Math 135

Instructor Pierre Simon, pierre.simon@berkeley.eduGSI Jacopo Di Bonito, jacopo.dibonito@berkeley.edu.

Office Hour Simon: Wed 4-5:30 + Th 2:30 - 4 in Evans 733Di Bonito: Mo 12-1, Fr 12-1, Tu 3:30-4:30 in 741 Evans

Grade 10% homework + 20% midterm 1 + 25% midterm 45% finalTextbook Enderton, Elements of Set TheoryThanks to Ning McKenzie and Varun Khurana, without whose help the notes cannot be complete, and thanks Professor

Simon for his teaching!

Chapter I: Introduction

What is this all about?• Basic set theory: foundations for (modern) mathematicsMathematics aims at being extremely precise and leaving no room for disagreement. for centuries, math did not have a

firm foundation. In the XIX century the need for one became contingent as mathematicians started to disagree especially onproperties of infinite sets.

To avoid disagreements, one need to agree on 3 things:

• a common language–for us English, first order logic (Math 125A)

• basic facts–axioms ZFC (”set theory”)–content of this course

• rules of logic–classical logic (Math 125A)

Historical Facts:

• Cantor 1870’s: studied arbitrary subsets of R the real line, sparked a controversy amongst mathematicians

• Hilbert 1900: prove the consistency of mathematics, hope shattered by Godel 1931, Hilbert’s basis theorem on polyno-mial ring over Noetherian ring, nonconstructive.

• Zermelo 1908: first axiomatisation of set theory

• Frenkel 1922, Von Neumann 1925: completed the axiomatization, forming the system ZFC that we use today, C standsfor choice

Proposition. There exists irrational numbers α, β such that αβ is rational

Proof. Consider√

2√

2. If it is rational, the proposition is proved by letting α = β =

√2. If it is irrational, let α =

√2√

2, β =√

2 then αβ =√

22

= 2 is rational

In classical logic the above proof is valid, but not in intuitionistic logic. elog 2 = 2Sets: Informally, a set is a collection of objects.

Example. set of students enrolled in Math 135

An object either belongs or doesn’t belong to a given set. In a set, elements are not ordered, not repeated.

Axiom. Extentionality: two sets which contain exactly the same elements are equal.

However, a same set can be defined in various ways.

Example.{x ∈ R : x2 ≥ 2

}={x ∈ R : x ≤ −

√2 or x ≥

√2}

Idea of set theory: code all mathematical objects using sets. (similar to a computer encoding everything into sequences ofbits). This encoding is arbitrary, a convention. For us, a number, a line in the plane, a function will all be sets. 2 = {∅, {∅}}.

Set of what? Set of sets (of sets (of sets ...))∅: empty set. {∅} is a set of set. In set theory, we have only one kind of objects: sets, two primitive relations: equality

and belongs to. ∅ 6= {∅}.

1

Sets

∅: the empty set (tehre is onluy one by extensionality). ∅ 6= {∅} ”is reflected in the fact that a man with an empty containeris better off than a man with nothing–at least he has the container.”–Enderton

If a is a set, {a} denotes the set whose only element is a; if a, b are sets, {a, b} denotes the set whose only elementsare a and b. If a 6= b, {a, b} has two elements; if a = b, we can still write {a, b}but the set contains only one element, i.e.,{a, a} = {a}.

Example. {0, 1, 5, 1, 3, 1, 0} = {0, 1, 3, 5} = {3, 1, 5, 0}

We can also define sets with conditions: E = {n : n is an even integer }, F = {n : n is an even prime number } = {2}.∪: union of two sets, ∩: intersection of two sets. For example, E ∩ F = {2} = FNote: For all set a, a ∩ ∅ = ∅, a ∪ ∅ = a.∅ ∪ {∅} ∪ {{∅}} = {∅, {∅}}.a ⊂ b: ”a is included in b” or ”a is a subset of b” iff all the members of a are also members of b. Note that any set is a

subset of itself and is a subset of every set. The fact (that ∅ ⊆ a) is ”vacuously true”. In particular, ∅ ⊆ ∅, but ∅ /∈ ∅, and{∅} ∈ {{∅}} but {∅} 6⊆ {{∅}}. If a set a ⊆ ∅ then a = ∅.

If a is a set, Pa or P(a) denotes the power set of a, for example:

• P∅ = {∅}

• P{∅} = {∅, {∅}}

• P{0, 1} = {∅, {0}, {1}, {0, 1}}

Pa = {x | x is a subset of a} = {x | x ⊆ a}

First Order Logic

is a precise language to write mathematical statements (for us, statements about sets).We give an informal presentation Symbols:

• variables: x, y, z, a, b, c, . . .

• ∈,=

• logical connections: and & (∧), or ∨, not ¬, implies =⇒ , is equivalent to ⇐⇒

• quantifiers: exists ∃, for all ∀

• parenthesis ()

Using these symbols we can write formulas:

Example. Ψ0 = ∀x(x = x),Ψ1 = ∀x∃y(y ∈ x),Ψ2 = ∀x∃y(x ∈ y),Ψ3 = ∀x∀y((∀t(t ∈ x ⇐⇒ t ∈ y)) =⇒ x = y).A formula (with no free variables) can be either true or false (in the universe of sets). truth value of the above: T,F(∅),T(y = Px), T (Ψ3 expresses extensionality)

A formula with a free variable x, φ(x) = ∃y∃z((y ∈ z)&(z ∈ x)), meaning x is not under the scope of a quantifier. Youcannot say whether φ(x) is true or false without knowing what x is. However, we can ask, given a set a if φ(a) is true orfalse. For example, φ(∅) is false, φ({∅}) is also false, but φ({{∅}}) is true: take y = ∅, z = {∅}.

Note: a ⊆ b is not a first order formula, but can be written as a first order formula: ? : ∀x(x ∈ a =⇒ x ∈ b). In practice,we do allow ourselves to write a ⊆ b as a first order formula, knowing that it is a shorthand for ?.

x = Py can be a shorthand for ∀u(u ∈ x ⇐⇒ u ⊆ y).A =⇒ B is true when either A is false or both A and B are true. A =⇒ B share the truth value of ¬A ∧BChapter 2:

1. Introduce the first axiom of set theory (others will be added later) and show how to use them to construct sets.

2. Formally, each axiom is a first order formula. We will state them in English and first order logic.

Axiom. Extentionality Axiom: If two sets have exactly the same members, then they are equal:

∀a∀b[∀x(x ∈ a ⇐⇒ x ∈ b) =⇒ a = b]

2

Axiom. Empty Set Axiom: There is a set that has no member:

∃a(∀x¬(x ∈ a))

or equivalently∃a(∀x(x /∈ a))

Theorem. There is a unique set which has no members.

Proof. By the empty set axiom, there is a set which has no member. Suppose a and b are two such sets, then a and b havethe same members. Hence by the extentionality axiom, a and b are equal. Therefore, there is a unique set which has nomembers.

Definition. The unique set without members is called the empty set and denoted ∅.

We cannot yet prove the existence of any other set. Need more axioms for that.

Axiom. Pairing Axom: For any sets u and v, there is a set having as members just u and v:

∀u∀v∃a∀x[x ∈ a ⇐⇒ (x = u or x = v)]

We can now prove the existence of other sets.

Proposition. Thee is a unique set whose only member is the empty set.

Proof. By the pairing axiom applied to u = ∅ and v = ∅, there is a set whose only element is ∅. By extensionality, there is aunique such set.

This set will be denoted {∅}.What other sets can we build? If we apply the pairing axiom with u = ∅ and v = {∅}, we obtain {∅, {∅}}. If we apply

the pairing axiom with u = v = {∅}, we obtain {{∅}}. If u = {∅, {∅}}, v = ∅, we obtain {{∅, {∅}}, ∅}.We can already construct infinitely many sets, but we cannot yet prove the existence of sets with more than two elements.

In particular, we would like to be able to construct the union of two sets.

Axiom. Union Axiom: For any set a, there is a set b whose members are precisely the members of the members of a.

∀a∃b∀x[x ∈ b ⇐⇒ ∃c(c ∈ a&x ∈ c)]

Idea: a is the set of sets which we want to take the union of, and b is that union. We write b =⋃a.

Exercise. Union of two sets: let c, d be two sets. Applying the pairing axiom, we construct the set a = {c, d}, then applyingthe union axiom to the set a gives us b =

⋃a = c ∪ d.

We can then iteratively construct the union of any finite number of sets.

Exercise. If a, b, c are sets, we can build a set {a, b, c}. Pairing axiom with the empty set to form {a, ∅}, {b, ∅}, {c, ∅} andtake union. Or, pairing axiom to get {a, b}, {b, c} and pairing axiom again to get {{a, b}, {b, c}} then union axiom applied tothe set above.

We can build all hereditary finite sets, i.e., sets that can be completely written with ∅, {}. We cannot yet prove theexistence of the intersection of any two sets. New axiom!

Introduction to lecture 4: We want to be able to construct sets given by any condition

x ∈ a ⇐⇒ · · ·

In other words we would like to construct sets by giving a condition that its elements satisfy, for example, being an integeror a finite set.

There are two caveats (”disasters”) to this that we have to be careful about:

• unclear definitions: ”the smallest integer that cannot be defined in less than 100 English words”

• unbounded definitions: {x : x /∈ x}.

{x : x /∈ x}: the set of sets which do not belong to themselves.Why is this a problem?Let a = {x : x /∈ x}. Do we have a ∈ a? Assume we do, that is, a ∈ a. By definition, any element x ∈ a satisfies x /∈ x.

In particular, this holds for a, so a /∈ a, contradicting the hypothesis. Hence a ∈ a is false, therefore a /∈ a. By definition, acontains all sets x such that x /∈ x, so a ∈ a. We again reach a contradiction.

This is called ”Russell’s paradox”. To avoid this problem, we will restrict the use of formulas to define sets, and allowonly to use formulas to define subsets of existing sets.

3

Breaking things up: subsets

Axiom. Subset Axioms (comprehension axioms): For every set c and first order formula φ(x, t1, . . . , tk), where t1, . . . , tk areany sets, there is a set b whose elements are precisely the elements of c that satisfy φ(x, t1, . . . , tk).

∀c,∀t1, . . . , tk∃b[∀x(x ∈ b ⇐⇒ x ∈ c&φ(x, t1, . . . , xk)]

For every formula φ, we get one axiom. In total, we have infinitely many subset axioms.We can now construct the intersection of two sets.

Proposition. If a and b are two sets, there is a unique set c whose elements are precisely the sets which are members of aand of b.

Proof. Using the subset axiom, we can construct the subset of a composed of elements which are elements of b. More formally,the following is a subset axiom:

∀a∀b∃c[∀x(x ∈ c ⇐⇒ x ∈ a&x ∈ b)]

In the definition of subset axioms, we have set c = a, t1 = b, b = c, φ(x, t1) = (x ∈ t1). The set c given by the axiom has therequired property. By extensionality, there is a unique such set.

This set c is called the intersection of a and b and is denoted a∩ b. As for unions, we can construct the intersection of anarbitrary set of sets.

Proposition. Let A be a nonempty set, then there is a set B whose elements are precisely the sets which are a member ofall members of A. We write B =

⋂A

Proof. As A is nonempty, let c be one of its elements. Then the following is a subset axiom:

∀c∀A∃B[∀x(x ∈ B ⇐⇒ x ∈ c&(∀u(u ∈ A =⇒ x ∈ u)))]

The set B given by this axiom has the required property.

Note:⋂∅ is not defined. It is all sets which is not a set, because false premises imply anything.

Notation: The set of elements of a that satisfy φ(t1, . . . , tk) is denoted {x ∈ a | φ(x, t1, . . . , tk)}.Warning: If we write b = {x ∈ a | φ(. . . )} then b should not appear in φ(. . . ). The textbook contains it explicitly in the

subset axiom.We can now for the first time prove that some set does not exist.

Theorem. There is no set which has all sets as elements.

Proof. Assume there were such a set, call it a. Then by a subset axiom, we can construct the set b = {x ∈ a : x /∈ x}. Thisleads to a contradiction as discussed before. Therefore, such a set a does not exist.

The subset axiom allows us to construct subsets of already existing sets. To do mathematics, it is often needed to considerall subsets of a set a. For this we need yet another axiom.

Axiom. Power Set Axiom: For any set a, there is a set b whose elements are precisely the subsets of a.

∀a∃b(∀x(x ∈ b ⇐⇒ x ⊆ a))

The set b is unique by extensionality and is called the the power set of a. It is denoted Pa or P(a).

Example. P∅ = {∅},P{∅} = {∅, {∅}}.

For finite sets, this axiom does not seem to give you new sets, but it will when applied to an infinite set a.

Algebra of Sets

We have seen two operations: ∩,∪. We define A − B or A \ B (relative complement of B in A) as the set of elements of Awhich are not element of B, that is:

A−B = {x ∈ A | x /∈ B}

Note we do not define the (absolute) complement of a set which would lead to Russell’s paradox.Note: A−B does not assume B is a subset of A.

4

Algebra of Sets

We have operations ∪,∩,P,−. They satisfy various identities listed in the textbook page 28 + 30.Distributive laws: A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C), A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C).De Morgan laws: C − (A∪B) = (C −A)∩ (C −B), C − (A∩B) = (C −A)∪ (C −B). There are corresponding identities

for infinite unions or intersections.distributive laws: A ∪

⋂B =

⋂{A ∪X : X ∈ B} , A ∩

⋃B =

⋃{A ∩X : X ∈ B}

de Morgan laws: C −⋂B =

⋃{C −X : X ∈ B} , C −

⋃B =

⋂{C −X : X ∈ B}

How to prove such identities?

• by listing all cases (only works for the finite series without infinite⋂,⋃

).

Example. A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C). Draw a Venn diagram.

we could take an x and enumerate all possibilities, x ∈ A or not, x ∈ B or not...(see textbook)

• with a proof either by equivalences or by double inclusion

Example. C −⋂B =

⋃{C −X : X ∈ B}. (You can put words in first-order formulas to make it more readable. The

textbook does it very often.) We have:

t ∈ C −⋂B ⇐⇒ (t ∈ C) ∧ ¬(t ∈

⋂B)

⇐⇒ (t ∈ C) ∧ ∃X(X ∈ B ∧ ¬(t ∈ X))

⇐⇒ ∃X(X ∈ B ∧ t ∈ C −X)

⇐⇒ t ∈⋃{C −X : X ∈ B}

Introduce a name for an element and turn equivalence to element-wise check, although sometimes you can directlyapply laws and obtain a chain of equality.

