The Unprovability of the Continuum Hypothesis Using the Method of Forcing

U.U.D.M. Project Report 2016:32

Alvar Bjerkeng van Keppel

Examensarbete i matematik, 15 hp Handledare: Vera Koponen Examinator: Jörgen Östensson Juni 2016

Department of Mathematics Uppsala University

Abstract The continuum hypothesis is the statement that no set has cardinality greater than N and smaller than R. We show that if ZFC is consistent, the continuum hypothesis is not provable from ZFC. This is done using the method of forcing, pioneered by Paul Cohen in 1963. Proofs and concepts are given with lots of detail to make reading as simple as possible.

1 1 Introduction

This thesis is meant to be a self contained, gentle and to the point introduction to the method of forcing. The reader is expected to have come in contact with mathematical logic and ZFC set theory. For ease of reading a lot of basic results in set theory are stated (but not proven) to refresh the readers memory. Forcing is a way of expanding a certain kind of model of set theory to a larger model. In this thesis forcing is used to prove that if ZFC is consistent, so is ZFC together with the negation of the continuum hypothesis(CH). We denote this by Cons(ZFC) ⇒ Cons(ZFC + ¬CH). This is one of the two parts in proving that the continuum hypothesis is independent of ZFC. The result was ﬁrst published by Paul Cohen in his article “The Independence of The Continuum Hypothesis” published 1963[1]. In this article Cohen refers to earlier work by Kurt G¨odelin 1940 proving (among other things) the other direction, Cons(ZFC) ⇒ Cons(ZFC+CH). We will only look at the Cons(ZFC) ⇒ Cons(ZFC+¬CH) direction. This thesis mainly follows the approach of Kunens excellent book [3].

2 The Logic of ZFC

The underlying logic of Zermelo-Fraenkel with the axiom of choice (ZFC) is classical ﬁrst order logic with equality and a binary relation symbol ∈. Thus the formulas will be of the form

(x = y), (x ∈ y), (φ ∧ ψ), (¬φ), (∃x φ) where x and y are variables and φ, ψ are formulas. In practice we drop parenthesis if no clarity is lost. We let φ∨ψ abbreviate the logically equivalent ¬(¬φ∧¬ψ), similarly for φ → ψ and φ ↔ ψ. In the same vein, ∀x φ abbreviates ¬∃x ¬φ. This is done to make proofs by induction on the structure of formulas have fewer cases. As a final convenience, the symbol ⊥ is used in a few places to denote falsehood in a formula. A possible definition of this symbol is thus ⊥ def⇔ ∃x(x ∈ x ∧ x∈ / x). At a glance, the language of first order logic does not seem expressive enough, or at least not terse enough, to formulate interesting notions in. We deal with this by defining new notation as shorthand for long formulas. Two examples of this are subset x ⊆ y def⇔ ∀z(z ∈ x → z ∈ y) and set comprehension x = {y ∈ Y | φ(y)} def⇔ ∀y(y ∈ x ↔ y ∈ Y ∧ φ(y)).

If the reader wants to refresh his or her memory about the axioms of ZFC they are included in an appendix at the back.

3 Metatheory and Formalism

As in the vast majority of mathematical works, we will be using mathematical prose as opposed to some formal (and therefore tedious) proof system. We will however implicitly assume that any of our proofs can be translated into a formal proof in some proof system for classical first order logic, except in a few special cases. These cases touch on the two separate types of logic used. On the one hand, we have the logic of ZFC and on the other hand, the logic of the metatheory. One problematic proof is of the form “for all formulas φ1, . . . , φn, X is true” where X depends on φ1, . . . , φn. Since we cannot quantify over formulas within first order logic, such statements and proofs cannot be translated as is. However, in the few cases where this pops up, we will see that for each concrete choice of formulas φ1, . . . , φn a formal proof in first order logic can be extracted for X in a straightforward way. A special case of this is proofs by induction on the structure of formulas. We can from such a proof extract a first order proof by manually unwinding the inductive steps until no reference to the induction hypothesis is needed. This is akin to proving P (83) without induction given a proof by induction for ∀n P (n). We then have a proof for P (0) and a proof for ∀n(P (n) → P (n + 1)) where we for simplicity assume no induction is used. By specializing ∀n(P (n) → P (n + 1)) for each n = 0,..., 82 and applying modus ponens 83 times, we get a proof for P (83) without using induction.

2 The underlying structure of the proof is as follows. To prove Cons(ZFC) ⇒ Cons(ZFC + ¬CH), we show the equivalent (ZFC + ¬CH ` ⊥) ⇒ (ZFC ` ⊥)., (?) where T ` φ stands for “there is a formal proof of φ using assumptions from T ”. To do this we assume the finite assumption property of the formal proof system, that if T ` ψ for some possibly infinite list of formulas T then there is a finite sublist φ1, φ2, . . . , φn of T such that φ1, φ2, . . . , φn ` ψ. We use this to reformulate (?) into cases, one for each finite sublist φ1, φ2, . . . , φn of ZFC where we let Φ ≡ φ1 ∧ φ2 ∧ ... ∧ φn: (Φ ∧ ¬CH ` ⊥) ⇒ (ZFC ` ⊥). (†) Let us consider a fixed Φ. We prove (†) by constructing a model1 of Φ ∧ ¬CH in ZFC using the method of forcing pioneered by Paul Cohen [1]. This is the meat of the thesis. This model M[G] is the expansion of a base model M which we augment with an extra set G, similar to how the field extension F (g) is the smallest subfield containing F ∪ {g} where F ⊂ G and g ∈ G. To make “a model of” more precise, we define in our metalogic a transformation (·)N on formulas, where φN is read as “φ relativized to N”. It simply bounds all quantifications to a set N and leaves everything else intact. As an example,

(∃x (x ∈ y → x = z))N becomes ∃x ∈ N (x ∈ y → x = z).

The construction of the model can then be stated as

ZFC ` ∃M[G] (Φ ∧ ¬CH)M[G].

We will gloss over something that is intuitively clear but requires a rigorous metalogical framework to deﬁne, namely that we can use deduction inside models N:

(ψ ` χ) ⇒ (ZFC ` ψN → χN ).

This is the last piece needed to prove (†).

ZFC ` ∃M[G] (Φ ∧ ¬CH)M[G] the constructed model Φ ∧ ¬CH ` ⊥ the supposition in (†) ZFC ` (Φ ∧ ¬CH)M[G] → ⊥M[G] deduction inside M[G] ZFC ` ∃M[G] ⊥M[G] by line one and three ZFC ` ⊥ simpliﬁcation

This concludes the rough sketch of the overarching structure of the proof. Our metatheory thus needs to support manipulation of the formal proof system, be able to prove the ﬁnite assumption property and support induction and recursion over formulas.

4 Well-founded Relations

Two concepts that are not really interesting to formalize outside of set theory are that of classes and of class functions. The intuition is simply that a class is a collection of sets that may be “bigger” than a set and a class function is a function whose domain and range are classes. Examples of classes are the class of all singleton sets and the class of all groups. An example of a class function is the Kuratowski operator that sends arbitrary sets x and y to {{x}, {x, y}}.

Deﬁnition 1. Let s1, . . . , sn be sets.

• A class A is deﬁned by a formula φA(x, z1, . . . , zn) and the sets si. We write a ∈ A if φA(a, s1, . . . , sn) holds.

• A class function F is deﬁned by a formula ψF(x1, . . . , xm, y, z1, . . . , zn) and the sets si where for sets r1, . . . , rm there is a unique set y such that ψF(r1, . . . , rm, y, s1, . . . , sn) holds. If F is a class function and ψF(a1, . . . , am, b, s1, . . . , sn) is true we use F(a1, . . . , am) as a shorthand for b. 1The exact meaning of “model” will be explained later.

3 The class of all sets, called the universe of sets is denoted by V. The choice of formula for a class or class function is usually implicit. For φV any tautology will do, x = x being the canonical choice. Every set is a class (fix z as the desired set in φ(x, z) def⇔ x ∈ z) but there are classes which are not sets. It should also be noted that classes and class functions are not objects in the underlying first-order logic, they are just an abstraction used to hide the formula they represent. Keep this in mind as certain operations on classes cannot be justified by ZFC (such as the collection of all subclasses) while others yield new classes without causing any trouble. If A is a class it is for example sane to speak of the class {a ∈ A : φ(a)} for some φ since this is represented by the formula φA(x) ∧ φ(x). Definition 2. A well-founded class W is a class W with a binary relation R (a class whose members are pairs of elements from W) where every nonempty subset of W has a R-minimal element; or expressed by a formula, ∀S ⊆ W(S 6= ∅ → ∃m ∈ S ∀s ∈ S ¬sRm). Note that for any well-founded class, ¬xRx (otherwise the set {x} has no minimal element). By using the axiom of choice there is an alternative characterization of well-founded relations that hopefully sheds more light on what structure well-foundedness imposes. Lemma 3. The following two statements are equivalent. • (W, R) is well-founded.

• Every function f : N → W, such that for all n ∈ N either f(n + 1)Rf(n) or f(n + 1) = f(n), satisﬁes ∃N ∈ N ∀n ≥ N f(N) = f(n). The second statement can be phrased as every R-descending N-sequence converges. Proof. Assume that (W, R) is well-founded and take a function f as above. Take a N ∈ N such that f(N) = min range(f). Then f(N) = f(N + 1) = ··· = f(N + n) = ... since by induction f(N + i + 1)Rf(N + i) is impossible by minimality of f(N + i) = f(N) for all i ∈ N. In the other direction, assume that every descending function f converges and towards a contradiction, assume that there is a nonempty set S ⊂ W with no R-minimal element. Let cS be a choice function of S and s ∈ S. Then

f(0) = s, f(n + 1) = cS({x ∈ S | xRf(n)}) defines a function since the set {x ∈ S | xRf(n)} being empty is equivalent to f(n) being a minimal element in S. This f contradicts our assumption since f(n) 6= f(n + 1) for all n ∈ N. This tells us that however we drop a bouncy ball down a well-founded staircase it will only bounce on a finite number of stairs. The canonical example of a wellfounded class is of course (V, ∈). That this is a wellfounded class follows directly from the axiom of foundation: ∀x [∃y(y ∈ x) → ∃y(y ∈ x ∧ ¬∃z(z ∈ x ∧ z ∈ y))] . Definition 4. The powerset of a set A denoted by PA. S S Definition 5. The union {x | ∃a ∈ A(x ∈ a)} of a set A is denoted by A or a∈A a. Definition 6. A relation R on a class A is set-like if for every a ∈ A the class {x ∈ A : xRa} is a set. Definition 7. Let R be a set-like relation on a class A and let x ∈ A. We define pred (as in predecessor) and closure by recursion as follows: pred(A, x, R) = {a ∈ A : aRx}

pred0(A, x, R) = pred(A, x, R) [ predn+1(A, x, R) = {pred(A, a, R): a ∈ predn(A, x, R)}

The set {predn(A, x, R): n ∈ N} exists by the replacement axiom so [ closure(A, x, R) = {predn(A, x, R): n ∈ N} is well deﬁned.

4 Proposition 8. If a relation R is set-like and well-founded on a class W then any nonempty subclass W0 of W has a R-minimal element. Proof. This is seen by ﬁxing a v ∈ W0 and taking a R-minimal element m of the set

S = W0 ∩ ({v} ∪ closure(W0, v, R)).

A m0 ∈ W0 such that m0Rm implies m0 ∈ S by construction of S. This contradicts the minimality of m in S so m must be a R-minimal element of W0. We now have all the machinery to prove that induction over well-founded sets is valid. To get a familiar special case of this general induction principle, take (W, R) = (N, <). This becomes

0 0 ∀w ∈ N (∀w < w φ(w )) → φ(w) → ∀w ∈ N φ(w), the total-induction principle on the natural numbers. If we fix w = 0 in the first quantifier above we get

(∀w0 < 0 φ(w0)) → φ(0) or equivalently φ(0),

the base case in the induction. Theorem 9. Given a well-founded set-like relation R on W and a formula φ then the following induction principle holds. ∀w ∈ W [∀w0 ∈ W w0Rw → φ(w0)] → φ(w) → ∀w ∈ W φ(w) In other words, to show that all elements of W have the property φ, it suffices to show that if all elements of W which are R-smaller than w satisfies φ then w satisfies φ.

Proof. Assume towards a contradiction that there is a w such that ¬φ(w) but ∀w ∈ W(∀w0Rw φ(w0)) → φ(w) holds. Without loss of generality assume that w is a minimal element not satisfying φ. Then for all w0Rw, φ(w0) by minimality of w. By assumption (∀w0Rw φ(w0)) → φ(w) so φ(w) holds, a contradiction.

It depends from case to case whether proof by induction or proof by a contradictory minimal element is preferable. Hand in hand with induction comes recursion, our next topic. For this we will need some notation. Deﬁnition 10. We deﬁne some standard notation concerning functions.

• The restriction of a (class) function f to a set A is f A = {ha, bi | a ∈ A ∧ f(a) = b}. • The image of a set A under a function f is f[A] = {f(a) | a ∈ A}. • The domain of a set R is dom(R) = {a | ∃baRb} if R is a binary relation and empty otherwise.

• The range of a set R is range(R) = {b | existsaaRb} if R is a binary relation and empty otherwise. Theorem 11 (Recursion Theorem). Let R be a well-founded and set-like relation on A and F : A×V → V a class function. Then there is a unique class function G : A → V such that ∀x ∈ A [G(x) = F x, G pred(A, x, R) ]. (∗) Proof. For the reader that has never seen this definition of a recursively defined function, a short expla- nation is in order. In a recursive definition we are allowed to make use of any previously defined values, in this case the values of G in points below x in the R-order. What values G takes in these points are encoded in G pred(A, x, R). Finally F decides what value G(x) has by looking at x itself and the previously defined values in G pred(A, x, R). But what about base cases? These are sneakily covered by the phrasing “previously defined values”, so for a base case b ∈ A, G = F b, G pred(A, b, R) = F(b, G ∅) = F(b, ∅).

