈ xˆyˆ. 3) Let r ∈ xˆyˆ. We have r < pq for some positive p ∈ xˆ and q ∈ yˆ. Since ∃s ∈ xˆ such that p < s and ∃t ∈ yˆ such that q < t we have r < pq < st, thus pq ∈ xˆyˆ. Since xˆyˆ satisfies the definition of a cut we conclude xˆyˆ is a real number. To show that xˆyˆ is positive we notice that xˆ > ˆ0 ⇒ ˆ0 ⊂ xˆ ⇒ 0 ∈ xˆ ⇒ ∃p > 0, p ∈ xˆ. Similarly ∃q > 0, q ∈ yˆ. Now let r ∈ xˆ and r > 0, s ∈ yˆ and s > 0 be arbitrary. We thus have 0 < rs ⇒ 0 ∈ xˆyˆ ⇒ ˆ0 ⊂ xˆyˆ ⇒ xˆyˆ > 0.

85 We complete the definition of multiplication by:   ˆ0  if xˆ = ˆ0 or yˆ = ˆ0      −(−ˆ x)ˆ y if xˆ < ˆ0 and yˆ > ˆ0 xˆyˆ =   x(−ˆ y )) if xˆ > ˆ0 and yˆ < ˆ0 −(ˆ     ((−ˆ x)(−ˆ y )) if xˆ < ˆ0 andˆ y < ˆ0 Corollary If xˆand yˆ are real numbers, then xˆyˆ is a real number. Associativity and commutativity of multiplication follows immediately from the definition of multiplication and the associativity and commutativity of rational numbers. Exercise For p, q ∈ Q, A ∈ N show i) pˆ · qˆ = pd · q. ii)

Y i∈A

pˆi =

Y [ pi . i∈A

Theorem 8.10 The real number ˆ1 = {p|p < 1} is the multiplicative identity. Proof First consider a positive real number xˆ. xˆ · ˆ1 = {r|r < pq, p ∈ xˆ, q ∈ ˆ1, p > 0, q > 0}. If t ∈ xˆ · ˆ1, then ∃p ∈ xˆ, p > 0 and q ∈ ˆ1, q > 0 such that t < pq < p ∈ xˆ, thus t ∈ xˆ. Now if t ∈ xˆ, then there exists q ∈ xˆ, q > 0, such that t < q. We t+q t+q t+q < q and < 1. We thus have t < q · , thus now note that t < 2 2q 2q t ∈ xˆ · ˆ1. Hence xˆ · ˆ1 = xˆ.

86 For xˆ = ˆ0 we always have ˆ0 · ˆ1 = ˆ0. For xˆ < ˆ0 we have xˆ · ˆ1 = −((−ˆ x) · ˆ1) = −(−ˆ x) = xˆ. Theorem 8.11 xˆ 6= ˆ0, if and only if ∃ˆ y such that xˆyˆ = ˆ1. Proof In the converse direction we simple note that ˆ0ˆ y = ˆ0 ∀ˆ y ∈ R. Thus, if xˆyˆ = ˆ1, then neither xˆ nor yˆ can be ˆ0. In the forward direction, let xˆ > 0, and let yˆ = {s|∃ p, q > 0, q ∈ xˆ, pq < 1 and s < p}. We first show that yˆ is a cut. 1. xˆ > ˆ0 ⇒ 0 ∈ xˆ ⇒ ∃q ∈ xˆ where q > 0. Since 0 · q = 0 < 1 ∀q we have 0 ∈ yˆ. Thus yˆ 6= ∅. To show that yˆ 6= Q, pick q ∈ xˆ, q > 0, then q −1 · q = 1, thus q −1 6∈ yˆ. 2. If s ∈ yˆ and t < s, then t < p, thus t ∈ yˆ. 3. If s ∈ yˆ, then s < p, thus s
0 where pq < 1 we have s < p, thus sq < pq < 1. Hence yˆxˆ = {r|r < sq < pq < 1, s, p, q > 0} = {r|r < 1} = ˆ1. To complete the proof, assume xˆ < 0, then ∃ˆ y such that (−ˆ x) · yˆ = 1, thus xˆ · (−ˆ y ) = −ˆ xyˆ = (−ˆ x) · yˆ = 1.

87

The set of Real Numbers is a Field To complete the verification that the real numbers form a field we demonstrate that multiplication distributes over addition. Theorem 8.12 xˆ · (ˆ y + zˆ) = xˆ · yˆ + xˆ · zˆ. To prove Theorem 8.12 we need the following lemma. Lemma If p ∈ xˆ and q ∈ yˆ where p, q > 0, then pq ∈ xˆyˆ. Proof ∃ r ∈ xˆ such that p < r, thus pq < rq, thus pq ∈ xˆyˆ Proof of Theorem 8.12 i) Assume xˆ, yˆ + zˆ > ˆ0, then either yˆ > ˆ0 or zˆ > ˆ0. xˆ · (ˆ y + zˆ) = {t|t < p · s where p ∈ xˆ, s ∈ yˆ + zˆ and p, s > 0} If yˆ and zˆ are positive, then for s ∈ (ˆ y + zˆ) we can write s = q + r where q ∈ yˆ, r ∈ zˆ and q, r > 0. Thus we have xˆ·(ˆ y +ˆ z ) = {t|t < p·s = p(q+r) = pq+pr, p, q, r > 0 with p ∈ xˆ, q ∈ yˆ, r ∈ zˆ}. Now xˆyˆ + xˆzˆ =

{t|t < u + v, u ∈ xˆyˆ and v ∈ xˆzˆ}

= {t|t < pq + p0 r, p, q, p0 , r > 0, p, p0 ∈ xˆ, q ∈ yˆ, r ∈ zˆ}. Thus xˆ · (ˆ y + zˆ) ⊆ xˆyˆ + xˆzˆ. Now let p, p0 ∈ xˆ, p, p0 > 0, then pq + p0 r = p(q + and either

p0 p r) = p0 ( 0 q + r) p p

p0 p p p0 ≤ 1 or 0 ≤ 1, thus either 0 q ∈ yˆ or r ∈ zˆ. p p p p

88 Thus xˆyˆ + xˆzˆ ⊆ xˆ · (ˆ y + zˆ), and thus xˆyˆ + xˆzˆ = xˆ · (ˆ y + zˆ). Now without loss of generality, assume yˆ > ˆ0 and zˆ ≤ ˆ0. Then we have xˆ · (ˆ y + zˆ) = {t|t < ps, p > 0, s > 0, p ∈ xˆ and s ∈ yˆ + zˆ}. Now, s = q − r, where q ∈ yˆ, −r ∈ zˆ, q, r > 0. Thus xˆ · (ˆ y + zˆ) = {t|t < ps = p(q − r) = pq − pr, p, q, r > 0}. Now xˆyˆ + xˆzˆ = xˆyˆ − xˆ(−ˆ z ) = {t|t < p − q, p ∈ xˆyˆ, and q ∈ xˆ(−ˆ z )}. We have p = su, q = rv, where s ∈ xˆ, u ∈ yˆ, r ∈ xˆ, v ∈ −ˆ z . Thus xˆyˆ + xˆzˆ = xˆyˆ − xˆ(−ˆ z ) = {t|t < su − rv}. Thus we again have xˆ · (ˆ y + zˆ) ⊆ xˆyˆ + xˆzˆ. s s r r z. And again either < 1 or < 1. Thus either u ∈ yˆ or v ∈ −ˆ s r r s r s Thus su − rv = s(u − v) = r( u − v), and one is in xˆ · (ˆ y + zˆ). s r Thus xˆyˆ + xˆzˆ ⊆ xˆ · (ˆ y + zˆ), and thus xˆyˆ + xˆzˆ = xˆ · (ˆ y + zˆ). ii) Now assume xˆ > ˆ0, yˆ + zˆ < ˆ0. Then xˆ · (ˆ y + zˆ) = =

