Math 145A - Set Theory I

Math 145a - Set Theory I

Taught by William Boney Notes 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 the course. 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 4 1.1 Three problems ...... 4 1.2 Functions and pairs ...... 5 1.3 Natural numbers to real numbers ...... 5

2 September 6, 2016 7 2.1 Zermelo-Fraenkel with Choice ...... 7 2.2 Equivalents of Replacement ...... 9 2.3 Basic notions ...... 9

3 September 8, 2016 10 3.1 Classes ...... 10 3.2 Axiom of Choice ...... 10 3.3 Cardinality of sets ...... 11

4 September 13, 2016 13 4.1 Well-founded and well-ordered ...... 13 4.2 Transﬁnite induction ...... 13

5 September 15, 2016 16 5.1 Ordinals for real ...... 16 5.2 Properties of ordinals ...... 17

1 Last Update: August 27, 2018 6 September 22, 2016 19 6.1 Ordinal induction and arithmetic ...... 19 6.2 Building the universe ...... 20 6.3 Cardinals ...... 21

7 September 27, 2016 23 7.1 Cardinal arithmetic ...... 23 7.2 Coﬁnality ...... 25

8 October 4, 2016 26 8.1 Successor cardinals are regular ...... 26

9 October 4, 2016, Make-up class 28 9.1 Regular limit cardinals ...... 28 9.2 Inaccessible cardinals and models of ZFC ...... 29

10 October 6, 2016 32 10.1 The measure problem ...... 32 10.2 The Baire space ...... 33

11 October 11, 2016 35 11.1 Midterm ...... 35

12 October 18, 2016 36 12.1 Topology of ωω ...... 36 12.2 Measures on the Baire space ...... 37

13 October 20, 2016 38 13.1 Borel sets ...... 38 13.2 Projective sets ...... 39 13.3 Filters and ultraﬁlters ...... 40

14 October 25, 2016 41 14.1 Ultralimits ...... 41 14.2 Ultraproducts ...... 42

15 October 25, 2016, Make-up class 43 15.1 Banach-Tarski Paradox ...... 43

16 October 27, 2016 45 16.1 Dense linear orderings ...... 45

17 November 1, 2016 47 17.1 Trees ...... 47 17.2 Branches of trees ...... 47

2 18 November 3, 2016 49 18.1 ω1-Aronszajn trees ...... 49 18.2 Other classes of tress ...... 50

19 November 8, 2016 51 19.1 Suslin trees ...... 51

20 November 10, 2016 53 20.1 The idea of L ...... 53 20.2 First-order logic ...... 53

21 November 15, 2016 56 21.1 Substructures and elementary substructures ...... 56 21.2 Models of set theory ...... 57 21.3 L´evyhierarchy ...... 57

22 November 17, 2016 60 22.1 G¨odelcodes ...... 60 22.2 Reﬂection ...... 61 22.3 Constructing the constructible universe ...... 62

23 November 22, 2016 64 23.1 L satisfies ZF ...... 64 23.2 Defining definablity ...... 65

24 November 29, 2016 67 24.1 L satisﬁes Choice ...... 67 24.2 Models satisfying V = L ...... 68

25 December 1, 2016 70 25.1 L satisﬁes GCH ...... 70 25.2 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 three motivating problems.

1.1 Three problems The cardinality of X is the number of elements in X. For instance we are going 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 greater cardinality than X.

An immediate corollary is that |N| < |P(N)| = |R|. This leads to a question that comes from Cantor: is there anything between them? This fundamental problem was left open until it was answered as not an answer. The statement is independent of ZFC, i.e., the foundations of ZFC is just not enough to answer this problem. The continuum hypothesis states that the answer to this is no, and the generalized continuum hypothesis deals with the existence of sets Y with |X| < |Y | < |P(X)| for an arbitrary set X. The second problem is about combinatorics. The ordered set (R, <) can be characterized by (1) dense linear ordering without endpoints, (2) connected in the order topology, and (3) separable in the order topology. Now Suslin’s question was whether one can replace the third condition (3) to (3’) every collection of disjoint open intervals is countable. The answer to this is again, who knows? The word combinatorics is used here because such questions are related to trees, or proper generalization of trees. Lemma 1.2 (König’slemma). If T is an infinite tree such that every vertex has 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 what happens when objects get larger and larger. It fails on the next one. This is basically because there is this nice countable dense set Q. But things get interesting when we take it to the next step. Then the independence phenomenon comes in to the picture. The third theme is measure theory. This is the measure theory in the actual measure theoretic sense. Theorem 1.3. (1) (Vitali) There are non-measurable sets. (2) (Banach-Tarski) You can decompose one sphere and reassemble them into two spheres of equal volume. One of my friends use the Banach-Tarski theorem to give a physical proof of the negate of the axiom of choice. But I don’t accept this proof. There is a good reason we can’t do this in nature, and it’s because they never occur in the physical 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 all set. 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 can think 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 in this 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 define n < m if and only if n ∈ m. Likewise we can say that n ≤ m if and only if n ⊆ 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 define n+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 k to n × m. Now we get to the integers Z. Now there is a standard way to make a monoid into a group, in algebra. For (m, n) ∈ N2, we want this to represent m − n. But because we don’t know subtraction yet, we set

(m, n) ∼ (m0, n0) if and only if m + n0 = m0 + n.

This is in fact an equivalence relation. So we deﬁne

2 Z = {[(m, n)]∼ :(m, n) ∈ N }. Math 145a Notes 6

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

0 0 0 0 [(m, n)]∼ + [(m , n )]∼ = [(m + m , n + n )]∼, 0 0 0 0 0 0 [(m, n)]∼ · [(m , n )]∼ = [(mm + nn , mn + m n)]∼.

In a similar way, we can construct the rationals. This is just turning a ring into the ﬁeld of fractions. For (a, b), (c, d) ∈ Z2 with b, d 6= 0, we deﬁne an equivalence 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 problem about the reals in the problem set. You could give a metric and then deﬁne the Cauchy sequences as real numbers, but the way I am going to do is through Dedekind cuts. Say that L ⊆ Q is proper downward closed if ∅ 6= L 6= Q and for any x < y ∈ Q if y ∈ L then x ∈ L. Deﬁne L0 ∼ L1 if they have the same supremum, which can be written as “for any x ∈ L0 and n ∈ N there exists 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 be two types of axioms. The ﬁrst will be that certain sets exist. The second type will be that we can get some sets from other sets. To deal with sets, we are going to 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 S x. S This is a bit diﬀerent from what we are used to, because Xi is what we S i∈I 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 topological space. This also says that you can get all of the subsets of x, no matter how huge 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 out here. We note that we haven’t actually justiﬁed that the set of natural numbers is a set. Now this was what Zermelo came along when Russel found a problem. People were happy for a while but there were still problems and Fraenkel added this single 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 is first 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 first is when there are a finite number of sets. The second is when the sets have some additional structure so that you can find a certain element easily. You need the Axiom of Choice when you are picking among an infinite pair of socks but you don’t need it when you are picking among an infinite pair of shoes, because you can 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 to talk 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 satisﬁes something 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 contains the 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 the actually image using Comprehension. It is also easy to see that Replacement implies Collection. Proposition 2.1. Under the other axioms of ZF, the Replacement Schema implies the Comprehension Schema.

Proof. Take a and φ(x, a1, . . . , an). If ¬ϕ(e, a1, . . . , an) for all e ∈ a, then we are 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 used to 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 S S By Union, we know that x = y∈x y is a set. So if {Xi : i ∈ I} is a set, then [ [ Xi = {Xi : i ∈ I} i∈I is a set.

