Math 145a - Set Theory I

Taught by William BoneyNotes by Dongryul Kim

Fall 2016

This course was taught by William Boney. We met on Tuesdays and Thurs-days from 1:00pm to 2:30pm. We used Set Theory: The Independence Phe-nomenon by Peter Koellner as a textbook. There were 15 people enrolled in thecourse. Grading was based on 60% weekly homeworks, 20% in-class midterm,and 20% three-hour final. The course assistance was Justin Cavitt.

Contents

1 September 1, 2016 41.1 Three problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Functions and pairs . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Natural numbers to real numbers . . . . . . . . . . . . . . . . . . 5

2 September 6, 2016 72.1 Zermelo-Fraenkel with Choice . . . . . . . . . . . . . . . . . . . . 72.2 Equivalents of Replacement . . . . . . . . . . . . . . . . . . . . . 92.3 Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 September 8, 2016 103.1 Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Axiom of Choice . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.3 Cardinality of sets . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 September 13, 2016 134.1 Well-founded and well-ordered . . . . . . . . . . . . . . . . . . . 134.2 Transfinite induction . . . . . . . . . . . . . . . . . . . . . . . . . 13

5 September 15, 2016 165.1 Ordinals for real . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.2 Properties of ordinals . . . . . . . . . . . . . . . . . . . . . . . . 17

1 Last Update: August 27, 2018

6 September 22, 2016 196.1 Ordinal induction and arithmetic . . . . . . . . . . . . . . . . . . 196.2 Building the universe . . . . . . . . . . . . . . . . . . . . . . . . . 206.3 Cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

7 September 27, 2016 237.1 Cardinal arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . 237.2 Cofinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

8 October 4, 2016 268.1 Successor cardinals are regular . . . . . . . . . . . . . . . . . . . 26

9 October 4, 2016, Make-up class 289.1 Regular limit cardinals . . . . . . . . . . . . . . . . . . . . . . . . 289.2 Inaccessible cardinals and models of ZFC . . . . . . . . . . . . . 29

10 October 6, 2016 3210.1 The measure problem . . . . . . . . . . . . . . . . . . . . . . . . 3210.2 The Baire space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

11 October 11, 2016 3511.1 Midterm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

12 October 18, 2016 3612.1 Topology of ωω . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3612.2 Measures on the Baire space . . . . . . . . . . . . . . . . . . . . . 37

13 October 20, 2016 3813.1 Borel sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3813.2 Projective sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3913.3 Filters and ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . 40

14 October 25, 2016 4114.1 Ultralimits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4114.2 Ultraproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

15 October 25, 2016, Make-up class 4315.1 Banach-Tarski Paradox . . . . . . . . . . . . . . . . . . . . . . . 43

16 October 27, 2016 4516.1 Dense linear orderings . . . . . . . . . . . . . . . . . . . . . . . . 45

17 November 1, 2016 4717.1 Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4717.2 Branches of trees . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2

18 November 3, 2016 4918.1 ω1-Aronszajn trees . . . . . . . . . . . . . . . . . . . . . . . . . . 4918.2 Other classes of tress . . . . . . . . . . . . . . . . . . . . . . . . . 50

19 November 8, 2016 5119.1 Suslin trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

20 November 10, 2016 5320.1 The idea of L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5320.2 First-order logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

21 November 15, 2016 5621.1 Substructures and elementary substructures . . . . . . . . . . . . 5621.2 Models of set theory . . . . . . . . . . . . . . . . . . . . . . . . . 5721.3 Levy hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

22 November 17, 2016 6022.1 Godel codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6022.2 Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6122.3 Constructing the constructible universe . . . . . . . . . . . . . . 62

23 November 22, 2016 6423.1 L satisfies ZF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6423.2 Defining definablity . . . . . . . . . . . . . . . . . . . . . . . . . . 65

24 November 29, 2016 6724.1 L satisfies Choice . . . . . . . . . . . . . . . . . . . . . . . . . . . 6724.2 Models satisfying V = L . . . . . . . . . . . . . . . . . . . . . . . 68

25 December 1, 2016 7025.1 L satisfies GCH . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7025.2 Forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3

Math 145a Notes 4

1 September 1, 2016

Because it’s hard to start from the very bottom, I am going to start with threemotivating problems.

1.1 Three problems

The cardinality of X is the number of elements in X. For instance we aregoing to say that |X| = |Y | if there is a bijection f : X → Y .

Theorem 1.1. For every X, its power set P(X) = {Y : Y ⊆ X} has greatercardinality than X.

An immediate corollary is that |N| < |P(N)| = |R|. This leads to a questionthat comes from Cantor: is there anything between them? This fundamentalproblem was left open until it was answered as not an answer. The statement isindependent of ZFC, i.e., the foundations of ZFC is just not enough to answerthis problem. The continuum hypothesis states that the answer to this isno, and the generalized continuum hypothesis deals with the existence of setsY with |X| < |Y | < |P(X)| for an arbitrary set X.

The second problem is about combinatorics. The ordered set (R, <) can becharacterized by (1) dense linear ordering without endpoints, (2) connected inthe order topology, and (3) separable in the order topology. Now Suslin’s ques-tion was whether one can replace the third condition (3) to (3’) every collectionof disjoint open intervals is countable. The answer to this is again, who knows?

The word combinatorics is used here because such questions are related totrees, or proper generalization of trees.

Lemma 1.2 (Konig’s lemma). If T is an infinite tree such that every vertexhas finitely many neighbors, then there is an infinite path through T .

One thing set-theorists like to do is to take a true statement and ask whathappens when objects get larger and larger. It fails on the next one. This isbasically because there is this nice countable dense set Q. But things get inter-esting when we take it to the next step. Then the independence phenomenoncomes in to the picture.

The third theme is measure theory. This is the measure theory in the actualmeasure theoretic sense.

Theorem 1.3. (1) (Vitali) There are non-measurable sets.(2) (Banach-Tarski) You can decompose one sphere and reassemble them intotwo spheres of equal volume.

One of my friends use the Banach-Tarski theorem to give a physical proofof the negate of the axiom of choice. But I don’t accept this proof. There is agood reason we can’t do this in nature, and it’s because they never occur in thephysical world. This is related to the hierarchy of sets in descriptive set theory.

Math 145a Notes 5

1.2 Functions and pairs

“All the world’s a set, and all mathematicians and theorems merely elements.”—William Setspeare

Every mathematical object is built out of sets, and those elements are allset. So in fact, every thing can be written in curly braces and commas.

As an example, how do we encode a function f : A → B as sets? We canthink it as a set of ordered pairs

f = {(a, f(a)) : a ∈ A}.

This is also called the graph of f but is actually just f .But to do this, we need to know what ordered pairs are. Given x, y, we want

a set (x, y) such that (x, y) = (a, b) if and only if x = a and y = b. We define

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

You will work this out in you homework. One thing to watch out is when x = y.

1.3 Natural numbers to real numbers

We are not going to construct the natural numbers N, which starts with 0 inthis course. We define

0 = {} = ∅.

The next number is 1. We’re going to define 1 to be

1 = {∅} = {0}.

Then we define 2 = {∅, {∅}} = {0, 1} and so on. Given n, we are going to define

n+ 1 = n ∪ {n}.

Then the natural numbers is N = {n : n ∈ N} although it looks silly. We definen < m if and only if n ∈ m. Likewise we can say that n ≤ m if and only ifn ⊆ m. Here, X ⊆ Y is defined as ∀x(x ∈ X → x ∈ Y ).

To define n + m, we need more than just elements. We are going to definen+m as the unique k such that there is a bijection from k to ({0}×m)∪({1}×n).Likewise, we define nm to be the unique k such that there is a bijection from kto n×m.

Now we get to the integers Z. Now there is a standard way to make a monoidinto a group, in algebra. For (m,n) ∈ N2, we want this to represent m−n. Butbecause we don’t know subtraction yet, we set

(m,n) ∼ (m′, n′) if and only if m+ n′ = m′ + n.

This is in fact an equivalence relation. So we define

Z = {[(m,n)]∼ : (m,n) ∈ N2}.

Math 145a Notes 6

If you have in mind what (m,n) represents, it is not hard to extend additionand multiplication. We define

[(m,n)]∼ + [(m′, n′)]∼ = [(m+m′, n+ n′)]∼,

[(m,n)]∼ · [(m′, n′)]∼ = [(mm′ + nn′,mn′ +m′n)]∼.

In a similar way, we can construct the rationals. This is just turning a ringinto the field of fractions. For (a, b), (c, d) ∈ Z2 with b, d 6= 0, we define anequivalence relation

(a, b) ∼ (c, d) if ad = bc.

Then we can define the equivalences classes as Q. We end up with

[(a, b)]∼ + [(c, d)]∼ = [(ad+ bc, bd)]∼.

Lastly I want to talk about the reals mainly because I have put a problemabout the reals in the problem set. You could give a metric and then define theCauchy sequences as real numbers, but the way I am going to do is throughDedekind cuts. Say that L ⊆ Q is proper downward closed if ∅ 6= L 6= Qand for any x < y ∈ Q if y ∈ L then x ∈ L. Define L0 ∼ L1 if they have thesame supremum, which can be written as “for any x ∈ L0 and n ∈ N thereexists a y ∈ L1 such that x− 1/n < y and vice versa.” Then we let

R = {[L]∼ : L ⊆ Q proper downwards closed}.

The equivalence relation is to take care of (−∞, 0) ∼ (−∞, 0].

Math 145a Notes 7

2 September 6, 2016

We’re going to define Zermelo-Fraenkel set theory with Choice. There will betwo types of axioms. The first will be that certain sets exist. The second typewill be that we can get some sets from other sets. To deal with sets, we are goingto use a single binary relation ∈ (and =), which intuitively means “member of”.

2.1 Zermelo-Fraenkel with Choice

This is just a list of axioms.

Axiom of Extensionality.

∀x, y, [x = y ↔ ∀z(z ∈ x↔ z ∈ y)]

That is, sets are made up of elements.

We are going to define x ⊆ y to mean

∀z(z ∈ x→ z ∈ y).

So we can restate the Axiom of Extensionality as

∀x, y, [x = y ↔ (x ⊆ y ∧ y ⊆ x)].

Axiom of Empty set.∃a∀x(x /∈ a)

That is, there is an empty set, which we will denote ∅.

Axiom of Pairing.

∀x, y∃z∀w(w ∈ z ↔ w = x ∨ w = y)

If x, y are sets, so is {x, y}.

This implies that because ∅ is a set, {∅} is also a set.

Axiom of Union.

∀x∃a∀z(z ∈ a↔ ∃y(z ∈ y ∧ y ∈ x))

This a is denoted⋃x.

This is a bit different from what we are used to, because⋃i∈I Xi is what we

now defined as⋃{Xi : i ∈ I}. In a little bit, we will justify this. We will need

to think about why the set of Xi is a set.

Axiom of Power Set.

∀x∃P(x)∀x(z ∈ P(x)↔ z ⊆ x)

P(x) is the collection of all subsets of x.

Math 145a Notes 8

This is what justifies talking about its subsets, like in defining a topologicalspace. This also says that you can get all of the subsets of x, no matter howhuge x is.

Axiom of Infinity.

∃a(∅ ∈ a ∧ ∀y(y ∈ a→ y ∪ {y} ∈ a)

There is a set that contains all the “natural numbers”.

It would be nice to show that y 6= {y}. The next axiom will help us outhere. We note that we haven’t actually justified that the set of natural numbersis a set.

Now this was what Zermelo came along when Russel found a problem. Peoplewere happy for a while but there were still problems and Fraenkel added thissingle axiom on top of Zermelo’s system.

Axiom of Foundation/Regularity.

∀x[x 6= ∅ → ∃y(y ∈ x ∧ ∀z(z ∈ x→ z /∈ y))]

You can show that this is equivalent to ¬(∃{xn : n ∈ N}(xn+1 ∈ xn)).A formula is going to be ϕ(x1, . . . , xn) and built from

• xi ∈ xj and xi = xj ,

• using logical symbols →, ∧, ∨, ¬,

• using quantifiers ∃, ∀.

This is called first order logic, and the beauty of set theory is that everything isfirst order logic, because of the Axiom of the Power Set and the other axioms.

Replacement Schema. For every ϕ(x, y, z1, . . . , xn), include

∀a1, . . . , an∀a[∀y ∈ a∃!zϕ(y, z, a1, . . . , an)→ ∃b∀z(z ∈ b↔ ∃y ∈ aϕ(y, z, a1, . . . , an)]

If ϕ(x, y, a1, . . . , an) defines a function with domain a, then its image is a set.

We can’t write this as a single axiom for metamathematical reasons. Al-though ZFC looks like a finite list of axioms, it is not because this is a schema.It gives an axiom for every formula.

Axiom of Choice.

∀x[∀y(y ∈ x→ y 6= ∅)→ f a function with domain x(∀y ∈ x(f(y) ∈ y))]

There are instances when picking a choice function becomes easy. The firstis when there are a finite number of sets. The second is when the sets have someadditional structure so that you can find a certain element easily. You need theAxiom of Choice when you are picking among an infinite pair of socks but youdon’t need it when you are picking among an infinite pair of shoes, because youcan choose the left one.

Math 145a Notes 9

2.2 Equivalents of Replacement

There is an equivalent formulation of the Replacement Schema that I want totalk about.

Comprehension.

∀a1, . . . , an∀a∃b∀x(x ∈ b↔ (x ∈ a ∧ ϕ(x, a1, . . . , an))

This means that if you have a set, then the elements in there that satisfiessomething is also a set.

Collection.

∀a1, . . . , an∀a[y ∈ a∃!zψ(y, z, a1, . . . , an)→ ∃b∀z(y ∈ aψ(y, z, a1, . . . , an)→ z ∈ b)]

This means that if you have some function then there is the set that containsthe image.

Now if you have both Comprehension and Collection, then you can get Re-placement by just taking the set that contains the image and restricting it to theactually image using Comprehension. It is also easy to see that Replacementimplies Collection.

Proposition 2.1. Under the other axioms of ZF, the Replacement Schemaimplies the Comprehension Schema.

Proof. Take a and φ(x, a1, . . . , an). If ¬ϕ(e, a1, . . . , an) for all e ∈ a, then weare done since what we want is the empty set. If not, let e ∈ a be a witness.Define ψ(x, z, a, e, a1, . . . , an) to be

[x ∈ a ∧ ϕ(x, a1, . . . , an)→ x = z] ∧ [¬(x ∈ a ∧ ϕ(x, a1, . . . , an))→ z = e].

Then this ψ does what we want to do in Replacement.

2.3 Basic notions

We are still in a very clunky stage because we don’t have a lot of tools we usedto have.

Given x, y, we define the intersection as

x ∩ y = {a ∈ x : a ∈ y}.

We also can define the ordered tuple as

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

For n-tuples, we define (x1, x2, x3) = ((x1, x2), x3) and iterate this process.We can now define the product as

X × Y = {(x, y) ∈ P(P(X ∪ Y )) : x ∈ X ∧ y ∈ Y }.

Then we can define functions too.

Definition 2.2. A set f is a function if there exists X,Y such that f ∈P(X,Y ) such that for all x ∈ X there is a unique y ∈ Y such that (x, y) ∈ f .We are going to write this as f(x) = y because we are mathematicians.

Math 145a Notes 10

3 September 8, 2016

By Union, we know that⋃x =

⋃y∈x y is a set. So if {Xi : i ∈ I} is a set, then⋃

i∈IXi =

⋃{Xi : i ∈ I}

is a set.

3.1 Classes

Everything we talk about is a set. But we need to distinguish it from thefollowing proposition.

Proposition 3.1. There is no set of all sets.

Proof. Suppose x is the set of all sets. Then x ∈ x contradicts Foundation. Oralternatively {y : y /∈ y} is a set by Comprehension, and Russel’s paradox leadsto a contradiction.

But I’m talking about the collection of sets, by giving a quantifier. A classis simply any collection of sets. Clearly any set is a class. A proper classis a class that is not a set. Definable classes are classes with a first-orderdefinition, and all other classes are called undefinable classes.

Example 3.2. The formula x = x defines the class V , which is also called VonNeumann universe. Another class is Grp, which is the class of all groups. Ican’t give an example of an undefinable class, because if I can, then it would bedefinable.

The reason there are proper classes is because there can be collections thatare too big. Suppose a is a set and X is a class such that there is a bijectionf : a → X . Here f is not really a set but a definition of a function. Then byReplacement, X is a set.

3.2 Axiom of Choice

Let us recall that this axiom says that

∀x(∀y(y ∈ x→ y 6= ∅)→ ∃function f with dom. x s.t. ∀y(f(y) ∈ y)).

Definition 3.3. A relation < on X is called a well-ordering if and only if

• ∀x, x 6< x.

• ∀x, y, x < y or x = y or y < x.

• ∀x, y, z, (x < y and y < z → x < z).

• ∀∅ 6= Y ⊆ X,∃y ∈ Y s.t. ∀z ∈ Y (y < z or z = y).

Math 145a Notes 11

An example of a well-ordering is (N, <). This is actually equivalent to in-duction.

Theorem 3.4. The following are equivalent in ZF.

(1) The Axiom of Choice.

(2) Every set can be well-ordered.

(3) The product of nonempty sets is nonempty.

(4) For all x, y, either |X| ≤ |Y | or |Y | ≤ |X|.

Given a collection {Xi : i ∈ I}, what is∏i∈I Xi? Some element x ∈

∏i∈I Xi

should associate to a function that picks an element of xi for each i ∈ I.