−(ˆ x · (−(ˆ y + zˆ)) −(ˆ x(−ˆ y − zˆ))

= −(−ˆ xyˆ − xˆzˆ) = xˆyˆ + xˆzˆ. iii) Assume xˆ < ˆ0, yˆ + zˆ > ˆ0. Then xˆ · (ˆ y + zˆ) =

−(−ˆ x · (ˆ y + zˆ))

= −(−ˆ xyˆ − xˆzˆ) = xˆyˆ + xˆzˆ.

89 iv) Assume xˆ < ˆ0, yˆ + zˆ < ˆ0. Then xˆ · (ˆ y + zˆ) =

(−ˆ x · (−(ˆ y + zˆ)))

=

(−ˆ x · (−ˆ y − zˆ))

=

(−ˆ x(−ˆ y ) − xˆ(−ˆ z ))

=

−(ˆ x(−ˆ y ) + xˆ(−ˆ z)

= −(−(ˆ xyˆ + xˆzˆ)) = xˆyˆ + xˆzˆ v) If yˆ + zˆ = ˆ0, then zˆ = −ˆ y and we have xˆ · (ˆ y + zˆ) = xˆ · ˆ0 = ˆ0 and xˆyˆ + xˆzˆ = xˆyˆ + xˆ(−ˆ y ) = xˆyˆ − xˆyˆ = ˆ0. vi) Finally, if xˆ = ˆ0, then xˆ · (ˆ y + zˆ) = ˆ0 · (ˆ y + zˆ) = ˆ0 = ˆ0 · yˆ + ˆ0 · zˆ = xˆyˆ + xˆzˆ

The Least Upper Bound Property The real numbers enjoy a property that is not shared by the rational numbers, known as the least upper bound property. Definition If X is a set of real numbers and the real number u satisfies the condition that x ≤ u ∀x ∈ X, then u is said to be an upper bound for X. Equivalently if the real number l satisfies the condition that x ≥ l ∀x ∈ X, then l is said to be a lower bound for X. Any set that has an upper bound is said to be bounded above, if the set has a lower bound it is said to be bounded below. If a set has both an upper bound and a lower bound, we say it is bounded. Definition An upper bound, u of a set X, that satisfies the condition, if v is also upper bound, then u ≤ v, is said to be the supremum, or the least

90 upper bound of X. A similar definition can be made for the greatest lower bound or infimum. We can now state and prove the following theorem. Theorem 8.13 The Supremum Property

Every non-empty set of real

numbers bounded above has a supremum. Proof Let A be a non-empty collection of real numbers that is bounded [ above by uˆ. We will show that xˆ is a real number and is the supremum. x ˆ∈A

1. For any xˆ ∈ A ∃p ∈ xˆ ⇒ p ∈

[ x ˆ∈A

xˆ ≤ uˆ ∀ˆ x ∈ A and

∃q 6∈ uˆ ⇒ q 6∈ xˆ ∀ˆ x ∈ A ⇒ q 6∈

[ x ˆ∈A

2. If q < p where p ∈

[

[

xˆ ⇒

xˆ 6= ∅.

x ˆ∈A

[

xˆ ⇒

xˆ 6= Q.

x ˆ∈A

xˆ ⇒ p ∈ xˆ for some xˆ ⇒ q ∈ xˆ ⇒ q ∈

[

[

xˆ.

x ˆ∈A

x ˆ∈A

3. If p ∈

[

xˆ ⇒ p ∈ xˆ for some xˆ ⇒ ∃q > p such that q ∈ xˆ ⇒ q ∈

x ˆ∈A

xˆ.

x ˆ∈A

Thus

[

xˆ is a cut and hence a real number.

x ˆ∈A

Now we show that xˆ ≤

[ x ˆ∈A

xˆ. Hence

[

[

xˆ is the supremum. If xˆ ∈ A, then xˆ ⊂

x ˆ∈A

xˆ is an upper bound. Now let yˆ
m.

92 2. Let kn be a decreasing sequence of integers. We say kn is inessential if there exists N ∈ N such that kn+1 = kn − 1 ∀n ≥ N . We say kn is essential if it is not inessential. 3. We will say any rational number of the form 2k , k ∈ Z is a binary. Consider the set of all binary-valued sequences on α ⊆ ω of the form sn = 2kn where kn is an essential decreasing integer-valued sequence, i.e., B = {(2kn )|(kn ) is an essential decreasing integer-valued sequence}. We will construct a bijection, b, between B and the positive real numbers, i.e. we construct b : R+ ↔ B. If b(ˆ x) = (2kn ), we will say (2kn ) is the binary representation of xˆ. We construct b by demonstrating how to compute the image of xˆ ∈ R+ . xˆ > ˆ0 ⇒ 0 ∈ xˆ ⇒ ∃p > 0, p ∈ xˆ. Thus ∃k ∈ Z such that 2k < p. Now ∃q 6∈ xˆ and ∃m such that 2k+m ≥ q. Thus ∃n such that ˆ2k+n > xˆ. Consider {m|ˆ2k+m > xˆ} ⊆ ω. Pick the least element,n, thus 2ˆk+n > xˆ and ˆ2k+n−1 ≤ xˆ, thus 2k+n−1 is the largest element of the form 2m in xˆ. Set k + n − 1 = k0 . Now consider xˆ − ˆ2k0 . There exists a largest element of the form 2m in xˆ − ˆ2k0 , call that element 2k1 .