Example. Prove that for all set A,⋃

PA = A by double inclusion:

⊇: A ⊂ A =⇒ A ∈PA =⇒ A ⊂⋃

PA

or : ∀t(t ∈ A) =⇒ {t} ∈PA =⇒

∀t(t ∈ A) =⇒ (t ∈⋃

PA)

⊆: ∀t(t ∈⋃

PA =⇒ ∃B(B ∈PA ∧ t ∈ B)

=⇒ B ⊆ A =⇒ t ∈ A

=⇒⋃

PA ⊂ A

Chapter 3. Relations and Functions

Pairs

An (ordered) pair: 〈a, b〉 (or often written (a, b)) is different from the set {a, b} in at least two ways:

• 〈a, b〉 6= 〈b, a〉 where as {a, b} = {b, a}

• 〈a, a〉 6= 〈a〉 whereas {a, a} = {a}

We nonetheless want to encode pairs as sets.First attempt: 〈a, b〉2 = {{a}, b}. Does not work: {{∅}, {{∅}}}. It could be 〈∅, {{∅}}〉 or 〈{∅}, {∅}〉Second attempt works: {a, b} = {{a, ∅}, {b, {∅}}}. It works, only when a = {∅}, b = ∅ it collapses and in this case we

also know what are a and b. –indexing a, b by the hierarchy of sets due to Varun

Definition. If x, y are sets, the pair 〈x, y〉 is {{x}, {x, y}}.

Proposition. If 〈a, b〉 = 〈c, d〉 then a = c and b = d –indexing by cardinality of elements as sets.

You really need to split into cases since it can also collapse.

5

Proof. Assume a = b. Then 〈a, b〉 = {{a}}. If c 6= d, then {c} 6= {c, d} so 〈c, d〉 has two elements and cannot be equal to{{a}}. Therefore c = d and 〈c, d〉 = {{c}}. Since {{c}} = {{a}}, we have a = c and b = a = c = d.

Assume a 6= b, then 〈a, b〉 = {{a}, {a, b}} has two distinct elements. We must have c 6= d (otherwise 〈c, d〉 has oneelement). Then 〈a, b〉 has a unique element with one element {a}, similarly 〈c, d〉 has a unique element with one element {c}.Therefore {a} = {c} so a = c. We then have {a, b} = {c, d}. Since a = c and a 6= b, b = d.

Definition. If A and B are two sets, the Cartesian product of A and B denoted A × B is the set of pairs of the form〈a, b〉, a ∈ A, b ∈ B.

Note: For this to make sense, we have to prove that this set exists and is unique. Uniqueness follows from extensionality.How to show existence? We want to use a subset axiom. For this, we need to find a set PP(A ∪B) of which it is a subsetand a first-order formula defining it.

Lemma. If x ∈ A and y ∈ B, then 〈x, y〉 ∈PP(A ∪B).

Pairing axiom {A,B}, union axiom⋃{A,B} = A ∪B, and the power set axiom PP(A ∪B)

Proof. Let C = A∪B, so x ∈ C and y ∈ C. As x ∈ C, {x} ∈PC. As x, y ∈ C, {x, y} ∈PC. Therefore, {{x}, {x, y}} ⊂PC,hence {{x}, {x, y}} ∈PPC as was to be shown.

A formula used to define A×B is:

p ∈ A×B ⇐⇒ ∃x∃y(x ∈ A ∧ y ∈ B ∧ p = {{x}, {x, y}}

Cheating: p = {{x}, {x, y}} is not first order.

φ(p;x, y) = ∀z(z ∈ p ⇐⇒ z = {x} ∨ z = {x, y})= ∀z(z ∈ p ⇐⇒ ((∀u(u ∈ z ⇐⇒ u = x)) ∨ (∀v(v ∈ z ⇐⇒ (v = x ∨ v = y)))))

It now follows from the subset axioms that the set A×B exists.Notation: A2 = A×AWe define (ordered) n-tuples inductively3-tuples: 〈x, y, z〉 = 〈〈x, y〉, z〉.n-tuples: 〈x1, · · · , xn〉 = 〈〈x1, · · ·xn−1〉, xn〉.Note: the ”natural” definition 〈x, y, z〉 = {{x}, {x, y}, {x, y, z}} does not work. 〈x, y, x〉 = 〈x, y, y〉.

Definition. A (binary) relation is a set of ordered pairs.

Example. order on Z, <Z= {〈x, y〉 : x, y ∈ Z, x is less than y}.divisibility on N, |N = {〈a, b〉 : a, b ∈ N, a divides b}N× N is a relation (mere generally A×B for any sets A and B is a relation.RQ = {〈x, y〉 : x, y ∈ Q, x · y = 1}.

Definition. A relation are symmetric if for all x, y we have 〈x, y〉 ∈ R ⇐⇒ 〈y, x〉.

Example. In the four examples above, the last two are symmetric.

Notation. We usually write xRy instead of 〈x, y〉 ∈ R

Example. 2 <Z 4 is the same thing as 〈2, 4〉 ∈<Z

Definition. If R is a relation, its domain domR = {x|∃y(xRy)}; its range is rgR = {y|∃x(xRy)}; its field is fldR =domR ∪ rgR.

Example. dom(<Z) = Z = rg(<Z); dom(<N) = N, rg(<N) = N− {0}. Note: N = {0, 1, 2, . . . }. dom(RQ) = Q− {0}.

In less precise metaphysical definitions, you can say 0 is related to nothing so it is also in the domain, but our definitionexcludes it, because 0 does not appear on the left side (both sides de facto) of any pairs in RQ a priori.

Note: Why does domR exist? To apply subset axiom, we need to find a set of which this is a subset. In fact, we can take⋃⋃R.

If x ∈ domR, then for some y, {{x}, {x, y}} ∈ R. Then {x}, {x, y} ∈⋃R and x, y ∈

⋃⋃R.

Therefore, domR = {x ∈⋃⋃

R|∃y(xRy)}, hence this set exists by a subset axiom. Similarly, rgR, fldR ⊆⋃⋃

R

Remark. We define n-ary relations as subsets of n-tuples. We do not define the range of an n-ary relations. In this chapter,class, we mean binary relation. ∅ is a relation. We can define relation in first order logic, by saying every element of R is anordered pair.

6

Definition. A function is a relation F such that for every x ∈ domF there is exactly one y with 〈x, y〉 ∈ F . If F is afunction, x ∈ domF , the unique y such that 〈x, y〉 ∈ F is called the image of x by F or the value of F at x and we writey = F (x). We say that F is a function from A to B if F is a function and A = domF and rgF ⊆ B. We write F : A→ B.We say that F is onto B if rgF = B. We say that the function F is one-to-one or injective if for every y ∈ rgF , there is aunique x ∈ A such that xFy or F (x) = y.

Note: A function can have as domain a set of pairs or n-tuples. For example, addition + on R is a function from R2 → R.We define some operations on relations (though they are used especially with functions)

Definition. Let R,S be two relations, the inverse of R is the relation R−1 = {〈u, v〉 ∈ rg(R)×dom(R)|vRu}; the compositionR ◦ S = {〈u, v〉 ∈ dom(S) × rg(R)|∃t(uSt ∧ tRv)}; the restriction of R to a set A is R|A = {〈u, v〉 ∈ R|u ∈ A ∧ uRv}; theimage of A under R is F [[A]] = rg(F |A) or by the textbook notation ran(F |A) = {v ∈ ran(R)|∃u(u ∈ A ∧ uRv)}.

You can unpack everything but we have proved the existence of domain and range of relations.

Example. If R =<Z, R−1 =>Z, R ◦R = {〈u, v〉 ∈ Z2|v − u ≥ 2}. If S =<R then S ◦ S = S. R[[{−2, 0, 4}]] = {−1, 0, . . . } =

{a ∈ Z|a ≥ −1}.

Let F : R → R, x 7→ x2, F−1 = {〈u, v〉 ∈ R2|u ≥ 0, v =√u ∨ v = −

√u}. F−1 is well defined as a relation but not as a

function. However,(F |R≥0

)−1is a function, x 7→

√x.

Note: If F is a function, F−1 is a function if and only if F is one-to-one. The domain of F−1 is the range of F andvice-versa.

Note: If F is a relation but not a function, we never write F (a). However, we can consider F [[A]] or F [[{a}]]. ifF : R→ R, x 7→ x2, F−1[[{4}]] = {2,−2}. You cannot write F−1(4) because F−1 is not a function.

Proposition. If F and G are functions, then F ◦G is a function whose domain is {x ∈ domG | G(x) ∈ domF}. If x is in itsdomain, F ◦G(x) = F (G(x)).

Proof. Let us show domF ◦G = {x ∈ domG | G(x) ∈ domF} by double inclusion.(⊂): Let x ∈ domF ◦G, so there is y such that 〈x, y〉 ∈ F ◦G. Hence there is z such that 〈x, z〉 ∈ G and 〈z, x〉 ∈ F . This

implies that x ∈ domG and G(x) = z, and z is in the domain of F .(⊃): If x ∈ domG and G(x) ∈ domF , then F (G(x)) is well defined and we have 〈x,G(x)〉 ∈ G and 〈G(x), F (G(x))〉 ∈ F

hence 〈x, F (G(x))〉 ∈ F ◦G so x ∈ domF ◦G.

We show that F ◦ G is a function. If x ∈ domF ◦ G, then there are y, z such that 〈x, y〉 ∈ G and 〈y, z〉 ∈ F . As Gis a function we must have y = G(x) and as F is a function we must have z = F (y) = F (G(x)) so if 〈x, z〉 ∈ F ◦ G thenz = F (G(x)). Hence F ◦G is a function and F ◦G(x) = F (G(x)) for x ∈ domF ◦G.

Definition. If A is a set, we define the identity function of A : IA = {〈a, a〉 | a ∈ A}.

If F : A→ B is one to one and onto B then F−1 is a function and we have

F ◦ F−1 = IB

F−1 ◦ F = IA

Warning : To prove that G is the inverse of F : A→ B one needs to show BOTH.

Theorem. Assume that F : A→ B and A 6= ∅

(a) There is a function G : B → A such that G ◦ F = IA (a ”left-inverse”) if an only if F is one to one

(b) There is a function G : B → A such that F ◦G = IB (a ”right-inverse”) if and only if F is onto B

Proof. (a) Assume F has a left-inverse G, let y ∈ ranF , let x ∈ domF such that F (x) = y then G(F (x)) = G(y) hencex = G(y) so there is a unique such x and F is one-to-one.

Conversely, assume that F is one-to-one. As A is nonempty, we can choose some v ∈ A. We define a functionG : B → A as follows: for y ∈ ranA, there is a unique x ∈ A such that F (x) = y and we set G(y) = x. For y ∈ B−ranF ,set G(x) = v. One checks that for any x ∈ A,G(F (x)) = x.

(b) Assume there is a function G : B → A such that F ◦ G = IB . Let y ∈ B, set x = G(y) then x ∈ A and F (x) =F (G(y)) = IB(y) = y. Therefore F is onto B.

If F is onto B, then we do have F ◦F−1 = IB where F−1 is a relation, needing not to be a function. To build G witha subset axiom, we need a formula to define it: that formula would have to pick out one specific x suchthat F (x) = y for every y ∈ B. In general, we have no way to do that.

7

Axiom. of Choice (first version): For any relation R, there is a function G ⊂ R with domG = domR.

Now let G ⊂ F−1 with domG = domF−1 = ranF = B. Then for y ∈ B we have 〈y,G(y)〉 ∈ F−1 so 〈G(y), y〉 ∈ Fhence F (G(y)) = y.

Definition. A relation R is single-rooted if for every y ∈ ranR, there is a unique x ∈ domR with xRy.

Notation: If F is a function with domain I, we can consider⋃{F (i) | i ∈ I}. We will denote this by

⋃i∈I F (i). Same for⋂

.

Example.⋂{X ∪A | A ∈ A} =

⋂A∈AX ∪A.

Warning :⋂A∈AX ∪A is very different from

⋂X ∪A.

Theorem. Let R be a relation, then for any set A 6= ∅, we have

a) R[[⋃A]] =

⋃A∈AR[[A]] = ∪{R[[A]] | A ∈ A};

b) R[[⋂A]] ⊆

⋂A∈AR[[A]], equality holds if R is single-rooted;

c) R[[A−B]] ⊇ R[[A]]−R[[B]] with equality if R is single-rooted.

Proof. We have

t ∈⋂A∈A

R[[⋂A]] ⇐⇒ ∀A∃x(A ∈ A =⇒ x ∈ A&xRt)

t ∈ R[[⋂A]] ⇐⇒ ∃x ∈

⋂A(xRt)

⇐⇒ ∃x∀A(A ∈ A =⇒ x ∈ A&xRt)

∗ =⇒ ∀A∃x(A ∈ A =⇒ x ∈ A&xRt)

⇐⇒ t ∈⋂A∈A

R[[⋂A]]

We have shown R[[∩A]] ⊆⋂A∈AR[[A]]. Now assume that R is single-rooted, then ∀t ∈ ranR there is at most one x such

that xRt. If for all A ∈ A there is x ∈ A with xRt, then all those x’s have to be the same. Since A is nonempty, we dohave

∀A∃x(A ∈ A =⇒ x ∈ A&xRt) =⇒ ∃x∀A(A ∈ A =⇒ x ∈ A&xRt)

So we deduce the equality R[[⋂A]] =

⋂A∈AR[[A]].

Definition. Let R be a relation on a set A

(i) R is reflexive on A if for all x ∈ A (you have to specify A) we have xRx;

(ii) R is transitive if for all x, y, z if xRy and yRz hold, then xRz holds;

The relation R is an equivalence relation on A if R is a relation on A and R is reflexive on A, transitive and symmetric.If R is an equivalence relation on A, and x ∈ A, the set

[x]R = {t ∈ A|xRt}

is called the equivalence class of x (modulo R)

Note [x]R is a set and [x]R = {t ∈ A|tRx} by symmetry.

