pmath433/733 — model theory and set theory · theory, we look at set theory as a language of...

PMATH433/733 — Model Theory and

Set Theory

Classnotes for Fall 2018

by

Johnson Ng

BMath (Hons), Pure Mathematics major, Actuarial Science Minor

University of Waterloo

Table of Contents

List of Definitions 7

List of Theorems 9

I Set Theory

1 Lecture 1 Sep 06th 15

1.1 Introduction to Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.2 Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.2.1 Zermelo-Fraenkel Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15


2.1 Ordinals (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1.1 Zermelo-Fraenkel Axioms (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1.2 Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24


3.1 Ordinals (Continued 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.1.1 Cartesian Products and Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.1.2 The Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.1.3 Well-Orderings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32



4.1.1 Well-Orderings (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35



5.1.1 Transfinite Induction & Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

PMATH433/733 — Model Theory and Set Theory 3



6.1.1 Transfinite Induction & Recursion (Continued) . . . . . . . . . . . . . . . . . . . . . . . 47

6.1.2 Ordinal Arithmetric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

6.1.2.1 Ordinal Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

6.1.2.2 Ordinal Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6.1.2.3 Ordinal Exponentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53



7.1.1 Well-Orderings and Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7.2 Cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8 Lecture 8 Oct 02nd 63

8.1 Cardinals (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

8.1.1 Axiom of Choice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

9 Lecture 9 Oct 04th 69

9.1 Cardinals (Continued 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

9.1.1 Axiom of Choice (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

9.1.2 Hierarchy of Infinite Cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71


10.1 Cardinals (Continued 3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

10.1.1 Cardinal Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

10.1.1.1 Cardinal Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

10.1.1.2 Cardinal Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

10.1.2 An Interlude on the Continuum Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . 79

II Model Theory


11.1 First-order Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

11.1.1 Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

11.1.2 Language . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87


12.1 First-order Logic (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

12.1.1 Language (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

12.1.2 Terms & Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

13 Lecture 13 Oct 23rd 97

4 TABLE OF CONTENTS - TABLE OF CONTENTS

13.1 First-order Logic (Continued 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

13.1.1 Terms & Formulas (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

14 Lecture 14 Oct 25th 10514.1 First-order Logic (Continued 3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

14.1.1 Terms & Formulas (Continued 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

14.1.2 Elementary Embeddings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

15 Lecture 15 Oct 30th 11115.1 First-order Logic (Continued 5) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

15.1.1 Elementary Embeddings (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

15.1.2 Parameters and Definable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

16 Lecture 16 Nov 01st 11716.1 First-order Logic (Continued 6) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

16.1.1 Parameters and Definable Sets (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . 117

16.1.2 Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

17 Lecture 17 Nov 06th 12317.1 First-order Logic (Continued 7) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

17.1.1 Theories (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

17.2 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

17.2.1 A proof of compactness using ultraproducts . . . . . . . . . . . . . . . . . . . . . . . . . 125

18 Lecture 18 Nov 08th 13118.1 Compactness (Continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

18.1.1 A Proof of Compactness Using Ultraproducts (Continued) . . . . . . . . . . . . . . . . . 131

19 Lecture 19 Nov 13th 13719.1 Compactness (Continued 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

19.1.1 A Proof of Comapactness Using Ultraproducts (Continued 2) . . . . . . . . . . . . . . . 137

20 Lecture 20 Nov 15th 14320.1 Compactness (Continued 3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

20.1.1 First Applications of Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Bibliography 147

Index 149

� List of Definitions

1 � Definition (Successor) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 � Definition (Definite Condition) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 � Definition (Subsets) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 � Definition (Natural Numbers) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5 � Definition (Class) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

6 � Definition (Ordered Pairs) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

7 � Definition (Cartesian Product) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

8 � Definition (Definite Operation) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

9 � Definition (Functions) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

10 � Definition (Strict Partially Ordered Set) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

11 � Definition (Strict Totally Ordered Set) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

12 � Definition (Well-Order) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

13 � Definition (Ordinals) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

14 � Definition (Successor Ordinal) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

15 � Definition (Limit Ordinal) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

16 � Definition (α-function) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

17 � Definition (Ordinal Addition) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

18 � Definition (Ordinal Multiplication) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

19 � Definition (Ordinal Exponentiation) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

20 � Definition (Isomorphism) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

22 � Definition (Equinumerous) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

23 � Definition (Cardinal) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

24 � Definition (Finite & Countable) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

25 � Definition (Choice Function) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

26 � Definition (Cardinality) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

28 � Definition (Cardinal Numbers) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

6 � LIST OF DEFINITIONS - � LIST OF DEFINITIONS

29 � Definition (Cardinal Sum) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

30 � Definition (Cardinal Product) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

31 � Definition (I-sequence) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

32 � Definition (Generalized Cardinal Sum) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

33 � Definition (Generalized Cardinal Product) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

34 � Definition (Cardinal Exponentiation) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

35 � Definition (Structure) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

36 � Definition (Expansion & Reduct) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

37 � Definition (Language) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

38 � Definition (L-structure) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

39 � Definition (L-Embedding) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

40 � Definition (Substructure) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

41 � Definition (L-isomorphism) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

42 � Definition (L-terms) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

43 � Definition (Interpretation) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

44 � Definition (Atomic L-Formulas) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

45 � Definition (L-formulas) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

46 � Definition (Bound and Free Variables) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

47 � Definition (L-Sentences) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

48 � Definition (Satisfaction / Realization) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

49 � Definition (L-Definable) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

50 � Definition (Elementary Embeddings) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

51 � Definition (Parameters) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

52 � Definition (B-Definable) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

54 � Definition (L-Theory) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

55 � Definition (Model) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

56 � Definition (L-elementary / L-axiomatizable) . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

57 � Definition (Theories of a Structure) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

58 � Definition (L-elementarily Equivalent) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

59 � Definition (Implication) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

60 � Definition (Complete) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

61 � Definition (Partial L-elementary Map) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

62 � Definition (Filter) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

63 � Definition (Ultrafilter) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

64 � Definition (Ultraproduct) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128


65 � Definition (Ultrapower) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

66 � Definition (ℵ1-Compact) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

1 List of Theorems

1 � Axiom (Empty Set Axiom) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2 � Axiom (Pairset Axiom) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 � Axiom (Axiom of Extension) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 � Axiom (Union Set Axiom) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 � Axiom (Infinity Axiom) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

6 � Axiom (Powerset Axiom) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7 � Axiom ((Bounded) Separation Axiom) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

8 � Axiom (Replacement Axiom) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

9 � Axiom (Replacement Axiom (Restated)) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

10 1 Theorem (Induction Principle) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

11 + Lemma (Properties of the Natural Numbers) . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

12 0 Proposition (ω is Strictly Well-Ordered) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

13 + Lemma (Proper Subsets of an Ordinal Are Its Elements) . . . . . . . . . . . . . . . . . . . . 38

14 0 Proposition (Properties of Ordinals) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

15 0 Proposition (Properties of Ordinals 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

16 1 Theorem (Transfinite Induction Theorem v1) . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

17 1 Theorem (Transfinite Induction Theorem v2) . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

18 1 Theorem (Transfinite Recursion v1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

19 � Corollary (Transfinite Recursion v2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

20 0 Proposition (Properties of Ordinal Addition and Ordinal Multiplication) . . . . . . . . . . . 52

21 + Lemma (Rigidity of Well-Orderings) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

22 + Lemma (Strict Well-Ordering 6' Any of Its Proper Initial Segment) . . . . . . . . . . . . . . 56

23 1 Theorem (Strict Well-Ordered Sets are Isomorphic to a Unique Ordinal) . . . . . . . . . . . 56

24 + Lemma (Schröder-Bernstein Theorem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

25 0 Proposition (The Least Cardinality Not Equinumerous to Subsets of a Set) . . . . . . . . . . 63

26 � Axiom (Axiom of Choice) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

27 1 Theorem (Axiom of Choice and Its Equivalents) . . . . . . . . . . . . . . . . . . . . . . . . . 65


28 0 Proposition (Using Cardinals to Measure Sets) . . . . . . . . . . . . . . . . . . . . . . . . . . 69

29 0 Proposition (Lesser Cardinality) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

30 � Corollary (Cardinalities are Always Comparable) . . . . . . . . . . . . . . . . . . . . . . . . 70

31 0 Proposition (Functions are “Lossy Compressions”) . . . . . . . . . . . . . . . . . . . . . . . . 70

32 0 Proposition (Countable Union of Countable Sets is Countable) . . . . . . . . . . . . . . . . . 71

33 + Lemma (Cardinal Numbers are Cardinals) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

34 + Lemma (Ordinals Index the Cardinal Numbers) . . . . . . . . . . . . . . . . . . . . . . . . . 72

35 + Lemma (Infinite Cardinals are Distant) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

36 0 Proposition (All Infinite Cardinals are Indexed by the Ordinals) . . . . . . . . . . . . . . . . 73

37 1 Theorem (Dominance of the Larger Cardinal) . . . . . . . . . . . . . . . . . . . . . . . . . . 77

38 1 Theorem (Properties of Cardinal Sum) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

39 1 Theorem (Cantor’s Diagonalization) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

40 � Axiom (Continuum Hypothesis) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

41 � Axiom (Generalized Continuum Hypothesis) . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

42 0 Proposition (Structure Traversal with respect to Quantifiers) . . . . . . . . . . . . . . . . . . 100

43 � Corollary (Isomorphisms are Elementary Embeddings) . . . . . . . . . . . . . . . . . . . . 106

44 0 Proposition (� Tarski-Vaught Test) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

45 1 Theorem (Downward Löwenheim-Skolem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

46 � Corollary (Automorphisms on B-definable Sets) . . . . . . . . . . . . . . . . . . . . . . . . . 117

47 0 Proposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

48 + Lemma (Equivalence to a Complete Theory) . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

49 0 Proposition (Elementary Equivalence in Finite Stuctures) . . . . . . . . . . . . . . . . . . . . 123

50 � Corollary (Corollary of Compactness Theorem) . . . . . . . . . . . . . . . . . . . . . . . . . 125

51 + Lemma (Existence of Ultrafilters) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

52 + Lemma (Equivalent Characterization of an Ultrafilter) . . . . . . . . . . . . . . . . . . . . . . 127

53 1 Theorem (Łos) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

54 � Corollary (Corollary of Łos) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

55 0 Proposition (Ultraproducts with a Non-Principal Ultrafilter is ℵ1-compact) . . . . . . . . . . 139

56 1 Theorem (Compactness Theorem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Foreword

This course has a ratio of about 1:3 for naive set theory to model

theory.

Part I

Set Theory

1 Lecture 1 Sep 06th

1.1 Introduction to Set Theory

In this course, we shall focus only on practical set theory, which

is more commonly knowned as naive set theory. In practical set

theory, we look at set theory as a language of mathematics. Some of

the examples of which we look into in this flavour of set theory are

(transfinite) induction and recursion, and the measuring of the sizes

of sets.

Another approach to set theory, one that is deemed required in

order to learn set theory is a more formal way, is to look at set theory

as the foundations of mathematics. Such an approach is more ax-

iomatic, rigorous, and grounding as compared to practical set theory.

This course will try to work around going into these topics, as they

can take a life of their own, and within the context of this course, the

topics that will be explored using this approach are not required.

1.2 Ordinals

1.2.1 Zermelo-Fraenkel Axioms

We use the natural numbers, i.e.

0, 1, 2, 3, 4, ...

to count finite sets. There are two related meanings attached to the

word “count” here:

• enumeration; and

• measuring (of sizes)

16 Lecture 1 Sep 06th - Ordinals

In order to facilitate the introduction to certain axioms that we

shall need, let our current goal be to develop an infinitary general-

ization of the natural numbers, so as to be able to enumerate and

measure arbitrary sets. We have that

enumeration → ordinals

measuring → cardinals

where→ represents “leads to” here.To construct the natural numbers, we require 3 basic notions that

shall remain undefined but understood:

• a set;

• membership, denoted by ∈; and

• equality.

One such construction is

0 := ∅, the empty set

1 := {0} = {∅}, the set whose only member is 0

2 := {0, 1} = {∅, {∅}}, the set whose only members are 0 and 1.

� Definition 1 (Successor)

Given a natural number n, the successor of n is the natural number nextto n, which can be obtained by

S(n) := n ∪ {n}.

We can use the definition of a successor to construct the rest of the

natural numbers.

Example 1.2.1

Just to verify to ourselves that the definition indeed works, observe

that

S(1) = 2 = {∅, {∅}} = {∅} ∪ {{∅}}.

So to construct the natural number 3, we see that

S(2) = 3 = {0, 1, 2} = 2∪ {2} = {∅, {∅}} ∪ {{∅, {∅}}}

= {∅, {∅}, {∅, {∅}}}


Looking at these, we start wondering to ourselves: how do we

know that ∅ exists in the first place? How do we know that we can

use ∪ and what does it even mean? Now it is meaningless if we

cannot take that ∅ always exists, nor is it meaningful if we cannot

take the ∪ of sets. And so, to allow us to continue, or even start with

these notions, we require axioms.

� Axiom 1 (Empty Set Axiom)

There exists a set, denoted by ∅, with no members.

With this axiom, we can indeed construct 0. To get 1 from 0, we

have that 1 is a set whose only member is zero, and so if we take a

member from 1, that member must be 0.

� Axiom 2 (Pairset Axiom)

Given set x, y, there exists a set, denoted by {x, y}, whose only membersare x and y. In other words,

t ∈ {x, y} ↔ (t = x ∨ t = y)

Now note that in � Axiom 2, if x = y, then the set {x, y} has only

x as its member. For example, we realize that 1 = {0, 0} = {0}.But why exactly does this equality make sense? What exactly does

“realize” mean?

� Axiom 3 (Axiom of Extension)

Given sets x, y, x = y if and only if x and y have the same members.

Now, using the above 3 axioms, we are guaranteed that

0 = ∅ exists by the Empty Set Axiom

1 = {∅} exists by the Pairset Axiom

2 = {∅, {∅}} exists by the Pairset Axiom

18 Lecture 1 Sep 06th - Ordinals

Now we’ve constructed 3 to be the set whose only members

are 0, 1, and 2. So far, within our axioms, there is no such thing as

{0, 1, 2}, which is what our 3 is supposed to be. We now require the

following axiom:

� Axiom 4 (Union Set Axiom)

Given a set x, there exists a set denoted by ∪x, whose members are pre-cisely the members of the members of x, i.e.

t ∈ ∪x ↔ (t ∈ y for some y ∈ x)

So, by this axiom, we have that given any n, S(n) = ∪{n, {n}}, or

in other words,

t ∈ S(n)↔ t ∈ n ∨ t = n.

With all of the above axioms, we can iteratively construct each and

every natural number in a rigorous manner. However, our goal is to

construct infinitely many of them.

It is tempting to simply take the infinitude of natural numbers

simply as an axiom, i.e.

There exists a set whose members are precisely the natural numbers.

There is a certain rule to which we set down axioms, and that is,

axioms must be expressable in a “finitary” manner, i.e. they must be

expressible using first-order logic.

� Definition 2 (Definite Condition)

We define a definite condition as follows:

• x ∈ y and x = y are definite conditions, where x and y are bothindeterminants, standing for sets, or are sets themselves;

• if P and Q are definite conditions, then so are

– not P, denoted as ¬P;

– P and Q, denoted as P ∧Q;


– P or Q, denoted as P ∨Q;

– for all x, P, denoted as ∀xP; and

– there exists x, P, denoted as ∃xP.

Example 1.2.2

x ∈ 1, 0 ∈ 2, 2 ∈ 0

are all definite conditions. Note, however, that 2 ∈ 0 is false.

å Note“If P then Q”, which is also written as P → Q, is also a definite condi-tion since it is “equivalent”1 to the statement ¬P ∨Q. 1 We have yet to define what equivalent

statements are but we shall take this forgranted for now.Consequently, Pi f andonlyi f Q, which is also expressed as P↔ Q, can

be written as(¬P ∨Q) ∧ (¬Q ∨ P)

Now, with this definition, and first-order logic notations in mind,

we can write:

Exercise 1.2.1Write � Axiom 3 in first-order logicnotation.

Ò Solution

∀x ∀y

(x = y)↔ (∀t ((t ∈ x)↔ (t ∈ y)))

• Empty Set Axiom: ∃x ∀t ¬(t ∈ x)

• Pairset Axiom: ∀x ∀y ∃p ∀t (t ∈ p↔ ((t = x) ∨ (t = y)))

• Union Set Axiom: ∀x ∃z ∀t ((t ∈ z)↔ (∃y ((y ∈ x) ∧ (t ∈ y))))

Note that the statement that we proposed as an axiom for the set

of natural numbers in page 18 is not definite, although that itself is

not obvious.

For example, we may try to write

∃x (∀t ((t ∈ x)↔ ((t = 0) ∨ (t = 1) ∨ (t = 2) ∨ ...)))

and then notice that we do not have the notion of “...” within the

“tools” that we are allowed to use.


2.1 Ordinals (Continued)

2.1.1 Zermelo-Fraenkel Axioms (Continued)

We stopped at the discussion about allowing for an infinite set, so

that we can construct our set of infinite natural numbers. The idea

here is to take the smallest set that contains 0 and is preserved by thesuccessor function1 1 Q: Why the smallest set?

� Axiom 5 (Infinity Axiom)

There exists a set I that contains 0 and is preserved by the successorfunction. We may express this as

∃I((0 ∈ I) ∧ ∀x(x ∈ I → S(x) ∈ I))

where we have defined that S(x) ∈ I means

∃y(∀t(t ∈ y↔ (t ∈ x) ∨ (t = x)) ∧ (y ∈ I))

We call I the successor set.

Since we want the smallest of such successor sets, we can try tak-

ing the intersection of all successor sets. But before we can do that,

we require more axiomatic statements.

� Definition 3 (Subsets)

22 Lecture 2 Sep 11th - Ordinals (Continued)

x ⊆ y means that every element of x is an element of y, i.e.

∀t((t ∈ x)→ (t ∈ y))

With a definition of a subset, we can define the Powerset Axiom.

� Axiom 6 (Powerset Axiom)

Given a set x, there exists a set P(x) that contains all subsets of x, i.e.

∀t((t ⊆ P(x))↔ (t ⊆ x))

We also require the following axiom.

� Axiom 7 ((Bounded) Separation Axiom)

Given a set x and a definition condition P, there exists a set whose ele-ments are precisely the members of x that satisfies P, i.e. There are two important aspects to the

Bounded Separation Axiom:

• it is bounded by the set x; and

• P is a definite condition.

∀x ∃y ∀t ((t ∈ y)↔ ∀y ((t ∈ x) ∧ P(t)))

wherey = {z ∈ x | P(z)}.

Exercise 2.1.1 (Set Intersection)

Prove that given a non-empty set x, there exists a set ∩x satisfying

∀t ((t ∈ ∩x)↔ ∀y ((y ∈ x)→ (t ∈ y)))

Ò Proofto be solved

� Definition 4 (Natural Numbers)


Let I be a successor set. The set of natural numbers is2 2 We can also write J ⊆ I as J ∈ P(I)and invoke the Bounded SeparationAxiom.

ω := ∩{J ⊆ I : J is a successor set }

å NoteJ being a successor set can be expressed by the definite condition

(0 ∈ J) ∧ ∀x (x ∈ J → S(x) ∈ J),

so we can write the definite condition in the above definition by

ω := ∩{J ⊆ I : (0 ∈ J) ∨ ∀x (x ∈ J → S(x) ∈ J)}

Exercise 2.1.2

Show that the definition of ω does not actually depend on I, i.e. if given I1

and I2 such that we have

ω1 = ∩{J ⊆ I1 : J is a successor set }

ω2 = ∩{J ⊆ I2 : J is a successor set }

we haveω1 = ω2.

Ò Proofto be solved

Another useful axiom that we will use later is the following:

� Axiom 8 (Replacement Axiom)

Suppose P is a binary definite condition3 such that for every set x, there 3 A binary definite condition has onlytwo variables.is a unique y satisfying P(x, y). Given a set A, there is a set B such that

t ∈ B if and only if there is an a ∈ A with P(a, t).