3.1 Classes Everything we talk about is a set. But we need to distinguish it from the following 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. Or alternatively {y : y∈ / y} is a set by Comprehension, and Russel’s paradox leads to a contradiction. But I’m talking about the collection of sets, by giving a quantifier. A class is simply any collection of sets. Clearly any set is a class. A proper class is a class that is not a set. Definable classes are classes with a first-order definition, and all other classes are called undefinable classes. Example 3.2. The formula x = x defines the class V , which is also called Von Neumann universe. Another class is Grp, which is the class of all groups. I can’t give an example of an undefinable class, because if I can, then it would be definable. The reason there are proper classes is because there can be collections that are too big. Suppose a is a set and X is a class such that there is a bijection f : a → X . Here f is not really a set but a definition of a function. Then by Replacement, 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 induction. 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|. Q Q 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-order this set, and then there is a unique element we can pick, which is the minimal element. Set ∗ [ X = {Xi : i ∈ I}

∗ ∗ and let < be a well-order of X . Then define f by f(i) = min Xi ∈ Xi. This is a choice function. There are some weakenings of the Axiom of Choice, and one of this is the Axiom of Dependent Choice. Axiom of Dependent Choice. Let R be a relation on X such that for all x ∈ 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 of time 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 to P(X). Here, f : X → Y is a surjection if and only if for all y ∈ Y there exists an 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, then a∈ / 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 ( y if x = f(y) g(x) = 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 inverse image. 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 prototype for 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 linear ordering 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 such that xT C(R)y if and only if x1,...,n ∈ X such that x1 = x, xn = y, and xiRxi+1. (3) Define

extR(x) = {y ∈ X : yRx}, predR(x) = {y ∈ X : yT C(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 all x ∈ 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 your parents 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 a well-ordering. In fact, if you have any well-ordering on a set, you can throw in the 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 < j implies Xi ⊆ Xj and Ri ⊆ Rj and x ∈ Xi, y ∈ Xj − Xi implies xRjy. This is an increasing sequence of well-orderings, in the sense that you are adding a S S few 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. If Y ⊆ 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 theorem to (N, <). Then the predecessor of 0 is empty, and so 0 ∈ Y . Then the rest is the 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 every element 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 and G : X × V → V is a function. Then there exists a unique function F : X → V such that for each x ∈ X, F (x) = G(x, F | ). (Note that V is all of set predR(x) theory, so F | is also in V .) predR(x)

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, pred (x) ⊆ D and f(x) = G(x, f| ). If f : D → V R predR(x) 1 1 and 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 of domains with a good function is a set, and then use replacement. We now claim that dom F = X. This is because if f : D → X and x = min(X − D) then set g : D ∪ {x} → X by ( f(y) if y ∈ D g(y) = G(y, f| ) if y = x predR(y) and g is good. Then dom F = Xby induction. Deﬁnition 4.5. If (X,R) is a well-ordering, and x ∈ X then denote

R Ix = (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 any x0, x1 ∈ X, x0Rx1 if and only if f(x0)Rf(x1). Theorem 4.6 (Compatibility). Let (X,R) and (Y,S) be well-orderings. Then one of the following happens.

(a) (X,R) =∼ (Y,S). R ∼ (b) There exists an x ∈ X such that Ix = (Y,S). ∼ S (c) There exists an y ∈ Y such that (x, R) = Iy . Math 145a Notes 15

Proof. Let z∈ / Y . Deﬁne G : X × V → V by ( min Y − f 00 pred (x) if f : pred x → X and Y − f 00 pred (x) 6= ∅ G(x, f) = S R R R z otherwise, where 00 denote the image. By transﬁnite recursion, we get a F : X → V such that F (x) = G(x, F | ). Set A = {x ∈ X : F (x) 6= z} and let f = F | . predR(x) A 00 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) and f is an injection. Now I have three cases. Case 1. dom f = X and im f = f 00 dom f = Y . The f :(X,R) =∼ (Y,S) and so f is an isomorphism. Case 2. dom f ( X. 00 Let x0 = min(X − dom f). Then dom f = predR x0. If Y − f predR x0 6= ∅, then

00 G(x0, f) = min Y − f predR x0 6= z so x ∈ A = dom f. This shows that f : IR ∼ (Y,S). 0 x0 = Case 3. im f ( Y . Set y = min Y − im f. Then as before dom f = X. So f :(X,R) ∼ IS . 0 = y0 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 uniquely representative. Recall that X is ∈-transitive if and only if x ∈ y ∈ X implies x ∈ X. Definition 5.1. A set α is an ordinal if and only if (α, ∈) is a well-ordering and α is transitive. This is a very useful definition, because this is first-order and thus gives a definable 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 called the successor, and we will denote this α + 1. Also if X ∈ On then S 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 else might 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} unless x = ∅, and then write y = z ∪ {z} unless y = ∅, and so forth. If it hits ∅ at some 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 = T 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 that ω 3 n 7→ ω + n is a definition of a function. So we can use replacement to get X = {ω + n : n < ω} as a set. Then you can take S X as an ordinal, and we write ω + ω = S X. Similarly, we can add many ωs, and so write ω + ··· + ω = ωn. Then we can use the same trick, and define ω · ω = S{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 ∈ α implies 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 and α ≤ 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 ∈ On such 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 unique R ∼ β ∈ On such that Ia = (β, ∈). Suppose not, and let a be a R-minimal coun- R ∼ terexample. For every bRa, there exists a unique βb such that Ib = (βb, ∈). Then set β = {βb : b ∈ predR a} is a set by replacement, and in fact, it is R an ordinal. So we can take π : Ia → β by taking b to βb. This is an order isomorphism. So there always exists a unique βa ∈ On, and cal this map a 7→ βa. Then α = {βa : a ∈ X} is an ordinal, and this map is a order morphism between α and 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 get R ∼ R,S. Use comparability to get f : Ia = (Y,S) for some a ∈ X, or the other way round. Then f −1 : Y → X is an injection. To do the other direction, we need the fact that ordinals can be larger than any set. Consider any X and get α with an injection g : X → α, by the next theorem. Set R on X by xRy if and only if g(x) ∈ g(y). By injectivity, this is a well-order.

Theorem 5.7 (in ZF, Hartog). For every Y , there exists an α such that there is 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, and let

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

Then considering the map A 3 R 7→ ot(dom R,R). Then α ∈ On is the image of 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 T C ∈ C and T C is the minimal element of C. Proof. We ﬁrst show that T C is an ordinal. If a ∈ b ∈ T C then for all γ ∈ C, we have a ∈ b ∈ γ and so a ∈ γ by transitivity. Then a ∈ γ and so a ∈ T C. So T C is transitive. Also T C is linearly ordered under ∈ as T C is a subset of On. Now let us prove that T C ≤ γ for all γ ∈ C. This is just because T C ⊂ γ and so T C ≤ γ. Third, T C is actually in C. Suppose T C/∈ C. Then T C < γ for all γ ∈ C. So T C + 1 ≤ γ for all γ ∈ C, and hence T C + 1 ⊆ T C. This is clearly a contradiction.

This shows that any subclass of On always has a minimal element.√ Note that 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 hβiii∈κ with βi < α S for all i ∈ κ such that α = i∈κ βi. Note that in the limit, the sequence cannot be indexed by the natural numbers. This is because the least ordinal not equinumerous with ω is not a countable union of smaller ordinals because all smaller ordinals will have at most ω many unions. To prove that these cover all cases, let C be the class of ordinals that does not fall in these categories. Let α be the least element of C, which is α = S{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, S (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 operation is 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, S (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, S 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, λ S β (3) if λ is a limit ordinal then α = β<λ α . Theorem 6.2. Suppose β > 0. Then for all α there are unique α there are unique γ1, γ2 such that α = β · γ1 + γ2. S Proof. First let us prove existence. Let γ1 = {λ : β · λ ≤ α}. Then β · γ1 ≤ S 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 (by the existence part we have proved) and so ζ + τ = ζ + λ + σ > ζ + λ. ∗ ∗ ∗ To show uniqueness, suppose that α = βγ1 + γ2 with γ2 < β. By the ∗ ∗ ∗ definition of γ1, we have γ1 ≤ γ1 but if γ1 < γ1 then β(γ1 + 1) ≤ α. Hence ∗ ∗ ∗ βγ1 +β < α which is a contradiction. So γ1 = γ1. Similarly we get γ2 = γ2. Theorem 6.3 (Cantor’s normal form). Suppose α > 0. Then α can be written uniquely as

β1 β2 βn α = ω · k1 + ω · k2 + ··· + ω · 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 α0 < α can be written in such a form. Let β = S{β0 : β0 β β ω ≤ α}. Then there are unique γ1, γ2 < ω such that α = ω · γ1 + γ2 such β+1 that γ1 < ω as otherwise ω ≤ α. If γ2 = 0, we are done. If γ2 > 9 then β γ2 < ω < α so γ2 has the appropriate form. Then combining the representation with ωβ, we get the representation of β.

