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Home , Continuum (set theory), Continuum hypothesis, Generic filter, Transitive set

The Hypothesis and the of Theory

Rachel Minster November 2015

Advisor: Dr. Alan Dow Department Honors Program University of North Carolina at Charlotte

1 Introduction

This research project will address the independence of the Continuum Hypothe- sis from the axioms of . In order to do this, we will explore fundamental set theory, G¨odel’sconstructible , and Cohen’s method of .

1.1 Independence and A statement is independent if neither it nor its negation are provable from the same system. In practice, this is equivalent to both the statement and its negation being consistent. This involves two separate arguments, and if both are satisfied, the result is independence. The system of set theory, the ZFC axioms, is a basis for all proofs in set theory. We take the axioms to be true when proving statements in set theory, so we assume that the axioms together are consistent. With this, we are interested in what the ZFC system can or cannot prove. For instance, the axioms cannot prove that they are consistent. This is due to G¨odel’sSecond Incompleteness Theorem, which states that a sufficiently expressive system cannot be used to prove its own consistency.

1.2 Basic Definitions

We take a model of ZFC using ∈ and =. This is our set-theoretic universe, V. Definition 1.1. A set is an of V. Definition 1.2. A is a collection of all x that satisfy a formula φ(x, y1, y2, ..., yn). Classes can be sets if they occur in V, but some do not.

1 Definition 1.3. A set x is transitive if every element of x is a of x. If y ∈ z and z ∈ x, then y ∈ x. Definition 1.4. A well-ordering is a linear order where every nonempty subset has a least element.

1.3 The Axioms The Zermelo-Frankel Axioms with the , abbreviated to ZFC Axioms, are the basis for set theory. ZFC fulfills G¨odel’srequirements for a sufficiently rich axiom system. His First Incompleteness Theorem suggests that, in a good , there will be at least one statement true but not provable. Also, axioms should be ”obviously” true. [7] An interesting fact about the axioms is that they are an infinite system. The axioms stated below make up the standard list, but it has been proven that ZFC is not finitely axiomatizable, so there is no actual finite collection of axioms equivalent to the ZFC system. All the axioms provide a basis for the existence of various sets. Using these, we can build different sets and know they are sets instead of classes. The standard axioms are listed below, and a more intuitive definition is included after each axiom.

Axiom (Axiom of Existence). ∃z z = ∅ The is a set, so the universe is not empty.

Axiom ().

∀x ∀y [x = y ↔ ∀u(u ∈ x ↔ u ∈ y)]

Two sets are equal if they contain the same elements.

Axiom ().

∀x ∀y ∃z (x ∈ z ∧ y ∈ z)

For any two sets, another set exists which contains both.

Axiom ( Axiom).

∀x [x 6= ∅ → ∃z z = {w :(∃y ∈ x)(w ∈ y)}]

A set of sets exists and is a set itself. The class {w :(∃y ∈ t)(w ∈ y)} is referred to as the ”big union”for any term t, denoted as S t.

2 Axiom (Axiom of Foundation).

∀x [x 6= ∅ → (∃y ∈ x)(x ∩ y = ∅)]

Each set is well-founded. This means that each non-empty set has a minimal element under the ∈ relation.

Axiom (Replacement Axiom Scheme). For each formula φ(x, u, v, w1, ..., wk),

∀w1, ..., ∀wk ∀x [∀u ∈ x ∃!v φ → ∃z z = {v : ∃u ∈ x φ}]

The image of a set under any formula defining a function will also be in a set.

Axiom (Comprehension Axiom Scheme). For each formula φ(x, y, w1, ..., wk),

∀w1, ..., wk ∀x ∃z z = {y : y ∈ x ∧ φ(x, y, w1, ...wk)}

All elements in a set satisfying a certain property or formula comprise a set.

Axiom ( Axiom).

∀x ∃z z = {y : y ⊆ x}

We denote {y : y ⊆ x} as P(x). We call P(x) the power set of x, and the axiom states that P(x) is a set itself.

Axiom (Axiom of Infinity).

∃x[x 6= ∅ ∧ ∀y y ∈ x → ∃z ∈ x y ∈ z]

This allows for the existence of infinite sets.

Axiom (Axiom of Choice).

∀X [(∀x ∈ X ∀y ∈ X(x = y ↔ x ∩ y = ∅)) → ∃z (∀x ∈ X ∃!y y ∈ x ∩ z)]

A set can be created from one or more elements of pairwise disjoint sets. Impli- cations of this axiom include that every set can be well-ordered. This becomes important in the construction of new sets and models. Not every theorem in set theory needs the assumption of the Axiom of Choice, but we will frequently assume it for the existence of different types of sets.

3 1.4 Ordinals and Cardinals Definition 1.5. A set is an ordinal if it is transitive and well-ordered by the ∈ relation. We denote the class of all ordinals by ON. There are two types of ordinals. A successor ordinal is an ordinal α of the form β ∪ {β} = β + 1, which we call the successor of some ordinal β. If α = succ(α), then α is called a limit ordinal.