24 Lecture 2 Sep 11th - Ordinals (Continued)

å NoteThe slogan for the Replacement Axiom is:

The image of a set under a definite operation exists.

These eight axioms, along with another ninth axiom called the

Regularity Axiom4, constitutes the Zermelo-Fraenkel Set Theory. 4 We shall not discuss too much aboutthis. According to the lecture and thelecture notes, the Regularity Axiomstates that every set has a minimalelement. On Wikipedia, the axiomstates that every set has an element thatdoes not intersect with the set itself.

Note that all axioms, save the Extensionality, assert the existence of

sets.

2.1.2 Classes

There are times where we are interested in a collection of sets that do

not form a set themselves.

Example 2.1.1 (Russell’s Paradox)

There is no set containing all sets.

Ò ProofSuppose such a set exists, and call it U. Now consider the set

R := {x ∈ U : x /∈ x},

which exists by Bounded Separation. Observe that

R ∈ R =⇒ R /∈ R E

=⇒ R /∈ R =⇒ R ∈ R E

Thus such a set U cannot exist.

To talk about such collections, that may or may not be sets, we

define classes.

� Definition 5 (Class)

A class is any collection of sets defined by definite property, i.e. given anydifinite condition P,

J z | P(z) K


is the class of all sets satisfying P.

Here, instead of Bounded Separation, we have what is called un-bounded separation.

å NoteWe shall use J K rather than { } to emphasize that we are talking aboutclasses, i.e. we may be talking about non-sets.

Example 2.1.2

Set := J z | z = z K

is the universal class of all sets.

å Note• Every set is a class.

Ò ProofSuppose x is a set. We may write

x = J z | z ∈ x K .

�

• Some classes are not sets; these are called proper classes. E.g. theuniversal class of all sets, and

Russell := J z | z /∈ z K .


3.1 Ordinals (Continued 2)

3.1.1 Cartesian Products and Function

� Definition 6 (Ordered Pairs)

Given sets x, y, an ordered pair of x and y is defined as1 1 This invokes the Pairset Axiom thrice.Why did we not define an ordered pairas

(x, y) = {{x}, {y}}

instead?

(x, y) = {{x}, {x, y}}

å NoteNote that we must have

((x, y) = (x′, y′)) ⇐⇒ (x = x′ ∧ y = y′).

Ò ProofThe ( ⇐= ) direction is clear by Extensionality. For the other

direction, we shall break it into 2 cases:

Case 1: x = y. Then {x, y} = {x} by Extensionality, and so

(x, y) = {{x}}

Therefore, we have that

{{x}} = (x, y) = (x′, y′) = {{x′}, {x′, y′}}

28 Lecture 3 Sep 13th - Ordinals (Continued 2)

So we have

{x} = {x′} =⇒ x = x′

and

{x} = {x′, y′} =⇒ y′ = x = y.

Thus we have

x = x′ ∧ y = y′.

Case 2: Suppose x 6= y and x′ 6= y′ 2 We have 2 If any of them are equal, Case 1 wouldapply.

{{x}, {x, y}} = {{x′}, {x′, y′}}

Then

{x} = {x′} ∨ {x} = {x′, y′}

The latter leads to a contradiction, since it would imply

x′ = x = y′.

Thus x = x′. Also, we have

{x, y} = {x′} ∨ {x, y} = {x′, y′}

Now the former leads to a contradiction since it would imply

that

x = x′ = y.

Now since x = x′, it must be that y = y′, otherwise y = x′ = xwould contradict our assumption. Therefore, we have that

x = x′ ∧ y = y′.

�

With ordered pairs, we can build Cartesian products:

� Definition 7 (Cartesian Product)

Given classes X and Y, the Cartesian Product of X and Y is defined as

X×Y := J z : z = (x, y), x ∈ X, y ∈ Y K


å NoteWe can express this definition using definite conditions;

∀x, y

((x ∈ X) ∧ (y ∈ Y) ∧

(∃a, b(∀t(t ∈ a↔ t = x)) ∧ ∀t(t ∈ b↔ (t = x) ∨ (t = y))

)∧

∀t(

t ∈ z↔((t = a) ∨ (t = b)

)))

å Note• A Cartesian product is a class.

• If A is a set and B is a class, and B ⊆ A, then B is also a set. This iseasy to show: observe that by Extentionality,

B = {a ∈ A | a ∈ B}.

By Bounded Separation Axiom, B is a set3. 3 This statement can be rephrased as:subclasses of a set are subsets.

Consequently, Cartesian products of sets are sets themselves; if Xand Y are sets, we want to show that X × Y is a set so it is sufficient

to show that it is contained in one. Recall that

(x, y) = {{x}, {x, y}}

and {x, y} ⊂ X ∪ Y which means {x, y} ∈ P(X), and we observe

that {x} ∈ P(X ∪ Y). So (x, y) ∈ P(X ∪ Y). Therefore, X × Y ⊂P(P(X ∪Y)), and we show to ourselves that X×Y is indeed a set.

� Definition 8 (Definite Operation)

Given classes X and Y, a definite operation f : X → Y is a subclassΓ( f ) ⊆ X×Y such that

∀x ∈ X ∃!y ∈ Y (x, y) ∈ Γ( f ).


å NoteWe write f (x) = y to mean (x, y) ∈ Γ( f ). We also refer to Γ( f ) as thegraph of f .

Example 3.1.1

The successor function S : Set→ Set is a definite operation such that To show that S is a definite operation,we need to show that S is a definitecondition.S(x) = x ∪ {x}

This is true since is can be expressed as

∀t(t ∈ y↔ (t ∈ x ∨ t = x)).

å NoteIf X and Y are sets and f is a definite operation, then Γ( f ) ⊆ X × Y is aset. In such a case, we call f a function.

� Definition 9 (Functions)

A function is a definite operation f : X → Y where X and Y are both sets.

We can now restate the Replacement Axiom.

� Axiom 9 (Replacement Axiom (Restated))

If f : X → Y is a definite operation, and A ⊆ X is a set, then ∃B ⊆ Ythat is a set such that t ∈ B if and only if t = f (a) for some a ∈ A.

3.1.2 The Natural Numbers

1 Theorem 10 (Induction Principle)


Suppose J ⊆ ω, 0 ∈ J and whenever n ∈ J, S(n) ∈ J. Then J = ω.

Ò ProofBy assumption, J is a successor set, therefore ω ⊆ J by definition.

Thus, sinec J ⊆ ω, we have J = ω. �

+ Lemma 11 (Properties of the Natural Numbers)

Suppose n ∈ ω. We have

1. n ⊆ ω;

2. ∀m ∈ n m ⊆ n;

3. n /∈ n;

4. n = 0 Y 0 ∈ n; and

5. y ∈ n =⇒ S(y) ∈ n Y S(y) = n.

Ò Proof1. Let4 4 We construct this J and show that it is

a successor set. Note that if J = ω, ourproof is complete.J := {n ∈ ω : n ⊆ ω} ⊆ ω.

Note that ∅ ⊆ ω and so 0 ⊆ ω. By membership, 0 ∈ J.

Suppose m ∈ J. Consider S(m) = m ∪ {m}. Since J ⊆ ω, m ∈ ω.

Since m ∈ ω, {m} ⊆ ω. Therefore S(m) = m ∪ {m} ⊆ ω,

and so S(m) ∈ J. So J is a successor set. And thus by Induction

Principle, J = ω.

2. Let

J := {n ∈ ω : ∀m ∈ n, m ⊆ n}.

It is vacuously true that 0 ∈ J since ∅ is a subset of every n ∈ J.

Suppose n ∈ J. Then ∀m ∈ n, we have m ⊆ n. Consider

S(n) = n ∪ {n}. Note that n ∈ S(n) and n ⊆ S(n). For x ∈ S(n)such that x 6= n, we must have that x ∈ n. By assumption,

x ⊆ n ⊆ S(n). Therefore, S(n) ∈ J, and so J is a successor set. By

the Induction Principle, J = ω.


3. Let

J := {n ∈ ω : n /∈ n}.

We have 0 = ∅ /∈ ∅. So 0 ∈ J.

Let n ∈ J. Consider S(n) = n ∪ {n}. In particular, note that

n ∈ S(n). Suppose, for contradiction, that S(n) ∈ S(n). Then

S(n) = n or S(n) ∈ n.

S(n) = n =⇒ n ∈ S(n) = n En /∈ n.

S(n) ∈ n =⇒ S(n) ⊆ n by part 2 =⇒ n ∈ n En /∈ n.

Thus S(n) /∈ S(n) and so S(n) ∈ J. So J is a successor set, and so

by the Induction Principle, J = ω.

4. It suffices to show that

ω = {0} ∪ {n ∈ ω : 0 ∈ n}.

Let J = RHS. We have that 0 ∈ J. Suppose n ∈ J such that

n 6= 0. Then 0 ∈ n. Since n ⊆ S(n) = n ∪ {n}, we have that

0 ∈ S(n). Therefore, S(n) ∈ J. So J is a successor set, and so by

the Induction Principle, J = ω as required.

5. Let

J := {n ∈ ω : y ∈ n =⇒ S(y) ∈ n Y S(y) = n}.

0 ∈ J is vacuously true, since there are no y ∈ 0. Suppose n ∈ J.

Let y ∈ S(n) = n ∪ {n}. We have two choices: either y ∈ n or

y = n. If y ∈ n, then S(y) ∈ n Y S(y) = n, since n ∈ J. We have

that

S(y) ∈ n ⊆ S(n) in which case we are done; and

(sy) ⊆ n ∈ S(n).

Otherwise, if y /∈ n, then y = n. Then we simply have S(y) =

S(n). Thus J is a succesor set and so by the Induction Principle,

J = ω.

�

3.1.3 Well-Orderings


� Definition 10 (Strict Partially Ordered Set)

A strict partially ordered set (or strict poset5) is a set E together with 5 This is my unofficial terminology

R ⊆ E2 = E× E such that

1. (anti-reflexive) ∀a ∈ E (a, a) /∈ R;

2. (anti-symmetric) ∀a, b ∈ E (a, b) ∈ R ∧ (b, a) ∈ R =⇒ a = b;and

3. (transitivity) ∀a, b, c ∈ E (a, b), (b, c) ∈ R =⇒ (a, c) ∈ R.

� Definition 11 (Strict Totally Ordered Set)

A strict poset is total (or linear) if

∀a, b ∈ E (a, b) ∈ R Y (b, a) ∈ R

� Definition 12 (Well-Order)

A strict linear order is well-ordered if

∀X ⊆ E(X 6= ∅) ∃a ∈ X ∀b ∈ X(b 6= a) (a, b) ∈ R

i.e. every nonempty subset of E has a least element.

We shall prove the following next lecture.6 6 Anti-reflexivity and Anti-symmetrywere proven in this lecture, but Iam moving it to the next for ease ofreading.

0 Proposition (ω is Strictly Well-ordered)(ω,∈) is a strict well-ordering.



4.1.1 Well-Orderings (Continued)

+ Lemma (Lemma 11)Suppose n ∈ ω. We have

1. n ⊆ ω;

2. ∀m ∈ n m ⊆ n;

3. n /∈ n;

4. n = 0 Y 0 ∈ n; and

5. y ∈ n =⇒ S(y) ∈ n Y S(y) = n.

0 Proposition 12 (ω is Strictly Well-Ordered)

(ω,∈) is a strict well-ordering.

Ò ProofBy Lemma 11, we have that ∀n ∈ ω, n /∈ n. (anti-reflexivity Ë).

∀n, m ∈ ω, suppose, for contradiction, that n ∈ m and m ∈ n.

Again, by Lemma 11, we have n ⊆ m and m ⊆ n, which implies

that n = m. Thus, we have n ∈ m = n and m ∈ n = m, a

contradiction to the fact that n /∈ n and m /∈ m (anti-symmetry Ë).

∀x, y, z ∈ ω such that x ∈ y and y ∈ z, by Lemma 11, y ∈ z =⇒y ⊆ z =⇒ x ∈ z (transitivity Ë).

To show totality of the relation, let n ∈ ω. WTS for any m ∈ ω,

either

m ∈ n, m = n, or n ∈ m.

Let1 1 We construct J such that J will containall the possible cases, and use this factto prove that J = ω so these 3 cases arethe only scenarios that can happen.

J = n∈n∪ {n}

=n∪ {m ∈ ω : n ∈ m}

>n.

Case 1: n = 0. In this case, we have22 Note that 0 = ∅.

J = ∅ ∪ {∅} ∪ {m ∈ ω : 0 ∈ m}

As a consequence of Lemma 11 (4), we have that J = ω.


Case 2: n 6= 0. Again, by Lemma 11 (4), since n 6= 0, we must have

0 ∈ n ⊆ J and so 0 ∈ J. Now suppose that m ∈ J.

Case 2(a): m ∈ n. Then by Lemma 11 (5), S(m) ∈ n or S(m) = n.

S(m) ∈ n =⇒ S(m) ∈ JS(m) ∈ n =⇒ S(m) ∈ J

Case 2(b): m = n. Then S(m) = S(n) = n ∪ {n}. And so n ∈S(m), which implies S(m) ∈ J.

Case 2(c): n ∈ m Then since S(m) = m ∪ {m}, we have that

m ∈ m ⊆ S(m). Therefore S(m) ∈ J.

Therefore, J is a sucessor subset of ω. Thus by the Induction

Principle, J = ω. (totality Ë)

To prove that ∈ is a well-ordering, suppose X ⊆ ω is non-empty.

Suppose, for contradiction, that X has no ∈-least element. Now

consider

J = {n ∈ ω : S(n) ∩ X = ∅}

Claim: J is a successor set.3 3 Since we want to prove that ∈ is awell-ordering, we can suppose thatthere is a non-empty subset of ωthat is not empty, and has no ∈-leastelement. The core idea here is that, bythe construction of J, if J = ω, then allelements of ω would be disjoint fromX, forcing X to be the empty set.

By Lemma 11 (4), 0 is the ∈-least element of ω. If 0 ∈ X, then 0

would be ∈-least in X, contradicting our supposition. Thus 0 /∈ X,

And so

S(0) ∩ X = (0∪ {0}) ∩ X = {0} ∩ X = ∅

since 0 /∈ X. Thus 0 ∈ J.

Suppose n ∈ J. By construction of J, we have S(n) ∩ X = ∅.

Observe that

S(S(n)) ∩ X = (S(n) ∪ {S(n)}) ∩ X.

Now if RHS of the above is non-empty (aiming for contradiction),

then we may have S(n) ∈ X. Then S(n) would be the ∈-least

element in X, a contradiction. If m ∈ S(n), we have that m /∈ Xsince S(n) ∩ X = ∅. Thus SS(n) ∩ X = ∅ and so S(n) ∈ J.

Therefore, by the Induction Principle, J = ω.

We observe that ∀n ∈ ω,

∅ = S(n) ∩ X) = (n ∪ {n}) ∩ X

=⇒ n /∈ X, and so we must have X = ∅ (well-ordered Ë). �


å NoteGiven n, m ∈ ω, we often write n < m to mean n ∈ m.

� Definition 13 (Ordinals)

An ordinal is a set α satisfying:

1. x ∈ α =⇒ x ⊆ α;

2. (α,∈) is a strict well-ordering.

Example 4.1.1

ω is an ordinal: ∀n ∈ ω, by Lemma 11, n ⊆ ω, and ω is proven to

have a strict well-ordering under ∈.

Example 4.1.2

Every natural number is an ordinal (finite ordinals): by Lemma 11 (2),

the first property is satisfied; well-ordering follows from the property

of ω.

Let Ord denote the class of all ordinals. We shall show later that

Ord is a proper class.

Exercise 4.1.1

Verify that for a set to be an ordinal is a definite condition.

Observe that

∀t(t ∈ Ord↔ ((x ∈ t =⇒ x ⊆ t) ∧ ((t,∈) is a strict well-ordering )))

where (x ∈ t =⇒ x ⊂ t) is the definite condition

∀x(x ∈ t→ ∀a(a ∈ x → a ∈ t))

and (t,∈) is a strict well-ordering is the definite condition

∀s(s ⊆ t ∧ s 6= ∅→ ∃a(a ∈ s→ ∀b(b ∈ s ∧ b 6= a→ (a, b) ∈ (∈))))


+ Lemma 13 (Proper Subsets of an Ordinal Are Its Elements)

If α, β ∈ Ord and α ( β, then α ∈ β.

Ò ProofWe shall prove that α is the least element in β that is not in α it-

self.4 4 We shall construct a subset of β \ αand show that α is its element.

Let D := β \ α = {x ∈ β : x /∈ α} ⊂ β 5. Since α ( β, D 6= ∅. 5 Exists by Bounded Separation Axiom.

Since β ∈ Ord, (β,∈) has a strict well-ordering, and so D has a

least element, d. Note that d ∈ β, and since β ∈ Ord, d ⊆ β.

Claim: α = d.6 WTS α ⊆ d. ∀x ∈ α, we have x, d ∈ β. Then since 6 If α = d, then α is the said leastelement.

(β,∈) is a strict well-ordering, we have either

x < d, x = d, or d < x

Note that x 6= d, otherwise x = d ∈ D = β \ α.

7 If d < x, then d ∈ x (by our notation). Now since α ∈ Ord, 7 This is an errorneous proof.

Warning

d < x ∧ x < α

=⇒transitivity

d < α =⇒ d ∈ α

This argument is errorneous becausewe do not yet know if α ∈ β.

x < α =⇒ x ∈⊆, and so d ∈ α, which is yet another contradiction

(d ∈ D = β \ α).

Thus we must have x < d, i.e. x ∈ d. So α ⊆ d.

WTS d ⊆ α. Suppose not. Then let x ∈ d \ α. Then since d ∈ D =

β \ α, we have x ∈ β \ α, which then contradicts the minimality of d.

Therefore, d = α as required. �

0 Proposition 14 (Properties of Ordinals) Some of proofs of these properties areavailable in the course notes.Exercise 4.1.2Prove Item 3, Item 4, and Item 5 of0 Proposition 14.

1. Every member of an ordinal is an ordinal.

2. α ∈ Ord =⇒ α /∈ α.

3. α ∈ Ord =⇒ S(α) ∈ Ord.

4. α, β ∈ Ord =⇒ α ∩ β ∈ Ord.

5. α, β ∈ Ord =⇒ α ∈ β ∨ α = β ∨ β ∈ α.

6. E ⊆ Ord a subset =⇒ (E,∈) is a strict well-ordering.8 8 I think that such an E need notbe an ordinal itself. For example,E = {1, 5, 10} ⊂ Ord, but 4 ∈ 5 and4 /∈ E, and so 5 ∈ E but 5 * E.


7. Ord is a proper class.

Ò Proof1. Suppose x ∈ β ∈ Ord. WTS x ∈ Ord, and we shall show that x

satisfies � Definition 13.

Since β ∈ Ord, x ∈ β =⇒ x ⊆ β. Thus (x,∈) is a strict well-

ordering (through inheriting the property). So it suffices to show

that y ∈ x =⇒ y ⊆ x. So let y ∈ x, and let t ∈ x 9. Observe that 9 To show that y ⊆ x, we need to showthat ∀t ∈ y, t ∈ x.

t ∈ y =⇒ t < y

y ∈ x =⇒ y < x

and t, y, x ∈ β ∈ Ord. Therefore, by transitivity, we have t < y <

x =⇒ t ∈ x.

2. Suppose not, i.e. α ∈ α. Then α ⊆ α ∈ Ord, and so (α,∈) is a

strict well-ordering, i.e. α /∈ α, a contradiction.

6. Suppose A ⊆ E and A 6= ∅. Let α ∈ A.

Case 1: α ∩ A = ∅. Then ∀β ∈ α =⇒ β /∈ A. Therefore α is

∈-least in A.

Case 2: a ∩ A 6= ∅. Let A′ = α ∩ A ⊆ α. Since α ∈ A ⊆ E ⊆ Ord,

we have (α,∈) is a strict well-ordering, and so A′ has a strict

well-ordering as well, and thus it must have a ∈-least element, x.