One thing to note is that α ≥ β1 contains equality. This is because ordinals such that = ω, which is quite diﬀerent from cardinal arithmetic. Furthermore 0 there even is an ordinal 0 = 0 .

6.2 Building the universe S 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. T Lemma 6.5. {x : x = x} = α∈On Vα. Math 145a Notes 21

T S Proof. It suﬃces to show that x ⊆ Vα then x ∈ Vα. Note that α∈On S α∈On 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 by transfinite 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 as −1 −1 f is a choice function, hβ is a bijection. For x, y ∈ X, let xRy if h (x) < h (y) then R well orders X.

6.3 Cardinals Deﬁnition 6.8. We say that X and Y are equinumerous (X ≈ Y ) if there is a 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 and 00 00 T Y0 = y, and take Xn+1 = g (Yn) and Yn+1 = f (Xn). Let X∞ = Xn and T n∈N Y∞ = Yn. Let n∈N  h(x) if x ∈ X − X  2n 2n+1 −1 h(x) = g (x) if x ∈ X2n+1 − X2n+1  f(x) if x ∈ X∞.

−1 Note that f|X2n−X2n+1→Y2n+1−Y2n+2 is a bijection, and g |X2n+1−X2n+2→Y2n−1−Y2n+1 is a bijection. So h is a bijection when restricted to X − X∞ → Y − Y∞. Also h 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 a r :(−1)j 7→ 2a3b5j b 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 exists n n−1 a1, a2, . . . , an ∈ Q such that anx + an−1x + ··· + 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) ··· ,

n where x is the ith solution of an irreducible anx +···+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 least cardinal 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, κ is a cardinal, then κ+ is the next cardinal. Finally, let Card be the collection or cardinals. Lemma 7.1. If A ⊆ Card and A is a set, then S A ∈ Card. Proof. If A has a largest element κ, then S A = κ ∈ Card. Assume S A/∈ Card and A does not have a largest element to get a contradiction. There exists an α ∈ S A such that α ≈ S A. Let κ ∈ A such that κ < α. But then κ ≈ α as α = |S A|.

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

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

Proof. Let α be the least such that ℵα ≥ κ. If α = β + 1 then ℵβ < κ, so + κ = ℵβ = ℵα. If α is a limit then for all γ < α, ℵγ < κ so ℵα ≤ κ. So κ = ℵα. Continuum Hypothesis (CH). Every subset of P(ω) that can not be mapped invectively into ω has a bijection with P(ω). This was the ﬁrst natural question arose that is independent of ZFC. G¨odel’s incompleteness theorem says that the consistency of Peano arithmetic is independent 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 cardinal κ. 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 that iα · α ≈ iα because iα · iα = iα. (We haven’t prove this yet, but hopefully S get to it.) So γ<α Vγ ≤ iα · α and so Vα ≤ iα. But we know that iγ ≤ Vγ for all γ < α, and so [ [ iγ ≤ Vγ ≈ Vα. γ<α

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 the least ordinal α such that X ≈ α. The reason we need AC is because we should know that X is equinumerous with 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 infinite if |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 infinite α < 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)

|{(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 shows that (κ × κ, δ) =∼ (κ, ∈). Corollary 7.9. If κ, λ ∈ Card and κ, λ 6= 0, and at least is inﬁnite, then κ + 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 .

B C B Proof. Fix f ∈ (A ) , i.e., f : C → A . For c ∈ C, let fc : B → A be such ∗ ∗ that fc = f(c). Then deﬁne f : B × C → A be such that f (b, c) = fc(b). It is easy 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 an equivalence relation. What will happen if we don’t have AC? Still X ≈ Y will be an equivalence relation, and {Y : X ≈ Y } will be a class. But there might not be a function e : 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 the least β such that there exists an f : β → α such that for allα ¯ < α there exists an 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 is singular.

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’t write 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 countably many 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 α0 < α there exists β0 < β such that f(β0) > α0. If α is a limit ordinal, the cofinality cof(α) is cof(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 function f : 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 → V given by ( max{g(β), sup (h(γ) + 1) if h with β ⊆ dom h, F (β, h) = γ<β {{0}} otherwise.

The following all have coﬁnality ω: 2 ω, ℵω, ω + ω, ω , 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, then there exists an γ < cof α and a bijection f : γ → cof α. Then we can compose f which is cofinal (or its modification that is both increasing and cofinal) and cof 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 be written as a (< κ) sized union (< κ) sized sets. Proof. Let us first prove the forward direction. For contradiction, suppose that S κ is regular and κ = i<α Xi with α < κ and |Xi| < κ. Because κ is regular and |Xi| < κ, sup Xi < κ for each i < α. Then the map g : α → κ by g(i) = sup Xi S is cofinal, because κ = i<α Xi. 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. Set Xi = f(i) + 1. Then |Xi| < κ because f(i) < κ and κ is a cardinal. Also S i<α Xi = 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 = Xi i<α

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

[ + Xi ≤ κ · κ = κ < κ . i<α

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 ZFC cannot 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, then cof β = cof α. Proof. Let g : cof α → α and h : cof β → β be increasing and cofinal. Then f ◦ h : cof β → α is cofinal, and this implies cof α ≤ cof β. For the other direction, 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 γ0 < cof α such that g(γ0) > f(δ). Because f(k(γ0)) > g(γ0) > f(δ) and f is nondecreasing, k(γ0) > 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 limit needs to satisfy α ≥ ℵα.

Proposition 9.5. α ≤ ℵα.

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

α = sup β ≤ sup ℵβ = ℵα. β<α β<α

There are indeed limit ordinals α with α = ℵα. Take α0 = ω and αn+1 =

ℵαn . Then we can set

α∗ = sup αn. n<ω

This satisfy α∗ = ℵα∗ , because

ℵα∗ = sup ℵβ = sup ℵαn+1 = sup αn+1 = α∗. β<α∗ n<ω n<ω So there are ℵ-fixed points. But the fixed point we have built has cofinality ω. 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 cardinal. κ is a strong limit if λ < κ implies 2λ < κ. κ is strongly inaccessible if it is regular and strong limit.

Theorem 9.7 (König’stheorem). For any cardinal κ, κcof κ > κ. If λ < cof κ λ λ then κ = supµ<κ µ + κ. Proof. (1) Let f : cof κ → κ be cofinal and increasing. We know that κcof κ = cof κ |cof κ → κ|. If κ = κ, let {gα : α < κ} enumerate cof κ → κ. Define g : 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. Then g(β) 6= gα(β). (2) Now suppose λ < cof κ. Let f : λ → κ. Then f can’t be coﬁnal, so supα<λ f(α) < κ. Set βf = sup f“λ. Then actually f : λ → βf . This shows S that {λ → κ} = α<κ{λ → α}. Thus [ κλ = |λ → κ| = {λ → α} ≤ sup|λ → α| = sup αλ + κ. α<κ α<κ α<κ

We have a function ℵ : On → Card that is deﬁned by  ω if α = 0  + ℵα = ℵβ if α = β + 1  supβ<α ℵα if α a limit.

Now replacing the successor cardinal by taking exponentiation, we also get a function i : On → Card given by  ω if α = 0  iα = 2iβ if α = β + 1  supβ<α iα if α a limit.

The continuum hypothesis then says that ℵ1 = i1, and the generalized continuum hypothesis states that ℵα = iα for all α ∈ On. S Recall that we have the hierarchy V0 = ∅, Vα+1P(Vα), and Vα = Vβ S β<α 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 function n ∈ 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 point because κ is strongly inaccessible. Now suppose that ϕ, a, a1, . . . , an ∈ Vk such that

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

V Then use replacement on ϕ k , which is ϕ relativized to Vκ. This gives a b ∈ V such 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 bijection g : 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. Then b ∈ Vγ+1. This shows that Vκ satisfies Replacement. Now we need to say why this shows that ZFC cannot prove nor disprove the existence of strongly inaccessible cardinals. Theorem 9.10 (Gödel’sIncompleteness Theorem). No sufficiently powerful consistence 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 cardinal. 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 inaccessible cardinals, then ZFC proves the consistency of ZFC. We will see later that if ZFC proves the existence of weakly inaccessible cardinals, 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 talk about 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 talk about 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 with lengths is the this only works for intervals. This leads to the deﬁnition of a measure.

