7 The Axioms of Infinity, Replacement, and Choice Are “True”

It is now clear that the inequality in (6) holds term by term. In fact, forU ∈X this follows from (8), and forU∈Z it follows because

|U ∈Zα⇔U ∈Z= (U ∈Zα∨ U ∈Z)∧(U ∈Zα ∨ U ∈Z)

andU ∈ZU ∈Zα for allU.

6.8. Proposition.The regularity axiom

∀x

∃y (y∈x)⇒ ∃y (y∈x∧y∩x=∅) is “true.”

Proof.We fix X ∈V^B. The axiom with the “constant” X in place of xhas the form R ⇒S. We must show that R ⇒S = 1. It suffices to prove that R ∧ S= 0, where

R= '

Y∈V^B

Y ∈X, (9)

S= %

Y∈V^B

Y ∈X∨ '

Z∈V^B

Z∈Y ∧ Z∈X

. (10)

We suppose thatR∧S=a= 0, and show that this leads to a contradiction.

It follows from (9) and (10) that there exists a Y ∈V^B such that Y ∈X ∧ a= 0. We chooseY to have the least rank of any element with this property.

It is again clear from (9) and (10) that Y ∈X ∧a '

Z∈V^B

Z ∈Y ∧ Z ∈X.

On the right we may sum only over Z ∈D(Y), without changing the value of the sum. Hence, there must exist aZ ∈D(Y) such that

Z ∈X ∧ Y ∈X ∧ a= 0,

so that Z ∈ X ∧a = 0. But the rank of Z is less than the rank of Y,

contradicting the choice ofY.

7 The Axioms of Infinity, Replacement, and

Xi with probability ai, but such anX might not exist. However, it turns out that there always exists anXwithXi∈Xaifor alli∈I; moreover, there exists a leastX with this property.

7.2. Lemma.

(a) Under the conditions in7.1, the function XofY Y ∈X='

i∈I

ai∧ Y ∈Xi (1)

is a random class X that is equivalent to a random set. In addition, Xi∈Xai, and ifXis any random class such thatXi∈Xai for eachi, thenY ∈XY ∈X for allY.

We shall say that X (or the equivalent random set) collects the Xi with probabilities a_i.

(b) Under the same conditions, the functionZ ofY Y ∈Z='

i

ai∧ Y ∈Xi (2)

is a random class Z that is equivalent to a random set. If we also have a_i∧a_j = 0 for all i =j, then Z =X_i a_i, and for any random class Z such that Z =X_i a_i for each i, we have Y ∈ Z Y ∈ Z for allY.

We shall say thatZ glues together theXi with probabilities ai.

Proof.It is easily verified that the functionsZ andX defined by formulas (1) and (2) are extensional.

There exists an ordinalαsuch thatXi∈V_α^Bfor alli. We show thatX ∼Xα

andZ ∼Z_α. For anyY ∈V^B we have Y ∈X_α= '

U∈V_α^B

Y =U ∧ U ∈X_α

= '

U∈V_α^B

'

i

Y =U ∧a_i∧ U =X_i

= '

U∈V_α^B

'

i

a_i∧ Y =X_i ∧ U =X_i.

If we consider the term withU =Xi on the right, we obtainai∧ Y =Xi Y ∈ Xα, so that Y ∈ X Y ∈ Xα by (1), and the assertion follows by 6.3(c).

Similarly, for anyY ∈V^B we have Y ∈Zα= '

U∈V_α^B

'

i

Y =U ∧ai∧ U ∈Xi

= '

U∈V_α^B

'

i

ai∧ Y ∈Xi ∧ Y =U.

Since Y ∈ Xi = ,

U∈V_α^BY = U ∧ Y ∈ Xi, it follows that a_i∧ Y ∈X_iY ∈Z_α, andY ∈ZY ∈Z_α by (2).

Now letX and Z be any random sets with the properties in (a) and (b).

