MATH 220C Set Theory

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1 MATH 220C Set Theory L A TEX by Kevin Matthews Spring 2017 (Updated January 7, 2018) Continuum Hypothesis Definition (Same Cardinality). Two sets A, B have the same cardinality iff there is a bijection from A to B. Theorem (Cantor). The natural numbers N and the real numbers R do not have the same cardinality. Suppose we have an infinite subset A R of the real numbers. Must A have either the same cardinality as the real numbers R or the same cardinality of the natural numbers N? The Continuum Hypothesis (CH) says that the answer to this question is yes. It turns out that CH is independent of the axioms of set theory. We will show in the second part of this course that there is no proof of CH. This uses Gödel s constructible universe. In some offerings of MATH 223S, it is shown that there is no proof of CH. This uses forcing. 1 Basics 1.1 The Language of Set Theory Definition (Language of Set Theory). The language of set theory consists of one binary relation symbol, which denotes set membership. Remark. We will often use other symbols. These are not symbols in the language of set theory, but rather abbreviations which may be rewritten in terms of the language of set theory. Example. We will write A B to denote that A is a subset of B. The symbol is not a symbol in the language of set theory. It is an abbreviation for ( x) x A x B. 1.2 Zermelo-Fraenkel Axioms with Choice Which axioms do sets satisfy? In this section we will list the axioms of Zermelo-Fraenkel with Choice (ZFC), and begin to convince ourselves that they are reasonable things for sets to satisfy and that we can do basic things with them which we would expect. Some of these axioms are a single sentence, but others are actually axiom schemata, which means that we actually have a collection of different axioms, usually one for each formula in the language of set theory. Axiom 1 (Set Existence). ( x)x = x Set Existence says that a set exists (and satisfies the additional condition that it is equal to itself, which we can ignore). Axiom 2 (Extensionality). ( x)( y)[(( z)(z x z y)) x = y] 1

2 Extensionality says that if x, y have the same elements then they are equal. Axiom 3 (Pairing). ( x)( y)( z)(x z y z) Pairing says that if we have two sets then we can find a set which has them as elements. Axiom 4 (Union). ( x)( z)[( y)y x y z] Union says that given a collection of sets we can find a set which contains all of these sets as subsets. Axiom 5 (Powerset). ( x)( z)[( y)y x y z] Powerset says that if we have a set then we can find a set which has all its subset sets as elements. Axiom 6 (Comprehension). For each formula φ(x, z, w 1,..., w n ): ( z)( w 1 ) ( w n )( y)[( x)x y (x z φ)] Before continuing with the rest of the axioms, we go over some applications of Comprehension. Claim. If a, b are sets then there is a set who elements are exactly a, b. Proof. By Pairing, we have a set z such that a z and b z. Take φ to be x = w 1 x = w 2. By Comprehension, taking z = z, w 1 = a, and w 2 = b, we have a set y such that ( x)(x y (x z (x = a x = b))). Then the elements of y are precisely a, b. Definition Given sets a, b, the set above is denoted {a, b}. Remark. We only claimed the existence of such a set. For this definition to make sense, we need to first show that this set is unique. That it is unique is a simple consequence of Extensionality. This same remark holds for the next few claims, and won t be mentioned explicitly. Claim. There is a set with no elements. Proof. By Set Existence, we have a set z. Let φ be x x. By Comprehension, we get a set y such that ( x)(x y (x z x x)). This set y has no elements. Definition (Emptyset). The set above is denoted. Claim. Given two sets a, b there is a set c such that x c (x a x b). Proof. By our work above, we have the set {a, b}. By Union, we have a set z such that a z, b z. Take φ to be x w 1 x w 2. By Comprehension, with w 1 = a and w 2 = b, we get a set c such that ( x)(x c (x z (x a x b))). This set c is as desired. Definition Given sets a, b, the set c above is denoted a b. Claim. Given a set a, there is a set y such that x y x a. Proof. Similar to the above; exercise. 2

3 Definition Given a set a, the set above is denoted P(a). Definition (Successor). Given a set x we define S(x) = x {x}. successor of x, and makes sense by the above work. This is often called the Axiom 7 (Infinity). ( x)( x ( y)(y x S(y) x)) Infinity says that we have a set which is infinite, although note that we have not defined the concept of an infinite set yet. Claim. Let φ(x) be x ( y)(y x S(y) x). There is a set ω such that u ω if and only if ( x)(φ(x) u x). Proof. By Infinity, there is a set z such that φ(z). By Comprehension, there is a set ω such that ( u)(u ω (u z ( x)(φ(x) u x))). This set satisfies the required condition. Definition (Natural Numbers). The set ω above is the natural numbers. Remark. For things like this, we will write ω = {u z ; ( x)(φ(x) u x)}, in place of ( u)(u ω (u z ( x)(φ(x) u x))). Claim. We have ω. Proof. As before, let φ(x) be x ( y)(y x S(y) x). Fix x such that φ(x). We must check that x. But this is clear because the definition of φ says x. Claim. The set ω is closed under the successor S. Proof. As before, let φ(x) be x ( y)(y x S(y) x). Fix y ω. We must show that S(y) ω. Fix x such that φ(x). We must show that S(y) x. As y ω, we have y x. By definition of φ, we get S(y) x, as desired. Our work shows that ω is a witness for Infinity. There are three more axioms (some of which are actually axiom schemata). We will see Foundation and Choice later. Axiom 8 (Replacement). For each formula φ(x, y, z, w 1,..., w n ): ( w 1 ) ( w n )[(( x z)(!y)φ) ( u)( x z)( y u)φ]. Replacement says that if φ defines a function on a set z, then there is a set which contains the image of z under this function. Of course, similarly to above, we can use Comprehension to get that the image itself is a set. We omit the proof because it is the same idea as before. Claim. For any φ as above and any sets w 1,..., w n, z, if ( x z)(!y)φ then there is a set u such that y u ( x z)φ. 3