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

[ X µ Xn = µ(Xn). n<ω n<ω

We say that µ is translation invariant if for all X ∈ Γ and r ∈ R, its translate X + 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 things like point measures. For X ⊆ R, deﬁne ∗ P µ (X) = inf{ `(Ui): Ui is an open cover of X}, P µ∗(X) = sup `(Ci): Ci are closed and disjoint in X}. {

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

Theorem 10.3 (Vitali). There is no translation measure on all of R (under AC). Math 145a Notes 33

Proof. Set a relation ∼ on R by r ∼ s if s − r ∈ Q. This is an equivalence relation. 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 S q∈Q0 V + q. This satisﬁes [ [0, 1] ⊆ V + q ⊆ [1, 2]. q∈Q0

Also V + q0 ∩ V + q1 = ∅. Then X 1 ≤ µ(V ) ≤ 3. n<ω 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 can be 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. For instance suppose there is a dictionary of all words using a and b. If you tear this into halves and erase all the ﬁrst letters, you get two copies of the dictionary without 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 the measurable sets. On the other hand, there are the Vitali sets and the set coming from the Banach-Tarski paradox on the complex and nonmeasurable side. So where 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. There is 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 of these combined have a tree structure, by looking at the initial segments, which we denote by ⊇. The Baire space lives over all of this. Given two topological spaces X and Y , a map f : X → Y is a homeomorphism if and only if it is a bijection such that if U ⊆ X is open then f“U ⊆ Y is 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 enough because Q has measure zero and nobody cares about Q. To show this, we ﬁrst need a topology on ωω. Math 145a Notes 34

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

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

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

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

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

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

ω ω [ ω − Ns = {x ∈ ω : s 6⊆ x} = Nt t∈nω,t6=s 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

n x n 6= y n = s ∈ ω.

ω ω Then y ∈ Ns and x ∈ ω − Ns will be a partition of ω 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 the exam.

The goal of this problem is to prove Mostowski’s Collapsing Theorem using Transﬁnite Recursion. Theorem 11.1 (Mostowski). If E is a well-founded and extensional relation on a set P , then there is a transitive set M and an isomorphism π :(P,E) =∼ (M, ∈).

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

x = y ⇔ ∀z ∈ X(zRx ⇔ zRy).

You can prove this theorem however you want, but here’s a nice outline: (i) We wish to deﬁne π on P so that π(x) = {π(z): zEx}. Find a 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 transitive. (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 ω Ns = {x ∈ ω : 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 are going to do is to use continued fractions. Consider 26 1 1 1 = 317 = 5 = 1 . 317 26 12 + 26 12 + 1 5+ 5 Then we can associate to 26/317 the sequence [12, 5, 5]. r r Fix a r ∈ (0, 1)\Q. Deﬁne the sequence xn ∈ (0, 1)\Q and kn ∈ ω+ = ω\{0} for n < ω as

r r j 1 k r 1 r r j 1 k x0 = r, k0 = r and xn+1 = r − kn, kn+1 = r . x0 xn xn+1 This process never terminates because r is irrational. If this process never terminates, then r is in fact irrational because of the Euclidean algorithm. Then deﬁne ω r r ω f : (0, 1) \ Q → ω+; f(r) = hk0, k1,...i ∈ ω. 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 inverse is continuous. We ﬁrst show that f is injective. For contradiction, r < s ∈ (0, 1) \ Q r t s and f(r) = f(s). We claim that if r < t < s, then kn = kn = kn and n r n t s s (−1) xn < (−1) xn < (−1) xn. For n = 0, this is easy. Then

r j 1 k t j 1 k sj 1 k r k0 = r ≥ k0 = t ≥ k0 s = k0. x0 x0 x0 t r s So k0 = k0 = k0 = 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 and l < ω be the ﬁrst index the expansions diﬀer at. Then r < s if and only if l r s (−1) (kl − kl ) > 0. This shows that 1 1 > 1 > ··· converges to x from above, k0 k0 + 1 k1+ k2 1 1 1 < 1 < ··· converges to x from below. k0 + k0 + 1 k1 k1+ 1 k2+ k3 Math 145a Notes 37

Going back to what we are trying to show, given x ∈ U there exists an > 0 such that (x − , x + ) ⊆ U. Now ﬁnd and N so that both sequences get in x x x <ω (x − , x + ). Letting s = hk0 , k1 , . . . , kN i ∈ ω. Then you can show that Ns ⊆ 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 and µ(A) = 0.

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

[ 1 1 A ⊆ U = a − , a + . n 2n n 2n n<ω

∗ Then µ (A) ≤ µ(U) = 4 for every > 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. ω ω Now let us give a measure on ω. Because ω = N∅ corresponds to (0, 1)\Q, we are going to make the measure of the whole space 1. Then for each ﬁnite sequence s, we have

N = N ∪ N ∪ N ∪ · · · . s sa(0) sa(1) sa(2) We make each N get a 2−k−1 proportion of N . sa(k) s <ω Deﬁnition 12.4. Let τ be the topology generated by B = {Ns : s ∈ ω}. Deﬁne µ : B → [0, 1] by

Y 1 µ(N ) = , s 2s(i)+1 i

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. 0 0 0 ω Definition 13.1. For α ∈ On define Σα, Πα, ∆α ⊆ P( ω) in the following way. For α = 1, 0 0 0 0 0 Σ1 = {open sets}, Π1 = {closed sets}, ∆1 = Σ1 ∩ Π1 = {clopen sets}. For α > 1, we define

0 ω [ [ 0 Σα = {A ⊆ ω : A = Bn for Bn ∈ Πβ}, n<ω β<α 0 ω ω 0 0 0 0 Πα = {A ⊆ ω : ω \ A ∈ Σα}, ∆σ = Πα ∩ Σα. 0 S 0 We call ΣOn = α∈On Σα the Borel sets. Actually, you are going to see in your homework that Σ0 = Σ0 . We also ω1 On have this hierarchy:

0 0 0 Σ1 Σ2 Σ3 ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ 0 0 0 0 ∆1 ∆2 ∆3 ∆4 ··· ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ 0 0 0 Π1 Π2 Π3

0 This actually comes up in analysis. Σ2 are the countable unions of closed sets, 0 and Π2 are the countable intersections of open sets. 0 Theorem 13.2. ΣOn is the smallest σ-algebra containing closed the open sets and closed under complement and countable union. 0 0 Proof. Let A ∈ ΣOn. Then for some α ∈ On, A ∈ Σα. For countable union, let 0 0 S An ∈ ΣOn for n < ω. Let αn ∈ On such that An ∈ Σα. Set β = n αn. Then 0 0 S 0 0 An ∈ Σβ and so An ∈ Πβ+1. Thus n<ω An ∈ Σβ+2. This shows that ΣOn is closed under countable union and complement. Now let us consider and arbitrary σ-algebra m. We show by induction that 0 0 Σα ⊆ m. For α = 1, m contains the open sets, which is Σ1. Let α > 1 0 0 S and suppose that Σβ ⊆ m for β < α. Let A ∈ Σα,r write A = n Bn for S 0 ω S Bn ∈ β Πβ. Then ω − Bn ∈ β Σβ0 ⊆ m and so

[ ω ω A = ( ω \ ( ω \ Bn)) ∈ m. n 0 This shows that Σα ⊆ m. Then Borel sets are measurable, because measurable sets are closed under complement 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 projection of a measurable set in R2 not being measurable. In fact, sets obtained by projection lie somewhere on the line between the measurable and the non-measurable side. Consider the product (ωω)n. This space is the set of n-tuples of sequences of natural numbers, and the topology is given by the basis

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

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

ω G : x ∈ ω 7→ (¯x1,..., x¯n) wherex ¯i(k) = F (x(k))(i).

This now gives us a measure, and Borel sets for (ωω)n. Now we are ready to define projective set. Definition 13.4. Given A ⊆ (ωω)n+m, define

ω n ω m projn(A) = {x ∈ ( ω) : ∃y ∈ ( ω) such that (x, y) ∈ A}. This is nice from a model theoretic point of view, because we can describe using a existential quantiﬁer.

1 Definition 13.5. Define Σ0 to be the Borel sets. We define

1 ω n ω n+m Σ1 = {A ⊆ ( ω) : A = projn(B) for Borel B ⊆ ( ω) }, 1 1 Π1 = {complements of elements of Σ1}. Then deﬁne similarly,

1 ω m+n 1 Σk+1 = {projn(A): A ⊆ ( ω) is Πk} 1 1 Πk+1 = {complements of elements of Σk+1}.