Lemma. Let R be an equivalence relation on A, let x, y ∈ A then:

(i) x ∈ [x]R

(ii) if xRy then [x]R = [y]R

(iii) if ¬(xRy) then [x]R ∩ [y]R = ∅

Proof. (i) follows from reflexivity of R(ii) Assume xRy, let z ∈ [x]R. Then we have xRz. We also have yRx by symmetry, so by transitivity, yRz holds. Thus

z ∈ [y]R and therefore [x]R ⊂ [y]R. By symmetry, [x]R ⊃ [y]R. Thus, [x]R = [y]R.(iii) We prove the contrapositive: if [x]R ∩ [y]R 6= ∅ then xRy. Assume z ∈ [x]R ∩ [y]R, then xRz and yRz hold. By

transitivity xRy holds.

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The set of all equivalence classes (modulo R) is denoted A/R: ”the quotient of A by R (or modulo R) or ”A modulo R”.

Example. R on R∗, xRy if xy > 0 (iff x and y have the same sign)n ∈ N. Consider ≡n on Z defined by x ≡n y if n divides y − x.If H ≤ G as subgroup, the relation ≡H : x ≡H y if xH = yH is an equivalence relation on G.If f : A→ B then we define Ef : xEfy if f(x) = f(y). Then Ef is an equivalence relation on A. Actually any equivalence

relation can be written in this form. Let R be an equivalence relation on A. Let πR : A → A/R, x 7→ [x]R (”the canonicalprojection”). We then have R = EπR . Why? It follows from the lemma that for x, y ∈ A, we have [x]R = [y]R =⇒ xRy.Hence πR(x) = πR(y) =⇒ xRy so xEπRy =⇒ xRy.

Definition. A partition Π of a nonempty set A is a set of subsets of A such that:

(i) ∅ ∈ Π

(ii) ∀B,C ∈ Π, B 6= C =⇒ B ∩ C = ∅

(iii) ∀x ∈ A,∃B ∈ Π, x ∈ B

Note Π ⊂PA

Definition. If R is an equivalence relation on A, then A/R is a partition on A.

Proof. We have that x ∈ [x]R for any x ∈ A, hence no element of A/R is empty. This also shows condition (iii). If [x]R 6= [y]R,then ¬(xRy) and therefore [x]R ∩ [y]R = ∅ so (ii) also holds.

Conversely, let Π be a partition of A. Note that by (ii) and (iii), for any x ∈ A, there is a unique B ∈ Π such that x ∈ B.We define the relation RΠ on A by xRΠy if x and y belong to the same element of Π. Then RΠ is an equivalence relationand A/RΠ = Π.

3 ways to build an equivalence relation R:

• as a relation on A

• as the partition A/R

• as the canonical projection A→ A/R

Let R be an equivalence relation on A. Let G : A/R → B be a function, then we can ”lift” G to A that is, we can defineG : A→ B by G = G ◦ πR.

A

A/R B

πR G

G

A

A/R B

FπR

F

More frequent situation, we are given F : A→ B we want to construct F : A/R→ B such that F = F ◦ πR.

Lemma. Given A,B,R as above, there is F : A/R→ B such that F = F ◦ πR if and only if

∀x, y ∈ A, xRy =⇒ F (x) = F (y)

“F is constant on each equivalence class.”

Proof. Assume there is F : A/R → B such that F = F ◦ πR. Let x, y ∈ A and assume xRy. Then [x]R = [y]R, soπR(x) = πR(y). Therefore F ◦ πR(x) = F ◦ πR(y), hence F (x) = F (y).

Conversely, assume that xRy =⇒ F (x) = F (y). Then define F : A/R→ B, [x]R 7→ F (x) or F = {〈[x]R, F (x)〉 : x ∈ A}.This is well-defined or this is a function as: if [x]R = [y]R then F (x) = F (y).

In particular, if F : A → A is compatible with R, that is xRy =⇒ F (x)RF (y) then we can construct F : A/R → A/R

such that F ([x]R) = [F (x)]R.

A A

A/R A/R

F

πRπR◦F

πR

F

Let F ′ = πR ◦ F and build F as above with F ′.

We need to represent natural numbers as sets. A priori, we can do this in various ways. Perhaps, the most natural waywould be to set 0 = ∅, 1 = {∅}, 2 = {{∅}} and in general n + 1 = {n}. However, the representation would make it difficultto define the set of natural numbers. We are actually going to use the Von-Neumann representation: 0 = ∅, 1 = {∅}, 2 ={∅, {∅}} = 1 ∪ {1}, 3 = 2 ∪ {2} = {∅, {∅}, {∅, {∅}}}. In general, n+ 1 = n ∪ {n}.

Note, with this representation, we have:

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• n has exactly n elements;

• n = {0, 1, 2, . . . , n− 1};

• in particular, if m < n then m ∈ n and m ⊂ n.

Definition. If a is a set, the successor of a, defined a+ is a+ = a ∪ {a}.

Example. ∅+ = ∅ ∪ {∅} = {∅}, ∅++ . . .

Definition. A set A is inductive if for any a ∈ A we have a+ ∈ A and ∅ ∈ A.

Example. If A is inductive, ∅ ∈ A so + ∈ A also ∅++ ∈ A and so on. Hence A contains all the sets that we want to take asrepresenting the natural numbers, but A could contain other sets.

Note: we cannot yet prove the existence of an inductive set. (note that such a set is necessarily infinite although wecannot define “infinity” yet without the natural numbers so the note is informal). All the axioms do not bring hereditaryfinite set to infinite set.

Axiom. Infinity Axiom: There exists an inductive set.

∃A[∅ ∈ A&(∀a ∈ A)a ∪ {a} ∈ A]

Definition. A natural number is a set that belongs to all inductive sets. We let ω denote the set of natural numbers.

Why does it exists? We know there is an inductive set A, we can define ω as a subset of A

ω = {n ∈ A | ∀B(“B is inductive” =⇒ n ∈ B}

Proposition. ω is inductive.

Proof. ∅ belongs to all inductive sets, so ∅ ∈ ω.If a ∈ ω then for any inductive set A, a ∈ A, hence a+ ∈ A as A is inductive, so a+ belongs to all inductive set, hence

a+ ∈ ω.

Notation: we set 0 = ∅, n = (n−1)+. Note 0, 1, 2, 3, . . . are all elements of ω. (caveat: we do not know whether ω containsnonstandard numbers but our theory never knows. “Compactness theorem”???)

Induction Principle

Any inductive subset of ω is equal to ω.

Proof. By definition of ω, any inductive set A satisfies ω ⊂ A. In particular, if A ⊂ ω is inductive, then A = ω.

Proposition. Every natural number 6= 0 is a successor of a natural number.

Proof. Let P = {n ∈ ω | n = 0 ∨ ∃m ∈ ω, n = m+}. Then P is inductive:

• 0 ∈ P ;

• if n ∈ P then n ∈ P then n ∈ ω so n+ ∈ P .

By the induction principle, P = ω which proves the result.

Transitive sets A set A is transitive if∀x ∈ A∀y ∈ x(y ∈ A)

In English, “A is transitive if an element of an element of A is an element of A”. (somehow contradicts the late feudal systemaround medieval Europe, “vassal of my vassal is not my vassal” seemingly to be reflected also in board games.)

Example. ∅ is transitive, {∅} is transitive but {{∅}} is not. However {∅, {∅}} is.

Note: A set A is transitive

• if and only if⋃A ⊂ A;

• if and only if A ⊂PA;

• if and only if ∀a ∈ A, a ⊂ A.

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Lemma. If a is transitive, then⋃a+ = a

Proof. ⋃a+ =

⋃(a ∪ {a})

= (⋃a) ∪ (

⋃{a})

= (⋃a) ∪ a

= a, as⋃a ⊂ a

Proposition. Every natural number is a transitive set.

Proof. Let P = {n ∈ ω | n is transitive }.

• ∅ ∈ P ;

• if n ∈ P then⋃n+ = n ⊂ n+ so n+ is transitive, hence n+ ∈ P

Therefore P is inductive. By the induction principle, P = ω.

Proposition. ω is a transitive set.

Proof. Let P = {n ∈ ω | n ⊂ ω}. We have

• ∅ ∈ P ;

• if n ∈ P then n+ = n ∪ {n} ⊂ ω as n ⊂ ω and {n} ∈ ω

Therefore P is inductive and P = ω. By one of the characterizations of transitive sets, ω is transitive.

Proposition. The function ω → ω, n 7→ n+ is one-to-one.

Proof. If m,n ∈ ω satisfy m+ = n+ then⋃m+ =

⋃n+. As m,n are transitive

⋃n+ = n,

⋃m+ = m so m = n.

Recursion

Theorem. Let A be a set, a ∈ A and F : A → A. There exists a unique function h : ω → A such that h(0) = a, h(n+) =F (h(n)) for all n ∈ ω.

Example. h(0) = 2, h(n+ 1) = 3h(n)− 1

Proof. Uniqueness: Let h, h′ be two functions satisfying the condition. Define T = {k ∈ ω | h(k) = h′(k)}. We want to showT = ω, for this it is enough to show that T is inductive.

• h(0) = h′(0) = a =⇒ 0 ∈ T

• Assume that n ∈ T, h(n) = h′(n). We have h(n+) = F (h(n)) = F (h′(n)) = h′(n+). Therefore n+ ∈ T .

Hence T is inductive and therefore T = ω. It follows that h = h′.Existence: Call a function v acceptable if:

• dom(v) ⊂ ω

• ran(v) ⊂ A

• 0 ∈ dom(v) and v(0) = a

• if n+ ∈ dom(v) then n ∈ dom(v) and v(n+) = F (v(n))

Let A be the set of all acceptable functions. It is a subset of Aω the set of functions from ω → A and define h =⋃A. Note

that a priori, we only know that h is a relation between ω and A, i.e. h ⊂ ω ×A.

Example. {〈∅, a〉} ∪ {〈∅, b〉}

We will show that h is de facto a function and satisfies the requirements.To show dom(h) = ω:

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• Let T = dom(h). We need to show that T is inductive. The function v0 = {〈0, a〉} is acceptable. so dom(v0) ⊂dom(h) = T

• Assume n ∈ T , there is an acceptable function v such that n ∈ dom(v). If n+ ∈ dom(v) then n+ ∈ T ; otherwise,consider v′ = v ∪ {〈n+, F (v(n))〉}. We have to check that v′ is acceptable.

The first 3 conditions are clear.

The 4th condition: Let k+ ∈ dom(v′).

– If k+ = n+ (note: n ∈ dom(v) ⊂ dom(v′)) then V ′(k+) = F (v(n)) = F (v′(n)) = F (v′(k))

– If k+ 6= n+, k+ ∈ dom(v) hence k ∈ dom(v) as v is acceptable, so k ∈ dom(v′), v′(k) = v(k) and v′(k+) = v(k+) =F (v(k)) = F (v′(k)).

As v′ ∈ A, n+ ∈ dom(v′), n+ ∈ T .

Therefore T is inductive and dom(h) = ω.To show h is a function: Consider S = {n ∈ ω | there is at most oney ∈ A〈n, y〉 ∈ h}

• If 〈0, y〉 ∈ h then there is v ∈ A such that 〈0, y〉 ∈ v i.e. v(0) = y. By definition of acceptability, v(0) = a so y = a.Therefore 0 ∈ S.

• Assume n ∈ S then ∃b ∈ A∀v ∈ A(v(n) = b). If n+ ∈ dom(h) then ∃v ∈ A(n+ ∈ dom(v)) then by the 4th condition ofacceptability n ∈ dom(v) and v(n+) = F (v(n)) = F (b) so n+ ∈ S.

Therefore S is inductive, and h is a function.It also follows from the discussion above that 〈0, a〉 ∈ h and if 〈n, b〉 ∈ h then 〈n+, F (b)〉 ∈ h so h satisfies the required

properties.

Arithmetic

Addition: For m ∈ ω, we use the recursion theorem to define the function Am : ω → ω with

{Am(0) = mAm(n+) = Am(n)+ . By the

recursion theorem such function exists and is unique. We define for m,n ∈ ω,m+ n = Am(n).

In other words we have

{m+ 0 = mm+ n+ = (m+ n)+

Multiplication: For m ∈ ω, we define (using the recursion theorem) Mm : ω → ω by

{Mm(0) = 0Mm(n+) = Mm(n) +m

. For

m,n ∈ ω,m · n = Mm(n).

Example. 1 + 1 = A1(1) = A1(0+) = A1(0)+ = 1+ = 2

Theorem. • Addition is

– associative: m+ (n+ p) = (m+ n) + p for all m,n, p ∈ ω– commutative: m+ n = n+m

• Multiplication is associative and commutative

• Addition is distributive over multiplication m · (n+ p) = m · n+m · p

Proof. Addition is associative: prove by induction on p. Fix m,n ∈ ω and consider T = {p ∈ ω | m+ (n+ p) = (m+n) + p}.

• m+ (n+ 0) = m+ n = (m+ n) + 0 = Am+n(0) so 0 ∈ T

• Assume p ∈ T,m+(n+p+) = m+(n+p)+ = (m+(n+p))+ = ((m+n)+p)+ as p ∈ T . Then ((m+n)+p)+ = (m+n)+p+

as desired.

Then T = ω and m+ (n+ p) = (m+ n) + p for all m,n, p ∈ ωAddition is commutative:

• first show by induction on n that 0 +m = m+ 0 = m for all m ∈ ω.

– 0 + 0 = 0

– Assume 0 +m = m then 0 +m+ = (0 +m)+ = m+ by induction hypothesis

Hence by induction 0 +m = m for all m ∈ ω.

12

• next show by induction that n+ +m = (n+m)+ for all n,m ∈ ω. Fix n ∈ ω and we prove it by induction on m:

– n+ + 0 = n+ = (n+ 0)+

– assume n+ + m = (n + m)+ and we compute n+ + m+ = (n+ + m)+ = ((n + m)+)+ by induction hypothesis.Then ((n+m)+)+ = (n+m+)+

• Now fix m ∈ ω, we prove by induction on n that m+ n = n+m for all n ∈ ω

– n = 0 : m+ 0 = m = 0 +m

– assume m+n = n+m then m+n+ = (m+n)+ = (n+m)+ by induction hypothesis. As above (n+m)+ = n+ +m.