4 1.3 Ordered Pairs, Functions, and Relations Definition (Ordered Pair). For sets x, y we define the ordered pair x, y := {{x, y}, {x}}. Note that this makes sense by our previous work. We must check that this ordered pair behaves the way that we expect it to. We leave the next claim as an exercise. It is straight-forward. Claim. For sets x, y, u, v, if x, y = u, v then x = u and y = v. Definition (Cartesian Product). For sets a, b we define the cartesian product a b := { x, y ; x a y b}. It is not immediate from our previous work that the cartesian product of two sets is necessarily a set. We will give two proofs that it is indeed a set. The first will use Powerset but not Replacement, and the second will use Replacement but not Powerset. This is because later we will sometimes want to assume only one (not both) of these axioms in particular settings. Proof. Notice that for any x a and any y b we have x, y P(P(a b)). By Comprehension, a b = {u P(P(a b)) ; ( x a)( b y) u = x, y } is a set. Proof. Fix u a and define Q u := {u} b. We claim that Q u is a set. Let φ(x, y, z, w 1 ) be y = w 1, x. Taking w 1 = b and z = b, noting that for each x z there is a unique y such that y = u, x, by Replacement (technically, the claim just after it) the image of b under the function v u, v is a set. But this set is precisely Q u. Similarly to what we just did, we can show that Q := {Q u {u} a} is a set. It is the image of a under the function u Q u. The details are left as an exercise. Finally, by Union (together with Comprehension) Q is a set. Notice that this is precisely the set we are looking for. With the cartesian product established, we can now define several other familiar notions. Definition (Set Function). Given sets a, b, f we say f is a (set) function from a to b, and write f : a b, iff f a b ( x a)(!y b) x, y f. Definition (Set Relation). Given a sets a, R we say R is a (set) relation on a iff R a a. Definition (Order). Given a relation R on a set a, we say R is a (non-strict) order on a iff R is reflexive: ( x a) x, x R; R is transitive: ( x a)( y a)( z a)( x, y R y, z R) x, z R; and R is antisymmetric: ( x a)( y a)( x, y R y, x R) x = y. We say R is a (strict) order on a iff R is anti-reflexive: ( x a) x, x / R; R is transitive; and R is antisymmetric. There is a correspondence between non-strict and strict orders. Usually there is not much reason to distinguish between a strict order and its corresponding non-strict order, although it will often be more convenient to work with one or the other in various definitions. For an order R, we will sometime write xry in place of x, y R, as is standard with orders denoted by < or and so on. 4

5 Definition (Linear/Total Order). An order R on a set a is called total (or linear) iff ( x a)( y a) x, y R x = y y, x R. Definition (Equivalence Class). A relation R on a set a is called an equivalence relation on a iff R is reflexive; R is transitive; and R is symmetric: ( x a)( y a) x, y y, x. Definition (Well-founded). A strict order R on a set a is called well-founded iff ( u a)(u ( x u)( y u) y, x / R). A non-strict order will be referred to as well-founded if its corresponding strict order is wellfounded. In this definition above, we only specified that R was a strict because if it were non-strict then we would have to deal with the case that y = x. So an order is well-founded if every non-empty subset of the underlying set has a minimal (not minimum) element. Example. The standard strict linear order < on N is well-founded. Definition (Well-ordering). A well-ordering of a set a is a linear order R on a which is wellfounded. In many cases, if we want to verify that we have a well-founded order, we will simply check that it is transitive, antisymmetric, and well-founded. We may not bother checking whether our order is strict or non-strict. 1.4 Ordinals The ordinals will be the isomorphism types of well-orderings; each well-ordering will be isomorphic to exactly one ordinal. See the homework for part of this. To start our development of the ordinals, first recall which it means for an order to be transitive, and compare to the following definition. Definition (Transitive). A set u is called transitive iff ( x)( y)(x y y u) x u. This is an instance of transitivity for a restriction of the relation in the case that the third element is the fixed set u. To speak more precisely about restrictions, we have the following definition. Definition (Restriction). Given a set z, the restriction of to z is z := { a, b z z ; a b}. Definition (Ordinal). A set α is an ordinal if α is transitive and α is a (strict) well-ordering. Claim. The empty-set is an ordinal. Proof. That is transitive holds vacuously, because it has no elements to verify the condition for. That is a well-ordering is also easy, because = and the empty-set again is vacuously a linear order and is vacuously well-founded because it has no non-empty subsets to check. Claim. For any set α, if α is an ordinal then S(α) is an ordinal. 5

6 Proof. First we need to check that S(α) is transitive. Suppose y S(α) and x y. We must show x S(α). Either y α or y = α. In the former case, transitivity of α gives that x α, and so x α S(α). In the latter case, x y = α S(α). Next we need to check that S(α) is a well-order. To do this, we must check that S(α) is a linear order and is well-founded. We start by showing that it is a (strict) linear order. We start with anti-reflexivity. Suppose x α. We show x / x. Well, {x} is a non-empty subset of α, so it has a ɛ-minimal element. This element must clearly be x. In particular, x / x. Next we do transitivity. Suppose x, y, z S(α) and x y, y z. Either z α or z = α. In the former case, by transitivity of α, y z α means y α, and x y α means x α. By transitivity of α we see that x z. In the latter case, x y and y α. By transitivity of α, x α = z. All that is left is antisymmetric. Suppose x, y S(α) with x y and y x. If x, y α then this follows inductively. If x, y = α then this is immediate. If x α and y = α then α = y x α and so α α. Similarly to above, {α} is now a non-empty subset of α, yet we can t find an -minimal element. If x = α and y α then the same contradiction arises. Finally we check that S(α) is well-founded. Suppose x S(α). If x = {α} then α is an -minimal element (because α α as we saw above). Otherwise, x \ {α} is a non-empty subset of α, so inductively has an -minimal element y. Then y x α so y α and therefore y is an -minimal element of x. We will use induction to simplify some of our proofs, instead of needing to recall the formula φ pertaining to the definition of ω each time. The proof of induction will in fact be the final time which we reference this specific formula φ. Claim (Induction). For any set y, if y ω, y, and ( n)n y S(y) y, then y = ω. Proof. It suffices to check that ω y. Suppose n ω. Then ( x)(φ(x) n x). So to conclude that n y, we need only show that φ(y) holds. But these are precisely the assumptions on y. Claim. Every natural number n ω is an ordinal. Proof. Let y := {n ω ; n is an ordinal}. Certainly y ω. We already saw that y. And we already saw that for x ω, if x y then S(x) y. By induction, we conclude ω = y. Definition (Finite). A set x is finite if there exists a natural number n ω and a bijection between x and n. Theorem The set ω is an ordinal. Proof. We must show that ω is transitive, and that ω is a well-ordering. We start by showing that it is transitive. We prove by induction that a := {n ω ; n ω} = ω. Certainly a ω. As ω, we have a. Now suppose n a. Then n ω and {n} ω, hence S(n) = n {n} ω, which is to say S(n) ω. By induction, a = ω, and thus ω is transitive. To check that ω is a well-ordering, we must check that it is a linear order, and that it is well-founded. Given n ω, prove by induction that for all m ω, n m or n = m or m n? INCOMPLETE To see that is it well-founded, suppose that x ω is non-empty. Let n x. Consider x S(n). This is a non-empty subset of the ordinal S(n) (it contains n) and hence has a minimal element. This will also be a minimal element of x. 6