The class of ordinals, ON, is well-ordered by the Axiom of Foundation. Thus there is no infinite descending sequence of ordinals, and there is a least element of any non-empty set of ordinals. Theorem 1.6. Let A be a well-ordereable set. Then A, under its well-ordering, is order-isomorphic to a unique ordinal α. Due to the transitive nature of ordinals, we can define α < β ↔ α ∈ β for any two ordinals α, β. We can also define , multiplication, and exponentiation on the ordinals through transfinite recursion. [7] Let α, β ∈ ON and δ be a limit ordinal. Addition: α + 0 = α α + 1 = succ(α) α + succ(β) = succ(α + β) α + δ = sup{α + β : β < δ}

Multiplication:

α · 0 = 0 α · succ(β) = (α · β) + α α · δ = sup{α · β : β < δ}

Exponentiation:

α0 = 1 αsucc(β) = (αβ) · α αδ = sup{αβ : β < δ}

Definition 1.7. The α such that a well-orderable set A is order-isomorphic to α ∈ ON is called the of A, denoted by |A|. |A| is defined for every A under the Axiom of Choice. Mentioning |X| assumes that X can be well-ordered. Informally and traditionally, cardinality refers to the size of a set. Two sets have the same size if there is a between them. Definition 1.8. An ordinal α is a cardinal if |α| = α.

4 The ordinal referring to the set of natural is denoted by ω, which we use frequently as it is one of the smallest infinite sets. ω is also a cardinal as |ω| = ω. Definition 1.9. A set is countable if its cardinality is at most ω, and uncountable if otherwise. There are many countable finite sets, but we are more interested in the infinite ones. Countable infinite sets all have cardinality ω. Cantor’s discovery of uncountable sets has profound implications, as there are thus infinite sets of different sizes. Notation. We denote the least cardinal greater than an ordinal α by α+. We tend to denote the least cardinal greater than ω as ω1, which is the first uncount- able cardinal. The second is ω2, and so on. Now we can state the , abbreviated CH, which proposed. It seems very plausible, but we cannot prove it to be true or false. [6] Both the statement and its negation are consistent with ZFC, which we will show.

Theorem 1.10 (Continuum Hypothesis). |P(ω)| = ω1

Cantor’s theorem, which we discuss later, states that |P(ω)| ≥ ω1. The hypothesis goes a step further, but this extra step cannot be verified in set theory.

1.5 History Every branch of mathematics has a foundation of axiomatic statements assumed to be true. These express properties and rules for that branch. All mathematical concepts can be expressed in terms of sets and the membership relation. The axiom system of set theory, ZFC, can then be used to derive all mathematics, which makes set theory the foundation of mathematics. The only problems come from infinite sets, as they behave differently than finite sets. This leads to questions about the behavior of infinite sets. Once the behavior of certain finite sets is determined, the succeeding question asks if the same is true for a larger set. This continues into the infinite. If something is true for countable sets, is it then true for uncountable sets? Set theorists strive to answer these questions. Major questions in set theory spark philosophical debates among mathemati- cians. There are multiple theorems and statements that, if true or false, would have huge implications. Unfortunately, they are still undecidable, so we cannot know whether they are true or false. To see the implications, set-theorists take a model of ZFC with whatever statement or combination of statements they wish to use. For example, ZFC +¬CH can be used to see what would happen if the Continuum Hypothesis was not true. Another important statement, Martin’s Axiom, is used the same way. It’s interesting to see that different results come from ZFC + MA than ZFC+ CH. [5].

5 There are also many parts of set theory that lead us to consider mathematics with a philosophical perspective. The nature of infinity, for example, inherently requires philosophical thought. Also, formal proofs are finite and cannot go on infinitely. This is not a usual problem in other branches of mathematics.

1.5.1 Cantor Georg Cantor is considered the founder of modern set theory. He was the first to prove that there are different levels of infinity. He did this by showing that the real numbers are not countable, which implies there are more real numbers than natural numbers. The method he used is commonly known as the diagonal argument, which shows that the natural numbers and real numbers cannot be put in a one-to-one relation. To summarize the proof, he lists all the natural numbers in correspondence with the real numbers. He thens creates another real that differs from all the others by changing at least one decimal place. This does not have a corresponding since they were all used already. Thus the cardinality of the reals is greater than ω. The reals are then uncountable. This result was shocking to the mathematical community because up until this point, many mathematicians tended to believe that infinity belonged to the realms of philosophy or theology. Limits in Calculus may tend towards infinity, but this abstract concept had no precise definition. Cantor’s argument made it possible to define different levels of infinity. Another important theorem Cantor proposed and proved, known as Cantor’s Theorem, is |x| < |P(x)|. He proved that there was no surjection from |x| to P(x). It is sufficient to show that any function from a set to its power set cannot be surjective. Theorem 1.11 (Cantor’s Theorem). There is no bijection mapping x to P(x).

Proof. Given a function f : x → P(x), let y = {z ∈ x : z∈ / f(z)}. Pick a z ∈ x such that f(z) = y. If z ∈ y, then z ∈ f(z), implying z∈ / y. If z∈ / y, then z∈ / f(z), implying z ∈ y. Then there is no z ∈ x where f(z) = y, so f is not surjective.