1 Theorem 13.6 (Lusin). Σ1 are measurable. 1 The question is, are Σ2 measurable? The amazing thing is that this is one of those maybe questions. Theorem 13.7 (Solovay). If there exists an inaccessible cardinal, then then Con(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 above and 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 point measure, which is like µϕ(A) = 1 if ϕ ∈ A and µϕ(A) = 0 if ϕ∈ / A. This measure is {0, 1}-valued, and we can represent µϕ by a set Uϕ ⊆ P(R) such that

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 ( 1 if A ∈ F µF (A) = 0 if not is a finitely additive measure. Definition 13.10. A filter F on I is an ultrafilter if for every X ⊂ I, either X ∈ F or I \ X ∈ F . A filter (or ultrafilter) F is principal if there exists an x0 ∈ 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 .A filter 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 0 ⊇ F is a filter then F 0 = F . T Lemma 14.2. If F0 ⊆ P(I) satisfies for each X1,...,Xn ∈ F0, i≤n Xi 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( Xi ⊆ X)}. i≤n

Then F is a filter. Proof of Proposition 14.1. I need to prove two directions. Suppose that F is an ultrafilter, and let F 0 ⊇ F . Assume that X ∈ F 0 but X/∈ F . Then I \ X ∈ F , and so I \ X ∈ F 0. This contradicts that F 0 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 lemma and maximality of F , either X ∈ F or I \ X ∈ F . This shows that F is an ultrafilter.

14.1 Ultralimits Theorem 14.3 (Bolzano-Weierstrass). Every bounded sequence has a convergent 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 to one on all sequences (with reasonable restrictions)? These restrictions will be stuff like, keeping sums and products, and the output should be a limit point. Fix an ultrafilter U on ω. Define the ultralimit of han ∈ R : n < ωi as

L = lim an U if and only if for every > 0 there exists an X ∈ U such that for every n ∈ X, |an − L| < . Theorem 14.4. Every bounded sequence has an ultralimit. Math 145a Notes 42

Proof. Let hani ⊆ [−B,B]. Deﬁne by induction bn,Cn, and Xn ∈ U such that 1−n bn ≤ bn+1 < cn+1 ≤ cn with cn − bn ≤ 2 B and

Xn = {k < ω : ak ∈ [bn, cn]} ∈ U.

We construct the sequence by letting b0 = −B, c0 = B, X0 = ω, dividing the interval [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 rigorous and were called “ghosts of departed quantities”. Later Weierstrass and other people came up with the -δ definitions. But as ultrafilters were developed, Robbinson came up with a way of making it real math. Q ω Fix a non-principal ultrafilter on ω. Set n<ω R = R = {f : ω → R}. But this is now not a field. Q 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 and multiplication behaves well under this order.

∗ Q 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 know that 2π/1 is irrational. Then for all n 6= m, ρn(l) and ρm(l) are disjoint. Let

∞ [ A = C ∪ ρn(l). n=0

S∞ n Then ρ(A) = C ∪ n=1 ρ (l) and so A = ρ(A) ∪ l. This kind of thing is what we are going to do. Denote by Fn the free group on n generators. The elements will be words −1 −1 −1 made up of letters x1, . . . , xn and also x1 , . . . , xn , such that no xi and xi are adjacent. Multiplication of two elements will be concatenating two words and canceling everything out. Consider the free group F2 with generators x, y. Let w(x) be the set of starting with x, and likewise deﬁne w(y), w(x−1), w(y−1). Then there is a partition

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

n Denote by SOn the group of rotations of R about the origin, with multiplication 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 that are the rotation ϕ about the x axis by arccos(1/3) and the rotation ψ about the z axis by arccos(1/3). Let hϕ, ψi be the subgroup of SO3 generated by ϕ and ψ. The proof is a bunch of computation, but the point is that there is a copy of F2 in SO3. Let S2 be the sphere of radius 1 in R3, and let B3 be the closed ball of radius 1 in R3. Then the group hϕ, ψi acts on S2 and gives gives an equivalence relation p ∼ q if and only if σ(p) = q for some σ ∈ hϕ, ψi. 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 ∈ hϕ, ψi may ﬁx some stuﬀ. So set

D = {p ∈ S2 : ∃1 6= σ ∈ hϕ, ψi, σ(p) = p}.

Then D is countable because F2 is countable and a rotation ﬁxes exactly two points. Also hϕ, ψi(S2 \ D) ⊆ (S2 \ D). Math 145a Notes 44

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

S2 \ D = hϕ, ψiM = 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 under any iterations of σ. (First fixed the axis and then pick the angle.) Then for every m 6= n, σm(D) ∩ σn(D) 6= ∅. If we set [ D = σn(D), n<ω 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 an equivalence relation. We first make S2 into S2 \ D. Then using the fact that ϕ does not fix M, we make S2 \ D into (S2 \ D) \ M. Using the decomposition

hϕ, ψi = 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 back and get to two pieces of S2. Finally, we can extend this way of making two S2s from S2, to get a way of making two B3 \{0} from one B3 \{0}. The way to ﬁll this last center is to use the same thing: consider a circle passing through 0 and use an irrational rotation. Theorem 15.1. B3 can be decomposed into ﬁnitely many pieces and then these pieces 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 an Q ultrafilter 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 that xy = 1. Proof. Then {n < ω : f(n) 6= 0} ∈ U. Define ( 1 f(n) 6= 0 g(n) = f(n) 0 f(n) = 0.

Then f · g = 1. This can be generalized.

Theorem 16.2 (Lo´s) . 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))} is in U.

So R and ∗R are exactly the same. But they are diﬀerent, because the map j : R → ∗R is injective but not surjective. For instance, f(n) = n−1 is an inﬁnitesimal 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 (without endpoints) 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,

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

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

(a) dom(fn) ⊆ L and ran(fn) ⊆ Q,

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

(d) for every x, y ∈ dom(fn), x

You can build this function by throwing in an or bn in the function using that the orderings < and

Proof. Let X0 ⊆ L be a dense countable subset. Then (X0, <) is a countable ∼ ∗ dense linear ordering without endpoints and so (X0, <) = (Q, <). Deﬁne f : L → R by ∗ f (x) = sup{f(y): y ∈ X0, y < x}. R This is an isomorphism. But mathematicians didn’t stop here and wanted other characterizations.

Deﬁnition 16.7. A linear ordering (L,

17 November 1, 2016 17.1 Trees We kind of know trees. The set of ﬁnite sequences <ωω has branches that keep branching. This is sort of the setup that we want to deal with. Note that there are no loops.

Deﬁnition 17.1. (T,

Deﬁnition 17.2. A partial order (T,

Deﬁnition 17.3. Let (T,

Tα = {x ∈ T : htT (x) = α}.

For example, for <ωω, the n-level sets are the sequences of length n and htT = ω. Example 17.4. Fix (inﬁnite) cardinals κ, λ and set T = (<κλ, ⊆), which is the sequences in λ with length less than κ. This is a tree, that has height κ. Every tree is a subtree of this tree.