Proof of (2) → (1). Let {Xi : i ∈ I} with Xi 6= ∅. We are going to well-orderthis set, and then there is a unique element we can pick, which is the minimalelement. Set

X∗ =⋃{Xi : i ∈ I}

and let <∗ be a well-order of X∗. Then define f by f(i) = minXi ∈ Xi. Thisis a choice function.

There are some weakenings of the Axiom of Choice, and one of this is theAxiom of Dependent Choice.

Axiom of Dependent Choice. Let R be a relation on X such that for allx ∈ X there exists a y ∈ X such that xRy. Then there exists a set {xn ∈ X :n ∈ N} such that xnRxn+1.

You can of course do this finitely many times with ZF, but you run out oftime as you try to do it infinitely many times. That is why you need this axiom.This is sometimes used in other areas of mathematics like analysis.

3.3 Cardinality of sets

We want to use the cardinal |X| is an emblem representing the size of X. Like-wise, we want to use ordinals to represent a specific order type.

Theorem 3.5 (Cantor). For every set X, thre is not surjection from X toP(X). Here, f : X → Y is a surjection if and only if for all y ∈ Y there existsan x ∈ X such that f(x) = y.

Proof. Suppose f : X → P(X) was a surjection. Define

D = {x ∈ X : x /∈ f(x)} ∈ P(X).

By surjectivity, there exists an a ∈ X such that f(a) = D. If a ∈ D, thena /∈ f(a), so a /∈ D. If a /∈ D, then a /∈ f(a), so a ∈ D.

This leads us to talking about sizes of sets.

Math 145a Notes 12

Definition 3.6. Fix X and Y that are nonempty. We write:

• |Y | ≥ |X| if there exists a surjection f : X → Y .

• |X| ≤ |Y | if there exists an injection f : X → Y .

• |X| = |Y | if there exists a bijection f : X → Y .

For example, |X| ≤ |P(X)| and |X| 6= |P(X)|. Also, we have

|N| = |Z| = |Q| = |Q[X]|.

These are called countable sets. Also,

|R| = |C| = |P(N)|

and this is called the continuum.The ≤ and ≥ symbols are not order relations. The whole thing |X| ≥ |Y | is

just one symbol. So we may want to ask a few questions.

(1) Is |Y | ≤ |X| equivalent to |X| ≥ |Y |?(2) Does |X| ≤ |Y | and |Y | ≤ |X| imply |X| = |Y |?(3) What is |X|?(4) For every X,Y , is it true that either |X| ≤ |Y | or |Y | ≤ |X|?

Proposition 3.7. The answer to (1) is yes!

Proof. Suppose f : Y → X is injective. Let y0 ∈ Y and set g : X → Y by

g(x) =

{y if x = f(y)

y0 else.

This is a surjective function.Now suppose that g : X → Y is a surjection. Put a well-ordering < on

X and define f : Y → X by f(y) = min g−1{y}, where g−1{y} is the inverseimage. This requires the Axiom of Choice.

Math 145a Notes 13

4 September 13, 2016

4.1 Well-founded and well-ordered

As I said, we are going to use ordinals as some kind of emblems. As a prototypefor well-foundedness, we use (V,∈).

Definition 4.1. let R be a binary relation on X (possibly definable classes).

(1) R is strict if and only if ¬(xRx) for all x ∈ X. Also R is a linearordering if and only if it is transitive, antisymmetric, and for all x, y ∈ X,xRy, yRx, or x = y.

(2) The transitive closure TC(R) of R is the binary relation on X suchthat xTC(R)y if and only if x1, . . . ,n ∈ X such that x1 = x, xn = y, andxiRxi+1.

(3) Define

extR(x) = {y ∈ X : yRx}, predR(x) = {y ∈ X : yTC(R)x} = extTC(R)(x).

(4) Given Y ⊆ X, Y is R-transitive if and only if x ∈ Y implies extR(x) ⊆ Y .

(5) We call that R is well-founded if and only if extR(x) is a set for allx ∈ X, and every ∅ 6= Y ⊆ X has an R-minimal element.

(6) R is a well-ordering if and only if it is a strict linear order that is well-founded.

As an example, if yRx means that y gave birth to x, then extR(x) are yourparents and predR(x) is the ancestors going up in your family tree.

Example 4.2. The universe (V,∈) is well-founded. Also (N, <) is well-ordered.Also if we throw in ∞, then (N ∪ {∞}, <∗) with n < ∞ for all n ∈ N is awell-ordering. In fact, if you have any well-ordering on a set, you can throw inthe biggest element to get a new well-ordering.

Suppose I is a set and for all i ∈ I, (Xi, Ri) is a well-ordering such that i < jimplies Xi ⊆ Xj and Ri ⊆ Rj and x ∈ Xi, y ∈ Xj − Xi implies xRjy. Thisis an increasing sequence of well-orderings, in the sense that you are adding afew elements that is bigger than every other thing. Then (

⋃i∈I Xi,

⋃i∈I Ri) is

well-ordered.

4.2 Transfinite induction

Theorem 4.3 (Transfinite induction). Suppose R is well-founded on X. IfY ⊆ X such that for every x ∈ X, predR(x) ⊆ Y implies x ∈ Y , then Y = X.

Note that this generalizes the induction we already know. Apply the theoremto (N, <). Then the predecessor of 0 is empty, and so 0 ∈ Y . Then the rest isthe same as what we learn as “strong induction”.

Math 145a Notes 14

Proof. Suppose not. Then Y ( X. Let x = minR(X − Y ). Then because everyelement R-below x is in Y . Thus predR(x) ⊆ Y , and x ∈ Y .

Theorem 4.4 (Transfinite recursion). Suppose R is well-founded on X andG : X × V → V is a function. Then there exists a unique function F : X → Vsuch that for each x ∈ X, F (x) = G(x, F |predR(x)). (Note that V is all of settheory, so F |predR(x) is also in V .)

Proof. We first prove uniqueness. Suppose F1 and F2 both work. Let

Y = {x ∈ X : F1(x) = F2(x)}.

If predR(x) ⊆ Y then

F1(x) = G(x, F1|pred(x)) = G(x, F2|pred(x)) = F2(x),

so x ∈ Y . Then by induction, Y = X.Let us now prove existence. For D ⊆ X, we call f : D → V good if and only

if for every x ∈ D, predR(x) ⊆ D and f(x) = G(x, f |predR(x)). If f1 : D1 → Vand f2 : D2 → V , then f1|D1∩D2 = f2|D1∩D2 are equal by uniqueness. Then

F =⋃{f : f is good}

is a function, because we can first use Comprehension to show that the set ofdomains with a good function is a set, and then use replacement. We now claimthat domF = X. This is because if f : D → X and x = min(X −D) then setg : D ∪ {x} → X by

g(y) =

{f(y) if y ∈ DG(y, f |predR(y)) if y = x

and g is good. Then domF = Xby induction.

Definition 4.5. If (X,R) is a well-ordering, and x ∈ X then denote

IRx = (predR(x), R ∩ (predR(x)× predR(x))).

We also say that f : (X,R) ∼= (Y, S) if f : X → Y is a bijection and for anyx0, x1 ∈ X, x0Rx1 if and only if f(x0)Rf(x1).

Theorem 4.6 (Compatibility). Let (X,R) and (Y, S) be well-orderings. Thenone of the following happens.

(a) (X,R) ∼= (Y, S).

(b) There exists an x ∈ X such that IRx∼= (Y, S).

(c) There exists an y ∈ Y such that (x,R) ∼= ISy .

Math 145a Notes 15

Proof. Let z /∈ Y . Define G : X × V → V by

G(x, f) =

{minS Y − f ′′ predR(x) if f : predR x→ X and Y − f ′′ predR(x) 6= ∅z otherwise,

where ′′ denote the image. By transfinite recursion, we get a F : X → V suchthat F (x) = G(x, F |predR(x)). Set A = {x ∈ X : F (x) 6= z} and let f = F |A.

Then by induction f ′′ predR(x) = predS(f(x)). Then x0Rx1 → f(x0)Sf(x1).Since R is linear and strict, we get that x0Rx1 if and only if f(x0)Sf(x1) andf is an injection. Now I have three cases.

Case 1. dom f = X and im f = f ′′ dom f = Y .The f : (X,R) ∼= (Y, S) and so f is an isomorphism.

Case 2. dom f ( X.Let x0 = min(X − dom f). Then dom f = predR x0. If Y − f ′′ predR x0 6= ∅,then

G(x0, f) = minY − f ′′ predR x0 6= z

so x0 ∈ A = dom f . This shows that f : IRx0∼= (Y, S).

Case 3. im f ( Y .Set y0 = minY − im f . Then as before dom f = X. So f : (X,R) ∼= ISy0 .

Math 145a Notes 16

5 September 15, 2016

5.1 Ordinals for real

As “emblems for well-ordering”, we want them to be canonical, and uniquelyrepresentative. Recall that X is ∈-transitive if and only if x ∈ y ∈ X impliesx ∈ X.

Definition 5.1. A set α is an ordinal if and only if (α,∈) is a well-orderingand α is transitive.

This is a very useful definition, because this is first-order and thus gives adefinable class On of ordinals. There is a order on On given by ∈.

Example 5.2. Clearly ∅ ∈ On. If α ∈ On, then α ∪ {α} ∈ On. This is calledthe successor, and we will denote this α + 1. Also if X ∈ On then

⋃X ∈ On.

Using this, we see that 0 = ∅, 1 = {∅}, 2 = {∅, {∅}}, and so on, are all ordinals.

We set ω = N. We haven’t proved that this is a set, so we need to show this.Heuristically, a collection is not a set only when it is too big. N is not that big,so it is going to be a set. Recall that the Axiom of Infinity states that

∃∞(∅ ∈ ω ∧ ∀x(x ∈ ∅ → x ∪ {x} ∈ ∞)).

This set ∞ will contain all the natural numbers, but we don’t know what elsemight be inside. So we use comprehension on the expression

ψ(x) = “x 6= ∅ ∨ (∃x = y ∪ {y} ∧ ∀y ∈ x∃z ∈ y(y 6= ∅ ∨ y = z ∪ {z})”.

Then we can define

ω = N = {x ∈ ∅ : ψ(x)}.

Why is this what we want? It is clear that if n is a natural number, n ∈ ω.Conversely, let x ∈ ω. By the definition of ω, we can write x = y ∪ {y} unlessx = ∅, and then write y = z ∪ {z} unless y = ∅, and so forth. If it hits ∅ atsome point, then this contradicts foundation.

I have another definition for ω. Consider the set

A = {Y ∈ P(∞) : ∀x ∈ Y (x = ∅ ∨ ∃y ∈ Y (x = y ∪ {y}))}

and define ω =⋂A.

This is the first limit ordinal. Let us now keep going. We see that ω + 1 =ω ∪ {ω} is an ordinal, and also ω + 2 = (ω + 1) ∪ {ω + 1} = {0, 1, . . . , ω, ω + 1}is an ordinal. This way, we get a ω + n for all natural numbers n.

Next we would want to construct ω + ω. How do we do this? We see thatω 3 n 7→ ω + n is a definition of a function. So we can use replacement to getX = {ω + n : n < ω} as a set. Then you can take

⋃X as an ordinal, and we

write ω + ω =⋃X.

Similarly, we can add many ωs, and so write ω + · · · + ω = ωn. Then wecan use the same trick, and define ω · ω =

⋃{ω · n : n < ω}. We can go on still

further and get a whole hierarchy.

Math 145a Notes 17

5.2 Properties of ordinals

Proposition 5.3. Let α, β ∈ On. The following are equivalent:

(1) α ∈ β (= α < β)

(2) α ( β

Proof. Let us first prove (1) ⇒ (2). Assume α ∈ β. Then by transitivity, γ ∈ αimplies γ ∈ β, and so α ⊆ β. Also, α 6= β by foundation.

Now we prove (2) ⇒ (1). Set γ = min(β − α). By your homework, γ ∈ On,and so α = pred∈ γ. Again by your homework, α = γ ∈ β.

Theorem 5.4 (Linearity). For all α, β ∈ On, α ≤ β or β ≤ α.

Proof. By another homework problem, α∩β ∈ On. If α∩β = α then α ⊆ β andα ≤ β. Similarly, if α ∩ β = β then β ≤ α. If α ∩ β ( α, β, then α ∩ β ∈ α, β.Thus α ∩ β ∈ α ∩ β, which contradicts Foundation.

Theorem 5.5. For every well-ordering (X,R) there exists a unique α ∈ Onsuch that (X,R) ∼= (α,∈). This α is called the order type of (X,R).

Proof. We fix X and R and claim that for every α ∈ X, there exists a uniqueβ ∈ On such that IRa

∼= (β,∈). Suppose not, and let a be a R-minimal coun-terexample. For every bRa, there exists a unique βb such that IRb

∼= (βb,∈).Then set β = {βb : b ∈ predR a} is a set by replacement, and in fact, it isan ordinal. So we can take π : IRa → β by taking b to βb. This is an orderisomorphism.

So there always exists a unique βa ∈ On, and cal this map a 7→ βa. Thenα = {βa : a ∈ X} is an ordinal, and this map is a order morphism between αand X.

Theorem 5.6 (in ZF). The following are equivalent:

(1) Well-ordering principle.

(2) For all X,Y , either |X| ≤ |Y | or |Y | ≤ |X|.

Proof. For (1) ⇒ (2), let X and Y be any sets and use well-ordering to getR,S. Use comparability to get f : IRa

∼= (Y, S) for some a ∈ X, or the otherway round. Then f−1 : Y → X is an injection.

To do the other direction, we need the fact that ordinals can be larger thanany set. Consider any X and get α with an injection g : X → α, by the nexttheorem. Set R on X by xRy if and only if g(x) ∈ g(y). By injectivity, this isa well-order.

Theorem 5.7 (in ZF, Hartog). For every Y , there exists an α such that thereis no injection from α into Y .

Math 145a Notes 18

Proof. Set

α = {β ∈ On : ∃injection : β → Y }.

We need to verify that this is a set. First look at P(Y × Y ) which is a set, andlet

A = {R ∈ P(Y × Y ) : R well orders some Z ⊆ Y }.

Then considering the map A 3 R 7→ ot(domR,R). Then α ∈ On is the imageof this map.

By Foundation, α /∈ α and so there is no injection α→ Y . We are done.

Math 145a Notes 19

6 September 22, 2016

Lemma 6.1. If C ⊂ On then⋂C ∈ C and

⋂C is the minimal element of C.

Proof. We first show that⋂C is an ordinal. If a ∈ b ∈

⋂C then for all γ ∈ C,

we have a ∈ b ∈ γ and so a ∈ γ by transitivity. Then a ∈ γ and so a ∈⋂C.

So⋂C is transitive. Also

⋂C is linearly ordered under ∈ as

⋂C is a subset of

On.Now let us prove that

⋂C ≤ γ for all γ ∈ C. This is just because

⋂C ⊂ γ

and so⋂C ≤ γ.

Third,⋂C is actually in C. Suppose

⋂C /∈ C. Then

⋂C < γ for all γ ∈ C.

So⋂C + 1 ≤ γ for all γ ∈ C, and hence

⋂C + 1 ⊆

⋂C. This is clearly a

contradiction.

This shows that any subclass of On always has a minimal element. Notethat this is not always true of linear orders, i.e., {a ∈ R : a >

√2} does not

have a minimal element.

6.1 Ordinal induction and arithmetic

Note all ordinals α fall into one of these three categories:

(1) α = 0,

(2) α is a successor, i.e., α = β + 1 for some β,

(3) α is a limit, i.e., there exists an increasing sequence 〈βi〉i∈κ with βi < αfor all i ∈ κ such that α =

⋃i∈κ βi.

Note that in the limit, the sequence cannot be indexed by the natural num-bers. This is because the least ordinal not equinumerous with ω is not a count-able union of smaller ordinals because all smaller ordinals will have at most ωmany unions.

To prove that these cover all cases, let C be the class of ordinals that doesnot fall in these categories. Let α be the least element of C, which is α =

⋃{β :

β < α}, because α is not a successor.Fix α ∈ On. We would like to define addition inductively. Define

(1) α+ 0 = α,

(2) α+ (β + 1) = (α+ β) + 1,

(3) if γ is a limit, then α+ γ =⋃β<γ α+ β.

This can be visualized as just attaching β at the large end of α. This operationis not commutative, as ω + 2 6= 2 + ω = ω.

Likewise we define multiplication of the ordinals in the following way:

(1) α · 0 = 0,

(2) α · (β + 1) = α · β + α,

(3) if γ is a limit ordinal then α · γ =⋃β<γ α · β.

Math 145a Notes 20

The way to look at it is to lay down α and copy that β times. For instance,ω · 2 = (ω · 1) + ω = ω + ω. On the other hand, 2 · ω =

⋃n<ω 2 · n = ω. So

multiplication is also not commutative.Then we define exponentiation as

(1) α0 = 1,

(2) αβ+1 = αβ · β,

(3) if λ is a limit ordinal then αλ =⋃β<λ α

β .

Theorem 6.2. Suppose β > 0. Then for all α there are unique α there areunique γ1, γ2 such that α = β · γ1 + γ2.

Proof. First let us prove existence. Let γ1 =⋃{λ : β · λ ≤ α}. Then β · γ1 ≤

α < β(γ1 + 1). Once again, let γ2 =⋃{τ : βγ1 + τ ≤ α}. Then βγ1 + γ2 ≤ α <

βγ1 + γ2 + 1.To prove uniqueness, we first show that λ < τ then ζ + λ < ζ + τ for every

ζ. This is true because we know that there exists a σ such that τ = λ+ σ (bythe existence part we have proved) and so ζ + τ = ζ + λ+ σ > ζ + λ.