93 We show 2k1 < 2k0 . ˆ2k1 ≤ xˆ − ˆ2k0 ⇒ ˆ2k1 + ˆ2k0 ≤ xˆ If 2k0 ≤ 2k1 , then we have ˆ2k0 + ˆ2k0 ≤ ˆ2k1 + ˆ2k0 < xˆ. Thus ˆ2k0 +1 = ˆ2 · ˆ2k0 < xˆ. Since 2k0 is the largest binary in xˆ we have a contradiction, thus 2k1 < 2k0 . We pick 2k2 to be the largest binary in xˆ − ˆ2k0 − ˆ2k1 = xˆ − (ˆ2k1 + ˆ2k1 ). Again 2k2 < 2k1 . We continue in this fashion to construct the sequence. This construction process defines our map b. We now need to show that b is well defined, one to one and onto. To demonstrate that b is well defined we need only show that the maximal binary in any real number is unique. If 2m and 2n are maximal binaries in xˆ, then 2m = 2n ⇒ m = n. Thus the maximal binary is unique and b is well defined. To show that b is onto, we need to define some notation and prove a lemma. Let (an ) be an α-sequence, where α ⊆ ω. ³X ´ X Let ak = a0 + a1 + · · · + an for n ∈ α. Then we have ak is an k∈n

k∈n

α-sequence. Which we call the sequence of partial sums of (an ). Now let (an ) be an α-sequence of non-negative rational numbers, α ⊂ ω. ³X ´ Each element of ak is a rational number since each sum is finite. Also X k∈n

ak ≤

X k∈m

k∈n

ak if n < m, since each ai ≥ 0.

94 If the associated sequence of real numbers

³X ´ \ ak is bounded above, k∈n

then we define

nX o X \ \ an = sup an . n∈α

k∈n

Lemma If (an ) = (2kn ) where kn ∈ Z, n ∈ α ⊂ ω and kn < km if n > m, ³X ´ \ki then 2 is bounded above. i∈n

Proof

X i∈n

Thus

X

2ki ≤

X

X

2k0 −i ≤

2k0 −i = 2k0 +1

i∈ω

i∈n

2ki < 2k0 +1 ∀n ∈ α ⇒

X \ki k0 +1 ∀n ∈ α 2 ≤ 2[

i∈n

i∈n

³X ´ \ki Thus 2 is bounded above. i∈n

Now let (2kn ) be a sequence where (kn ) is a decreasing essential sequence X 2km+n < 2km ∀m ∈ N. Thus we see of integers. Since kn is essential n∈N−{0}

P k that (2 ) is the binary representation of the real number \ 2 n. kn

To show that b is one to one, if b(ˆ x) = b(ˆ y ), then {2kn } = {2jn } only if 2kn = 2jn ∀n. Theorem 8.14 The real numbers are uncountable. To prove this theorem we need two lemmas and a corollary. Lemma 1 The set of all bi-valued ω-sequences is uncountable. That is S = {s : ω → {0, 1}} is uncountable. Proof The demonstration is by contradiction.

95 Assume there exists a bijection b : ω → S. Define a bi-valued ω-sequence s by

 ¡ ¢  0 if b(n) (n) = 1 s(n) = ¡ ¢  1 if b(n) (n) = 0.

Since b is a bijection, ∃ k where b(k) = s. Then we have  ¡ ¢  0 if b(k) (k) = s(k) = 1 s(k) = ¡ ¢  1 if b(k) (k) = s(k) = 0 which is a contradiction. Lemma 2 The set of all bi-valued ω-sequences where s(k) = 0 for only a finite number of times is countable. That is, A = {s|∃N ∈ ω such that s(n) = 1 ∀n ≥ N } is countable. Proof There is a reasonably obvious bijective map [ j:A↔ {s : n → {0, 1}|n ∈ ω}. n∈ω Ã ! [ Thus C(A) ≤ C {s : n → {0, 1}|n ∈ ω} , and the countable union of n∈ω

finite sets is countable. Corollary The set of all bi-valued ω-sequences where s(k) = 0 infinitely often is uncountable. Proof of Thoerem 8.14 Let A = {s : ω → {0, 1}, where sn = 0 infinitely often}. There exists a bijection

( b:A↔

X\ sn · 2−n |s ∈ A

) ⊂R

n∈ω

defined by b(s) =

X\ sn · 2−n . Since A is uncountable, so is R. n∈ω

96 Exercise Show that {s : ω → {0, 1}, where sn = 0 infinitely often} = {x|0 ≤ x ≤ 1}.

The Cauchy Sequence Construction of the Real Numbers We conclude the chapter with a set of exercises that leads to an alternate construction of the real numbers. Definition The functions Abs : R → R defined by   x if x ≥ 0 Abs(x) =  −x if x < 0 is the absolute value function. We abbreviate Abs(x) by |x|, i.e. Abs(x) = |x|. Definition A rational-valued sequence, sn , that satisfies the following: ∀² > 0, ² ∈ Q, ∃N ∈ N such that ∀n, m > N ⇒ |sn − sm | < ² is called a Cauchy Sequence. We define an equivalence relation on the set of all rational-valued Cauchy Sequences by sn ≡ tn iff ∀² > 0 ∃N ∈ N such that ∀n ≥ N ⇒ |sn − tn | < ². We define an order on the equivalence classes of Cauchy Sequences by S ≥ T iff ∃N ∈ N such that ∀n > N ⇒ sn − tn ≥ 0 ∀(sn ) ∈ S and ∀(tn ) ∈ T.

97 Exercises 1. Show that the relation as defined above is in an equivalence relation. 2. Show that the order defined for the equivalence classes of Cauchy Sequences is a linear order. 3. Show that the set of equivalence classes of rational-valued Cauchy sequences is order isomorphic to the Real Numbers.

98

IX

Complex numbers, Quaternions and Octonions

Since the product of two positive real numbers is positive, and the product of any two negative real numbers is also positive, the solution to the equation x2 + 1 = 0 is vacuous in the Real numbers. However by extending the real numbers to a larger set of numbers we can create solutions to equations such as the example given above. The construction of this larger set of numbers from the Real numbers is far easier than the construction of the Reals from the Rationals.

Complex Numbers Definition The Complex Numbers, C, is the cartesian product of the Real Numbers with themselves, R × R, with the following arithmetic. 1. (a, b) + (c, d) = (a + c, b + d). 2. (a, b) · (c, d) = (ac − bd, ad + bc). The real numbers are embedded into the Complex numbers by the following injection map J : x ,→ (x, 0). Theorem 9.1 The complex numbers are a field. Proof From the definition of addition and multiplication, the results of the binary operation produces an ordered pair of real numbers hence the Complex numbers are closed with respect to the binary operations of addition and multiplication.