17.2 Branches of trees

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

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

A = {hn, n, . . . , ni(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 < κ immediate successors. “Finitely branching” means “< ω-branching”. Theorem 17.8 (König). Any finitely splitting tree with a root of height ω has a branch. Math 145a Notes 48

Proof. We are going to build hsn : n < ωi such that Sn ∈ Tn and sn

k [ succ(sn) = succ(ti) ∪ {t1, . . . , tk}. i=1

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 < κ many 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. α To construct this, for α < β < κ define sα,β ∈ κ that is constant β. Then T = {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. This is because you can construct the same tree but use the cofinal sequence instead of all of κ. Anyways, all ω-trees always have branches, by König’stheorem. We are going 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,

18.1 ω1-Aronszajn trees Definition 18.1. T is called a κ-Aronszajn tree if it is a κ-tree with no cofinal branch. So König’stheorem says that there are no ω-Aronszajn trees. On the other hand 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 the increasing sequences of length n for n < ω. Then Qω is the increasing sequences of order type ω + 1. For ω ≤ α < ω1, Qα is the increasing sequences of order type α + 1. We claim that htQ = ω1. To show this, it suffices to show that for each α < ω1, there exists an Aα such that otp(Aα) = α + 1. This is true because there is an order preserving morphism from any countable ordinal to Q. On the other hand, htQ ≤ ω1 because Q is countable. This shows that Q has no cofinal branch. If there is one, then you can glue them to get a function on ω1 to Q. Are the levels of Q countable? This is not ω+1 ℵ0 true because Qω ∼ Q ∼ 2 . So our first try doesn’t really work. Now build a subtree T ⊆ Q with countable levels and htT = ω1 but not changing the height notes. Define ∀β < γ < α and s ∈ Tβ and q ∈ Q such that max s < q, (∗)α = . ∃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 Tα satisfies (∗)α. We define T0 = ∅ and if Tα is defined and (∗)α+1 holds,

Tα+1 = {s a hqi ∈ Qα+1 : s ∈ Tα, q ∈ Q, q > max(s)}. S If α is a limit, then denote T<α = β<α Tβ. Because α is countable, cof(α) = ω. Find an increasing sequence hαn < α : n < ωi such that supn αn = α. Let x ∈ 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 S yx,q = n<ω xn a hqi ∈ 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 → κ such that x

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

• if x ∈ Tα, then there exist y1, y2 ∈ Tα+1 such that x

19 November 8, 2016 19.1 Suslin trees

Recall that Q is the countable dense linear ordering and R is the separable complete dense linear ordering. We asked if we can replace separability with the 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 uncountable antichain. Our goal is to prove that there exists a Suslin tree if and only if there exists a 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 a common root. Then the new tree T 0 is still Suslin. Suppose x ∈ T 0 cannot be arbitrarily extended, i.e., have a successor on all x 0 1 higher levels. Then T = {y ∈ T : x <0 y} is countable. Let T = {x ∈ T 0 : |T x| > ω}. This is a subset of T 0 so T 1 has no uncountable branches or antichains. Also htT 1 = ω1 because if htT 1 = β < ω1 then node in T0 at level β + 1 has countably many successors and there are also countably many nodes below so T0 is countable. This shows that T1 is still Suslin. Also by a similar argument, every node is arbitrarily extendible. Now we want to ensure every node has at least 2 immediate successors. We 1 1 claim that if α < ω1 and x ∈ Tα, then there exists an β > α and y1 6= y2 ∈ Tβ such that x < y1 and x < y2. This is because if such an x exists, then we can 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 clear that T 2 has no uncountable branches and antichains. Because level sets are antichains, this includes the fact that level sets are countable. Then everything works 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 that two or more nodes at a limit level have same predecessors, add a node there right below the limit level and make T 3. This additional node will take care of the 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 countable L0 ⊆ 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 set C = {aβ, bβ : β < α} is countable. Find Iα = (aα, bα)L whose closure is disjoint from 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-sized antichain 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 branch because 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 , x succ(x) = {y ∈ Tht (x)+1 : x

z ∈ T = max(pred x ∩ pred y) and x < y ⇔ x < y or ∗ x0 6= y0 ∈ succ(z), x0 < x, y0 < y and x0

Then (T, <∗) is a dense linear ordering. If X is a collection of disjoint intervals in (T, <∗), then we can pick aI ∈ I for each I ∈ X to get an antichain {aI }I∈X in T . Finally, if D ⊆ T is countable, then α = sup{htT x : x ∈ D} < ω1 and so we 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 statement ϕ is independent of axioms T if and only if there is a model T satisfying ϕ and a model T failing ϕ. Example 20.1. “Being abelian” is independent of the group axioms. This simply means that you can’t neither prove nor disprove “being abelian” from just the group axioms. On the other hand, “product of two elements have an inverse” is not independent from the group axioms.

ℵ0 We’ve been saying that 2 = ℵ1 is independent from ZFC. To prove this, we have to construct two models of ZFC, one that satisfies the continuum hypothesis 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 to find a model M ⊆ V , that satisfies ZFC. Or we can try to find V ⊆ M outside V . For instance, if κ is a strongly inaccessible cardinal, then Vκ is a model of ZFC. 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 V only as our universe/starting model of ZFC. But then we run into class/definablity issues. The other way is to start with some set V 0 that is a model of ZFC and treat V 0 as V . But then we have to assume Con(ZFC) and is also less satisfying because all we are talking about is not all of math but a small chunk of math that is indistinguishable from all of math from the inside. We are mainly going to work in the second viewpoint.

20.1 The idea of L S Recall that we have deﬁned V0 = ∅, Vα+1 = P(Vα), and Vδ = Vα if δ is S α<δ 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

∃ ﬁrst-order formula ϕ(x, y , . . . , y ) and a , . . . , a ∈ X Def(X) = A ⊆ X : 1 n 1 n . such 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 = Lα for limit δ, L = Lα. α<δ α∈On Doing this is much subtle than just constructing abelian groups and non-abelian groups, because this L is going to be our new “universe” and there are metamathematical 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 semantics that tells us how sentences connect to the real world. Let us describe the syntax of L. We define the terms Terms(L) as the smallest 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 made by 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 sentences like ∀y(¬y ∈ x). Deﬁne the free variable function FV inductively. For terms, we set

n [ FV(c) = ∅, FV(x) = {x}, FV(F (t1, . . . , tn)) = FV(ti). i=1 Then for formulas, we let

n [ FV(R(t1, . . . , tn)) = FV(ti), FV(¬ϕ) = FV(ϕ), i=1 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}. In short:

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-model M consists of Math 145a Notes 55

• a universe/underlying set |M|, • for F ∈ L with arity n, a map F M : M n → M, • for n-ary relation R ∈ L, RM ⊆ M n, • for constant c ∈ L, an element cM ∈ L.

If we have a term t and an L-structure M with FV(t) = {x1, . . . , xn}, build tM : M n → M as follows: • (c)M : M 0 → M has value cM , • xM : M → M is the identity, n M • F (t1,...,Fn): M → M is (m1, . . . , mn) 7→ F (t1(m•), . . . , tm(m•)).

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

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

M M M (M ϕ(a1, . . . , an)) = ((t1 (a•), . . . , tm (a•)) ∈ R ).

• 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 of graphs, 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 = x and xn = y and (xi, xi+1) ∈ E”. You can’t quantify over a varying number of variables. Math 145a Notes 56

21 November 15, 2016

Last time we set up the notion of ﬁrst-order logic. For set theory, the language is τ = {E}. The syntax is built from xEy nd logical operations like negation, conjunction, and quantiﬁcation. On the semantic side, there is a structure M = (M,EM) consisting of a set M and EM ⊆ M 2. Finally, we have semantics, that is, 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 of N ) if • M ⊆ N, and N n M • for every n-ary F ∈ τ, F M = F , • for every n-ary R ∈ τ, RN ∩ M n = RM, • for every c ∈ τ, cN = cM. This comes up in every parts of mathematics. But by what we have done so far, we have a tighter connection between the two structures. Now this substructure axioms implies that M aEb if and only if N aEb for every a, b ∈ M. Given τ-structures M and N , we say M 4 N (M is an elementary 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}, ∈) and N = (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 such that • f is a bijection, M n • if F ∈ τ and a1, . . . , an ∈ F , then f(F (a1, . . . , an)) = F (f(a1, . . . , an)), M • if R ∈ τ and a1, . . . , an ∈ M, then (a1, . . . , an) ∈ R if and only if N (f(a1), . . . , f(an)) ∈ R , • if c ∈ τ, then f(cM) = cN .

Proposition 21.1. Let M, N be {E}-structures and let M 4 N . If M is a model 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 of 4. Theorem 21.2 (Downward L¨owenheim–Skolem–Tarski theorem). Given a τ- structure M and A ⊆ M, there is a τ-structure N such that

(1) N 4 M, (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 satisfy some 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-foundedness axiom 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 the ∗ Q ∗ ultraproduct N = (N,E)/U so that N = {[f]U :(f : ω → U)} and

∗ [f]U E [g]U ⇔ {n < ω : N f(n)Eg(n)} ∈ U. It turns out that N ∗ also satisﬁes ZFC including well-foundedness. But for k < ω, we get take ( 0 if n ≤ k fk : n 7→ n − k if n > k

∗ and this is a infinite decreasing sequence, i.e., N [fk+1]U E[fk]U . This is why we want transitivity. Theorem 21.3 (Mostowski). Suppose M is a {E}-structure that satisfies extensionality and such that EM is well-founded (actually well-founded, not satisfies the axiom). Then there is a unique transitive set X and an isomorphism π : M =∼ (X, ∈). This theorem allows us to check axioms more easily.