To show uniqueness, suppose that α = βγ∗1 + γ∗2 with γ∗2 < β. By thedefinition of γ1, we have γ∗1 ≤ γ1 but if γ∗1 < γ1 then β(γ∗1 + 1) ≤ α. Henceβγ∗1 +β < α which is a contradiction. So γ∗1 = γ1. Similarly we get γ∗2 = γ2.

Theorem 6.3 (Cantor’s normal form). Suppose α > 0. Then α can be writtenuniquely as

α = ωβ1 · k1 + ωβ2 · k2 + · · ·+ ωβn · kn

where n ∈ N and α ≥ β1 > β2 > · · · > βn and all ki are positive elements of N.

Proof. Let us first show existence. If α = 1, then it has a representation α =ω0 · 1. Now assume every α′ < α can be written in such a form. Let β =

⋃{β′ :

ωβ′ ≤ α}. Then there are unique γ1, γ2 < ωβ such that α = ωβ · γ1 + γ2 such

that γ1 < ω as otherwise ωβ+1 ≤ α. If γ2 = 0, we are done. If γ2 > 9 thenγ2 < ωβ < α so γ2 has the appropriate form. Then combining the representationwith ωβ , we get the representation of β.

One thing to note is that α ≥ β1 contains equality. This is because ordinals εsuch that ε = ωε, which is quite different from cardinal arithmetic. Furthermorethere even is an ordinal ε0 = εε00 .

6.2 Building the universe

Definition 6.4. Define V0 = ∅, Vα+1 = P(Vα), and Vλ =⋃α<λ Vα if λ is a

limit ordinal.

It turns out that these contain everything.

Lemma 6.5. {x : x = x} =⋂α∈On Vα.

Math 145a Notes 21

Proof. It suffices to show that x ⊆⋂α∈On Vα then x ∈

⋃α∈On Vα. Note that

by replacement, there exists a β such that x ⊆⋃α<β Vα. So x ⊆ Vβ+ω, so

x ∈ P(Vβ+ω) = Vβ+ω+1.

Definition 6.6. The rank of x is the least α such that x ∈ Vα.

Theorem 6.7. Assuming the axiom of choice, any set can be well-ordered.

Proof. Let f : P(X) − {∅} → X be a choice function. Let h be defined bytransfinite recursion using

h(α) = f(X − range(h|α))

when range(h|α) 6= X. Let β be the least α such that range(h|α) = X. Note asf is a choice function, hβ is a bijection. For x, y ∈ X, let xRy if h−1(x) < h−1(y)then R well orders X.

6.3 Cardinals

Definition 6.8. We say that X and Y are equinumerous (X ≈ Y ) if there isa bijection from X to Y . We say X 4 Y if there is an injection from X into Y .

Theorem 6.9 (Cantor-Bernstein thereom). If X 4 Y and Y 4 X then X ≈ Y .

Proof. Suppose f : X → Y and g : Y → X are injections. Let X0 = X andY0 = y, and take Xn+1 = g′′(Yn) and Yn+1 = f ′′(Xn). Let X∞ =

⋂n∈NXn and

Y∞ =⋂n∈N Yn. Let

h(x) =

h(x) if x ∈ X2n −X2n+1

g−1(x) if x ∈ X2n+1 −X2n+1

f(x) if x ∈ X∞.

Note that f |X2n−X2n+1→Y2n+1−Y2n+2is a bijection, and g−1|X2n+1−X2n+2→Y2n−1−Y2n+1

is a bijection. So h is a bijection when restricted to X −X∞ → Y − Y∞. Alsoh is an injection on X∞ as f is an injection, and also it is surjection onto Y∞.So h : X → Y is a bijection.

For example, we can easily show that N ≈ Q. Clearly N 4 Q since N ⊆ Q.Next, we can set r : Q→ N so that

r : (−1)ja

b7→ 2a3b5j

where j ∈ {0, 1} and gcd(a, b). This is an injection, so N ≈ Q.

Definition 6.10. We say that a real number x is algebraic if there existsa1, a2, . . . , an ∈ Q such that anx

n + an−1xn−1 + · · · + a0 = 0. The set of

algebraic numbers is denoted by A.

Math 145a Notes 22

Then A ≈ N. Clearly N 4 A, and by the injection s : A→ N

s(x) = 2i3r(a0)5r(a1)7r(a2) · · · ,

where x is the ith solution of an irreducible anxn+ · · ·+a1x+a0 = 0. So A 4 N

and thus A ≈ N.

Definition 6.11. An ordinal α is a cardinal if β 6≈ α for every β < α.

For instance, ω+ω is not a cardinal. For each cardinal κ, let κ+ be the leastcardinal that is greater than κ.

Math 145a Notes 23

7 September 27, 2016

Recall that α is an ordinal, then α is a cardinal if α 6≈ β for all β < α. Also, κis a cardinal, then κ+ is the next cardinal. Finally, let Card be the collectionor cardinals.

Lemma 7.1. If A ⊆ Card and A is a set, then⋃A ∈ Card.

Proof. If A has a largest element κ, then⋃A = κ ∈ Card. Assume

⋃A /∈ Card

and A does not have a largest element to get a contradiction. There exists anα ∈

⋃A such that α ≈

⋃A. Let κ ∈ A such that κ < α. But then κ ≈ α as

α = |⋃A|.

Definition 7.2. Let ℵ0 = ω and ℵα+1 = ℵ+α for all ordinals α, and let ℵλ =⋃α<λ ℵα for λ a limit.

Theorem 7.3. If κ ∈ Card then either κ is finite or κ = ℵα for some ordinalα.

Proof. Let α be the least such that ℵα ≥ κ. If α = β + 1 then ℵβ < κ, soκ = ℵ+β = ℵα. If α is a limit then for all γ < α, ℵγ < κ so ℵα ≤ κ. Soκ = ℵα.

Continuum Hypothesis (CH). Every subset of P(ω) that can not be mappedinvectively into ω has a bijection with P(ω).

This was the first natural question arose that is independent of ZFC. Godel’sincompleteness theorem says that the consistency of Peano arithmetic is inde-pendent of Peano arithmetic, but this is kind of ad hoc.

7.1 Cardinal arithmetic

Generalized Continuum Hypothesis (GCH). If κ is a cardinal then κ+ ≈P(κ).

Theorem 7.4. GCH implies AC.

Proof. Assume GCH. It suffices to show that for all α, Vα ≈ κ for some cardinalκ. Assume not, and let α be the least one such that Vα 6≈ κ for any κ ∈ Card.Note that α 6= 0. If α is a limit ordinal then for all γ < α, Vγ is well-ordered.Let iγ be the cardinality of Vγ , and let iα = sup{iγ : γ < α}. Note thatiα · α ≈ iα because iα · iα = iα. (We haven’t prove this yet, but hopefullyget to it.) So

⋃γ<α Vγ ≤ iα · α and so Vα ≤ iα. But we know that iγ ≤ Vγ

for all γ < α, and so ⋃γ<α

iγ ≤⋃Vγ ≈ Vα.

So iα ≈ Vα, which is a contradiction.Now consider the case α = β + 1. Then Vβ ≈ iβ so by GCH, P(Vβ) ≈

Vβ+1 ≈ iβ . This is again a contradiction.

Math 145a Notes 24

Definition 7.5 (Assuming AC). The cardinality of X, written as |X|, is theleast ordinal α such that X ≈ α.

The reason we need AC is because we should know that X is equinumerouswith at least one ordinal.

Definition 7.6. If κ, λ ∈ Card, then we define

κ+ λ = |({0} × κ) ∪ ({1} × λ)|, κ · λ = |κ× λ|.

Definition 7.7 (Assuming AC). A set is finite if |X| < ω, and a set is infiniteif |X| ≥ ω.

Theorem 7.8. If κ is an infinite cardinal then κ · κ = κ.

Proof. Base case ω ·ω = ω is straight forward. Assume this holds for all infiniteα < κ. We want to define a well-ordering on κ × κ of order type (κ,∈). For(α, β), (γ, δ) ∈ κ× κ, let

(α, β)C (γ, δ) ⇔ max(α, β) < max(γ, δ) or max(α, β) = max(γ, δ) ∧ (α, β) <lex (γ, δ).

We claim (but not prove) that this is a well-ordering.Note that for any (α, β) ∈ κ× κ,

|{(γ, δ) : (γ, δ)C (α, β)}| ≤ |(max(α, β) + 1)× (max(α, β) + 1)|.

But by the inductive hypothesis,

|(max(α, β) + 1)× (max(α, β) + 1)| = |max(α, β) + 1| < κ.

So for (α, β) ∈ (κ, κ), (α, β) has at most κ many C-predecessors. This showsthat (κ× κ, δ) ∼= (κ,∈).

Corollary 7.9. If κ, λ ∈ Card and κ, λ 6= 0, and at least is infinite, thenκ+ λ = κ · λ = max{κ, λ}.

Definition 7.10 (Also under AC). If κ, λ ∈ Card, let κλ = |{f : λ→ κ}|.

Lemma 7.11. If A,B are sets then AB = {f : A→ B}.

Theorem 7.12. For all sets A,B,C, (AB)C ≈ AB×C .

Proof. Fix f ∈ (AB)C , i.e., f : C → AB . For c ∈ C, let fc : B → A be suchthat fc = f(c). Then define f∗ : B ×C → A be such that f∗(b, c) = fc(b). It iseasy to check that f 7→ f∗ is a bijection.

Corollary 7.13 (Assuming AC). If κ, λ, µ ∈ Card, then (κλ)µ = κλ·µ.

Lemma 7.14. If 2 ≤ κ ≤ λ and λ is infinite, then 2λ = κλ.

Proof. This is because 2λ ≤ κλ ≤ (2κ)λ = 2κ·λ = 2λ.

Math 145a Notes 25

If λ < κ then things are more interesting. Without AC, X ≈ Y is anequivalence relation.

What will happen if we don’t have AC? Still X ≈ Y will be an equivalencerelation, and {Y : X ≈ Y } will be a class. But there might not be a functione : V → V such that e(X) ≈ X for all X and e(X) = e(Y ) if and only if X ≈ Y .There is a trick.

Scott’s trick. For all X ∈ V consider the least αX such that {Y : Y ≈ X} ∩VαX 6= ∅. We let {Y : Y ≈ X} ∩ VαX “represent” |X|.

This works quite well as a representative.

7.2 Cofinality

Definition 7.15. Let α be a limit ordinal. The cofinality of α, cof(α), is theleast β such that there exists an f : β → α such that for all α < α there existsan β < β such that f(β) > α.

For example, cof(ℵω) = ω, because f : n 7→ ℵn is cofinal in ℵω.

Definition 7.16. A limit ordinal α is regular if cof(α) = α. Otherwise, α issingular.

If κ is a limit and κ is regular, then Vκ � ZFC.

Math 145a Notes 26

8 October 4, 2016

The countable union of countable sets is countable. This shows that I can’twrite ℵ1 as a less than ℵ1 sized union of less that ℵ1 sized sets. For this reason,we are going to call ℵ1 regular.

On the other hand, ℵω = supn<ω ℵn and so ℵω is the union of countablymany less thab ℵω sized sets. Then we are going to call ℵω singular.

Definition 8.1. A function f : β → α is cofinal if for every α′ < α there existsβ′ < β such that f(β′) > α′. If α is a limit ordinal, the cofinality cof(α) iscof(α) = min{β : ∃cofinal f : β → α}.

Definition 8.2. A limit ordinal is regular if cof α = α and singular if cof α <α. A limit ordinal α is Godzilla if cof α > α.

Proposition 8.3. There are no Godzilla ordinals.

Proof. The map id : α→ α is cofinal.

Lemma 8.4. The composition of two increasing cofinal functions is cofinal.

Lemma 8.5. Let α be a limit ordinal. There is an increasing cofinal functionf : cof α→ α.

Proof. Let g : cof α → α be cofinal. Set f(β) = max{g(β), supγ<β(f(γ) + 1)}.To define this function, we have to use transfinite recursion on F : cof α×V → Vgiven by

F (β, h) =

{max{g(β), supγ<β(h(γ) + 1) if h with β ⊆ domh,

{{0}} otherwise.

The following all have cofinality ω:

ω, ℵω, ω + ω, ω2, κ+ ω, ℵω2 .

8.1 Successor cardinals are regular

Theorem 8.6. For every cardinal κ, κ+ is regular.

Proposition 8.7. If α is a limit ordinal,

(1) cof(cof(α)) = cof α.

(2) cof α is a regular cardinal.

Proof. (1) Suppose f : cof α → α and g : cof(cof α) → cof α are increasing.Then their composite f ◦g : cof(cof α)→ α is also cofinal, and since cof(cof α) ≤cof α, we are done by minimality.

(2) We need only prove that cof α is a cardinal. If cof α is not a cardinal, thenthere exists an γ < cof α and a bijection f : γ → cof α. Then we can composef which is cofinal (or its modification that is both increasing and cofinal) andcof α→ α so that γ → α is cofinal. This contradicts the minimality of cof α.

Math 145a Notes 27

Lemma 8.8. If κ is a cardinal then κ is regular if and only if κ cannot bewritten as a (< κ) sized union (< κ) sized sets.

Proof. Let us first prove the forward direction. For contradiction, suppose thatκ is regular and κ =

⋃i<αXi with α < κ and |Xi| < κ. Because κ is regular and

|Xi| < κ, supXi < κ for each i < α. Then the map g : α→ κ by g(i) = supXi

is cofinal, because κ =⋃i<αXi. This contradicts that κ is cofinal.

Let’s now show the other direction. Suppose κ is singular and α = cof κ <κ. This means that there is a increasing cofinal function f : α → κ. SetXi = f(i) + 1. Then |Xi| < κ because f(i) < κ and κ is a cardinal. Also⋃i<αXi = κ because given β < κ there exists an i < α with f(i) > β.

Proof of Theorem 8.6. Suppose not. Then there exists a decomposition

κ =⋃i<α

Xi

for α < κ+ and |Xi| < κ+. Because only the cardinality of α matter, we mayassume that α ≤ κ, and also |Xi| ≤ κ. Then∣∣∣⋃

i<α

Xi

∣∣∣ ≤ κ · κ = κ < κ+.

This is a contradiction. And so κ+ is regular.

Math 145a Notes 28

9 October 4, 2016, Make-up class

9.1 Regular limit cardinals

Can limit cardinals be regular? I spoiled the answer by telling you that ZFCcannot decide.

Theorem 9.1. If α is a limit ordinal, then cof ℵα = cof α.

Lemma 9.2. If f : β → α is a function that is cofinal and nondecreasing, thencof β = cof α.

Proof. Let g : cof α → α and h : cof β → β be increasing and cofinal. Thenf ◦ h : cof β → α is cofinal, and this implies cof α ≤ cof β. For the otherdirection, define k : cof α→ β by

k(γ) = min{δ < α : f(δ) > g(γ)}.

I want to show that k is cofinal. Let δ < β. There exists and γ′ < cof α suchthat g(γ′) > f(δ). Because f(k(γ′)) > g(γ′) > f(δ) and f is nondecreasing,k(γ′) > δ. Thus k is cofinal. So cof β ≤ cof α.

Proof of Theorem 9.1. The map f : α → ℵα given by f(β) = ℵβ is cofinal.Then by the lemma, cof α = cof ℵα.

Corollary 9.3. There is no cofinal nondecreasing f : ℵ2 → ℵ1.

Corollary 9.4. If α < ℵα, then ℵα is singular.

Proof. cof ℵα = cof α ≤ α < ℵα.

This proves that a lot of limit cardinals are singular. So the regular limitneeds to satisfy α ≥ ℵα.

Proposition 9.5. α ≤ ℵα.

Proof. We use induction. Clearly 0 ≤ ℵ0, and α ≤ ℵα implies α+ 1 ≤ ℵα. If αis a limit,

α = supβ<α

β ≤ supβ<αℵβ = ℵα.

There are indeed limit ordinals α with α = ℵα. Take α0 = ω and αn+1 =ℵαn . Then we can set

α∗ = supn<ω

αn.

This satisfy α∗ = ℵα∗ , because

ℵα∗ = supβ<α∗

ℵβ = supn<ωℵαn+1

= supn<ω

αn+1 = α∗.

So there are ℵ-fixed points. But the fixed point we have built has cofinalityω.

Math 145a Notes 29

9.2 Inaccessible cardinals and models of ZFC

Definition 9.6. A cardinal κ is weakly inaccessible if it is a regular limit car-dinal. κ is a strong limit if λ < κ implies 2λ < κ. κ is strongly inaccessibleif it is regular and strong limit.

Theorem 9.7 (Konig’s theorem). For any cardinal κ, κcof κ > κ. If λ < cof κthen κλ = supµ<κ µ

λ + κ.

Proof. (1) Let f : cof κ → κ be cofinal and increasing. We know that κcof κ =|cof κ → κ|. If κcof κ = κ, let {gα : α < κ} enumerate cof κ → κ. Defineg : cof κ→ κ by

g(β) = min(κ− {gα(β) : α < f(β)}).

Now I claim that for all α < κ, gα 6= g. Find β < cof κ so that f(β) > α. Theng(β) 6= gα(β).