7 1.5 Definition by Recursion and by Transfinite Induction Definition (Classes). Given a formula φ(x, w 1,..., w n ) and sets a 1,..., a n, is a class. {x ; φ(x, a 1,..., a n )} Classes are just collections of sets, which may or may not be sets themselves. If a class is not a set, we call it a proper class. Definition (Universe). The class V := {x ; x = x} is the universe. The universe is the class consisting of all sets (since any set x satisfies x = x). Theorem (Schema of Definition by Recursion). Given a class function F : V V, there exists a unique (set) function g : ω V satisfying ( n ω) g(n) = F (g n). Proof. (long) INCOMPLETE Theorem (Second Schema of Definition by Recursion). Given a class function F : V V and a set u, there exists a unique (set) function g : ω V satisfying g(0) = u and ( n ω) g(s(n)) = F (g(n)). Proof. Define for any set z { F (z(n)) F (z) := u if there is an n ω such that z is a function with domain S(n), otherwise. By (1.5.3), there is a (set) function g : ω V satisfying ( n ω)g(n) = F (g n). Then we see that and for any n ω, g(0) = F (g 0) = F ( ) = u, g(s(n)) = F (g S(n)) = F ((g S(n))(n)) = F (g(n)). For uniqueness, suppose we have another such function h. Then we can show by induction that g = h. Indeed, h(0) = u = g(0), and for n ω, if h(n) = g(n) then h(s(n)) = F (h(n)) = F (g(n)) = g(s(n)). Claim. For every set x there is a set y such that x y and y is transitive. Proof. For a set a, let F (a) = a. Let u = x. By the second schema of definition by recursion, there is a set function g : ω V such that g(0) = x and ( n ω) g(s(n)) = g(n). Let y := range(g). We must check that x y and y is transitive. The former holds because x = g(0) range(g) = y. To show the latter, suppose w y. We must show w y. By definition of y, there is an n ω such that w g(n). Then w g(n) = g(s(n)) range(g) = y. Definition (Class of Ordinals). The class of ordinals is denoted ON. We ve seen in the homework that ON is a (class) strict linear order. We show now that it is well-founded. What does it mean for a class relation to be well-founded? It could mean that every non-empty subset of the ordinals has an ON-minimal element. But it could also reasonably mean that every non-empty class of ordinals has an ON-minimal element. The latter is clearly a stronger statement, and in this case we will take well-founded to be mean this. 7

8 Theorem ( ON is Well-Founded). Every non-empty class of ordinals has an ON-minimal element. Proof. Fix non-empty A ON. Since A is non-empty, there is an ordinal α A. Consider A S(α). This is a non-empty subset of the ordinal S(α). Therefore it has an S(α)-minimal element γ. We claim that γ is ON-minimal in A. Assume, for a contradiction, that it is not. Then there is δ A with δ γ. Since δ γ S(α) and S(α) is an ordinal, we have δ S(α). Hence δ A S(α). This contradicts the S(α)-minimality of γ. In light of the fact that we now know that ON is a well-ordering, we will adopt a more suggestive notation. Definition We will denote ON by <. Recall section 1.4, where we proved the principle of induction, a method for showing that a property holds for all natural numbers. We now present a similar idea which gives a method for showing that a property holds for all ordinals. Theorem (Transfinite Induction). For a class of ordinals A, if ( α ON) α A α A then ON A. Proof. Assume, for a contradiction, that ON A. Then ON \A is a non-empty class of ordinals, and therefore has a <-minimal element γ. By definition of <-minimal, this means that for any ordinal α, if α < γ then α A. Equivalently, γ A (since < is just ON). By assumption, γ A. Contradiction. Definition Let α ON be an ordinal. We say α is a successor ordinal if there is an ordinal β ON such that α = S(β). We say α is a limit ordinal if α is non-zero and is not a successor ordinal. Theorem (Transfinite Induction 2). For a class of ordinals A, if 0 A, ( α ON) α A S(α) A, and for every limit ordinal γ, if ( δ < γ) δ A then γ A, then ON A. Proof. Assume, for a contradiction, that ON A. Then ON \A is a non-empty class of ordinals, and therefore has a <-minimal element γ. Note that γ 0 by assumption. If γ is a successor ordinal, then there is some ordinal β with S(β) = γ. Since β < γ and γ is <-minimal, we must have β A, yet S(β) / A. This contradicts our assumption. Hence γ is a limit ordinal. By <-minimality, every δ < γ must be in A. Then by assumption, γ A, a contradiction. Previously we proved two schemata of definition by recursion, which gave (set) functions ω V. Now we prove two schemata of definition by transfinite recursion, which will give (class) functions ON V. Theorem (Schema of Definition by Transfinite Recursion). Given F : V V, there is a unique G : ON V such that ( α ON) G(α) = F (G α). Proof. (long) INCOMPLETE Theorem (Second Schema of Definition by Transfinite Recursion). Given F : V V and a set u, there is a unique G : ON V such that 8

9 G(0) = u, ( α ON) G(S(α)) = F (G(α)), and for every limit ordinal γ, G(γ) = α<γ G(α). Proof. Define for any z, F (z(α)) if there is an ordinal α such that z is a function with domain S(α), F (z) := α<γ z(α) if there is a limit ordinal γ such that z is a function with domain γ, u otherwise. By (1.5.11), there is a (set) function G : ON V such that for all ordinals α, G(α) = F (G α). Then we see that for any ordinal α ON and any limit ordinal γ, we have G(0) = F (0 0) = F ( ) = u, G(S(α)) = F (G S(α)) = F ((G S(α))(α)) = F (G(α)), G(γ) = F (G γ) = (G γ)(α) = G(α). α<γ α<γ Uniqueness is straight-forward using (the second form of) transfinite induction. 1.6 Ordinal Arithmetic We can use the schema of definition by transfinite recursion to define many (class) functions on the ordinals, including ordinal arithmetic which we do now. In general, we will not explicitly write down a function F or F, but it will always be easy enough to figure out what F or F is if we wanted. Definition (Ordinal Addition). Given an ordinal α ON, define by transfinite recursion: α + ON 0 := α, for β ON, α + ON S(β) := S(α + ON β), for limit γ ON, α + ON γ := β<γ α + ON β. Definition (Ordinal Multiplication). Given an ordinal α ON, define by transfinite recursion: α ON 0 := 0, for β ON, α ON S(β) := α ON β + ON α, for limit γ ON, α ON γ := β<γ α ON β. Definition (Ordinal Exponentiation). Given an ordinal α ON, define by transfinite recursion: α 0 := 1, for β ON, α S(β) := α β ON α, for limit γ ON, α γ := β<γ αβ. Definition (von Neumann Hierarchy). Define, by transfinite recursion: V :=, for α ON, V S(α) := P(V α ), for limit γ ON, V γ := α<γ V α. It would be nice if α ON V α were the whole universe V. It turns out that we need an additional axiom to show this. We now present the second last axiom of ZFC. 9