99 (a, b) + (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) + (a, b), and (a, b) · (c, d) = (ac − bd, ad + bc) = (ca − db, cb + da) = (c, d) · (a, b). Hence addition and multiplication is commutative. ((a, b) + (c, d)) + (e, f ) = (a + c, b + d) + (e, f ) = ((a + c) + e, (b + d) + f ) = (a + (c + e), b + (d + f )) = (a, b) + (c + e, d + f ) = (a, b) + ((c, d) + (e, f ), and ((a, b) · (c, d)) · (e, f ) = (ac − bd, ad + bc) · (e, f ) = ((ac − bd)e − (ad + bc)f, (ac − bd)f + (ad + bc)e) = (ace − bde − adf − bcf, acf − bdf + ade + bce) = (a(ce − df ) − b(de + cf ), a(cf + de) + b(ce − df )) = (a, b) · (ce − df, cf + de) = (a, b) · ((c, d) · (e, f )). Hence addition and multiplication is associative. (a, b) · ((c, d) + (e, f )) = (a, b) · (c + e, d + f ) = (ac + ae − bd − bf, ad + af + bc + be) = (ac − bd, ad + bc) + (ae − bf, af + be) = (a, b) · (c, d) + (a, b) · (e, f ). Hence multiplication distributes over addition. (a, b) + (0, 0) = (a + 0, b + 0) = (a, b). Thus (0, 0) is the additive identity. (a, b)·(1, 0) = (a·1−b·0, a·0+b·1) = (a, b). Thus (1, 0) is the multiplicative identity. (a, b) + (−a, −b) = (a − a, b − b) = (0, 0). Thus any complex number (a, b) has an additive inverse (−a, −b). −b a2 −b2 −ab ab a , ) = ( − , + )= a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2 ab − ab , ) = (1, 0). Thus for any complex number (a, b) 6= (0, 0) the ( 2 a + b2 a2 + b2 a −b complex number ( 2 , 2 ) is the multiplicative inverse. 2 a + b a + b2 (a, b) · (

We have thus verified that the complex numbers forms a field. We note that the product (0, 1) · (0, 1) = (−1, 0) Since (−1, 0) is the embedding of the real number −1, we have (0, 1) as the square root of −1, and we have a solution to the equation x2 + 1 = 0.

100 The simplest and standard way to represent Complex numbers is to represent them as the formal expression a + bi, where i2 = (0, 1)(0, 1) = (−1, 0) = −1. Now standard arithmetic on binomial expressions yield the appropriate sums and products.

(a + bi) + (a0 + b0 i) = (a + a0 ) + (b + b0 )i (a + bi)(a0 + b0 i) = aa0 + ab0 i + a0 bi + bb0 i2 = (aa0 − bb0 ) + (ab0 + a0 b)i . For the complex number (a, b) we say (a, −b) is its complex conjugate. We denote the complex conjugate of a complex number c by c∗ , thus if c = (a, b), then c∗ = (a, b)∗ = (a, −b). s Definition For x = (a1 , a2 , · · · , an ) ∈ Rn = Πi∈n R, the real value

n X

a2i

i=1

is the norm of x. We say that the norm of a complex number is the norm of the pair of real numbers that represents it. We see that for any complex number c, we have c · c∗ = (a + bi)(a − bi) = a2 + b2 which is the norm squared of c.

Quarternions Definition The Quaternions is the set C × C with the following arithmetic: 1. (a, b) + (c, d) = (a + c, b + d) 2. (a, b) · (c, d) = (ac − d∗ b, ad + bc∗ ).

101 We designate the Quaternions with the blackboard bold face capital H, H. Let h ∈ H, then h = (a + bi, c + di). But again it is common practice to write h = a + bi + cj + dk. We leave it to the reader to verify i2 = j 2 = k 2 = −1. We also leave it to the reader to verify ij = −ji = k, jk = −kj = i, ki = −ik = j. Thus we see ij 6= ji, hence quaternions fail to have the commutative property.

Octonions The quaternion conjugate of a quaternion h = (a, b) is h∗ = (a∗ , −b). Definition The Octonions is the set H × H with the following arithmetic: 1. (a, b) + (c, d) = (a + c, b + d) 2. (a, b) · (c, d) = (ac − d∗ b, ad + bc∗ ). We designate the Octonions with a bold face capital K, K. Again it is common practice to designate Octonions by k = a + be1 + ce2 + de3 + ee4 + f e5 + ge6 + he7 . We leave it again to the reader to verify e2n = −1 and that for n 6= m we have anticommutivity, en em = −em en . We could continue constructing numbers in this fashion but more algebraic properties fail. In particular in the next constructions not every non zero number has a multiplicative inverse.

102

X

TRANSFINITE AND INFINITESIMAL NUMBERS

In his development of the Calculus, Isaac Newton introduced the concept of a “fluxion”, a number that was smaller than any positive number and yet was greater than zero. This concept seemed to be contrary to the Archimedian principle, and led one of the critics of early analysis, Bishop Berkeley, to comment “And what are these fluxions? The velocities of evanescent increments. And what are these same evanescent increments? They are neither finite quantities, nor quantities infinitely small, nor yet nothing. May we not call them ghosts of departed quantities?” At this time in history, when Newton’s genius is greatly admired we may be inclined to dismiss Bishop Berkeley’s comments as those of a pompous fool. However this is not true, his criticisms were well founded as the rigorous structure of Newton’s Calculus had not been well established. Criticism such as Bishop Berkeley, is good for the development of any scholarly field of study, as it forces those involved in that study to construct a firm basis for their claims. Of course the concept of fluxions has been replaced with the limit, and all seems well with the world, and Calculus. With Cantor’s development of transfinite ordinal and cardinal numbers it has become possible to make the concept of fluxions rigorous, and without contradicting Archimedes. We will not use the term fluxion, but will call such numbers infinitesimal. In this chapter we will construct transfinite Integers, Rational and Real Numbers. The construction will be a simple extension of our development of finite numbers. We begin our construction of transfinite and infinitesmals by noting that our construction of the integers, rational and real numbers used the ordinal

103 number ω as a basis on which to build the appropriate collection of sets that would be our numbers. An attempt to use a larger ordinal than ω will fail, as ordinal arithmetic is not commutative, and commutativity is crucial in the construction of the equivalence classes that represent the different numbers. Consider the equivalence relation defined on ω × ω that yielded the integers. (a, b) ≡ (c, d) ⇔ a + d = b + c. This relation is not an equivalence relation on ordinal numbers greater than ω as (ω, 1) 6≡ (ω, 1) since ω + 1 6= 1 + ω, hence the relation does not satisfy the reflexive property.