5 To prove uniqueness of G, assume we have G1, G2 satisfying the above and that a is minimal in A 0 satisfying G1(a) 6= G2(a). By R-minimality G1 pred(A, a, R) = G2 pred(A, a, R) since if a Ra and 0 0 G1(a ) 6= G2(a ) then a is not minimal with the property G1(a) 6= G2(a). This yields the contradiction

G1(a) = F(x, G1 pred(A, a, R)) = F(x, G2 pred(A, a, R)) = G2(a). The proof of existence is done by deﬁning a G that satisﬁes the equation. Call a function g an x-approximation of G if

dom(g) = {x} ∪ closure(A, x, R) and ∀y ∈ dom(g)[g(y) = F(y, g pred(A, y, R))].

The definition of G(a) is simply G(a) = ga(a) where ga is an a-approximation. Thus we need to prove that this definition is well-defined by showing uniqueness and existence of these approximations.

Uniqueness Given an x-approximation gx, a y-approximation gy and a z ∈ dom(gx) ∩ dom(gy) then in the same fashion as in the uniqueness argument about G, gx(z) = gy(z). In particular, given two x-approximations, they coincide everywhere and are thus equal. Existence By induction on x ∈ A. Assume that approximations exist for all x such that xRa. We deﬁne ga as follows: ( F(a, {hy, gy(y)i : y ∈ pred(A, a, R)}) if x = a ga(x) = gx(x) if x ∈ closure(A, a, R).

Then ga is an approximation since for any x ∈ {a} ∪ closure(A, a, R),

ga(x) = gx(x) = F(x, {hy, gy(y)i : y ∈ pred(A, x, R)}) = F(x, gx pred(A, x, R)) (∗) = F(x, ga pred(A, x, R))

where (∗) holds since pred(A, x, R) ⊆ dom(gx) and gx ⊆ ga.

The only part left to prove is that G satisﬁes the equation ∀x ∈ A [G(x) = F(x, G pred(A, x, R)]. To see that this is the case, ﬁx an x ∈ A and chase the equalities.

G(x) = gx(x) = F(x, {hy, gy(y)i : y ∈ pred(A, x, R)}) = F(x, {hy, G(y)i : y ∈ pred(A, x, R)}) = F(x, G pred(A, x, R)).

In practice we do not mention F at all and just define G in terms of base cases and earlier defined values. We sometimes define functions by recursion on some well-founded class (U, R) instead of V, the class of all sets. In this case we do not care about the values of G outside of U and to appease the recursion theorem we define G(x) = ∅ for x∈ / U. Note that if R well-founds U it also well-founds V. In a few places we will define relations on well-founded classes. To see that this is well defined, we formally view the relation Q as a function GQ with the same domain as Q:: ( 1 if the definition of Q with every occurence of Q(~y) replaced with GQ(~y) = 1 holds GQ(~x) = 0 otherwise.

Then the recursion theorem tells us GQ is well deﬁned and hence, the relation Q is as well.

5 Well-Orders, Ordinals, Cardinals And Cardinality

Apart from K¨onigstheorem, the rank function and a few results about coﬁnality at the end, this section is a refresher on the ordinal and cardinal numbers. As ordinals and cardinals are assumed to be familiar, basic results are only stated, not proven. For proofs, curious, oblivious and forgetful readers alike are directed to [2] or [4].

6 5.1 Ordinal Numbers Definition 12. • A partially ordered set or poset is a binary relation R on a class A such that for a, b, c ∈ A, R is: transitive: aRb and bRc implies aRc. antisymmetric: If aRb and bRa then a = b. irreflexive: ¬aRa. • A relation R on a class A is called a well-order if it is both well-founded and totally ordered. • A class A is transitive if x ∈ y ∈ A implies x ∈ A or equivalently, if y ∈ A implies y ⊆ A. • An ordinal or an ordinal number is a transitive set α that is totally ordered by ∈. Due to the axiom of foundation, this is equivalent to being well-ordered. The class of ordinals are denoted by Ord. We use the convention that the first greek letters α, β, γ and δ always denote ordinals while λ is sometimes used to denote a limit ordinal. The symbols < and ∈ are used interchangably when dealing with ordinals and α ≤ β has the usual meaning of a < b or a = b.

• The successor function SW on a well-order W is deﬁned at every non-maximal element in W by SW (x) = min{y ∈ W : x

• If β ∈ α and β is not maximal in α then Sα(β) = β ∪ {β} = S(β). Definition 14. The first ordinal ∅ is denoted as 0. An ordinal α is called a successor if α = S(β) for some other ordinal β. It is called a limit if it is neither a successor nor 0. Proposition 15. If A is a transitive subset of an ordinal α then A is an ordinal and either A = α or A ∈ α. Proposition 16. For ordinals α and β exactly one of α ∈ β, α = β or α 3 β is true. Theorem 17. The class of ordinals is well-ordered and set-like under the ∈-order. Lemma 18. There exists a limit ordinal. In particular, there is a smallest limit ordinal. Definition 19. The smallest limit ordinal is called ω. All ordinals smaller than ω are called finite and the rest are called infinite. In ZFC it is common practice to identify the finite ordinals with natural numbers. The usual operations (addition, multiplication and so on) can then be defined by recursion on N = ω. Proposition 20. If α and β are ordinals and x is a set of ordinals then • α ∪ β = max(α, β), • α ∩ β = min(α, β) and • For any x ⊂ Ord the union S x is the least upper bound of x in Ord. This is sometimes denoted as sup x. The next theorem together with the well-ordering principle justifies indexing an arbitrary set with some ordinal. Definition 21. A function f : V → W between two partial orders is an order isomorphism if it is 0 0 bijective and satisfies v

7 Theorem 22. Every well-ordered set is order-isomorphic to a unique ordinal. Deﬁnition 23. The class function rank : V → Ord is deﬁned with recursion on ∈ by

rank(x) = sup{rank(y) + 1 | y ∈ x}.

The rank function simply counts how “deep” a set is. Proposition 24. Every ordinal α has rank α. Proof. By induction. Whether α is 0, a successor or a limit,

rank(α) = sup{rank(γ) + 1 | γ < α} IH= sup{γ + 1 | γ < α} = α.

Deﬁnition 25. The class function R(α) is deﬁned on all ordinals by

R(0) = ∅ R(α + 1) = P(R(α)) [ R(λ) = R(γ) for limit ordinals λ. γ<λ

The “depth” of a set in R(α + 1) \ R(α) is then reasonably enough α + 1 since the newly generated subsets in R(α + 1) must contain sets in R(α) that are not in any R(β) where β < α. The next theorem shows this connection formally. Proposition 26. For all α ∈ Ord, R(α) = {x | rank(x) < α}. Proof. By induction over α where we assume the proposition holds for β < α. α = 0: In this case R(0) = ∅ = {x | rank(x) < 0}. α = S(β): To show R(α) ⊆ {x | rank(x) < α}, note that any subset of {x | rank(x) < β} has rank at most β: {x | rank(x) < α} ⊇ P({x | rank(x) < β}) = P(R(β)) = R(α). In the R(α) ⊇ {x | rank(x) < α} direction, take a set y such that rank(y) < α. For any x ∈ y, rank(x) < β so y ⊆ {x | rank(x) < β} = R(β), hence y ∈ P(R(β)) = R(β + 1) = R(α). S IH S α is a limit: R(α) = γ<α R(γ) = γ<α{x | rank(x) < γ} = {x | rank(x) < α}.

5.2 Cardinality Deﬁnition 27. A set X has the same cardinality as a set Y if there is a bijection between X and Y . This is written as X =c Y . A set X have cardinality lesser than or equal to Y if there is an injection from X to Y . This is written as X ≤c Y . If X ≤c Y and X 6=c Y we say X has a strictly smaller cardinality than Y and we write X

Theorem 28 (Cantor-Schr¨oder-Bernstein). If X ≤c Y and Y ≤c X then X =c Y . This theorem can be proven in a weak subtheory of ZFC, notably without the axiom of choice and replacement. Cantor showed that cardinality among the inﬁnite sets is a nontrivial concept. Theorem 29 (Cantor). The cardinality of PX is always greater than the cardinality of X.

Proof. Since x 7→ {x} is an injection, X ≤c PX. It follows that if PX ≤c X there is a bijection f : X → PX. Assume f is such a bijection and let f(y) = {x ∈ X | x∈ / f(x)} for some y ∈ X. Then y ∈ f(y) ⇔ y∈ / f(y), so no such f can exist and X

8 As a consequence of the well-ordering principle and the fact that every well-order is is isomorphic with some ordinal we can define the notion of cardinal number. Definition 30. A cardinal or cardinal number is an ordinal such that there is no bijection to any smaller ordinal. Cardinals are denoted by the letter κ and sometimes λ, although λ may also stand for a limit ordinal. The cardinality of a set X, denoted by |X|, is defined as the least ordinal bijective with X. For every ordinal α, we denote the αth infinite cardinal by ℵα. A cardinal κ is a successor cardinal if it is the smallest cardinal larger than some cardinal λ. A cardinal is a limit cardinal if it is not 0 nor the successor to any smaller cardinal. There is a natural correspondence between the ordering of cardinals and existence of injections, surjections and bijections between sets. Proposition 31. • There is an injection f : X → Y if and only if |X| ≤ |Y |. • There is a bijection f : X → Y if and only if |X| = |Y |.

• There is a surjection f : X → Y if and only if |X| ≥ |Y |.

Proposition 32. Either X c Y . Proof. Follows from the total order of the ordinals.

We deﬁne the usual arithmetic operations on cardinals and state some of their properties. Deﬁnition 33. • κ + λ := |κ ] λ| = |{hα, ii | (α ∈ κ ∧ i = 0) ∨ (α ∈ λ ∧ i = 1)}| • κ · λ := |κ × λ|

• κλ := |{f | f : λ → κ}|

Proposition 34. Let max(κ, λ) ≥ ℵ0. Then κ + λ = κ · λ = max(κ, λ). S Proposition 35. If X and its members are of at most cardinality κ ≥ ℵ0 then | X| ≤ κ. Proposition 36. For cardinals κ, λ and µ,(κλ)µ = κλ·µ).

Deﬁnition 37. The product of sets Bi indexed by i ∈ I is deﬁned as Y Bi = {f | (f is a function) ∧ dom(f) = I ∧ ∀i ∈ I(f(i) ∈ Bi)}. i∈I

Theorem 38 (K¨onig). Let I be a nonempty index set and Ai and Bi be sets for every i ∈ I. If Ai

9 5.3 Cofinality The results and definitions in this section will only be used in the last section and so can thus be skipped at first and returned to when needed. Definition 39. A set A ⊂ Ord is unbounded in α if ¬∃β ∈ α∀γ ∈ A(γ < β), that is, no strict bound of all of A exists in α. Definition 40. A function f : α → β is cofinal if the range of f is unbounded in β. Definition 41. The cofinality of an ordinal β, denoted cf β is defined as the least ordinal α such that there exists a cofinal function f : α → β. As an immediate consequence of the definition, the identity map is cofinal so cf β ≤ β and the cofinality of an ordinal is always defined. cf 0 = 0 and Also, by the function ∅, cf 0 = 0 and by the function (0 7→ α) from 1 to α + 1, cf(α + 1) = 1. Lemma 42. For every ordinal β, there is a nondecreasing cofinal map f : cf β → β. Proof. Let f : cf β → β be a cofinal map. We define the nondecreasing function g : cf β → Ord as follows: g(γ) = sup range f S(γ) . Since f(γ) ≤ g(γ) for every γ ∈ cf β, range(g) is clearly unbounded in β, but is it a subset of β? Assume that it is not and that α < cf β is the least ordinal such that g(α) ≥ β. Then α can not be 0 or a successor because in that case g(α) = f(α) < β so further assume α is a limit ordinal. But if sup range f S(α) ≥ β, either f : α → β is cofinal, contradicting the minimality of cf β or f(α) > β, contradicting the range of f. Thus range(g) ⊆ β and g : cf β → β is cofinal and nondecreasing. Proposition 43. If f : α → β and g : β → γ are nondecreasing cofinal maps then their composition g ◦ f : α → γ is nondecreasing and cofinal as well. Definition 44. An ordinal β is regular if cf β = β and singular otherwise. Proposition 45. Every regular ordinal is a cardinal. Proof. A bijection b : α → β is always cofinal so no regular β can be bijective with a smaller α. Lemma 46. For every β, cf β is a regular cardinal. Proof. First of all, cf cf 0 = cf 0 = 0 and cf cf S(α) = cf 1 = 1 are both regular. For limit ordinals β, let f : cf β → β and g : cf cf β → cf β be nondecreasing cofinal maps. The composition f ◦ g : cf cf β → β is cofinal so by the minimality of cf β, cf cf β = cf β, meaning cf β is regular. The previous proposition then tells us cf β is a cardinal. Proposition 47. The cofinality of every infinite limit ordinal is a limit ordinal. In particular, cf ω = ω, so ω is regular. Proof. Let λ be a limit ordinal. The cofinality of an ordinal is itself regular and since cf(α + 1) = 1, the only regular nonlimit ordinals are 0 and 1. All we have to do is exclude those possibilities. The cofinality of λ can clearly not be 0 or 1 since λ is nonempty and α ∈ λ is strictly bounded by α + 1 ∈ λ, respectively. Since ω is the least limit ordinal and cf α ≤ α, cf ω = ω. Lemma 48. If there is a nondecreasing cofinal map from α to β then cf α = cf β. Proof. The composition of nondecreasing cofinal maps f 0 : cf α → α and f : α → β shows that cf β ≤ cf α. To show that cf α ≤ cf β, let g : cf β → β be a nondecreasing cofinal function and consider the mapping

h : cf β → α γ 7→ {δ ∈ α | f(δ) ≤ g(γ)}.

First of all, it maps ordinals to ordinals since f is nondecreasing and every transitive subset (δ < δ0 ∧δ0 ∈ A ⇒ δ ∈ A) of an ordinal is an ordinal. Second, for every δ ∈ α, there is some γ ∈ cf β such that f(δ) ≤ g(γ) by cofinality of g, so h : cf β → α is cofinal. By definition of cofinality, cf α ≥ cf β.