10 Axiom 9 (Foundation). ( x)(x ( w x) ( z)(z x z w)) Foundation says that the class relation on the universe V is well-founded. That is, every set in V has an -minimal element. With this, we can show that the V α cover the whole universe V. This will be one of the few times where we actually use Foundation in this course. We will need to recall the fact that every set is contained in a transitive set (see subsection (1.5)). Theorem For every set x, there is an ordinal α ON such that x V α. Equivalently, V = V α. α ON Proof. Assume, for a contradiction, that the result is false. Then A := {x ; x / α ON V α} is nonempty, so pick y A. Let t := TC({y}) be the transitive closure of the set containing y. Note that y A t, and hence by Foundation A t has a minimal -element x. Suppose w x. Then w x and x t, so by transitivity of t we have w t. By -minimality of x, w / A t. Thus x / A. This means that w α ON V α. Define the class function F given by w { α if α is the least ordinal such that w V α, 0 otherwise. By Replacement, the image Q of x under F is a set. Note Q ON. Therefore sup Q is an ordinal. Now for any w x, by our work above we have w V F (w). As F (w) Q we have F (w) < sup Q. By the monotonicity of the von Neumann hierarchy we see w V F (w) V sup Q. Thus we have shown x V sup Q. Therefore x P(V sup Q ) = V S(sup Q). This contradicts that x A. In light of this fact, we can define the von Neumann rank of an arbitrary set. Definition (von Neumann Rank). Given a set x, the von Neumann rank of x is the least ordinal α ON such that x V α. Remark. The von Neumann rank of a set is always a successor ordinal. Indeed, V = so the rank of a set cannot be zero, and for a limit ordinal γ ON, V γ = α<γ V α so any element must have appeared at some earlier level. 1.7 Cardinals Definition Given sets x and y, we write x y if there exists an injection x y, and we write x y if there exists a bijection x y. Remark. It is straight-forward to check that is transitive and reflexive, and is an equivalence relation (each as class relations). Theorem (Schröder-Bernstein). For sets x and y, if x y and y x then x y. Proof. Let f : x y and g : y x be injections. By recursion on n ω, define x 0 := x y 0 := y x S(n) := g (y n ) y S(n) := f (x n ). (Formally, we actually are defining a function n x n, y n : ω V using one of the schema of definition by recursion.) 10

11 We claim that for all n < ω, x S(n) x n and y S(n) y n. We prove this by induction. For the base case, we have x S(0) = g (y 0 ) = g (y) x, y S(0) = f (x 0 ) = f (x) y. Inductively assuming that x S(n) x n and y S(n) y n, we have Now we can define x S(S(n)) = g ( y S(n) ) g (yn ) = x S(n), y S(S(n)) = f ( x S(n) ) f (xn ) = y S(n). x := {x n ; n < ω}, y := {y n ; n < ω}. These are sets because n x n, y n is a (set) function. Next note that f (x ) = ( f {xn ; n < ω}) = { f (x n ) ; n < ω} = {y S(n) ; n < ω} = y. As f is an injection, this means that f x : x y is a bijection. Finally, we claim that for all n < ω, f n := f (x n \ x S(n) ) : x n \ x S(n) y S(n) \ y S(S(n)) and g n := g (y n \ y S(n) ) : y n \ y S(n) x S(n) \ y S(S(n)) are bijections. Since f is an injection, it distributes over set-subtraction: f n ( (xn \ x S(n) ) ) = f (x n ) \ f ( x S(n) ) = ys(n) \ y S(S(n)). Again because f (and hence f n ) is an injection, this shows that f n is a bijection as claimed. The calculation for g n is similar. Now we can define h : x y by h(u) := (f x ) {f n ; n < ω even} {gn 1 ; n < ω even}. Check that h is a bijection, using the previous claims. Definition (Cardinal). An ordinal κ ON is a cardinal if for all ordinals α < κ we have α κ. Definition (Cardinality). Given a cardinal κ, a set x has cardinality κ if x κ. In this case we write card(x) = κ. Lemma If the cardinality of a set is defined, then it is unique. Proof. Suppose x is a set, and κ 1, κ 2 are cardinals. Assume that x has cardinality κ 1 and also has cardinality κ 2. Then κ 1 x κ 2. We cannot have κ 1 < κ 2, because κ 2 being a cardinal would imply κ 1 κ 2. We cannot have κ 2 < κ 1, because κ 1 being a cardinal would imply κ 2 κ 1. As < is a total order, we must have κ 1 = κ 2. Claim. Given a set x and an ordinal γ, if x γ then card(x) is defined. Proof. Let α ON be least such that x α. We need only show that α is a cardinal. Suppose β < α. If we were to have β α, then we would have β x by transitivity, which could contradict the minimality of α. Thus β α, and α is a cardinal. Lemma The cardinality of a set is defined if and only if the set can be well-ordered. Proof. Let x be a set. First suppose that we have an ordinal γ ON with x γ. Let f : x y be a bijection. Define an order on x by urv if and only if f(u) < ON f(v). Then R is a well-ordering of x. Conversely, suppose that R is a well-ordering of x. Since we saw (homework?) that every wellordering is isomorphic to a unique ordinal, there is a γ ON so that (x, R) = (γ, <). In particular, γ x, so we are done by the previous claim. 11