(2) Now suppose λ < cof κ. Let f : λ → κ. Then f can’t be cofinal, sosupα<λ f(α) < κ. Set βf = sup f“λ. Then actually f : λ → βf . This showsthat {λ→ κ} =

⋃α<κ{λ→ α}. Thus

κλ = |λ→ κ| =∣∣∣ ⋃α<κ

{λ→ α}∣∣∣ ≤ sup

α<κ|λ→ α| = sup

α<καλ + κ.

We have a function ℵ : On→ Card that is defined by

ℵα =

ω if α = 0

ℵ+β if α = β + 1

supβ<α ℵα if α a limit.

Now replacing the successor cardinal by taking exponentiation, we also get afunction i : On→ Card given by

iα =

ω if α = 0

2iβ if α = β + 1

supβ<α iα if α a limit.

The continuum hypothesis then says that ℵ1 = i1, and the generalized contin-uum hypothesis states that ℵα = iα for all α ∈ On.

Recall that we have the hierarchy V0 = ∅, Vα+1P(Vα), and Vα =⋃β<α Vβ

for limit α. Then V =⋃α∈On Vα is our universe. As we climb up, we would like

to see how much of ZFC does the Vα satisfy.What does it mean for Vα to satisfy an axiom of ZFC? We say that Vα

satisfies an axiom ϕ if ϕ is true if we bind all quantifiers to Vα. For example,the Empty Set states that

∃x, ∀y(y /∈ x).

Math 145a Notes 30

Now we restrict this to the set Vα and consider

∃x ∈ Vα,∀y ∈ Vα(y /∈ x).

If α > 0, this is going to be true since ∅ ∈ Vα.

Proposition 9.8. For an ordinal α,

(1) Vα satisfies Extensionality and Foundation.

(2) if α > 0 then Vα satisfies Emptyset.

(3) if α > ω then Vα satisfies Infinity.

(4) if α is limit then Vα satisfies Union, Pairing, and Powerset.

We need Replacement, which states that

∀a, a1, . . . , an(∀x ∈ a∃!y ϕ(x, y, a1, . . . , an)→ ∃b∀x∀y ∈ b ϕ(x, y, a1, . . . , an)).

This is not always possible, because for instance the image of the functionn ∈ ω 7→ ω + n does not give a set in Vω+ω.

Now this problem arise because ω + ω has small cofinality.

Theorem 9.9. If κ is strongly inaccessible then Vκ satisfies Replacement.

Proof. First note that |Vω+α| = βα by induction. Also, if κ is a i-fixed pointbecause κ is strongly inaccessible.

Now suppose that ϕ, a, a1, . . . , an ∈ Vk such that

∀x ∈ a∃!y ∈ Vk, ϕ(x, y, a1, . . . , an).

Then use replacement on ϕVk , which is ϕ relativized to Vκ. This gives a b ∈ Vsuch that

∀x ∈ a∃!y ∈ b, ϕ(x, y, a1, . . . , an).

We then need to check that the image in b lives in Vκ.We know that a ∈ Vκ, so a ∈ Vω+α for all α < κ. Thus |a| ≤ |Vω+α| = iα <

κ. Replacement gives a function f : a → b given by ϕVκ . Now get a bijectiong : λ→ α for some cardinal λ.

Define h : λ→ κ

h(β) = rank f(g(β)).

Since λ < cof κ = κ, h cannot be cofinal. So we get γ = supβ<λ h(β) + 1. Thenb ∈ Vγ+1. This shows that Vκ satisfies Replacement.

Now we need to say why this shows that ZFC cannot prove nor disprove theexistence of strongly inaccessible cardinals.

Theorem 9.10 (Godel’s Incompleteness Theorem). No sufficiently powerfulconsistence theory with a recursive axiomization can prove its consistency.

Math 145a Notes 31

Corollary 9.11. ZFC cannot prove that ZFC is consistence.

Corollary 9.12. ZFC cannot prove the existence of a strongly inaccessible car-dinal.

Proof. Suppose it could. Then ZFC proves ∃ strongly inaccessible. But ZFC +∃ strongly inaccessible proves Vκ satisfies ZFC. If a set satisfies some axioms,then they are consistent. So if ZFC proves existence of strongly inaccessiblecardinals, then ZFC proves the consistency of ZFC.

We will see later that if ZFC proves the existence of weakly inaccessiblecardinals, then ZFC proves the existence of strongly inaccessible cardinals.

Math 145a Notes 32

10 October 6, 2016

So last lecture finishes introducing the basic notions. Now we are going to talkabout problems we mentioned in first class.

10.1 The measure problem

Let’s do a bit of introductory measure theory. We like R and would like to talkabout the sizes of subsets of R. One thing we can use is the cardinality of sets,but this is a very coarse notion.

Proposition 10.1. For any r < s in R, |(r, s)| = |R|.

Proof. The function tan : (−π/2, π/2)→ R is a bijection.

What about its length? We can define λ((r, s)) = s− r. Then problem withlengths is the this only works for intervals. This leads to the definition of ameasure.

Definition 10.2. Given Γ ⊆ P(R), a measure µ : Γ→ R≥0∪{∞} is a functionsuch that

(1) µ(∅) = 0, µ([0, 1]) = 1,

(2) if X,Y ∈ Γ and X ⊆ Y then µ(X) ≤ µ(Y ),

(3) if Xn ∈ Γ for n < ω are pairwise disjoint, then

µ( ⋃n<ω

Xn

)=∑n<ω

µ(Xn).

We say that µ is translation invariant if for all X ∈ Γ and r ∈ R, its translateX + r ∈ Γ and µ(X) = µ(X + r).

It follows from translation invariance and some other properties that µ((r, s)) =s−r. Another good thing about translation invariance is that it excludes thingslike point measures.

For X ⊆ R, define

µ∗(X) = inf{∑`(Ui) : Ui is an open cover of X},

µ∗(X) = sup{

∑`(Ci) : Ci are closed and disjoint in X}.

One is approximating X from the outside and one is approximating from theinside. Then X is called measurable if µ∗(X) = µ∗(X).

Theorem 10.3 (Vitali). There is no translation measure on all of R (underAC).

Math 145a Notes 33

Proof. Set a relation ∼ on R by r ∼ s if s − r ∈ Q. This is an equivalencerelation. Now use the axiom of choice to pick one thing in each class in [0, 1].This means a function f : R/ ∼→ [0, 1] such that f([r]∼) ∼ r. Now set

V = {f([r]∼) : r ∈ R} ⊆ [0, 1].

Assume V is measurable by some µ. Set Q = [−1, 1] ∩ Q and consider⋃q∈Q′ V + q. This satisfies

[0, 1] ⊆⋃q∈Q′

V + q ⊆ [1, 2].

Also V + q0 ∩ V + q1 = ∅. Then

1 ≤∑n<ω

µ(V ) ≤ 3.

Now we have a problem.

Once you get to R3, things get a bit worse.

Theorem 10.4 (Banach-Tarski paradox). For n ≥ 3, the unit ball in Rn canbe decomposed into finitely many pieces and reassembled (by congruent actions)into two copies of itself.

The problem is that the free group with 2 generators is non-amenable. Forinstance suppose there is a dictionary of all words using a and b. If you tear thisinto halves and erase all the first letters, you get two copies of the dictionarywithout doing anything too crazy. This is the key idea of the proof.

There are the open and closed sets, and there are also the Borel sets on themeasurable sets. On the other hand, there are the Vitali sets and the set comingfrom the Banach-Tarski paradox on the complex and nonmeasurable side. Sowhere is the line? This is what descriptive set theory is about.

10.2 The Baire space

Consider the Baire space ωω, which is the set of functions from ω to ω. Thereis a bijection f : ωω → R. But the real number has some more structure on it,and we need to keep some of this structure.

For n < ω, the set nω is the functions from n = {0, . . . , n − 1} to ω. All ofthese combined have a tree structure, by looking at the initial segments, whichwe denote by ⊇. The Baire space lives over all of this.

Given two topological spaces X and Y , a map f : X → Y is a homeomor-phism if and only if it is a bijection such that if U ⊆ X is open then f“U ⊆ Yis open, and if V ⊆ Y is open then f−1V ⊆ X is open.

We will build a homeomorphism between ωω and R−Q. This is good enoughbecause Q has measure zero and nobody cares about Q. To show this, we firstneed a topology on ωω.

Math 145a Notes 34

Definition 10.5. For s ∈ <ωω, let NS = {x ∈ ωω : S ⊆ X}.

Proposition 10.6. The set {Ns : s ∈ <ωω} forms a basis for the topology

τ = {X ⊆ ωω : ∀x ∈ X∃s ∈ <ωω such that x ∈ Ns ⊆ X}.

Proof. We want to show that τ is a topology. Clearly ∅, ωω ∈ τ because N∅ =ωω. It is also trivially closed under unions. Now let X,Y ∈ τ and a ∈ X ∩ Y .Then there exists s, t ∈ <ωω such that a ∈ Ns ⊆ X and a ∈ Nt ⊆ Y . Theneither s ⊆ t or t ⊆ s. If t ⊆ s then a ∈ Ns = Ns ∩Nt ⊆ X ∩ Y .

Proposition 10.7. For all s ∈ <ωω, Ns is clopen.

Proof. Ns is open by the definition of τ . Let s ∈ nω. Then

ωω −Ns = {x ∈ ωω : s 6⊆ x} =⋃

t∈nω,t6=s

Nt

is open.

This shows that ωω cannot be homeomorphic to R.

Proposition 10.8. ωω is totally disconnected.

Proof. Let x 6= y ∈ ωω. There exists a minimal n < ω such that

x � n 6= y � n = s ∈ nω.

Then y ∈ Ns and x ∈ ωω −Ns will be a partition of ωω into open sets.

Math 145a Notes 35

11 October 11, 2016

11.1 Midterm

There was a midterm. Here is an interesting question that appeared in theexam.

The goal of this problem is to prove Mostowski’s Collapsing Theoremusing Transfinite Recursion.

Theorem 11.1 (Mostowski). If E is a well-founded and extensionalrelation on a set P , then there is a transitive set M and an isomor-phism π : (P,E) ∼= (M,∈).

We say that a binary relation R on X is extensional iff for everyx, y ∈ X, we have

x = y ⇔ ∀z ∈ X(zRx⇔ zRy).

You can prove this theorem however you want, but here’s a niceoutline:

(i) We wish to define π on P so that π(x) = {π(z) : zEx}. Finda function G so that the output of well-founded recursion is π(and show that this is the case).

(ii) Show that π is injective and xEy ⇔ π(x) ∈ π(y).

(iii) Set M = π”P . Show that π is surjective and that M is transi-tive.

(iv) Conclude the theorem.

In fact this isomorphism is unique, but you don’t need to show this.If you do show this, 5 bonus points.

Math 145a Notes 36

12 October 18, 2016

Recall that the topology on ωω is given by the clopen basis

Ns = {x ∈ ωω : s ≤ x}

where s = <ωω.

12.1 Topology of ωω

Now we want to show that ωω is homeomorphic to (0, 1) \ Q. The way we aregoing to do is to use continued fractions. Consider

26

317=

131726

=1

12 + 526

=1

12 + 15+ 1

5

.

Then we can associate to 26/317 the sequence [12, 5, 5].Fix a r ∈ (0, 1)\Q. Define the sequence xrn ∈ (0, 1)\Q and krn ∈ ω+ = ω\{0}

for n < ω as

xr0 = r, kr0 =⌊ 1

xr0

⌋and xrn+1 =

1

xrn− krn, krn+1 =

⌊ 1

xrn+1

⌋.

This process never terminates because r is irrational. If this process neverterminates, then r is in fact irrational because of the Euclidean algorithm. Thendefine

f : (0, 1) \Q→ ωω+; f(r) = 〈kr0, kr1, . . .〉 ∈ ωω.

Proposition 12.1. The map f is a homeomorphism between (0, 1)\Q and ωω.

To show this, we have to show that f is bijective, continuous, and its inverseis continuous.

We first show that f is injective. For contradiction, r < s ∈ (0, 1) \ Qand f(r) = f(s). We claim that if r < t < s, then krn = ktn = ksn and(−1)nxrn < (−1)nxtn < (−1)sxsn. For n = 0, this is easy. Then

kr0 =⌊ 1

xr0

⌋≥ kt0 =

⌊ 1

xt0

⌋≥ ks0

⌊ 1

xs0

⌋= kr0.

So kt0 = kr0 = ks0 = k0. The other parts can be worked out.That f is surjective and continuous will be a homework.So now let us prove that f−1 is continuous. Let U ⊆ R be an open set and

we want to show that f”U ⊆ ωω is open. In general, if r < s ∈ (0, 1) \ Q andl < ω be the first index the expansions differ at. Then r < s if and only if(−1)l(krl − ksl ) > 0. This shows that

1

k0>

1

k0 + 1k1+

1k2

> · · · converges to x from above,

1

k0 + 1k1

<1

k0 + 1k1+

1

k2+ 1k3

< · · · converges to x from below.

Math 145a Notes 37

Going back to what we are trying to show, given x ∈ U there exists an ε > 0such that (x − ε, x + ε) ⊆ U . Now find and N so that both sequences get in(x − ε, x + ε). Letting s = 〈kx0 , kx1 , . . . , kxN 〉 ∈ <ωω. Then you can show thatNs ⊆ f”U .

12.2 Measures on the Baire space

Knowing a measure on the irrationals would give us a measure on R.

Proposition 12.2. If A ⊆ R is countable, then A is Lebesgue measurable andµ(A) = 0.

Proof. Write A = {an : n < ω}. If we prove that the outer measure is 0, thenwe are done. Take any ε > 0, and look at

A ⊆ Uε =⋃n<ω

(an −

1

2n, an +

1

2n

).

Then µ∗(A) ≤ µ(Uε) = 4ε for every ε > 0.

Corollary 12.3. Q has measure 0.

If λ is a measure on R \Q, then λ+(A) = λ(A \Q) is a measure on R.Now let us give a measure on ωω. Because ωω = N∅ corresponds to (0, 1)\Q,

we are going to make the measure of the whole space 1. Then for each finitesequence s, we have

Ns = Nsa(0) ∪Nsa(1) ∪Nsa(2) ∪ · · · .

We make each Nsa(k) get a 2−k−1 proportion of Ns.

Definition 12.4. Let τ be the topology generated by B = {Ns : s ∈ <ωω}.Define µ : B → [0, 1] by

µ(Ns) =∏i<n

1

2s(i)+1,

where s ∈ nω.

Math 145a Notes 38

13 October 20, 2016

We have defined the measure on the Baire space ωω.

13.1 Borel sets

We are going to define Borel sets by a hierarchy.

Definition 13.1. For α ∈ On define Σ0α,Π

0α,∆

0α ⊆ P(ωω) in the following way.

For α = 1,

Σ01 = {open sets}, Π0

1 = {closed sets}, ∆01 = Σ0

1 ∩Π01 = {clopen sets}.

For α > 1, we define

Σ0α = {A ⊆ ωω : A =

⋃n<ω

Bn for Bn ∈⋃β<α

Π0β},

Π0α = {A ⊆ ωω : ωω \A ∈ Σ0

α}, ∆0σ = Π0

α ∩ Σ0α.

We call Σ0On =

⋃α∈On Σ0

α the Borel sets.

Actually, you are going to see in your homework that Σ0ω1

= Σ0On. We also

have this hierarchy:

Σ01 Σ0

2 Σ03

∆01 ∆0

2 ∆03 ∆0

4 · · ·

Π01 Π0

2 Π03

⊆ ⊆ ⊆⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆⊆ ⊆ ⊆

This actually comes up in analysis. Σ02 are the countable unions of closed sets,

and Π02 are the countable intersections of open sets.

Theorem 13.2. Σ0On is the smallest σ-algebra containing closed the open sets

and closed under complement and countable union.

Proof. Let A ∈ Σ0On. Then for some α ∈ On, A ∈ Σ0

α. For countable union, letAn ∈ Σ0

On for n < ω. Let αn ∈ On such that An ∈ Σ0α. Set β =

⋃n αn. Then

An ∈ Σ0β and so An ∈ Π0

β+1. Thus⋃n<ω An ∈ Σ0

β+2. This shows that Σ0On is

closed under countable union and complement.Now let us consider and arbitrary σ-algebra m. We show by induction that

Σ0α ⊆ m. For α = 1, m contains the open sets, which is Σ0

1. Let α > 1and suppose that Σ0

β ⊆ m for β < α. Let A ∈ Σ0α,r write A =

⋃nBn for

Bn ∈⋃β Π0

β . Then ωω −Bn ∈⋃β Σβ0 ⊆ m and so

A =⋃n

(ωω \ (ωω \Bn)) ∈ m.

This shows that Σ0α ⊆ m.

Then Borel sets are measurable, because measurable sets are closed undercomplement and countable unions. So Borel sets are on the measurable side.

Math 145a Notes 39

13.2 Projective sets

I have told you the story of some graduate student worrying about the projectionof a measurable set in R2 not being measurable. In fact, sets obtained by projec-tion lie somewhere on the line between the measurable and the non-measurableside.

Consider the product (ωω)n. This space is the set of n-tuples of sequencesof natural numbers, and the topology is given by the basis

{Ns1 × · · · ×Nsn : s1, . . . , sn ∈ ωω}.

Proposition 13.3. The two spaces ωω and (ωω)n are homeomorphic.

Proof. Fix some bijection F : ω → nω. Define G : ωω → (ωω)n by

G : x ∈ ωω 7→ (x1, . . . , xn) where xi(k) = F (x(k))(i).

This now gives us a measure, and Borel sets for (ωω)n. Now we are readyto define projective set.