Transfinite Arithmetic Our choices of binary operations were made from our experiences with tangible collection of objects (e.g. boxes of apples), and these tangible collections must be finite by their very nature. The extension of these binary operations to abstract and infinite sets, although well motivated is quite arbitrary. So our strategy is to define a new arithmetic, that agrees with ordinal arithmetic on finite sets, but satisfies the appropriate field axioms on infinite sets. As noted in Chapter 3, the ordinal number ω ω can be expressed as {x | x =

n X

ω i αi where n, αi ∈ ω}.

i=0

Thus every element of ω ω is of the form of a polynomial of indeterminate ω with coefficients and exponents in ω. We thus define an arithmetic on ω ω to be polynomial arithmetic. n X i=0

i

ω αi +

m X i=0

max{n,m} i

ω βi =

X i=0

ω i (αi + βi )

104 and

n X i=0

i

ω αi ·

m X

j

ω βj =

j=0

n X m X

ω i+j (αi · βj ).

i=0 j=0

Where the arithmetic on ω is ordinal arithmetic. We may now easily verify that these two operations are commutative and associative, and that multiplication distributes over addition. Thus we may repeat our construction of integers, rational and real numbers with this arithmetic to form transfinite integers, transfinite rational, and transfinite real numbers. Both the transfinite rational numbers and transfinite real numbers contain elements that satisfy the conditions of Newton’s fluxions. We will call those numbers infinitesimal. If we let n ∈ ω be an arbitrary element , and we let ω represent the transfinite integer (ω, 0), 1 represent the transfinite integer (1, 0) and n represent the transfinite integer (n, 0) we observe that 1 · n < 1 · ω ⇒ [1, ω] < [1, n], where [1, ω] and [1, n] are transfinite rational numbers. Since the restriction of the construction of transfinite rational numbers is the construction of the rational numbers, we may say that [1, n] is a rational number. Thus for any positive real number ², by the Archimedian property there exists an integer n such that [1, n] < ², and since [1, ω] < [1, n] we have [1, ω] < ². And thus [1, ω] satisfies the condition of Newton’s fluxions.

Transfinite Numbers We define the ω-Transfinite Natural Numbers, Nω to be ω ω , with polynomial arithmetic. We define the ω-Transfinite Integers, Zω , to be the collection of equivalence classes {[x, y]|x, y ∈ Nω where (x, y) ≡ (z, w) ⇔ x + w = y + z}.

105 We define the ω-Transfinite Rational numbers, Qω , to be the collection of equivalence classes {[x, y]|x, y ∈ Zω where (x, y) ≡ (z, w) ⇔ xw = yz}. And we define the ω-Transfinite Real Numbers, R, to be the collection of all Dedekind cuts of transfinite rational numbers. We now observe that the ordinal number ω (ω

ω)

can be expressed as

n X {x | x = (ω ω )i αi where n, αi ∈ ω ω }. i=0

Thus every element of ω ω is of the form of a polynomial of indeterminate ω ω with coefficients and exponents in ω ω . We again define arithmetic as polynomial arithmetic where the arithmetic on ω ω is the arithmetic previously defined. We now define the ω ω -Transfinite Natural Numbers, ω ω -Transfinite Integers, ω ω -Transfinite Rational Numbers, and ω ω -Transfinite Real Numbers in the analogous fashion. We may continue indefinitely constructing Transfinite numbers in this fashion, We thus define any number constructed in this fashion to be a Transfinite Number. Thus a Transfinite Integer is an α-Transfinite Integer for some appropriate ordinal number α. And equivalently for Transfinite Natural Numbers, Transfinite Rational Numbers, and Transfinite Real numbers. We will adopt the convention, that when referring to an element of the transfinite real numbers that is the embedded image of either a transfinite natural number, transfinite integer, or transfinite rational number, we will simply refer to that number as being an element of that respective embedded

106 set, that is, a transfinite natural number, transfinite integer or transfinite rational number. Let ξ be an arbitrary transfinite real number. Consider the cut {α|α2 < ξ}. √ We may consider this cut to represent the transfinite real number ξ. Questions What is

√

ω ? What is

√

ω + 1 ? What is

√ 3

ω?

107

XI

SURREAL NUMBERS

The origin of Surreal numbers is credited to John Conway, however the name was coined by Don Knuth. For consistency we may use the symbol S for surreal numbers, however the need to do so seldom arises.

The Constructive Definition of Surreal Numbers John Conway defined Surreal numbers by formulating two simple rules for the construction of Surreal numbers plus the definitions for addition and multiplication. With these two rules and two definitions, a collection of numbers is constructed that includes all Real numbers, and all Ordinal numbers. The collection although not a set, but rather a proper class, forms a field. Rule 1: Every number is represented by a pair of sets of previously constructed numbers, a left set and a right set, where no number in the left set is greater than or equal to any number of the right set. Rule 2: A number, a, is less than or equal to a number, b, if and only if no member of a0 s left set is greater than or equal to b, and no member of b0 s right set is less than or equal to a. The first rule tells us how to construct new Surreal numbers from previously constructed numbers. The second rule defines the order relation of the collection of surreal numbers that is necessary for the construction. We will develop the surreal numbers using an alternate definition, that uses some of the principles already developed in Set Theory.

108

The Function Definition of Surreal Numbers

Definition A Surreal number is a function from an ordinal number to a two point space. The two point space is designated by {+, −}. The domain of a surreal number will be called its length. The function from the ordinal number 0 = {} is of course the empty function,{}, and is considered to be a surreal number, and we call it 0.

Example 5 = {0, 1, 2, 3, 4}. Thus

0

1

2

3

4

↓

↓

↓

↓

↓

+ +

-

+

-

Is a surreal number Recall that we defined a sequence as a function whose domain is an ordinal number. If the domain of a sequence is the ordinal number α then we say that the sequence is an α-sequence. Thus a surreal number, a, is a binary valued α-sequence for some ordinal α, and its length is α, which we indicate by l(a) = α. The 0-sequence is of course the empty set. Definition If a is an α-sequence, and b is a β-sequence, such that a ∩ b = b, then b is an initial segment of a. If b 6= a, then b is a proper initial segment. We see that if b is a proper initial segment of a, then l(b) < l(a). Two surreal numbers are equal if they are equal as sets. We define a linear order on the class of surreal numbers by the following: Let a and b be surreal numbers, and let c be the maximal initial segment

109 that is in both a and b, where c is a γ-sequence. We say    a(γ) = + and b(γ) = − or    a > b if a(γ) = + and b(γ) is undefined or     a(γ) is undefined and b(γ) = − .

The Canonical Representation of Surreal Numbers Let a be a surreal number. If A0 = {a0 |a0 is an initial segment of a and a0 < a}, and A00 = {a00 |a00 is an initial segment of a and a < a00 }, then we say A0 |A00 is the canonical representation of a. Example (+ + − − +) = 1 83 = {0, 1, 1 14 }|{2, 1 21 }

Addition of Surreal Numbers We define addition in the following way. If a = A0 |A00 and b = B 0 |B 00 are in canonical form, then a + b = {a + b0 , b + a0 }|{a + b00 , b + a00 } ∀ a0 ∈ A,0 b0 ∈ B 0 , a00 ∈ A00 , b00 ∈ B 00 . We now verify that the elements of the left set are truly less than the elements of the right set. But first we must verify that addition is commutative. a + b = {a + b0 , b + a0 }|{a + b00 , b + a00 } = {b + a0 , a + b0 }|{b + a00 , a + b00 } = b + a To complete the verification we induct on the ordinal sum of the lengths of a and b. We make the inductive hypothesis that if l(c) + l(d) < l(a) + l(b), l(c0 ) + l(d0 ) < l(a) + l(b), c < c0 , and d0 < d0 , then c + d < c0 + d0 .