12 The following claim is vacuously true, as there are no ordinals less than zero. Claim. The ordinal 0 is a cardinal. Theorem (Pigeonhole principle). Given a natural number n ω and a function f : n n, if f is injective then f is surjective. Proof. We proceed by induction. There is only one (injective) function, and it is surjective. Now inductively assuming that this holds for n, suppose f : S(n) S(n) is injective. Let h : S(n) S(n) be the function which swaps n and f(n), and is the identity everywhere else. Then h f : S(n) S(n) satisfies (h f)(n) = h(f(n)) = n by choice of h. Hence (h f) n : n n. Inductively, (h f) n is surjective. If m < n, then there is a k < n such that (h f)(k) = m. So h(f(k)) = m. If f(k) / {n, f(n)}, then m = h(f(k)) = f(k). If f(k) = n then m = h(f(k)) = h(n) = f(n). If f(k) = f(n), then since f is injective we have k = n and therefore m = h(f(k)) = h(f(n)) = n, contradicting that m < n. Finally, either f(n) = n, or f(n) < n and thus there is a j such that (h f)(j) = f(n). In the latter case, f(j) = n by definition of h. In either case, we see that f is surjective. From the pigeonhole principle we quickly get two simple consequences. Theorem Every natural number is a cardinal. Proof. Assume, for a contradiction, that n < ω is not a cardinal. Then there is some ordinal m < n with m n. Let f : n m be a bijection. Then f : n n is injective. By the pigeonhole principle, f : n n is surjective. Yet it misses m n, so we have a contradiction. Theorem The ordinal ω is a cardinal. Proof. Assume, for a contradiction, that we have some n < ω with n ω. Let f : ω n be a bijection. Then f n n is an injection. By the pigeonhole principle, it must be a surjection. Yet it misses f(n), so we have a contradiction. Theorem Infinite successor ordinals are not cardinals. Proof. Let α ω be an ordinal. Define f : S(α) α by { 0 if ξ = α, f(ξ) := 1 + ξ if ξ < α. Check that f is a bijection. Then S(α) α and so S(α) cannot be a cardinal. 1.8 Cardinal Arithmetic Claim. Given cardinals κ, δ, the cardinality of the set {0} κ {1} δ is defined. Proof. It suffices to show this set is equinumerous with an ordinal. Define f : {0} κ {1} δ κ+ ON δ by f(0, ξ) := ξ f(1, µ) := κ + ON µ. Check that f is a bijection. Definition (Cardinal Addition). Given cardinals κ, δ, the cardinal sum of κ and δ is κ + CR δ := card({0} κ {1} δ). Theorem For any natural numbers n, m ω we have n + ON m = n + CR m. 12

13 Proof. INCOMPLETE (easy) Corollary For any natural numbers n, m ω we have n + ON m = m + ON n. Theorem Given cardinals κ, δ, if κ / ω or δ / ω then κ + CR δ = max{κ, δ}. Proof. INCOMPLETE (not long) Claim. Given cardinals κ, δ, the cardinality of the set κ δ exists. Proof. INCOMPLETE (says see homework) Definition (Cardinal Multiplication). Given cardinals κ, δ, the cardinal product of κ and δ is κ CR δ := card(κ δ). Theorem For any natural numbers n, m ω we have n CR m = m ON n. Proof. INCOMPLETE (short and easy) Corollary For any natural numbers n, m ω we have n ON m = m ON n. Theorem For any cardinal κ we have 0 CR κ = 0 and κ CR 0 = 0. Proof. Simply note and 0 CR κ = card( κ) = card( ) = 0 κ CR 0 = card(κ ) = card( ) = 0. Theorem Given non-zero cardinals κ, δ, if κ ω or δ ω then κ CR δ = max{κ, δ}. Proof. We will assume that κ δ, as the argument for the case δ κ is analogous. This means that κ ω. Since δ 0, we have κ κ δ. And certainly κ δ κ κ. So it suffices to show that for any κ ω we have κ κ κ. We proceed to show this by transfinite induction. So assume that for all infinite cardinals τ < κ, τ τ τ (and hence τ τ τ). Define an order on κ κ as follows: τ, µ τ, µ max{τ, µ} < max{τ, µ } or max{τ, µ} = max{τ, µ } and max{τ, µ} = τ and µ < µ max{τ, µ} = µ and max{τ, µ } = µ and τ < τ. or Label both axis of a plot by κ and then check which squares the two pairs lie one. The inner square is smaller. If they are on the same square, then the vertical side of the square is smaller. If they are on the same side of the same square, then compare using the induced order by the ordinal which labels that side of the square. 13

14 We claim that is well-founded. To see this, suppose that A κ κ is non-empty. Pick α ON least so that A ({α} S(α) S(α) {α}) (the smallest square which meets A), and then take an element of A which minimizes the order in the vertical component if possible, and otherwise take an element of A which minimizes the order in the horizontal component. Make this precise, and check that this is a -minimal element in A. Next we claim that is a linear order. This simply involves checking many cases. This means that there is an ordinal γ ON such that (κ κ, ) is isomorphic to (γ, <), say via f : κ κ γ. We claim that γ κ, which would show that κ κ κ. Assume, for a contradiction, that κ < γ. Then κ f (κ κ). So let ˆξ, ˆµ κ κ such that f( ˆξ, ˆµ ) = κ. Let τ = S(max{ˆξ, ˆµ}). Then f 1 (κ) τ τ. Hence f 1 κ witnesses κ τ τ. Since κ ω, we must have τ ω, for otherwise the infinite set κ injects into the finite set τ τ. Note that τ κ by definition, and we cannot have equality because τ is a successor ordinal and κ is an infinite cardinal, therefore τ κ. But then by our inductive assumption, τ τ τ. Combining this with κ τ τ above, we see κ τ. Contradiction. Definition Given sets x and y, define y x := {f P(x y) ; f is a function from y to x}. Definition (Cardinal Exponentiation). Given cardinals κ, δ, if the cardinality of the set δ κ exists then define κ δ := card( δ κ). Claim. For any ordinal α ON and any set z we have S(α) z α z z. Theorem For any natural numbers n, m ω, the cardinality of m n exists and m n n m. Proof. A simple induction, using the previous claim. Theorem Given an infinite cardinal κ ω and a natural number m ω, the cardinality of m κ exists and we have κ m = κ. Proof. This is also a simple induction, using the previous claim together with the fact that κ κ κ as shown in the proof of(1.8.9). Definition Given an ordinal α ON and a set x, define <α x := β x. Definition Given a cardinal κ and an ordinal α ON, if the cardinality of the set <α κ exists then define κ <α := card( <α κ). Theorem For an infinite cardinal κ ω, the cardinality of the set <ω κ exists and κ <ω = κ. Proof. Fix a bijection θ : κ κ κ. Define a function ψ : ω (m κ) κ (that is, for each m, ψ m : m κ κ) as follows: ψ 0 :=, 0 ψ 1 := f f(0) β<α ψ S(m) := f θ( ψ m (f m), f(m) ). That is, we first break up f : S(m) κ as f m : m κ and f(m), then use the previous map ψ m to get an element of κ κ out of the restricted map, and finally apply the bijection θ to what remains. Check that each ψ m is a bijection. Now define φ : <ω κ κ ω by φ := f ψ m (f), m 14