10 Proposition 49. Every inﬁnite successor cardinal is regular. Proof. Let κ be the successor cardinal of the cardinal µ and assume towards a contradiction that cf(κ) < S κ. Let f : cf κ → κ be coﬁnal. Then for every γ ∈ κ, f(γ)

κ Proposition 51. For any inﬁnite cardinal κ, κ

κ κ Proof. Assume that λ = cf 2 ≤c κ and let f : λ → 2 be coﬁnal. By applying K¨onigstheorem and a bit of cardinal arithmetic we get the following contradiction:

κ [ Y κ κ λ κ·λ κ 2 = f(i)

By applying proposition 50 to the map f(n) = ℵn, dom(f) = ω we see it is coﬁnal in ℵω and by ℵ0 proposition 47, cf ℵω ≤ ω. Thus we see that by the last proposition, 2 = ℵω is not a possibility.

cf κ Proposition 52. For every inﬁnite cardinal κ, κ

Proof. Let f : cf κ → κ be a coﬁnal map. Since f(γ)

6 Relativization

The the only mathematical structures we are interested in are the ones with a single binary relation, that is to say the ones that possibly could be models of ZFC. But as it turns out we can get by using only structures where the binary relation is ∈. We therefore define what it means for a formula to hold in a transitive class M. Recall that a class is transitive if y ∈ A implies y ⊆ A. Definition 53. Let M be a transitive class. The relativization of a formula φ to M, denoted by φM is defined by recursion (in the metalogic) on the structure of φ as follows. 1.( x = y)M is x = y.

2.( x ∈ y)M is x ∈ y. 3.( ¬ψ)M is ¬ ψM. 4.( ψ ∧ χ)M is ψM ∧ χM.

5.( ∃x ψ)M is ∃x(x ∈ M ∧ ψM).

11 The rest of the logical connectives and the universal quantifier are as mentioned defined to be shorthands containing only ¬,∧ and existential quantifiers, i.e φ ∨ ψ := ¬(¬φ ∧ ¬ψ) and ∀x φ := ¬∃x(¬φ). We abbreviate ∃x(x ∈ y ∧ φ) by ∃x ∈ y φ and ∀x(x ∈ y → φ) by ∀x ∈ y φ for our common mental well being’s sake. Note that the defined universal quantifier works as expected under relativization in the following sense:

(∀x φ)M def⇔ (¬∃x ¬φ)M def⇔ ¬(∃x ¬φ)M def⇔ ¬∃x(x ∈ M ∧ ¬φM) ⇔ ¬∃x¬(x ∈ M → φM) def⇔ ∀x(x ∈ M → φM) def⇔ ∀x ∈ M φM.

M The reason for deﬁning relativization and using the notation φ instead of the more familiar M φ is that [3] does so. We read φM as “φ is true in M”. Similarly, if S is a list of sentences in the metalogic, we say that “M is a model of S” or “S is true in M” if every sentence φ in S is true in M 2.

Definition 54. For any constant c uniquely defined by some formula φc(y) (such as ω and ∅) and any M class function F(x1, . . . , xn) defined by some formula φF(x1, . . . , xn, y), let c be the object defined by M M M φc (y) and F (x1, . . . , xn) be defined as the y making φF (x1, . . . , xn, y) hold. This notation will only be used when such a y is guaranteed to exist and is unique.

Deﬁnition 55. Let M be a transitive class. A formula φ with free variables among x1, . . . , xn is called absolute for M if M ∀x1, . . . , xn ∈ M(φ(x1, . . . , xn) ↔ φ (x1, . . . , xn)) holds. A formula φ is called absolute if M is given by the context or if for any transitive class M, φ is absolute for M.

Lemma 56. Let M be a transitive class and let φ and ψ be absolute for M. Then ¬φ, φ ∧ ψ, ∃x ∈ y φ and ∀x ∈ y φ are absolute for M. Proof. The first two cases are true by definition of relativization on negations and conjunctions. In the existential case, fix values in M for y and the free variables x1, . . . , xn of ∃x ∈ y φ.

def def (∃x ∈ y φ)M ≡ (∃x(x ∈ y ∧ φ))M ≡ ∃x(x ∈ M ∧ (x ∈ y)M ∧ φM) (?) (†) def ⇔ ∃x(x ∈ M ∧ x ∈ y ∧ φ) ⇔ ∃x(x ∈ y ∧ φ) ≡ ∃x ∈ y φ

The (?)-equivalence follows by absoluteness of φ and x ∈ y. The left to right of the (†)-equivalence is clear, for the right to left direction, consider that y ∈ M and that M is transitive yields x ∈ y ⇒ x ∈ M. The universal quantiﬁer case follows from the existential case and that ∀x ∈ y φ is equivalent to ¬∃x ∈ y ¬φ.

Deﬁnition 57. A formula is ∆0 if it is of the form 1. x ∈ y or x = y,

2. ¬φ or φ ∧ ψ where φ and ψ are ∆0.

3. ∃x (x ∈ y ∧ φ) or ∀x (x ∈ y → φ) where φ is ∆0.

Note that ∀x ∈ y φ is short for ¬∃x(x ∈ y ∧ ¬φ) and is therefore also ∆0 if φ is. As it happens, this type of formula with only bounded quantiﬁers is relevant in other topics also. In arithmetic, where (<) bounds variables instead of (∈) we can for such a formula φ(x1 . . . xn) replace the free variables with numbers n1 . . . nm and check if φ(n1, . . . nm) is true by mechanically checking for existence up to the given bounds for each quantiﬁer. All primitive recursive functions can also be described by ∆0 formulas. The curious reader is referred to [6].

Corollary 58. ∆0 formulas are absolute. Alternatively, any formula whose quantiﬁers are all bounded is absolute. 2By true, we here mean derivable from ZFC.

12 Deﬁnition 59. An ordered pair hx, yi is coded as the set {{x}, {x, y}}. The class function taking x, y to {{x}, {x, y}} is called the Kuratowski operator. With this in mind we let

Qha, bi ∈ x : φ abbreviate Qw ∈ x∃p ∈ w∃a, b ∈ p(w = ha, bi ∧ φ)

where Q is one of the quantiﬁers ∀ or ∃. Note that if φ is ∆0, so is Qha, bi ∈ x : φ.

To demonstrate what can be deduced from the absoluteness of ∆0 formulas, here is a list of formulas accompanied with ∆0 equivalents. That these actually are equivalent is “easily seen” by inspection. Note that the “deﬁnition” of intersection has a quirk, to make it well deﬁned for all sets we say that T ∅ has the arbitrary value ∅.

x ∈ y ⇔ x ∈ y x = y ⇔ x = y x ⊆ y ⇔ ∀w ∈ x(w ∈ y) z = {x, y} ⇔ x ∈ z ∧ y ∈ z ∧ ∀w ∈ z(w = x ∨ w = y) z = {x} ⇔ z = {x, x} z = hx, yi ⇔ z = {{x}, {x, y}} z = ∅ ⇔ ∀w ∈ z(w 6= w) z = x ∪ y ⇔ x ⊆ z ∧ y ⊆ z ∧ ∀w ∈ z(w ∈ x ∨ w ∈ y) z = x ∩ y ⇔ z ⊆ x ∧ z ⊆ y ∧ ∀w ∈ x(w ∈ y → w ∈ z) z = x \ y ⇔ z ⊆ x ∧ ∀w ∈ x(w∈ / y ↔ w ∈ z) z = S(x) ⇔ x ⊆ z ∧ x ∈ z ∧ ∀w ∈ z(w ∈ x ∨ w = x) (x is transitive) ⇔ ∀y ∈ x ∀z ∈ y(z ∈ x) x ∈ Ord ⇔ ∀y, z ∈ x(y ∈ z ∨ y = z ∨ y 3 z) ∧ (x is transitive) x = 0 ⇔ x = ∅ (x is a successor ordinal) ⇔ ∃y ∈ x(S(y) = x) ∧ x ∈ Ord (x is a limit ordinal) ⇔ ¬∃y ∈ x(S(y) = x) ∧ x ∈ Ord ∧ x 6= 0 z = ω ⇔ (z is a limit ordinal) ∧ ¬∃y ∈ z(y is a limit ordinal) [ z = x ⇔ ∀y ∈ x(y ⊆ z) ∧ ∀w ∈ z∃y ∈ x(w ∈ y) \ z = x ⇔ (∀y ∈ x(z ⊆ y)) ∧ (x = ∅ → z = ∅) ∧ ∀y ∈ x∀w ∈ y((∀y0 ∈ x(w ∈ y0)) → w ∈ z) z = A × B ⇔ ∀w ∈ z∃hx, yi ∈ z(x ∈ A ∧ y ∈ B ∧ w = hx, yi) ∧ ∀x ∈ A∀y ∈ B∃w ∈ z(w = hx, yi) R is a relation ⇔ ∀w ∈ R∃hy, zi ∈ R(w = hy, zi) z = dom R ⇔ (R is a relation) ∧ (∀x ∈ z∃ha, bi ∈ R(a = x)) ∧ (∀ha, bi ∈ R ∃x ∈ z(a = x)) z = range R ⇔ (R is a relation) ∧ (∀x ∈ z∃ha, bi ∈ R(b = x)) ∧ (∀ha, bi ∈ R ∃x ∈ z(b = x)) R is a function ⇔ (R is a relation) ∧ ∀ha, bi, ha0, b0i ∈ R(a = a0 → b = b0) z = R(x) ⇔ (R is a function) ∧ ∃ha, bi ∈ R(a = x ∧ b = z) R is an injective function ⇔ (R is a function) ∧ ∀ha, bi, ha0, b0i ∈ R(b = b0 → a = a0)

Proposition 60. The formulas to the right of the ‘⇔’ signs are ∆0 so their equivalent counterparts on the left are therefore absolute.

Lemma 61. The formula z = {w ∈ y | φ(w, x1, . . . , xn)} where z is not a parameter of φ is ∆0 if φ is ∆0 and absolute if φ is absolute.

Proof. The formula is really a notational convenience instead of writing ∀w ∈ y(w ∈ z ↔ φ(w, x1, . . . , xn)). This formula is clearly ∆0 if φ is and by lemma 56 it is absolute if φ is absolute.

13 7 Model Construction

The goal of this section is to show that for every ﬁnite sublist φ1, . . . , φn of ZFC,

M M ZFC ` ∃M (M is transitive) ∧ (M ≤c ℵ0) ∧ φ1 ∧ · · · ∧ φn .

At a glance, this looks a lot like the prerequisites for the compactness theorem of first order logic but differs somewhat. Let “M s” be defined (in ZFC) in the usual way in the spirit of Tarski’s truth definition by recursion over encoded formulas3. For sets S of encoded formulas we use the standard abuse of notation M S to mean ∀s ∈ S(M s). The set of all encoded axioms of ZFC is denoted by pZFCq. The compactness theorem specialized to the theory pZFCq can then be stated as the sentence

∀S ∈ P(pZFCq)[S

M It is provable for transitive models M that M pφq is equivalent to φ so this is only a syntactic difference. In order to deduce the consequent from the compactness theorem we would need to prove the antecedent in a single first order proof while showing that there are models for all finite subsets of pZFCq. This is in stark contrast to the section goal of proving that for a fixed finite sublist of ZFC we can prove that there is a countable transitive model of these axioms where we can tailor our use of axioms in the proof to suit the required axioms. This somewhat motivates why the section goal is achieveable and why the antecendent of the compactness theorem would be harder, if possible at all, to prove. In fact, the consequent is not provable in ZFC! This follows from Gödelssecond incompleteness theorem which tells us that ZFC 0 4 ¬ ProvablepZFCq(p⊥q) and by standard results from mathematical logic , this is equivalent to ∃M pZFCq. But how is this possible? The section goal and the antecendent looks so similar. Do they not say the same thing? Not necessarily: If we take a model theoretic approach and assume that we in the metatheory have a model M of ZFC then ωM might be nonisomorphic to the natural numbers in the metatheory5. If this is the case, there are so called nonstandard numbers in ωM that have no corresponding natural number in the metatheory. This implies that “finite” inside of M is not the same thing as “finite” in the metatheory. Together with the encodings of formulas from the metatheory, the set of encoded formulas in M will contain codes for formulas whose length are nonstandard. The encoded proofs will similarly include the encodings of proofs from the metatheory as well as proofs that include nonstandard formulas or are of nonstandard length. Definition 62. A formula is defined to have the following subformulas:

• x ∈ y and x = y have no subformulas •∃ x φ and ¬φ have φ as a subformula and • φ ∧ ψ has φ and ψ as subformulas.

Deﬁnition 63. A list φ1, . . . , φn (in the metalogic) is subformula closed if, for any φi, all subformulas of φi are in the list. The following lemma is not the usual form of the Tarski-Vaught test6 but the core idea is the same. Lemma 64 (Tarski-Vaught Test). Let M, N be transitive classes such that M ⊂ N and let the list φ1, . . . φn be subformula closed. Then the following clauses are equivalent:

(a) For every φi, M N ∀x1, . . . , xn ∈ M(φi (x1, . . . , xn) ↔ φi (x1, . . . , xn)) is true. 3An example of an encoding is the G¨odel coding of arithmetic formulas down to natural numbers. In the text we refer to some unspeciﬁed way of mapping formulas in the metatheory to sets. 4If a theory is consistent it has a model and vice versa. 5This is similar to nonstandard models of Peano arithmetic. 6The usual Tarski-Vaught test can be found in [5]

14 (b) Whenever φi(y1, . . . , ym) ≡ ∃x φj(x, y1, . . . , ym) (with all free variables displayed),

N N ∀y1, . . . , ym ∈ M ∃x ∈ N φj (x, y1, . . . , ym) → ∃x ∈ M φj (x, y1, . . . , ym) .

As a special case, if N = V this test gives a criteria for when a formula is absolute. Whenever we refer to this result we will be referring to (b)⇒(a).

Proof. To show (a)⇒(b), take a φi(y1, . . . , yn) ≡ ∃x φj(x, y1, . . . , ym) and apply (a). For (b)⇒(a), we proceed by induction on φi taking as induction hypothesis that (a) is true for any subformula of φ. All cases except for the quantiﬁer case are trivial:

(x ∈ y)M def⇔ x ∈ y def⇔ (x ∈ y)N (for x, y ∈ M) (x = y)M def⇔ x = y def⇔ (x = y)N (for x, y ∈ M) (¬ψ)M def⇔ ¬ψM ⇔IH ¬ψN def⇔ (¬ψ)N (ψ ∧ χ)M def⇔ ψM ∧ χM ⇔IH ψN ∧ χN def⇔ (ψ ∧ χ)N.