The other statements in the theorem are similarly proven by induction.

Definition. For m,n ∈ ω, define m <ω n if m ∈ n. We write m ≤ω n if (m <ω n or m = n)

Note: the textbook does not use<ω (it has always ∈ instead), and writes ∈ instead of≤ω so<ω= {〈m,n〉 | m,n ∈ ω,m ∈ n}is a (binary) relation on ω.

Lemma. a) For m,n ∈ ω we have m ∈ n ⇐⇒ m+ ∈ n+

b) For m ∈ ω,m /∈ m

By far we have not put axiom forbidding m /∈ m because integers do not need it. Later we will.

Proof. ⇐= : Assume m+ ∈ n+. Recall n+ = n ∪ {n}, so either m+ ∈ n or m+ = n. If m+ = n then m ∈ m+ = n.If m+ ∈ n, then m ∈ m+ ∈ n. As n is transitive, this implies m ∈ n.=⇒ by induction on n. T = {n ∈ ω | ∀m(m ∈ n =⇒ m+ ∈ n+)}

• 0 ∈ T is true as there is no m ∈ 0

• assume k ∈ T let m ∈ k+. As above either m = k or m ∈ k. If m ∈ k,m+ ∈ k+ as k ∈ T , so m+ ∈ k+ ∈ k++. As k++

is transitive, m+ ∈ k++ so k++ ∈ T .

If m = k,m+ = k+ ∈ k++.

Thus k+ ∈ T and T = ω.

We prove by induction that n /∈ n for all n ∈ ω.

• 0 /∈ 0 as ∅ /∈ ∅

• Assume n /∈ n. We can rewrite the above as the contrapositive m /∈ n ⇐⇒ m+ /∈ n+. Applying it for m = n.

Theorem. The relation <ω is a linear ordering on ω, i.e.:

• <ω is transitive

• ∀m,n ∈ ω exactly one of the following holds:

m <ω n,m = n, n <ω m

Proof. If a ∈ b, b ∈ c are natural numbers, a ∈ c as c is a transitive set. Therefore <ω is transitive.We show that no two of

m <ω n,m = n, n <ω m

can hold at onceAssume m <ω n,m = n, then m ∈ m which is impossible by the lemma. Same for m = n, n <ω m. Finally if m <ω n

and n <ω m then m ∈ n ∈ m as n is transitive then m ∈ m again a contradiction.We show that at least one holds:Let T = {n ∈ ω | (∀m ∈ ω)m <ω n ∨m = n ∨ n <ω m}.

• 0 ∈ T . We show by induction that ∀n ∈ ω(n 6= 0 =⇒ 0 ∈ n). Induction step 0 ∈ n ∈ n+ by transitivity.

• Assume n ∈ T , let m ∈ ω. As n ∈ T , either m <ω n or m = n or n <ω m. If m = n then m ∈ n+ so m <ω n+. If

m ∈ n ∈ n+,m <ω n. If n ∈ m then n+ ∈ m+ by the lemma. Either n+ = m or n+ ∈ m =⇒ n+ <ω m.

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Corollary. For all m,n ∈ ω

• m <ω n iff m ( n

• m ≤ω n iff m ⊆ n

From now we write < instead of <ω when there is no confusion.

Proposition. For all m,n, p ∈ ω we have

(i) m < n ⇐⇒ m+ p < n+ p

(ii) if p 6= 0,m < n ⇐⇒ m · p < n · p

Proof. =⇒ we do an induction on p.

• m = m+ 0 < n+ 0 = n so p = 0 satisfies

• (m+ p)+ = m+ p+ < n+ p+ = (n+ p)+ ⇐⇒ m+ p < n+ p by the previous lemma

⇐= : Assumem+p < n+p. To showm < n it suffices to showm 6= n and n 6< m by the trichotomy. m = n =⇒ m+p = n+pwhich contradicts m+ p < n+ p.

n < m then n+ p < m+ p which again contradicts m+ p < n+ p.(ii) Induction step: m · p+ = m · p+m < n · p+m < n · p+ n = n · p+

Corollary. For all m,n, p ∈ ω,m+ p = n+ p =⇒ m = n. If p 6= 0,m · p = n · p =⇒ m = n

Proof by contradiction.

Theorem. Well-ordering property for ω. Let A ⊂ ω be non-empty, then A has a least element, that is, there is a ∈ A suchthat ∀b(b ∈ A =⇒ b ≤ a).

ω is well-ordered. Any set with such an order is called ”well-ordered”.

Proof. Assume for a contradiction that A does not have a least element.Let B = {n ∈ ω | ∀m(m < n =⇒ m /∈ A)}. We show that B is inductive.

• 0 ∈ B because ¬∃m(m < 0)

• Assume n ∈ B. Let m < n+ either m < n or m = n. If m = n and n ∈ A then n ∈ A ∩ B is a least element of A,contradicting our assumption. Therefore n /∈ A and m /∈ A so n+ ∈ B.

Thus B = ω and A = ∅.

Strong Induction Principle Let A ⊆ ω and assume that:

∀a ∈ ω[(∀b(b < a&b ∈ A)) =⇒ a ∈ A]

Then A = ω. ”Strong” means stronger requirements: ∀b < aYou do not need 0 because implicitly this is encoded in ¬∃m(m < 0)

Proof. Let A be as in the statement. Then ω−A cannot have a least element. If b ∈ ω−A is the least element, (∀c < b)c /∈ ω−Aso (∀c < b)c ∈ A then b ∈ A by hypothesis.

Dealing with negativity: the integers

Goal: build Z A natural way to construct Z would be to consider for example ω×{0, 1} where +n is represented by the pair〈n, 0〉 and −n by 〈n, 1〉. (we would need to quotient out 〈0, 0〉, 〈0, 1〉.)

Note that with this representation, defining addition would take a little bit of work.We are going to do differently by a construction that generalizes naturally to other situation (in particular to construct

Q.) We are going to represent an integer a by the set of all pairs 〈m,n〉 ∈ ω2 such that m− n = a.

Definition. Define the relation ∼ on ω2 by 〈m,n〉 ∼ 〈p, q〉 if m+ q = n+ p.

Proposition. The relation ∼ is an equivalence relation on ω2.

Proof. transitivity: assume 〈m,n〉 ∼ 〈p, q〉 and 〈p, q〉 ∼ 〈r, s〉. m + q = n + p and p + s = r + q. Similarly we getm+ q + p+ s = n+ p+ q + r which gives 〈m,n〉 ∼ 〈r, s〉.

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Definition. Let Z, the set of integers, be the quotient ω2/ ∼.

Notation: the class of 〈m,n〉 is denoted [〈m,n〉]∼ or [〈m,n〉]

Example. [〈0, 0〉] = {〈0, 0〉, 〈1, 1〉, . . . } = 0Z(6= 0)

We now want to define the operations on Z. Addition: we want to define +Z so that [〈m,n〉] +Z [〈p, q〉] = [〈m+ p, n+ q〉].For this to make sense, we need to check compatibility with ∼.

Lemma. If 〈m,n〉 ∼ 〈m′, n′〉 and 〈p, q〉 ∼ 〈p′, q′〉 then 〈m+ p, n+ q〉 ∼ 〈m′ + p′, n′ + q′〉.

Consider F : ω2 × ω2 → ω2, (〈m,n〉, 〈p, q〉) 7→ 〈m + p, n + q〉. By the lemma, F is compatible with ∼, hence there is aunique function +Z: ω2/ ∼ ×ω2/ ∼→ ω2/ ∼ such that [〈m,n〉] +Z [〈p, q〉] = [F (〈m,n〉, 〈p, q〉)].

Example. 1Z +Z (−2)Z = [〈1, 0〉] +Z [〈0, 2〉] = [〈1 + 0, 0 + 2〉] = [〈1, 2〉] = [〈0, 1〉] = (−1)Z.

Theorem. (Z,+Z, 0Z) is an abelian group:

• the opration +Z is associative, commutative

• 0Z is an identity (neutral) element: 0Z +Z a = a+Z 0Z = a for all a ∈ Z

• every a ∈ Z has an inverse for +Z: there is b ∈ Z such that a+Z b = b+Z a = 0Z.

associativity: definition of +Z twice and associativity of + on ωThe inverse of [〈m,n〉] is [〈n,m〉]The additive inverse of a will be denoted −a. We write a− b to mean a+Z (−b).Multiplication: we want to define +Z so that [〈m,n〉] ·Z [〈p, q〉] = [〈mp+ nq,mq+ np〉]. We need to check that this makes

sense:

Lemma. If 〈m,n〉 ∼ 〈m′, n′〉 and 〈p, q〉 ∼ 〈p′, q′〉 then 〈mp+ nq,mq + np〉 ∼ 〈m′p′ + n′q′,m′q′ + n′p′〉

This allows us to define ·Z as the unique operation on Z× Z satisfying [〈m,n〉] ·Z [〈p, q〉] = [〈mp+ nq,mq + np〉].

Theorem. (Z,+Z, ·Z, 0Z, 1Z) is an integral domain.

• (Z,+Z) is an abelian group

• ·Z is associative, commutative, distributive over +Z has a neutral (identity) element 1Z 6= 0Z

• tehre are no zero divisors: if a ·Z b = 0Z then a = 0Z or b = 0Z.

Proof. 1Z is an identity: [〈1, 0〉] ·Z [〈m,n〉] = [〈1 ·m+ 0 · n, 1 · n+ 0 ·m〉] = [〈m,n〉].No zero divisors: assume a ·Z b = 0Z, a = [〈m,n〉], b = [〈p, q〉] so mp + nq = mq + np. Assume m > n, p > q. Write

m = n+m′,mp+ nq = mq + np ⇐⇒ n(p+ q) +m′p = n(p+ q) +m′q ⇐⇒ m′p = m′q ⇐⇒ p = q as m′ 6= 0. Similarly ifn > m. So if m 6= n then p = q hence [〈p, q〉] = 0Z.

Ordering on ZWe define a relation <Z by [〈m,n〉] <Z [〈p, q〉] if m+ q < p+ n. For this to make sense, we need to check compatibility with∼, that is, if 〈m,n〉 ∼ 〈m′, n′〉 and 〈p, q〉 ∼ 〈p′, q′〉 then m+ q < p+ n =⇒ m′ + q′ < p′ + n′

Proposition. The relation <Z is a linear ordering on Z, that is:

• <Z is transitive

• for all a, b ∈ Z exactly one of the following holds:

a <Z b, a = b, b <Z a

Proposition. (a) for all a, b, c ∈ Z, a <Z b ⇐⇒ a+ c <Z b+ c

(b) if 0Z <Z c then a <Z b ⇐⇒ a ·Z c <Z b ·Z c

ω as a “subset” of Z We can define a map E : ω → Z, n 7→ [〈n, 0〉].

Proposition. E is one-to-one and for all m,n ∈ ω

• E(m+ n) = E(m) +Z E(n)

• E(m · n) = E(m) ·Z E(n)

• m < n ⇐⇒ E(m) <Z E(n)

We can identify ω with its image.

15

The rational numbers

The rational ab for a, b ∈ Z will be encoded by the pair 〈a, b〉, or rather by its class under some equivalence relation. Let

Z′ = Z− {0}

Definition. Define the relation ∼ on Z× Z′ by 〈a, b〉 ∼ 〈c, d〉 if a ·Z d = c ·Z b.

Proposition. The relation ∼ is an equivalence relation on Z× Z′.

Definition. The set Q of rationals is the quotient Z× Z′/ ∼

Example. 0Q = [〈0Z, 1Z〉]∼ = [〈0Z, 2Z〉]∼, ( 12 )Q = [〈1Z, 2Z〉]∼.

In the rest of this section, we drop the index ∼ and (most) of the index Z. We define +Q as the binary operation on Qthat satisfies [〈a, b〉] +Q [〈c, d〉] = [〈ad+ cb, bd〉] (note bd 6= 0 because we already know that Z is an integral domain.) We needto check that this makes sense:

Lemma. If 〈a, b〉 ∼ 〈a′, b′〉 and 〈c, d〉 ∼ 〈c′, d′〉 then 〈ad+ cb, bd〉 ∼ 〈a′d′ + c′b′, b′d′〉.

Example. 2Q +Q 3Q = [〈2, 1〉] +Q [〈3, 1〉] = [〈2 · 1 + 1 · 3, 1 · 1〉] = [〈5, 1〉] = 5Q.(1/2)Q +Q (1/2)Q = 1Q.

We define ·Q as the binary operation on Q that satisfies: [〈a, b〉] ·Q [〈c, d〉] = [〈ac, bd〉]. This is allowed as: if 〈a, b〉 ∼ 〈a′, b′〉and 〈c, d〉 ∼ 〈c′, d′〉 then 〈ac, bd〉 ∼ 〈a′c′, b′d′〉.

Theorem. (Q,+Q, ·Q, 0Q, 1Q) is a field, that is,

• it is an integral domain

• for any a ∈ Q if a 6= 0Q then there is b ∈ Q, a ·Q b = 1Q.

Proof. We need to check

• +Q is associative, commutative, 0Q + a = a for all a ∈ Q

• every a ∈ Q has an additive inverse. The additive inverse of [〈p, q〉] is [〈−p, q〉].

[〈p, q〉] +Q [〈−p, q〉] = [〈p · q + (−p) · q, q · q〉] = [〈0, q2〉] = 0Q

• ·Q is associative, commutative, distributive over +Q.

• 1Q ·Q a = a for all a ∈ Q

• every non-zero rational has a multiplicative inverse. The multiplicative inverse of [〈p, q〉] is [〈q, p〉] (assuming p 6= 0 aswe only consider nonzero rational).