15 where m is the unique value such that f m κ. Since each ψ m is a bijection (hence injection) it is easy to see that φ is an injection. Note that image(φ) κ ω κ κ = dom(θ). Thus we may consider the map <ω κ κ given by f θ(φ(f)). Since φ and θ are injections, so is this map. Thus <ω κ κ. As the other direction is easy to check, the desired result follows. So we have κ <ω = κ. In particular, we showed <ω κ κ. However, we do not necessarily have ω κ κ in general. Claim. For any cardinal κ we have κ 2 P(κ). Proof. Check that the map z χ z is a bijection, where χ z is the indicator function of z on κ. Theorem For any cardinal κ we have P(κ) κ. Proof. Assume, for a contradiction, that P(κ) κ. In particular, since P(κ), this means that there is a surjection f : κ P(κ). Let z := {ξ κ ; ξ / f(ξ)}. Then z P(κ), so as f is surjective, there is a ξ κ such that f(ξ) = z. But then by definition of z, ξ z if and only if ξ / f(ξ) = z. Contradiction. So we have κ κ 2. Theorem There is no largest cardinal. Proof. Let κ be a cardinal. We will find a cardinal which is strictly greater than κ. Note that by Comprehension and Powerset, A := {r P(κ κ) ; r is a well-order of κ} is a set. For each r A, there is an ordinal γ r ON such that r = (γ r, <), namely the order-type of r. By Replacement, Q := {S(γ r ) ; r A} is a set. Certainly sup(q) ON. We will show that κ < sup(q) and κ sup(q), which would mean card(sup(q)) is a cardinal strictly greater than κ. To see that κ < sup(q), note that κ A, and hence S(γ κ ) Q. But γ κ = κ, so in fact S(κ) Q. Therefore sup(q) S(κ) > κ. Finally, we check that κ sup(q). Assume, for a contradiction, that f : κ sup(q) is a bijection. Define a relation r on κ by u, v r if and only if f(u) < f(v). Then f witnesses an isomorphism (κ, r) (sup(q), <). This means r A and γ r = sup(q). Hence S(sup(Q)) = S(γ r ) Q. Contradiction. In fact, sup(q) is a cardinal, and moreover is the smallest cardinal which is greater than κ, but we didn t need to show these facts to complete the proof. These are needed, however, justify the following definitions. Definition Given a cardinal κ, κ + is the smallest cardinal which is greater than κ. Definition Define by transfinite recursion: ℵ 0 := ω; ℵ S(α) := ℵ + α ; ℵ λ := {ℵ α ; α < λ}. Check that each ℵ α is a cardinal, and that every cardinal is of this form. 15

16 1.9 Axiom of Choice and Cardinals We know consider the final axiom of ZFC. Axiom 10 (Choice). ( x)[( y 1 x)( y 2 x)(y 1 = y 2 y 1 y 2 = ) ( y x)y ( z)( y x)(!w)w z y] Choice says that we x is a non-empty collection of non-empty sets, we can pick a set z which meets each member of x exactly once. Claim. Given a set x, if x can be well-ordered then Choice holds for this particular x. Proof. Let r be a well-ordering on x. Check that { z := u } x ; ( y x) such that u is the r-least element of y satisfies the condition in Choice. Claim. The ordinals ON are a proper class. Proof. We must show that ON is not a set. Suppose u ON is a set. We will show u ON. Well u = sup(u) ON. Therefore S( u) ON. But S( u) / u. Thus u ON. Lemma Given a set z, there is no class function ON z which is injective. Proof. Assume, for a contradiction, that there is such a function G : ON z. Define F : z ON by { α G(u) = {α}, F := u 0 G(u) =. Since G is injective, G(u) cannot contain more than a single element. Hence F is well-defined. So F is a class function. By Replacement, F (z) is a set. But it s easy to check F (z) = ON. This contradicts the previous claim. Theorem (Uses AC). Every set can be well-ordered. Proof. It suffices to show that every set is equinumerous with an ordinal. So fix a set E. We will essentially just start enumerating elements of E, indexed by larger and larger ordinals, until we exhaust it. Choice comes in when we make sure that we can always pick a next element to enumerate. We start by showing that there is a function f : P(E) \ {E} E such that for all z E, f(z) E \ z. This will be the function which allows us to always pick the next element to enumerate. Let x := {{z} (E \ z) ; z E}. Then each member of x is non-empty, since E \ z is non-empty whenever z E. And any two distinct members of x are disjoint, since all their respective members differ on the first coordinate. Therefore, by Choice, there is a set w which intersects each member of x precisely once. Check that f := w ((P(E) \ {E}) E) satisfies our requirements. Now we can define the function which essentially enumerates E, via transfinite recursion: { ( G(α) ) G(α) f E, G := α E otherwise. (It is not important what the result is in the second case, only that it is different from everything possible in the first case.) Suppose α < β and G(α) ( E. If G(β) = E then G(α) G(β). Otherwise, by definition of G(β) ) G and f we have G(β) = f E \ G(β). In this case, G(α) G(β) (as α β) and hence G(α) G(β) regardless. Now if there is no γ ON such that G(γ) = E, then G is, by the previous paragraph, an injection ON E, which is impossible. So let γ ON be least so that G(γ) = E. Then the set function G γ is, again by the previous paragraph, an injection γ E. By definition of G we must have G(γ) E. Check, using transfinite induction, that G(ON) E. Thus we must have G(γ) = E, and hence G γ is a bijection. 16