1.5.2 Hilbert Cantor and his arguments about infinity became much more prominent be- cause of . In an address in 1900 about the unsolved mathematical problems of the day, the first thing Hilbert mentioned was the Continuum Hy- pothesis. He suggested that it was a very plausible theorem although no one had managed to prove it. The plan Hilbert laid out to work towards proving it did contribute to the developing studies on the continuum. Also in this address, Hibert focused on another set theoretical problem. This was the well-ordering of the reals, which proved using the Axiom of Choice. [6]

6 1.5.3 The Axiom of Choice The Axiom of Choice is a controversial axiom because of its critically useful as well as non-intuitive consequences. It allows us to choose objects we cannot fully describe and use them in proofs. The axiom asserts the existence of choice sets without defining their elements, making it different than the other ZF Axioms. This is useful in all branches of mathematics, but in order to use the Axiom of Choice, it must be taken as an axiom. Zermelo first stated the axiom as an explicit assumption. Mathematicians debated whether it was acceptable to assume it, though G¨odelproved that, taking ZF to be consistent, the other axioms did not disprove it. later proved that ZF did not prove the Axiom of Choice either. So the Axiom of Choice is independent from the other axioms. [3] There is a Well-Ordering Theorem which states that every set can be well- ordered. Cantor proposed it and thought it fundamentally true though he never proved it. This is another statement hotly debated by mathematicians as it seems counter-intuitive. How can uncountable sets, for example, have a well- ordering? Zermelo proved it, but needed the Axiom of Choice to do so, so the theorem remained controversial. We take the Well-Ordering Theorem as true, as we take the Axiom of Choice as true. An important result of the theorem is that the real numbers can be well-ordered. [7]

1.5.4 Cohen Paul Cohen received the for his contributions to set theory. He discovered the technique of forcing, which directly led to independence results for both the Continuum Hypothesis and the Axiom of Choice. Through forc- ing, Cohen was able to prove CH and ¬CH consistent from ZF. As G¨odelhad already shown CH to hold in the L, only the negation was necessary, but both can be shown through forcing. His techniques were also used to discover interesting about the continuum. [5]

2 CH is Consistent 2.1 The Universe of ZF Notation. The class {x : x = x}, or all the sets equal to themselves, is denoted by V, which we refer to as the universe. In contrast, the empty set ∅ corresponds with the set {x : x 6= x}.

The set-theoretical universe is denoted as V for Von Neumann, who con- tributed a great deal to a better understanding of it. He came up with a way to break it down into a hierarchy of smaller sets, which allows us to better analyze the axioms. V is a class because the universe is not an element of our universe, which we took as our definition of a set. V ∈ V would violate several axioms. Suppose

7 for instance that x ∈ x. If such an x existed, then {x} exists from the Axiom of Pairing. The only element in this set is x. If x ∈ x, then y ∩ {x} 6= ∅ as the only y ∈ {x} is x. So the Axiom of Foundation is not satisfied and V ∈ V, or V being a set, is not possible.

2.1.1 Russell’s Paradox The paradox considers the class of all sets that are not elements of themselves, mathematically represented by z = {x : x∈ / x}. Is z a set? If z ∈ z, z is by definition not an element of itself, which contradicts our original assumption. If z∈ / z, then z is an element of itself, which again gives us a contradiction. To resolve this paradox, we must conclude that z does not exist as a set in V. Going further, we can say that nothing is an element of itself. So the set {x : x∈ / x} = {x : x = x} = V. This does not exist, so the mathematical universe does not exist as a set. We can, however, construct different sets in V because the axioms we assume provide for that. [7]

2.1.2 The Basics Definition 2.1. A subset S of an ordinal α is cofinal if ∀β ∈ α ∃σ ∈ S β ∈ σ.

Definition 2.2. The cofinality of an ordinal α, denoted by cf(α), is the smallest cardinality of a cofinal subset of α. Intuitively, the cofinality of an ordinal is the number of steps that go up towards the ordinal. This is another way to the relative size of a set, and provides a way to classify infinite cardinals into two categories.

Definition 2.3. An infinite cardinal α is regular if its cofinality is itself, so cf(α) = α. Otherwise, the cardinal is singular. The ordinal ω, for example, is regular, as it takes ω ”steps” to get to ω. + Also, for each infinite ordinal κ, κ is a . So the cardinal ω1 is regular.

2.1.3 Breaking Down the Universe

We can break down V into a hierarchy, which allows us to exploit the well- founded nature of ∈. Definition 2.4. The smallest containing a set x as a subset is referred to as the transitive of x, denoted as trcl(x). is useful to establish induction and recursion schemes for V. Recursion in particular allows us to define a new way to construct sets. Definition 2.5. An ordinal rank can be associated to each set x, denoted by rank(x), by the recursive rule rank(x) = sup {rank(y) + 1 : y ∈ x}, with rank(∅) = 0 and rank(α) = α ∀α ∈ ON.

8 Rank allows us to view each set at a particular level. A smaller rank between two sets implies that the larger set contains the smaller. Using this, we can define a hierarchy of sets in V.

Definition 2.6. The cumulative hierarchy defines Vα for different α ∈ ON. Let δ be a limit ordinal. [7]

V0 = ∅ Vα+1 = P(Vα) [ Vδ = {Vα : α < δ}

This recursively defined cumulative hierarchy allows us to break V into dif- ferent Vα sets. There are several important results of this hierarchy, especially in combination with rank. These include that every Vα is transitive. For any two ordinals α and β with β < α, Vβ ⊆ Vα. Also, if rank(x) < α ∈ ON, then x ∈ Vα. Putting these results together gives us the important theorem that S V = {Vα : α ∈ ON}. Every set has a rank and thus a corresponding Vα, so when every Vα is put together, the resulting class is the universe.

2.1.4 and Reflection Certain formulas can be interpreted differently by different classes. A formula may not hold in one class, but could still be true in another. The idea of relativization defines the of a formula in a particular class. Definition 2.7. Let M be any class and φ any formula. The relativization of φ to M, denoted by φM , is a formula defined as follows. [4] a) (x = y)M is x = y b) (x ∈ y)M is x ∈ y c) (φ ∧ ψ)M is φM ∧ ψM d) (¬φ)M is ¬(φM ) e) (∃xφ)M is ∃x (x ∈ M ∧ φM ) Logical deductions keep their intended meaning through different relativiza- tions. This idea of a formula meaning the same thing in different classes is known as absolutness.