21.3 Lévyhierarchy There is also the Lévyhierarchy that allows to rank the complexity of formulas. Of course you can define the complexity as the number of quantifiers, but there are 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

If M ⊆ N but M4 6 N , then the failure must start at some quantification. But this first failure cannot happen at a bounded quantifier because if y ∈ M then automatically x ∈ M. Let us know define the Lévyhierarchy. We let

∆0 = Σ0 = Π0 as formulas with just bounded quantiﬁcation. So this is the minimal set such that 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 equivalence classes of formulas.

Proposition 21.4. If ϕ(x1, . . . , xn) is ∆0 and M ⊆ N have transitive underlying sets, then for all a1, . . . , an ∈ M, M ϕ(a1, . . . , an) if and only if N ϕ(a1, . . . , an). Proof. From the substructure, we know they agree on the atomic formulas. So we 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), 0 such that for all a, a , a1, . . . , an ∈ M,

0 0 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. So M 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 are transitive 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(∀z0 ∈ y(z ∈ z0 ∨ z = z0)) ∃z ∈ x∀z0 ∈ x(z ∈ z0 ∨ z = z0) and other stuﬀ. Math 145a Notes 60

22 November 17, 2016

Last time, we defined the Lévyhierarchy, where ∆0 = Σ0 = Π0 are the ones using only bounded quantification.

Proposition 22.1. ∆0 formulas are absolute for transitive models. That is, if ϕ(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 even some 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 L¨owenhiem–Skolem, there is a countable transitive model M (M, ∈) ZFC. Let ω1 ∈ M be the element such that

M 00 M “ω1 is the ﬁrst uncountable cardinal .

M M Then V countable(ω1 ) but M ¬countable(ω1 ). This is Skolem’s paradox. That is, if ZFC is consistent, then there are countable models. But ZFC implies that there are uncountable sets. There is also a philosophical argument by Hilary Putnam, about the question “am I a brain in the box?” If I am indeed a brain in the box, then that is from the perspective of the “real” world. But if we ask the question on the “brain in the box” world, the two notions of being a brain in the box doesn’t match. So the answer is always no, in a very unsatisfying way.

22.1 G¨odelcodes In the beginning of the course, we have said that everything is a set. But we are now talking about formulas. Are these sets? There is a way to code formulas as sets. We can assign to each symbol a diﬀerent number, to each variable xn a pair (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 Gödelcodes in the following way. Denote by dϕe the Gödel 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 h(Mα,Eα): α < κi be a sequence of {E}-structures such that

(a) α < β implies (Mα,Eα) ⊆ (Mβ,Eβ), S S (b) if α is a limit then Mα = β<α Mβ and Eα = β<α Eβ,

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. n For each ϕ(x, y1, . . . , yn), deﬁne fϕ : M → 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, if there exists such a x, we are choosing that x, and if there doesn’t exist such a x then we are choosing anything. This requires choice. In the homework, you proved the following fact: S If X = α<κ Xα, |Xα| < κ, and Xα is increasing and continuous, kn and {(fn : X → X): n < ω, kn < ω} is a sequence of functions, then

kn {α < κ : fn”Xα ⊆ Xα for all n < ω}

is a club. 1Recall that C being a club means that, if X ⊆ C is not coﬁnal then sup X ∈ C, and C itself 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 at M ∃xϕ(x, a1, . . . , an). If it is false, then fϕ(a1, . . . , an) ∈ M0 ⊆ Mα by minimality. 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. For atomic formulas, it follows immediately from (Mα,Eα) ⊆ (M,E). Negation and conjunction are more straightforward. For existential quantifiers, this is what the Skolem functions are designed for. If Mα ∃xϕ(x, a1, . . . , an), so does M. If M ∃xϕ(x, a1, . . . , an) then M ϕ(fϕ(a1, . . . , an), a1, . . . , an). So Mα ϕ(fϕ(a1, . . . , an), a1, . . . , an). So this is a set-theoretical version of reflection. There is a more generalized reflection, called the generalized reflection.

Theorem 22.5 (Generalized reﬂection). Let hωα : α ∈ Oni be a deﬁnable S transitive 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 set as ∃ϕ(x, y , . . . , y ) and a , . . . , a ∈ M DefM (X) = Y ⊆ X : 1 n 1 n . such that y ∈ Y ↔ (M,E) ϕ(y, a1, . . . , an) We now define

Lα [ [ L0 = ∅,Lα+1 = Def Lα,Lδ = Lα for a limit δ, L = Lα. α<δ α∈On

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”  f is a function with domain y + 1∧   f(0) = ∅∧   S  = ord(y) ∧ ∃f  (∀β ∈ y + 1(lim(β) → f(β) = {f(γ): γ ∈ β}))∧  .  0  (∀β, β0 ∈ y + 1(β = β0 + 1 → f(β) = Deff(β ) f(β0)))∧ 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 0 0 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 order formula “Lev(x, α) = x ∈ Lα”. We now want to show that L ZF and also L V = L.

23.1 L satisﬁes ZF L Proposition 23.1 (L ZF). For each axiom ϕ of ZF, ZF proves ϕ (relativisation of ϕ to L). Proof. By one of your homeworks, L is transitive. Then FoundationL and ExtensionalityL holds. Now let us now show that the other axioms are also satisﬁed. 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 show that P(ω) ∩ L ⊆ Lω1 but P(ω) ∩ L 6⊂ Lα for α < ω1. So we have to go up far up into the hierarchy. For X ⊆ a, let ( 0 if X/∈ L f(X) = β 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 use the general reﬂection principle. There exists a α < γ ∈ On such that for all 0 0 0 0 b1, . . . , bn, bn+1 ∈ Lγ ,

0 0 0 0 Lγ ϕ(b1, . . . , bn+1) ↔ L ϕ(b1, . . . , bn+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 M Theorem 23.2. The formula DEF(x, M) = “x = Def M” is a Σ1-formula. Corollary 23.3. (1) If W ⊆ V is a transitive, definable model of ZFC with W On ⊂ W , then L = L (= {x : W ∃α Lev(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 that W “∃y(x ∈ y ∧ Lev(y, α))”. Because Σ1-formulas are absolutely upward, we W have ∃y(x ∈ y ∧ Lev(y, α)). Then x ∈ Lα. This shows that Lα ⊆ Lα. Now let W us show that Lα ⊆ Lα by induction. If x ∈ Lα+1, then

x = {z ∈ Lα : Lα ϕ(x, a1, . . . , an)} W W W = {z ∈ Lα : Lα ϕ(x, a1, . . . , an)} ∈ Lα+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 formula x = DefM M. To prevent the formula from climbing up the hierarchy, we are going to set W to be the “master domain” for quantification. So our strategy is to first quantify everything over W and then at the end set W to be the things we want to actually quantify over. First suppose ω ∈ W . Note that omega(x) = “x = ω” is ∆0. So we may use ω 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”  ATOM(F(m), M, W)   ∨∃k ∈ m(F (m) = d¬F (k)e)  ∧ ∀m ∈ n + 1    ∨∃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 variables in {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 M r satisfy the formula)). Here, satisfying the formula can be written as

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

∃i ∈ r, k ∈ m(F (m) = d∃xiF (k)e) → S(m) = {s ∈ M r : ∃s¯ ∈ S(k)∀i0 < r(i0 = i → s(i) =s ¯(i0))} 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 ∈ M r(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 ∈ M r(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 that y DEF(x, y), which codes x = Def 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 ⊆-minimal inner model of ZF containing all ordinals. Today we are going to show that λ + L Choice, and also L ∀λ(2 = 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 formula ϕ(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 no parameters) is a sequence ϕ0, . . . , ϕk such that each term in the sequence is an atomic formula or the negation or conjunction or existential quantified formula of something before. Now define <τ on formulas by:

sequence of ϕ is shorter than the sequence of ψ or ϕ <τ ψ ↔ lengths are equal and ϕ

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

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

βx = min{β < α : x ⊆ Lβ}, ϕx = min{ϕ : ∃a1, . . . , an ∈ Lβ(x = {y ∈ Lβ : Lβ ϕ(y, a1, . . . , an)})}, <τ (ax, . . . , ax) = min{(a , . . . , a ) ∈ Ln : x = {y ∈ L : L ϕ(y, a , . . . , a )}}. 1 n 1 n βx β β 1 n <βx