[〈p, q〉] ·Q [〈q, p〉] = [〈pq, qp〉] = 1Q

Note: every element of Q can be written as [〈p, q〉] with q >Z 0 as if q <Z 0 then −qZ >Z 0 and [〈p, q〉] = [〈−p,−q〉].We define <Q as a binary relation on Q by [〈p, q〉] <Q [〈r, s〉] if ps < qr where q > 0 and s > 0.This is well-defined by the following lemma:

Lemma. If 〈p, q〉 ∼ 〈p′, q′〉 and 〈s, t〉 ∼ 〈s′, t′〉 and q, q′, s, s′ are positive then ps < qt ⇐⇒ p′s′ < q′t′.

Example. (1/2)Q <Q 1Q because 1 · 1 < 2 · 1 in Z.

Proposition. The relation <Q is a linear ordering on Q and satisfies

• r <Q s ⇐⇒ r + t <Q s+ t for all t ∈ Q

• r <Q s ⇐⇒ r · t <Q s · t for all r, s ∈ Q, 0 <Q t

Thus, (Q,+Q, ·Q, 0Q, 1Q, <Q) is an ordered field.

16

Note that Q does not contain Z however we have a natural embedding (morphism of commutative rings) E1 : Z→ Q, n 7→[〈n, 1〉] which is one-to-one and for all m,n ∈ Z we have

E1(m+Z n) = E1(m) +Q E1(n)

E1(m ·Z n) = E1(m) ·Q E1(n)

m <Z n ⇐⇒ E1(m) <Q E1(n)

One can identify Z with its image in Q under E. Now, 1 could mean 3 things (i.e. name 3 different sets)

1 = {∅} ∈ ω1Z = [〈1, 0〉]∼ω = {{{1}, {0, 1}}, {{2}, {1, 2}}, · · · } = {{{{∅}}, {∅, {∅}}} , · · · }1Q = [〈1, 1〉]∼Z = {〈1Z, 1Z〉, 〈2Z, 2Z〉, 〈(−1)Z, (−1)Z〉, · · · }

In the same way, 1 + 1 could mean 3 things:

1 + 1 ∈ ω1Z +Z 1Z

1Q +Q 1Q

2− 3 could mean two things:2Z − 3Z, 2Q − 3Q

1/2 could mean only one thing:1Q ÷ 2Q

In practice, this ambiguity does not cause problems: because of the embeddings E and E1, we can do computations in anyof ω,Z or Q and obtain the same answers.

The real numbers

The jump from Q to R is in some sense wider than the cases from ω to Z and Z to Q: there is no way to encode elements ofR by tuples of elements of Q.

We could use decimal expansions to construct R from (possibly infinite) sequences of {0, 1, · · · , q}. careful: 0.12999 · · · =0.13000 · · · .

Another construction is as a quotient of converging sequences of elements of Q. RQ

Namely, R = {convergent sequences of elements of Q} /equivalence relation of having the same limit. One needs to definethe set on top and the relation down without using R. This can be done using the notion of Cauchy sequence.

This construction generalizes to the construction of the completion of a metric space.

Definition. A Dedekind cut is a subset x of Q such that:

(a) ∅ 6= x 6= Q

(b) x is “closed downward”, i.e.,q ∈ x&r < q =⇒ r ∈ x

(c) x has no largest member.

R ⊂PQ. There is no equivalence relation here; a real is a cut.

Ordering For x, y ∈ R, definex <R y ⇐⇒ x ⊂ y.

Theorem. The relation <R is a linear ordering on R

Proof. Transitivity follow from transitivity of (. We have to show that for x, y ∈ R, exactly one of

x <R y, x = y, y <R x

holds. From the definition it is clear that no two of them can hold. To see that at least one holds, take x, y ∈ R, assumex 6⊆ y. Then there is r ∈ x − y ∩ Q. Let t ∈ y. If t < r then t ∈ x as x is downward closed. If t ≥ r then r ∈ y as y isdownward closed, contradiction. Therefore y ( x.

17

Definition. Let A ⊂ R. An upper bound for A is an element a ∈ R such. that for all x ∈ A, x ≤R a. A is bounded if it hasan upper bound. A least upper bound for A is an upper bound for A which is less <R than any other upper bound.

{r ∈ Q | r2 < 2

}has the least upper bound

√2 in R but has no least upper bound in Q.

Theorem. Let ∅ 6= A ⊂ R be bounded, then A has a least upper bound.

Proof. Assume A is bounded and z ∈ R is an upper bound. Let b =⋃A and we show that b is the least upper bound for A.

• b ∈ R:

– b 6= ∅ as A 6= ∅; b 6= Q as b ⊆ z and z ( Q– b is downward closed: let r ∈ b and s < r, then ∃a ∈ A such that r ∈ a then s ∈ a as a is downward closed, sos ∈ b.

– b has no greatest element: let r ∈ b there is a ∈ A such that r ∈ a. Then there is s ∈ a such that r < s and s ∈ bso r is not a greatest element of b.

• b is an upper bound: If a ∈ A then a ⊆ b so a ≤R b.

• b is the least upper bound: Let c ∈ R be an upper bound for A then for all a ∈ A, a ⊆ c hence⋃A ⊆ c that is b ≤R c.

Addition For reals x and y, define:x+R y = {q + r | q ∈ x&r ∈ y}

We let0R = {r | r < 0}

Proposition. For x, y ∈ R, we have

• x+R y ∈ R

• (x+R y) +R z = x+R (y +R z)

• x+R y = y +R x

• 0R +R x = x

Additive inverse For x ∈ R, let−x = {r ∈ Q | (∃s > r)− s /∈ x} .

Theorem. For every x ∈ R

(a) −x ∈ R,

(b) x+R (−x) = 0R.

Proof. Verify −x is a cut.

• −x 6= ∅: there is t ∈ Q− x then −t− 1 ∈ −x (take r = −t); −x 6= Q: take t ∈ x, then −t /∈ x: if r > −t then −r < tso −r ∈ x

• x is downward closed: if s ∈ −x and s′ < s as s ∈ −x there is r > s,−r /∈ x then r > s′,−r /∈ x so s′ ∈ −x.

• x has no largest element: if s ∈ −x there is r > s,−r /∈ x. Let s′ ∈ Q, s < s′ < r then s′ ∈ −x (take the same r)

x+R (−x) = {q + s | q ∈ x&(∃r > s)− r /∈ x}

If q+ s ∈ x+R (−x) with r as above, then s < r and as −r /∈ x, q < −r, so q+ s < 0 hence q+ s ∈ 0R. Conversely, let p ∈ 0Rso p < 0 and −p2 > 0. By exercise 19 ∃q ∈ x such that q− p

2 /∈ x. Let r = p2 − q, s = p− q so −r /∈ x and s < r hence s ∈ −x,

and p = q + s ∈ x+R (−x). Thus, 0R ⊂ x+R (−x) and 0R = x+R (−x).

Definition. For x ∈ R, define |x| = x ∪ −x.

so |x| ≥ 0R for all x ∈ R

18

Multiplication

Definition. If x, y ∈ R with x, y ≥R 0

x ·R y = 0R ∪ {rs | 0 ≤ r ∈ x&0 ≤ s ∈ y} .

If x, y <R 0, thenx ·R y = |x| ·R |y|

If one of the x, y is negative and one is nonnegative, then

x ·R y = −(|x| ·R |y|)

Theorem. (R,+R, ·R, 0R, 1R, <R) is an ordered field which is moreover complete.

We have a natural map E2 : Q → R, p 7→ {r ∈ Q | r < p} which is one-to-one and respects addition, multiplication andthe order. It is a morphism of ordered fields.

Measuring sizes of sets: Cardinals

Idea: If two sets have the same size, then there is a one-to-one correspondence (bijection) between the two sets.

Definition. Two sets A and B are equinumerous (write A ≈ B) if ∃ a bijection from A onto B.

Note: This is a symmetric condition: if A→ B is a bijection, f−1 : B → A is a bijection.

Example. • ω is equinumerous with the set E of even natural numbers. A bijection f : ω → E,n 7→ 2n.

• ω is equinumerous with Z, g : Z→ ω, n 7→

0 n = 02n n > 0

−2n− 1 n < 0

• ω is equinumerous with Q, 0, 1,−1, 2,−2, 12 ,−

12 , 3,−3, 1

3 ,−13 ,

23 ,−

23 ,

32 ,−

32 · · ·

• (−1, 1) is equinumerous with R, x 7→ tan(π2x) is a bijection.

Proposition. For any A,B,C

(a) A ≈ A

(b) If A ≈ B, then B ≈ A

(c) If A ≈ B and B ≈ C then A ≈ C

≈ is not an equivalence relation because there is no set of all sets.

Proof. (a) Consider id: A→ A a bijection

(b) said

(c) If f : A→ B and g : B → C are two bijections, then g ◦ f : A→ C is a bijection.

Theorem (Cantor). a) ω is not equinumerous with R

b) A is not equinumerous with P(A) for any set A.

Proof. diagonalisation trickmidterm 1

Finite Sets

Definition. A set A is finite if it is equinumerous to a natural number. Otherwise, we say A is infinite.

19

Pigeonhole Principle No natural number is equinumerous with a proper subset of itself.

Proof. DefineT = {n ∈ ω | any one-to-one function n→ n is onto n}

We show T is inductive.

• 0 ∈ T : there is one function from 0→ 0, namely ∅ and it is onto 0.

• Assume n ∈ T , want to show n+ ∈ T . Let f : n+ → n+ be one-to-one. Case 1 fJnK ⊆ n. As n ∈ T, f �n: n → n isonto n. As f is one-to-one, f(n) cannot be in n (n+ = n ∪ {n}), hence f(n) = n. Therefore f is onto n+. Case 2

n ∈ ran(f �n), let a ∈ n such that f(a) = n. Let k = f(n), define u :

n+ → n+

x 7→ x x 6= n&x 6= kk 7→ nn 7→ k

. Then u is a bijection

and u ◦ f is one-to-one and u ◦ fJnK ⊆ n. By Case 1, u ◦ f is a bijection n+ → n+. Then so is f = u−1 ◦ u ◦ f as u is abijection.

Therefore, T = ω. It follows that if f : n→ A is a bijection with A ⊆ n, then f is onto n and A = n.

Corollary. If m < n are natural numbers, then m 6≈ n

Proof. Follows from the pigeonhole principle that m 6⊆ n.

Corollary. • If A is a finite set, then A is not equinumerous with a proper subset of itself.

• ω is an infinite set.

Proof. Let A be finite, then for some n ∈ ω,∃f : A → n a bijection. Assume we have bijection g : A → B whereB ⊆ A. Let C = fJBK ⊆ n, then f ◦ g ◦ f−1 : n → C is a bijection from n to C. By pigeonhole principle, C = n, hence

A = f−1JnK = f−1JCK = B.

A n

B C

g

f

f

It also follows that if A is finite, then it is equinumerous with a unique natural number. This number is called thecardinality (or cardinal number) of A, written card(A).

Example. card({1, 2, 3}) = 3, card(n) = n, card(∅) = 0.

Infinite Cardinalities

Let ℵ0 = cardω. We have seen ℵ0 = cardZ = cardQ. We can define operations on cardinal numbers:

Definition. Let κ and λ be any cardinal numbers.

(a) κ+ λ = card(K ∪ L) where K,L are any sets such that card(K) = κ, card(L) = λ and K ∩ L = ∅.

(b) κ · λ = card(K × L) where K,L are any sets such that card(K) = κ, card(L) = λ.

(c) κλ = card(LK) where K,L are any sets such that card(K) = κ, card(L) = λ,LK ⊂ P(K × L) are functions fromL→ K.

We have to check these definitions make sense: well-defined.

Proposition. Assume K ≈ K ′, L ≈ L′

(a) If K ∩ L = K ′ ∩ L′ = ∅, then K ∪ L ≈ K ′ ∪ L′.

(b) K × L ≈ K ′ × L′.

(c) LK ≈L′ K ′.

20

Proof. Since K ≈ K ′,∃f : K → K ′ one-to-one; similarly ∃g : L → L′ one-to-one. Define h : K ∪ L → K ′ ∪ L′, x 7→{f(x) x ∈ Kg(x) x ∈ L . Since K ∩ L = ∅, h is well-defined as a function; since K ′ ∩ L′ = ∅, h is one-to-one.

Define h : K × L→ K ′ × L′, 〈x, y〉 7→ 〈f(x), g(y)〉 for x ∈ K and y ∈ L.

Define H :L K →L′ K ′, x 7→ f ◦ x ◦ g−1.L′ K ′

L K

g−1

x

f H is one-to-one: Suppose ∃x 6= x′ ∈L K then ∃t ∈ L with

x(t) 6= x′(t). Then H(x)(g(t)) = f(x(t)) 6= f(x′(t)) = H(x′)(g(t)) since f is one-to-one. Hence H(x) 6= H(x′). Define theinverse L′K ′ →L K, y 7→ f−1 ◦ y ◦ g.

Example. n+ ℵ0 = card({n, n− 1, . . . , 1} ∪ ω) = card(ω) = ℵ0,ℵ0 + ℵ0 = ℵ0, n · ℵ0 = ℵ0 think the zigzag for rationals. Bydefinition of exponentiation,

2cardA = card(A2).

We have shown that A2 ≈PA,A ⊃ X 7→ χX , f 7→ f−1J{1}K, and hence

2cardA = card(A2) = cardPA.

In particular, cardPω = 2ℵ0 . (The term “power set” is rooted in the fact that cardPA equals 2 raised to the power cardA.)

Theorem. For any κ, λ, µ

1. κ+ λ = λ+ κ and κ · λ = λ · κ

2. κ+ (λ+ µ) = (κ+ λ) + µ and κ · (λ · µ) = (κ · λ) · µ

3. κ · (λ+ µ) = κ · λ+ κ · µ

4. κλ+µ = κλ · κµ

5. (κ · λ)µ = κµ · λµ

6. (κλ)µ = κλ·µ

Proof. Choose, for convenience sake, pairwise disjoint sets K,L,M with cardK = κ, cardL = λ, cardM = µ. exercise formost, but we show ∃H :M (LK)→(L×M) K, f 7→ (〈l,m〉 7→ f(m)(l)) and S :(L×M) K →M (LK), g 7→ (l 7→ (m 7→ g(〈l,m〉)))bijections.

Ordering Cardinal Numbers

Definition. A set A is dominated by a set B (written A 4 B) iff there is a one-to-one function from A into B.

cardA ≤ cardB ⇐⇒ A 4 B

Check it is well-defined by composing three one-to-one maps.