Definition 2.8. Let φ(x1, ..., xn) be a formula. Given two classes M,N with M N M ⊂ N, φ is absolute for M,N if ∀x1, ..., xn ∈ M, φ (x1, ..., xn) ↔ φ (x1, ..., xn). Also, φ is absolute for M if φ is absolute for M and V, meaning ∀x1, ..., xn ∈ M, M φ (x1, ..., xn) ↔ φ(x1, ..., xn). If the formula holds in a given model, we would like to say that the truth reflects to the original class. We can do this through the Reflection Theorem.

Theorem 2.9 (Levy Reflection Theorem). Given any formulas φ1, ..., φn, ∀α ∈ ON ∃β ∈ ON where β ≥ α such that φ1, ..., φn are absolute for Vβ].

9 This theorem states that we can find a big enough β so that any formula is absolute in Vβ. It holds for any cumulative hierarchy. An interesting con- sequence of this theorem is that ZF is not finitely axiomatizable. ZF proves the Levy Reflection Theorem, so if ZF could be expressed in a finite number of axioms, we could find a model of them. This would imply that ZF is consistent, and ZF cannot prove its own consistency. [3]

2.2 The Constructible Universe 2.2.1 Definability We want to precisely define formulas using in order to clearly show their truth relativized to any set. G¨odeloperations allow us to define a function equivalent to any formula. For any set A, Df(A, n) for definable function, takes an A and produces a subset of An for some n ∈ ω. Every formula relativized to A is an element of Df(A, n) for some n. The different types of definable functions can be boiled down to logical operators, which allows us to assert absoluteness. There are three types, listed with their respective operator: Projection (∃), Diagonalization with ∈, and Diagonalization with =. Each definable function can be expressed as one of a doubly enumerated list of functions from ω.

Definition 2.10. En(m, A, n) is defined by recursion on m ∈ ω. [4] i j (1) If m = 2 · 3 with i, j < n, then En(m, A, n) = Diag∈(A, n, i, j). i j (2) If m = 2 · 3 · 5 with i, j < n, then En(m, A, n) = Diag=(A, n, i, j). (3) If m = 2i · 3j · 52, then En(m, A, n) = An r En(i, A, n). (4) If m = 2i · 3j · 53, then En(m, A, n) = En(i, A, n) ∩ En(j, A, n). (5) If m = 2i · 3j · 54, then En(m, A, n) = Proj(A, En(i, A, n + 1), n). (6) If m is not one of these forms, then En(m, A, n) = ∅.

Theorem 2.11. Let φ(x0, ..., xn−1). Then for some m ∈ ω and any set A, {s ∈ An : φA(s(0), ..., s(n − 1))} = En(m, A, n). In other words, for every formula φ, there is an m, n where φA and En(m, A, n) mean the same thing. [4] Definition 2.12. Given a set A, the definable power set of A, denoted by D(A), is the set of of A that are definable from a finite number of elements of A by a formula relativized to A.

2.2.2 Elementary Submodels Definition 2.13. M is a model of a formula φ, denoted by M |= φ, if φM holds. φM is the relativisation of φ to M, or M’s interpretation of φ. Definition 2.14. A is an elementary submodel of B, denoted A ≺ B, if A ⊂ B and ∀n, m[En(m, A, n) = En(m, B, n) ∩ An].

10 Intuitively, this definition can be stated that for all formulas φ with pa- rameters a1, a2, ..., an ∈ M, M |= φ(a1, a2, ..., an) ↔ N |= φ(a1, a2, ..., an). The formal definition, which is legitimate in ZF, is the same as seemingly quantifying over all formulas. We assume the Axiom of Choice for the next theorem, which in particular says that given a B, there is a countable A such that A ≺ B. [7] Theorem 2.15 (L¨owenheim-Skolem Theorem). Let X ⊆ N. Then there is an M such that 1) M ≺ N 2) X ⊆ M 3) |M| ≤ max{ω, |X|}.

We can extract elementary submodels of sets, but not of V as we are working within V. This would violate G¨odel’sSecond Incompletness Theorem.

2.2.3 L L is G¨odel’sconstructible universe, which is, as we will discuss, a model of ZFC and CH. The sets in L are similar to the sets in the cumulative hierarchy with one major difference. Instead of taking the unrestricted power set of the previous set to get the successor, we take only the definable subsets. [4] We can construct each level of L through recursion. Let δ be a limit ordinal. L(0) = ∅ L(α + 1) = D(L(α)) [ L(δ) = {L(α): α < δ}

S Then L = L(α) ∀α ∈ ON. In order to use L, we want the axioms to hold there. It turns out that they do hold, so L is a model of ZFC. As an example, we will show that the Axiom of Comprehension holds in L. It will be important to note that the Levy Reflection Principle also holds for this hierarchy.

Theorem 2.16. Comprehension holds in L.

Proof. Let φ(x, z, v1, ..., vn) be a formula with z, v1, ..., vn ∈ L. To satisfy Com- prehension, we will show that y = {x ∈ z : φ(x, z, v1, ..., vn) holds} is in L. The Levy Reflection Principle allows us to find a big enough β so that φ is absolute for L(β) and z, v1, ..., vn ∈ L(β). Then φ corresponds to some En(m, L(β), n), so y ∈ D(L(β)).