Then x is the ∈-least element in A.

7. If Ord is a set, then by Item 6, (Ord,∈) is a strict well-ordering.

Also, by Item 1, every element of Ord is a subset of Ord. There-

fore, Ord satisfies � Definition 13, and so Ord ∈ Ord, which

contradicts Item 2. Therefore Ord /∈ Set.

�



å NoteIf A, B ∈ Ord, we will write A < B to mean A ∈ B.

0 Proposition 15 (Properties of Ordinals 2)

1. If α ∈ Ord, then α < S(α), and there is nothing in between.

2. Let E ⊆ Ord, where E 6= ∅ is a set, and sup E := ∪E. Thensup E ∈ Ord, and it is a least upper bound for E.1 1 I noted down from the lectures that

this is "not necessarily strict", but Ido not remember what it means now.(Clarification required.)

Perhaps this related to my question;can E = sup E?

This is not necessarily true. If E /∈ Ord,then E 6= sup E.

3. If E ⊆ Ord is a subset, then there is a least ordinal that is not in E.

Exercise 5.1.1

Prove 0 Proposition 15 Item 3.

Recommended strategy: α ∈ Ordsuch that E ( α and take the leastelement of α \ E (which is non-empty).Prove that this least element is the leastordinal that is not in E.

You can take α = SS(sup E). Verifythat this works.

Ò Proof1. Since S(α) = α ∪ {α}, α ∈ S(α) and so α < S(α).

It suffices to show that ∀x < S(α), we have x ≤ α. Let x < S(α),i.e. x ∈ S(α). So x ∈ α or x = α, i.e. x < α or x = α.

2. By definition, ∀x ∈ E ⊆ Ord, we have that x ⊆ Ord. Since ∪E ⊆E, we have that ∪E ⊆ Ord is a subset. Thus by 0 Proposition 14

Item 6, (∪E,∈) is a strict well-ordering.

2 Suppose α ∈ ∪E, then ∃e ∈ E such that α ∈ e ⊆ E ⊆ Ord. So e

2 This part shows that ∪E is also anordinal.

is an ordinal and so α ⊆ e. 3 ∀x ∈ α, we have x ∈ e ∈ E, and so

3 Now we show that α ⊆ ∪E.x ∈ ∪E by definition. Therefore α ⊆ ∪E.

And so, we have shown that ∪E = sup E ∈ Ord.


Claim 1: sup E is an upper bound for E.

Suppose, for contradiction, that ∃e ∈ E such that sup E < e.

Then since sup E and e are both ordinals, we have sup E ∈ e ∈ E.

Then by definition of ∪, we have that sup E ∈ ∪E = sup E, but

by 0 Proposition 14 Item 2, sup E /∈ sup E, a contradiction.

Thus sup E is an upper bound as claimed.

Claim 2: sup E is the supremum (least upper bound).

∀α < sup E, we have that α ∈ sup E = ∪E, and so ∃e ∈ E such

that α ∈ e. Then α < e ∈ E, i.e. α is not an upper bound of E.

� Definition 14 (Successor Ordinal)

The successor ordinal is an ordinal of the form S(α) for some α ∈ Ord.

� Definition 15 (Limit Ordinal)

A limit ordinal is an ordinal that is not a successor.

Example 5.1.1

0 and ω are both limit ordinals; 0 is vacuosly a limit ordinal, and ω is

not a successor of any α ∈ Ord 4. 4 Need a more careful proof, which Icannot do. The idea is to show that anysuch ordinal α will be an element ofω, and so will its successor S(α), andω /∈ ω.

On the other hand, for n ∈ ω such that n 6= 0, ∃ ∪ n ∈ ω such that

S(∪n) = n 5.5 See A1.

Exercise 5.1.2

Prove that S(ω) is a successor ordinal.

Ò SolutionWe have that ω ∈ Ord, and so S(ω) is a successor ordinal.

5.1.1 Transfinite Induction & Recursion

1 Theorem 16 (Transfinite Induction Theorem v1)


Suppose P is a definite condition, with the property

∀α ∈ Ord∧(∀β < α P(β)) =⇒ P(α). (5.1)

Then P is true of all ordinals.

Ò ProofP(0) is vacuously true, since there are no elements that are less

than 0. Suppose P(α) is false for some α ∈ Ord such that α > 0. By

the Bounded Separation Axiom,

D := {β ≤ α : ¬P(β)}

is a set 6. Note that D 6= ∅, since α ∈ D. Since α ∈ Ord, we have 6 Note that β ≤ α ⇐⇒ β < α ∨ β =α ⇐⇒ β ∈ S(α)D ⊆ α ⊆ Ord, and so (D,∈) has a strict well-ordering. Let α0 ∈ D

be ∈-least. Then ∀β < α0, we have that ¬P(β), which contradicts

the assumption Equation (5.1). Thus P(α) is true for all ordinals. �

1 Theorem 17 (Transfinite Induction Theorem v2)

Suppose P is a definite condition satisfying This statement strongly resembles theInduction Princple that we have learntin the earlier years of university. Incontrast, v1 resembles Strong InductionPrinciple. It can be shown that v1 ⇐⇒v2. v1 =⇒ v2 is proven in this lecture.

Exercise 5.1.3Prove that 1 Theorem 17 =⇒1 Theorem 16.

1. P(0);

2. ∀β ∈ Ord P(β) =⇒ P(S(β)); and

3. If α ∈ Ord is a limit ordinal and ∀β < α, P(β), then P(α).

Then P is true of all ordinals.

Ò ProofIt suffices to show that P satisfies Equation (5.1), i.e. ∀α ∈ Ord, we

want to prove that ∀β < α, if P(β), then P(α).

When α = 0, we have P(0) and so Equation (5.1) is satisfies.

When α > 0 is a limit ordinal, our assumption immediately satis-

fies Equation (5.1). Now suppose α > 0 is a successor ordinal, and

suppose that α = S(γ) for some γ ∈ Ord. By the assumption in


Equation (5.1), we have that

∀β < γ P(β) =⇒ P(γ)

and so by condition (2), we have P(S(γ)) since γ ∈ Ord. Thus we

have P(α) = P(S(γ)). �

We shall prove the following in the next lecture:

1 Theorem (Transfinite Recursion)Let X be a class of all definite operations whose domain is an ordinal.Given a definite operation

G : X → Set

∃!F : Ord → Set, a definite operation, such that F(α = F(F �α)), for allα ∈ Ord.

We want to use Transfinite Recursion to construct definite opera-

tions on ordinals such that they have properties that we are familiar

with (and hence desire).

å Note (Notation - Restriction)Let H : U → Y be a definite operation on classes U, Y, and Z ⊆ U asubclass. H �Z is the definite operation

H �Z : Z → Y

obtained by restricting H onto Z.

å NoteIn the theorem, we stated that F has its domain on Ord. We know that forα ∈ Ord, α ⊆ Ord, and so F �α makes sense; in particular,

F �α: α→ Set .

Note that F �α∈ X, and so G(F �α) is valid and makes sense.



6.1.1 Transfinite Induction & Recursion (Continued)

1 Theorem 18 (Transfinite Recursion v1)

Let X be a class of all definite operations whose domain is an ordinal.Given a definite operation

G : X → Set

∃!F : Ord→ Set, a definite operation, such that F(α) = G(F �α)), for allα ∈ Ord.

Before proving the theorem, we shall note the following definition.

� Definition 16 (α-function)

Using definitions in 1 Theorem 18, a function t with domain in theordinals is called an α-function defined by G if

∀β < α t(β) = G(t �β).

Ò ProofWe shall first prove for uniqueness. Suppose F and F′ are two


definition operations such that

F : Ord→ Set F′ : Ord→ Set

F(α) = G(F �α) F′(α) = G(F′ �α)

1Suppose ∀β < α ∈ Ord, we have F(β) = F′(β). Note that 1 Here, we want to use TransfiniteInduction v1 to show that they areunique.

F(β) = F′(β) ⇐⇒ F �α= F′ �α

=⇒ F(α) = G(F �α) = G(F′ �α) = F′(α)

Thus, uniqueness of F is guaranteed.

To prove existence, firstly, we note that the α-functions defined

in � Definition 16 are approximations to the F that we want.

However, before going further, we need to show that they are

also unique and that we can form a chain of extensions on these

functions over Ord.

Uniqueness of tα Let t, t′ be a α-functions defined by G. WTS ∀β <

α, t(β) = t′(β). If an α-function defined by G exists, we shall

denote it as tα.

Consider the definite condition

P(x) := (x ≥ α) ∨ (t(x) = t′(x)).

Suppose that ∀γ < β, P(γ) holds, i.e. γ ≥ α or t(γ) = t′(γ), which

implies that t �β= t′ �β. Therefore t(β) = t′(β), i.e. P(β) holds.

Thus tα is unique if it exists by Transfinite Induction.

tα as a chain of extensions Now ∀β < α ∈ Ord, we have that β ⊆ α.

If tα and tβ exist, then

Γ(tβ) ⊆ Γ(tα),

or in other words

tα �β= tβ.

We shall denote this relation as tβ ⊆ tα.

Existence of tα The existence of tα is a definite condition: by the

Replacement Axiom, a function that maps α 7→ tα is definite, and

by Bounded Separation Axiom, the set

Γ(tα) = {(β, G(tα �β) | β < α, tα(β) = G(tα �β)} ⊆ Ord× Set


exists. 2Now t0 = t∅ is vacuously true. Suppose for any successor 2 We use Transfinite Induction v2 toCTP.ordinal α, tα exists. Since α ∈ Ord, we have that there is nothing

between α and its successor S(α), and α < S(α). Thus

tS(α) = tα ∪ {(α, G(tα))},

which exists by Union Set Axiom. We see that tS(α) extends tα onto

α itself.

Suppose α > 0 is a limit ordinal. Since tα is a definite condition

by Replacement, by Bounded Separation, we have that3 3 Verify the motivation in using or thereason behind getting this.

tα =⋃

β<α

tβ = ∪{tβ : β < α}.

Note that tα is, indeed, an α-function defined by G:

• tα is a function on α: we have that ∀β < α,the tβ’s form a chain

of extensions;

• tα is an α-function defined by G: ∀β < α, since α is a limit

ordinal, S(β) < α, and so

tα(β) = tS(β)(β) = G(tS(β) �α) = G(tβ �α).

And so by Transfinite Induction, ∀α ∈ Ord, tα exists as required.

Construction of F Now for any β < α, we have a chain of exten-

sions

t0 ⊆ t1 ⊆ t2 ⊆ . . . ⊆ tβ ⊆ tα ⊆ . . .

Let4 4 I will leave the proof unfinished here.Need to verify my understandingF :=

⋃α∈Ord

tα = ∪ J tα | α ∈ Ord K

� Corollary 19 (Transfinite Recursion v2)

Given G1 ∈ Set, G2 : Set → Set a definite operation, G3 : X → Set

a definite operation, where X is the class of all definite opeartion whosedomain is an ordinal. Then

∃!F : Ord→ Set

such that


1. F(0) = G1;

2. ∀α ∈ Ord F(S(α)) = G2(F(α)); and

3. ∀β > 0 a limit ordinal, F(β) = G3(F �β).

Ò ProofThe result is clear by 1 Theorem 16 with G : X → Set defined by

G( f ) =

G1 f = ∅

G2( f (α)) Dom( f ) = S(α)

G3( f ) Dom( f ) > 0 a limit ordinal

�

6.1.2 Ordinal Arithmetric

6.1.2.1 Ordinal Addition

� Definition 17 (Ordinal Addition)

Let β ∈ Ord. For any α ∈ Ord, we define

β + α

using Transfinite Recursion5 on α as follows: 5 Note that we are using� Corollary 19 with

G1 = β

G2 = S : Set→ Set

G3 : X → Set by G3( f ) = sup Img( f )

• β + 0 := β;

• if α is a successor ordinal, then β + S(α) := S(β + α); and

• if α > 0 is a limit ordinal, then β + α := sup{β + γ : γ < α}.

Exercise 6.1.1

Using both the Induction Principle and Transfinite Induction, prove thatβ + α ∈ Ord.

Example 6.1.1


We have that

0, 1, 2, ..., ω, ω + 1, ω + 2, ..., ω + ω

Observe that

ω + 1 = ω + S(0) = S(ω + 0) = S(ω)

and so ω + 1 is a successor to ω. In general, we have that ∀α ∈ Ord,

α + 1 = S(α). For example,

ω + 2 = ω + S(1) = S(ω + 1).

On the other hand, note that

ω + ω = sup{ω + n : n ∈ ω}.

Unlike regular addition, ordinal addition is not commutative. For

instance, while ω + 1 = S(ω),

1 + ω = sup{1 + n : n ∈ ω} = ω.

Exercise 6.1.2

Prove that ordinal addition is only commutative for “finite” ordinals.

6.1.2.2 Ordinal Multiplication

� Definition 18 (Ordinal Multiplication)

Let β ∈ Ord. For any α ∈ Ord, we define

β · α

using Transfinite Recusion6 as follows: 6 Here, we use

G1 = 0

G2 : Set→ Set by G2(x) = x + β

G3( f ) = sup Img( f )

• β · 0 := 0;

• if α is a successor ordinal, β · S(α) := βα + β

• if α > 0 is a limit ordinal, β · α := sup{β · γ : γ < α}.

Example 6.1.2

We have

ω · 1 = ω · S(0) = ω · 0 + ω = 0 + ω = ω.


Exercise 6.1.3

Prove that in general, ∀β ∈ Ord, we have β · 1 = β.

Exercise 6.1.4

Prove that ∀α, β ∈ Ord, α · β ∈ Ord.

Example 6.1.3

We have

ω · 2 = ω · S(1) = ω · 1 + ω = ω + ω.

Exercise 6.1.5

Prove that in general, ∀β ∈ Ord, we have β · 2 = β + β.

Note that ordinal multiplication, like its addition counterpart, is

not necessarily commutative.

Example 6.1.4

While we have

1 ·ω = sup{1 · n | n ∈ ω} = ω = ω · 1,

observe that

2 ·ω = sup{2 · n | n ∈ ω} = ω 6= ω + ω = ω · 2.

0 Proposition 20 (Properties of Ordinal Addition and OrdinalMultiplication)

Let α, β, δ ∈ Ord.

Exercise 6.1.6Prove 0 Proposition 20.

• α < β ⇐⇒ δ + α < δ + β;

• α = β ⇐⇒ δ + α = δ + β;

• ((associativity)) (α + β) + δ = α + (β + δ);

• if δ 6= 0, then α < β ⇐⇒ δα < δβ;

• if δ 6= 0, then α = β ⇐⇒ δα = δβ;

• (αβ)δ = α(βδ).


6.1.2.3 Ordinal Exponentiation

� Definition 19 (Ordinal Exponentiation)

Let β ∈ Ord. For any α ∈ Ord, define

βα

using Transfinite Recursion by

• β0 := 1;

• if α is a successor ordinal, then βS(α) := βα · β;

• if α > 0 is a limit ordinal, then βα := sup{βγ | γ < α}.

In the next lecture, we shall study the following theorem:

1 Theorem (Strict Well-Ordered Sets are Isomorphic to a Unique Ordinal)Every strict well-ordering is isomorphic to an ordinal. Both the ordinaland the isomorphism are unique.

The definition of isomorphism is:

� Definition 20 (Isomorphism)

Let E, F be sets and R, S be relations defined on each set respectively so.We say that (E, R) and (F, S) are isomorphic, which we denote by

(E, R) ' (F, S),

if ∃ f : E→ F, a bijection, such that

e1Re2 ⇐⇒ f1S f2.

Such an f is called an isomorphism.



7.1.1 Well-Orderings and Ordinals

Before proving the theorem stated at the end of last lecture, we re-

quire the following 2 lemmas:

+ Lemma 21 (Rigidity of Well-Orderings)

Well-orderings are rigid, i.e. the only automorphism1 is the identity. 1 An automorphism is an isomorphismfrom a set to itself.

Ò ProofSuppose (E,<) is a well-ordering, and f : E → E an automor-

phism. Let2 2 We look at the fellas that were‘moved’.D = {x ∈ E | f (x) 6= x}.

Suppose for contradiction that f is not the identity map, i.e. D 6=∅. Then, since D ⊆ E, D has a well-ordering, and so we can pick

α ∈ D to be the least element.

Case 1: f (a) < a. Then f (a) /∈ D since a is least. But then that

would mean

f ( f (a)) = f (a) =⇒ f (a) = a

since f is a bijection. This contradicts the choice that a ∈ D.

Case 2: a < f (a). Since f is a bijection, its inverse exists, and so

f−1(a) < a =⇒ f−1(a) /∈ D

⇐⇒ f f−1(a) = f−1(a) ⇐⇒ a = f−1(a)


which contradicts the choice that a ∈ D, yet again. Thus there is no

such element a ∈ D, forcing D = ∅, and so f must be the identity

map. �

+ Lemma 22 (Strict Well-Ordering 6' Any of Its Proper InitialSegment)

A strict well-ordering is not isomorphic to any proper initial segment3 of 3

� Definition 21 (Initial Segment)For a strict well-order (E,<), an initialsegment is a subset of the form

{x ∈ E : x < b}

for some b ∈ E.

itself.

Ò ProofLet (E,<) be a strict well-ordering. ∀b ∈ E,

Ib := {x ∈ E : x < b}

has an induced well-order, in particular (Ib,<).

Suppose for contradiction that there exists an isomorphism

f : E→ Ib. Let

D = {x ∈ E | f (x) 6= x}

b /∈ Ib and so b /∈ D. Proof is left incomplete until I verify theproof with the prof.

1 Theorem 23 (Strict Well-Ordered Sets are Isomorphic to aUnique Ordinal)

Every strict well-ordering is isomorphic to an ordinal. Both the ordinaland the isomorphism are unique.

Ò ProofUniqueness

Let (E, R) be a strict well-ordering. Suppose that

(E, R) ' (α,∈) and (E, R) ' (β,∈) (7.1)


where α, β ∈ Ord. Then, either

α < β, α = β or β < α.

Suppose α < β (this argument also works for β < α, by simply

swapping the inequality on α and β). Then α is a proper initial

segment of β. From Equation (7.1), we have that (α,∈) ' (β,∈) via

(E, R), but this contradicts Lemma 22. Thus we must have α = β.

Existence

If E = ∅, then we can choose 0 ∈ Ord to be the ordinal of which

E is isomorphic to. Suppose E 6= ∅. Denote an initial segment of Eby

Ix := {y ∈ E | y < x}.

Let

A = {x ∈ E | ∃β ∈ Ord (Ix, R) ' (β,∈)}.

Notice that (Ix, R) ' (β,∈) is a definite condition: the both Ix and

β are sets, and so by Replacement, there is a graph, Γ( f ), from Ix to

β; injectivity of an element in Γ( f ) is expressible as

∀y1∀y2∀β1∀β2(y1, y2 ∈ Ix ∧ β1, β2 ∈ Ord((y1, β1) = (y2, β2)↔ y1 = y2));

surjectivity is expressible as

∀β(β ∈ α→ ∃y(y ∈ Ix → (y, β) ∈ Γ( f ))).

Thus by Bounded Separation, A is a set. Also, A is nonempty, since

the least element of E will be isomorphic to 0.

By our uniqueness proof above, let f be a function on A, where

f (x) is the unique ordinal that is isomorphic to (Ix, R). By Replace-

ment,

Img( f ) = { f (x) ∈ Ord | x ∈ A} ⊂ Ord

is a set. By 0 Proposition 15 Item 3, ∃α ∈ Ord \ Img( f ) that is ∈-

least. We want to show that f : A → Img( f ) is an isomorphism

between (E, R) and (α,∈).4 4 This is why we need to show that

1. f is order-preserving, which isone of the requirements of anisomorphism (by our definition in� Definition 20);

2. f is injective;

3. α = Img( f );

4. A = E,

where the last 2 items will force f to besurjective.

f is order-preserving: We shall also show here that A is down-ward closed. ∀x, y ∈ E, we want to show that xRy ∧ y ∈ A =⇒x ∈ A. By the assumption, we have that

f (x)h' IX ( Iy

h' f (y)


Since f (x), f (y) ∈ Ord, we have either

f (x) < f (y), f (x) = f (y), or f (y) < f (x).