Fix y1, . . . , ym ∈ M and assume φi ≡ ∃x φj(x, y1, . . . , ym); then

M def M IH N (b) N def N φi ⇔ ∃x ∈ M φj (x, y1, . . . , ym) ⇔ ∃x ∈ M φj (x, y1, . . . , ym) ⇔ ∃x ∈ N φj (x, y1, . . . , ym) ⇔ φi .

This is all the preparation we need to create the initial model of our ﬁnite fragment.

Theorem 65 (The Reﬂection Theorem). Given any list (in the metatheory) of formulas φ1, . . . , φn and an ordinal α there is a β > α such that φ1, . . . , φn are absolute for R(β).

Proof. We assume that φ1, . . . , φn is subformula closed, if it is not, append the missing formulas and get a longer ﬁnite list of formulas. We will use the Tarski-Vaught test on R(β) and V to show the absoluteness of φ1, . . . , φn. For each φi ≡ ∃x φj(x, y1, . . . , ym), deﬁne Fi and Gi as ( m γ if γ is the smallest ordinal where ∃x ∈ R(γ) φj(x, y1, . . . , ym) Gi : V → Ord, Gi(y1, . . . , ym) = 0 if no such γ exists.

Fi : Ord → Ord, Fi(γ) = sup{Gi(y1, . . . , ym) | y1, . . . ym ∈ R(γ)}

For all other φi, let Fi(γ) = 0 and Gi(x1, . . . , xn) = 0 for every γ ∈ Ord and x1, . . . , xn ∈ V. We now recursively deﬁne an ω-sequence by

β0 = α βk+1 = max(βk + 1, F1(βk),..., Fn(βk))

and let β = supk∈ω βk. The sequence is strictly increasing so β must be a limit ordinal and R(β) = S k∈ω R(βk). Assume that for y1, . . . , ym ∈ R(β),

φi(y1, . . . , ym) ≡ ∃x φj(x, y1, . . . , ym) holds.

For p from 1 to m, let kp ∈ ω be minimal such that yp ∈ R(βkp ). Let k = max(k1, . . . , km). By construction of βk+1, there is an x ∈ R(βk+1) ⊂ R(β) such that φj(x, y1, . . . , ym). This is all that the Tarski-Vaught test requires, so the formulas are absolute for R(β).

Corollary 66. Any finite fragment φ1, . . . , φn of ZFC has a model. It is worth pointing out that m,n and p in the proof of the reflection theorem lives in the metalogic. This is why this argument cannot be expanded to all of ZFC, the proposition would not be expressible nor provable in first order logic. Next we prove that we can cut out a small chunk of the model while preserving the absoluteness of the formulas. The proof uses the same technique as the Löwenheim-Skolem downward theorem7 from mathematical logic and is therefore called the same thing.

7For the usual L¨owenheim-Skolem downwards, see [5]

15 Theorem 67 (L¨owenheim-Skolem Downwards). Given formulas φ1, . . . , φn and a set X there is a set A ⊇ X where |A| ≤ max(ℵ0, |X|) such that φ1, . . . , φn are absolute for A. Proof. We once again assume that the list is subformula closed. Apply the reﬂection theorem to φ1, . . . , φn with α ≥ max(1, rank(X)) to get a β > α. The reason we require that α is positive is that the construction assumes that ∅ ∈ R(β). Then X ⊆ R(α) ⊆ R(β) and the formulas are absolute for R(β). We will then construct a set A such that X ⊆ A ⊆ R(β) and |A| ≤ max(ℵ0, |X|) where A R(β) A and R(β) passes the Tarski-Vaught test. It then follows that φi(... ) ⇔ φi (... ) ⇔ φi (... ) and the formulas are thus absolute for A.

Pick a choice function on R(β). For each φi ≡ ∃x φj(x, y1, . . . , yli ) with all free variables shown where we assume that li ≥ 1 (otherwise ∃x (x = x) and other sentences become problematic), deﬁne the li function Hi : R(β) → R(β) by

( R(β) x where x ∈ R(β) is the chosen element such that φj (x, y1, . . . , yli ) Hi(y1, . . . , yl ) = i ∅ if no such x ∈ R(β) exists.

li For the remaining φi, let Hi[Ak ] denote ∅. Let

l1 ln [ A0 = X,Ak+1 = H1[Ak ] ∪ · · · ∪ Hn[Ak ],A = Ak. k∈ω

Then all the Ak have (by induction) cardinality at most |X| so the union has cardinality at most ℵ0 · |X| = max(ℵ0, |X|). Furthermore, since A and R(β) by construction passes the Tarski-Vaught test for φ1, . . . , φn and since these formulas are absolute for R(β) they are also absolute for A. Definition 68. A relation R is extensional on A (or simply extensional when A is clear from the context) if ∀x, y ∈ Ax = y ↔ ∀z ∈ A(zRx ↔ zRy). Definition 69. Given a class A and a set-like, well-founded relation R, the Mostowski collapsing function is defined as

F : A → V, F(a) = {F(a0) | a0Ra}.

We call {F(a) | a ∈ A} the Mostowski collapse of A. Lemma 70. If A is well-founded by a set-like, well-founded, extensional relation R then the Mostowski collapsing function is bijective to the Mostowski collapse of A and for a, b ∈ A, aRb ⇔ F(a) ∈ F(b). Further, the Mostowski collapse of A is transitive. Proof. If a0 6= b0 there is by extensionality of R some x ∈ A such that either xRa0 and ¬(xRb0) or ¬(xRa0) and xRb0. Either way, F(x) will be a member of F(a0) or F(b0) but not both; thus F(a0) 6= F(b0) and F is injective. It is by definition surjective and thus also bijective. That aRb implies F(a) ∈ F(b) follows from the definition. The converse is true by injectivity of F: If F(a) ∈ F(b) there is some a0Rb where F(a) = F(a0) and thus aRb. The Mostowski collapse is transitive by definition: For any x ∈ y ∈ {F(a) | a ∈ A} there are a0, a00 ∈ A such that F(a0) = x, F(a00) = y, hence x ∈ {F(a) | a ∈ A}.

Proposition 71. If X is a set where ∈X = {hx, yi | x ∈ y ∈ X} is extensional, the Mostowski collapse M of X will for any formula φ satisfy

X M ∀x1, . . . , xn ∈ X φ (x1, . . . , xn) ↔ φ (F(x1),..., F(xn)) .

Proof. The proof is by induction on formulas with the induction hypothesis that the theorem holds for any subformulas. Let x0 = F(x) and y0 = F(y). It is true for the atomic formulas:

(x ∈ y)X ≡ x ∈ y ⇔ x0 ∈ y0 ≡ (x0 ∈ y0)M (previous proposition) (x = y)X ≡ x = y ⇔ x0 = y0 ≡ (x0 = y0)M . (injectivity of F)

16 Let y1, . . . , yn ∈ X and let ~y denote y1, . . . , yn and ~z denote F(y1),..., F(yn). It then follows that

(ψ(~y) ∧ χ(~y))X ≡ ψX (~y) ∧ χX (~y) ⇔IH ψM (~z) ∧ χM (~z) ≡ (ψ(~z) ∧ χ(~z))M (¬ψ(~y))X ≡ ¬ψX (~y) ⇔IH ¬ψM (~z) ≡ (¬ψ(~z))M . Finally, when the formula is an existential quantiﬁcation, (∃x ψ(x, ~y))X ≡ ∃x ∈ X ψX (x, ~y) ⇔IH ∃x ∈ X ψM (F(x), ~z) ⇔ ∃x ∈ M ψM (x, ~z) ≡ (∃x ψ(x, ~z))M .

It should be noted that ∈X is not always extensional. Consider the set X = {∅, {{∅}}}: ∅ ∈/ ∅, {{∅}} ∈/ ∅, ∅ ∈/ {{∅}}, {{∅}} ∈/ {{∅}}.

X If ∈X were extensional then (∅ = {{∅}}) would be true. We can at last stitch these steps together and get a countable transitive model.

Theorem 72 (Countable Transitive Model Theorem). For any list φ1, . . . , φn of axioms of ZFC there is a countable transitive set M such that φ1, . . . , φn are true in M. Proof. Start by adding the axiom of extensionality to the list if it is not already present. Apply the L¨owenheim-Skolem Downwards theorem with X = ∅. The resulting set A is then countable and our formulas are true in A. The formulas are by the last proposition also true in the Mostowski collapse of A. By the previous lemma the Mostowski collapse is transitive and bijective with A and thus also countable. An odd consequence of this theorem is that if we add a constant symbol N to our formal ﬁrst order N language then the theory TN = ZFC +ZFC +(N is transitive)+(N ≤c ℵ0) is a conservative extension of ZFC. To clarify, ZFCN means the axioms of ZFC relativized to N and “conservative extension” means that if TN ` φ and φ does not contain the symbol N then ZFC ` φ. An informal argument N N for conservativity goes like this: Let p be a proof of φ from TN and let φ1 , . . . , φn be the assumptions from ZFCN used in p. The countable transitive model theorem then says there exists a model M of M M N φ1 , . . . , φn that has the needed properties to replace N in the proof p. Every occurence of φi can be M replaced by a derivation of φi , and similarly for (N is transitive) and (N ≤c ℵ0). With this in mind we will from now on work in the theory TM . As stated in section 3, Metatheory and Formalism we will assume that if φ ` ψ then T ` φM → ψM . A proof of this assumption would be of a proof theoretic nature and is therefore outside the scope of this thesis. We end the section by proving a few properties of M.

Lemma 73. Let u1, . . . , un ∈ M and φ(x, u1, . . . , un) be absolute for M. If φ(x, u1, . . . , un) deﬁnes a M unique x ∈ V and φ (y, u1, . . . , un) for some y ∈ M then x = y. M Proof. Let x be the unique set satisfying φ(x, u1, . . . , un). If y ∈ M satisﬁes φ (y, u1, . . . , un) then by absoluteness, φ(y, u1, . . . , un) so x = y. This lemma highlights a mental shortcut in reasoning about objects in M. The implicit reasoning in a lot of places goes as follows: A certain construct C can be made by appealing to ZFC. Some formula M φc(x, u1, . . . , un) describes the construct uniquely. By reasoning inside M, for some x ∈ M, φC (x). If the formula happens to be absolute for M, φC (x) also holds. Examples of this pattern are in order: Absolute set comprehensions If φ(x) is an absolute formula then by lemma 61, so is z = {w ∈ y | φ(w, x1, . . . , xn)}. Thus

M (z = {w ∈ y | φ(w, x1, . . . , xn)}) ⇔ z = {w ∈ y | φ(w, x1, . . . , xn)}

M or equivalently, ({w ∈ y | φ(w, x1, . . . , xn)}) = {w ∈ y | φ(w, x1, . . . , xn)}.

17 Empty set The empty set is absolute by proposition 60 (the gigantic list of equivalent formulas) so ∅M = ∅). Domain and range The domain and range of a binary relation is by the same proposition absolute so domM R = dom R and rangeM R = range R.

Union Again, our friend proposition 60 tells us that unions are absolute, so (x ∪ y)M = x ∪ y and (S x) = S x. Lemma 74. If X ⊆ M and X has a ﬁnite cardinality then X ∈ M.

Proof. The proof is by induction on the cardinality of X. |X| = 0 : If |X| = 0 then X = ∅ and by lemma 73 and lemma 60, ∅ ∈ M. |X| = n + 1 : Pick x ∈ X. Then x ∈ M and |X \{x}| = n. The induction hypothesis yields X \{x} ∈ M. By lemma 73 and lemma 60, {x} ∈ M and by the same lemmas X = (X \{x}) ∪ {x} ∈ M.

Some of the functions and relations used, notably the forcing relation (that will be defined later) and the rank function are defined by recursion on some well ordering. The following result gives us a criteria when such a function or relation is absolute. Theorem 75. Let A be well-founded by R and F : A × V → V a defined function. If A, R and F are absolute for M,(R is set-like)M and

∀x ∈ AM (pred(A, x, R) ⊆ M)(†) then the deﬁned function G characterized by

∀x ∈ A [G(x) = F(x, G pred(A, x, R)] is absolute for M. Proof. We start by proving that (R is well-founded)M . By absoluteness RM = R ∩ (M × M). For any nonempty S ∈ M, S ⊆ AM we see that there is an R-minimal element in S. This is also an RM -minimal element so (R is well-founded)M . We can therefore use the recursion theorem (theorem 11) to deﬁne G inside of M such that

M (∀x ∈ A [G(x) = F(x, G pred(A, x, R)]) M M M M M M ⇔ ∀x ∈ A [G (x) = F (x, G pred(A , x, R ))].

We note that pred(A, x, R) = pred(AM , x, RM ) by (†) and by absoluteness of A and R in M. Assume there is an R-minimal x ∈ AM such that G(x) 6= GM (x). Then G(y) = GM (y) for all yRx so M G pred(A, x, R) = G pred(A, x, R) and we get the contradiction

M M M M M G(x) = F(x, G pred(A, x, R)) = F (x, G pred(A , x, R )) = G (x).

Lemma 76. The rank function is absolute for M. Proof. First of all, rank(x) = sup{rank(y) + 1 | y ∈ x}. The well-founded relation is ∈, which is set-like in M since M is transitive. The domain A = V is clearly absolute. The function F uses supremum (i.e. union), successor and function application, all shown to be absolute. Therefore according to theorem 75, rank is absolute for M.

18 8 Forcing

As in the end of the last section, M will be assumed to be a countable transitive model of ZFC. That is to say, M is countable and transitive and for each axiom φ of ZFC, φM is true.