110 We have a0 < a < a00 , b0 < b < b00 . Thus a + b0 < a + b00 , a + b0 < a00 + b = b + a00 and b + a0 < b + a00 , b + a00 < b00 + a = a + b00 . Thus a + b does represent a Surreal number. However we notice that a + b is not in canonical form. Is it well defined? If A and B are sets of surreal numbers such that a < b ∀a ∈ A and b ∈ B, then A|B is the “first” surreal number c, such that a < c < b ∀a ∈ A and b ∈ B. That is, c is the surreal number with the minimal domain such that a < c < b ∀a and ∀b. Hence the sum is well defined. The way to think of why this is true is this way. Start at 0 and try to get into the gap between the two sets as quickly as possible. If there is an element of the lower set greater or equal to zero step towards the upper set, if the opposite is true step toward the lower set. If you arrive between the two sets then you have defined that “youngest” surreal number. If you are still “amid” one of the two sets, step towards the other set and continue until you arrive between them. If at any stage you were to step away from the set in which you are not amid, then you can never arrive between the sets, thus there is only one way to arrive between the sets, and when you first get there you stop. Example Compute 1+1. To add 1+1 we need to know 0+1, to add 0+1 we need to know 0+0. We have 0 = ∅|∅, thus 0 + 0 = {0 + b0 , 0 + a0 }|{0 + b00 , 0 + a00 }, where a0 , a00 , b0 , b00 ∈ ∅ →←. Since there are no elements in the empty set to add to

111 0 each of those sums does not exist and thus 0 + 0 = ∅|∅ = 0. We now have 1 = {0}|∅, thus 0 + 1 = {0 + 0, 1 + a0 }|{0 + b00 , 1 + a00 } where a0 , a00 , b00 ∈ ∅ →← . Thus we have 0 + 1 = {0}|∅ = 1. Finally 1 + 1 = {1 + 0, 1 + 0}|{1 + b00 , 1 + a00 } where a00 , b00 ∈ ∅ →← . Thus 1 + 1 = {1}|∅ = 2. We can see that 0 = ∅|∅ is the additive identity, since if we let a = A0 |A00 , then a + 0 = {0 + a0 }|{0 + a00 } = a. Now we can define the additive inverses of surreal numbers. Let a be a surreal number. Define −a as:   + if a(α) = − −a(α) =  − if a(α) = +. Example Let a + (+ − +) =

3 4

= {0, 12 |{1},

thus −a = (− + −) = − 34 = {−1}|{0, − 12 }. 1 3 3 1 a + b = {− , − }|{ , } = 0. 4 4 4 4 The canonical representation of the opposite of a given surreal number a = A0 |A00 has the opposites of A00 as its lower set and the opposites of A0

112 as its upper set. Thus if a = A0 |A00 , then −a = −A00 | − A0 where −A00 = {−a00 |a00 ∈ A00 } and −A0 = {−a0 |a0 ∈ A0 }. Now a + (−a) = {a + (−a00 ), (−a) + a0 }|{a + (−a0 ), −a + a00 }

Since a0 < a < a00 ⇒ −a00 < −a < −a0 We have a + (−a00 ) < a00 + (−a00 ) = 0 −a + (a0 )

< a0 + (−a0 ) = 0

a + (−a0 )

> a0 + (−a0 ) = 0

−a + a00

> −a00 + a00 = 0.

Thus

a + (−a) = 0.

We can verify that addition of surreal numbers is associative and thus satisfy the axioms for an abelian group under addition. Let a = A0 |A00 , b = B 0 |B 00 , and c = C 0 |C 00 . (a + b) + c = {a + b0 , b + a0 }|{a + b00 , b + a00 } + c =

{(a + b) + c0 , c + (a + b0 ), c + (b + a0 }|{(a + b) + c00 , c + (a + b00 ), c + (b + a00 )}

= {a + (b + c0 ), a + (c + b0 ), (b + c) + a0 }|{a + (b + c00 ), a + (c + b00 ), (b + c) + a00 } =

a + {b + c0 , c + b0 }|{b + c00 , c + b00 }

=

a + (b + c).

The Multiplication of Surreal Numbers We now define multiplication. As a motivation for the following definition consider real numbers a, b, c, d where a < b, and c < d. We then have b − a > 0 and d − c > 0. Thus (b−a)(d−c) > 0 ⇒ bd+ac−ad−bc > 0 ⇒ bd > ad+bc−ac, and bc < bd+ac−ad.

113 Definition Let a = A0 |A00 ,

b = B 0 |B 00 , then

ab = {a0 b + ab0 − a0 b0 , a00 b + ab00 − a00 b00 }|{a0 b + ab00 − a0 b00 , a00 b + ab0 − a00 b0 } where a0 ∈ A0 , b0 ∈ B 0 , a00 ∈ A00 and b00 ∈ B 00 . Exercise Consider ² = (+ − − − · · · ) = {0}|{1, 12 , 41 , · · · } and ω = (+ · · · ) = {0, 1, · · · }|∅. Find ² · ω.

The Field Properties of Surreal Numbers We have confirmed that Surreal numbers with the binary operation of addition forms an abelian group. We now want to complete our verification that Surreal numbers with the operations of addition and multiplication form a field. Of course we must verify that the definition of multiplication does indeed define a Surreal number, to do that we must first verify the properties of commutativity and associativity of multiplication, and the distributive property. We again make the inductive hypothesis that the inequalities necessary are valid for the product of surreal numbers whose sum of lengths are less than the sum of the lengths of a and b, and thus the definition of multiplication will be well defined. Let a = A0 |A00 , b = B 0 |B 00 and c = C 0 |C 00 . We now have ab =

{a0 b + ab0 − a0 b0 , a00 b + ab00 − a00 b00 }|{a0 b + ab00 − a0 b00 , a00 b + ab0 − a00 b0 }

= {b0 a + ba0 − b0 a0 , b00 a + ba00 − b00 a00 }|{b0 a + ba00 − b0 a00 , b00 a + ba0 − b00 a0 } = ba Where a0 ∈ A0 , a00 ∈ A00 , b0 ∈ B 0 , and b00 ∈ B 00 . Thus multiplication is

114 commutative. (ab)c =

{(ab)0 c + (ab)c0 − (ab)0 c0 , (ab)00 c + (ab)c00 − (ab)00 c00 }| {(ab)0 c + (ab)c00 − (ab)0 c00 , (ab)00 c + (ab)c0 − (ab)00 c0 }