If f (x) = f (y), then Ix ' Iy which contradicts Lemma 22. If

f (y) < f (x), then

h(Ix) ⊆ h(Iy) = f (y) < f (x) = h(Ix),

which is a contradiction. Thus we must have f (x) < f (y), i.e. fpreserves order as claimed, and f (x) is an initial segment of f (y),i.e. f (x) ∈ Ord, which implies that x ∈ A.

α = Img( f ): Suppose β ∈ α =⇒ β < α =⇒ β ∈ Img( f ) by

choice of α being the least. Therefore, α ∈ Img( f ).

Now suppose that β ∈ Img( f ). Then ∃x ∈ A such that (Ix, R) '(β,∈). Again, we have 3 possibilities; either

α < β, α = β or β < α.

Now α < β =⇒ α ∈ β =⇒ α = h(Iy) where yRx and Iy ⊂ IX .

Since A is downward closed, α ∈ Img( f ), a contradiction. We also

have that α 6= β since β ∈ Img( f ) and α ∈ Ord \ Img( f ), i.e.

α /∈ Img( f ). Thus β < α, and so β ∈ α. Therefore α = Img( f ).

f is injective: Suppose that f (x) = f (y). xRy =⇒ Ix is an

initial segment of Iy, which contradicts Lemma 22. The argument is

similar for if yRx. Thus we must have x = y.

A = E: It suffices to show that E \ A = ∅. Suppose for contradic-

tion that E \ A 6= ∅, i.e. E = A. Then ∃x ∈ E \ A, since E \ A ⊂ Ewhich has a strict well-ordering. Since f preserves order, for any

y ∈ E, xRy =⇒ y /∈ A. On the other hand, yRx =⇒ y ∈ A. Thus

Ix = A. However, since f is an isomorphism between (A, R) and

(α,∈) (since we assume that A = E), we have that Ix = A ' α, i.e.

x ∈ A, which is a contradiction to the choice that x ∈ E \ A.

This completes the proof. �


7.2 Cardinals

While ordinals allow us to enumerate, we cannot use it to “measure”.

For example, ω 6= ω + 1, but they have the same size.

� Definition 22 (Equinumerous)

Two sets A and B have the same size, or equinumerous, if there is abijection from A to B. We denote this relation by |A| = |B|.5 5 Note that we have yet to define |·|.

The following is a well-known theorem that makes proving equinu-

merosity a lot easier.

+ Lemma 24 (Schröder-Bernstein Theorem)

Given two sets A and B, |A| = |B| if and only if there exists injections inboth directions, i.e. an injection from A to B, and an injection from B toA.

Ò ProofThe ( =⇒ ) direction is easy, since a bijection exists. So it suffices to

show the ( ⇐= ) direction. We shall use A ↪→ B to say that there is

an injection from A to B.

Suppose

A ↪→ Bg↪→ A

Then ∃ f : A→ A an injective map, and we would have

f (A) ⊆ g(B) ⊆ A. (7.2)

From here, it suffices to show that for an injective ap f : X → X, if

we have

f (X) ⊆ Y ⊆ X,

then |Y| = |X|. From our observation in Equation (7.2), we have

X ⊇ Y ⊇ f (X) ⊇ f (Y) ⊇ f 2(X) ⊇ f 2(Y) ⊇ f 3(X) ⊇ . . .

60 Lecture 7 Sep 27th - Cardinals

Let

Z = X \Y·∪ f (X) \ f (Y)

·∪ f 2(X) \ f 2(Y)

·∪ . . .

and

W = X \ Z

Then

X = Z·∪ W

Claim: For sets A and B such that B ⊆ A, we have that6 6 Forgive me if the proof of this claim isa little sloppy.

f (A \ B) = f (A) \ f (B).

Note that since f is injective, f : B → f (B) is a bijective map.

Suppose ∃x ∈ A \ B such that f (x) ∈ f (B). Since f : B → f (B) is

bijective, ∃b ∈ B such that f (b) = f (x), but f is injective. Thus the

claim is true.

Using a similar argument, it can be shown that f (A·∪ B) =

f (A)·∪ f (B).

Observe that

f (Z) = f (X \Y)·∪ f ( f (X) \ f (Y))

·∪ f ( f 2(X) \ f 2(Y))

·∪ . . .

= f (X) \ f (Y)·∪ f 2(X) \ f 2(Y)

·∪ f 3(X) \ f 3(Y)

·∪ . . .

and note that

f (Z) = Z \ (X \Y).

Since W = X \ Z, we have W ⊆ Y. Since Z ∩W = ∅, we still have

f (Z) ∩W = ∅. Also, note that (X \Y) ∩W = ∅. Thus, we have

Y = X \ (X \Y) = (Z·∪ W) \ (X \Y) = f (Z)

·∪ W

Let g : X → Y such that

g(A) =

A A ⊆W

f (A) A ⊆ Z

Clearly so, g is bijective. �

Example 7.2.1

We claimed that |ω| = |ω + 1|.


We can simply use the identity map from ω → ω + 1. For ω + 1→ω, consider the mapping

f (α) =

S(α) α ∈ ω

0 α = ω

This map is clearly injective by properties of elements of ω.

� Definition 23 (Cardinal)

A cardinal is an ordinal α with the property that ∀β < α, |α| 6= |β|

In the next lecture, we shall see that the collection of cardinals is a

proper class, and is a subclass of the ordinals.

� Definition 24 (Finite & Countable)

A set A is finite if |A| = |n| for some n ∈ ω. A is countable if A isfinite or |A| = |ω|.

8 Lecture 8 Oct 02nd

8.1 Cardinals (Continued)

å NoteIf κ ∈ Card and κ is infinite, then κ is a limit ordinal. In other words,successor ordinals are either finite or are not Cardinals. This is true since∀α ∈ Ord such that α ≥ ω, clearly we have α ↪→ S(α), and we candefine a function f : S(α)→ α such that

f (β) =

S(β) β < ω

β ω ≤ β < α

0 β = α

,

which is injective, and so by Schröder-Bernstein, |S(α)| = |α|.

0 Proposition 25 (The Least Cardinality Not Equinumerous toSubsets of a Set)

∀E ∈ Set ∃α ∈ Ord ∀e ⊆ E

|e| 6= |α|

and there exists a least such h(E) ∈ Ord.

å NoteNote that h(E) ∈ Card.

64 Lecture 8 Oct 02nd - Cardinals (Continued)

Ò ProofSuppose not, i.e. ∃α ∈ Ord such that |h(E)| = |α| and α < h(E).Since h(E) is the least, we must have |α| = |A| or some A ⊆ E.

Then |h(E)| = |A|, which is a contradiction to the definition of

h(E). �

In particular, we have that h(ω) is an uncountable cardinal.

Now to prove 0 Proposition 25.

Ò ProofIt suffices to prove the existence of h(E). Consider the class

H = J α ∈ Ord | ∃e ⊆ E |e| = |α| K .

If we can show that H is a set, then the least element not in H shall

be our h(E). Consider

W := {(A, R) | A ⊆ E, (A, R) is a strict well-ordering }

which is a set by Replacement, i.e.

W ⊆ P(E)×P(E× E).

Note that W 6= ∅, since ∅ ⊆ E and the empty relation would be

a well-ordering of ∅. By 1 Theorem 23, ∃ f : W → Ord such that

f (A, R) is a unique ordinal. By Replacement, Img( f ) ⊆ Ord is a

set.

Claim: Img( f ) = H: It is clear that Img( f ) ⊆ H, since all ele-

ments of Img( f ) are isomorphic to some subset of E by definition.

Now let α ∈ H. Then ∃g : α → A a bijection, for some A ⊆ E. Then

define ≺ on A by

a ≺ b ⇐⇒ g−1(a) < g−1(b).

Then (α,<)g→'

(A,≺). Therefore, (A,≺) ∈ W and f (A,≺) = (α,<

), i.e. H ⊆ Img( f ). Thus H = Img( f ) and so H is a set as required.

�


RemarkThis revelation tells us that Card is a proper class.

To use Card to measure all sets, we need every set to be equinu-

merous with a cardinal. In particular, every set would then have to

be equinumerous to an ordinal. This would then require evrey set to

have a strict well-ordering, which is something that we cannot prove

with our axioms thus far.

8.1.1 Axiom of Choice

� Definition 25 (Choice Function)

Suppose F is a set. A choice function on F is a function

c : F → ∪F such that ∀F ∈ F c(F) ∈ F.

å NoteIf ∅ ∈ F , then F has no choice function, since nothing belongs in ∅.

� Axiom 26 (Axiom of Choice)

Every F ∈ Set such that ∅ /∈ F admits a choice function.1 1 This is, again, an existential axiom.

å NoteUnlike the other axioms, while the Axiom of Choice asserts the existenceof choice functions on sets, the choice function need not be unique. Re-call that in other axioms, the sets of which we assert their existence areunique.

1 Theorem 27 (Axiom of Choice and Its Equivalents)

TFAE


1. Axiom of Choice

2. Well-ordering Principle: Every set admits a well-ordering.

3. Zorn’s Lemma: If (E, R) is a strict poset with the property that everytotally ordered subset of E has an upper bound, i.e.

∀A ⊂ E(∀a, b ∈ A aRb∨ bRa∨ a = b) ∀a ∈ A∃e ∈ E(aRe∨ a = e).

Then (E, R) has a maximal elements, i.e. ∃z ∈ E such that ∀x ∈ E,¬zRx.

Ò Proof(1) =⇒ (2): Let A ∈ Set. If A = ∅, then there is nothing to

do and the statement is vacuously true. So suppose A 6= ∅. By

the asusmption, fix a choice function c on F := P(A) \ {∅}. Let

θ ∈ Ord \A. Define a definite operation F : Ord→ Set such that

F(α) =

c(A \ Img(F �α)) A \ Img(F �α) 6= ∅

θ otherwise

Note that F exists by Transfinite Recursion2. 2 I am not sure how.

Claim 1: F halts, i.e. ∃α ∈ Ord such that F(α) = θ and ∀β ∈ Ord

such that α < β, F(β) = θ.

Suppose not. Then F must be injective and has codomain ∪F =

∪(P(A) \ {∅}) = A, i.e. we have that F : Ord → A injective.

Then by 0 Proposition 25, there exists h(A) that is the least ordinal

that is not equinumerous with any subset of A. We may consider

F �h(A): h(A) → A since h(A) ⊂ Ord. Now ∀α < β ∈ h(A), by

our supposition, F(β) 6= θ, and so F(β) = c(A \ Img(F �β)) ∈(A \ Img(F �β)). Thus F(β) /∈ Img(F �β). But since α < β, it must

be that F(α) ∈ Img(F �β). Then F(α) 6= F(β). Consequently, since

∀β ∈ h(A), we have that F(β) 6= θ, and so we created an injection

from h(A) to A. This is impossible by the definition of h(A). Thus

it must be the case that ∃β ∈ h(A) such that F(β) = θ.

Let α be the least such β. The previous paragraph showed that

F �α is an injection from α to A. It remains to show that the map is


surjective. Suppose Img(F �α) 6= A. Then A \ Img(F �α) 6= ∅. Then

F(α) = c(A \ Img(F �α)) ∈ A

which is a contradiction since F(α) = θ. So A = Img(F �α) and so

F � α is surjective.

Therefore F �α is a bijection from α to A, and so A has an in-

duced strict well-ordering from α.

(3) =⇒ (1): Let F be a set such that ∅ /∈ F . Let Λ be the set of

all partial choice functions on F , identified with their graphs, i.e.

∀ f ∈ Λ, f : G → ∪G such that f (G) ∈ G for all G ∈ G, and G ⊆ F .

Note that Λ is indeed a set since the graphs exist by Replacement,

and Λ is therefore a set from Bounded Separation. Λ 6= ∅, since

the function f (F) = x exists for F ∈ F 6= ∅, and F 6= ∅.

Now (Λ,⊆) is a poset, where we order the functions by exten-

sions. For every Θ that is a totally ordered subset of Λ, we have

that the union, ∪Θ is the upper bound of Θ. Thus the assump-

tions for Zorn’s Lemma are satisfied, and so there exists a maximal

function

f : G → ∪G in Λ.

To prove that this f is a choice function on F , we want to prove

that f : F → ∪F , i.e. we need to show that Dom( f ) = F . Suppose

not, i.e. ∃F ∈ F such that F /∈ Dom( f ). Then since F 6= ∅,

∃x ∈ F, and so f ∪ {(F, x)} is a larger partial choice function on

F , contradicting the maximality of f in Λ. Thus Dom( f ) = F as

claimed, and so f is a choice function on F , proving the Axiom of

Choice.

(2) =⇒ (3): Suppose (E, R) is a strict poset. By (2), let < be a

strict well-ordering on E. Now by 1 Theorem 23, ∃!α ∈ Ord such

that (E, R) ' (α,∈) through a unique isomorphism. Therefore,

we may assume that E ∈ Ord. Suppose that (E, R) satisfies the

assumptions of Zorn’s Lemma.

Assume, to the contrary, that (E, R) does not have an R-maximal

element. From 0 Proposition 25, ∃h(E) ∈ Ord such that ∀β < h(E),∀A ⊆ E, |β| 6= |A|3. Recursively so, define F : h(E)→ E by 3 The strategy here is to use, once

again, 0 Proposition 25 to arrive at acontradiction that is similar to when wewere proving (1) =⇒ (2).F(0) = e for some e ∈ E

F(S(β)) = < -least element γ of E such that F(β)Rγ


and for β > 0 a limit ordinal,

F(β) =

< -least γ such that F(ζ)Rγ for all ζ < β if such γ exists

e otherwise

Note that this function is well-defined, since F(β)Rγ properly

distinguishes the ordinals, for there are no R-maximal element in

E, and < is a strict well-ordering on E.

Now since h(E) ∈ Card, it is a limit ordinal, to show that F is

injective, it suffices to show that ∀β < h(E), F �β is strictly order-

preserving, i.e. ∀x < y < β, F(x)RF(y). We shall prove this by

Transfinite Induction.

β = 0 is vacuously true. If β > 0 is a limit ordinal, then β =⋃γ<β γ, and so the strict ordering is preserved as given by the

induction hypothesis. For β a successor ordinal, consider F �S(β).

∀x < y < S(β), if y 6= β, then we are done by the induction

hypothesis. Suppose y = β. Since β is a successor ordinal, ∃γ ∈Ord such that S(γ) = β. Since γ < S(γ) (by 0 Proposition 15

Item 1), either x = γ or x < γ. If x < γ, then our proof is complete

by the induction hypothesis. If x = γ, then since x = γ < S(γ) = β,

regardless if γ is a limit ordinal or successor ordinal, we have that

F(γ)RF(β).

Thus, by Transfinite Induction, we have that F �β is strictly

order-preserving for any β < h(E), implying that F : h(E) ↪→ E,

hence contradicting the definition of h(E). Thus, an R-maximal

must exist. �

9 Lecture 9 Oct 04th

9.1 Cardinals (Continued 2)

9.1.1 Axiom of Choice (Continued)

From hereon, unless stated otherwise, we shall assume AC.

0 Proposition 28 (Using Cardinals to Measure Sets)

Assume AC. Every set is equinumerous with a cardinal.

Ò ProofLet A ∈ Set. By the Well-Ordering Principle, A is well-orderable,

i.e. there is a strict well-ordering on A. By 1 Theorem 23, ∃!α ∈Ord such that (A,<) ' (α,∈). Let

S = {β ≤ α | |β| = |α|},

which is a set by Bounded Separation. Note that S 6= ∅ since

α ∈ S. Let β be the least such ordinal in S. By this minimal choice,

β ∈ Card, |β| = |α| = |A|. �

And now our notation of |A| = |B| makes sense provided the

following definition.

� Definition 26 (Cardinality)

Let A ∈ Set. |A|, in which we shall call the cardinality of A, is the(unique) cardinal which is equinumerous with A.

70 Lecture 9 Oct 04th - Cardinals (Continued 2)

0 Proposition 29 (Lesser Cardinality)

∀A, B ∈ Set |A| ≤ |B| ⇐⇒ ∃ f : A ↪→ B.

Ò ProofLet κ = |A| and λ = |B|. If κ ≤ λ, then

Abijection→ κ

id↪→ λ

bijection→ B

By composition of the 3 functions, A ↪→ B.

Conversely, suppose ∃h : A ↪→ B. Suppose to the contrary that

λ < κ. Then λ ⊆ κ. Then

κbijection→ A

h↪→ B

bijection→ λ

and so there exists an injection from κ → λ. By Schröder-Bernstein,

we have |κ| = |λ|, which contradicts the fact that κ ∈ Card. Thus

κ ≤ λ. �

� Corollary 30 (Cardinalities are Always Comparable)

∀A, B ∈ Set A ↪→ B ∨ B ↪→ A.

Ò ProofWLOG, suppose ¬(B ↪→ A). Then ¬(|B| ≤ |A|), i.e. |A| < |B|, i.e.

(not by 0 Proposition 29), ∃ f : A ↪→ B. �

0 Proposition 31 (Functions are “Lossy Compressions”)

Suppose f : A→ B. Then |Img( f )| ≤ |A|.


Ò ProofLet

F = { f−1(y) | y ∈ Img( f )}

be the set of fibres1 of f . Note that ∅ /∈ F , as otherwise we 1

� Definition 27 (Fibres)Let f : A → B. For y ∈ Img( f ), thefibre of y is defined as

f−1(y) = {x ∈ A | f (x) = y}.

The fibres are also commonly called thepullback.

would be saying that ∃y ∈ Img( f ) that has no pre-image. Let

h : Img( f )→ A by

h(y) := c( f−1(y)) ∈ A

where c : F → ∪F is a choice function that exists by AC. Clearly

so, h is injective, and so by 0 Proposition 29, |Img( f )| ≤ |A|. �

0 Proposition 32 (Countable Union of Countable Sets is Count-able)

Let A ∈ Set be countable, and every a ∈ A is also countable. Then ∪A iscountable.

Ò Proofto be added

9.1.2 Hierarchy of Infinite Cardinals

å Note (Notation)Let κ ∈ Card. Let κ+ := h(κ), which is the least ordinal not equinu-merous with any subset of κ. We proved that ∀E ∈ Set, h(E) ∈ Card.Therefore, κ+ ∈ Card.

Remarkκ+ is the least cardinal that contains κ. This follows immediately from thedefinition of h(κ).

And so we observe that κ 7→ κ+ is a “successor” operation on

cardinals.2 2 Note that κ + 1 is not necessarily κ+.


� Definition 28 (Cardinal Numbers)

Using Transfinite Recursion, define the following ordinal-enumeratedcollection of cardinals

• ℵ0 = ω;

• ∀α ∈ Ord that is a successor ordinal, ℵS(α) = ℵα+1 = ℵ+α ; and

• ∀α > 0 that is a limit ordinal, ℵα := sup{ℵβ | β < α}.

The ℵα’s are called cardinal numbers.

+ Lemma 33 (Cardinal Numbers are Cardinals)

If α ∈ Ord, then ℵα ∈ Card.

Ò ProofWe shall use Transfinite Induction. The result is clear for α = 0,

since ℵ0 = ω and ∀n ∈ ω, |n| 6= |ω|. For successor ordinals

α, since h(α) is an ordinal, we have that ℵ+α = h(ℵα) is also a

cardinal. Now for α > 0 a limit ordinal, suppose that β < ℵα. Since

ℵα = sup{ℵγ | γ < α}, ∃γ < α such that β < ℵγ. Since ℵγ is a

cardinal by the Inductive Hypothesis, and β < ℵγ < ℵα, we have

that

|β| < |ℵγ| ≤ |ℵα| .

Thus ℵα is not equinumerous with any lesser ordinal, i.e. it is a

cardinal. �

+ Lemma 34 (Ordinals Index the Cardinal Numbers)

∀α < β ∈ Ord, we have ℵα < ℵβ.