Deﬁnition 77. A forcing poset is a partial order (P, ≤P ) with a greatest element 1P and for every p ∈ P there are distinct q, r ∈ P smaller than p such that no s ∈ P satisﬁes both s ≤P q and s ≤P r. From

now on P = (P, ≤, 1P) always denotes some forcing poset such that P ∈ M. To make the forcing construction more intuitive we will interpret the elements of P as some sort of “information”. This interpretation has no technical impact on the proofs but will hopefully make the arguments less alien. We interpret p ≤ q for p, q ∈ P as “p contains as much or more information than q”. That more information is smaller in P is a notational inconvenience inherited from Kunens book. The existence of a greatest element tells us that there is a smallest amount of information. The last criterion for a forcing poset tells us that it is always possible to get more information in at least two ways that are incompatible with each other. This interpretation motivates some of the following deﬁnitions. Deﬁnition 78. Let a, b ∈ P and S ⊆ P. (a) a is called an extension of b (or a extends b) if a ≤ b. (b) a and b are said to be compatible if there is a common extension e of a and b (e ≤ a, b) and incompatible if no such extension exists. That a and b are incompatible is written as a ⊥ b. We interpret a ⊥ b as a and b being contradictory information.

(c) S is dense in R ⊆ P (or simply dense when R = P) if for every p ∈ R there is an extension of p in S: ∀p ∈ R ∃s ∈ S s ≤ p

(d) S is dense below p if for every q ≤ p there an extension of q in S. An alternative formulation is that S is dense in {q ∈ P | q ≤ p}. (e) S is closed upwards if s ∈ S, p ∈ P and s ≤ p implies p ∈ S. In our information interpretation this corresponds to S being closed under subtraction of information. (f) S is a filter if it is closed upwards and all pairs of elements in S have a common extension in S. A filter is similar to a theory in logic: All pairs a, b ∈ S have a “conjunction” in S that “implies” a and b and all p “weaker” than something in S also lies in S. 8 We note that all of these concepts can be defined by ∆0-formulas and are therefore absolute. The simplest possible example of a forcing poset are strings over the alphabet {0, 1} where the ordering a ≤P b is defined by “a has b as a prefix” and the greatest element is the empty string. The empty string is then a prefix of every string in P and every string s ∈ P can be extended to two incompatible strings by appending a zero or a one at the end. We can view these strings as the result of flipping a coin several times, zeroes for heads and ones for tails. Definition 79. A filter G ⊆ P is P-generic over A if G ∩ D 6= ∅ for every dense D ∈ A. Lemma 80 (Generic Filter Existence). For every p ∈ P there is a filter G ⊆ P that is P-generic over M where p ∈ G.

Proof. Let D0,D1,...,Dn,... be the countably many dense subsets of P in M and cP a choice function on P. The descending sequence

p0 = p, pn+1 = cP({q ∈ P | q ≤ pn} ∩ Dn) then cuts every dense subset of P in M. To expand this sequence into a ﬁlter, let [ G = {q ∈ P | pn ≤ q}. n∈N

For any q, r ∈ G, there is piq , pir such that piq ≤ q, pir ≤ r. Thus pmax(iq ,ir ) ∈ G extends both q and r. Clearly G is closed upwards so G is a ﬁlter.

8 These formulas would then have P as a formal parameter.

19 That a set D ⊆ P is dense tells us that if we carefully walk down P like so:

p0 ≥ p1 ≥ · · · ≥ pn,

a single careless misstep pn+1 ≤ pn can cause us to step on D regardless of how we walked to begin with. Let us view an arbitrary subset S ⊆ P as an event and say that S occurs to a filter F ⊂ P if S ∩ F 6= ∅. If we view dense sets as accidents then a generic G is maximally unlucky in the sense that every never surely avoidable accident happens to G. Let us call this the Murphy’s law interpretation of generic sets. A generic set G is in fact so unlucky that it cannot exist in the model M that it is generic over. We prove this by proving no P-generic set over V can exist and apply the same reasoning inside of M. Lemma 81. There is no P-generic filter over V. Proof. Let G be the supposedly generic set. We construct a dense set D that does not intersect G. For each p ∈ P, let q and r be incompatible extensions of p. Because G is a filter, G can at most contain one of q and r, lets say q∈ / G. We define D to be a set of all such q’s. Then D is dense and G ∩ D = ∅. Definition 82. The poset Fn(I,J) is defined as the finite partial functions from I to J

Fn(I,J) = {p | p

0 0 i) Clearly, 1Fn(I,J) = ∅. For any f ∈ Fn(I,J), f ∪ {hi, ji} and f ∪ {hi, j i} are incompatible if j, j ∈ J are distinct and i ∈ I \ dom(f). ii) If f, g ∈ G there is a function h ∈ G where f, g ⊆ h by the definition of filter; S G is in other words a function. To prove surjectivity, consider the set Dj = {f | ∃i ∈ I : hi, ji ∈ f}. It is dense and by lemma 61 and 73, Dj ∈ M. it is always possible to extend a function in an undefined point of I S with the value j. Then G ∩ Dj 6= ∅ so the function ( G) must be defined in every point j ∈ J.

The next subject of our attention are the P-names. To understand the definition of P-names we explore a strained parable in which the mathematician Zermela modifies the definition of the cumulative hierarchy (the R(α) below) several times to finally arrive at the definition of P-names. Hopefully this transition will make P-names more intuitive than just stating the definition. Zermela has taken a fancy in characteristic functions and wishes to modify the cumulative hierarchy such that at each level there are characteristic functions instead of sets. Furthermore, the domain of these functions should only contain characteristic functions, so that there are characteristic functions all the way down9. She comes up with the C(α) hierarchy.

R(0) = ∅ C(0) = ∅ C0(0) = ∅ R(α + 1) = P(R(α)) C(α + 1) = {f : C(α) → {0, 1}} C0(α + 1) = {f ⊆ C0(α) × {0, 1}} [ [ [ R(λ) = R(γ) C(λ) = C(γ) C0(λ) = C0(γ) γ<λ γ<λ γ<λ She is pleased with this deﬁnition, a set can be extracted from a characteristic function by “peeling away” the characteristic functions like so:

peel(f) = {peel(g) | g ∈ dom(f) ∧ f(g) = 1}.

9As opposed to turtles.

20 It is even possible to deﬁne an explicit mapping χγ : R(γ) → C(γ) such that peel(χγ (x)) = x:

χ0 : ∅ → ∅ ( ! 1 if ∃y ∈ x : g = χα(y) χα+1(x) = g 7→ 0 otherwise

χλ(x) = χrank(x)+1(x) She proves this by induction on γ by assuming that the statement holds for ordinals smaller than γ. For γ = 0 and γ = λ this is trivially true. When γ = α + 1 we see that ( ! 1 if ∃y ∈ x : f = χα(y) peel(χα+1(x)) = peel f 7→ 0 otherwise

= {peel(g) | g ∈ C(α) ∧ ∃y ∈ x : g = χα(y)}

= {peel(χα(y)) | y ∈ x} IH= {y | y ∈ x} = x. On a second glance she sees that that the peel function suggests a generalization of the definition. By relaxing the criterion in the successor case from f : C(α) → {0, 1} to f ⊆ C(α) × {0, 1} we get a new hierarchy C0(α), of what she calls characteristic relations. We still interpret peel(f) = x as f representing x and since the change from C(α) is a relaxation, C(α) ⊆ C0(α) and we still have representatives for all sets. What happens if we replace {0, 1} with the poset of coin tossing outcomes P in C0(α)? If we interpret hx, pi ∈ y as “the object w x represents is a member of the object that y represents, if p is true” then membership depends on information outside of y. Until all coin tosses are done, we cannot know for sure whether or not object a is a member of object b. We could, if we knew the first n outcomes, determine some membership relations and perhaps deduce some properties of the objects. Of a whim, she decides to name these structures P -names since they name the objects before all the coin tosses are done. Looking more closely, 1P ∈ P corresponds to 1 ∈ {0, 1} but 0 has no substitute in P since every event in P is possible. In hindsight, a characteristic relation does not need any zeroes, any and all pairs with a zero can be removed without changing what the relation represents, all the way down. This convinces her that any set can be represented by a P -name. In fact, every characteristic relation with no zeroes can be transformed into a in terms of peel equivalent P -name by replacing 1 with 1P . She thus feels confident that something similar to χγ is definable for the P -names. With this Zermela has satisfied her thirst for math for the day and goes on her merry way.

Deﬁnition 84. The class of P-names is deﬁned by recursion over the ordinals and is denoted by VP.

P0 = ∅ Pα+1 = {f ⊆ Pα × P} [ Pλ = Pα (for limit ordinals λ) α<λ [ VP = Pα α∈Ord To show that being a P-name is absolute we will deﬁne a predicate by recursion and show that this predicate is absolute and that it is equivalent to being a member of VP.

def Deﬁnition 85. P-name(x) ⇔ [(x is a binary relation) ∧ ∀hy, zi ∈ x (P-name(y) ∧ z ∈ P)]

Lemma 86. If α ≤ β then Pα ⊆ Pβ. Proof. By induction on β. When β = 0, α = β or when β is a limit, the statement is trivially true. The case of interest is thus β = γ + 1. Let f ∈ Pα where α ≤ γ. By deﬁnition of Pα there is then a δ ≤ α such that IH f ⊆ Pδ × P ⊆ Pγ × P, and by deﬁnition of Pγ+1 = Pβ, f ∈ Pβ.

21 Proposition 87. For all sets x, P-name(x) is equivalent to x ∈ VP. Proof.

P-name(x) ⇒ x ∈ VP : By induction on the rank of x, take as induction hypothesis that the implication holds for all y with smaller rank than x. Assume P-name(x). We then know that every element of x is a pair hy, pi with P-name(y) and p ∈ P. The induction hypothesis then gives us an

αy such that y ∈ Pαy . Let β = suphy,pi∈x αy. By the previous lemma, all such y’s then lie in Pβ and thus x ⊆ Pβ × P, or equivalently x ∈ Pβ+1. x ∈ VP ⇒P-name(x): Assume that the implication holds for P-names with rank smaller than τ ∈ VP. For τ ∈ VP we know τ is a binary relation where any hπ, pi ∈ τ satisﬁes p ∈ P. By the induction hypothesis P-name(π), thus P-name(τ).

Proposition 88. The predicate P-name(x) is absolute. Proof. Recall the comment at the end of section 4, saying that formally, predicates deﬁned by recursion are really class functions taking the value 1 if the predicate is true for x and 0 if it is false. Theorem 75 then states that if 1. R well-founds A, 2. A, R and F : A × V → V are absolute for M,

3.( R is set-like)M and 4. ∀x ∈ AM (pred(A, x, R) ⊆ M) then G such that ∀x ∈ A [G(x) = F(x, G pred(A, x, R)] is absolute for M. We simply prove that these conditions are met. Here A = V, aRb ⇔ a ∈ dom(b) and our F is implicit in the definition of the predicate P-name(x). 1. According to lemma 3 a relation R is well-founded if and only if there is no infinite descending R-sequence. An infinite descending R-sequence implies an infinite descending ∈-sequence since if x ∈ dom(y) there is a z such that x ∈ {x} ∈ {{x}, {x, z}} = hx, zi ∈ y. Hence R is well-founded by the axiom of foundation.

2. The class V, the class relation R and the class function F are all deﬁned using only ∆0-concepts and are therefore absolute. 3. Clearly, pred(V, x, R) = dom(x) = domM (x). Since M is a model of ZFC, for every x ∈ M, domM (x) ∈ M.

4. By transitivity of M, dom(x) ⊆ M.

Deﬁnition 89. The set of P-names inside M are denoted by M P.

We go on to deﬁne the valuation and encoding functions that are to P-names as peel and χγ are to characteristic relations.

Deﬁnition 90. The valuation function val : VP × P(P) → V is deﬁned by val(τ, S) = {val(π, S) | ∃hπ, pi ∈ τ : p ∈ S}.

When we apply val on a syntacticly small expression (like a variable with a possible subscript) we will use the terse notation τS instead of val(τ, S). In the few cases of a syntactically large expression (such as a set comprehension) the notation val({...... },S) will be used.

22 Deﬁnition 91. The encoding function encode : V → VP is deﬁned by recursion over ∈ as

encode(x) = {hencode(y), 1Pi | y ∈ x}. As with the valuation function we use an alternative notation on syntacticly small expressions, namely xˆ instead of encode(x).

When we writex ˆS we mean (ˆx)S.

Proposition 92. For any x ∈ V and S ⊆ P where 1P ∈ S,x ˆS = val(encode(x),S) = x.

Proof. By induction over ∈ with the hypothesis that for y ∈ x,y ˆS = y we get

IH xˆS = val({hy,ˆ 1Pi | y ∈ x},S) = {yˆS | y ∈ x} = {y | y ∈ x} = x.

Lemma 93. For every x ∈ M,x ˆ ∈ M P. Proof. As any pedantic or doubtful reader will show, the class function ‘encode’ satisﬁes theorem 75 so for x ∈ M, encode(x) = encodeM (x) ∈ M.

Furthermore,x ˆ is a P-name by construction, hencex ˆ ∈ M P. We are now ready to construct the forcing model.

Definition 94. Our forcing model M[G] is defined as M[G] = {πG | π ∈ M P}. Our forcing model M[G] is then the concrete set we get if we say that the information that is true in P is that in our generic filter G.10 We can immediately prove a few properties of M[G]. Proposition 95. M ⊆ M[G] Proof. By lemma 93. Proposition 96. The set M[G] is transitive.

Proof. If x ∈ y ∈ M[G] there are by the deﬁnition of M[G] some P-names π, τ ∈ M P such that

hπ, pi ∈ τ, πG = x, τG = y and p ∈ G.

Hence x = πG ∈ M[G]. Proposition 97. The set M[G] is countable.

Proof. It is at least countable since M ⊆ M[G]. By the surjection val(·,G): M P → M[G] we get M ⊇ M P ≥c M[G] so it is also at most countable.

Theorem 98. The forcing model M[G] is a strict extension of M in the sense that M ( M[G]. Speciﬁcally, G ∈ M[G] but G/∈ M.

Proof. We have seen that M ⊆ M[G] in proposition 95 and G/∈ M in lemma 81. To prove G ∈ M[G] is thus what is left. The set ∆ = {hp,ˆ pi | p ∈ P} is deﬁnable by a ∆0-formula so by the axiom of comprehension inside M and lemma 73, ∆ ∈ M P. Using lemma 92 we see that

∆G = {pˆG | p ∈ G} = {p | p ∈ G} = G,

hence G ∈ M[G]. We can at this point prove that M[G] models some of ZFC, but ﬁrst we deﬁne two helpful shorthands.