=

{(a0 b + ab0 − a0 b0 )c + abc0 − (a0 b + ab0 − a0 b0 )c0 , (a00 b + ab00 − a0 b00 )c + abc0 − (a00 b + ab00 − a00 b00 )c0 , (a0 b + ab00 − a0 b00 )c + abc00 − (a0 b + ab00 − a0 b00 )c00 , (a00 b + ab0 − a00 b0 )c + abc00 − (a00 b + ab0 − a00 b0 )c00 }| {(a0 b + ab0 − a0 b0 )c + (ab)c00 − (a0 b + ab0 − a0 b0 )c00 , (a00 b + ab00 − a00 b00 )c + (ab)c00 − (a00 b + ab00 − a00 b00 )c00 , (a0 b + ab00 − a0 b00 )c + (ab)c0 − (a0 b + ab00 − a0 b00 )c0 , (a00 b + ab0 − a00 b0 )c + (ab)c0 − (a00 b + ab0 − a00 b0 )c0 }

= {a0 bc + ab0 c − a0 b0 c + abc0 + abc0 − a0 bc0 − ab0 c0 + a0 b0 c0 , a00 bc + ab00 c − a00 b00 c + abc0 − a00 bc − ab00 c + a00 b00 c, a0 bc + ab00 c − a0 b00 c + abc00 − a0 bc00 − ab00 c + a0 b00 c00 , a00 bc + ab0 c − a00 b0 c + abc00 − a00 bc00 − ab0 c00 + a00 b0 c00 }| {a0 bc + ab0 c − a0 b0 c + abc00 − a0 bc00 − ab0 c00 + a0 b0 c00 , a00 bc + ab00 c − a00 b00 c + abc00 − a00 bc00 − ab00 c00 + a00 b00 c00 , a0 bc + ab00 c − a0 b00 c + abc0 − a0 bc0 − ab00 c0 + a0 b00 c0 , a00 bc + ab0 c − a00 b0 c + abc0 − a00 bc0 − ab0 c + a00 b0 c0 =

{a0 (bc) + a(b0 c + bc0 − b0 c0 ) − a0 (b0 c + bc0 − b0 c0 ), a0 (bc) + a(b00 c + bc00 − b00 c00 ) − a0 (b00 c + bc00 − b00 c00 ), a00 (bc) + a(b0 c + bc0 − b0 c0 ) − a00 (b0 c + bc − b0 c0 ), a00 (bc) + a(b00 c + bc00 − b00 c00 ) − a00 (b00 c + bc00 − b00 c00 )}| {a0 (bc) + a(b0 c + bc00 − b0 c00 ) − a0 (b0 c + bc00 − b0 c00 ), a0 (bc) + a(b00 c + bc0 − b00 c0 ) − a0 (b00 c + bc0 − b00 c0 ), a00 (bc) + a(b0 c + bc0 − b0 c0 ) − a00 (b0 c + bc0 − b0 c0 ), a00 (bc) + a(b00 c + bc00 − b00 c00 ) − a00 (b00 c + bc00 − b00 c00 )}

115

= {a0 (bc) + a(bc)0 − a0 (bc)0 , a00 (bc) + a(bc)00 − a00 (bc)00 }| {a0 (bc) + a(bc)00 − a0 (bc)00 , a00 (bc)0 − a00 (bc)0 } = a(bc) Where a0 ∈ A0 , a00 ∈ A00 , b0 ∈ B 0 , b00 ∈ B 00 , c0 ∈ C 0 , and c00 ∈ C 00 . Thus multiplication is associative. a(b + c) =

{a0 (b + c) + a(b + c)0 − a0 (b + c)0 , a00 (b + c) + a(b + c)00 − a00 (b + c)00 }| {a0 (b + c) + a(b + c)00 − a0 (b + c)00 , a00 (b + c) + a(b + c)0 − a00 (b + c)0 }

=

{a0 (b + c) + a(b + c0 ) − a0 (b + c0 ), a0 (b + c) + a(b0 + c) − a0 (b0 + c), a00 (b + c) + a(b + c00 ) − a00 (b + c00 ), a00 (b + c) + a(b00 + c) − a00 (b00 + c)}| {a0 (b + c) + a(b + c00 ) − a0 (b + c00 ), a0 (b + c) + a(b00 + c) − a0 (b00 + c), a00 (b + c) + a(b + c0 ) − a00 (b + c0 ), a00 (b + c) + a(b0 + c) − a00 (b0 + c)}

=

{a0 b + a0 c + ab + ac0 − a0 b0 − a0 c0 , a0 b + a0 c + ab0 + ac − a0 b0 − a0 c, a00 b + a00 c + ab + ac00 − a00 b − a00 c00 , a00 b + a00 c + ab00 + ac − a00 b00 − a00 c}| {a0 b + a0 c + ab + ac00 − a0 b − a0 c00 , a0 b + a0 c + ab00 + ac − a0 b00 − a0 c, a00 b + a00 c + ab + ac0 − a00 b − a00 c0 , a00 b + a00 c + ab0 + ac − a00 b0 − a00 c}

=

{ab + a0 c + ac0 − a0 c0 , ac + a0 b + ab0 − a0 b0 , ab + a00 c + ac00 − a00 c00 , ac + a00 b + ab00 − a00 b00 }| {ab + a0 c + ac00 − a0 c00 , ac + a0 b + ab00 − a0 b00 , ab + a00 c + ac0 − a00 c0 , ac + a00 b + ab0 − a00 b0 }

=

{ab + (ac)0 , (ab)0 + ac}|{ab + (ac)00 + (ab)00 + ac} = ab + ac

Where a0 ∈ A0 , a00 ∈ A00 , b0 ∈ B 0 , b00 ∈ B 00 , c0 ∈ C 0 , and c00 ∈ C 00 . Thus multiplication distributes over addition.

116 We now verify that the Surreal number 1 = {0}|{} is the multiplicative identity. First we must verify that any surreal number multiplied times 0 is 0. Let a = A0 |A00 , and we have 0 = {}|{}, then we have a · 0 = {a0 b + ab0 − a0 b0 , a00 b + ab00 − a00 b00 }|{a0 b + ab00 − a0 b00 , a00 b + ab0 − a00 b0 } where a0 ∈ A0 , b0 ∈ {}, a00 ∈ A00 and b00 ∈ {} Since there are no elements in {} we conclude that the left and right sets of the product are empty and thus is the Surreal number 0. Now consider a · 1. a · 1 = {a0 · 1 + a · 0 − 1 · 0}|{a00 · 1 + a · 0 − a00 · 0} = {a0 }|{a00 } = a. Hence 1 is the multiplicative identity. To demonstrate that all non-zero surreal numbers have multiplicative inverses is considerably more technical than the calculation arguments used for the associative and distributive properties. A complete development of inverses is given in the book “An introduction to the theory of Surreal Numbers” by Harry Gonshor. We give here an intuitive argument as to why inverses should exist. Given a non-zero surreal number x we naively pick a candidate y0 = B 0 |B 00 for its inverse. y0 > 0 if x > 0 and y0 < 0 if x < 0. If x · y0 > 1, then we construct our next candidate by y1 = B 0 |B 00 ∪ {y0 } if x > 0 or y1 = B 0 ∪ {y0 }|B 00 if x < 0. If x · y0 < 1, then we construct our next candidate by y1 = B 0 ∪ {y0 }|B 00 if x > 0 or y1 = B 0 |B 00 ∪ {y0 } if x < 0. We proceed in this fashion until the product is in fact 1. We naively believe that the procedure will eventually end as we will exhaust all possible

117 numbers that could lie between the sets of x · yα except for 1. Then yα will be the multiplicative inverse of x.