L is then a model of ZFC. Also, in every transitive model, ”being L(α)” is absolute. This is because each L(α) is built from the definable power sets, which consist only of absolute functions. Axiom (Axiom of Constructibility).

V = L

11 This axiom states that every set belongs to L. It is consistent with ZFC, as no finite collection of axioms can disprove it. More importantly, it holds in L, which we will show.

Theorem 2.17. L |= ZF + V = L

Proof. We have already discussed that L is model of ZF. We need to show (V = L)L. Let x ∈ L, and fix α ∈ ON. Since L(α) is absolute for transitive models, and L is a transitive model of ZF, x ∈ L(α) → (x ∈ L(α))L. This means that ∀x ∈ L ∃α ∈ ON so that (x ∈ L(α))L. [4]

The Axiom of Choice holds in L, so L is a model of ZFC. This allows us to prove the following theorem. Theorem 2.18. If α ≥ ω, then |L(α)| = |α|. Proof. We will prove by induction on α. Assume α ≥ ω and ∀β < α (β ≥ ω → |L(β)| = |β|). ∀β < α, |L(β)| ≤ |α| because whenever β < α, |L(β)| = |β| ≤ |α|. If α is a successor ordinal of the form α = β + 1, then |L(β)| = |β| by induction, and |β| = |α| as β is infinite. L(α) = D(L(α)), so |L(α)| = |α|. If α is a limit ordinal, then L(α) = S{L(β): β < α}. Then |L(α)| = max(|α|, sup{|L(β)| : β ∈ α}. By induction, this is max(|α|, sup{|β| : β ∈ α}, which is equal to |α|. Then |L(α)| = |α|.

2.3 CH holds

CH follows when we take V = L because, as we will show, each subset of ω can ω be constructed at a countable level, or P(ω) ⊂ L(ω1). This implies 2 ≤ ω1. This can be strengthened to imply GCH, which is the statement |P(κ)| = κ+ for any cardinal κ. [4] Lemma 2.19 (Mostowski Collapsing Theorem). Let R be a well-founded, ex- tensional relation on a set X. Then there is a unique transitive set M and a unique ∈-isomorphism f : X → M. A useful fact is that any ∈-isomorphism between transitive sets is the iden- tity.

Notation. We will use o(M) to denote ON ∩ M.

Lemma 2.20. Let M be a transitive set. Then o(M) ∈ ON. It is not hard to see that o(M) is an ordinal, as every x ∈ o(M) is an ordinal and therefore transitive. In fact, o(M) is the first ordinal not in M.

Lemma 2.21. There is a finite conjunction χ of axioms of ZF +V = L such that ∀M, if M is transitive and M |= χ, then M = L(o(M)).

12 Proof. There is a finite conjunction of axioms of ZF such that ”being an ordinal, rank, or L(β) is absolute. [4] Let χ be those axioms in conjunction with V = L. Suppose M is transitive and M |= χ. Then (∀x(x ∈ L))M , so M = LM . Now M M S L = {x ∈ M :(∃α(x ∈ L(α))) } = {L(α): α ∈ M}. Because M is transitive and models χ, o(M) is a limit ordinal. This implies that L(o(M)) = S{L(α): α ∈ M}. Then L(o(M) = LM = M. Theorem 2.22. V = L → GCH holds Proof. We first prove that P(L(α)) ⊆ L(α+) ∀α ∈ ON. Take X ∈ P(L(α)). Let A = L(α)∪{X}. We pick α ≥ ω, and A is transitive because L(α) is transitive for every α ∈ ON, so |A| = |α|. Let χ be the finite conjunction of axioms of ZF+ V = L described in Lemma 2.21. By the Levy Reflection Principle, ∃β ∈ ON such that A ⊆ L(β), where L(β) |= χ. We can find an elementary submodel K ≺ L(β) using the L¨owenheim-Skolem Theorem such that A ⊆ K and |K| = |A|. Then because L(β) |= χ, K |= χ. Then using the Mostowski Collapsing Theorem, we can find a transitive class M such that there is an isomorphism from K to M so K =∼ M. A is transitive, so the isomorphism is the identity on A, which implies that A ⊆ M. Thus M |= χ and |M| = |α|. M is transitive and M |= χ, implying M = L(o(M)). |M| = |α|, so |o(M)| = |α|. Then o(M) < |α+|. Now A ⊆ M = L(o(M)) ⊆ L(α+). Thus A ⊆ L(α+), so X ∈ L(α+). Now given that P(L(α)) ⊆ L(α+), GCH follows. For every cardinal κ, |P(κ)| ≤ |P(L(κ))| ≤ |L(κ+)| = κ+. Then |P(κ)| = κ+. Also, 2κ = κ+. [2] CH naturally follows from this as well, taking κ = ω. Thus CH is consistent with ZFC.

3 ¬CH is Consistent

ω ω ¬CH is equivalent to the statement 2 ≥ ω2 where 2 is the set of functions from ω into 2. Our plan to show that ¬CH is consistent is to find a countable, transitive model of a finite fragment of ZFC and add enough subsets of ω to it so that ¬CH holds. We will do this through forcing. [5]

3.1 Posets

Definition 3.1. A poset P is a pair hP, ≤i where ≤ is a relation on P that is reflexive, transitive, and antisymmetric, which means that if p, q ∈ P, p ≤ q and q ≤ p, then p = q. Poset stands for partially ordered set. We call their elements conditions, and each poset has a unique maximum element.