Ò ProofWe shall use Transfinite Induction on β. β = 0 is true vacuously


so. For β a successor ordinal, suppose ∀α < β, ℵα < ℵβ. Now

ℵβ+1 = ℵ+β = h(ℵβ), which we have that ℵβ < h(ℵβ) as proven

before.

Now suppose that β > 0 is a limit ordinal. Then ℵβ = sup{ℵα |α < β} and ∀γ < α < β, ℵγ < ℵα. Since β is a limit ordinal,

∃ζ < β such that α < ζ. By the Induction Hypothesis, we have that

ℵα < ℵζ , and ℵζ ≤ ℵβ. Thus ℵζ ⊆ ℵβ, and so ℵα < ℵβ. �

+ Lemma 35 (Infinite Cardinals are Distant)

∀α ∈ Ord, α ≤ ℵα. The inequality is strict if α is a successor ordinal.

Ò ProofAgain, we shall use Transfinite Induction on α. For α = 0, we have

0 < ℵ0 = ω, and so α = 0 holds. For α a successor ordinal, suppose

α ≤ ℵα. Thus α + 1 ≤ ℵα + 1. By definition of h(ℵα) and definition

of cardinal numbers

α + 1 ≤ ℵα + 1 < ℵ+α = ℵα+1.

Thus the statement holds for α + 1, and indeed, the inequality is

strict for successor ordinals. For α > 0 a limit ordinal, suppose that

∀β < α, we have β ≤ ℵβ. By Lemma 34, we have

β ≤ ℵβ < ℵα.

Thus

α = sup{β | β < α} < ℵα

as required. �

0 Proposition 36 (All Infinite Cardinals are Indexed by the Ordi-nals)

Every infinite cardinal is of the form ℵα for some α ∈ Ord.


Ò ProofLet κ ∈ Card. By Lemma 35, we have κ ≤ ℵκ < ℵκ+1. And so we

can show that ∀β ∈ Ord, ∀κ < ℵβ, ∃α < β such that κ = ℵα. We

shall use Transfinite Induction on β.

Since there are no infinite cardinals strictly below ω, β = 0 is

trivially true. Suppose β = γ + 1 is a successor ordinal, where

γ ∈ Ord, and suppose κ < ℵβ. Since γ < β, Lemma 34 implies that

ℵγ < ℵβ = ℵγ+1 = ℵ+γ , and by definition, there are no cardinals

between ℵγ and ℵβ. Thus κ ≤ ℵγ. We thus have that either κ = ℵγ

or, by the Induction Hypothesis, ∃α < γ such that κ = ℵα. Thus the

statement holds for successor ordinals.

Let β > 0 be a limit ordinal and κ < ℵβ = sup{ℵγ | γ <

β}. Then ∃γ < β such that κ < ℵγ, and so by the Induction

Hypothesis, ∃α < γ such that κ = ℵα. �

Consequently, we have ourselves an ordinal-valued, order-preservingcomplete indexing of the infinite cardinals.

Exercise 9.1.1

For the inequality in Lemma 35, show that the equality can occur. In partic-ular, consider the sequence of ordinals defined recursively by α0 = 0, andαn+1 = ℵαn , and verify that α = ℵα. In fact, this works if we start with anyordinal α0, not just 0.


10.1 Cardinals (Continued 3)

10.1.1 Cardinal Arithmetic Contents in this lecture should beextended upon. A lot of the contentsneed to be explored for it was presentedtersely so without all the details ofwhich we need.

10.1.1.1 Cardinal Summation

� Definition 29 (Cardinal Sum)

∀κ1, κ2 ∈ Card. Let the cardinal sum

κ1 + κ2 := |X1 ∪ X2|

where X1, X2 ∈ Set such that

|X1| = κ1 and |X2| = κ2

and X1 ∩ X2 = ∅.

RemarkThis definition does not depend on the choice of X1 and X2, i.e. if we haveanother X′1 and X′2 such that

Exercise 10.1.1Prove this remark.

∣∣X′i ∣∣ = |Xi| = κi i = 1, 2

and X′1 ∩ X′2 = ∅, then ∣∣X′1 ∪ X′2∣∣ = |X1 ∪ X2| .

This shows that the cardinal summation is well-defined.


å NoteIf X, Y ∈ Set are arbitarily chosen, then

|X ∪Y| ≤ |X|+ |Y|

WarningThe cardinal summation is different from ordinal summation. For ex-ample, ℵ0 + ℵ0, where the + represents the ordinal summation, is not acardinal, as we have shown that |ω + ω| = |ω|.

Therefore, there is a need to explicitly mention the context of which +

is used.

10.1.1.2 Cardinal Product

� Definition 30 (Cardinal Product)

∀κ1, κ2 ∈ Card, the cardinal product

κ1κ2 := |X1 × X2|

where X1, X2 ∈ Set with |Xi| = κi, for i = 1, 2.

Exercise 10.1.2

Prove that � Definition 30 is well-defined.

RemarkWe may as well choose

X1 = κ1 and X2 = κ2,

and so κ1κ2 = |κ1 × κ2|.1 1 We cannot do so for cardinal sums, forκ1 ∩ κ2 6= ∅.

Exercise 10.1.3

Prove that the cardinal sum and product agrees with ordinal sum and prod-ucts on the finite ordinals. This is the usual arithmetic on natural numbers.


1 Theorem 37 (Dominance of the Larger Cardinal)

Let κ1, κ2 ∈ Card not both finite. Then

Exercise 10.1.4Prove/read 1 Theorem 37.

1. κ1 + κ2 = max{κ1, κ2}; and

2. if neither κ1 or κ2 is 0, then κ1κ2 = max{κ1, κ2}.

We can generalize the notions of cardinal sum and cardinal prod-

uct. But first, a definition.

� Definition 31 (I-sequence)

Let I ∈ Set. By an I-sequence of sets, we mean a definite operation2 2 Note that f : I → Img( f ) is a functionby the Replacement Axiom.

f : I → Set .

We write such sequences as

(xi : i ∈ I)

where xi := f (i).

� Definition 32 (Generalized Cardinal Sum)

Suppose (κi : i ∈ I) is a sequence of cardinals. We define the (general-ized) cardinal sum to be

Σi∈Iκi :=

∣∣∣∣∣⋃i∈I

Xi

∣∣∣∣∣where (Xi : i ∈ I) is a sequence of pairwise disjoint sets with |Xi| =κi.

Exercise 10.1.5

Check that � Definition 32 is well-defined.


å NoteObserve that ⋃

i∈IXi = ∪{Xi | i ∈ I} = ∪ Img( f )

where (Xi : i ∈ I) is a sequence of functions and f : I → Set is given byf (i) = Xi.

1 Theorem 38 (Properties of Cardinal Sum)

Let I ∈ Set be infinite, and (κi : i ∈ I) a sequence of cardinals not allzero. Then

Exercise 10.1.6Prove/read 1 Theorem 38

1. supi∈I κi ∈ Card; and

2. ∑i∈I κi = max{|I| , supi∈I κi}.

Example 10.1.1

We have that

∑0<n∈ω

n = max{|ω \ {0}| , sup0<n∈ω

n} = max{ℵ0, ω} = ℵ0.

� Definition 33 (Generalized Cardinal Product)

Suppose (κi : i ∈ I) is a sequence of cardinals. Then the cardinalproduct is defined as

∏i∈I

κi := |X1 × X2 × . . .| =∣∣∣∣×i∈I

Xi

∣∣∣∣where (Xi : i ∈ I) is a sequence of sets with |Xi| = κi.

Exercise 10.1.7

Check that � Definition 33 is well-defined.


å NoteNote that ×

i∈IXi is properly defined, as

×i∈I

Xi := {(ai : i ∈ I) | ∀i ∈ I, ai ∈ Xi}

=

{f : I →

⋃i∈I

Xi, f (i) = Xi

}.

RemarkOnce again, we might as well take Xi = κi, just as we did when we definedpair products.

Example 10.1.2

Suppose κi = 2 for all i ∈ I. Define a function such that

×i∈I

2→ P(I)

given by

(ai : i ∈ I) 7→ {i ∈ I : ai = 1}.

Note that 2 = {0, 1}, and so each ai = 0 or 1. Clearly so, this is a

bijection3. Consequently, we have 3 This is the usual correspondencebetween subsets and characteristicfunctions.

∏i∈I

2 =

∣∣∣∣×i∈I2∣∣∣∣ = |P(I)|

We claim that

|I| < |P(I)| .

10.1.2 An Interlude on the Continuum Hypothesis

1 Theorem 39 (Cantor’s Diagonalization)

∀I ∈ Set, we have |I| < |P(I)|.

Ò ProofClearly we have that I ↪→ P(I) through the map i 7→ {i}. Thus

|I| ≤ |P(I)|.

Suppose to the contrary that there exists a bijection f : I → P(I).


Let4 4 This definition looks awfully familiarto Russell’s Paradox.∆ = {i ∈ I : i /∈ f (i)} ⊆ I.

which is a set by Bounded Separation. Thus ∆ ∈ P(I). Then

∃i0 ∈ I such that f (i0) = ∆. Now if i0 ∈ ∆, then i0 ∈ f (i0) = ∆,

but this contradicts the membership condition which states that

i0 /∈ f (i0). If i0 /∈ ∆, then i0 /∈ f (i0) = ∆, but by the membership

condition, it must be that i0 ∈ ∆, yet another contradiction. Thus

such a bijection does not exist. �

This proves our earlier claim. In fact, so long as |I| > 2,

∏i∈I

2 6= max{|I| , supi∈I

2} = max{|I| , 2},

since RHS = |I| while LHS = |P(I)|.

� Definition 34 (Cardinal Exponentiation)

∀κ, λ ∈ Card, the cardinal exponentiation is defined as

κλ := |Fun(λ, κ)| ,

where Fun(λ, κ) is the set of all functions from λ to κ.

Exercise 10.1.8

Prove that

∏i<λ

κ = κλ.

As a consequence, we have that 2λ = |P(λ)|. Then what is

|P(ℵ0)|?

� Axiom 40 (Continuum Hypothesis)

We have thatℵ0 < 2ℵ0 = |P(ℵ0)| ,

i.e. 2ℵ0 = ℵ1.


More generally,

� Axiom 41 (Generalized Continuum Hypothesis)

∀κ ∈ Card, we have2κ = κ+.

In this course, we will not assume the Continuum Hypothesis (nor

for the general case).

It has been proven by Paul Cohen (1963)5 that the Continuum 5 Cohen, P. J. (1963). The indepen-dence of the continuum hypothesis.https://www.ncbi.nlm.nih.gov/pmc/

articles/PMC221287/

Hypothesis is independent from ZFC.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC221287/


Part II

Model Theory


The course will not cover for cofinality and coregularity, but it may

be helpful to read through the material.

We shall now venture into Model Theory.

11.1 First-order Logic

11.1.1 Structure

� Definition 35 (Structure)

A structureM consists of the following data:

1. A non-empty set M, called the universe ofM;

2. A sequence (ci : i ∈ Icon), where Icon is an I-sequence of constants, ofdistinguished elements of M, called the constants ofM;

3. A sequence of M-valued functions on powers of M,

( fi : Mni → M : i ∈ Ifun),

called the basic functions ofM, and for each i ∈ Ifun, ni < ω iscalled the arity of fi;

4. A sequence of subsets of powers of M,

(Ri ⊂ Mmi : i ∈ Irel)

called the basic relations ofM, and for each i ∈ Irel, mi < ω iscalled the arity of Ri.

86 Lecture 11 Oct 16th - First-order Logic

å Note1. Note that we defined M 6= ∅, as there is nothing too interesting to

study from an empty universe.

2. Also, note that we shall always use the corresponding capital alphabetas the universe of the structure (e.g. the universe of the structureM isM).

3. While M 6= ∅, we allow Icon, Ifun and Irel to be ∅. In such a case, thestructure that we have is about a pure set.

4. If f is a 0−ary basic function, then we have

f : M0 → M.

By convention, we have that M0 = 1 = {0}, and so we have f (0) ∈M. Thus, such an f is “essentially” a basic constant. Therefore, weshall usually assume that the arity of f is strictly positive.

Example 11.1.1

Consider R, the set of real numbers1. The following are some of the 1 We did not construct R in our sectionon Set Theory. I may write up a fullconstruction from N to R on my site.structures that we can study on R:

1. as a pure set: R = R, with Icon = Ifun = Irel = ∅.

2. as an ordered set: R = (R,<), with the basic binary relation <,

and no basic functions or constants.

3. as an additive group: R = (R, 0,+,−), with

• 0 as a constant;

• + as a basic binary function; and

• − as a basic unary function.

4. as a ring: R = (R, 0, 1,+,−, ·).

5. as a Q-vector space: R = (R, 0,+,−, (λq)q∈Q), where λq : R → R

is scalar multiplication by q.

6. as an ordered ring: R = (R, 0, 1,+,−, ·,<).


� Definition 36 (Expansion & Reduct)

SupposeM and N are structures with the same universe. We say thatM is an expansion of N or that N is a reduct ofM if the basic con-stants, basic functions, and basic relations of N are contained in those ofM.

Example 11.1.2

• (R, 0,+,−) is a reduct of (R, 0, 1,+,−, ·);

• (R, 0, 1,+,−, ·,<) is an expansion of (R, 0, 1,+,−, ·).

11.1.2 Language

Example 11.1.3

Consider the following 2 structures

R = (R, 0,+,−)

Z =(

Z�2Z, 0,+,−)

Notice that the 0,+ and − do not not actually share the same mean-

ing, since, e.g., + in R is addition on the reals, while + in Z is addi-

tion on the integers modulo 2.

� Definition 37 (Language)

A language L consists of 3 sets of symbols:

1. a set Lcon of constant symbols;

2. a set Lfun of function symbols, where each function symbol comes withan arity, which is a natural number;

3. a set Lrel of relation symbols, where each relation symbol comes withan arity, which is a natural number.

� Definition 38 (L-structure)


An L-structure is a structureM with bijections between

Lcon → Icon c 7→ cM

Lfun → Ifun f 7→ fM

Lrel → Irel R 7→ RM

that preserves arity. We call

cM, fM, and RM

the Interpretation of c, f , R inM.

So in our previous example, we have that R and Z are both L-

structures, where L = {0,+,−} is the language with

• one constant symbol, 0;

• one binary function symbol, +;

• one unary function symbol, −.

This particular language is often referred to as the language of addi-tive groups, for obvious2 reasons. 2 Obvious, if you have studied Group

Theory.Remark• To be precise, we should really write

R = (R, 0R,+R,−R),

so as to not obfuscate the symbols themselves and their interpretations inR, but we shall forgive ourselves for this abuse of notation, for we shallleave this for the context to resolve this confusion.

• Every group is naturally an L-structure, where L = {0,+,−}. Theconverse is definitely false. E.g. the following L-structure

M = (Z, 0M,+M,−M)

given by

0M = 12

+M : Z2 → Z (a, b) 7→ −2

−M : Z→ Z a 7→ 22

https://tex.japorized.ink/PMATH347S18/classnotes.pdf

https://tex.japorized.ink/PMATH347S18/classnotes.pdf


is not a group, since there are no identities nor inverses. But it is indeedan L-structure.

Example 11.1.4

Let F be a field, and let L be the language of F-vector spaces, i.e.

Lcon = {0}

Lfun = {+,−, (λ f ) f∈F}

Lrel = ∅

where + is a binary function, − a unary function, and λ f a unary

function for each f ∈ F. Then any F-vector spaces V is an L-

structure:

0 is interpreted as the zero vector

+ is interpreted as vector addition

− is interpreted as the additive inverse of a vector

λ f is interpreted as scalar multiplication by f

The converse is, however, not true, and we shall study the reasons

behind why this is not true later on.

� Definition 39 (L-Embedding)

Suppose L is a language, andM and N are L-structures. An L-embedding, j : M → N , is an injective function j : M → N suchthat

1. ∀c ∈ Lcon j(cM) = cN ;

2. ∀ f ∈ Lfun each with arity n f , and ∀a1, . . . , an f ∈ M, we have

j(

fM(a1, . . . , an f ))= fN

(j(a1), . . . , j(an f )

)3. ∀R ∈ Lrel each with arity mR, and ∀(a1, . . . , amR) ∈ MmR , we have

(a1, . . . , amR) ∈ RM ⇐⇒(

j(a1), . . . , j(amR))∈ RN .

� Definition 40 (Substructure)


Continuing with the above assumptions and notation, if M ⊆ N, then wesay thatM is a substructure on N if the identity map id : M → N isan L-embedding. We denote this notion, without confusion, byM⊂ N .

We may also say that N is an extension ofM.


12.1 First-order Logic (Continued)

12.1.1 Language (Continued)

å NoteNotice that L-structuresM⊆ N iff

• M ⊆ N;

• ∀c ∈ Lcon cM = cN ;

• ∀ f ∈ Lfun, each with arity n f , fM = fM �Mn f ;

• ∀R ∈ Lrel, each with arity mR, RM ⊆ RN ∩MmR .

Exercise 12.1.1

Suppose N is an L-structure with A ⊂ N, where A 6= ∅. Show that A isthe universe of some (unique) substructure of N iff

• ∀c ∈ Lcon cN ∈ A;

• ∀ f ∈ Lfun that is n-ary, fN (An) ⊂ A.

The choice of the language determine what the substructures are.

Example 12.1.1

The following is table showing an example of a substructure in the

given structures:

92 Lecture 12 Oct 18th - First-order Logic (Continued)

Structure Substructures

R all non-empty subsets

(R, 0,+) all submonoids of R 1

(R, 0,+,−) all subgroups

(R, 0, 1,+,−, ·) all subrings

(R,<) all non-empty subsets with induced order

(R, 0,+,−, (λq)q∈Q) all Q-subspaces1 Not that substructures of (R, 0,+) isnot necessarily a group, and this is whywe have always specified for − for thelanguage of additive groups. However,this is not necessary for the uniqueinverse in the language of fields, ofwhich we shall clarify in a later section.

� Definition 41 (L-isomorphism)

An L-isomorphism is a surjective L-embedding.

Exercise 12.1.2

Let j : M → N be an L-embedding. Then j induces an L-isomorphismbetweenM and a unique substructureM′ ⊆ N .

12.1.2 Terms & Formulas

From hereon, we shall have the following countable infinite set,

Var = {x0, x1, . . .},

which we shall call as our set of variables2. 2 This is not to be confused with thevariance in probability theory.

� Definition 42 (L-terms)

Let L be a language. The set of L-terms is the smallest set of strings ofsymbols from

L ∪Var∪{(, )} ∪ {, } 3

satisfying 3 By {(, )} and {, }, we mean the punc-tuation symbols “(”, “,”, “)”.

• every variable is an L-term;

• every constant of Lcon is also an L-term;

• 4if t1, t2, . . . , tn are L-terms and f ∈ Lfun is n-ary, then f (t1, . . . , tn) 4 This is a recursive formula.

is also an L-term.

https://en.wikipedia.org/wiki/Monoids


We often write a term t as

t = t(x1, . . . , xn)

to mean that the variables in t come from {x1, . . . , xn}.

RemarkAll x1, . . . , xn need not always appear in t.

We shall use the following example to show another abuse of

notation that we shall gladly do in this course:

Example 12.1.2

Let L = {0, 1,+,−,×}. The following is an L-term:

×(+(x0,−(x1)),×(1, x2)).

It is fairly straightforward to verify that the above is indeed an L-

term: using items (1), (2) and (3) in our definition above, by the

given order,

1. all x0, x1, 1, x2 are L-terms by (1) and (2);

2. − is a unary function, and so by (3), we have that −(x1) is an

L-term;

3. + is a binary function, and by (3), we have that +(x0,−(x1)) is an

L-term;

4. × is a binary function, and so by (3), we have that ×(1, x2) is an

L-term;

5. finally, by (3), ×(+(x0,−(x1)),×(1, x2)) is an L-term.

We shall “informally” write this as

(x0 + (−x1))(1x2).

Note that we do not simply write x2 = 1x2, since we may not have

that 1 acts as a “multiplicative identity” of sorts, as we would have in

a ring.