118

APPENDIX 1 The Continuum Hypothesis A major question arising from the very beginning of Set Theory is one regarding the relationship between ordinal numbers and cardinal numbers. Namely, which ordinal is the first uncountable ordinal? We know that ordinal numbers and cardinal numbers are identical until the ordinal ω + 1 which has the same cardinality as ω, namely ℵ0 . Since we know that 2ℵ0 has cardinality greater than ℵ0 could that be the next cardinal number? I.e. does there not exist an ordinal number whose cardinality is strictly greater than ℵ0 and strictly less than 2ℵ0 ? We call ℵ1 the next cardinal greater than ℵ0 , and we formally state the hypothesis ℵ1 = 2ℵ0 , which is known as the continuum hypothesis. Recall that cardinal numbers are ordinal numbers so the existence of the next larger cardinal number is guaranteed by the well ordering of ordinal numbers. If we let ℵ be an arbitrary cardinal number we designate the next larger cardinal number by ℵ+ . We can thus generalize the continuum hypothesis by ℵ+ = 2ℵ ∀ℵ ≥ ℵ0 . Paul Cohen demonstrated in 1963 that the answer to the above question cannot be decided with the Zermelo-Fraenkel axioms. That is the continuum hypothesis is independent of ZF Axioms! A sketch of the argument that establishes this fact is given in Keith Devlin’s book The Joy of Sets, and a rigorous account is given in Bell’s book Boolean-Valued Models and Independence Proofs in Set Theory.

119

APPENDIX 2 The number 1 The symbol that we use to represent a set is the two braces { and }. The symbols should be considered a single symbol, since in representing sets the left or right brace alone is meaningless. If we should replace these symbols with a simple closed curve, that is a circle, then we can essentially illustrate numbers. Here we illustrate the various numbers 1. The ordinal number 1 is also the cardinal number 1 and the natural number 1. The integer 1 and rational 1 are represented by their canonical representative from their respective equivalence classes. The real 1 is an infinite collection of rational numbers and does not lend itself to a concise picture, and hence we do not represent it here.

120 ORDINAL 1

121 INTEGER 1

122 RATIONAL 1

123 SURREAL 1

124

APPENDIX 3

Quantifiers ∀ ∃ !

for all or for any there exists Unique

Logical Connectives not ∨ or ∧ and ⇒ implies a ⇒ b; if a, then b ⇔ if and only if a ⇔ b; if a, then b and if b, then a 6⇒ does not imply a 6⇒ b ≡ −(a ⇒ b) Logical Contradiction →← The preceding statement is self contradictory. e.g. (a ⇒ b) ∧ (a 6⇒ b) →←

Truth Tables a −a T F F T

a T T F F

b a⇒b T T F F T T F T

a T T F F

b a∨b T T F T T T F F

a T T F F

a T T F F

b a⇔b T T F F T F F T

b a∧b T T F F T F F F

Bibliography [1] Halmos, Paul R. Naive Set Theory D. Van Nostrand Company, Inc., Princeton, NJ, 1960 [2] Stoll, Robert R. Set Theory and Logic W. H. Freeman and Company, San Francisco Ca, 1963 [3] Rudin, Walter The Principles of Mathematical Analysis, third edition McGraw-Hill, inc., New York, NY, 1964 [4] Landau, Edmund Foundations of Analysis Chelsea Publishing Company, New York, NY, 1951 [5] Gonshor, Harry An Introduction to the Theory of Surreal Numbers Cambridge University Press, Cambridge, U.K., 1986 [6] Knuth, Donald Surreal Numbers Addison-Wesley Publishing Company, Menlo Park, CA., 1974. [7] Devlin, Keith The Joy of Sets, Second Edition Springer-Verlag, New York, NY, 1993 [8] Bell, J.L. Boolean-Valued Models and Independence Proofs in Set Theory Oxford University Press, London, 1977 [9] Websters New World Dictionary Warner Books Inc., New York, 1990

Index Term Absolute Value addition additive identity additive inverse Aleph antisymmetric Archimedes Archimedian Property Arithmetic Axiom of Choice Axiom of extension Axiom of infinity Axiom of pairing Axiom of power sets Axiom of regularity Axiom of unions Axiom schema of replacement Axiom schema of restriction Axiom schema of specification Bijection binary operation Binary representation Bishop Berkeley bound C canonical representation Cantor Cantor’s Theorem Cardinal Arithmetic Cardinal Number cardinality Cartesian Product Cauchy Sequence chain Choice Function codomain comparable complete Complex Numbers Compliment Composition continuation continuum hypotheses Conway countable countably infinite counting numbers Counting Theorem Cut De Morgan laws Dedekind cut

Page 96 51 67 67 33 20 102 73 50 16 3 10 6 9 17 7 17 17 3 30 50 92 102 39 98 109 102 33 50 32 30 13 96 41 15 13 43 40 98 8 30 45 118 107 32 32 59 47 78 8 78

Term disjoint disjoint union division algorithm domain element embedding Empty Set equivalence relation exponentiation Field fluxion Function greatest lower bound H image induction infimum initial segment injection integer addition integer multiplication Integers Integral Domain Intersection Isaac Newton K Knuth least upper bound Limit Ordinal linear order lower bound Mathematical Induction map multiplication N natural numbers Negative Integers Octonians One to One Onto order order isomorphic order preserving order type Ordinal Arithmetic Ordinal Number partial order Partition Positive Integers Power Power set

Page 8 51 74 13 1 65 4 31 52,54 71 102 13,30 40,90 101 13 72 40,90 108 65 62 62 60 67 7 102 75 101 40,89 27 21 40 72 30 51 59 59 66 101 30 30 20 41 40 48 52 22 20 31 66 52 9

Term preimage product Projection Map Proper Class proper initial segment Proposition Q Quaternions R range Rational numbers Real Number reflexive relation Russel’s Paradox S Schroder-Bernstein Theorem section sequence Set simple closed curve spaces subset

Page 13 51 15 19 108 3 68 100 78 13 68 78 20 20 5 107 34 21 91 1 119 1 8

Term successor successor set sum supremum Supremum Property Surreal numbers total order tower Transfinite Induction Transfinite numbers Transfinite Recursion Theorem transitive Trichotomy Trichotomy Property uncountable union upper bound weak section well ordered Well ordering theorem Z Zermelo Fraenkel Zorn’s Lemma

Page 10 11 51 26,40,89 90 107 21 43 24 102 27 20 21 80 32 7 26,39 22 21 45 59 1 41