10The reason to pick a generic ﬁlter as our chosen “truth” is as far as the author can see technical and unsatisfying.

23 Deﬁnition 99. Let up(x, y) (short for unordered pair) denote {hx, 1Pi, hy, 1Pi} and op(x, y) (short for ordered pair) denote {hup(x, x), 1Pi, hup(x, y), 1Pi}.

Proposition 100. For P-names τ and π, val(up(τ, π),G) = {τG, πG} and val(op(τ, π),G) = hτG, πGi.

Proposition 101. The operations up and op are ∆0 and hence absolute. Lemma 102. If π, τ ∈ M P then op(π, τ) ∈ M P and up(π, τ) ∈ M P.

Proof. By lemma 73, both up(π, τ) and op(π, τ) are elements of M. The set up(π, τ) is clearly a P-name and therefore the set up(π, τ) is also a P-name. Theorem 103. The axioms of extensionality, inﬁnity, pairing and foundation are true in M[G]. Proof. This is an excellent time to skim through these axioms in the appendix. Extensionality We have shown that M[G] is transitive. If (∀z(z ∈ x ↔ z ∈ y))M[G] is true for x, y ∈ M[G] then for any z ∈ x, z ∈ M[G] so z ∈ y. This means that x ⊆ y and similarly for x ⊇ y so x = y. Pairing

Let x, y ∈ M[G] and σ, τ ∈ M P be such that σG = x and τG = y. By propositions 101 and 102, {x, y} = val(up(σ, τ),G) ∈ M[G]. Inﬁnity

Let φ(x) be the ∆0-formula ∅ ∈ x ∧ ∀y ∈ x(S(y) ∈ x). The axiom of inﬁnity is the statement Inf ≡ ∃x φ(x). By construction of M (theorem 72), we know that the axiom of inﬁnity is true in M and since M ⊆ M[G],

InfM ⇔ ∃x ∈ Mφ(x) ⇒ ∃x ∈ M[G]φ(x) ⇔ InfM[G] .

Union We wish to show that for any set x ∈ M[G] there is a set U ∈ M[G] containing all x00 where x00 ∈ x0 ∈ x for some x0, or equivalently, that S x ⊆ U.11 Pick an x ∈ M[G] and an accompanying S τ ∈ M P such that x = τG. We shall see that the valuation of the P-name σ = dom(τ) is such a U. To see that σ is a P-name, note that for every hπ, pi ∈ σ, π is a P-name and p ∈ P so σ satisﬁes the P-name predicate. By lemma 73 and absoluteness of union and dom(·), σ ∈ M P. Recall that 0 00 00 0 0 0 00 00 0 τG = x, so for arbitrary x , x where x ∈ x ∈ x, we can pick hτ , p i ∈ τ and hτ , p i ∈ τ such 0 00 0 0 00 00 00 00 00 00 that p , p ∈ G, x = τG and x = τG. Then hτ , p i ∈ σ and therefore x = τG ∈ σG. Hence σG is an acceptable U. Foundation By ﬁxing x ∈ M[G] in the foundation axiom

∀x [∃y(y ∈ x) → ∃y(y ∈ x ∧ ¬∃z(z ∈ x ∧ z ∈ y))]

we see that the inner formula only contains bounded quantiﬁers and is therefore absolute by lemma 58. Since the inner formula is true for all sets x it is speciﬁcally true for all x ∈ M[G].

To prove that the rest of the axioms also are modelled by M[G] we need a technical aid, the forcing relation. The forcing relation, written as p φ(τ1, . . . , τn) for some p ∈ P, a formula φ(x1, . . . , xn) and P-names τ1, . . . , τn is a formula which roughly states that “if p is true in the strange world where membership is determined by what events in P occurs and Murphys law is true then we can deduce that φ(τ1, . . . , τn) is true”. More concretely, we will prove the forcing theorem that states

M M[G] ∃p ∈ G(p φ(τ1, . . . , τn)) ⇔ φ (τ1G, . . . , τnG) 11This is not a typo, we are indeed searching for a superset of the union. If you wish, look up the axiom of union in the appendix.

24 where τ1G means (τ1)G, and so on. In other words, this theorem tells us that every true statement (in M[G]) is forced (in M) and every forced statement (in M) is true (in M[G]). If we view the forcing relation as a kind of deduction relation where an event is guaranteed to occur if it is a dense set then the forcing theorem is akin to a combination of the soundness theorem (everything forced is true) and the completeness theorem (everything true is forced) for first order logic. The forcing theorem will make it almost trivial to prove that M[G] models the rest of ZFC. In the definition of many sets will be required to be dense below p. The relevance of this is given by clause (b) in the following lemma. If this discussion and the following lemma is not enough to get a grip on the definition of the forcing relation, it may help to suspend disbelief and assume that the forcing theorem is true while reading the definition.

Lemma 104. Let E ⊆ P and E ∈ M. (a) Either G ∩ E 6= ∅ or ∃p ∈ G∀e ∈ E(p ⊥ e). (b) If p ∈ G and E is dense below p then G ∩ E 6= ∅. Proof. Let D = {p ∈ P | ∃e ∈ E(p ≤ e)} ∪ {p ∈ P | ∀e ∈ E(p ⊥ e)} . | {z } | {z } D1 D2 D is dense since given p ∈ P either

• p is compatible with something in E and an extension of p lies in D1, or

• p is incompatible with all e ∈ E and p ∈ D2.

By lemma 73 and absoluteness of dense below, D ∈ M and thus G ∩ D 6= ∅. If there is p ∈ G ∩ D1 then e ∈ E such that p ≤ e exists, yielding e ∈ G ∩ E since G is closed upwards. If on the other hand p ∈ G ∩ D2 we have p which satisfies ∀e ∈ E(p ⊥ e), proving (a). To prove (b), assume towards a contradiction that there is q ∈ G ∩ D2. Let r be an extension of p and q in G. E is dense below p so there is some e ≤ r ≤ q contradicting that q is incompatible with all of E. Since G ∩ D 6= ∅ and G ∩ D2 = ∅, there must be some q ∈ G ∩ D1, showing that for some e ∈ E, q ≤ e. Moreover G is a filter, so e ∈ G and G ∩ E 6= ∅. In one of the clauses of the forcing relation we need a well-founded relation to make the definition by recursion well-defined.

Deﬁnition 105. The relation R is deﬁned on V × V as  P-name(σ) for σ = π1, π2, τ1, τ2 def  hπ1, π2iR hτ1, τ2i ⇔ π1 ∈ dom(τ1)  π2 ∈ dom(τ2).

Lemma 106. The relation R is well-founded and absolute.

Proof. It is absolute because it is deﬁned using only the absolute predicate P-name and ∆0-concepts. To see that it is well-founded, assume towards a contradiction that there is a sequence where hπn+1, τn+1iR hπn, τni for every n ∈ N. For well chosen pi, i ∈ N, it then holds that

π0 3 hπ1, p1i = {{π1}, {π1, p1}} 3 {π1} 3 π1 3 hπ2, p2i 3 · · · 3 πn 3 · · · .

This contradicts the well-foundedness of ∈ according to lemma 3.

Deﬁnition 107 (The Forcing Relation). The recursive deﬁnition in clause (a) is over R . Clause (a) should be compared to the axiom of extensionality.

(a) p τ1 = τ2 if = (α) for all hπ1, s1i ∈ τ1, Forcingα (p, τ2, π1, s1) is dense below p, where

= Forcingα (p, τ2, π1, s1) = {q ≤ p | q ≤ s1 → ∃hπ2, s2i ∈ τ2(q ≤ s2 ∧ q π1 = π2)},

25 = (β) and for all hπ2, s2i ∈ τ2, Forcingβ (p, τ1, π2, s2) is dense below p, where

= Forcingβ (p, τ1, π2, s2) = {q ≤ p | q ≤ s2 → ∃hπ1, s1i ∈ τ1(q ≤ s1 ∧ q π1 = π2)}.

(b) p τ1 ∈ τ2 if the following set is dense below p:

{q ∈ P | ∃hπ, si ∈ τ2 (q ≤ s ∧ q π = τ1)}.

p φ(τ1, . . . , τn) and p ψ(τ1, . . . , τn).

(d) p ¬φ(τ1, . . . , τn) if there is no q ≤ p such that q φ(τ1, . . . , τn).

(e) p ∃x φ(x, τ1, . . . , τn) if the following set is dense below p:

{q ∈ P | ∃σ ∈ VP (q φ(σ, τ1, . . . , τn)}.

This definition is not only complicated but metalogically somewhat troublesome. The problem is as usual that first order logic does not allow definitions by recursion on formulas. We get around this by having a first order formula ForcesM, ,≤ (v, φ) representing (v φ) over the model M and forcing poset P P for each formula φ and variable v. As an example, the formula ForcesM, ,≤ (v, ψ ∧χ) should be defined P P P as ForcesM, ,≤ (v, ψ) ∧ ForcesM, ,≤ (v, χ). The rest is defined in a similar fashion. P P P P

Lemma 108. The formula φ(p, τ1, τ2) ≡ p τ1 = τ2 is absolute. Proof. Theorem 75 states that if 1. R well-founds A, 2. A, R and F : A × V → V are absolute for M, 3.( R is set-like)M and 4. ∀x ∈ AM (pred(A, x, R) ⊆ M)

then G such that ∀x ∈ A [G(x) = F(x, G pred(A, x, R)] is absolute for M. We simply prove that P P these conditions are met. Here A = V × V , R = R and our F is implicit in (a).

1. That R well-founds A has already been proven in lemma 106. 2. By proposition 87 and 88, x ∈ VP is absolute and by absoluteness of z = hx, yi, so is VP × VP.

R is absolute by lemma 106. F is absolute because

• (D is dense below p) is ∆0, • the sets in (α) and (β) are deﬁned by ∆0-formulas (the occurence of q π1 = π2 is translated into a function application (∆0) of a variable that should contain the graph of G up to but not including hτ1, τ2i), • the quantiﬁcations in (α) and (β) are bounded. 3. It is from ZFC provable that

P P pred(V × V , hτ1, τ2i, R ) = {hπ1, π2i ∈ τ1 × τ2|π1 ∈ τ1 ∧ π2 ∈ τ2} M is a set. By inference inside M,(R is set-like) . M 4. Let hπ1, π2i ∈ A, hτ1, τ2i ∈ A and hπ1, π2iR hτ1, τ2i. Then π1 ∈ dom(τ1) and π2 ∈ dom(τ2) so by transitivity of M, π1, π2 ∈ M. A few applications of the axiom of pairing yields hπ1, π2i ∈ M.

Lemma 109. The formula φ(p, τ1, τ2) ≡ p τ1 ∈ τ2 is absolute.

26 Proof. The set {q ∈ P | ∃hπ, si ∈ τ2 (q ≤ s ∧ q π = τ1)} is absolute since p π1 = π2 is absolute and the rest of the formula inside of the set comprehension is deﬁned using bounded quantiﬁers. A given set being dense below p in the implicit poset P is also absolute. Hence, p τ1 ∈ τ2 is absolute.

Lemma 110 (Monotonicity of ). If p φ(τ1, . . . , τn) and r ≤ p then r φ(τ1, . . . , τn). Proof. The proof is by induction on formulas. = φ(τ1, τ2) ≡ τ1 = τ2: Fix a hπ1, s1i ∈ τ1 and consider Forcingα (p, τ2, π1, s1) in the deﬁnition of . This set is by assumption dense below p and for r ≤ p,

= = Forcingα (r, τ2, π1, s1) = Forcingα (p, τ2, π1, s1) ∩ {q ∈ P | q ≤ r}. = By the deﬁnition of dense below, Forcingα (r, τ2, π1, s1) is thus dense below r. The (a)(β) case is similar. Thus r τ1 = τ2.

φ(τ1, τ2) ≡ τ1 ∈ τ2: The set in (b) is unchanged by substituting p for r so denseness below p implies denseness below r, hence r τ1 ∈ τ2. φ ≡ ψ ∧ χ: True by the induction hypothesis and the deﬁnition of . φ ≡ ¬ψ: There is no q ≤ p such that q ψ, in particular no such q ≤ r. φ ≡ ∃x ψ(x): If the set in (e) is dense below p it is also dense below r ≤ p.

Once again, by τ1G we mean (τ1)G. M M[G] Lemma 111. If p ∈ G, τ1, τ2 ∈ M P and (p τ1 = τ2) then (τ1G = τ2G) . Proof. Since (p σ = π) and (x = y) are absolute we can drop the relativizations and show that if (p τ1 = τ2) then τ1G = τ2G. The proof is by induction on R . It is enough to prove τ1G ⊆ τ2G, then τ1G ⊇ τ2G follows by symmetry. We need to show that for each hπ1, s1i ∈ τ1 where s1 ∈ G there is a 0 hπ2, s2i ∈ τ2 such that s2 ∈ G and π1G = π2G. Take an extension p ∈ G of p and s1. By monotonicity, 0 p τ1 = τ2 so by the deﬁnition of forcing the set = 0 0 Forcingα (p , τ2, π1, s1) = {q ≤ p | q ≤ s1 → ∃hπ2, s2i ∈ τ2(q ≤ s2 ∧ q π1 = π2)} 0 0 is dense below p . Call this set D. Lemma 104 then tells us there is a r ∈ G ∩ D. Since r ≤ p ≤ s1 and r ∈ D there is a hπ2, s2i ∈ τ2 such that r ≤ s2 and r π1 = π2. By the induction hypothesis, π1G = π2G. Finally, because r ∈ G, π2G ∈ τ2 so τ1G ⊆ τ2G.

M[G] M Lemma 112. If (τ1G = τ2G) then there exists a p ∈ G such that (p τ1 = τ2) . Proof. As in the previous proof the formulas are absolute so we may drop the relativizations and we do

induction over R . The induction is carried out by deﬁning a set S ⊆ P and proving that (a) S lies in M, (b) S is dense and

Assuming (a), (b) and (c) are true, genericity of G guarantees a p ∈ G ∩ S and by (c) p τ1 = τ2 as 0 0 desired. S is deﬁned as the set of all r ∈ P where one of ( ), (α ) or (β ) below holds.