13 Definition 3.2. Two elements p, q of a poset P are compatible if ∃r ∈ P such that r ≤ p, q. When two elements are compatible, we say they have a common extension r, or that r extends p and q. If there is no common extension, we say p and q are incompatible.

Definition 3.3. An antichain in P is a set A ⊂ P such that any two elements are incompatible.

Definition 3.4. If every antichain in a poset P is countable, we say that P has the countable chain condition, abbreviated ccc.

Definition 3.5. A subset D ⊆ P is dense if for any p ∈ P, ∃q ∈ D such that q extends p.

Definition 3.6. A subset G ⊆ P is a filter if 1) ∀p, q ∈ G ∃r ∈ G such that r extends both p and q. 2) Whenever q ∈ G and q ≤ p, then p ∈ G. Using these definitions, we can create a poset with certain characteristics that will allow us to use forcing.

3.2 Countable Transitive Models Another piece we need to use forcing is a countable, transitive model. Since we are trying to work within ZFC to prove ¬CH, we would like a countable, transitive model of ZFC. Unfortunately, this is not provable from ZFC. To get around this, we can take any finite fragment of ZFC axioms. We can show that there is a M that models this fragment. [5] If we take a model M that satisfies extensionality, then by using the Mostowski Collapsing Theorem we can find a transitive N that is isomorphic to M. This is useful for when our original M is not automatically transitive. To get a specific transitive model of finite fragements of ZFC, we can use the cumulative hierarchy. Vα is always transitive, which gets us a step closer. Lemma 3.7. If φ is a finite conjuction of ZFC axioms, then there is an α ∈ ON such that Vα |= φ. Now we can find a transitive model, so all we need is a countable one coming from it. Taking a finite fragment of ZFC φ, we have a Vα that models φ. We can take a countable elementary submodel M ⊆ Vα, which gives us a countable model M such that M |= φ. If our M is not transitive, we can use the Mostowski Collapsing Theorem to find a transitive N isomorphic to M. Then as M is countable, so is N. Thus we have a countable transitive model of a finite fragment of ZFC, satisfying the following theorem.

Theorem 3.8. For any finite fragment of ZFC φ1, φ2, ...φn, there is a countable transitive model M such that M |= φ1 ∧ φ2 ∧ ... ∧ φn.

14 3.3 M[G] Definition 3.9. Let M be a countable transitive model of a finite fragment of ZFC, and let P be a poset with P ∈ M. A filter G is P-generic if for every set D ∈ M with D dense in P, we have that G ∩ D 6= ∅. We would like to be able to form a new model containing both M and G, thereby adding G into our model. This new model, M[G], can be recursively constructed, which is done through ”names.” The first issue is the existence of a generic filter, but we can show this by inductively choosing an element in each set dense in a poset P. Theorem 3.10. Let M be a countable transitive model of a finite fragment of ZFC with P ∈ M. Then for any condition p ∈ P, there is a P-generic filter G with p ∈ G. [4]

Proof. Let D be the set of all subsets Dn of M which are dense in P, with n ∈ ω. We can inductively choose a sequence qn with n ∈ ω so that p = q0 ≥ q1 ≥ ... and qn+1 ∈ Dn. Take G to be the filter generated by this sequence {qn : n ∈ ω}.

Definition 3.11. τ is a P-name if τ is a relation and ∀hσ, pi ∈ τ, σ is a P-name and p ∈ P. The P-names together form a class. P-names need to be interpreted to have any meaning. We use a generic filter G to interpret all the P-names in a model M, the class of which is denoted by M P.

Definition 3.12. Let τ be a P-name and G a filter on P. The value of τ under G, denoted τ[G], is the set τ[G] = {σ[G]: hσ, pi ∈ τ and p ∈ G}.

τ[G] is the interpretation of the P-name τ under a filter G. hσ, pi ∈ τ can be thought of as σ[G] having at least p of belonging to τ[G]. The set of all the interpretations of the P-names in M under G makes up the set M[G]. Definition 3.13. Let M be a countable transitive model of a finite fragment of ZFC. Let P ∈ M and G be a filter. Then M[G] is defined as {τ[G]: τ ∈ M P}. Now that we have formed M[G], we want to show that it is a countable tran- sitive model of ZFC so we can use it for forcing. The countable and transitive part is not too difficult. Sending a of names to their interpre- tations yields a countable set, so M[G] is countable. M[G] is transitive, also coming from the fact that M is transitive and elements of τ[G] are of a similar form σ[G]. This shows the first part of the following theorem.

Theorem 3.14. If G is a P-generic filter, then M[G] is a countable transitive model of a finite fragment of ZFC such that M ⊆ M[G] and G ∈ M[G]. We have already discussed that M[G] is countable and transitive. To show that M ⊆ M[G], we take an element in M and find a name that has an in- terpretation of that element. There is a name because that is the definition of

15 M[G]. Then every element of M is in M[G], so M ⊆ M[G]. We find a name to show that G ∈ M[G] as well. To show that M[G] |= ZFC , we can show that the usual axioms hold in M[G]. [5] We will show an example after stating the fundamental forcing theorems.

3.4 Forcing The general idea of forcing is the construction of an M[G] that adds new sets from a filter G into a model M. In this process, sets may be added that would change interpretations of key elements. M[G]’s interpretation of countable and uncountable might be different than M’s, for example. Absoluteness is helpful here as structures such as L would not change from M to M[G].