Definition 13.4. Given A ⊆ (ωω)n+m, define

projn(A) = {x ∈ (ωω)n : ∃y ∈ (ωω)m such that (x, y) ∈ A}.

This is nice from a model theoretic point of view, because we can describeusing a existential quantifier.

Definition 13.5. Define Σ10 to be the Borel sets. We define

Σ11 = {A ⊆ (ωω)n : A = projn(B) for Borel B ⊆ (ωω)n+m},

Π11 = {complements of elements of Σ1

1}.

Then define similarly,

Σ1k+1 = {projn(A) : A ⊆ (ωω)m+n is Π1

k}Π1k+1 = {complements of elements of Σ1

k+1}.

Theorem 13.6 (Lusin). Σ11 are measurable.

The question is, are Σ12 measurable? The amazing thing is that this is one

of those maybe questions.

Theorem 13.7 (Solovay). If there exists an inaccessible cardinal, then thenCon(ZFC + projective sets are measurable).

Theorem 13.8 (Shelah). If projective sets are measurable, then Con(ZF +∃inaccessible cardinal).

There is somehow a connection between something that happens way aboveand something that happens on the ground.

Math 145a Notes 40

13.3 Filters and ultrafilters

There is another approach to measure theory. Consider for example, the pointmeasure, which is like µϕ(A) = 1 if ϕ ∈ A and µϕ(A) = 0 if ϕ /∈ A. Thismeasure is {0, 1}-valued, and we can represent µϕ by a set Uϕ ⊆ P(R) suchthat

A ∈ Uϕ ←→ µϕ(A) = 1.

This is exactly what a filter is.

Definition 13.9. Fix as set I. A set F ⊆ P(I) is a filter if

(a) ∅ ∈ F , I /∈ F .

(b) If X ⊂ Y ⊂ I and X ∈ F , then Y ∈ F .

(c) If X,Y ∈ F , then X ∩ Y ∈ F .

If F is a filter on I then µF on {A ⊆ I : A ∈ F or I \A ∈ F} given by

µF (A) =

{1 if A ∈ F0 if not

is a finitely additive measure.

Definition 13.10. A filter F on I is an ultrafilter if for every X ⊂ I, eitherX ∈ F or I \X ∈ F . A filter (or ultrafilter) F is principal if there exists anx0 ∈ I such that for all A ⊆ I, A ∈ F implies x0 ∈ A.

Every ultrafilter on a finite set is principal.

Math 145a Notes 41

14 October 25, 2016

Recall that if we fix a set I, a filter F on I satisfies the axioms (1) ∅ /∈ F , I ∈ F ,(2) X ⊆ Y ⊆ I and X ∈ F implies Y ∈ F , (3) X,Y ∈ F implies X ∩ Y ∈ F . Afilter U is an ultrafilter if and only if for all X ⊆ I, either X ∈ U or I \X ∈ U .

Given a filter F , we can define a finitely additive measure µF on {X ⊆ I :X ∈ F or I \X ∈ F}.

Proposition 14.1. A filter F is an ultrafilter if and only if it is maximal, i.e.,if F ′ ⊇ F is a filter then F ′ = F .

Lemma 14.2. If F0 ⊆ P(I) satisfies for each X1, . . . , Xn ∈ F0,⋂i≤nXi 6=,

then there is a filter F ⊇ F0.

Proof. Without loss of generality, assume F0 6= ∅. Define F by

F = {X ⊆ I : ∃X1, . . . , Xn ∈ F0(⋂i≤n

Xi ⊆ X)}.

Then F is a filter.

Proof of Proposition 14.1. I need to prove two directions. Suppose that F is anultrafilter, and let F ′ ⊇ F . Assume that X ∈ F ′ but X /∈ F . Then I \X ∈ F ,and so I \X ∈ F ′. This contradicts that F ′ is a filter.

Now suppose F is a maximal filter. Let X ⊆ I. Then either F0 = F ∪{X} or F1 = F ∪ {I \ X} has the finite intersection property. By the lemmaand maximality of F , either X ∈ F or I \ X ∈ F . This shows that F is anultrafilter.

14.1 Ultralimits

Theorem 14.3 (Bolzano-Weierstrass). Every bounded sequence has a conver-gent subsequence.

This is a great theorem, but it is annoying that there are many limit points.So, is there a way of extending the “limit function” on convergent sequence toone on all sequences (with reasonable restrictions)? These restrictions will bestuff like, keeping sums and products, and the output should be a limit point.

Fix an ultrafilter U on ω. Define the ultralimit of 〈an ∈ R : n < ω〉 as

L = limUan

if and only if for every ε > 0 there exists an Xε ∈ U such that for every n ∈ Xε,|an − L| < ε.

Theorem 14.4. Every bounded sequence has an ultralimit.

Math 145a Notes 42

Proof. Let 〈an〉 ⊆ [−B,B]. Define by induction bn, Cn, and Xn ∈ U such thatbn ≤ bn+1 < cn+1 ≤ cn with cn − bn ≤ 21−nB and

Xn = {k < ω : ak ∈ [bn, cn]} ∈ U.

We construct the sequence by letting b0 = −B, c0 = B, X0 = ω, dividing theinterval [bn, cn] by half and choosing the large side. Let L = limn→∞ bn =limn→∞ cn. Then L = limU an.

If U is a principal filter, which only cares only m, then limU an = am.

14.2 Ultraproducts

Newton and Leibniz did calculus using infinitesimals, but they were not rigorousand were called “ghosts of departed quantities”. Later Weierstrass and otherpeople came up with the ε-δ definitions. But as ultrafilters were developed,Robbinson came up with a way of making it real math.

Fix a non-principal ultrafilter on ω. Set∏n<ω R = ωR = {f : ω → R}. But

this is now not a field.

Definition 14.5. For f, g ∈∏

R, let f =U g if and only if {n < ω : f(n) =g(n)} ∈ U .

Then this is an equivalence relation and respects + and =. We say that[f ]U < [g]U if and only if {n ∈ w : f(n) < g(n)} ∈ U . The addition andmultiplication behaves well under this order.

Definition 14.6. Define the hyperreals ∗R =∏

R/ =U with +, ·, <, etc.

You can define any function, like cos, on the hyperreals in a similar way.

Proposition 14.7. The ring ∗R has multiplicative inverses, i.e., is a field.

Math 145a Notes 43

15 October 25, 2016, Make-up class

15.1 Banach-Tarski Paradox

Let us first work in R2. Let C be the circle of radius 1 and let l be the line(0, 1) × {0}. Let ρ be the rotation of R about the origin by 1 rad. We knowthat 2π/1 is irrational. Then for all n 6= m, ρn(l) and ρm(l) are disjoint. Let

A = C ∪∞⋃n=0

ρn(l).

Then ρ(A) = C ∪⋃∞n=1 ρ

n(l) and so A = ρ(A) ∪ l. This kind of thing is whatwe are going to do.

Denote by Fn the free group on n generators. The elements will be wordsmade up of letters x1, . . . , xn and also x−11 , . . . , x−1n , such that no xi and x−1iare adjacent. Multiplication of two elements will be concatenating two wordsand canceling everything out.

Consider the free group F2 with generators x, y. Let w(x) be the set ofstarting with x, and likewise define w(y), w(x−1), w(y−1). Then there is apartition

F2 = {e} ∪ w(x) ∪ w(y) ∪ w(x−1) ∪ w(y−1)

= x−1w(x) ∪ w(x−1)

= y−1w(y) ∪ w(y−1).

Denote by SOn the group of rotations of Rn about the origin, with multi-plication being composition. Note that there is no embedding of F2 into SO2.However, there is an embedding of F2 into SO3, given by the generators thatare the rotation ϕ about the x axis by arccos(1/3) and the rotation ψ about thez axis by arccos(1/3). Let 〈ϕ,ψ〉 be the subgroup of SO3 generated by ϕ andψ. The proof is a bunch of computation, but the point is that there is a copyof F2 in SO3.

Let S2 be the sphere of radius 1 in R3, and let B3 be the closed ball ofradius 1 in R3. Then the group 〈ϕ,ψ〉 acts on S2 and gives gives an equivalencerelation p ∼ q if and only if σ(p) = q for some σ ∈ 〈ϕ,ψ〉.

We can try to set M∗ to be the set of representatives. Then

S2 = M∗ ∪ w(ϕ)M∗ ∪ w(ψ)M∗ ∪ w(ϕ−1)M∗ ∪ w(ψ−1)M∗

but it is not necessarily disjoint, because 1 6= σ ∈ 〈ϕ,ψ〉 may fix some stuff. Soset

D = {p ∈ S2 : ∃1 6= σ ∈ 〈ϕ,ψ〉, σ(p) = p}.

Then D is countable because F2 is countable and a rotation fixes exactly twopoints. Also 〈ϕ,ψ〉(S2 \D) ⊆ (S2 \D).

Math 145a Notes 44

So just as we had before, let ∼ be the equivalence relation on S2 \D and letM be a set of representatives. Then

S2 \D = 〈ϕ,ψ〉M = M ∪ w(ϕ)M ∪ w(ψ)M ∪ w(ϕ−1)M ∪ w(ψ−1)M

= ϕ−1w(ϕ)M ∪ w(ϕ−1)M = ψ−1w(ψ)M ∪ w(ψ−1)M.

are partitions.How do we get rid of D? Since D is countable, we can find a rotation

σ ∈ SO3 that fixes no element of D and sends no element of D to another underany iterations of σ. (First fixed the axis and then pick the angle.) Then forevery m 6= n, σm(D) ∩ σn(D) 6= ∅. If we set

D =⋃n<ω

σn(D),

then σ(E) = E \D. So you can rearrange S2 \D to get S2.Now let us bring everything together. Note that “being able to decompose

A into finitely many pieces an rearrange to make B by rigid motions” is anequivalence relation. We first make S2 into S2 \D. Then using the fact that ϕdoes not fix M , we make S2 \D into (S2 \D) \M . Using the decomposition

〈ϕ,ψ〉 = ϕ−1w(ϕ) ∪ w(ϕ−1) = ψ−1w(ψ) ∪ w(ψ−1),

we can make (S2 \D) \M into two pieces of S2 \D. Then we can track backand get to two pieces of S2. Finally, we can extend this way of making two S2sfrom S2, to get a way of making two B3 \ {0} from one B3 \ {0}. The way tofill this last center is to use the same thing: consider a circle passing through 0and use an irrational rotation.

Theorem 15.1. B3 can be decomposed into finitely many pieces and then thesepieces can be translated and rotated to get two copies of B3.

Math 145a Notes 45

16 October 27, 2016

Last time we talked about the hyperreals, which is constructed by fixing anultrafilter U on ω and putting an equivalence relation on

∏R = {f : ω → R}.

Then we can extend functions R→ R to ∗R→ ∗R.

Proposition 16.1. For any x ∈ ∗R with x 6= 0, there is an y ∈ ∗R such thatxy = 1.

Proof. Then {n < ω : f(n) 6= 0} ∈ U . Define

g(n) =

{1

f(n) f(n) 6= 0

0 f(n) = 0.

Then f · g = 1.

This can be generalized.

Theorem 16.2 ( Los). If ϕ(x1, . . . , xn) is any first order formula and [f1]u, . . . , [fn]u ∈∗R, then ϕ([f1]u, . . . , [fn]u) holds if and only if {k < ω : ϕ(f1(k), . . . , fn(k))} isin U .

So R and ∗R are exactly the same. But they are different, because the mapj : R → ∗R is injective but not surjective. For instance, f(n) = n−1 is aninfinitesimal that is greater than 0 but is smaller than any positive real number.

16.1 Dense linear orderings

Definition 16.3. A relation (L,≤) is a dense linear ordering (withoutendpoints) if

(1) ≤ is a linear ordering.

(2) For every x < y ∈ L there exists a z ∈ L such that x < z < y.

(3) For every y 3 L, there exist x, z ∈ L such that x < y < z.

For example, (Q, <), (R, <), and (α×Q, <lex) are dense linear orderings.

Theorem 16.4 (Cantor’s theorem). If (L,≤L) is a countable dense linear or-dering, then (L,<L) ∼= (Q, <).

Proof. Write L = {an : n < ω} and Q = {bn : n < ω}. Now we are going to doa “back and forth” definition. Build ⊆-increasing functions fn for n < ω suchthat

(a) dom(fn) ⊆ L and ran(fn) ⊆ Q,

(b) {a1, . . . , an} ⊆ dom(fn) and {b1, . . . , bn} ⊆ ran(fn),

(c) dom(fn) is finite,

(d) for every x, y ∈ dom(fn), x <L y if and only if fn(x) < fn(Y ).

Math 145a Notes 46

You can build this function by throwing in an or bn in the function using thatthe orderings < and <L are dense and without endpoints. Once you have thissequence, you can glue it as f =

⋃n<ω fn to get an isomorphisms between

(Q, <) and (R, <).

You can ask the same thing for R. Is any dense linear ordering with thesize |R| isomorphic to R? There is set of the form α × Q that has the samecardinality with R, and it is not isomorphic to R.

Definition 16.5. A dense linear ordering (L,<) is separable if and only if ithas a countable dense subset. It is complete if and only if every ∅ 6= X ( Lthat is bounded above has a least upper bound.

Note that these are not first-order formulas because they involve the exis-tence of sets.

Theorem 16.6. If (L,<) is a dense linear ordering without endpoints, separableand complete, then (L,<L) ∼= (R, <).

Note that we are not requiring anything about the cardinality of L.

Proof. Let X0 ⊆ L be a dense countable subset. Then (X0, <) is a countabledense linear ordering without endpoints and so (X0, <) ∼= (Q, <). Define f∗ :L→ R by

f∗(x) = supR{f(y) : y ∈ X0, y < x}.

This is an isomorphism.

But mathematicians didn’t stop here and wanted other characterizations.

Definition 16.7. A linear ordering (L,<L) satisfies the countable chain con-dition if and only if every collection of disjoint open intervals is countable.

Because every open set in R contains some rational number and Q is count-able, (R, <) satisfies the countable chain condition. In general, if a dense linearordering is separable, then it satisfies the countable chain condition.

Why is this called the countable chain condition? Given a partial order(P,<), two elements p, q ∈ P are called incomparable if ¬(p ≤ q ∨ q ≤ p). Aset C ⊆ P is an antichain if every p 6= q ∈ C are incomparable, and we saythat C ⊆ P is a strong (upward) antichain if for every x 6= y ∈ C there isno z ∈ P such that x <P z and y <P z. We say that P satisfies the κ-chaincondition if and only if every strong antichain of P has size at most κ. In thecase of the linear order (L,<L), we can form a partial order (P,<P ) by lettingP = {(a, b)L : a < b ∈ L} and <P is ⊇. Then satisfying the countable chainconditions is equivalent to P satisfying the ℵ0-chain condition.

Suslin’s hypothesis. Every complete dense linear ordering satisfying the count-able chain condition is isomorphic to R.

A Suslin line is a complete dense linear ordering that does not satisfy thecountable chain condition.

Math 145a Notes 47

17 November 1, 2016

17.1 Trees

We kind of know trees. The set of finite sequences <ωω has branches that keepbranching. This is sort of the setup that we want to deal with. Note that thereare no loops.

Definition 17.1. (T,<T ) is a partial order if <T is a transitive binary rela-tion. For an x ∈ T , define pred<T (x) = {y ∈ T : y <T x}. Then pred<T (x) is apartially ordered by the restriction of <T .

Definition 17.2. A partial order (T,<T ) is a tree if for every x ∈ T , pred<T (x)is well-ordered.

These trees can go infinitely wide, and also infinitely tall.

Definition 17.3. Let (T,<T ) be a tree. For x ∈ T , define its height as htT (x)as the order type of predT (x). Define the height of T as htT = sup{htT (x)+1 :x ∈ T}. Also, for α < htT , the α-level of T is defined by

Tα = {x ∈ T : htT (x) = α}.

For example, for <ωω, the n-level sets are the sequences of length n andhtT = ω.

Example 17.4. Fix (infinite) cardinals κ, λ and set T = (<κλ,⊆), which is thesequences in λ with length less than κ. This is a tree, that has height κ. Everytree is a subtree of this tree.

17.2 Branches of trees

Definition 17.5. A branch thorugh a tree (T,<) is a sequence b = 〈bα ∈ T :α < htT 〉 such that bα ∈ Tα and α < β implies bα <T bβ .

Branches in <κλ are given by f ∈ κλ.

Example 17.6. Let T = (A,⊆) for A ⊆ <ωω by

A = {〈n, n, . . . , n〉(k-many) : n < ω, k ≤ n}.

This tree does not have any branches.

Definition 17.7. Say that a tree T is < κ-branching if every x ∈ T has < κimmediate successors. “Finitely branching” means “< ω-branching”.

Theorem 17.8 (Konig). Any finitely splitting tree with a root of height ω hasa branch.

Math 145a Notes 48

Proof. We are going to build 〈sn : n < ω〉 such that Sn ∈ Tn and sn <T sn+1

and {x ∈ T : sn <T x} is infinite. Start with a root s0. Then {x ∈ T :s0 <T x} = T −{s0} is infinite. Suppose we have sn with immediate successorst1, . . . , tk. Then

succ(sn) =

k⋃i=1

succ(ti) ∪ {t1, . . . , tk}.

So there exists an i0 such that succ(ti0) is infinite.

Definition 17.9. T is a κ-tree if htT = κ and T is < κ-splitting (and has < κmany roots). This is the same as saying that for every α < κ, |Tα| < κ.

Note that there is a tree of height κ that is < κ+-splitting and has no branch.To construct this, for α < β < κ define sα,β ∈ ακ that is constant β. ThenT = {sα,β : α < β < κ} is a tree with no branch. So we can restrict to κ-trees.