17 Combining this result with previous work, we now know that the cardinality of any set is defined. In particular, we have the following result. Corollary (Uses AC). For every cardinal κ, the cardinality 2 κ = card( κ 2) is defined. We previously saw that κ 2 κ. So we now have two ways to get a cardinal greater than κ: κ + = sup{order-type(r) ; r is a well-ordering of κ}, 2 κ = card(p(κ)). As κ + is the least cardinal which is greater than κ, we certainly have κ + 2 κ. DPick λ such that κ = ℵ λ. Definition (Continuum Hypothesis). The continuum hypothesis, denoted CH, is 2 ℵ 0 = ℵ 1. Definition (Generalized Continuum Hypothesis). The generalized continuum hypothesis, denoted GCH, is ( κ ω) 2 κ = κ +. Definition Let δ be a limit ordinal, let γ be an ordinal, and let f : γ δ. We say f is cofinal if ( α < δ)( ξ < γ) f(ξ) > α. Definition Let δ ω be a limit ordinal. The cofinality of δ, denoted cof(δ), is the least ordinal γ ON for which there is a cofinal function γ δ. Note that id : δ δ is cofinal, and hence the cofinality of δ is well-defined. In particular, this shows that cof(δ) δ. It s also straightforward to check that cof(δ) ω. Example. Define f : ω ℵ ω by f := n ℵ n. Then f is cofinal, so cof(ℵ ω ) ω. Since the reverse inequality always holds, cof(ℵ ω ) = ω. Definition Let δ be a limit ordinal. We say δ is regular if cof(δ) = δ. We say δ is singular if cof(δ) < δ. By our remarks above, singular ordinals are precisely those which are not regular. 17

18 2 Consistency Proofs Recall that ZF is ZF minus Foundation. Our first aim is to prove Con(ZF ) Con(ZF). Note that Con(Γ) is actually defined as Incon(Γ), and in basically all consistency proofs we will actually be proving something of the form Incon(Γ) Incon(Λ), though we present the result as the consequence Con(Λ) Con(Γ). All of this is happening in the metatheory. 2.1 Relativization Definition (Relativization). Given a class M and a formula φ, define the relativization of φ to M, denoted φ M, as follows: (x = y) M is x = y; (x y) M is x y; ( φ) M is φ M ; (φ ψ) M is φ M ψ M ; (( x)φ) M is ( x M)φ M. We present two proofs of the following lemma; one semantic in ZF and one syntactic in PA. Lemma Let Γ and Λ be sets of formulas in the language of set theory. Let θ(x) be a formula in the language of set theory with one free variable. If Λ ( x)θ(x) and for each σ Γ, Λ σ {x ; θ(x)}, then Con(Λ) Con(Γ). Proof (Semantic in ZC ). Suppose A = (A; A ) is a model of Λ. Let B = ({a A ; A = θ[a]}; A B). Then for any sentence τ of set theory we have B = τ if and only if A = τ {x ; θx}. And if σ Γ then since Λ = σ {x ; θ(x)} we get A = σ {x ; θ(x)}. Combining these we see that B = σ. Also, since Λ = ( x)θx, B. Thus B = Γ and so Con(Γ). Proof (Syntactic in PA). Assume, for a contradiction, that φ 1 φ n is a deduction of a contradiction τ τ in Γ. Let σ 1,..., σ k be the members of Γ which occur in this deduction. By assumption, for each σ i there is a deduction ψ i of σ {x ; θ(x)} {x ; θ(x)} {x ; i in Λ. Then ψ 1 ψ k φ 1 φ θ(x)} n is a deduction of τ {x ; θ(x)} τ {x ; θ(x)} in Λ. Thus Incon(Λ), a contradiction. As mentioned, we actually showed that Incon(Γ) Incon(Λ) and not just Con(Λ) Con(Γ). We think of θ as defining the universe of a class model of Γ from a class model of Λ. 2.2 Class of Well-Founded Set Recall that our first goal is to show Con(ZF ) Con(ZF). By (2.1.2), it suffices to find a formula θ (which defines a class {x ; θ(x)}) and check for each axiom σ of ZF that ZF σ {x ; θ(x)}. This class will be the class of well-founded sets. Definition (In ZF ). The class of well-founded sets are WF := V α. α ON We showed V = WF in ZFC. Of course, we are technically actually working with the formula which defines WF, which says that there is an α so that α is an ordinal and x V α (and this last statement is actually rewritten in terms of the definitin of V α and so on). Instead of just showing that we have ZF σ WF for each axiom σ of ZF, we will prove some more general results. For the remainder of this section, we will be working in ZF, that is, we will not use foundation in any of our arguments. This is because we ultimately want to conclude ZF proves each σ WF. 18

19 Lemma Given a class M, if M then ZF (Set Existence) M. Proof. This is immediate. Lemma Given a class M, if M is transitive then ZF (Extentionality) M. Proof. The sentence (Extentionality) M is ( x M)( y M)(( z M)(z x z y) x = y). Suppose x, y M, and assume that ( z M)(z x z y), that is, x M = y M. Since M is transitive, x, y M. Hence x = x M = y M = y. Lemma Given a class M, if M WF then ZF (Foundation) M. Proof. The sentence (Foundation) M is ( x M)(( z M)z x ( z M)(z x ( y M)y x y / z)). Suppose x M is such that x M. Then (x M) WF. By definition of M, there is some β ON such that (x M) V β. So a least such β ON exists, and note that it is not a limit ordinal. Therefore let α ON be least such that (x M) V α+1. Let z (x M) V α+1. We will show that this z works. Suppose y x M. If y z then y z V α, hence y V α. Thus y (x M) V α = by minimality of α, a contradiction. Thus y / z. 2.3 Layered Classes Definition (Layered class). A class M is layered if there is a class function α M α : ON V such that: 1. for every α ON, M α is transitive; 2. for every α, β ON, α β M α M β ; 3. for every α ON, M α M S(α) ; 4. M = α ON M α. Example. The class WF is layered, as witnessed by α V α. homework. Parts of this were justified in the Lemma Given a class M, if M is layered then ZF (Pairing) M. Proof. The sentence (Pairing) M is ( x M)( y M)( z M)(x z y z). Suppose x, y M. As M is layered, there are α, β ON such that x M α and y m β. Then by the definition of layered, x, y M max{α,β}. Moreover, M max{α,β} M S(max{α,β}) M, so M max{α,β} M. Hence we may take z := M max{α,β}. Lemma Given a class M, if M is layered then ZF (Union) M. Proof. The sentence (Union) M is ( x M)( u M)( y x M)( z y M)z u. Suppose x M. Since M is layered, there is an α ON such that x M α. Suppose y x M and z y M. So z y x M α. By transitivity of M α, z M α. Thus we may take u := M α M. 19