Example. 0 ≤ κ for all κ. For any finite cardinal n we have n < ℵ0. For any two finite cardinals m and n, we havem∈n =⇒ m ⊆ n =⇒ m ≤ n. κ < 2κ for any cardinal κ. In particular, there is no largest cardinal number.

Definition. A set A is dominated by a set B (written A 4 B) iff there is a one-to-one function from A into B.

Theorem (Schroder-Bernstein Theorem). a) If K 4 L and L 4 K, then K ≈ L

b) if κ ≤ λ and λ ≤ κ then κ = λ

Proof. Let f : K → L and g : L → K be one-to-one. We want to construct a bijection h : K → L. The idea is to use f onsome part of K and g on the rest.

Define C0 = K − ran(g) and define by induction.

Cn+1 = gJfJCnKK

Let C∗ =⋃n∈ω Cn. We define h : K → L by h(x) =

{f(x) x ∈ C∗g−1(x) x ∈ K − C∗

. Note if x ∈ K − C∗ then x ∈ ran(g) so g−1

makes sense.

• h is one-to-one: As f and g−1 are one-to-one, h �C∗ and h �K−C∗ are one-to-one. If x ∈ C∗, y ∈ K − C∗, h(x) = h(y)then we have f(x) = g−1(y) so y = g(f(x)). If x ∈ Cn then y ∈ Cn+1 contradiction to y /∈ C∗.

21

• h is onto L: Let y ∈ L. If g(y) /∈ C∗, then h(g(y)) = g−1(g(y)) = y. If g(y) ∈ C∗, note g(y) /∈ C0, so there isn ∈ ω such that g(y) ∈ Cn+1. Hence there is x ∈ Cn such that g(y) = g(f(x)) so y = f(x) as g is one-to-one. Asx ∈ Cn, h(x) = f(x) = y.

Example. To prove A ≈ B, it is often easier to prove A 4 B and B 4 A

• ω ≈ ω × ω, n 7→ (n, 0), (a, b) 7→ 2a3b. Decomposition into prime numbers is unique.

• R ≈P(ω), f : R→P(Q), a 7→ a as Dedekind cut. g :ω 2→ R, η 7→ 0.η(0)η(1)η(2) · · · ,ω 2 is the set of functions fromω to 2. We have to justify decimal expansion but we omit it. We may not be able to avoid analysis at this point. Weneed P(ω) ≈ω 2 4 R.

Proposition. Let κ, λ, µ be cardinal numbers. Then:

a) κ ≤ λ =⇒ κ+ µ ≤ λ+ µ

b) κ ≤ λ =⇒ κ · µ ≤ λ · µ

c) κ ≤ λ =⇒ κµ ≤ λµ

d) κ ≤ λ =⇒ µκ ≤ µλ (if (κ, µ) 6= (0, 0))

To prove point c) of the theorem, we need some form of the axiom of choice (AOC). We will use Zorn’s Lemma. Equivalencebetween Zorn’s Lemma and AOC is proven through ordinals and transfinite induction (induction on infinite set)

Lemma (Zorn’s Lemma). Let A be a nonempty set such that for every chain B ⊂ A , we have⋃

B ∈ A . (B is called achain iff for any C,D ∈ B, either C ⊆ D or D ⊆ C.) Then A contains an element M (a “maximal” element) such that Mis not a subset of any other set in A .

Proof. By Zorn’s Lemma. Let K,L be two sets. We want to show that either K 4 L or L 4 K. We consider A to be theset of functions f : X → L where X ⊆ K and f is one-to-one.

A is nonempty as ∅ ∈ A .Let B ⊆ A be a chain, then for any f, g ∈ B, either f �dom(g)= g or g �dom(f)= f .Then

⋃B is a function and is one-to-one. Its domain is the union of the domains of functions in B. By Zorn’s Lemma,

there is h ∈ A which is maximal. If dom(h) = K then h : K → L is one-to-one and K 4 L. Otherwise, let a ∈ K − dom(h).If ran(h) = L, then h−1 : L → K is one-to-one and L 4 K. Otherwise, let b ∈ L − ran(h), h ∪ {〈a, b〉} ∈ A , contradictingmaximality of h.

Countable Sets

Definition. A set A is countable if A 4 ω, equivalently cardA ≤ ℵ0.

Proposition. A countable union of countable sets is countable.If A is countable, the set of finite sequences of elements of A is countable.

Idea:⋃n∈ω A

n is a countable union of countable sets.Consequences: the set of (finite) words, sentences in a finite alphabet is countable. Think about it: for π, e ∈ R you can

describe in natural language with a sentence. However, the set of real numbers that can be defined by an English sentence(or book!) is countable. “Most” real numbers cannot be described. The set of all proofs is also countable. . . There is alsothe paradox if you try to use minimum argument. There is no good way to define what it means to be defined. Some moviesome ET has continuous alphabet.

There are countably many algebraic numbers, i.e., roots of nonzero polynomials in Z[x] or Q[x]. “Most” real numbers arenot algebraic.

Arithmetic of Infinite Cardinals

Absorption Law of Cardinal Arithmetic Let κ and λ be cardinal numbers, the larger of which is infinite and thesmaller of which is nonzero. Then

κ+ λ = κ · λ = max(κ, λ)

Observation: ℵ0 + ℵ0 = ℵ0,ℵ0 × ℵ0 = ℵ0.

Lemma. For κ infinite cardinal, κ+ κ = 2× κ = κ

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Proof. Let B has cardB = κ. We want a bijection B → B tB = B × 2. Define

H = {f | ∃A ⊆ B, bijection f : A↔ A× 2}

Claim 1: H 6= ∅. Let A ⊆ B be countably infinite, then cardA+ cardA = cardA as ℵ0 + ℵ0 = ℵ0 so there is a bijectionf : A→ A× 2. This f ∈H .

Claim 2: If C ⊆H is a chain then⋃C ∈H .

• union of a chain of functions is a function.

• union of a chain of 1-1 functions is 1-1. If 〈a1, b〉 ∈ f1, 〈a2, b〉 ∈ f2 then WLOG f1 ⊂ f2 and 〈a1, b〉 ∈ f2, too. Thena1 = a2 because f2 is 1-1.

• dom⋃C =

⋃{domf | f ∈ C}, ran

⋃C =

⋃{ranf | f ∈ C} =

⋃{domf × 2 | f ∈ C} =

⋃{domf | f ∈ C} × 2 =

dom⋃C × 2 requires some arguments.

By Zorn’s Lemma there is a maximal f0 ∈H . Let A0 = domf0 then ranf0 = A0 × 2.Now we know 2 · λ = λ.Claim: B \ A0 is finite. Otherwise, let D be a countably infinite subset of B \ A0. Let g0 be bijection D → D × 2. We

know g0 exists because ℵ0 = ℵ0 × 2. Let h = f0 ∪ g0,domh = A0 ∪D and h ∈H .Since B \A0 is finite, say card(B \A0) = n. Let λ = cardA0, then κ = λ+ n ≤ 2 · λ = λ ≤ κ so λ = κ.

Lemma. For any infinite cardinal κ, we have κ · κ = κ

Proof. Let cardB = κ. It suffices to show that B ×B ≈ B. Define

H = {f | ∃A ⊆ B, f : A↔ A×A}

. . . , now we know λ · λ = λ.Let λ = cardA0. Suppose λ ≤ card(B \A0). Then ∃S ⊆ B \A0 with cardD = λ. card((A0×D)∪ (D×A0)∪ (D×D)) =

3 ·λ ≤ λ ·λ = λ. Hence there is g0 : D ↔ (A0×D)∪ (D×A0)∪ (D×D). Let h = f0∪ g0 ∈H , contradicting the maximalityof f0. Hence λ > card(B \A0).

Finally, κ = cardA0 + card(B \A0) ≤ 2 · λ = λ ≤ κ so λ = κ.

Proof. By the symmetry we may suppose that λ ≤ κ. Then

κ ≤ κ+ λ ≤ κ+ κ = κ · 2 = κ

andκ ≤ κ · λ ≤ κ · κ = κ

Hence equality holds throughout.

Recall: for infinite κ, λ, κ + λ = κ · λ = max(κ, λ). Addition and multiplication is not very interesting. However,exponentiation is much more complicated to understand in general.

We know κλ ≤ 2κ·λ as a function is a set of pairs so we have an injection from LK ↪→P(K × L).

Example. κκ ≤ 2κ·κ = 2κ. Also κκ ≥ 2κ by monotonicity so κκ = 2κ. In general, if λ ≥ κ, κλ = 2λ.

Axiom of Choice

Theorem. tfae:

1. For any relation R, there is a function F ⊆ R with domF = domR.

2. The product of non-empty sets is non-empty, i.e., if H is a function with domH = I and H(i) 6= ∅ for all i ∈ I, thenthere is s ∈

∏i∈I H(i), i.e. s is a function with domain I and s(i) ∈ H(i) for all i ∈ I.

3. For any set A, there exists a “choice function” for A, i.e., a function F : PA \ {0} → A such that F (X) ∈ X for eachX.

4. Zorn’s lemma

Proposition. (uses AC) If A is infinite, then ω 4 A

Proof. Take choice function F and build by recursion f :

ω → A0 7→ F (A)

n+ 1 7→ F (A \ fJnK)A \ fJnK 6= ∅ because A if not finite.

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Ordinals

Idea is to allow you to keep counting. ε0 = ωωω.. .

Well-Orderings

Definition. A well-ordering on a set A is a linear ordering on A with the property that any nonempty subset of A has asmallest element.

Example. <ω is a well-ordering on ω. <Z is not a well-ordering as Z has not smallest element.Define the ordering / on Z by: 0 / 1 / 2 / 3 · · · /−1 /−2 /−3 · · · . This is a well-ordering: if there is positive numbers...if

there is not...This ordering has order type “ω + ω” whatever that means.

Lemma. Let < be a linear ordering on a set A, then this is a well-ordering iff there is no infinite decreasing sequence ofelements of A, i.e., there is no function f : ω → A with f(n+) < f(n) for all n ∈ ω.

Proof. We prove the contrapositive. Assume there is a decreasing function f : ω → A. Let X = ranf then X ⊆ A isnon-empty. If x ∈ X, there is n ∈ ω such that x = f(n) then f(n+) ∈ X is smaller than x so X has no smallest element.

(uses AC) We prove the contrapositive. Let X ⊆ A nonempty with no smallest element. Let F : P(X) \ {∅} → X bea choice function on X. For any x ∈ X, let X<x = {y ∈ X | y < x} 6= ∅ as X has no smallest element. Define by recursion

g :

ω → Af(0) = F (X)

f(n+ 1) = F (X<f(n))so we have f(n+ 1) < f(n) for all n ∈ ω.

Transfinite induction

Definition. Let A be linearly ordered by < and t ∈ A. The initial segment up to t, denoted segt is the set {x ∈ A : x < t}

Example. In ω, segn = n is a very special property

Definition. A subset B of A is <-inductive if for all t ∈ A, segt ⊆ B =⇒ t ∈ B.

Theorem (Transfinite induction principle). Assume that < is a well ordering on A. Then any <-inductive subset of A isequal to A.

Proof. Let B ⊆ A be <-inductive. Assume B 6= A. Then C := A \ B 6= ∅. As < is a well ordering, there is t ∈ C a leastelement in C. Then segt ⊆ B. As B is <-inductive, t ∈ B which is a contradiction.

Notation: <AB = {f : f is a function from segt for some t ∈ A, to B}

Theorem (Transfinite recursion theorem, preliminary form). Assume that < is a well ordering on A and let G :<A B → B.Then there is a unique function F : A→ B such that for all t ∈ A,

F (t) = G(F �segt)

The limitation of this theorem is that we need to know B in advance.

Example. Hereditarily finite sets: ∅∪P∅∪PP∅ · · · . We want to build a function n 7→P◦· · ·◦P∅ where G : f 7→P(ranf).With the axioms we have so far, we cannot build a function without specifying its domain and range a priori. If we canspecify the domain we have already built the set. Here is a paradox.

Do we want to build this set? Does it lead to Russell’s paradox? Nah. . . This set is countable so it is not too big. It isreasonable for us to build it.

Idea: G(f) = P(ranf) is not a function, as we have not specified its domain. It is a FORMULA.

We are going to think of G as a function-class, i.e., a rule which to each set x assigns some set y and in fact this rule canbe represented by a first order formula φ(f, y) = (y = P(ranf)).

Axiom (Replacement axiomS). “The image of a set by a function class is a set.” For any formula φ(x, y) not containing theletter B, the following is an axiom

∀A [(∀x ∈ A)∀y1∀y2 (φ(x, y1)&φ(x, y2) =⇒ y1 = y2) =⇒ ∃B∀y(y ∈ B ⇐⇒ ∃x ∈ Aφ(x, y))]

Left says φ is a function-class; Right says B is the image.

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Theorem (Transfinite recursion, general case, it is a theorem for each function class, so there are theoremS). For anyfunction-class φ(x, y), assume < is a well ordering on A and ∀x there is exactly one y such that φ(x, y), then there is a uniquefunction F with domF = A such that

φ(F �segt, F (t)),∀t ∈ A

Example. Take (A,<) = (ω,<ω), φ(f, y) =

(y =

{P(ranf)) f is a function∅ otherwise

)F (0) = P∅, F (1) = PP∅, F (2) =

P(P∅ ∪PP∅) . . .Now ∃ranF =

⋃n∈ω(◦nP)∅

sketch of proof. 1. Definition: A function v is constructed up to t for t ∈ A if domv = {x : x ≤ t} and for all x ∈domv, φ(v �segt, v(x)).

2. ∀t ∈ A there is at most one function constructed up to t. Proof by contradiction.