( ) r τ1 = τ2 0 (α ) ∃hπ1, s1i ∈ τ1 r ≤ s1 ∧ ∀hπ2, s2i ∈ τ2∀q ∈ P ([q ≤ s2 ∧ q π1 = π2] → q ⊥ r) 0 (β ) ∃hπ2, s2i ∈ τ2 r ≤ s2 ∧ ∀hπ1, s1i ∈ τ1∀q ∈ P ([q ≤ s1 ∧ q π1 = π2] → q ⊥ r)

27 (a) S is absolute since all quantiﬁers are bounded and subformulas such as q ⊥ r and p π1 = π2 are absolute. Thus S ∈ M.

(b) To show denseness we take a p ∈ P and ﬁnd an extension of p in S. If p τ1 = τ2 then p will do. Otherwise p 1 τ1 = τ2 so either (α) or (β) in the deﬁnition of the forcing relation does not hold. Assume (α) is false, i.e. for some hπ1, s1i ∈ τ1 the set

{q ≤ p | q ≤ s1 → ∃hπ2, s2i ∈ τ2(q ≤ s2 ∧ q π1 = π2)} is not dense below p. By deﬁnition of dense below there is an r ≤ p such that

∀t ≤ r(t ≤ s1 ∧ ∀hπ2, s2i ∈ τ2 ¬(t ≤ s2 ∧ t τ1 = τ2)). 0 To see that r fulfills (α ), note that r ≤ s1 and that there is no q ∈ P that fulfills q ≤ s2 ∧q τ1 = τ2. The implication in (α0) is thus vacuously true, r fulfills (α0) and r ∈ S extends p as desired. The case where (β) fails is symmetrical.

(c) In order to show that if r ∈ G ∩ S then r τ1 = τ2, assume towards a contradiction that r ∈ G and 0 0 that r satisﬁes (α ). Take π1, s1 as in (α ). Since r ≤ s1 and r ∈ G, s1 ∈ G holds, hence π1G ∈ τ1G. Fix

•h π2, s2i ∈ τ2 such that s2 ∈ G and π1G = π2G (exists since τ1G = τ2G), • q0 ∈ G such that q0 π1 = π2 (by the induction hypothesis) and • q ∈ G extending q0 and s2 (by G being a filter). According to (α0), q and r are incompatible but by construction they are members of the filter G and therefore compatible. The case where r satisfies (β0) is similarly contradictory. Since r ∈ G ∩ S 0 0 cannot satisfy (α ) nor (β ) it must fulfill ( ), namely that r τ1 = τ2, as desired.

M M[G] M[G] Lemma 113. If p ∈ G and (p τ1 ∈ τ2) then (τ1G ∈ τ2G) . Conversely, if (τ1G ∈ τ2G) then M ∃p ∈ G(p τ1 ∈ τ2) Proof. As previously mentioned, the relativization can be dropped since forcing for ∈ is absolute. Ac- cording to the deﬁnition of p τ1 ∈ τ2, the set

D = {q | ∃hπ, si ∈ τ2(q ≤ s ∧ q π = τ1)}

is dense below p. Take a q ∈ G ∩ D and a hπ, si ∈ τ2 where q ≤ s and q π = τ1. Lemma 111 then tells us that πG = τ1G. Since G is a ﬁlter and q ∈ G, q ≤ s we conclude that s ∈ G. Thus hπ, si ∈ τ2 implies πG ∈ τ2G, hence τ1G = πG ∈ τ2G as desired. For the second part, assume τ1G ∈ τ2G. Then there is a hπ, si ∈ τ2 such that πG = τ1G. According to lemma 112 there is a r ∈ G where r π = τ1. Let p be an extension of r and s in G. We show that, as the deﬁnition of p τ1 ∈ τ2 states, the set

{q | ∃hπ, si ∈ τ2(q ≤ s ∧ q π = τ1)}

is dense below p. Let q ≤ p. Then q π = τ1 by monotonicity since q ≤ p ≤ s. Thus the above set is not merely dense below p, it contains every element smaller than p.

Theorem 114 (The Forcing Theorem). Let φ(x1, . . . , xn) be a formula with all free variables shown and let τ1, . . . , τn ∈ M P. M M[G] 1. If p ∈ G and (p φ(τ1, . . . , τn)) then φ (τ1G, . . . , τnG), M[G] M 2. If φ (τ1G, . . . , τnG) then ∃p ∈ G(p φ(τ1, . . . , τn)) . Proof. The proof is by induction on the structure of φ. The induction hypothesis is that the theorem holds for every subformula. The parameters of subformulas are not written (ψ and χ instead of ψ(τ1, . . . , τn) and χ(τ1, . . . , τn)) when they do not play an active role in the proof and would only clutter the argument.

28 φ ≡ (τ1 = τ2) This is exactly lemma 111 and 112.

φ ≡ (τ1 ∈ τ2) Proven in lemma 113.

M[G] M (ψ ∧ χ) ⇒ ∃p ∈ G(p ψ ∧ χ) Assume (ψ ∧ χ)M[G]. Pulling apart the conjunction results in ψM[G] and χM[G], applying the 0 00 0 M 00 M induction hypothesis yields some p , p ∈ G such that (p ψ) and (p χ) . An extension 0 00 M M p ∈ G of p and p will by monotonicity satisfy (p ψ) and (p χ) . By definition of M therefore also (p ψ ∧ χ) . M M[G] ∃p ∈ G(p ψ ∧ χ) ⇒ (ψ ∧ χ) M Let (p ψ ∧ χ) for some p ∈ G. The definition of p ψ ∧ χ and relativization yields that M M M[G] M[G] (p ψ) and (p χ) . By the induction hypothesis we get ψ and χ and by definition of relativization also (ψ ∧ χ)M[G].

M[G] M (¬ψ) ⇒ ∃p ∈ G(p ¬ψ) Assume (¬ψ)M[G] and consider the set

M M M Dψ = {r ∈ P | (r ψ) ∨ (r ¬ψ) }.

M M M Dψ is dense: for r ∈ P,(r ¬ψ) is defined as (∀q ≤ r (q 1 ψ)) ; either such a q exists and M M M M M (q ψ) , q ∈ Dψ or there is no such q and r ∈ Dφ . Fix a p ∈ G ∩ Dψ . Assuming (p ψ) and M[G] M using the induction hypothesis gives us ψ , contrary to our assumption. The definition of Dψ M therefore guarantees that (p ¬ψ) . M M[G] ∃p ∈ G(p ¬ψ) ⇒ (¬ψ) M M[G] Assume r ∈ G and (r ¬ψ) . Towards a contradiction, also assume that ψ . By the induction M hypothesis there is a q ∈ G such that (q ψ) . An extension p ∈ G of q and r will then fulfill M M M M[G] (p ψ) , contradicting the definition of (p ¬ψ) , namely ∀s ≤ p (s 1 ψ) . Thus (¬ψ) must hold.

M[G] M (∃x ψ(x)) ⇒ ∃p ∈ G(p ∃x ψ(x)) M[G] M[G] Assume (∃x ψ(x)) . Fix a σ ∈ M P where ψ (σG) and apply the induction hypothesis to get M M a p ∈ G such that (p ψ(σ)) . By monotonicity all q ≤ p fulﬁll (q ψ(σ)) . The set

M {q | ∃σ ∈ M P (q φ(σ, τ1, . . . , τn) }

from the deﬁnition of p ∃x ψ(x) (but relativized to M) is not only dense below p as required by the deﬁnition, it even contains all q ≤ p.

M M[G] ∃p ∈ G(p ∃x ψ(x)) ⇒ (∃x ψ(x)) M Let (p ∃x ψ(x)) for some p ∈ G, that is to say the set

M P M D∃ = {r ∈ P | ∃σ ∈ M (r ψ(σ)) }

M M M is dense below p. Fix a r ∈ G ∩ D∃ and let σ be as in the deﬁnition of D∃ . Then (r ψ(σ)) M[G] M[G] and the induction hypothesis yields ψ (σG), witnessing that (∃x ψ(x)) .

We now have the tools needed to prove that M[G] models the remaining axioms of ZFC not proven in theorem 103.

Theorem 115. The axioms of ZFC are true in M[G]. Proof. In theorem 103 we proved that M[G] satisﬁes extensionality, pairing, inﬁnity, union and foundation.

29 Comprehension Schema 12 We need to show that for z, w1, . . . , wn ∈ M[G] and for a formula φ(x, z, w1, . . . , wn), the set M[G] S = {x ∈ z | φ (x, z, w1, . . . , wn)} is a member of M[G]. Let σ, τ1, . . . , τn ∈ M P be such that σG = z, τ1G = w1, . . . , τnG = wn and let M ρ = {hπ, pi ∈ dom(σ) × P | p (π ∈ σ ∧ φ(π, σ, τ1, . . . , τn)) }.

Clearly ρ ∈ M P so ρG ∈ M[G]. We show that ρG = S by proving ρG ⊆ S and ρG ⊇ S.

ρG ⊆ S Assume that x ∈ ρG. By deﬁnition of ρ there is a hπ, pi ∈ ρ such that π ∈ dom(σ), x = πG M and p (π ∈ σ ∧ φ(π, σ, τ1, . . . , τn)) . By the forcing theorem it follows that πG ∈ σG and M[G] φ (πG, σG, τ1G, . . . , τnG), so ρG ⊆ S. M[G] 0 ρG ⊇ S Let x ∈ σG and φ (x, σG, τ1G, . . . , τnG). It is immediate that there is a hπ, p i ∈ σ 0 such that p ∈ G and πG = x. The forcing theorem tells us there is a p ∈ G such that M (p π ∈ σ ∧ φ(π, σ, τ1, . . . , τn)) . By construction of ρ, hπ, pi ∈ ρ and x = πG ∈ ρG. Replacement Schema

We wish to show that for every formula φ(x, y, s, z1, . . . , zn) and σG, τ1G, . . . , τnG ∈ M[G] where all P-names are in M, when M[G] ∀x ∈ σG ∃y ∈ M[G] φ (x, y, σG, τ1G, . . . , τnG) there is a ρ ∈ M P such that M[G] ∀x ∈ σG ∃y ∈ ρG φ (x, y, σG, τ1G, . . . , τnG). 0 We ﬁrst note that for every x ∈ σG there is a π ∈ dom(σ) such that x = πG ∈ σG and a π ∈ M P M[G] 0 such that φ (πG, πG, σG, τ1G, . . . , τnG). The forcing theorem then tells us there is a p ∈ G such 0 M that p φ(π, π , σ, τ1, . . . , τn) . The idea behind the proof is to ﬁnd an α inside of M large enough that RM (α) contains all the needed P-names π0. To construct our ρ, we step inside of M. Consider the class function F :

F : dom(σ) × P → Ord, ( α if α is minimal such that there is a π0 ∈ R(α) ∩ VP where p φ(π, π0, σ, τ , . . . , τ ) F (π, p) = 1 n 0 if no such α exists. The axiom of replacement tells us that the image of F is a set. Let α = sup(range(F )) and P ρ = (R(α) ∩ V ) × {1P}. We now step back out of M. For x ∈ σG with π ∈ dom(σ) such that x = πG ∈ σG there is M[G] 0 by our initial assumption y ∈ M[G] such that φ (πG, y, σG, τ1G, . . . , τnG). Let π ∈ M P be 0 of minimal rank with the property y = πG. By the forcing theorem, some p ∈ G then satisﬁes 0 M 0 (p φ(π, π , sigma, τ1, . . . , τn)) and it follows that hπ , 1Pi ∈ ρ by construction of ρ. Hence 0 y = πG ∈ ρG. Powerset Fix an x ∈ M[G]. We wish to prove that there is a set z ∈ M[G] such that any subset y ∈ M[G] of x is a member of z. Pick a σ ∈ M P such that x = σG. Let ρ = S × P where M S = {τ ∈ M P | dom(τ) ⊆ dom(σ)} = P (dom(σ) × P).

We will show that ρG contains all subsets of x in M[G]. To see that ρ ∈ M, note that dom and cartesian products are absolute and that M is closed under Pm. That ρ is a P-name is seen by applying the P-name(x)-predicate. Fix a µ ∈ M P such that µG ⊆ σG. The P-name M τ = {hπ, pi ∈ dom(σ) × P | (p π ∈ µ) }

is deﬁnable inside M and therefore a member of S. We prove that τG = µG; it then follows that µG = τG ∈ ρG. 12The formula is technically required to not have y as a free variable because in the axiom schema, y plays the role of the set comprehension.

30 M τG ⊆ µG If hπ, pi ∈ τ and p ∈ G then (p π ∈ µ) by construction of τ. By the forcing theorem πG ∈ µG.

τG ⊇ µG Assume x ∈ µG. There is then some π ∈ dom(σ) such that x = πG. Since πG ∈ µG, M by the forcing theorem some p ∈ G satisﬁes (p π ∈ µ) . Then hπ, pi ∈ τ and therefore x = πG ∈ τG. Choice

We wish to show that for any σ ∈ M P, there is a well-order R ∈ M[G] of σG. Pick a bijection γπ 7→ π in M from some ordinal α to dom(σ). Recall that val(op(π, τ),G) = hπG, τGi. Let

ρ = {hop(γcπ, π), pi | hπ, pi ∈ σ}.

Since ρ is deﬁned in absolute terms and with parameters in M, ρ ∈ M P. It follows that ρG is a surjection ρG : A → σG where A = {γπ | hπ, pi ∈ σ} ⊆ α. Because we have already proven that M[G] models all of ZFC except for the axiom of choice we deﬁne the following injective function f inside M[G]:

−1 f : σG → A, f(x) = min(ρG {x}).

Recall that being an ordinal is absolute. Thus σG is bijective with the subset range(f) ⊆ α of an ordinal. The induced order x R y ⇔ f(x) < f(y) on σG then has for any nonempty S ⊂ σG the least element f −1(min f[S]). This order is also deﬁned using only absolute concepts, so R ∈ M[G] and the axiom of choice is true in M[G].