Definition 3.15. Let φ(τ1, τ2, ..., τn) be a formula with τ1, ...τn ∈ M P. Let p ∈ P. We say p φ(τ1, ..., τn), or p forces φ, if for every P-generic filter G with p ∈ G, we have M[G] |= φ(τ1[G], ..., τn[G]). There are two major theorems we use in forcing, particularly when proving consistency.

Theorem 3.16 (Forcing Theorem A). If M[G] |= φ(τ1[G], ..., τn[G]), then there is a p ∈ P such that p φ(τ1, ..., τn).

Theorem 3.17 (Forcing Theorem B). For any formula φ, p ∈ P, τ1, ..., τn ∈ M P, there is a formula ψ such that for all p ∈ P and τ1, ..., τn ∈ M P we have that whenever p φ(τ1, ...τn),M |= ψ(p, τ1, ..., τn). The proofs of the forcing theorems, which we omit, need another relation that works inside the model, as a generic G works outside of M [4]. This can ∗ be resolved by defining , a recursively defined relation which lines up with the forcing relation but does not mention M[G] at all, instead working entirely ∗ M ∗ within M. It turns out that p φ(τ1, ..., τn) iff (p φ(τ1, ..., τn)) . is used to prove the two forcing theorems, which then allow us to prove important consistency results. As an example of M[G] modeling ZFC, we will show that the Axiom of Comprehension holds in M[G]. Theorem 3.18. Comprehension holds in M[G].

Proof. Let φ be a formula and τ0, τ1, ..., τn be names such that τ0[G], τ1[G], ..., τn[G] ∈ M[G]. To satisfy Comprehension, we would like the set X = {σ[G] ∈ τ0[G]: M[G] |= φ(σ[G], τ0[G], ..., τn[G])} to be in M[G]. To show this, we will construct a name for X. M Choose a big enough β ∈ ON so P, τ0, ..., τn are all elements of Vβ . Define M χ = {hσ, pi :(σ ∈ Vβ ∧ p (σ ∈ τ0) ∧ φ(σ, τ0, ...τn))}. By Forcing Theorem B and Comprehension in M, χ ∈ M P. We will show χ[G] = X. Let ρ ∈ χ[G]. By definition of names, there is some p ∈ G such that p (ρ ∈ τ0 ∧ φ(ρ, τ0, ...τn)). By definition of , ρ[G] ∈ τ0[G] and M[G] |= φ(ρ[G], τ0[G], ..., τn[G]). Thus ρ[G] ∈ X, so χ[G] ⊆ X.

16 Let π[G] ∈ X. Then M[G] |= φ(π[G], τ0[G], ..., τn[G]). By Forcing Theorem A, there is some p ∈ G such that p (π ∈ τ0 ∧ φ(π, τ0, ..., τn). By definition M 3.12, there is some p0 ∈ G so that hπ, p0i ∈ τ0. Then π ∈ Vβ . So hπ, p0i ∈ χ, so π[G] ∈ χ[G] and X ⊆ χ[G]. Thus χ[G] = X.

3.5 ¬CH holds To show that ¬CH is consistent, we would like to construct an M[G] to satisfy ω ZFC+ ¬CH. This M[G] should add subsets of ω to M so M[G] |= 2 ≥ ω2.A possible problem with this forcing could be that through the process of making P(ω) bigger, we might unintentionally change the idea of cardinality. For exam- ple, if we take P to be the set of finite functions from Z into R, the uncountable reals in M turn out to be countable in M[G]. A series of lemmas are needed to prove that the M[G] we will construct models ¬CH. They will discuss useful characteristics about posets in general that we want our poset to have. The proofs are from Palumbo [5]. Definition 3.19. A family A of sets is called a 4-system if there is some fixed set r such that if A, B ∈ A and distinct, then A ∩ B = r. Lemma 3.20 (4-system Lemma). Suppose A is an uncountable family of finite sets, there is an uncountable B ⊆ A that forms a 4-system. Lemma 3.21. Let I be an arbitrary set and J be a countable set. Then Fn(I,J) is ccc. Proof. Let X be an uncountable subset of Fn(I,J). We want to show that there are compatible p, q ∈ X such that p and q are distinct. Let D = {dom(p): p ∈ X}. If D is countable, then each dom(p) is finite implies that S D is countable. There are countably many possible p with dom(p) contained in S D as J is countable. Then since X is uncountable, D must be as well. Assuming then that D is uncountable, then D is an uncountable family of finite sets. By the 4-system Lemma, there is an uncountable B ⊆ D such that B is a 4-system with a root r. Let Y = {p ∈ X :dom(p) ∈ B}. If p ∈ Y , then r ⊆dom(p). As J is countable, there are only countably many possibilities for p  r. Y is uncountable, so there must be distinct p, q ∈ Y such that p  q = q  r. Because p, q ∈ Y , dom(p) and dom(q) are elements of B, so dom(p)∩dom(p) = r. Then p ∪ q is well-defined, so p and q are compatible. Then there are no uncountable antichains, so Fn(I,J) is ccc.

Lemma 3.22. Suppose P ∈ M is ccc in M, and let A, B ∈ M. Let a filter G be P-generic and let f ∈ M[G] with f : A → B. Then there is a function F : A → B with F ∈ M, f(a) ∈ F (a) ∀a ∈ A, and F (a) is countable in M ∀a ∈ A.