Now given x, y ∈ Lα, we deﬁne the order as   βx < βy or x <α y ↔  βx = βy and ϕx <τ ϕy or  β = β , ϕ = ϕ and (ax, . . . , ax)

Then <α is a well-ordering of Lα because it is a lexicographic product of well orderings. 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 a well-ordering. I haven’t made a big deal out of this yet, but we haven’t used Choice in V to 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 models think 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. In proving this, we are going to use the fact that if α < δ and δ is a limit, then Lδ 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 for some 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 because given α ∈ δ, we have α ⊆ δ by transitivity. We now claim that δ is a limit ordinal. If not, suppose δ = α + 1 so that α ∈ M is the largest ordinal. Apply V = L to α to get y ∈ M and β ∈ M such that M ord(β) ∧ α ∈ y ∧ Lev(y, β). But note that this is a Σ1 formula. So M is actually right about everything, i.e., y ∈ Lβ and α ∈ Lβ and so β > α. This contradicts 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 ∈ M such that y = Lβ and α < β. Then Lα ⊆ Lβ ⊆ M. This shows that Lδ = S 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 limit ordinal then Lδ (V = L).

Lemma 24.5. If δ is a limit and α < δ, then hLβ : β < αi ∈ Lδ.

This allows us to transfer the Σ1 formula Lev(x, α) from L to Lδ. Although Σ1 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 = δ0 + ω for some limit δ0, and δ a limit of limit ordinals. First look at the case δ = ω. Then α < δ is 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 exists 0 0 a limit δ < δ so that α < δ . Then hLβ : β < αi ∈ Lδ0 ⊆ Lδ. Finally assume that δ = δ0 + ω. If α < δ0, then we are done. Otherwise, there is some n < ω such that α = δ0 + n. Suppose first that n = 0. Then 0 0 α = δ . By induction, for every β < α we have hLβ0 : β < βi ∈ Lδ0 . This shows 0 that (β ,Lβ0 ) ∈ Lδ0 , and thus

hLβ : β < αi = {(β, Lβ): β < α} ⊆ Lδ0 .

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

 function with domain α0∧  0 0 “x = (α , y) where ord(α ) ∧ ∃f  f(0) = ∅ ∧ DEF(f(β + 1), f(β))∧  ” S 0 β < α limit → f(β) = β0<β f(β ) So

hLβ : β < αi = {y ∈ Lδ0 : Lδ0 ϕ(y)} ∈ Lδ0+1.

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

0 hLβ : β ∈ δ + ni ∈ 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. Now it suﬃces to show Lδ Lev(Lα, α). Now use the previous lemma with α + 1. So f = hLβ : β < α + 1i ∈ Lδ. We now want to show that

 f is a function with domain α + 1∧  Lδ  f(0) = ∅ ∧ DEF(f(α + 1), f(α))∧  . S 0 (β limit → f(β) = β0<β f(β )) ∧ f(α) = Lα

But everything here is ∆0, because of the things you are going to prove in the homework. Thus L thinking that all these are true implies Lδ thinking that all these 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. Last time we showed that L ZFC +(V = L). Also, we showed that M (V = L) if and 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 singletons that show up later in the hierarchy.

Proof. Find a limit δ0 such that X ∈ Lδ. We have a structure (Lδ, ∈). Use the Downward L¨owenheim–Skolem–Tarksi to ﬁnd an elementary substructure M (M,E ) = 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 transitive. But we know that EM is an extensional well-founded relation. Use Mostowski’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 a + limit 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 that X ∈ 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 is because π(∅) = ∅ and π(x) = {π(y): y ∈ x} = {y : y ∈ x} = x. This shows that π(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} ∪ λ is checked to be transitive. By condensation, we have {X} ∪ λ ∈ Lλ) and so + X ∈ 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 to do 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. It will lack details and ignore metamathematical issues. Assume we have a countable, transitive model M of ZFC. We want to build a 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-generic filter 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 element 1P ∈ P. (This is a ∆0 statement.) We say that a subset D ⊆ P is dense if and only if for every p ∈ P, there is a q ∈ D such that q ≤ p. We also say that G ⊆ P is 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 there exists 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 as a black box for now. A Take P = Add(ω, 1) ∈ M. Then Dn,α, Diﬀα,β,Dα ∈ M for n < ω, α = 0, S and A ∈ P(ω)∩M. If G is M-generic, it meets all of these. Then G : 1×ω → 2 is a function. Deﬁne

G [ X = X0 = {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 an M α ∈ M such that M “α > ω1 is a cardinal”. Set P = Add(ω, α) ∈ M. Find a M-generic G ⊆ P and form M[G]. From G, we get M[G] “∃ injection from α to P(ω)”. ω So M[G] “2 ≥ α”. We would like to show that M[G] “α > ω1”. But this statement is Σ2 or something, and so we can’t say this directly. But it turns out that M[G] still thinks this is true. This is because Add(ω, α)M has the countable chain condition. If P ∈ M has the countable chain condition and G ⊇ P is M-generic, then M and M[G] have the same cardinalities and the same cofinalities. This means that M[G] “α > ω1 is a cardinal”. These two combined shows that M[G] ¬ CH. You can run the same argument by using κ instead of ω although it becomes more 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 that for all κ, λ ∈ dom F , (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, κ M[G] “∀κ ∈ dom F (2 = 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-namex ˙ and a M-generic filter G, we define x˙ G = {y˙G : (y, ˙ p) ∈ x,˙ p ∈ G} and M[G] = {x˙ G :x ˙ ∈ M is a P-name}. For each x ∈ M, we can define a canonical namex ˇ by

xˇ = {(ˇy, 1P): y ∈ x}. Thenx ˇG = x and so M ⊆ M[G]. Also Γ˙ = {(ˇp, p): p ∈ P}. Then for all G ⊆ P, Γ˙ G = G ∈ M[G]. Forcing black box. Let P ∈ M be a p.p. Then there is a deﬁnable relation such that ∀a˙ 1,..., a˙ n ∈ Name and p ∈ P and ϕ(x1, . . . , xn), the following are equivalent:

(1) M “p ϕ(˙a1,..., a˙ n)”. G G (2) For any M-generic G with p ∈ G, M[G] ϕ(˙a1 ,..., a˙ n ). The book has more detailed exposition to forcing. Index

< κ-branching, 47 homeomorphism, 33 V = L, 62 hyperreals, 42 κ-embeddable tree, 50 inaccessible cardinals, 29 P-name, 72 Lo´s’stheorem, 45 independence, 53 König’stheorem, 29, 47 Aronszajn tree, 49 Löwenheim–Skolem–Tarski Banach-Tarski Paradox, 43 theorem, 57 Banach-Tarski paradox, 33 Lévyhierarchy, 58 beth numbers, 29 language, 53 bounded quantifier, 57 limit, 19 branch, 47 measure, 32 Cantor normal form, 20 translation invariant, 32 Cantor’s theorem, 45 Mostowski’s theorem, 57 Cantor-Bernstein theorem, 21 natural numbers, 16 cardinal, 22 normal tree, 50 cardinality, 24 class, 10 order type, 17 club, 61 ordinal, 16 cofinality, 25, 26 complete, 46 partial order, 47 condensation lemma, 70 principal, 40 constructible universe, 62 product, 9 countable chain condition, 46 rank, 21 countable set, 12 reflection, 61 regular ordinal, 25 definable power set, 62 dense linear ordering, 45 separable, 46 singular ordinal, 25 elementary substructure, 56 Skolem function, 61 equinumerous, 21 Skolem’s paradox, 60 strong limit cardinals, 29 filter, 40, 71 substructure, 56 formula, 8, 54 successor, 19 free variable, 54 Suslin line, 46 function, 9 Suslin tree, 51 syntax, 54 Gödelcodes, 60 Gödel’sincompleteness theorem, 30 term, 54 Godzilla ordinal, 26 transfinite induction, 13 transitive closure, 13 height, 47 tree, 47

73 Math 145a Notes 74 ultraﬁlter, 40 well-founded relation, 13 ultralimit, 41 well-ordering, 10, 13

Von Neumann universe, 10 ZFC, 7