Likewise, if κ is singular, then there also exists a κ-tree with no branch. Thisis because you can construct the same tree but use the cofinal sequence insteadof all of κ.

Anyways, all ω-trees always have branches, by Konig’s theorem. We aregoing to prove next time that there is an ω1-tree with no cofinal branches.

Math 145a Notes 49

18 November 3, 2016

Recall that (T,<T ) is a tree if for every x ∈ T , pred<T (x) is well-ordered. Atree T is a κ-tree if and only if htT = κ and |Tα| < κ for every α < κ.

18.1 ω1-Aronszajn trees

Definition 18.1. T is called a κ-Aronszajn tree if it is a κ-tree with nocofinal branch.

So Konig’s theorem says that there are no ω-Aronszajn trees. On the otherhand there is an ω1-Aronszajn tree.

Theorem 18.2 (Aronszajn). There is an ω1-Aronszajn tree.

Proof. As a first try, set

Q = {s : ∃α ∈ On such that s : α→ Q is increasing and s has a maximum}.

Lemma 18.3. If α < ω1 then there exists an Aα ⊆ Q such that ot(Aα, <) = α.

Because s has a max, either α = 0 or α = β + 1. Note that Qn is theincreasing sequences of length n for n < ω. Then Qω is the increasing sequencesof order type ω + 1. For ω ≤ α < ω1, Qα is the increasing sequences of ordertype α+ 1.

We claim that htQ = ω1. To show this, it suffices to show that for eachα < ω1, there exists an Aα such that otp(Aα) = α + 1. This is true becausethere is an order preserving morphism from any countable ordinal to Q. On theother hand, htQ ≤ ω1 because Q is countable.

This shows that Q has no cofinal branch. If there is one, then you can gluethem to get a function on ω1 to Q. Are the levels of Q countable? This is nottrue because Qω ∼ ω+1Q ∼ 2ℵ0 . So our first try doesn’t really work.

Now build a subtree T ⊆ Q with countable levels and htT = ω1 but notchanging the height notes. Define

(∗)α =

{∀β < γ < α and s ∈ Tβ and q ∈ Q such that max s < q,

∃t ∈ Tγ such that s ⊆ t and max t < q

}.

We are going to define T , and at the same time, prove that every level Tαsatisfies (∗)α. We define T0 = ∅ and if Tα is defined and (∗)α+1 holds,

Tα+1 = {s a 〈q〉 ∈ Qα+1 : s ∈ Tα, q ∈ Q, q > max(s)}.

If α is a limit, then denote T<α =⋃β<α Tβ . Because α is countable, cof(α) = ω.

Find an increasing sequence 〈αn < α : n < ω〉 such that supn αn = α. Letx ∈ T<α and q ∈ Q such that max(s) < q. Set m = min{n : αn > htT (x)}.Build an increasing sequence {xn : n < ω} such that x0 = x, htT xn = αm+n−1,and max(xn) < q, using (∗). Then set

yx,q =(⋃

n<ω xn)a 〈q〉 ∈ Qα.

Setting Tα = {yx,q} works.

Math 145a Notes 50

18.2 Other classes of tress

Definition 18.4. A tree is κ-embeddable if there exists a function f : T → κsuch that x <T y implies f(x) 6= f(y).

Definition 18.5. A tree is κ-special if it is the union of κ-many antichains.

Proposition 18.6. If a κ+-tree is κ-embeddable, then it has no cofinal branch.

Proof. Restrict the function to a branch.

Definition 18.7. A tree T is normal if κ = htT and

• |T0| = 1,

• for every limit α < x and x, y ∈ Tα, if predT (x) = predT (y) then x = y,

• if α < β < κ and x ∈ Tα then there exists a y ∈ Tβ such that x <T y,

• if x ∈ Tα, then there exist y1, y2 ∈ Tα+1 such that x <T y1 and x <T y2.

Math 145a Notes 51

19 November 8, 2016

19.1 Suslin trees

Recall that Q is the countable dense linear ordering and R is the separablecomplete dense linear ordering. We asked if we can replace separability withthe Suslin property, that every collection of disjoint open intervals is countable.A Suslin line is a complete dense linear ordering that is not separable.

Definition 19.1. A Suslin tree is an ℵ1-Aronszajn tree that has no uncount-able antichain.

Our goal is to prove that there exists a Suslin tree if and only if there existsa Suslin line.

Lemma 19.2. If there is a Suslin tree, then there is a normal Suslin tree.

We are again going to be doing some pruning.

Proof. Start with a Suslin tree T . We first make T have one root, by adding acommon root. Then the new tree T 0 is still Suslin.

Suppose x ∈ T 0 cannot be arbitrarily extended, i.e., have a successor on allhigher levels. Then T x = {y ∈ T 0 : x <0 y} is countable. Let T 1 = {x ∈T 0 : |T x| > ω}. This is a subset of T 0 so T 1 has no uncountable branches orantichains. Also htT 1 = ω1 because if htT 1 = β < ω1 then node in T0 at levelβ + 1 has countably many successors and there are also countably many nodesbelow so T0 is countable. This shows that T1 is still Suslin. Also by a similarargument, every node is arbitrarily extendible.

Now we want to ensure every node has at least 2 immediate successors. Weclaim that if α < ω1 and x ∈ T 1

α, then there exists an β > α and y1 6= y2 ∈ T 1β

such that x < y1 and x < y2. This is because if such an x exists, then wecan extend it to an branch by the arbitrary extendibility. Set T 2 = {x ∈ T 1 :x has ≥ 2 immediate successors} with the restricted orders. Again, it is clearthat T 2 has no uncountable branches and antichains. Because level sets areantichains, this includes the fact that level sets are countable. Then everythingworks out fine and because splitting occurs cofinally, T 2 is Suslin.

Finally we need to take care of the splitting at limit levels. In the case thattwo or more nodes at a limit level have same predecessors, add a node thereright below the limit level and make T 3. This additional node will take care ofthe problem. Finally, T 3 is a normal Suslin tree.

Theorem 19.3. There is a Suslin tree if and only if there is a Suslin line.

Proof. First consider a Suslin line L. Because it is not separable, no countableL0 ⊆ L is dense. Note that disjoint unions corresponded to strong (upward)antichains. We are going to build intervals {Iα = (aα, bα) : α < ω1} so that({Iα},⊇) is a Suslin tree. Let I0 be an arbitrary interval. For α > 0, the setC = {aβ , bβ : β < α} is countable. Find Iα = (aα, bα)L whose closure is disjointfrom C. Then for β < α, either Iβ ∩ Iα = ∅ or Iα ( Iβ . Then {Iα} form an

Math 145a Notes 52

tree under ⊇ because every subset has a ⊇-minimal element. It has no ω1-sizedantichain because L satisfies the Suslin property. Also its height is at most ω1

but it has ω1 elements, so its height is exactly ω1. Finally, there is no branchbecause a branch will give rise to a ω1-many collection of disjoint open intervals.So T is a Suslin tree.

Now consider a normal Suslin tree T . We may assume that for x ∈ T ,succ(x) = {y ∈ ThtT (x)+1 : x <T y} is countably infinite. For each x, let <x bea linear order on succ(x) so that succ(x) ∼= Q. Order T by <∗, that is definedas,

x <∗ y ⇔ x < y or

{z ∈ T = max(predx ∩ pred y) and

x′ 6= y′ ∈ succ(z), x′ < x, y′ < y and x′ <z y′.

}Then (T,<∗) is a dense linear ordering. If X is a collection of disjoint intervalsin (T,<∗), then we can pick aI ∈ I for each I ∈ X to get an antichain {aI}I∈Xin T . Finally, if D ⊆ T is countable, then α = sup{htT x : x ∈ D} < ω1 and sowe can pick to stuff above level α. This shows that this is a Suslin line.

Math 145a Notes 53

20 November 10, 2016

What does it mean for something to be independent of something? A statementϕ is independent of axioms T if and only if there is a model T satisfying ϕand a model T failing ϕ.

Example 20.1. “Being abelian” is independent of the group axioms. Thissimply means that you can’t neither prove nor disprove “being abelian” fromjust the group axioms. On the other hand, “product of two elements have aninverse” is not independent from the group axioms.

We’ve been saying that 2ℵ0 = ℵ1 is independent from ZFC. To prove this, wehave to construct two models of ZFC, one that satisfies the continuum hypoth-esis and one that fails the continuum hypothesis. But how do we build models?We already have V as a model, but we don’t know anything else. We can try tofind a model M ⊆ V , that satisfies ZFC. Or we can try to find V ⊆M outsideV . For instance, if κ is a strongly inaccessible cardinal, then Vκ is a model ofZFC. We are soon going to construct L that has the same ordinals as in V .

There are two metamathematical perspectives we can take. We can use Vonly as our universe/starting model of ZFC. But then we run into class/definablityissues. The other way is to start with some set V ′ that is a model of ZFC andtreat V ′ as V . But then we have to assume Con(ZFC) and is also less satisfyingbecause all we are talking about is not all of math but a small chunk of maththat is indistinguishable from all of math from the inside. We are mainly goingto work in the second viewpoint.

20.1 The idea of L

Recall that we have defined V0 = ∅, Vα+1 = P(Vα), and Vδ =⋃α<δ Vα if δ is

a limit. Then V =⋃α∈On Vα. But if we need only to get a model of ZFC, we

don’t really need to include everything at each step. Define

Def(X) =

{A ⊆ X :

∃ first-order formula ϕ(x, y1, . . . , yn) and a1, . . . , an ∈ Xsuch that A = {x ∈ X : ϕ(x, a1, . . . , an)}

}.

If M satisfies ZFC and X ∈M , then we expect Def(X) ⊆M .Now if this is true, we are going to define L as

L0 = ∅, Lα+1 = Def(Lα), Lδ =⋃α<δ

Lα for limit δ, L =⋃α∈On

Lα.

Doing this is much subtle than just constructing abelian groups and non-abeliangroups, because this L is going to be our new “universe” and there are meta-mathematical issues. So we need to be really precise in working with these.

20.2 First-order logic

Logic starts with a language (that always have =). A language L consists of

Math 145a Notes 54

• function symbols F with some arity nF < ω (i.e., binary, ternary, n-ary,etc.),

• relations symbols R with some arity nR < ω,

• constant symbols c.

We also need some syntax that tells us how to construct sentences and semanticsthat tells us how sentences connect to the real world.

Let us describe the syntax of L. We define the terms Terms(L) as thesmallest set containing

(1) every constant c of L,

(2) every variable x (from a fixed infinite set),

(3) if t1, . . . , tn ∈ Terms(L) and F ∈ L is an n-ary function, then F (t1, . . . , tn).

For instance, in the language of rings, the terms are things that can be madeby multiplying and adding variables.

The formulas Formulas(L) is the smallest set containing

(1) if t1, . . . , tn ∈ Terms(L) and R ∈ L is an n-ary relation, then R(t1, . . . , tn),

(2) if ϕ,ψ ∈ Formulas(L), then ¬ϕ, ϕ ∧ ψ, ϕ ∨ ψ, ϕ→ ψ,

(3) if ϕ ∈ Formulas(L) and x is a variable, then ∃xϕ and ∀xϕ.

In relation to set theory, in the language of set theory, L = {∈,=} and Terms(L) ={x : x is a variable}. So far, we don’t know how to interpret these sentences.But before this, we need to make sure we are not trying to interpret sentenceslike ∀y(¬y ∈ x).

Define the free variable function FV inductively. For terms, we set

FV(c) = ∅, FV(x) = {x}, FV(F (t1, . . . , tn)) =

n⋃i=1

FV(ti).

Then for formulas, we let

FV(R(t1, . . . , tn)) =

n⋃i=1

FV(ti), FV(¬ϕ) = FV(ϕ),

FV(ϕ ∧ ψ) = FV(ϕ) ∪ FV(ψ), FV(∃xϕ) = FV(ϕ) \ {x}.

Write ϕ(x1, . . . , xn) to mean ϕ ∈ Formulas(L) and FV (ϕ) ⊆ {x1, . . . , xn}. Inshort:

Terms: combine constants, variables, and functions.Atomic formulas: relations between terms.Formulas: closure of atomic formulas under logical operations.

Now let us get to semantics. Fix a language L. An L-structure or L-modelM consists of

Math 145a Notes 55

• a universe/underlying set |M |,• for F ∈ L with arity n, a map FM : Mn →M ,

• for n-ary relation R ∈ L, RM ⊆Mn,

• for constant c ∈ L, an element cM ∈ L.

If we have a term t and an L-structure M with FV(t) = {x1, . . . , xn}, buildtM : Mn →M as follows:

• (c)M : M0 →M has value cM ,

• xM : M →M is the identity,

• F (t1, . . . , Fn) : Mn →M is (m1, . . . ,mn) 7→ FM (t1(m•), . . . , tm(m•)).

For ϕ(x1, . . . , xn) ∈ Formulas(L), M an L-structure, and a1, . . . , an ∈ M , weare going to define M � ϕ(a1, . . . , an) again inductively as:

• if ϕ = R(t1(x•), . . . , tm(x•)), then

(M � ϕ(a1, . . . , an)) = ((tM1 (a•), . . . , tMm (a•)) ∈ RM ).

• if ϕ = ¬ψ, then

(M � ϕ(a1, . . . , an)) = ¬(M � ψ(a1, . . . , an)).

• if ϕ = ψ1 ∧ ψ2, then

(M � ϕ(a1, . . . , an)) = (M � ψ(a1, . . . , an)) ∧ (M � ψ(a1, . . . , an)).

• if vϕ(x1, . . . , xn) = ∃yψ(y, x1, . . . , xn), then

(M � ϕ(a1, . . . , an)) = ∃b ∈M, (M � ψ(b, a1, . . . , an)).

Note that you are only able to quantify over elements. So in the language ofgraphs, the notion of being connected can’t naturally be described in first logic,because it is “there exists a n < ω and vertices x0, . . . , xn such that x0 = xand xn = y and (xi, xi+1) ∈ E”. You can’t quantify over a varying number ofvariables.

Math 145a Notes 56

21 November 15, 2016

Last time we set up the notion of first-order logic. For set theory, the languageis τ = {E}. The syntax is built from xEy nd logical operations like negation,conjunction, and quantification. On the semantic side, there is a structureM =(M,EM) consisting of a set M and EM ⊆M2. Finally, we have semantics, thatis, given an ϕ(x1, . . . , xn) and a1, . . . , an, M � ϕ(a1, . . . , an).

21.1 Substructures and elementary substructures

Given τ -structures M and N , we say that M ⊆ N (M is a substructure ofN ) if

• M ⊆ N , and

• for every n-ary F ∈ τ , FN �Mn = FM,

• for every n-ary R ∈ τ , RN ∩Mn = RM,

• for every c ∈ τ , cN = cM.

This comes up in every parts of mathematics. But by what we have done sofar, we have a tighter connection between the two structures.

Now this substructure axioms implies thatM � aEb if and only if N � aEbfor every a, b ∈ M . Given τ -structures M and N , we say M 4 N (M is anelementary substructure of N ) if and only if

(a) M⊆ N ,

(b) if ϕ(x1, . . . , xn) is a formula and a1, . . . , an ∈M , then

M � ϕ(a1, . . . , an) if and only if N � ϕ(a1, . . . , an).

Let us look at an example. Take τ = {<} and M = (ω \ {0},∈) andN = (ω,∈). Then M ⊆ N , but M 4 N . This is because if we let ϕ(y) =∀x(x = y ∨ y < x) then

M � ϕ(1) but N 2 ϕ(1).

However, M ∼= N with the isomorphism given by M 3 x 7→ x − 1 ∈ N .Here, two structures are called isomorphic if there exists an f : M → N suchthat

• f is a bijection,

• if F ∈ τ and a1, . . . , an ∈ F , then f(FM(a1, . . . , an)) = Fn(f(a1, . . . , an)),

• if R ∈ τ and a1, . . . , an ∈ M , then (a1, . . . , an) ∈ RM if and only if(f(a1), . . . , f(an)) ∈ RN ,

• if c ∈ τ , then f(cM) = cN .

Proposition 21.1. Let M, N be {E}-structures and let M 4 N . If M is amodel of ZFC, so is N .

Math 145a Notes 57

Proof. For each sentence ϕ in ZFC, M � ϕ if and only if N � ϕ by definition of4.

Theorem 21.2 (Downward Lowenheim–Skolem–Tarski theorem). Given a τ -structure M and A ⊆M , there is a τ -structure N such that

(1) N 4M,

(2) A ⊆ N ,

(3) |N | ≤ |τ |+ |A|+ ℵ0.

Proof. This is going to be a homework.

21.2 Models of set theory

From now on, we are going to focus on τ = {E} and typically M will satisfysome fragment of ZFC. Recall that X is transitive if z ∈ y ∈ X implies z ∈ X.We also want to focus on structures M where M is transitive and EM =∈(restricted to M).

We might have this weird situation where M satisfies the well-foundednessaxiom but there are infinite decreasing EM-chains. Start with a structure N =(N,E) that satisfies ZFC. Take a non-principal ultrafilter U on ω, and form theultraproduct N ∗ =

∏(N,E)/U so that N∗ = {[f ]U : (f : ω → U)} and

[f ]UE∗[g]U ⇔ {n < ω : N � f(n)Eg(n)} ∈ U.