3. Let K = {v : ∃t ∈ A, v is constructed up to t}. We need a replacement axiom to define K.

4. F =⋃K is a function and domF = A.

Epsilon Images Application of transfinite induction: construct ordinal numbers. Assume < is a well ordering on A andtake for γ(x, y) = (y = ranx). The transfinite recursion theorem gives a unique function E with domain A such that ∀t ∈ A

E(t) = ran(E � segt) = EJsegtK = {E(x) | x < t}

Let α = ranE; we will call α the ∈-image of the well-ordered structure 〈A,<〉. (Later α will be called an ordinal number)

Example. Let A = {r, s, t}, r < s < t, E(r) = ∅ = 0, E(s) = {∅} = 1, E(t) = {∅, {∅}} = 2, and the ∈-image of 〈A,<〉 is 3.ω,<ω has epsilon image ω; the even natural numbers with restriction of<ω has epsilon image ω; A = {0, 1, 2, 3, . . . ,−1,−2,−3, . . . }

has epsilon image ω + ω.In fact, α 4 A,∈α= {〈x, y〉 ∈ α× α | x ∈ y} provided a well ordering on α.

Theorem. Let < be a well ordering on A and let E,α be as above.

1. E(t) /∈ E(t) for all t ∈ A

2. E maps A one-to-one onto α

3. For any s and t in A,s < t ⇐⇒ E(s) ∈ E(t)

4. α is a transitive set

Proof. If it is false, take the least t ∈ A such that E(t) ∈ E(t), then E(t) = {E(x) : x < t} so ∃x < t with E(x) = E(t).Then E(x) ∈ E(t) contradicts the minimality of t.

WTS E is injective. (onto is by definition α = ran(E)). Let s < t in A, then E(s) ∈ E(t) so E(t) 6= E(s) by a).If s = t, we cannot have E(s) ∈ E(t) by a). If s < t, then E(s) ∈ E(t). If s > t, then E(t) ∈ E(s) so if E(s) ∈ E(t) we

arrive at the contradiction E(s) ∈ E(s).α is transitive: If u ∈ v ∈ α then ∃t such that v = E(t) = {E(x) : x < t} so ∃s, u = E(s) so u ∈ α.E is an isomorphism from 〈A,<〉 to 〈α,∈〉. In particular, ∈ is a well-ordering on α.

Definition. An ordinal number is a set α which is transitive and on which ∈ is a well-ordering.

Theorem. A set α is an ordinal number iff it is the epsilon image of some well-ordered set.

Proof. We have seen that the epsilon image of a well-ordered set is an ordinal number.Conversely, let α be an ordinal number, then ∈ is a well-ordering and we show that it is its own epsilon image.Proof by transfinite induction that E(t) = t for all t ∈ α. Assume E(s) = s for all s < t i.e. s ∈ t. then E(t) = {E(x) :

x < t} = {x : x < t} = t by induction hypothesis.

If 〈A,<〉 is a well-ordered set, the ordinal number of 〈A,<〉 is its epsilon image.

Proposition. If α, β are ordinals such that 〈α,∈〉 and 〈β,∈〉 are isomorphic, then α = β.

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Proof. Let f : α→ β a bijection such that s ∈ t ⇐⇒ f(s) ∈ f(t),∀s, t ∈ α. Proof by transfinite induction that f(t) = t forall t ∈ α. Assume f(s) = s for all s ∈ t, t ∈ α. Then f(t) = {u : u ∈ f(t)}. If u ∈ f(t) then u ∈ β then u = f(s) for somes ∈ α (as f is onto β) so f(s) ∈ f(t) implies s ∈ t so f(s) = s ∈ t therefore f(t) ⊆ t.

For the other inclusion, let s ∈ t, then f(s) = s by inductive hypothesis and f(s) ∈ f(t) as f is an isomorphism so s ∈ f(t)and t ⊆ f(t). Therefore t = f(t).

Corollary. Let 〈A,<A〉 and 〈B,<B〉 be two well-ordered sets with ordinal numbers α, β resp. Then 〈A,<A〉 ∼= 〈B,<B〉 iffα = β.

Proof. 〈A,<A〉 ∼= 〈α,∈〉 and 〈B,<B〉 ∼= 〈β,∈〉 so if 〈A,<A〉 ∼= 〈B,<B〉 then 〈α,∈〉 ∼= 〈β,∈〉 hence α = β.Conversely, if α = β, then 〈A,<A〉 ∼= 〈α,∈〉 ∼= 〈β,∈〉 ∼= 〈B,<B〉.

Lemma. An initial segment of an ordinal is an ordinal.

Proof. Let α be an ordinal, t ∈ α and β = segt = {x ∈ α | x ∈ t} = t.

1. t is transitive: let u ∈ v ∈ t, then u ∈ t as ∈ is an ordering

2. t is well ordered by ∈ as it is a subset of α and a subset of a well ordering is a well ordering.

We also have that an element of an ordinal is an ordinal.

The class of all ordinal numbers is a transitive class well ordered by epsilon

Lemma. If α, β are two ordinals, then either α = β or α or one is isomorphic to a segment of the other.

Proof. If α ⊆ β, let t ∈ β \ α be the least element. Then if s ∈ t, we have s ∈ α so t ⊆ α. If u ∈ α then u ∈ β. If u ≥ t, thent ∈ u ∈ α so t ∈ α, contradiction. If t = u then t ∈ α so u ∈ t.

Therefore t = α and α = segt.If α 6⊆ β, let u ∈ α \ β be the least element. If β ⊆ α we show as above β = segu = u. Otherwise, let t ∈ β \ α be the

least element, we show that t = u. Contradiction.

Theorem. Let α, β, γ be ordinals, then

1. Any element of α is an ordinal

2. If α ∈ β ∈ γ then α ∈ γ

3. α /∈ α

4. Exactly one of the following holds:α ∈ β, α = β, β ∈ α

Proof. Seen in the previous proof. γ is transitive. Already saw. “At least one” follows from the previous lemma, “at mostone” by c).

So ∈ defines an “ordering” on the class of ordinals. We will sometimes write, α < β for α ∈ β, α ≤ β for α ∈ β or α = β,equivalently, α ⊆ β.

Theorem. 1. Any non-empty set of ordinals has a least element

2. Any transitive set of ordinals is an ordinal

3. 0 is an ordinal and if α is an ordinal, so is α+

4. If A is a set of ordinals, then⋃A is an ordinal

Proof. Let A be a nonempty set of ordinals. Pick any α ∈ A, let β be the least element of α∩A. Claim β is the least elementof A. Indeed, if we had γ < β for some γ ∈ A, then we would have γ < β < α so γ < α hence γ ∈ α ∩ A. This contradictsminimality of β.

If A is a transitive set of ordinals, it is well-ordered by ∈ by a), so it is an ordinal by definition of ordinals.exercise.As elements of ordinals are ordinals,

⋃A is a set of ordinals. If β ∈ α ∈

⋃A then there is γ ∈ A such that α ∈ γ so β ∈ γ

as γ is transitive. Hence β ∈⋃A so

⋃A is transitive and d) follows from b).

Note:⋃A is the least ordinal α such that a ⊆ α for each a ∈ A. In other words,

⋃A is the supremum (least upper

bound) of A.

Example.⋃ω = ω,

⋃{0, 1, 4} = 4.

Theorem (Burali-Forti). There is no set that contains all ordinals.

Proof. Assume A contains all ordinals. Using a subset axiom, let A′ ⊆ A be the set of all ordinals. Let α =⋃A′. Then α is

an ordinal and we have β ≤ α for all ordinals β. Then taking β = α+ gives a contradiction.

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Transfinite Recursion on all ordinals Want to define F : Ord → Sets which is a function class so F (α) = G(F �α).Let G be a function class (this means G(x) = y is given by a formula γ(x, y) so that ∀x there is exactly one y with γ(x, y)).For any ordinal α, we can define by transfinite recursion a function Fα with domain α such that Fα(β) = G(Fα �β) for allβ < α. And there is a unique such Fα. Now if α < α′, then Fα′ �α= Fα as it satisfies the recursion condition.

For any ordinal β let F (β) = Fα(β) for any α > β. F is a function class (you have to write F (β) = Fα(β)∀α > β in firstorder formula) defined on all ordinals and satisfies F (β) = G(F �β) for all ordinals β.

Thus, we can define a function (really a function classes) by transfinite recursion on all ordinals.

Theorem (AC). Every set A admits a well-ordering.

Proof. Let G : PA \ {∅} → A be a choice function on A. Pick e a set not in A (exists by Russell’s paradox). Define a

function class F on ordinals by transfinite recursion, F (α) =

{e A ⊆ ran(F �α)

G(A \ ran(F �α)) otherwise

• If there is an ordinal α with F (α) = e. Take the least such α (every class of ordinals has a least element, because youcan truncate it to obtain a set of ordinals). Then F �α: α→ A is a bijection.

– F �α is onto A as F (α) = e

– F �α is one-to-one: let β < γ < α, then F (γ) = G(A \ ran(F �γ)) ∈ A \ ran(F �γ) so F (β) 6= F (γ).

Define <A by a <A b ⇐⇒ (F �α)−1(a) < (F �α)−1(b). Then Fα is an isomorphism from 〈α,∈〉 to 〈A,<A〉. Inparticular, <A is a well-ordering.

• For no α do we have F (α) = e. Then F is a function class from ordinals to A which is one-to-one, so F−1 is a functionclass from a subset of A (map things outside of A to anywhere) to ordinals. By the replacement axiom, the image ofthat subset under F−1 is a set. But this set contains all ordinals; contradiction.

In fact, the statement of the theorem is equivalent to the axiom of choice.

Proposition. If every set has a well ordering, then there is choice function on every set.

Proof. Let A be a set and <A a well ordering on A. Define F : PA \ {∅} → A by F (B) =least element of B for allB ⊆ A,B 6= ∅. This is a choice function.

Proposition. Assume all sets have a well ordering, then Zorn’s lemma holds.

Proof. Let A be a set such that for every chain B ⊆ A ,⋃

B ∈ A .

Let<A be a well-ordering on A . Build by transfinite induction F : A → {0, 1}, F (a) =

{1 b <A a&F (b) = 1 =⇒ b ⊆ a0 otherwise

.

Let C = F−1J{1}K then C is a chain. Indeed, if a, b ∈ C , say b <A a by construction.Let c =

⋃C ∈ A by assumption. Claim c is a maximal element. If ∃d ∈ A , c ⊆ d, then for all b ∈ A if F (b) = 1 then

b ⊆ d so F (d) = 1. Thus d ⊂ c so d = c and c is maximal under inclusion.

Cardinals

Definition. A cardinal is an ordinal number α which is not equinumerous with any β < α. The cardinal number of a set Ais the least ordinal equinumerous with A.

Note:

• we saw that any set A is equinumerous to some ordinal, so the definition makes sense.

• the cardinal number of A is a cardinal.

Example. 0, 1, 2, 3 are cardinals. More generally all natural numbers are cardinals. ω is a cardinal, as ω 6≈ n for n ∈ ω.ω+ = ω ∪ {ω} is countable: ω+ ≈ ω so ω+ is not a cardinal. ω + ω is also countable (as the union of two countable sets - wehave seen a bijection ω + ω → Z not a cardinal. ω · ω =

⋃n∈ω nω is again countable as a countable union of countable sets,

not a cardinal.

In fact, we cannot explicitly construct an uncountable ordinal. We let ℵ1 be the smallest uncountable ordinal. Bydefinition, it is a cardinal.

Recall that ℵ0 was defined as the cardinal number of ω, which is ω. (In general, the cardinal number of a cardinal isitself.)

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Definition. Define ℵα for all ordinals α by transfinite recursion. ℵα =the least infinite cardinal different from ℵβ for β < α,i.e., ...ordinal not equinumerous with some ℵβ for β < α.

Note: By construction, ℵα < ℵβ for α < β.

Lemma. For any α, ℵα ≥ α.

Proof. We prove it by transfinite induction. Let α be an ordinal and assume ℵβ ≥ β for all β < α. Assume ℵα < α, thenℵα = β for some β < α. Then β ≤ ℵβ < ℵα, contradiction.

Proposition. Every infinite cardinal has the form ℵα for some cardinal α.

Proof. Let κ be a cardinal, then ℵκ ≥ κ. If = we are done. If ¿, by definition of ℵκ as κ is infinite, there is β < κ withℵβ = κ.

Proposition. If κ, λ are cardinals, then

• κ ≈ λ iff κ = λ

• κ 4 λ iff κ ≤ λ

Proof. If κ 6= λ, say κ > λ, then κ 6≈ λ by definition of κ being a cardinal.If κ ≤ λ, then κ ⊆ λ so κ 4 λ. If κ > λ, then we cannot have κ 4 λ, otherwise since we have λ 4 κ, we would have κ ≈ λ,

contradicting κ being a cardinal.

so the ordering that we defined previously on cardinals coincides with the ordering coming from ordinals.

Successors and limit ordinals There are these sorts of ordinals:

• 0

• successor ordinals: ordinals of the form α+ for some α an ordinal (1,2,3,ω+...)

• limit ordinals: all the other ordinals (ω, ω + ω...)

Proposition. If A is a set of cardinals, then⋃A is a cardinal.

Proof. We already know that⋃A is an ordinal. Let α ∈

⋃A, we have to show α ≈

⋃A. There is κ ∈ A such that α ∈ κ. If

α ≈⋃A, then since α 4 κ 4

⋃A, we would have α ≈ κ. This contradicts κ being a cardinal.

Consequence: If λ is a limit ordinal, then ℵλ =⋃α<λ ℵα. Why? We know that

⋃α<λ ℵα is a cardinal. Furthermore,

since λ is a limit ordinal, we cannot have⋃α<λ ℵα = ℵβ for some β < λ.

Proof. Assume⋃α<λ ℵα ≈ ℵβ for some β < λ. As λ is limit, β+ < λ so ℵβ+ 4

⋃α<λ ℵα but ℵβ+ 64 ℵβ , contradiction.

Now⋃α<λ ℵα is the least ordinal greater than all ℵα, α < λ so a fortiori, the least cardinal such. By definition, it is

ℵλ.

ω + ω is a limit ordinal but not a cardinal.Question: Is there an ordinal κ such that ℵκ = κ. Define by recursion a function f on ω by f(0) = ℵ0, f(n+ 1) = ℵf(0).

Let κ =⋃n<ω f(n). Each f(n) is a cardinal, hence κ is a cardinal. We always have κ ≤ ℵκ. Let α < κ, then there is n < ω

such that α < f(n) so ℵα < ℵf(n) = f(n+ 1) hence ℵα < κ. By definition of ℵκ, we have ℵκ ≤ κ. Therefore ℵκ = κ.