20 Lemma Given a class M, if M is layered then for every formula φ, ZF (Replacement φ ) M. Proof. INCOMPLETE (a bit long and heavy on notation) Lemma Given a class M, if M is layered and for each formula ψ(x, w 1,..., w n+1 ), for every ordinal α ON, and for each w 1,..., w n+1 M α we have {x M α ; ψ(x, w 1,..., w n+1 ) M } M, then for every formula φ, ZF (Comprehension φ ) M. Proof. (not too long or difficult) INCOMPLETE Note that M = WF satisfies the assumptions of the above lemma, for the simple reason that all subsets of V α belong to WF. 2.4 Absoluteness and Reflection It remains to check Powerset and Infinity in order to conclude that Con(ZF ) Con(ZF). To do this, we will use the concept of absoluteness. Definition (Absolute). Let φ(x 1,..., x n ) be a formula and let M N be classes. We say φ is absolute for (M, N) if ( x 1,..., x n M) φ M (x 1,..., x n ) φ N (x 1,..., x n ). We say φ is absolute for M if it is absolute for (M, V ). Example. Check that the formula x y is absolute for M whenever M is a transitive class. Lemma Let φ, ψ be formulas and let M N be classes. If φ and ψ are absolute for (M, N) then φ and φ ψ are also absolute for (M, N). Lemma Let φ(x 1,..., x n, y, z) be a formula and let M N be transitive classes. absolute for (M, N) then ( y z)φ is also absolute for (M, N). If φ is Proof. Fix x 1,..., x n, z M. Since M is transitive, z M and hence z M = z. Thus (( y z)φ) M is ( y z)φ M. Similarly, (( y z)φ) N is ( y z)φ N. Now since φ is absolute for (M, N), any witness y for ( y z)φ M is also a witness for ( y z)φ N, and vice-versa. Thus ( y z)φ M ( y z)φ N. Remark. Another way to show that x y is absolute for transitive classes is to note that it is defined to be ( z x) z y. Now z y is absolute for any classes, and so by the previous two lemmas, ( z x) z y is absolute. Definition ( 0 formula). A formula is 0 if it can be obtained from atomic formulas using,, and bounded quantifiers (( x y), ( x y)). The previous lemmas immediate give the following corollary. Corollary Let φ be a formula and let M N be transitive classes. absolute for (M, N). If φ is 0 then φ is Now we are able check Powerset. Lemma Let M be a class. If M is layered then ZF (Powerset) M. 20

21 Proof. The sentence (Powerset) M is ( w M)( u M)( x M)((x w) M x u). Since we showed x w is absolute for M, this is just ( w M)( u M)( x M)(x w x u). So fix w M. For each x P(w) M, let α x ON be least such that x M αx. By Replacement, let γ := sup{α x ; x P(w) M}. Check that u := M γ works. The following lemma shows that, as we would expect, we do not need to stick to the exact definition of a formula; we can replace it by an equivalent formula. This helps, for example, when we replace formulas by equivalent 0 formulas. Lemma Let Γ be a fragment of ZFC, let φ, ψ be formulas, and let M N be classes. Γ φ ψ and Γ M, Γ N, then φ is absolute for (M, N) if and only if ψ is absolute for (M, N). If The following is a weakening of the concept of absoluteness. Definition (Reflects). Let φ(x 1,..., x n ) be a formula and let M N be classes. We say φ reflects from M to N if ( x 1,..., x n M)φ M (x 1,..., x n ) φ N (x 1,... x n ). Definition (Σ 1 formula). A formula is Σ 1 if it can be obtained from a 0 formula using existential quantifiers, that is, is of the form ( x 1 )... ( x n )ψ where ψ is a 0 formula. For this more general class of formulas, we cannot guarantee absoluteness for transitive classes, but we can get the weaker reflection. Lemma Let φ be a formula and let M N be transitive classes. If φ is Σ 1 then φ reflects from M to N. We also have the same assurance that we can replace formulas by equivalent formulas. Lemma Let Γ be a fragment of ZFC, let φ, ψ be formulas, and let M N be classes. If Γ φ ψ and Γ M, Γ N, then φ reflects from M to N if and only if ψ reflects from M to N. Definition (Provably 1 ). Let Γ be a fragment of ZFC and let ψ be a formula. We say ψ is provably 1 in Γ if there are Σ 1 formulas φ 1 and φ 2 such that Γ ψ φ 1 and Γ ψ φ 2. With this less general class of formulas, we once again can guarantee absoluteness. Lemma Let Γ be a fragment of ZFC, let ψ be a formula, and let M N be transitive classes. If ψ is provably 1 in Γ and both Γ M and Γ N hold, then ψ is absolute for (M, N). Lemma For each of the following formulas φ, φ is provably 1 in ZF Powerset Replacement Infinity: 1. x y; 2. x = y; 3. x y; 4. z = {x, y}; 5. z = {x}; 6. z = x, y ; 7. z = x y; 21

22 8. z = x y; 9. z = x \ y; 10. z = S(x); 11. x is transitive; 12. z = x; 13. z = x. Hence. if (ZF Powerset Replacement Infinity) M and (ZF Powerset Replacement Infinity) N hold then each of the above formulas is absolute for (M, N). Definition (Class operation). Let Γ be a fragment of ZFC, let φ(x 1,..., x n, y) be a formula, and let D be a class. We say φ defines an operation on D provably in Γ if Γ ( x 1 ) ( x n ) x 1,..., x n D!yφ. Let M N be classes. The defined operation is absolute for (M, N) if φ is absolute for (M, N) and both (( x 1 ) ( x n ) x 1,..., x n D!yφ) M and (( x 1 ) ( x n ) x 1,..., x n D!yφ) N hold. Lemma Let Γ be a fragment of ZFC, let φ be a formula, let D be a class, and let M N be transitive classes. If φ is provably 1 in Γ, φ defines an operation on D provably in Γ, and both Γ M and Γ N hold, then the defined operation is absolute for (M, N). Lemma For each of the following operations (on V ), there is a formula φ such that φ is provably 1 in ZF Powerset Replacement Infinity and φ defines the original operation on V provably in ZF Powerset Replacement Infinity: 1. x, y {x, y}; 2. x {x}; 3. x, y x, y ; 4. x, y x y; 5. x, y x y; 6. x, y x \ y; 7. x S(x); 8. x x; 9. x x; 10.. Finally, we can use these results about absoluteness of defined operations to check the infinity axiom. Recall that this is our last step in proving our first consistency result. Lemma Given a class M, if M is layered, ω M, and (ZF Powerset Replacement Infinity) M holds, then ZF Infinity M. Proof. The sentence Infinity M is ( u M)( M u ( x u)(s(x)) M u). By our results above on defined operations, this is equivalent to ( u M)( u ( x u)s(x) u). Now it is immediate to see that this formula is witnessed by ω, which we are assuming to be in M. 22