� Definition 43 (Interpretation)


SupposeM is an L-structure, and t = t(x1, . . . , xn) is an L-term. Wedefine the itnerpretation of t inM to be the function

tM : Mn → M,

defined recursively by

• If t is xi for some 1 ≤ i ≤ n, then

tM : Mn → M (a1, . . . , an) 7→ ai.

• If t is c for some c ∈ Lcon, then

tM : Mn → M (a1, . . . , an) 7→ cM.

• If t is f (t1, . . . , tl), where t1, . . . , tl are L-terms and f is an l-aryfunction symbol, then tM : Mn → M

(a1, . . . , an) 7→ f (tM1 (a1, . . . , an), tM2 (a1, . . . , an), . . . , tMl (a1, . . . , an))

å NoteNote that the function tM depends not just on t but on its presentationt = t(x1, . . . , xn).

Example 12.1.3

Let L = {0,+,−} and R = (R, 0,+,−) an L-structure. Let t be the

term x ∈ Var. Then for y ∈ Var, the following are some interpreta-

tions of t in R:

1. if t = t(x), then we may interpret tR : R → R as the identity map,

i.e. r 7→ r;

2. if t = t(x, y), then we may interpret tR : R2 → R as taking the first

component of the tuple, i.e. (a, b) 7→ a;

3. if t = t(x, y), then we may interpret tR : R2 → R as taking the

second component of the tuple, i.e. (a, b) 7→ b.

Exercise 12.1.3


Prove the following: supposeM is an L-structure, and A ⊆ M such thatA 6= ∅, then A is the universe of the substructure iff ∀t = t(x1, . . . , xn)

that are L-terms, we have tM(An) ⊆ A.

Example 12.1.4

Let L = {0, 1,+,−,×} and Z = (Z,=, 1,+,−,×). Let P ∈ Z[x]where P is x2 + 2y− 1. Then let tP = tP(x, y) be the L-term

x2 + (y + y) + (−1).

Then the interpretation of tP is

tZP : Z2 → Z (a, b) 7→ a2 + 2b + 1.

� Definition 44 (Atomic L-Formulas)

An atomic L-formula is a (finite) string of symbols from

L ∪Var∪{(, )} ∪ {, } ∪ {=}

of the form:

1. (t = s) where t, s are L-terms;

2. R(t1, . . . , tl) where t1, . . . , tl are L-terms, and R ∈ Lrel with arityl ∈ ω.

� Definition 45 (L-formulas)

The set of L-formulas is the smallest set of (finite) strings of symbolsfrom

L ∪Var∪{(, )} ∪ {, } ∪ {=} ∪ {∧,∨,¬, ∃, ∀}

satisfying:

1. every atomic L-formula is an L-formula;

2. if φ, ψ are L-formulae, then so is (φ ∧ ψ), ¬φ, and (φ ∨ ψ);

3. if φ is an L-formula and x ∈ Var, then ∃xφ and ∀xφ are L-formulae.


Remark1. We shall use the following abbreviations:

• φ→ ψ for (¬φ ∨ ψ);

• φ↔ ψ for (φ→ ψ) ∧ (ψ→ φ).

2. We shall gleefully make the following abuse of notation: for the languageL = {×,<}, we write x < y2 instead of < (x,×(y, y)).

� Definition 46 (Bound and Free Variables)

Suppose φ is an L-formula. An occurrence of a variable x in φ is calledbound if it appears inside the scope of a quantifer (i.e. in the existence of∃ and ∀). Otherwise, they are called free.

Example 12.1.5

Let L = {∈}, and x, y, z ∈ Var. The following is an L-formula:

(x ∈ y) ∧ ∀z((z ∈ y)→

((z ∈ x) ∨ (z = x)

))We have that x, y are the free variables while z is bound.

We write φ = φ(x1, . . . , xn) to mean that the free variables of φ

come from {x1, . . . , xn}, but it is not necessary that all of the variables

appear in the L-formula.

For the sake of convenience, we hall assume that no variable can

be simultaneously bound and free in an L-formula.

13 Lecture 13 Oct 23rd

13.1 First-order Logic (Continued 2)

13.1.1 Terms & Formulas (Continued)

Example 13.1.1

Let L = {0, 1,+,−, ·}, and x, y ∈ Var. We have that x2 = ·(x, x) and

y are L-terms. Given R = (R, 0, 1,+,−, ·), we can interpret x2, using

(x, y) as (x2)R

: R→ R such that (r, s) 7→ r2

and interpret y, using (x, y) as

yR : R→ R such that (r, s) 7→ s.

Now since x2 and y are L-terms, we have that x2 = y is an atomicformula1. In the L-formula ∃x(x2 = y), the only free variable is y, 1 Atomic formulas are quantifier-free

and so all the variables are free.which is not “specified” by a quantifier.

RemarkNotice that an L-formula with free variablesay something about the free variable. Inour example of ∃x(x2 = y), the L-formula“says” that y has a square root.

In the L-formula ∀y ∃x(x2 = y), themeaning is different: it “says” that everyelement (in R) has a square root. Thisis an important observation for the nextdefinition.

On the other hand, for the L-formula ∀y; ∃x(x2 = y), both x and yare bound.

� Definition 47 (L-Sentences)

An L-formula with no free variable is called an L-sentence.

å NoteAn L-sentence has a truth value.� Definition 48 (Satisfaction / Realization)

Given a language L and an L-structureM, for some n < ω, let

φ = φ(x1, . . . , xn)

98 Lecture 13 Oct 23rd - First-order Logic (Continued 2)

be an L-formula2, a = (a1, . . . , an) ∈ Mn and x = (x1, . . . xn). 2 If n = 0, then φ is simply an L-sentence.

We define that a satisfies (or realizes) the formula φ(x) inM 3, 3 We may also say that φ(a) is true inM.of which we denote byM |= φ(a), through the following recursive

definition, which we iterate on the complexity of the formula φ(x): Notice that in the recursive definition,we start with atomic L-formulas firstbefore going to the L-formulas.1. If φ(x) is of the form t1 = t2, where

t1 = t1(x) and t2 = t2(x)

are L-terms, then

M |= φ(a) if tM1 (a) = tM2 (a),

where tM1 and tM2 are realizations of t1 and t2 inM, respectively.

2. If φ(x) is of the form

R(t1, . . . , tl), where ti = ti(x) are L-terms, R ∈ Lrel

with arity l < ω, then we defineM |= φ(a) if(tM1 (a), tM2 (a), . . . , tMl (a)

)∈ RM ⊆Ml

3. If φ(x) is of the formφ1(x) ∧ φ2(x),

where φ1(x), φ2(x) are known L-formulas, then we define

M |= φ(a) ⇐⇒ M |= φ1(a) ∧M |= φ2(a).

4. If φ(x) is of the formφ1(x) ∨ φ2(x),

where φ1(x), φ2(x) are known L-formulas, then we define

M |= φ(a) ⇐⇒ M |= φ1(a) ∨M |= φ2(a).

5. If φ(x) is of the form¬ψ(x),

where ψ(x) is a known L-formula, then we define4 4 Here, we conveniently assumed theLaw of Excluded Middle.

M |= φ(a) ⇐⇒ it is not the case thatM |= φ(a),

of which the latter shall be denoted asM 6|= ψ(a).

https://en.wikipedia.org/wiki/Law_of_excluded_middle


6. If φ(x) is of the form∃yψ(x, y),

where ψ(x, y) is an L-formula, and y a (single) variable, then wedefine

M |= φ(a) ⇐⇒ there exists b ∈ M such thatM |= ψ(a, b).

7. If φ(x) is of the form∀yψ(x, y),

where ψ(x, y) is an L-formula, and y a (single) variable, then wedefine

M |= φ(a) ⇐⇒ for everyb ∈ M such thatM |= ψ(a, b).

å NoteThe set of all realizations of an L-formula φ in the L-structureM isdenoted as

φM := {a ∈ Mn :M |= φ(a)}.

We may also call this set the subset of Mn defined by φ.

Finally, we can define definability.

� Definition 49 (L-Definable)

A subset S ⊂ Mn is said to be L-definable if S = φM for some L-formula φ = φ(x) = φ(x1, . . . , xn).

RemarkIf n = 0, we have that φ is an L-sentence, and since M0 = 1 = {0}, thereare two possible scenarios and that is

eitherM |= φ orM |= ¬φ.

Notice that φM is either 1 = {0} or 0 = ∅.

Example 13.1.2


Let L = {0, 1,+,−,×}, ψ(x, y) be the atomic L-formula x2 = y, and

φ(y) be the L-formula ∃x(x2 = y). It is clear that we can also write

φ(y) = ∃xψ(x, y).

Let R = (R, 0, 1,+,−,×). We have that

R |= ¬φ(−1) and R |= φ(2).

In fact, φR = R≥0 ⊆ R.

On the other hand, for the L-structure Q = (Q, 0, 1,+,−,×), we

have that

Q |= ¬φ(−1) and Q |= ¬φ(2).

It is worth noting that

φQ =

{n2

m2 : n, m < ω, m 6= 0}⊆ Q.

For the L-structure C = (C, 0, 1,+,−,×), we have that φC = C, or

we may also write that C |= ∀y φ(y) 5. 5 We may further simplify our notationsby letting

σ = ∀y∃x(x2 = y),

which we can then simply write C |= σ.

Example 13.1.3

Consider the same language as in the last example, and let φ(y, z) =∃x(x2 = y− z). In the L-structure R = (R, 0, 1,+,−,×), it is clear

that

φR ={(a, b) ∈ R2 : b ≤ a

}⊆ R2,

and we observe that the set of solutions is also the binary relation ≥,

i.e. ≥ is definable in the L-structure R.

Similarly so, we can show that ≤ is L-definable in R. 6 6 In other words, this example shows tous that we can define ≤ and ≥ usingjust the symbols = and −.

0 Proposition 42 (Structure Traversal with respect to Quantifiers)

LetM ⊆ N be L-structures, whereM is the L-substructure, andvariables x = (x1, . . . , xn). Let φ(x) be an L-formula, and a =

(a1, . . . , an) ∈ Mn. Then

1. if φ is a quantifier-free formula, then

M |= φ(a) ⇐⇒ N |= φ(a).


2. if φ is universal, i.e. φ(x) is of the form

∀y1∀y2 . . . ∀ymψ(x, y1, y2, . . . , ym),

where ψ is quantifier-free, then

M |= φ(a) ⇐= N |= φ(a).

3. if φ is existential, i.e. φ(x) is of the form

∃y− 1∃y2 . . . ∃ymψ(x, y1, y− 2, . . . , ym),

where ψ is quantifer-free, then

M |= φ(a) =⇒ N |= φ(a).

I shall directly quote from the coursenotes from Professor Moosa:

This proposition has a very typicalproof. In order to prove somethingabout all formulas one usually hasto begin by proving somethingabout terms and then proceedingby induction on the complexityof the formula. The result aboutterms is itself usually proved byinduction on the complexity of theterm.

Ò Proof1. Claim : We shall first show that if t(x) is an L-term, then tM =

tN �Mn . We shall prove this claim by using induction on the

complexity of t.

• If t is a variable xi, then we have

tM : Mn → M given by (a1, . . . , an) 7→ ai (13.1)

tN : Nn → N given by (b1, . . . , bn) 7→ bi

Since Mn ⊆ Nn, we have that the restriction of tN to Mn will

give us (13.1).

• If t is a constant symbol, i.e. some c ∈ Lcon, then we simply

have cM = cN , for M ⊆ N.

• If t is f (t1, . . . tl), where f ∈ Lfun, where t1, . . . , tl are L-terms,

then

tM(a) = fM(tM1 (a), . . . , tMl (a)

)= fM

(tN1 (a), . . . , tNl (a)

)∵ IH

= fN(tN1 (a), . . . , tNl (a)

)∵ note on page 91

= tN (a).

Therefore tN �Mn= tM as claimed a.


existential, since our assumption is that φ is quantifier-free.

This completes the proof.

�

Exercise 13.1.1

Prove (2) and (3) of 0 Proposition 42.



14.1.1 Terms & Formulas (Continued 2)

Example 14.1.1

Let L be the language of rings1. We have that in Z , where 1 From hereon, for the sake of simplic-ity, I shall gleefully use such decla-rations for a language wherever it isconvenient, and when the context isclear that I am merely using the com-mon symbols that are used in the saidtheory.

(Z, 0, 1,+,−,×) = Z ⊆ Q = (Q, 0, 1,+,−,×),

for the L-formula φ(x) = ∃y(x = 2y), we observe that

Z |= ¬φ(1) but Q |= φ(1).

This shows to us that existential formulas that

• are satisfied in an extension may not be satisfied in the substruc-

ture;

• is not satisfied in the substructure may be satisfied in the exten-

sion.

14.1.2 Elementary Embeddings

Considering 0 Proposition 42, it is interesting for us to consider

substructures that satisfies all the formulas of its extension, including

formulas that are either universal or existential.

� Definition 50 (Elementary Embeddings)

SupposeM ⊆ N are L-structures, andM is a substructure of N . AnL-embedding j : M → N is an elementary embedding if for any

106 Lecture 14 Oct 25th - First-order Logic (Continued 3)

L-formula φ and a ∈ Mn,

M |= φ(a) ⇐⇒ N |= φ(j(a)).

We call such anM an elementary substructure of N , and we denoteM� N .

In simpler words, any formula that is true in N is true in its ele-

mentary substructureM,

� Corollary 43 (Isomorphisms are Elementary Embeddings)

Every isomorphism is an elementary embedding.

Ò ProofWe already know the result for quantifier-free statements from

0 Proposition 42, and so it suffices to prove this statement by in-

duction on the number of quantifiers, which we shall call n. In fact,

it suffices to prove for the case of an existential L-formula, since we

can write ∀ as ¬∃¬.

Suppose g : M → N is an L-isomorphism, whereM ⊆ N .

There is nothing to show for n = 0. Suppose that the statement

holds for n = m. Consider an L-formula of the form φ(x) =

∃yψ(x, y), where x ∈ Mm+1, and ψ an L-formula with lower com-

plexity than φ. Then for a ∈ Mm+1,

M |= φ(a) ⇐⇒ M |= ∃bψ(a, b)

⇐⇒ N |= ∃bψ(g(a), g(b)) ∵ IH

⇐⇒ N |= ∃cψ(g(a), c) ∵ f is bijective

⇐⇒ N |= φ(g(a)).


Example 14.1.2

Let Q = (Q,<, 0) be an L = {<, 0}-structure. Show that the graph of

addition is not definable in L.


Ò Solution2We need make this inference from automorphisms on Q. Let σ : 2 Question: why did we look at

the automorphisms? Answer: By� Corollary 46, automorphisms thatfixes, in this case, ∅ pointwise, which istrivially true, should remap the definedset into an element of its own. We usethis fact to obtain a contradiction toshow that addition is not definable.

Q→ Q be an automorphism. Then we have that

x < y ⇐⇒ σ(x) < σ(y)

σ(0) = 0

and σ is a bijection

For example, the map x 7→ ax, for some a > 0, is an automorphism

on Q.

Consider the function

φ(x) =

2x + 1 x < −1

x x ∈ [−1, 1]

2x− 1 x > 1

which has a graph as in Figure 14.1.x

φ(x)

Figure 14.1: Graph of φ(x) in Exam-ple 14.1.2

It is rather clear that φ is indeed an automorphism, i.e. it has all

the properties that we listed above. However, such an automorphism

does not preserve addition itself, since

φ(1 + 2) = φ(3) = 5

φ(1) + φ(2) = 4.

0 Proposition 44 (� Tarski-Vaught Test)

SupposeM⊆ N . TFAE

1. M� N

2. For every L-formula φ(x, y) and all n-tuples a ∈ Mn, if N |=∃yφ(a, y), then there exists b ∈ M such that N |= φ(a, b).

Ò Proof(1) =⇒ (2) : SupposeM � N . Then ∃j :M→ N an elementary

embedding such that for all L-formula φ and a ∈ Mn,

M |= φ(a) ⇐⇒ N |= φ(j(a)).


In particular, the identity map is such an embedding. Let ψ(x) =

∃yφ(x, y). Then

N |= ψ(a) defn 50⇐⇒ M |= ψ(a)

⇐⇒ M |= ∃bφ(a, b)defn 50⇐⇒ N |= ∃bφ(a, b)

This completes ( =⇒ ). a

(2) =⇒ (1) : We shall use induction on the complexity of the

L-formula φ.

• If φ(a) is quantifier-free, then sinceM⊆ N , by 0 Proposition 42,

we haveM |= φ(a) ⇐⇒ N |= φ(a). This case also covers the

atomic formulas.

• If φ(a) is of the form φ1(x) ∨ φ2(x), where φ1(x), φ2(x) are L-

formulas whose results are known, then

M |= φ(a) ⇐⇒ M |= φ1(a) orM |= φ2(a)

⇐⇒ N |= φ1(a) or N |= φ2(a) ∵ IH

⇐⇒ N |= φ(a).

• If φ(x) is of the form φ1(x) ∧ φ2(x), the proof is the same as the

previous item.

• If φ(x) is of the form ¬ψ(x) for some L-formula ψ, then

M |= φ(a) ⇐⇒ M |= ¬ψ(a)

⇐⇒ N |= ¬ψ(a) ∵ IH

⇐⇒ N |= φ(a).

• As stated before in another proof, it suffices to prove for the

existential case, since ∀ can be written as ¬∃¬. Suppose φ(x) is

of the form ∃yψ(x, y), for some L-formula ψ(x, y) whose result

we already know. We know thatM |= φ(a) =⇒ N |= φ(a) by


0 Proposition 42. For the converse,

N |= φ(a) ⇐⇒ N |= ∃yψ(a, y)

=⇒ there exists b ∈ M N |= ψ(a, b)IH=⇒ there exists b ∈ MM |= ψ(a, b)

=⇒ M |= ∃yψ(a, y)

⇐⇒ M |= φ(a).


RemarkThe Tarski-Vaught Test gives us an alternate mechanism to check if a sub-structure is elementary.

1 Theorem 45 (Downward Löwenheim-Skolem)

SupposeM is an L-structure, and A ⊆ M. Then there exists an ele-mentary substructure ofM that contains A and is of cardinality at mostmax{|A| , |L| ,ℵ0}.

In particular, if L is countable, then every L-structure has a countableelementary substructure.

Ò ProofLet κ = max{|A| , |L| ,ℵ0}. Define, recursively so, a countable

chain of subsets of M,

A = A0 ⊆ A1 ⊆ . . . ⊆ An ⊆ . . . ,

such that each of their cardinality is less than κ, such that for each

n ≥ 0, if φ(x, y) is an L-formula, and a ∈ An such thatM |=∃yφ(a, y), then it is clear that there exists b ∈ An+1 withM |=φ(a, b).

Given an An, we shall show a way to construct An+1. Requires clarification

Example 14.1.3


Consider the language L of rings and the L-structure C = (C, 0, 1,+,−,×).We know that Q = (Q, 0, 1,+,−,×) is an L-substructure of C. It can

be shown that the structure that is the algebraic closure of Q is an

elementary L-substructure of C.3 Why? How can we showthis?

3



15.1.1 Elementary Embeddings (Continued)

Example 15.1.1

Q = (Q, 0,+,−) has no proper elementary subgroups.

Ò ProofLet G � Q. For a fixed n > 0, we know that

Q |= ∀x ∀y (y + y + . . . + y︸︷︷︸n times

= x)

Q |= ∃x (x 6= 0)

So G must be some non-trivial divisible subgroup of Q. Let nm ∈ G,

with n 6= 0. WMA1 n, m > 0. Then n ∈ G by n divisibility in G. 1 Short for We May Assume. We mayindeed assume that both n and m arestrictly positive, or we can just factorout −1.

Thus 1 ∈ G Thus Z ≤ G ≤ Q, and so2 G = Q by divisibility.

How does this follow?2Remark

By Downward Löwenheim-Skolem, (R, 0,+,−) has many proper elemen-tary subgroups, e.g. Q.3

How so? Is it simply bytaking q + r for q ∈ Q andr ∈ R \Q.15.1.2 Parameters and Definable Sets

Example 15.1.2

In (R,<), the subset (0, 1) is not L-definable: we could have said that

(0 < x) ∧ (x < 1), but 0, 1 /∈ L.