We define by transfinite recursion: i0 = ℵ0,iα+ = 2iα (cardinal exponentiation), iλ =⋃α<λ iα for λ a limit ordinal.

Example. i0 = ℵ0,i1 = cardR,i2 = cardPR. One can prove by induction iα ≥ ℵα. Continuum hypothesis: i1 = ℵ1 isundecidable. You may add an axiom saying iα = ℵα for all α but it may cause trouble. You may add without much troublean axiom that i1 = ℵ? for almost anything...cannot be ω? at some point ℵ1 or 2...? what is the best value?

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Ordinal Operations

Note: α 7→ ℵα is not a function because the domain is not a set. We will call it an “operation on ordinals”.

Definition. Let α 7→ tα be an operation on ordinals. Say that it is:

• monotone if α < β =⇒ tα < tβ , (why cannot be descending? you cannot even go over ω then!)

• continuous if tλ =⋃{tα : α < λ} for λ a limit ordinal,

• normal iff it is both monotone and continuous

Notation: Let S 6= ∅ is a set of ordinals, then supS :=⋃S: it is the least upper bound of S.

Lemma (8C). If α 7→ tα is a continuous operation such that tα < tα+ for each α then it is monotone.

Proof. Fix α and prove by transfinite induction on β that

α < β =⇒ tα < tβ

Case 1: β is zero. The above condition is vacuously true, since α 6< βCase 2: β is a successor ordinal γ+. Then

α < β =⇒ α ≤ γ=⇒

Case 3: β is a limit ordinal.

Example. α 7→ ℵα, α 7→ iα are normal operations.

Lemma (8E). Let S be a nonempty set of ordinals, α 7→ tα a normal operation then tsupS = sup{tα | α ∈ S}.

Proof. ≥ by monotonicity: α ∈ S =⇒ α ≤ supS =⇒ tα ≤ tsupS =⇒ RHS ≤ LHS.

• S has a greatest element γ = supS. Then tsupS = tγ ≤ sup{tα | α ∈ S}.

• λ := supS for sure have to be a limit ordinal. By continuity, tsupS = sup{tα | α < λ}. If α < λ, α ≤ β ∈ S for someβ ∈ S then tα ≤ tβ . Therefore, tλ = sup{tα | α < λ} = sup{tα | α ∈ S}.

Theorem (Veblen Fixed-Point). Let α 7→ tα be a normal operation. Then for any ordinal β, there is γ ≥ β such that tγ = γ.

Proof. f : ω into ordinals such that f(0) = β and f(n+) = tf(n). By monotonicity, tnβ < tn+

β . Define λ = sup ranf = sup{tnβ |n ∈ ω}. Claim tλ = λ.

λ has no largest element, and hence is a limit ordinal. Let S = ranf so λ = supS. By the previous theorem,

tλ = sup{tα | α ∈ S} = sup{tn+

β | n ∈ ω} = λ

Ordinal Arithmetic

This week we go by recursion. Next week we go by textbook.

Addition Define + on ordinals by transfinite recursion:

α+ 0 = α, α+ β+ = (α+ β)+

When λ is a limit ordinal, for continuity’s sake,

α+ λ = sup{α+ β | β < λ}

Note: For fixed ordinal α, the operation β 7→ α + β is a normal operation. In particular, by lemma 8C, it is monotone:β < β′ =⇒ α+ β < α+ β′.

Example. • n,m < ω, n+m is the same in the sense of ordinals as in the sense of natural numbers.

• α+ 1 = α+ 0+ = (α+ 0)+ = α+.

• ω + ω = sup{ω + n : n ∈ ω} = sup{ω, ω+, ω++, ω+++, . . . }.

• 1 + ω = sup{1 + n : n ∈ ω} = ω!!! 1 + ω 6= ω + 1. Addition is in general not commutative on ordinals.

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Multiplication Define · on ordinals by transfinite recursion:

α · 0 = 0, α · β+ = α · β + α, α · λ = sup{α · β : β < λ}

when λ is a limit ordinal.Note: for a fixed α > 0, β 7→ α · β is a normal operation. In particular, if α > 0, β < β′ then α · β < α · β′. Indeed, if

α > 0 then α · β + α > α · β + 0 by monotonicity on right of addition.

Example. α · 1 = α · 0+ = α · 0 + α = 0 + α = α the last step has to be proven by induction on α. this is the disadvantageof recursive definition.

1 · α = α can be proven by induction on α.

• 1 · 0 = 0

• 1 · β+ = 1 · β + 1 = β + 1 = β+

• If λ is limit, 1 · β for all β < λ then 1 · λ = sup{1 · β : β < λ} = sup{β : β < λ} = λ.

0 · α = 0 (induction on α). ω · 2 = ω · 1 + ω = ω + ω. 2 · ω = sup{2 · n : n ∈ ω} = ω. !!! ω · 2 6= 2 · ω. Multiplication onordinals is not commutative.

For n ∈ ω, ω+ω ·n = ω · (n+ 1) = ω ·n+ω 6= ω ·n. In general, α ·n = α+ α+ · · ·+ α︸ ︷︷ ︸n

. +, · are associative, so the above

is justified.Claim ω + ω · ω = ω · ω.

Proof. ω + ω · sup{n : n < ω} = ω + sup{ω · n : n < ω} = sup{ω + ω · n : n < ω} = sup{ω · (n+ 1) : n < ω} = ω · sup{n+ 1 :n < ω} = ω · ω using either the definition of ω · x or Lemma 8E where α 7→ tα = ω · α is a continuous operation.

Thus, ω + ω · ω 6= ω · ω + ω because + on the right is monotone.

Proposition. Let α, β, γ be ordinals. We have:

(α+ β) + γ = α+ (β + γ)

(α · β) · γ = α · (β · γ)

0 + α = α+ 0 = α

0 · α = α · 0 = 0

1 · α = α · 1 = α

α · (β + γ) = α · β + α · γ

Proof. All those statements are proven by transfinite induction on the rightmost variable. Let us do the last one: distributivityon the right, by transfinite induction on γ.

• γ = 0, α · (β + 0) = α · β

• α · (β + γ+) = α · (β + γ)+ = α · (β + γ) + α = α · β + α · γ + α = α · β + α · (γ + 1) = α · β + α · γ+

• α · (β + γ) = α · (β + sup{δ : δ < γ}) = α · sup{β + δ : δ < γ} = sup{α · (β + δ) : δ < γ} = sup{α · β + α · δ : δ < γ} =α · β + sup{α · δ : δ < γ} = α · β + α · sup{δ : δ < γ} = α · β + α · γ.

Exponentiation we define by transfinite recursion

α0 = 1

αβ+

= αβ · ααλ = sup{αβ : β < λ}

If α > 1, β 7→ αβ is a normal operation. In particular, monotone.

Example. ω2 = (ω1)+ = ω1 · ω. (α1 = α0+

= α0 · α = 1 · α = α).2ω = sup{2n : n ∈ ω} = ω.

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!!! Ordinal arithmetic is not cardinal arithmetic. ω = ℵ0. However, 2ω 6= 2ℵ0 , ω + ω 6= ℵ0 + ℵ0

If α ≥ 2, then β 7→ αβ is a normal operation. We only check αβ+

> αβ .

αβ+

= αβ · α≥ αβ · 2, · is monotone

= αβ + αβ

≥ αβ + 1 > αβ

Proposition. For any ordinals α, β, γ,

1. If 2 ≤ α, then β < γ =⇒ αβ < αγ

2. αβ+γ = αβ · αγ

3. (αβ)γ = αβ·γ

Proof. 1 normality (monotone). 2&3 fix α, β and prove by transfinite induction on γ as in the textbook.

Proposition. For any ordinals α, β, γ, we have

1. β ≤ γ =⇒ β + α ≤ γ + α

2. β ≤ γ =⇒ β · α ≤ γ · α

3. β ≤ γ =⇒ βα ≤ γα

These are false if ≤ is replaced by <, e.g., 1 < 2 but 1 + ω = 2 + ω = ω

Arithmetic of Order Type

Addition Idea: α+ β is obtained by concatenating (α,∈) with (β,∈), placing β after α.

Example. 2 = ••, ω = · · · , ω + 2 = · · · • • 6= ω because ω + 2 has greatest element but ω does not. 2 + ω = • • · · · =ω, ω + ω = • · · · • · · · 6= ω because ω + ω has an element which is not the least element and has no predecessor.

Let C = {0} × α ∪ {1} × β. Define <∈ be the lexicographic ordering on C (so we have (0, γ) <C (1, δ),∀γ ∈ α, δ ∈ β).Then <C is a well-ordering on C with ordinal number α+ β (proof by induction on β.

Multiplication Idea: α · β is obtained by taking β & replacing each of its element by a copy of α.

Example. ω · 2 = • · · · • · · · , 2 · ω = || || || . . . , ω2 = ω · ω = • · · · • · · · • · · · • · · · • · · · . . .

Define C = α×β and<C as the “inverse” (“Hebrew”) lexicographic ordering, i.e., (γ, δ) <C (γ′, δ′) ⇐⇒{

δ < δ′

or δ = δ′&γ < γ′

Then <C is a well-ordering on C and its ordinal number is α · β. Indeed, the isomorphism between (C,<C) and (α · β,∈) is

f :

{C → α · β

(γ, δ) 7→ α · δ + γ

Example. (ω + ω + 1)2 = (ω2 + ω + 1) · (ω2 + ω + 1) = (ω2 + ω + 1)ω2 + (ω2 + ω + 1)ω + (ω2 + ω + 1) = ω4 + · · ·

Theorem (Subtraction). Let α, β be 2 ordinals with α ≥ β. Then there is a unique δ such that α = β + δ.

Theorem (Division). Let α, δ be ordinals with δ 6= 0 then ∃!(β, γ) such that α = δ · β + γ and γ < δ.

They can be proven by induction on α.

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Ranks and the Vα hierarchy

Recall we started constructing sets from ∅ using power set axiom

∅,P∅,PP∅, . . .

Going on like this, we can build all hereditary finite sets (i.e. elements of⋃n∈ω P(n)∅. Now that we have ordinals, it is

natural to continue the sequence along the ordinals. We define Vα by transfinite recursion.

V0 = ∅Vα+ = PVα

Vλ =⋃β<λ

Vβ

Vω is the set of all hereditary finite sets.

Proposition. • For α ≤ β, Vα ⊆ Vβ .

• For all α, Vα is a transitive set.

• α ≤ Vα so α ∈ Vα+1

Proof. We prove the first two by induction on β:

• β = 0, V∅ is transitive.

• β = γ + 1, Vβ = PVγ is transitive as Vγ is transitive. Also Vγ ⊆ PVγ because Vγ is transitive. Thus, if δ < β thenδ ≤ γ so Vδ ⊆ Vγ ⊆PVγ = Vβ .

• If β = λ is a limit ordinal, Vβ =⋃α⊂β Vα so (1) follows. The union of transitive sets is transitive so Vβ is transitive.

We prove the last one by induction on α:

• α = 0, ∅ ⊆ V∅.

• If α = β + 1, then α = β ∪ {β}. By induction β ⊆ Vβ ⊆ Vα. Also β ∈ Vβ+1 = Vα so α ≤ Vα.

• If α = λ is a limit ordinal, then for β ∈ λ we have by induction β ⊆ Vβ+1 ≤ Vλ so λ ≤ Vλ.

Definition. A set A is grounded if A ⊆ Vα for some α. The minimal such α is called the rank of A.

Example. rk∅ = 0, rk{∅} = 1, rkω = ω. More generally if α is an ordinal, rkα = α. We have seen α ≤ Vα =⇒ rkα ≤ α.Prove by induction that rkPω = ω + 1 where Pω is not an ordinal. rkZ = ω + 1 (note that elements of Z are hereditaryfinite).

rkR = ω + 5.

Cardinality of Vα For any ordinal α, cardVα+1 = 2cardVα so for n ∈ ω, Vn is finite (of size 222...

so cardVω = ℵ0 (since it isa countable union of finite sets). cardVω+1 = 2ℵ0 = i1, cardVω+2 = i2 = 2i1 .

Proposition. • If A is grounded and a ∈ A, then a is grounded and rka < rkA.

• If every element of A is grounded, then A is grounded and rkA = sup{rk(a) + 1 | a ∈ A}.

Proof. Assume A is grounded, say A ⊆ Vα, then if a ∈ A we have a ∈ Vα. If α = β + 1, then a ⊆ Vβ so rka ≤ β < α. If α isa limit ordinal, then there is β < α such that a ⊆ Vβ so a ⊆ Vβ is transitive so rka ≤ β < α.

Assume all a ∈ A are grounded, let α = sup{rk(a) + 1 | a ∈ A} (exists, as a set of ordinals is always bounded). Theneach a ∈ A is in Vrk(a)+1 ⊆ Vα hence A ⊆ Vα so rkA ≤ α. (rkA = α is left as an exercise).

Theorem. The following two statements are equivalent:

• Every set is grounded.

• (Axiom of regularity) every nonempty set A has an element m such that m ∩A = ∅.

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Proof. Assume that every set is grounded, and let A be a nonempty set. Let µ := min{rka | a ∈ A} and select m ∈ A withrkm = µ.

If x ∈ m, then by a previous proposition, rka ∈ µ. We cannot have x ∈ A due to the minimality of µ. Hence m ∩ A = ∅as desired.

Suppose ∃c a set that is not grounded. Then {c} has a nongrounded member. Let B be the transitive closure of {c}.Let A = {x ∈ B | x is not grounded}. Since A 6= ∅, by regularity there is m ∈ A such that m ∩ A = ∅. We claim that everymember of m is grounded. If x ∈ m, then x ∈ B since B is transitive, but x /∈ A because m∩A = ∅, so x must be grounded.Since every member of m is grounded, by a previous proposition m is also grounded, contradicting the fact that m ∈ A.

Axiom (Regularity). Every nonempty set A has a member m with m ∩A = ∅.

The regularity axiom is also known as the foundation axiom or the “Fundierungsaxiom”. The idea first appeared in a1917 paper by Mirimanoff; the axiom was listed explicitly in von Neumann’s 1925 paper.

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