Proof. f ∈ M[G] implies the existence of a name τ ∈ M P such that τ[G] = f. ˇ ˇ Forcing Theorem A gives us a p ∈ G such that p τ : A → B. Define F : A → B as F (a) = {b ∈ B :(∃q ≤ p)q τ(a) = b}.

17 Forcing Theorem B implies that is defined in M, and F is defined using on objects in M. Thus F ∈ M. Suppose f(a) = b. f ∈ M[G], so M[G] |= f(a) = b. Then there is an r ∈ G such that r τ(a) = b. Both r, p ∈ G, so they are compatible. We can then find a q ≤ r, p. Then q τ(a) = b, and by definition, b ∈ F (a). For every b ∈ F (a), there is a qb ≤ p such that qb τ(a) = b. qb and qc are incompatible if b 6= c, we can show by assuming the contrary. If we take a q ≤ qb, qc, then q τ(a) = b and q τ(a) = c. But p ”τ is a function,” and q ≤ p, so we have a contradiction. So {qb : b ∈ B} is an antichain. P is ccc by assumption, so B is countable. Thus F (a) is countable for every a ∈ A.

Lemma 3.23. If a poset P ∈ M has the ccc in M, then P preserves cardinals. Proof. We need to show that if κ ∈ M and M |= ”κ is a cardinal,” then M[G] |= ”κ is a cardinal” as well. Suppose to the contrary that κ is not a cardinal in M[G]. Then ∃α ∈ ON with α < κ and a surjective function f : α → κ. From the previous lemma, we can get a function F ∈ M where F : α → P(κ), f(β) ∈ F (β) for every β < α, and F (β) is countable in M for every β < α. Because f is surjective, if a γ ∈ κ, then there is a β such that f(β) ∈ F (β). Then κ = S{F (β): β < α}. The union is definable in M because F ∈ M, so M |= |κ| ≤ |α| · |ω|. This implies that M |= |κ| ≤ |α|. But α < κ and κ is a cardinal in M, which is a contradiction. Thus M[G] |= ”κ is a cardinal.” Theorem 3.24. Let M be a countable transitive model of ZFC. Then there is a poset P ∈ M such that for any P-generic filter G over M, M[G] |= ¬CH.

Proof. We define our poset P = {p : X → 2 : X is a finite subset of ω2 × ω}, M M M or all the partial functions from ω2 × ω into 2, denoted by Fn(ω2 × ω, 2) . S M M Let g = G. g is then a function from ω2 × ω into 2. We can use g to M define ω2 subsets of ω. For each α < ω2 , define Aα = {n ∈ ω : g(α, n) = 1}. Each Aα belongs M[G], as well as the set containing every Aα. If every Aα M M[G] is distinct, f : ω2 → P(ω) is an injection, given by f(α) = Aα. Then ω M M[G] |= 2 ≥ ω2 . Suppose α 6= β, and let Dαβ = {p ∈ P : ∃n ∈ ω(hα, ni, hβ, ni ∈ dom(p) and p(α, n) 6= p(β, n)}. The domain of p is finite, so ∃n such that neither hα, ni nor hβ, ni belongs to dom(p). Then there is an extenstion of p in Dαβ, namely p∪{hhα, ni, 1i, hβ, ni, 0i} Thus Dαβ is dense, and Dαβ ∩G 6= ∅. Suppose p(α, n) 6= p(β, n). If p(α, n) = 1, then p(β, n) = 0, and n ∈ Aα, but not Aβ. Similarly for p(α, n) = 0. Thus Aα 6= Aβ ω M M M[G] Then M[G] |= 2 ≥ ω2 . The only thing left to show is that ω2 = ω2 so there are no unexpected results with the forcing. Our P is ccc in M by Lemma 3.2, and thus preserves cardinals by Lemma 3.4. Then M’s interpretation of ω M ω2 is the same as M[G]’s interpretation of ω2, so M[G] |= 2 ≥ ω2 . Thus M[G] |= ¬CH. [1] A possible problem with forcing this many subsets into M[G] include that the values of ω2 relativised to M and M[G] could be different. However, the

18 poset we constructed allows for the preservation of cardinals as it is ccc, so this does not end up being a problem. Thus ¬CH is consistent with ZFC.

4 Conclusion

We have shown that the Continuum Hypothesis is consistent with the ZFC Axioms under the assumption that V = L. We have also shown that the negation of the Continuum Hypothesis is consistent with ZFC through forcing. Together, these results imply independence.

References

[1] Cohen, Paul J. ”The Independence of the Continuum Hypothesis.” Proceed- ings of the National Academy of Sciences of the United States of America 50.6 (1963): 1143-1148. Print. [2] G¨odel,Kurt. ”The Consistency of the Axiom of Choice and of the General- ized Continuum Hypothesis.” Proceedings of the National Academy of Sciences of the United States of America 24.12 (1938): 556-557. Print. [3] Jech, Thomas. Set Theory. 3rd Millennium Rev and Expanded ed. Berlin: Springer, 2002. Print. Springer Monographs in Mathematics. [4] Kunen, Kenneth. Set Theory: An Introduction to Independence Proofs. Ed. Barwise et al. Amsterdam: Elsevier, 1983. Print. Vol 102 of Studies in Logic and the Foundations of Mathematics. [5] Palumbo, Justin. ”Forcing and Independence in Set Theory.” N.d. MS. [6] Steprans, Juris. ”History of the Continuum in the 20th Century.” 6 (2011). [7] Weiss, William. An Introduction to Set Theory. 5 Sept. 2014. MS.

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