It turns out that N ∗ also satisfies ZFC including well-foundedness. But fork < ω, we get take

fk : n 7→

{0 if n ≤ kn− k if n > k

and this is a infinite decreasing sequence, i.e., N ∗ � [fk+1]UE[fk]U . This is whywe want transitivity.

Theorem 21.3 (Mostowski). Suppose M is a {E}-structure that satisfies ex-tensionality and such that EM is well-founded (actually well-founded, not sat-isfies the axiom). Then there is a unique transitive set X and an isomorphismπ :M∼= (X,∈).

This theorem allows us to check axioms more easily.

21.3 Levy hierarchy

There is also the Levy hierarchy that allows to rank the complexity of formulas.Of course you can define the complexity as the number of quantifiers, but thereare certain bounded quantifier that is of the form

∃x ∈ yϕ(x, y, z1, . . . , xn) or ∀x ∈ yϕ(x, y, z1, . . . , zn).

Math 145a Notes 58

IfM⊆ N butM 64 N , then the failure must start at some quantification. Butthis first failure cannot happen at a bounded quantifier because if y ∈ M thenautomatically x ∈M .

Let us know define the Levy hierarchy. We let

∆0 = Σ0 = Π0

as formulas with just bounded quantification. So this is the minimal set suchthat when we quantify, ϕ(x, y, z1, . . . , zn) ∈ ∆0, then ∃x ∈ yϕ(x, y, z1, . . . , zn) ∈∆0. We then let ϕ(y1, . . . , yn) ∈ Σk+1 if it is

∃x1 · · · ∃xlψ(x1, . . . , xl, y1, . . . , yn).

We say that ϕ ∈ Πk+1 if it is

∀x1 · · · ∀xlψ(x1, . . . , xl, y1, . . . , yn).

We finally define ∆k+1 = Σk+1 ∩ Πk+1. Here, we are looking at equivalenceclasses of formulas.

Proposition 21.4. If ϕ(x1, . . . , xn) is ∆0 and M ⊆ N have transitive un-derlying sets, then for all a1, . . . , an ∈ M , M � ϕ(a1, . . . , an) if and only ifN � ϕ(a1, . . . , an).

Proof. From the substructure, we know they agree on the atomic formulas. Sowe check the other cases of ∆0. Negation and conjugation follows. For instance,

M � ¬ϕ ⇔ ¬(M � ϕ) ⇔ ¬(N � ϕ) ⇔ N � ¬ϕ.

For bounded quantification, consider ϕ(y, z1, . . . , zn) = ∃x ∈ yψ(x, y, z1, . . . , zn),such that for all a, a′, a1, . . . , an ∈M ,

M � ψ(a, a′, a1, . . . , an) ⇔ N � ψ(a, a′, a1, . . . , an).

If M � ϕ(a, a1, . . . , an), then there exists b ∈M such that

M � ψ(b, a, a1, . . . , an) ∧ b ∈ a.

Then M ⊆ N implies

N � ψ(b, a, a1, . . . , an) ∧ b ∈ a,

and thus N � ϕ(a, a1, . . . , an).Now suppose that N � ϕ(a, a1, . . . , an). Then there exists b ∈ N such that

N � ψ(b, a, a1, . . . , an) ∧ b ∈ a.

Because b ∈ a and a ∈M , by transitivity, we get b ∈M . SoM � ψ(b, a, a1, . . . , an)∧b ∈ a and by induction we get the result.

Math 145a Notes 59

We can extend this theorem a bit. We have just shown that if M ⊆ N aretransitive and ϕ is ∆0, then M � ϕ if and only if N � ϕ. We also have:

• if ϕ is Σ1, then M � ϕ implies N � ϕ.

• if ϕ is Π1, then N � ϕ implies M � ϕ.

Proposition 21.5. “x is the empty set” is ∆0.

Proof. We can write it as ∀a ∈ x¬(a ∈ x).

Proposition 21.6. “x is an ordinal” is ∆0.

Proof. x is an ordinal if and only if x is transitive and x is well-ordered by ∈.Transitivity can be written as

∀y ∈ x∀z ∈ y(z ∈ x)

and being well-ordered can be written as a conjunction of things like

∀y ∈ x∃z ∈ y(∀z′ ∈ y(z ∈ z′ ∨ z = z′))∃z ∈ x∀z′ ∈ x(z ∈ z′ ∨ z = z′)

and other stuff.

Math 145a Notes 60

22 November 17, 2016

Last time, we defined the Levy hierarchy, where ∆0 = Σ0 = Π0 are the onesusing only bounded quantification.

Proposition 22.1. ∆0 formulas are absolute for transitive models. That is, ifϕ(x1, . . . , xn) is ∆0, M ⊆ N is transitive, and a1, . . . , an ∈M , then

M � ϕ(a1, . . . , an)↔ N � ϕ(a1, . . . , an).

We have proved that empty(x) and ord(x) are ∆0, in models of ZF (or evensome fragment of ZF). Also countable(x) is Σ1, because we can write it as

∃f(f is an injective function and im(f) = ω and dom(f) = x).

We can even see that it is not ∆0.

Proposition 22.2 (Under Con(ZFC)). countable(x) is not ∆0.

Proof. By downward Lowenhiem–Skolem, there is a countable transitive model(M,∈) � ZFC. Let ωM1 ∈M be the element such that

M � “ωM1 is the first uncountable cardinal′′.

Then V � countable(ωM1 ) but M � ¬countable(ωM1 ).

This is Skolem’s paradox. That is, if ZFC is consistent, then there arecountable models. But ZFC implies that there are uncountable sets. There isalso a philosophical argument by Hilary Putnam, about the question “am I abrain in the box?” If I am indeed a brain in the box, then that is from theperspective of the “real” world. But if we ask the question on the “brain in thebox” world, the two notions of being a brain in the box doesn’t match. So theanswer is always no, in a very unsatisfying way.

22.1 Godel codes

In the beginning of the course, we have said that everything is a set. But we arenow talking about formulas. Are these sets? There is a way to code formulasas sets. We can assign to each symbol a different number, to each variable xn apair (1, n) and encode like

∀xn(xn ∈ xm) as (0, (1, n), 1, (1, n), 2, (1,m), 3).

But this is not very satisfactory because this doesn’t really match the semantics.So let us define Godel codes in the following way. Denote by dϕe the Godel

code of ϕ. We first assign numbers to symbols as

d(e = 0, d)e = 1, d=e = 2, d∈e = 3,

d¬e = 4, d∧e = 5, d∃e = 6, dxie = (7, i).

Math 145a Notes 61

Next we assign codes for formulas inductively as

dxi ∈ xje = (d∈e, dxie, dxje), dxi = xje = (d=e, dxie, dxje), d¬ϕe = (d¬e, dϕe),dxi ∧ xje = (d∧e, dxie, dxje), d∃xiϕe = (d∃e, dxie, dϕe).

Because every code starts with some symbol, we can parse it easily.

Theorem 22.3. M � ϕ(a1, . . . , an) is a Σ1-formula in M , dϕe, and a1, . . . , an.

Proof. Later.

22.2 Reflection

Theorem 22.4 (Reflection). Let κ = cof κ > ω and 〈(Mα, Eα) : α < κ〉 be asequence of {E}-structures such that

(a) α < β implies (Mα, Eα) ⊆ (Mβ , Eβ),

(b) if α is a limit then Mα =⋃β<αMβ and Eα =

⋃β<αEβ,

(c) |Mα| < κ.

Set M =⋃α<κMα and E =

⋃α<κEα. Then

C = {α < κ : (Mα, Eα) 4 (M,E)}

is a club1 in κ.

Proof. To prove this, we are going to define the Skolem function for M ⊆ N .For each ϕ(x, y1, . . . , yn), define fϕ : Mn →M such that if a1, . . . , an ∈M then

M � ∃xϕ(x, a1, . . . , an) implies M � ϕ(fϕ(a1, . . . , an), a1, . . . , an)

and also min{β < κ : fϕ(a1, . . . , an) ∈ Mβ} is as small as possible. In short, ifthere exists such a x, we are choosing that x, and if there doesn’t exist such ax then we are choosing anything. This requires choice.

In the homework, you proved the following fact:

If X =⋃α<κXα, |Xα| < κ, and Xα is increasing and continuous,

and {(fn : Xkn → X) : n < ω, kn < ω} is a sequence of functions,then

{α < κ : fn”Xknα ⊆ Xα for all n < ω}

is a club.

1Recall that C being a club means that, if X ⊆ C is not cofinal then supX ∈ C, and Citself is cofinal in κ.

Math 145a Notes 62

Note that the number of formulas is ω. So we see that

C = {α < κ : fϕ”Mα ⊆Mα for all ϕ ∈ Formulas(∈)}

is a club.We now claim that α ∈ C if and only if (Mα, Eα) 4 (M,E). First assume

that (Mα, Eα) 4 (M,E). If a1, . . . , an ∈ Mα and ϕ is a formula, look atM � ∃xϕ(x, a1, . . . , an). If it is false, then fϕ(a1, . . . , an) ∈ M0 ⊆ Mα byminimality. If it true, then Mα � ∃xϕ(x, a1, . . . , an) and again by minimality,fϕ(a1, . . . , an) ∈Mα. Therefore fα”Mα ⊆Mα.

Now let us show that (Mα, Eα) 4 (M,E) assuming that fα”Mα ⊆Mα. Foratomic formulas, it follows immediately from (Mα, Eα) ⊆ (M,E). Negationand conjunction are more straightforward. For existential quantifiers, this iswhat the Skolem functions are designed for. If Mα � ∃xϕ(x, a1, . . . , an), sodoes M . If M � ∃xϕ(x, a1, . . . , an) then M � ϕ(fϕ(a1, . . . , an), a1, . . . , an). SoMα � ϕ(fϕ(a1, . . . , an), a1, . . . , an).

So this is a set-theoretical version of reflection. There is a more generalizedreflection, called the generalized reflection.

Theorem 22.5 (Generalized reflection). Let 〈ωα : α ∈ On〉 be a definabletransitive hierarchy. Set ω =

⋃α∈On ωα. For each formula ϕ(x1, . . . , xn) there

is some α ∈ On such that for every a1, . . . , an,

(ωα,∈) � ϕ(a1, . . . , an)↔ (ω,∈) � ϕ(a1, . . . , an).

22.3 Constructing the constructible universe

For a transitive structure (M,E) and X ⊆M , define the definable power setas

DefM (X) =

{Y ⊆ X :

∃ϕ(x, y1, . . . , yn) and a1, . . . , an ∈Msuch that y ∈ Y ↔ (M,E) � ϕ(y, a1, . . . , an)

}.

We now define

L0 = ∅, Lα+1 = DefLα Lα, Lδ =⋃α<δ

Lα for a limit δ, L =⋃α∈On

Lα.

Theorem 22.6 (In ZF).

(1) L is a definable class, i.e, there exists a formula ϕ(x) such that x ∈ L↔ϕ(x).

(2) L � (∀xϕ(x)). This “∀xϕ(x)” is called V = L.

(3) L � ZFC + GCH.

Math 145a Notes 63

Proof of (1). We are going to show that ∃y(x = Defy y) is a definable formula(and we will even see later that it is Σ1). Set

Lev(x, y) = “ ord(y) ∧ x = Ly”

= ord(y) ∧ ∃f

f is a function with domain y + 1∧

f(0) = ∅∧(∀β ∈ y + 1(lim(β)→ f(β) =

⋃{f(γ) : γ ∈ β}))∧

(∀β, β′ ∈ y + 1(β = β′ + 1→ f(β) = Deff(β′) f(β′)))∧

f(x) = y

.

This is made so that f(α) = Lα for all α ∈ y + 1.Now x ∈ L is ∃α, y(Lev(y, α) ∧ x ∈ y). This is in Σ1.

We are further going to show that L is canonical, in the sense that if W ⊆W ′are models of ZF, then LW = LW

′.

Math 145a Notes 64

23 November 22, 2016

Last time we showed that L is a definable class. We also gave a first orderformula “Lev(x, α) = x ∈ Lα”. We now want to show that L � ZF and alsoL � V = L.

23.1 L satisfies ZF

Proposition 23.1 (L � ZF). For each axiom ϕ of ZF, ZF proves ϕL(relativisationof ϕ to L).

Proof. By one of your homeworks, L is transitive. Then FoundationL andExtensionalityL holds. Now let us now show that the other axioms are alsosatisfied. If x, y ∈ Lα, then {x, y} ∈ Lα+1 because we have

{x, y} = {z ∈ Lα : Lα � (z = x) ∨ (z = y)} ∈ Lα+1.

Likewise, we can define the union as

∪a = {z ∈ Lα : Lα � ∃y ∈ a(z ∈ y)} ∈ Lα+1.

We have ω ∈ Lω+1 and so there is an infinite set.For Power Set, we need to show that for any a ∈ L, P(a)∩L ∈ L. This is an

interesting statement, because P(a) ∩ L 6= DefLα a. In fact, we will later showthat P(ω) ∩ L ⊆ Lω1

but P(ω) ∩ L 6⊂ Lα for α < ω1. So we have to go up farup into the hierarchy. For X ⊆ a, let

f(X) =

{0 if X /∈ Lβ minimal such that X ∈ Lβ .

Then if γ = sup{f(X) : X ⊆ a}, P(a) ∩ L ⊆ Lγ . Then

P(a) ∩ L = {z ∈ Lγ : Lγ � z ⊆ a} ∈ Lα+1.

The next axiom is Comprehension. If a, b1, . . . , bn ∈ Lα and ϕ(x, y1, . . . , yn),we want to show that {x ∈ a : L � ϕ(x, b1, . . . , bn)} ∈ L. We have

{x ∈ a : Lα � ϕ(x, b1, . . . , bn)} ∈ Lα+1,

but they are not the same set because of existential quantifiers. So we usethe general reflection principle. There exists a α < γ ∈ On such that for allb′1, . . . , b

′n, b′n+1 ∈ L′γ ,

Lγ � ϕ(b′1, . . . , b′n+1)↔ L � ϕ(b′1, . . . , b

′n+1).

Then

{x ∈ Lγ : Lγ � x ∈ a ∧ ϕ(x, b1, . . . , bn)} ∈ Lγ+1.

Collection can be proven similarly.

Math 145a Notes 65

23.2 Defining definablity

Theorem 23.2. The formula DEF(x,M) = “x = DefM M” is a Σ1-formula.

Corollary 23.3. (1) If W ⊆ V is a transitive, definable model of ZFC withOn ⊂W , then L = LW (= {x : W � ∃αLev(x, α)}).(2) L � V = L.(3) L is the ⊆-minimal model of ZFC containing all ordinals.

Proof. (1) Let x ∈ LW . Then there exists an α ∈ OnW = On such thatW � “∃y(x ∈ y ∧ Lev(y, α))”. Because Σ1-formulas are absolutely upward, wehave ∃y(x ∈ y ∧ Lev(y, α)). Then x ∈ Lα. This shows that LWα ⊆ Lα. Now letus show that Lα ⊆ LWα by induction. If x ∈ Lα+1, then

x = {z ∈ Lα : Lα � ϕ(x, a1, . . . , an)}= {z ∈ LWα : LWα � ϕ(x, a1, . . . , an)} ∈ LWα+1.

(2) We have L ⊆ V . So by (1), L = LV = LL. Then for any x ∈ LL,

∃α(L � ∃y(x ∈ y ∧ Lev(y, α))).

(3) This is clear.

Let us now prove Theorem 23.2. We are going to try an define the formulax = DefM M . To prevent the formula from climbing up the hierarchy, we aregoing to set W to be the “master domain” for quantification. So our strategy isto first quantify everything over W and then at the end set W to be the thingswe want to actually quantify over.

First suppose ω ∈W . Note that omega(x) = “x = ω” is ∆0. So we may useω as a parameter. Define

TERM(x,M,W ) = ∃n ∈ ω(x = dxne) ∨ ∃a ∈M(x = dae).

Then we define the atomic formulas as

ATOM(x,M,W ) = ∃y, z ∈W ( TERM(y,M,W ) ∧ TERM(z,M,W )

∧ (x = dy = ze ∨ x = dy ∈ ze)).

Now define a formula sequence

FSEQ(F, dϕe,M,W ) =

∃n ∈ ω(“F is a func. with dom. n+ 1”

∧ ∀m ∈ n+ 1

ATOM(F(m),M,W)

∨∃k ∈ m(F (m) = d¬F (k)e)∨∃k, l ∈ m(F (m) = dF (k) ∧ F (l)e)∨∃k ∈ m∃i ∈ ω(F (m) = d∃xiF (k)e)

∧ F (n) = dϕe).

Math 145a Notes 66

We now encode the statement of a formula F being built only using variablesin {xi : i < n}. Let

VLEN(n, F,M,W ) =

∃i ∈ dom(F )( ATOM(F (i),M,W )∧(F (i)(1) = dxn+1e ∨ F (i)(w) = dxn−2e))

∧∀m ∈ ω(∃i ∈ dom(F )( ATOM(F (i),M,W )∧(F (i)(1) = dxn+1e ∨ F (i)(w) = dxn−2e))→ m ∈ n).

Now we define a satisfaction sequence.

SSEQ(S, F,dϕe,M,W ) =

∃n, r ∈ ω(“F, S are func. with dom. n+ 1”

∧ FSEQ(F, dϕe,M,W ) ∧VLEN(r, F,M,W )

∧ ∀m ∈ n+ 1(Based on how F (m) is formed

→ Form S(m) to be all tuples in Mr satisfy the formula)).