We would now like to rid ourselves of such a restriction.

� Definition 51 (Parameters)

Suppose L is a language,M an L-structure, and B ⊆ M. Let

LB := L ∪ {b : b ∈ B}

be the language L extended with new constant symbols b ∈ B. Theseadditional symbols are called parameters.

We can consequently talk about LB-structures,

MB = (M, bM = b)b∈B.

RemarkSince each of the bM = b ∈ M, we will often just writeM instead ofputting a subscript of B for the LB-structure.4 3

Example 15.1.3

With parameters, in (R,<){0,1}, we can use the L{0, 1}-formula

(0 < x) ∧ (x < 1),

or more easily written as

(0 < x) ∧ (x < 1),

to define the interval (0, 1).

Remark1. If φ(x1, . . . , xn) is an LB-formula, then there exists an L-formula

ψ(x1, . . . , xn, y1, . . . , ym)

for some m < ω, and b1, . . . , bm ∈ B such that

φ(x1, . . . , xn) = ψ(x1, . . . , xn, b1, . . . bm).

Moreover, given a1, . . . , an ∈ M,

M |= φ(a1, . . . , an) ⇐⇒ M |= ψ(a1, . . . , an, b1, . . . , bm).


As mentioned in an earlier comment, we will often write

ψ(~x,~b) as ψ(~x,~b).

2. Suppose N ⊆ M. Then N � M iff for every LN-sentence σ =

φ(b1, . . . , bm), for some b1, . . . , bm ∈ N,

N |= σ ⇐⇒ M |= σ.

In particular, notice that

N |= σ ⇐⇒ NN |= σ ⇐⇒ NN |= φ(b1, . . . , bm)

M |= σ ⇐⇒ M |= φ(b1, . . . , bm).

� Definition 52 (B-Definable)

LetM be an L-structure for some language L, and B ⊆ M a subset. Alet X ⊆ Mn, for some n < ω, is D-definable (or definable over B) ifthere is an LB-formula φ(x1, . . . , xn) such that

X = {(a1, . . . , an) ∈ Mn :MB |= φ(a1, . . . , an)}.

We say that X is 0-definable if it is ∅-definable. We say that X is de-finable if it is M-definable, i.e. definable with parameters from anywherein the universe. We say that X is quantifier-free definable (respec-tively existentially definable or universally definable) if there is aquantifier-free (respectively existential or universal) formula φ such thatX = φM.

å Note• If X is definable, then it is B-definable for some B ⊆ M.

• A function f : X → Y is B-definable if X ⊆ Mn, Y ⊆ Mm areB-definable, and

Γ( f ) ⊆ X×Y ⊆ Mn+m

is B-definable.

Example 15.1.4


Let L = {0, 1,+,−,×}, and R = (R, 0, 1,+,−,×)). We shall as-

sume (in this course, “perversely” so) that a ring is commutative and

unitary. What are the definable sets in R?

Example 15.1.5 (Zero Sets of Polynomials)

SupposeM = (F, 0, 1,+,−,×) where R is a field. Suppose P1, . . . , Pl ∈R[X1, . . . , Xn]. The zero set of {P1, . . . , Pl}

V(P1, . . . , Pl) = {a ∈ Rn : Pi(a) = 0, i = 1, . . . , l}

is called a variety. Such sets are called algebraic sets of Rn. They

form the closed sets of a topology on Rn, called a Zariski topology,

and the closed sets are called Zariski’s closed subsets. We can define

such a set using the LR-formula

l∧i=1

(Pi(x1, . . . , xn) = 0) .

It is clear that Zariski’s closed subsets of Rn are quantifier-free defin-

able in R, with parameters from R.

More generally, any finnite boolean combination5 of Zariski’s 4 Is this saying that from hereon, wewill somewhat be more loose on theconstants of the language, in that wecan use any of the constants from theunderlying universe?

closed subsets are quantifier-free definable. Sets that can be express-

ible as such are said to be Zariski-constructible.

In fact, Zariski’s closed subsets are exactly all of the quantifier-free

definable sets.

Exercise 15.1.1

Let B ⊆ R and S a subring generated by B. The LB-terms of R are preciselythe polynomial functions over S, i.e. for any LB-term, t(x1, . . . , xn), there isPt ∈ S[X1, . . . , Xn] such that tR = Pt, as functions on Rn.

In fact, for any ring A ⊇ S, tA = Pt, where A = (A, 0, 1,+,−,×).

Suppoer φ is an atomic LR-formula. Then φ can be

t(x1, . . . , xn) = s(x1, . . . , xn),

where t, s are LR-terms. Then by the exercise above, we have that

φR =∨(Pt − Ps),

i.e. it is Zariski closed (or Z-closed). We see that quantifier-free for-


mulas are Zariski-constructible (Z-constructible).

The following is a “fact” of which we shall see again later on.

1 TheoremIf R = F is an algebraically closed field, then every definable set in(F, 0, 1,+,−,×) is Z-constructible.

This is a quantifier-elimination theorem, of which we shall study

slightly later on in the course.

RemarkWith regards to the unnumbered theorem above, the only definable subsetesof F are the finite and cofinite6 sets. 5 Boolean combinations include taking

unions, intersections, and complements.

Example 15.1.6

In (R, 0, 1,+,−,×), (0, 1) is definable by (0 < x) ∧ (x < 1) and

< is definable in the structure. However, we have that < is not a

quantifier-free definable set. This is because R as a field is not alge-

braically closed.

Fortunately, just so that we can have some sort of closure7, we 6

� Definition 53 (Cofinite)A cofinite set is a set whose complementis a finite set.

shall present the following theorem without proof.

1 Theorem (Macintyre)If F is a field and the definable sets in (F, 0, 1,+,−,×) are all Z-constructible,then F is algebraically closed.

This gives us some sort of comfort knowing that we have some

method of checking if a field is algebraically closed.

16 Lecture 16 Nov 01st


16.1.1 Parameters and Definable Sets (Continued)

� Corollary 46 (Automorphisms on B-definable Sets)

Suppose X ⊆ Mn be a B-definable set inM, where B ⊆ M. Let

j ∈ AutB(M) := {σ ∈ Aut(M) | σ �B= idB},

such that j acts on Mn coordinate-wise. Then j(X) = X.

Ò ProofSay X is defined by φ(~x,~b) where φ(~x,~y) is a L-formula, ~x =

(x1, . . . , xn), ~y = (y1, . . . ym), and~b = (b1, . . . bm) ∈ Bm. In par-

ticular, X = {~a ∈ Mn | M |= φ(~a,~b). Since j is elementary, we

have

~a ∈ X ⇐⇒ M |= φ(~a,~b)

⇐⇒ M |= φ(j(~a), j(~b))

⇐⇒ M |= φ(j(~a),~b) ∵ j fixes B pointwise

⇐⇒ j(~a) ∈ X

Therefore, j(X) = X as required. �

118 Lecture 16 Nov 01st - First-order Logic (Continued 6)

16.1.2 Theories

� Definition 54 (L-Theory)

An L-Theory is simply a set of L-sentences.

� Definition 55 (Model)

A model of an L-theory T is an L-structureM such thatM |= σ for allσ ∈ T. We denote this byM |= T. A theory is said to be consistent if ithas a model.

� Definition 56 (L-elementary / L-axiomatizable)

A class K of L-structures is said to be L-elementary (or L-axiomatizble)if there is an L-theory T such that

M ∈ K ⇐⇒ M |= T.

Given T, the models of T is

Mod(T) := J M |M |= T K ,

so K is L-elementary if K = Mod(T).

Example 16.1.1

Let L = {e, ·, inv}. The following classes are elementary:1 How do we know this?1

• groups

• abelian groups

• divisible groups (∀x ∃y (yn = x))(note: this is infinitely axiomatizable)

• torsian free groups (∀x (xn = e→ x = e))(note: this is also infinitely axiomatizble)

• for any n, the class of groups of exponent n is elementary (∀x (xn =

e))


The following are not elementary (which we shall prove later)

• all torsion groups

• all finite groups

� Definition 57 (Theories of a Structure)

IfM is an L-structure, the theories of this structure is

Th(M) := {σ :M |= σ},

where σ is an L-sentence.

� Definition 58 (L-elementarily Equivalent)

Given two L-structuresM and N , we sya thatM is L-elementarilyequivalent to N if

Th(M) = Th(N ).

We denote this byM≡ N .

å NoteIf j : M → N is an elementary embedding, thenM ≡ N . In particular,ifM' N , thenM≡ N .

However, the converse is not true: we know from the DownwardLöwenheim Skolem thatM≡ N but it is not necessary thatM' N .

0 Proposition 47

LetM and N be L-structures. There exists an L-elementary embeddingj : M → N iff N has an expansion N ′ to LM such that N ′ |=Th(MM).


Ò ProofGiven j : M → N , let N ′ be the LB-structure that expands N ,

given by

N ′ = (N , bN′= j(b)),

for all b ∈ M. Then by the definition of an elementary L-embedding,

we must have N ′ |= Th(MM).

Conversely, if there is an expansion N ′ of N on LM such that

N ′ |= Th(MM), then we simply need to consider the mapping

j :M→ N by b 7→ bN′,

which will give us the desired elementary L-embedding. �

Example 16.1.2 (⊆ + ≡6=�)

Let N = (ω,<) andM = (ω \ {0},<) be L = {<} structures.

Clearly so,M ⊆ N . They are also isomorphic: j : N → M by

j(n) = n + 1 is an isomorphism. However, notice that

M |= ¬∃x (x < 1)

N |= ∃x (x < 1),

soM 6� N .

� Definition 59 (Implication)

Suppose T is an L-theory, and σ an L-sentence. We say that T impliesσ (or T entails σ), denoted by T |= σ, if for everyM |= T, we haveM |= σ.

� Definition 60 (Complete)

An L-theory T is said to be complete if for every L-sentence σ,

either T |= σ or T |= ¬σ.


å NoteIn words, a theory is complete if every L-formula or its negation can bederived.

Example 16.1.3

Th(M) is complete for any L-structureM.

Ò ProofFor every L-sentence σ, we have either σ ∈ Th(M) or σ /∈ Th(M),

which impliesM |= σ andM |= ¬σ, respectively.

Example 16.1.4

Let’s go on a quest for a complete theory: Let L = {0, 1,+,−,×} and

T be the theory of rings. Then, let

σ = ∀x (x 6= 0→ ∃y (xy = 1)),

which asserts that every non-zero element has a multiplicative inverse.

However, in the models

Z = (Z, 0, 1,+,−,×)

Q = (Q, 0, 1,+,−,×)

we know that

Z |= ¬σ while Q |= σ.

So T 6|= σ and T 6|= ¬σ, i.e. T is incomplete.

Let’s add the property of multiplicative inverses to our theory

and call the new theory T1. This is the theory of fields. Consider the

L-sentence

σ = ∃x (x2 + 1 = 0),

which asserts that every element has a square root. However, we know

that this is true in C but not in R. Thus T1 is still incomplete.

Let T2 be the theory of algebraically closed fields (ACF), i.e. fields

where every element is a root of some polynomial, and in particular

every element will have a square root, mitigating the incompleteness


above. However, the L-sentence σ = 1 + 1 = 0 is true in Halg2 but not

true in C.

Let T3 be the theory of algebraically closed fields with character-

istic p, with p either being prime or 0. This is a complete theory, but

we are not ready to prove this.

+ Lemma 48 (Equivalence to a Complete Theory)

Suppose T is a consistent L-theory. TFAE

1. T is complete;

2. T := {σ | T |= σ}, called the set of consequences of T, is maximallyconsistent;

3. T = Th(M) for some (equivalently any)2 M |= T;What does this mean?

4. Any 2 models of T are elementarily equivalent.

Ò Proof(1) =⇒ (2) : Suppose T is not maximal, i.e. T ( S for some S an

L-theory. Let σ ∈ S \ T. Then T 6|= σ. But T is complete, and so

T |= ¬σ. Thus ¬σ ∈ T ( S. Then we have that {σ,¬σ} ⊂ S and so

S has no models, i.e. S is inconsistent.

(2) =⇒ (3) : IfM is a model of T, thenM is a model of T, thus

T ⊆ Th(M). Since T is maximal, we must have T = Th(M).

(3) =⇒ (4) : SupposeM |= T such that T = Th(M). Let N |= T.

Then N |= T, and as above, we have T = Th(N ). Therefore

Th(M) = Th(N ), i.e. M≡ N .

(4) =⇒ (1) : Suppose T 6|= σ, where σ is an L-sentence. Then

there existsM |= T such thatM |= ¬σ. Let N |= T. By assump-

tion, since any two models of T are elementarily equivalent, we

must also have N |= ¬σ. Therefore T |= ¬σ. �

17 Lecture 17 Nov 06th


17.1.1 Theories (Continued)

� Definition 61 (Partial L-elementary Map)

LetM and N be L-structures, A ⊆ M, and f : A → N a function. f isa partial L-elementary map (pem) if for all L-formulae φ(x1, . . . , xn),and for all a1, . . . , an ∈ A,

M |= φ(a1, . . . , an) ⇐⇒ N |= φ( f (a1), . . . , f (an).

Remark• f : ∅→ N is a pem iffM≡ N .

• f : M→ N is a pem iffM� N (i.e. f is an elementary embedding).

0 Proposition 49 (Elementary Equivalence in Finite Stuctures)

Suppose at least one ofM and N finite. Then

M≡ N ⇐⇒ M ' N .

Ò ProofWithout loss of generality, suppose M is finite. SupposeM ≡ Nand f : ∅ → N is a pem. We shall extend f , one-by-one, until

Dom( f ) = M, so that f remains a pem and is an isomorphism

124 Lecture 17 Nov 06th - First-order Logic (Continued 7)

(Note: We must have |M| = |N| = n for some n < ω, since one of

them is finite).

By induction, suppose that we have a pem

f : {a1, . . . , ak} → N

for 0 ≤ k < n, and {a1, . . . , ak} ⊆ M. Let Xk+1,1, . . . , Xk+1,mk+1⊆ M

be the {a1, . . . , ak}-definable sets that contains ak+1. Note that there

are finitely many such subsets since P(M) is finite. Let

φk+1, j(a1, . . . , ak, x) 1 ≤ j ≤ mk+1

be some (fixed) L-formula that defines Xk+1, j, respectively so. With

that, we have that

M |=mk+1∧i=1

φk+1, j(a1, . . . , ak, ak+1),

which implies that

M |= ∃x

(mk+1∧i=1

φk+1, j(a1, . . . , ak, x)

).

Thus by the induction hypothesis and f being a pem,

N |= ∃x

(mk+1∧i=1

φk+1, j ( f (a1), . . . , f (ak), x)

).

Let b ∈ N satisfy this formula1 and define f ′ : {a1, . . . , ak, ak+1} → What are these other prop-erties, and why should b notsatisfy these other proper-ties?

1 During lecture, it was mentioned thatb should not satisfy other properties.

N such that

a1 7→ f (ai) for 1 ≤ i ≤ k

ak+1 7→ b.

It is clear that f ′ is a pem and an isomorphism. Thus by induction,

we have thatM' N . �

RemarkThis gives us an excuse to “ignore” finite structures when it comes to el-ementary equivalence, since it is no different from studying about isomor-phisms.


17.2 Compactness

1 Theorem (Compactness Theorem)Let L be a lagnauge, and T an L-theory. T is consistent iff every finitesubset of T is consistent.

The following is an equivalent formulation.

� Corollary 50 (Corollary of Compactness Theorem)

Supppose T is an L-theory for some langauge L, and σ some L-sentence.We have T |= σ iff there exists a finite subset Σ ⊆ T such that Σ |= σ.

Ò ProofIt is clear that T |= σ iff T ∪ {¬σ} is inconsistent. �

In this course, we shall study a different proof for the Compact-

ness Theorem, one that uses ultraproducts.

17.2.1 A proof of compactness using ultraproducts

Motivation Observe that T =⋃

n<ω Σn, where Σn ⊆ T is finite, and

Σ0 ⊆ Σ1 ⊆ Σ2 ⊆ . . . .

Since each Σn is consistent, there exists a modelMn |= Σn. Ideally,

the “limit” of theseMn’s should be a model of T.

� Definition 62 (Filter)

Let I ∈ Set and F ⊆ P(I). We say that F is a filter on I if

1. I ∈ F and ∅ /∈ F ;

2. A, B ∈ F =⇒ A ∩ B ∈ F ;

3. A ∈ F ∧ A ⊆ B =⇒ B ∈ F .

126 Lecture 17 Nov 06th - Compactness

RemarkIt is reasonable to think of filters as a notion of “largeness” for subsets ofI; indeed, since any superset of a smaller set in the filter is also in the filteritself.

Example 17.2.1

1. {R \ X | X has Lebesgue measure 0} is a filter on R.

2. Let I ∈ Set, and κ ≤ |I| ∈ Card. Then

F = {A ⊂ I : |I \ A| < κ}

is a filter. In particular, if I = ω, and κ = ℵ0. Then the above filter,

which is the set of cofinite sets, is called a Fréchet filter.

3. Let I ∈ Set \{∅}, and a ∈ I. The set

Fa = {A ⊆ I | a ∈ A}

is called a principal filter on I.

� Definition 63 (Ultrafilter)

An ultrafilter on I is a filter that is maximal.

Example 17.2.2

It is not difficult to notice that principal filters are maximal: Suppose

Fa is a principle filter on I based on a ∈ I, and suppose to the con-

trary that Fa ( G for some filter G on I. Then let A ∈ G \ Fa. In

particular, a /∈ A. However, {a} ∈ Fa ⊆ G. Then {a} ∩ A = ∅ ∈ Gsince G is a filter, but a filter should not contain ∅.

In general, it is rather diffcult to describe a non-principal ultrafil-

ter. However, there is a tool that tells us the existence of ultrafilters at

our disposal: Zorn’s Lemma!

+ Lemma 51 (Existence of Ultrafilters)

Every filter on I extends to an ultrafilter on I.


Ò ProofLet F be a filter on I. Consider

Z = {H ⊆ P(I) | H a filter , F ⊆ H}.

It is rather clear that (Z ,⊆) is a poset, closed under unions of

chains, and the union of filters is a filter. Thus the chains have

an upper bound, and by Zorn’s Lemma, there is a maximal filter,

which is an ultrafilter. �

+ Lemma 52 (Equivalent Characterization of an Ultrafilter)

Let F be a filter on I. F is an ultrafilter iff

∀A ⊆ I (A ⊆ F ) ∨ (I \ A ∈ F ).

Ò Proof(⇐= ) Suppose to the contrary that F ( G is a filter. Then let

A ∈ F \ F . Then by assumption, we have I \ A ∈ F ∈ G, but that

implies

∅ = A ∩ (I \ A) ∈ G,

and so G cannot be a filter. Thus F is an ultrafilter.

( =⇒ ) Suppose F is an ultrafilter. Let A ⊆ I, and suppose

A /∈ F . WTS I \ A ∈ F . Let

G := {X ⊆ I | Y \ A ⊆ X, for some Y ∈ F}.

We shall verify that G is a filter: since I ⊇ Y \ A for any Y ∈ F , I ∈G. Since ∅ is not a superset of any set, ∅ /∈ G. Suppose B, C ∈ G.

Then B ⊇ Y1 \ A and C ⊇ Y2 \ A, for some Y1, Y2 ∈ F . It is clear

that B ∩ C = (Y1 ∩ Y2) \ A. Thus B ∩ C ∈ G. Suppose B ∈ G and

B ⊆ C. Then there exists Y1 ∈ F such that C ⊇ B ⊇ Y1 \ A. Thus

C ∈ G. This proves our claim that G is a filter.

Quite clearly so, since A /∈ F , F ⊆ G. Since F is an ultrafilter,

we have that F = G. Therefore I \ A ∈ G = F , as required. �

128 Lecture 17 Nov 06th - Compactness

We are now ready to define what an ultraproduct is.