9 The Cardinal Correspondence Between M And M[G]

Recall that definition 82 tells us that Fn(I,J) is defined as the poset of all finite partial functions from I to J ordered by reverse inclusion. Also recall that lemma 83 says that if I,J ∈ M, I is infinite, |J| ≥ 2 and G is Fn(I,J)-generic over M then Fn(I,J) is a forcing poset and S G is a surjective function. Let us consider a Fn(κ × ω, 2)-generic G over M for some κ that is a cardinal in M and let F = S G. Note that F ∈ M[G] because G ∈ M[G] and unions are absolute. We can see F as a κ-sequence of characteristic functions of ω, namely (χα)α∈κ where χα(n) = F (α, n). These are all distinct, since the set Dαβ = {f ∈ Fn(κ × ω, 2) | ∃n ∈ ω f(α, n) 6= f(β, n)} is dense in M (for a finite f it is always possible to find an i such that f is undefined in (α, i) and (β, i)). This means that there are at least κ subsets of ω in M[G]. But there is nothing so far that tells us that just because κ is a cardinal in M it is a cardinal in M[G]. This is not the case for all forcing posets: If M 0 S 0 M we take an Fn(ω, ℵα )-generic filter G over M then G : ω → ℵα is a surjective function by lemma M 0 83. Consequently, ℵα is not a cardinal in M[G ], just another countable limit ordinal. This section is M M[G] M dedicated to sufficient conditions on P for ℵα = ℵα for all α ∈ Ord . Definition 116. A set D is called a ∆-system if there is a set r, called the root of D, such that whenever a, b ∈ D are distinct, a ∩ b = r. Lemma 117 (∆-system lemma). If A is an uncountable family of finite sets there is an uncountable ∆-system D ⊆ A.

Proof. In this proof we use the convention that for any set X, Xn denotes the set {x ∈ X | n = |x|}. We start by reiterating that a countable union of at most countable sets is countable, so for A to be uncountable, there is some n ∈ N where An >c ℵ0. The proof is by induction over this n with 0 0 the induction hypothesis that if m < n, A is a set of ﬁnite sets and Am is uncountable there is an uncountable ∆-system D0 ⊆ A0.

n = 1: The set A1 is by assumption uncountable. It is a ∆-system with root ∅ because {x} 6= {y} implies ⇒ {x} ∩ {y} = ∅.

31 n = m + 1: S Let Ae∈ = {a ∈ A | e ∈ a} for e ∈ A. We divide the proof into the case when there is an e such that Ae∈ is uncountable and when no such e exists.

Ae∈ is uncountable 0 0 Let A = {a \{e} | a ∈ Ae∈}. Then be : Ae∈ → A , be(a) = a \{e} is a bijection and the induction hypothesis rewards us with an uncountable ∆-system D0 ⊆ A0 with a root r0. It −1 0 0 0 follows that be [D ] is a ∆-system with the same cardinality as D and r ∪ {e} as root.

All Ae∈ are at most countable In this case we can simply pick sets that are disjoint until we have uncountably many. To do this formally, let cA be a choice function on P(A) and consider the following function deﬁnition.

F : ℵ1 → AF (α) = cA({a ∈ A | ∀γ < α(a ∩ F (γ) = ∅)})

In order for this deﬁnition to be valid, we need to show that the set comprehension above S is nonempty. For each γ < α the sets a∈F (γ) Aa∈ that have a nonempty intersection with F (γ) are at most countable. Furthermore, every ordinal α < ℵ1 is countable so the set S S γ<α a∈F (γ) Aa∈ is then also at most countable. Since A is uncountable, removing countably many sets from A does not result in an empty set and F is well deﬁned. Finally |range(F )| = ℵ1 > ℵ0, all sets in range(F ) are disjoint and range(F ) ⊆ A is a ∆-system with root ∅.

Definition 118. A set S ⊆ P is an antichain if all elements in S are incompatible. A forcing poset has the countable chain condition (c.c.c) if all antichains are at most countable. Lemma 119. If Fn(I,J) is a forcing poset and J is at most countable then Fn(I,J) has the c.c.c. Proof. Towards a contradiction, assume that S ⊆ Fn(I,J) is an uncountable antichain. Apply the ∆- system lemma to get an uncountable ∆-system D ⊆ S with root r ∈ Fn(I,J). Distinct functions f, g ∈ D then agree in i ∈ dom(r) and, because they are incompatible, disagree in some i0 ∈ dom(f)∩dom(g). Fix an f ∈ D and let dom(f \ r) = {i1, . . . , in}. For arbitrary g, h ∈ D, in a point im ∈ {i1, . . . , in} ∩ dom(g) either g(im) 6= h(im) or h(im) is undefined because otherwise g ∩ h ) r. Thus, im and g(im) uniquely identifies g in D and we can therefore define a surjection ( g if g ∈ D \{f} where g(i ) = j F : J × n → D,F (j, m) = m f if no such g exists.

This gives us a surjection from the countable J ×n to the supposedly uncountable D, a contradiction.

Lemma 120. Let A, B, ∈ M,(P has the c.c.c)M , f : A → B and f ∈ M[G]. There is then a function M F : A → P(B) where F ∈ M, ∀a ∈ A (f(a) ∈ F (a)) and ∀a ∈ A (|F (a)| ≤ ℵ0) .

Proof. In this proof we use the notation p τ(π) = σ as a shorthand for p op(π, σ) ∈ τ. M Take a τ ∈ M P such that f = τG and a p ∈ G such that (p τ is a function) . The function ˆ M F : A → P(B),F (a) = {b ∈ B | ∃q ≤ p(q τ(ˆa) = b) }.

M is then deﬁnable in M and whenever f(a) = b, this is forced in M so b ∈ F (a). To see that (|F (a)| ≤ ℵ0)

for a given a ∈ A, pick a choice function cP : P(P) → P and deﬁne Qa inside M:

Qa : F (a) → P ˆ M b 7→ cP({q ∈ P | q ≤ p ∧ (q τ(ˆa) = b) })

0 0 We wish to show that if b, b ∈ F (a) are distinct, then Qa(b) and Qa(b ) are incompatible. Assume 0 instead that there is a common extension q ∈ P of Qa(b) and Qa(b ). By the generic ﬁlter existence

32 lemma (lemma 80) there is then a P-generic G0 over M containing q. By monotonicity and the forcing theorem we get M[G0] M[G0] 0 M[G0] (τG0 is a function) ∧ (τG0 (a) = b) ∧ (τG0 (a) = b ) . 0 0 Hence, distinct b, b ∈ F (a) means incompatible Qa(b),Qa(b ). Therefore Qa[F (a)] is an antichain and because incompatible elements are distinct, Qa is a bijection. By the c.c.c of P inside M the antichain Qa[F (a)] is at most countable in M. Hence F (a) is also at most countable in M. Finally it is time to use the results about coﬁnality in subsection 5.3!

Deﬁnition 121. A forcing poset P preserves cardinalities if for any P-generic G over M, ∀α ∈ OrdM (α is a cardinal)M ⇔ (α is a cardinal)M[G].

It preserves coﬁnalities if for any P-generic G over M and any limit ordinal γ in M cfM γ = cfM[G] γ.

As it turns out, it is a stronger criterion for an ordinal to be a cardinal in M[G] then in M. This is because being a bijection is absolute and that being a cardinal is deﬁned as not being bijective to any smaller ordinal. Thus, if an ordinal is a cardinal in M[G], it is also a cardinal in M. Hence it is enough to check that ∀α ∈ OrdM (α is a cardinal)M ⇒ (α is a cardinal)M[G]

M M[G] to see if P preserves cardinals. Moreover, since ω is absolute, ω = ℵ0 = ℵ0 and a simple proof by induction shows that regardless of P, all the ﬁnite cardinals are preserved so we only need to check for the uncountable cardinals in M.

Lemma 122. If P preserves coﬁnalites then P preserves cardinals. Proof. Let α ≥ ω be a regular cardinal in M. By assumption cfM[G] α = cfM α = α so α is a regular cardinal in M[G] as well. We remind the reader that all successor cardinals are regular. Proposition 50 tells us the successor cardinals in a limit cardinal is unbounded. This together with the absoluteness of union yields M [ M [ M[G] M[G] ℵλ = ℵγ+1 = ℵγ+1 = ℵλ . γ+1<λ γ+1<λ

M for limit cardinals ℵλ in M.

Lemma 123. Let P be such that for every P-generic G over M it holds that whenever κ is a regular uncountable cardinal in M, κ is regular in M[G]. Then P preserves cofinalities. Proof. We prove the lemma separately for 0, finite successor ordinals, ω and the uncountable regular ordinals. The cofinality of 0 and successor ordinals are trivially preserved. Since ω is absolute and cf ω = ω, cfM ω = ω = cfM[G] ω. According to lemma 46, all regular ordinals are cardinals so the regular uncountable ordinals are preserved by assumption. Let γ ∈ M be a limit ordinal, and (κ = cf γ)M . Note that κ is regular in M and therefore also in M[G]. According to lemma 42 inside M there is a nondecreasing cofinal map f : κ → γ in M and hence in M[G]. Then

cfM[G] γ = cfM[G] κ (lemma 48 inside M[G] using f) = κ (κ is regular in M and M[G]) = cfM γ so the coﬁnality is preserved in all cases.

Lemma 124. If (P has the c.c.c)M then P preserves coﬁnalities and cardinals.

33 Proof. Assume towards a contradiction that P does not preserve cofinalities. By the previous lemma there is then a P-generic G and a κ ∈ M uncontable in M that is regular in M but not in M[G]. By definition of regularity and cofinality there is a cofinal f : α → κ in M[G] where α < κ. By lemma 120 there is a function F : α → P(κ) in M such that

M ∀γ ∈ α (f(γ) ∈ F (γ)) and ∀γ ∈ α (|F (α)| ≤ ℵ0) .

S M M Then the set S = γ∈α F (γ) has cardinality at most |α × ℵ0| = |α| inside M. Furthermore, S is unbounded in κ since range(f) ⊆ S, so a bijection b : |S|M → S in M is also a cofinal map b : |S|M → κ. The domain of the cofinal b : |S|M → κ then contradicts (κ is regular)M since |S|M = α < κ. Thus, no such κ ∈ M exists so P preserves cofinalities and by lemma 122 also cardinalities. Theorem 125. The theory ZFC + ¬CH is equiconsistent with ZFC.

M Proof. Let P = Fn(ℵ2 × ω, 2). Since 2 is at most countable in M we can apply lemma 119 inside of M to see that P has the c.c.c in M. By the previous lemma it therefore preserves cardinalities. For any S M M[G] P-generic G over M, G encodes ℵ2 = ℵ2 subsets of ω in M[G]. This yields the wanted conclusion, M[G] namely (ℵ2 ≤ |P(ω)|) .

References

[1] Paul J. Cohen. The independence of the continuum hypothesis. Proc. Nat. Acad. Sci. USA, 50(6):1143–1148, 1963. [2] Thomas Jech. Set theory, the third millennium edition. Springer Monographs in Mathematics. Springer-Verlag, 2003. [3] K. Kunen. Set Theory: An Introduction to Independence Results. Elsevier Science Publishers BV (North-Holland), Amsterdam, The Netherlands, 1980. [4] Yiannis Moschovakis. Notes on set theory, 2006.

[5] Philipp Rothmaler. Introduction to Model Theory. Cambridge University Press, 2000. [6] Helmut Schwichtenberg and Stanley S. Wainer. Proofs and computations. Cambridge University Press, 2011.

34 A The Axioms of ZFC

Extensionality Two sets are equal if they have the same members.

∀x∀y[∀z(z ∈ x ↔ z ∈ y) → x = y]

Foundation All sets are well-founded or alternatively, every set has a ∈-minimal element.

∀x [∃y(y ∈ x) → ∃y(y ∈ x ∧ ¬∃z(z ∈ x ∧ z ∈ y))]

Comprehension schema For all sets z, w1, . . . , wn the collection of all x ∈ z where φ(x, z, w1, . . . , wn) holds is a set. This set is denoted by {x ∈ z | φ(x, z, w1, . . . , wn)}. Formally, for each formula φ with free variables among x, z, w1, . . . , wn not including y, the following is an axiom:

∀z∀w1, . . . , wn∃y∀x[x ∈ y ↔ x ∈ z ∧ φ(x, z, w1, . . . , wn)].

This is therefore not an axiom but an axiom schema. Pairing For any sets x and y there is a set containing both of them. An application of comprehension then yields that {x, y} is a set. ∀x∀y∃z(x ∈ z ∧ y ∈ z) S S Union For any set of sets F the union F = x∈F x is a subset of some set U. ∀F ∃U ∀x∀Y [(x ∈ Y ∧ Y ∈ F ) → x ∈ U]

Replacement Schema Fix sets A, w1, . . . , wn. Assume φ(x, y, w1, . . . , wn) behaves like a function on A, that is for every x ∈ A there is a unique y such that φ(x, y, w1, . . . , wn). Then {y | x ∈ A ∧ φ(x, y, w1, . . . , wn)}, the image of A under φ is a subset of a set. (An application of comprehension again makes the image itself a set.)

Formally, for all formulas φ with free variables among x, y, A, w1, . . . , wn not including Y , the following is an axiom:

∀A∀w1, . . . , wn[∀x ∈ A∃y φ → ∃Y ∀x ∈ A∃y ∈ Y φ]

Inﬁnity Let S(x) = x ∪ {x}. ∃x(∅ ∈ x ∧ ∀y ∈ x S(y) ∈ x)

Power Set For any set x there is a set y such that y contains all subsets of x.

∀x∃y∀z(z ⊂ x → z ∈ y)

Note that this y is not necessary the power set of x, but the set {z ∈ y | z ⊆ x} is and exists by comprehension. Choice The axiom of choice has under the previous axioms three common equivalent forms. S Choice Function If F is a set of nonempty sets there is a function cF : F → F such that for all x ∈ F , cF (x) ∈ x. Well-ordering Principle All sets have a well-order.

Zorns Lemma Let (P, ≤P ) be a poset. A directed set is a set S such that if x, y ∈ S there is a z ∈ S such that x, y ≤P z. A maximal element is an element that is not smaller than any other element. Zorns lemma states that if all directed subsets of P has an upper bound in P then there exists a maximal element in P .