Here, satisfying the formula can be written as

ATOM(F (m),M,W ) ∧ F (M) = dxi ∈ xje → “S(m) = {s ∈Mr : s(i) ∈ s(j)}”

plus

∃i ∈ r, k ∈ m(F (m) = d∃xiF (k)e)→ S(m) = {s ∈Mr : ∃s ∈ S(k)∀i′ < r(i′ = i→ s(i) = s(i′))}

plus some other things. Then finally we can define satisfaction as

SAT(M, dϕe, s,W ) = ∃S, F ∈W (SSEQ(S, F, dϕe,M,W ) ∧ s ∈ S).

Finally, using this, we can define definability as

DEF(x,M) = ∃W ( MASTER(W,M)

∧(∀z ∈ w∃dϕe ∈W, i ∈ ω(FVAR(dϕe, {i},M,W ) ∧ ∀y ∈M

(y ∈ z ↔ ∃s ∈Mr(s(i) = y ∧ SAT(M, dϕe, s,W ))))

)∧(∀dϕe ∈W∃i ∈ ω(FVAR(dϕe, {i},M,W )→ ∃z ∈ x∀y ∈M(y ∈ z ↔ ∃r ∈ ω, s ∈Mr(SAT(M, dϕe, s,W ) ∧ s(i) = z)))

)).

We still have to define what the master formula is. We are going to define

MASTER(W,M) = ∃f∃x ∈W (

omega(x) ∧ “f is a func. with dom. ω”

∧ ∀n ∈ x(f(n) ⊆W ∧ f(n) ⊆ f(n+ 1))

∧ (“everything to start recursive def.” ⊆ f(0))

∧ ∀n ∈ x(“if we have elements that could be combined in f(n)

→ their combination is in f(n+ 1)”).

This shows that the advertised complexity of the statements are correct.

Math 145a Notes 67

24 November 29, 2016

So last time we went through a couple of things. The main thing was thatDEF(x, y), which codes x = Defy y is Σ1. This gave a formula Lev(x, α) ∈ Σ1

that codes x = Lα. This implies that L � (V = L) and so is the ⊆-minimalinner model of ZF containing all ordinals. Today we are going to show thatL � Choice, and also L � ∀λ(2λ = λ+).

24.1 L satisfies Choice

There are many equivalent forms of Choice, and we are going to use the well-ordering one. Note that x ∈ L if and only if there exist α ∈ On, a formulaϕ(y, x1, . . . , xn) and a1, . . . , an ∈ Lα such that

x = {y ∈ Lα : Lα � ϕ(y, a1, . . . , an)}.

To order such elements, we first order formulas. Each ϕ(x1, . . . , xn) (with noparameters) is a sequence ϕ0, . . . , ϕk such that each term in the sequence is anatomic formula or the negation or conjunction or existential quantified formulaof something before. Now define <τ on formulas by:

ϕ <τ ψ ↔(

sequence of ϕ is shorter than the sequence of ψ orlengths are equal and ϕ <lex ψ.

)Note that two equivalent formulas can be technically different formulas.

Proposition 24.1. <τ well-orders Formulas(τ).

Proof. Given X ⊆ Formulas(τ), we can first fine a minimal length and then<lex-minimal of the minimal length formulas, which are going to be finite.

Now define <α to be a well-order on Lα by recursion, so that if α < β, thenLα is an initial segment of Lβ . For α = 0, we have L0 = 0. So we don’t needanything. Suppose α > 0. Given x ∈ Lα, set

βx = min{β < α : x ⊆ Lβ},ϕx = min

<τ{ϕ : ∃a1, . . . , an ∈ Lβ(x = {y ∈ Lβ : Lβ � ϕ(y, a1, . . . , an)})},

(ax1 , . . . , axn) = min

<βx{(a1, . . . , an) ∈ Lnβx : x = {y ∈ Lβ : Lβ � ϕ(y, a1, . . . , an)}}.

Now given x, y ∈ Lα, we define the order as

x <α y ↔

βx < βy orβx = βy and ϕx <τ ϕy or

βx = βy, ϕx = ϕy and (ax1 , . . . , axn) <lex

βx(ay1, . . . , a

yn).

Then <α is a well-ordering of Lα because it is a lexicographic product of wellorderings. Also it is clear that the orderings extend each other.

Math 145a Notes 68

Corollary 24.2. L � Choice.

Proof. Given x ∈ L, find α so that x ⊆ Lα. Then <α restricted to x is awell-ordering.

I haven’t made a big deal out of this yet, but we haven’t used Choice in Vto show that L satisfies Choice.

Corollary 24.3. Con(ZF)→ Con(ZFC).

Proof. In a model of ZF, build L. Then L � ZFC.

24.2 Models satisfying V = L

Before we do that, we have to examine the question of when do transitive modelsthink M � V = L. Note that V = L is just the formula

(V = L) = ∀x∃α(ord(α) ∧ xEy ∧ Lev(y, α)).

The answer is that this holds exactly when M = Lδ for some δ a limit. Inproving this, we are going to use the fact that if α < δ and δ is a limit, thenLδ � DEF(Lα+1, Lα). This is one of the homework due Thursday.

Proposition 24.4. If M � (V = L) for some transitive M , then M = Lδ forsome limit δ.

Proof. By definition of V = L, M contains some ordinals. (Recall that ord(α) ∈∆0.) Set δ = On∩M 6= ∅. Since M is transitive, δ is an ordinal. This is becausegiven α ∈ δ, we have α ⊆ δ by transitivity.

We now claim that δ is a limit ordinal. If not, suppose δ = α + 1 so thatα ∈M is the largest ordinal. Apply V = L to α to get y ∈M and β ∈M suchthat M � ord(β)∧ α ∈ y ∧ Lev(y, β). But note that this is a Σ1 formula. So Mis actually right about everything, i.e., y ∈ Lβ and α ∈ Lβ and so β > α. Thiscontradicts the maximality of α. So δ is indeed a limit ordinal.

Now we show that M = Lδ. Let α ∈ δ. There exists y ∈ M and β ∈ Msuch that y = Lβ and α < β. Then Lα ⊆ Lβ ⊆ M . This shows that Lδ =⋃α<δ Lα ⊆ M . For the other direction, consider an arbitrary x ∈ M . There

exist y, β ∈ M such that x ∈ Lβ = y. Then β ∈ M ∩ On = δ and so x ∈ Lδ.Thus M = Lδ.

We would now like to show that its converse holds. That is, if δ is a limitordinal then Lδ � (V = L).

Lemma 24.5. If δ is a limit and α < δ, then 〈Lβ : β < α〉 ∈ Lδ.

This allows us to transfer the Σ1 formula Lev(x, α) from L to Lδ. AlthoughΣ1 formulas don’t go downwards, this sequence will serve as a witness.

Math 145a Notes 69

Proof. We use induction on δ. We have three cases: δ = ω, δ = δ′ + ω for somelimit δ′, and δ a limit of limit ordinals. First look at the case δ = ω. Then α < δis finite, and so Lω contains all finite sequences of finite things (more precisely,hereditarily finite sets).

It is also easy when δ is the limit of limits. Let α < δ and then there existsa limit δ′ < δ so that α < δ′. Then 〈Lβ : β < α〉 ∈ Lδ′ ⊆ Lδ.

Finally assume that δ = δ′ + ω. If α < δ′, then we are done. Otherwise,there is some n < ω such that α = δ′ + n. Suppose first that n = 0. Thenα = δ′. By induction, for every β < α we have 〈Lβ′ : β′ < β〉 ∈ Lδ′ . This showsthat (β′, Lβ′) ∈ Lδ′ , and thus

〈Lβ : β < α〉 = {(β, Lβ) : β < α} ⊆ Lδ′ .

But we want this set as a definable subset. Set ϕ(x) to be

“x = (α′, y) where ord(α′) ∧ ∃f

function with domain α′∧f(0) = ∅ ∧DEF(f(β + 1), f(β))∧β < α limit→ f(β) =

⋃β′<β f(β′)

”

So

〈Lβ : β < α〉 = {y ∈ Lδ′ : Lδ′ � ϕ(y)} ∈ Lδ′+1.

Now consider the case n > 0, where α = δ′+n. Then 〈Lβ : β < δ′+n−1〉 ∈Lδ by induction. We have Lα−1 = {y ∈ Lα−1 : Lα−1 � (y − y)} ∈ Lα. So weget (α− 1, Lα−1) ∈ Lα+5 or something. Then adding this element in, we get

〈Lβ : β ∈ δ′ + n〉 ∈ Lδ.

This finishes the proof.

Theorem 24.6. If δ is a limit, then Lδ � (V = L).

Proof. Remember this says that ∀x∃y, α(x ∈ y∧ord(α)∧Lev(y, α)). Let x ∈ Lδ.There exists some α < δ such that x ∈ Lα, and by construction, Lα ∈ Lδ. Nowit suffices to show Lδ � Lev(Lα, α). Now use the previous lemma with α + 1.So f = 〈Lβ : β < α+ 1〉 ∈ Lδ. We now want to show that

Lδ �

f is a function with domain α+ 1∧f(0) = ∅ ∧DEF(f(α+ 1), f(α))∧

(β limit → f(β) =⋃β′<β f(β′)) ∧ f(α) = Lα

.

But everything here is ∆0, because of the things you are going to prove in thehomework. Thus L thinking that all these are true implies Lδ thinking that allthese are true.

We are going to prove that this implies L � GCH next time.

Math 145a Notes 70

25 December 1, 2016

The plan today is finish talking about L and talk about forcing briefly. Lasttime we showed that L � ZFC +(V = L). Also, we showed that M � (V = L) ifand only if M = Lδ for some limit δ. The goal is to prove L � GCH.

25.1 L satisfies GCH

Lemma 25.1 (Condensation lemma). If an infinite set X ∈ L is transitive,then X ∈ L|X|+ .

Note that being transitive is very important. There are lots of singletonsthat show up later in the hierarchy.

Proof. Find a limit δ0 such that X ∈ Lδ. We have a structure (Lδ,∈). Usethe Downward Lowenheim–Skolem–Tarksi to find an elementary substructure(M,EM) =M 4 (Lδ,∈) such that

• X ∪ {X} ⊆M ,

• |M | ≤ |X|.

Because L � (V = L), we have M � (V = L). We don’t know that M is tran-sitive. But we know that EM is an extensional well-founded relation. UseMostowski’s theorem to find

π : (M,EM) ∼= (A,∈)

with A transitive. And A � (V = L). So by last time, A = Lγ for γ = On∩A alimit ordinal. Then |γ| ≤ |A| = |M | = |X|. So γ < |X|+ and A ⊆ L|X|+ .

But we are not there yet, because we want X ∈ L|X|+ . We claim thatX ∈ A. Recall that X ∈ M is transitive and for all x ∈ M , π(x) = {π(y) : y ∈M,M � yEx}. By induction on rank, for all x ∈ X ∪ {X}, π(x) = x. This isbecause π(∅) = ∅ and π(x) = {π(y) : y ∈ x} = {y : y ∈ x} = x. This shows thatπ(X) = X ∈ A.

Theorem 25.2. L � GCH.

Proof. Let us work in L. Let λ be a cardinal (i.e., let λ ∈ L such that L �“λ is a cardinal”). We claim that P(λ) ⊆ Lλ+ . Let X ∈ P(λ). Then {X} ∪ λis checked to be transitive. By condensation, we have {X} ∪ λ ∈ Lλ) and soX ∈ Lλ+ . Then |P(λ)| ≤ |Lλ+ | = λ+.

Corollary 25.3. Con(ZF)→ Con(ZFC + GCH + V = L).

Math 145a Notes 71

25.2 Forcing

So far, we have done everything by constructing L. But we don’t know how todo the other direction. This is done by forcing.

Theorem 25.4. Con(ZF)→ Con(ZFC +¬GCH + V 6= L).

Because I have an hour, I can going to try to give at least a flavor of forcing. Itwill lack details and ignore metamathematical issues.

Assume we have a countable, transitive model M of ZFC. We want to builda transitive N ⊇M such that

• N � ZFC,

• On∩M = On∩N ,

• N is built in a “controlled fashion”.

Here is the outline. Start with a partial order P ∈ M and find a M -genericfilter G /∈M . Then form M [G] ⊇M ∪ {G}, and argue in general M [G] � ZFC.Finally use the properties P to conclude something.

Let P ∈ M be a partial order, and assume that P has a greatest element1P ∈ P. (This is a ∆0 statement.) We say that a subset D ⊆ P is dense if andonly if for every p ∈ P, there is a q ∈ D such that q ≤ p. We also say that G ⊆ Pis a filter if

• ∀p ∈ P, q ∈ G(q ≤ p→ p ∈ G),

• ∀p, q ∈ G(∃r ∈ G(r ≤ p, q)).

A filter G is called M-generic if for all dense D ∈M , G ∩D 6= ∅.

Proposition 25.5. If M is a countable transitive model and P ∈M , then thereexists a M -generic filter G /∈M .

Assume, given a M -generic G, we can build a countable transitive M [G] )M ∪{G} such that M [G] � ZFC and M ∩On = M [G]∩On. We assume this asa black box for now.

Take P = Add(ω, 1) ∈ M . Then Dn,α,Diffα,β , DAα ∈ M for n < ω, α = 0,

and A ∈ P(ω)∩M . If G is M -generic, it meets all of these. Then⋃G : 1×ω → 2

is a function. Define

X = XG0 = {n < ω :

⋃G(0, n) = 1} ⊆ ω.

But X 6= A for A ∈M . So M [G] contains a subset of ω that M doesn’t.

Theorem 25.6. M [G] � (V 6= L).

Proof. LM [G] = LM ⊆M . But X ∈M [G]− LM [G].

Math 145a Notes 72

Here we added only one element. But we can do more than this. Find anα ∈M such that M � “α > ω1 is a cardinal”. Set P = Add(ω, α)M ∈M . Finda M -generic G ⊆ P and form M [G]. From G, we get

M [G] � “∃ injection from α to P(ω)”.

So M [G] � “2ω ≥ α”. We would like to show that M [G] � “α > ω1”. Butthis statement is Σ2 or something, and so we can’t say this directly. But itturns out that M [G] still thinks this is true. This is because Add(ω, α)M hasthe countable chain condition. If P ∈M has the countable chain condition andG ⊇ P is M -generic, then M and M [G] have the same cardinalities and thesame cofinalities. This means that M [G] � “α > ω1 is a cardinal”. These twocombined shows that M [G] � ¬CH.

You can run the same argument by using κ instead of ω although it becomesmore complicated. Also use can do it more than one cardinal at a time.

Theorem 25.7 (Easton). Suppose M satisfies F is a (definable class) such thatfor all κ, λ ∈ domF ,

(1) κ < λ→ F (k) ≤ F (l),

(2) cof(F (κ)) > κ,

(3) κ regular.

Then there exists a P ∈M such that for all M -generic G ⊆ P,

M [G] � “∀κ ∈ domF (2κ = F (κ))”

and M and M [G] have the same cardinals and cofinalities.

We are going at least open the black box and look inside. So what is M [G]?Fix a P ∈M . Define a P-name as, x is a P-name if and only if

x = {(y, p) : p ∈ P, y is a P-name}.

Given a P-name x and a M -generic filter G, we define

xG = {yG : (y, p) ∈ x, p ∈ G}

and M [G] = {xG : x ∈M is a P-name}.For each x ∈M , we can define a canonical name x by

x = {(y, 1P) : y ∈ x}.

Then xG = x and so M ⊆M [G]. Also Γ = {(p, p) : p ∈ P}. Then for all G ⊆ P,ΓG = G ∈M [G].

Forcing black box. Let P ∈M be a p.p. Then there is a definable relation such that ∀a1, . . . , an ∈ Name and p ∈ P and ϕ(x1, . . . , xn), the following areequivalent:

(1) M � “p ϕ(a1, . . . , an)”.

(2) For any M -generic G with p ∈ G, M [G] � ϕ(aG1 , . . . , aGn ).

The book has more detailed exposition to forcing.

Index

< κ-branching, 47V = L, 62κ-embeddable tree, 50P-name, 72 Los’s theorem, 45

Aronszajn tree, 49

Banach-Tarski Paradox, 43Banach-Tarski paradox, 33beth numbers, 29bounded quantifier, 57branch, 47

Cantor normal form, 20Cantor’s theorem, 45Cantor-Bernstein theorem, 21cardinal, 22cardinality, 24class, 10club, 61cofinality, 25, 26complete, 46condensation lemma, 70constructible universe, 62countable chain condition, 46countable set, 12

definable power set, 62dense linear ordering, 45

elementary substructure, 56equinumerous, 21

filter, 40, 71formula, 8, 54free variable, 54function, 9

Godel codes, 60Godel’s incompleteness theorem, 30Godzilla ordinal, 26

height, 47

homeomorphism, 33hyperreals, 42

inaccessible cardinals, 29independence, 53

Konig’s theorem, 29, 47

Lowenheim–Skolem–Tarskitheorem, 57

Levy hierarchy, 58language, 53limit, 19

measure, 32translation invariant, 32

Mostowski’s theorem, 57

natural numbers, 16normal tree, 50

order type, 17ordinal, 16

partial order, 47principal, 40product, 9

rank, 21reflection, 61regular ordinal, 25

separable, 46singular ordinal, 25Skolem function, 61Skolem’s paradox, 60strong limit cardinals, 29substructure, 56successor, 19Suslin line, 46Suslin tree, 51syntax, 54

term, 54transfinite induction, 13transitive closure, 13tree, 47

73

Math 145a Notes 74

ultrafilter, 40ultralimit, 41

Von Neumann universe, 10

well-founded relation, 13well-ordering, 10, 13

ZFC, 7