� Definition 64 (Ultraproduct)

Let (Mi : i ∈ I) be a sequence of L-structures indexed by I 6= ∅,and U an ultrafilter. An ultrafilter is an L-structure which is defined asfollows: we denote the ultraproductM as

M = ∏UMi,

where

• its universe isM = ×

i∈IMi/E,

where E is an equivalence relation defined as

(ai : i ∈ I)E(bi : i ∈ I) ⇐⇒ {i ∈ I : ai = bi} ∈ U ;

• its constants arecM = [(cMi : i ∈ I)]

for c ∈ Lcon;

• its n-ary functions are

fM([α1], . . . , [αn]) = [( fMi (a1(i), . . . , an(i)) : i ∈ I)]

where [α1], . . . , [αn] ∈ M and αj = (aj(i) : i ∈ I), where aj(i) ∈ Mi;

• its k-ary relations are

([α1], . . . , [αn]) ∈ RM ⇐⇒ {i ∈ I : (a1(i), . . . , an(i)) ∈ RMi} ∈ U .

Exercise 17.2.1

Verify that the functions and relations of the ultraproductM is well-defined; in particular, the definition of the functions and relations are in-dependent of the elements [αi] ∈ M.


å NoteNote that in the definition of an ultraproduct, we may notice that, almostdeceivingly so, U need not be an ultrafilter.


18.1 Compactness (Continued)

18.1.1 A Proof of Compactness Using Ultraproducts (Continued)

Recall our last note in the last lecture.

Example 18.1.1

Suppose that F = {I} is a filter (which it trivially is), then E is

the trivial equivalence relation, andM would just be the “cartesian

product” of theMi’s.

For instance, suppose that we have the language L = {0, 1,+,−,×}and I = ω. Then F = {ω}. Consider the L-structureM =

(Z, 0, 1,+,−,×). Then in

R = ×i∈I

(Z, 0, 1,+,−,×),

we have

(1, 0, 1, 0, . . .) · (0, 1, 0, 1, . . .) = (0, 0, 0, 0, . . .)

but neither (1, 0, 1, 0, . . .) nor (0, 1, 0, 1, . . .) are (0, 0, 0, 0, . . .). There-

fore R is not an integral domain.

Suppose that we have, instead, constructed an ultraproduct

S = ∏UMi

with an ultrafilter U . Then in comparison, we have

[(1, 0, 1, 0, . . .)] · [(0, 1, 0, 1, . . .)] = [(0, 0, 0, 0, . . .)] = 0S .

Now if [(1, 0, 1, 0, . . .)] 6= 0S , then any of the elements that is zero on

the even entries would not be in U . If [(0, 1, 0, 1, . . .)] 6= 0S , then any

132 Lecture 18 Nov 08th - Compactness (Continued)

of the elements that is zero on the odd entries would not be in U . But

these sets are complements of each other, and so one of them must be

in U since U is an ultrafilter.

1 Theorem 53 (Łos)

Suppose we have a sequence of L-structures (Mi : i ∈ I), an ultrafilterU , an ultraproduct of theMi’sM = ∏U , an L-formula φ(x1, . . . , xn),and elements [α1], . . . , [αn] ∈ M, where α1, . . . , αn ∈ ×

i∈IMi. Then

M |= φ([α1], . . . , [αn]) ⇐⇒ {i ∈ I :Mi |= φ(α1(i), . . . , αn(i))} ∈ U .

In particular, if n = 0, i.e. for any L-sentence σ, we have

M |= σ ⇐⇒ {i ∈ I :Mi |= σ} ∈ U .

Ò ProofWe shall first prove the following statement about L-terms, of

which we will need for the rest of the proof.

Claim If t(x1, . . . , xn) is an L-term, then for any [α1], . . . , [αn] ∈M, we have

tM([α1], . . . , [αn]) = [(tMi (α1(i), . . . , αn(i)) : i ∈ I)].

Proof of claim This is clearly true for when our L-terms is either

constants for functions, simply from the way of which we have

defined them. If our L-term is a variable, i.e. if t(xi) = xi, then the

result follows as the xi’s are variables for exactly any of the [αj]’s. a

We shall prove Łos’s theorem by the complexity of φ(x1, . . . , xn).

Atomic formulae If φ is of the form

t(x1, . . . , xn) = s(x1, . . . , xn),


then we have

M |= φ([α1], . . . , [αn])

⇐⇒ tM([α1], . . . , [αn]) = sM([α1], . . . , [αn])

⇐⇒ [(tMi (α1(i), . . . , αn(i)) : i ∈ I)] = [(sMi (α1(i), . . . , αn(i)) : i ∈ I)]

⇐⇒ {i ∈ I : tMi (α1(i), . . . , αn(i)) = sMi (α1(i), . . . , αn(i))} ∈ U

⇐⇒ {i ∈ I :Mi |= φ(α1(i), . . . , αn(i))} ∈ U .

If φ is of the form (t1, . . . , tk) ∈ Rk, where tj’s are L-terms, then

M |= φ([α1], . . . , [αn])

⇐⇒ (tM1 ([α1], . . . , [αn]), . . . , tMk ([α1], . . . , [αn])) ∈ RM ⊆ Mk

⇐⇒ {i ∈ I : (tMi1 (α1(i), . . . , αn(i)), . . . tMi

n (α1(i), . . . , αn(i))) ∈ RMi} ∈ U

⇐⇒ {i ∈ I :Mi |= φ(α1(i), . . . , αn(i))} ∈ U ,

where the second ⇐⇒ is by the interpretation of R inM.

Logical Operators If φ is of the form ¬ψ for some L-formula

ψ(x1, . . . , xn), then

M |= φ([α1], . . . , [αn])

⇐⇒ M 6|= ψ([α1], . . . , [αn])

IH⇐⇒ {i ∈ I :Mi |= ψ(α1(i), . . . , αn(i))} /∈ U

⇐⇒ {i ∈ I :Mi 6|= ψ(α1(i), . . . , αn(i))} ∈ U ∵ U is an ultrafilter

⇐⇒ {i ∈ I :Mi |= φ(α1(i), . . . , αn(i))} ∈ U .

If φ is of the form φ1 ∨ φ2, where φ1(x1, . . . , xn), φ2(x1, . . . , xn)

are L-formulas, then

M |= φ([α1], . . . , [αn])

⇐⇒ M |= φ1([α1], . . . , [αn]) andM |= φ2([α1], . . . , [αn])

⇐⇒ {i ∈ I :Mi |= φ1(α1(i), . . . , αn(i))},

{i ∈ I :Mi |= φ2(α1(i), . . . , αn(i))} ∈ U

⇐⇒ {i ∈ I :Mi |= φ1(α1(i), . . . , αn(i)),

Mi |= φ2(α1(i), . . . , αn(i))} ∈ U

⇐⇒ {i ∈ I :Mi |= φ(α1(i), . . . , αn(i))} ∈ U .

There is no need to show for the case of ∧, as it is equivalent to

134 Lecture 18 Nov 08th - Compactness (Continued)

showing ¬ ∨ ¬.

Quantifiers As mentioned in a few of the previous proofs in

earlier propositions and theorems, it suffices to only prove for

the existential case. Suppose φ is of the form ∃yψ(y), for some

L-formula ψ(x1, . . . , xn, y). Then

M |= φ([α1], . . . , [αn])

⇐⇒ there exists [β] ∈ M M |= ψ([α1], . . . , [αn], [β])

⇐⇒ there exists [β] ∈ M

Aβ := {i ∈ I :Mi |= ψ(α1(i), . . . , αn(i))} ∈ U

Now our goal is to show that

Aβ ∈ U

⇐⇒

B := {i ∈ I : there exists bi ∈ Mi Mi |= ψ(α1(i), . . . , αn(i), bi)} ∈ U

We can, in fact, show that Aβ = B. It is clear that Aβ ⊂ B, since we

may just label bi = β(i). For B ⊆ Aβ, define β : I → ⋃i∈I Mi by

β(i) =

bi i ∈ B

ei i /∈ B

where ei is just some dummy constant. Then β ∈ ×i∈I

Mi.

Now if i ∈ B, we have that β(i) = bi such thatMi |= ψ(α1(i), . . . , αn(i), bi),

and so

Mi |= ψ(α1(i), . . . , αn(i), β(i)),

i.e. i ∈ Aβ. This completes the proof. �

Example 18.1.2

The class of finite groups is not axiomatizable.

Ò ProofLet Gn = (Z/nZ, 0,+,−). Let U be an ultrafilter that extends the

Fréchet Filter. Then, let

G = ∏U

Gn.


Since each of the Gn’s satisfies the axioms of groups, by Łos’ The-

orem, we have that G is also a group. Also, G is infinite. Let σn be

the L-sentence

σn = ∃x1, . . . , xn

∧1≤i<j≤n

xi 6= xj

,

which is the sentence that says that “I have n elements (and I may

have more)”. Then we have

{m < ω : Gm |= σn} ⊇ {m < ω : m ≥ n} ∈ U

since the set in the middle is cofinite. Thus the first set is in U . In

particular, we have that

G |= σn =⇒ |G| ≥ n

for any n < ω. Then there exists an L-sentence σ such that it says

that “I have at least ℵ0-many elements (and I may have more)”.

This sentence cannot be satisfied by any of the finite structures.


19.1 Compactness (Continued 2)

19.1.1 A Proof of Comapactness Using Ultraproducts (Continued 2)

We shall introduce one of the special cases of an ultraproduct:

� Definition 65 (Ultrapower)

Given I 6= ∅, U an ultrafilter on I, and a seqeunce (Mi : i ∈ I) ofL-structures such thatMi =M, some L-structure. Then

M′ = ∏UM

is called the ultrapower ofM.

� Corollary 54 (Corollary of Łos)

Given an L-structureM, the diagonal map

δ :M→∏UM given by a 7→ [(a, a, . . .)]

is an elementary L-embedding.

å NoteFrom this corollary, we see that ultrapowers are elementary extensions ofthe single L-structure.

Also, if U is principal, thenM' ∏UM 1 1 This may be an interesting or at least arelatively simple exercise.

Exercise 19.1.1Show that if U is principal in� Corollary 54, thenM' ∏UM.

138 Lecture 19 Nov 13th - Compactness (Continued 2)

Ò ProofTo show that δ is an elementary L-embedding, we shall show that

M � ∏UM. Let φ(x1, . . . , xn) be an L-formula, and let d : M →MI be defined by

d(ai) = (ai, ai, . . .).

Then fromM, we have

M |= φ(a1, . . . , an)

=⇒ {i ∈ I :M |= φ(d(a1)(i), . . . , d(an)(i))}(∗)= {i ∈ I :M |= φ(a1, . . . , an)}

(∗∗)= I ∈ U

Łos⇐⇒ ∏UM |= φ([d(a1)], . . . , [d(an)])

⇐⇒ ∏UM |= φ(δ(a1), . . . , δ(an)),

where (∗) is because the two realizations give the same i’s, and

(∗∗) is because the i’s do not matter in the condition, allowing

all i ∈ I to be in the set. This gives us that L-formulas from the

original structure is true in its ultrapower.

From ∏UM, we have

∏UM |= φ(δ(a1), . . . , δ(an))

Łos⇐⇒ {i ∈ I :M |= φ(d(a1)(i), . . . , d(an)(i))} ∈ U .

Then, for any i0 in the above set, we have

M |= φ(d(a1)(i0), . . . , d(an)(i0)) =⇒ M |= φ(a1, . . . , an).


ℵ1-compactness We shall take a little detour and visit an aside that

utilizes Łos’ Theorem.

� Definition 66 (ℵ1-Compact)

We say that a set A is ℵ1-compact if any countable collection of definable


non-empty sets in A, with the finite intersection property2, then the 2 The finite intersection property issuch that every finite non-emptysubcollection non-empty sets has anon-empty intersection. See also noteson PMATH 351 (Real Analysis).

collection has a non-empty intersection.

0 Proposition 55 (Ultraproducts with a Non-Principal Ultrafilteris ℵ1-compact)

Suppose (Mi : i < ω) is a sequence of L-structures, and U a non-principal ultrafilter on ω. Then ∏UMi is ℵ1-compact.

å NoteThe given proof is not one that I can fully agree with, since in theproof we may somehow assume that the sets are nested, which is not givenin our assumption, and may not actually be a necessary condition. Theprovided proof will be recorded for archiving purposes.

Ò ProofLetM = ∏UMi, and (Fi : i < ω) a sequence of definable non-

empty sets inM, such that Fi 6= ∅. Suppose that Fj ⊆ Fi for i < j.We may assume that F0 = Mn, given by the L-formula x = x. For

0 < i < ω, let φi(x) be the L-formula that defines Fi. For each i, let

ni = max{n ≤ i :Mi |= ∃xφn(x)},

where we know that the set above is non-empty as F0 is defined.

Then we have that 0 ≤ ni ≤ i. Also for each i, let ai ∈ Mi be such

thatMi |= φni (ai). Now let a = (ai : i < ω). We want to show that

for all n < ω,

M |= φn([a])

so that

[a] ∈⋂

n<ω

Fn.

Come back to this proof after

the exam.Leaving it behind for now to focuson the more important stuff. Seealso https://math.stackexchange.

com/questions/3027612/

ultraproducts-with-a-non-principal-ultrafilter-is-aleph-one-compact

[solved].

Example 19.1.1

https://tex.japorized.ink/PMATH351F18/classnotes.pdf

https://tex.japorized.ink/PMATH351F18/classnotes.pdf

https://math.stackexchange.com/questions/3027612/ultraproducts-with-a-non-principal-ultrafilter-is-aleph-one-compact



140 Lecture 19 Nov 13th - Compactness (Continued 2)

(Z,<) is not ℵ1-compact but ∏U (Z,<), where U is a non-principal

ultrafilter, is ℵ1-compact.

We are now ready to tackle on the problem which we have set out

to solve.

1 Theorem 56 (Compactness Theorem)

Let T be an infinite L-theory. If every finite subset of T is consistent, thenT is consistent.

Ò ProofLet I be the collection of finite subsets of T. Then by assumption,

we have that for i ∈ I,

Mi |= i.

Consider the collection of supersets of i

Xi := {i′ ∈ I : i ⊆ i′}.

The collection of these Xi’s gives us the structure of a lattice:

i ∪ j

i j

i ∩ j

∅

Figure 19.1: Lattice structure from theXi’s. Red lines are the parts captured byXi .

3 Consider

Need to show an example ofV failing to “capture” super-sets.

3 Now if we simply consider the collec-tion of these collections:

V = {Xi : i ∈ I},

we notice that V is almost a filter, butnot quite: we have

• I = X∅ ∈ V and ∅ /∈ V ; and

• for Xi , Xj ∈ V , we have Xi ∩ Xj =Xi∪j ∈ V .

F := {Y ⊆ I : Y ⊃ Xi, i ∈ I}.

Notice that F is a filter, as

• I ∈ F , ∅ ∈ F ;

• ∀Y, Z ∈ F Y ∩ Z ⊃ Xi∪j =⇒ Y ∩ Z ∈ F ; and

• ∀Y ∈ F ∧ Z ⊃ Y =⇒ Z ⊃ Y ⊃ Xi for some i ∈ I =⇒ Z ∈ F .

Let U ⊃ F be an ultrafilter4. LetM = ∏U Mi.

4

Exercise 19.1.2Show that U is a non-principal ultrafilter.

To show thatM |= T, let σ ∈ T. Then {σ} ∈ I and so X{σ} ∈ Uby construction. Then

X{σ} = {i ⊂ T finite : σ ∈ i}.

By assumption, for each i ∈ X{σ}, we haveMi |= i, and so in

particularMi |= σ. Then in particular,

X{σ} ⊂ Sσ := {i ∈ I :Mi |= σ} =⇒ Sσ ∈ U .


Thus by Łos we haveM |= σ and soM |= T as required. �


20.1 Compactness (Continued 3)

20.1.1 First Applications of Compactness

Example 20.1.1

Consider the empty language and K the class of all infinite L-structures

(as pure sets). Show that K is axiomatizable but not finitely so.

Ò Proof

Bibliography

Cohen, P. J. (1963). The independence of the continuum hypothesis.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC221287/.

Moosa, R. (n.d.). Set Theory and Model Theory. (n.p.), 5th edition.


Index

0-definable, 113

B-Definable, 113

I-sequence, 77

M-definable, 113

ℵ1-Compact, 138

α-function, 47

∅-definable, 113

L-Definable, 99

L-Embedding, 89

L-Sentences, 97

L-Theory, 118

L-axiomatizable, 118

L-elementarily Equivalent, 119

L-elementary, 118

L-formulas, 95

L-isomorphism, 92

L-structure, 88

L-terms, 92

Łos, 132

arity, 85

Atomic L-Formulas, 95

automorphism, 55

Axiom of Choice, 65

Axiom of Extension, 17

basic functions, 85

basic relations, 85

binary definite condition, 23

Bound, 96

Bounded Separation Axiom, 22

Cantor’s Diagonalization, 79

Cardinal, 61

Cardinal Exponentiation, 80

Cardinal Numbers, 72

Cardinal Product, 76

Cardinal Sum, 75

Cardinality, 69

Cartesian Product, 28

Choice Function, 65

Class, 24

Cofinite, 115

Compactness Theorem, 125, 140

Complete, 120

consistent, 118

constants, 85

Continuum Hypothesis, 80

Corollary of Łos, 137

Countable, 61

Definite Condition, 18

Definite Operation, 29

Downward Löwenheim-Skolem, 109

Elementary Embeddings, 105

elementary substructure, 106

Empty Set Axiom, 17

Equinumerous, 59

existentially definable, 113

Expansion, 87

extension, 90

Fibres, 71

Filter, 125

Finite, 61

Fréchet filter, 126

Free Variables, 96

Functions, 30

Generalized Cardinal Product, 78

Generalized Cardinal Sum, 77

Generalized Continuum Hypothesis,81

graph of f , 30

Implication, 120

Induction Principle, 31

Infinity Axiom, 21

Initial Segment, 56

Interpretation, 88, 94

isomorphic, 53

Isomorphism, 53

Language, 87

Limit Ordinal, 42

Model, 118

Natural Numbers, 23

Ordered Pairs, 27

Ordinal Addition, 50

Ordinal Exponentiation, 53

Ordinal Multiplication, 51

Ordinals, 37

Pairset Axiom, 17

Parameters, 112

Partial L-elementary Map, 123

Powerset Axiom, 22

principal filter, 126

proper classes, 25

pullback, 71

quantifier-free definable, 113

Realization, 97

Reduct, 87

Replacement Axiom, 23, 30

Rigid, 55

Russell’s Paradox, 24

Satisfaction, 97

Schröder-Bernstein Theorem, 59

Separation Axiom, 22

Set Intersection, 22

Strict Linearly Ordered Set, 33

Strict Partially Ordered Set, 33

Strict Poset, 33

Strict Totally Ordered Set, 33

Structure, 85

Subsets, 21

Substructure, 90

Successor, 16

Successor Ordinal, 42

successor set, 21

Tarski-Vaught Test, 107

the set of consequences of T, 122

Transfinite Induction Theorem, 43

Transfinite Recursion, 47, 49

Ultrafilter, 126

Ultrapower, 137

Ultraproduct, 128

Union Set Axiom, 18

universally definable, 113

universe, 85

variety, 114

Well-Order, 33

Zariski, 114

Zariski-constructible, 114

Todo list

Requires clarification . . . . . . . . . . . . . . . . . . . . . . . . . 109

Why? How can we show this? . . . . . . . . . . . . . . . . . . . . 110

How does this follow? . . . . . . . . . . . . . . . . . . . . . . . . . 111

How so? Is it simply by taking q + r for q ∈ Q and r ∈ R \Q. . 112

How do we know this? . . . . . . . . . . . . . . . . . . . . . . . . 118

What are these other properties, and why should b not satisfy theseother properties? . . . . . . . . . . . . . . . . . . . . . . . . . 124

Come back to this proof after the exam. . . . . . . . . . . . . . . . 139

Need to show an example of V failing to “capture” supersets. . . 140

pmath433/733 — model theory and set theory · theory, we look at set theory as a language of...

Documents