It is clear from (1) that X_i ∈ X a_i. If X_i ∈ X a_i, for eachi, then Y ∈X=,

UY =U ∧ U ∈X,

iY =Xi ∧ Xi∈XY ∈X by (1).

Similarly, if ai ∧ aj = 0 for i = j then it is clear from (2) that ai∧ Y ∈Z=ai∧ Y ∈Xi, so that

ai∧ Xi=Z='

Y

ai∧ Y ∈Xi⇔Y ∈Z=ai

andXi=Zai.Now ifXi=Zai for eachi, then Y ∈ZY ∈Z ∧ Z=X_i

=Y ∈Xi ∧ Z=Xiai∧ Y ∈Xi,

so thatY ∈ZY ∈Z.

Here is our first application of Lemma 7.2(a):

7.3. Proposition.The axiom of infinity

∃x

∅∈x∧ ∀u(u∈x⇒ {u} ∈x) is “ true.”

Proof. When we proved that the axiom of pairing is “true,” we constructed for any U, W ∈ V^B an element Z ∈ V^B (unique up to equivalence) with the property that Y ∈ Z = Y = U ∨Y = W for all Y. It is natural to let {U, W}^B denote this elementZ, and let{U}^B={U, U}^B.

We now verify the axiom of infinity. We setX0=∅, X1={∅}^B, . . . , Xn = {Xn−1}^B, . . . .Further, we letX ∈V^B be the element that collects all the Xi

with probabilities 1. We show that

∅∈X∧ ∀u(u∈X ⇒ {u} ∈X)= 1.

It is obviously sufficient to prove that for all U ∈ V^B we have U ∈ X {U}^B∈X, that is, by (1);

'_∞

i=0U =X_i'_∞

i=0{u}^B=X_i.

In fact, since the formula u = x ⇔ {u} = {x} is “true,” and since Xi+1 = {Xi}^B, it immediately follows that

U =Xi={U}^B=Xi+1. 7.4.Lemma.Let W be a random class. Then there exists an elementX ∈V^B

such that '

U∈V^B

W(U) =W(X).

The left-hand side may be represented in the form ∃x(x ∈ W) = W =∅. Hence, intuitively, the lemma says that the probability that a given class is nonempty coincides with the probability that a suitable element occurs in it.

Proof.We first show that there exists an ordinalβ such that,

U∈VBW(U) = ,

U∈V_β^BW(U). In fact, let aγ = ,

U∈VB

γ W(U), and for any a ∈ B set γ(a) = min(γ|aγ > a) (or γ(a) = 0 if aγ > a for all γ). Finally, set β = sup_a∈Bγ(a). This is an ordinal, becauseB is a set. Ifγ > β, thenaγ aβ

by monotonicity, but we cannot haveaγ > aβ because of the choice ofβ.

Thus, let,

UW(U) =,

U∈V_β^BW(U). We index all the elements in V_β^B by an initial segment of ordinals (by the axiom of choice!):V_β^B={Uα}α∈I. We set

aα=W(Uα)∧ '

γ<α

W Uγ)

, α∈I.

Obviouslya_α∧a_γ= 0 forα=γ. Using Lemma 7.2(b), we glue together the sets U_α with probabilitiesa_α(α∈ I). We obtain a set X satisfying the conditions X =U_αa_αW(U_α). Using the extensionality ofW, we obtain

W(X) '

α∈I

X=Uα ∧W(Uα) = '

α∈I

W(Uα) = '

U∈VB

W(U).

7.5. Proposition.The replacement axiom

∀¯z ∀u ∀x

x∈u⇒ ∃!y P(x, y,z))¯

⇒ ∃w ∀y(y∈w⇔ ∃x(x∈u∧P(z, y,z))¯ is “ true” (here z¯= z1, . . . , zn).

Proof. We fix a “vector” Z = Z1, . . . , Zn with Zi ∈ V^B and an element U ∈ V^B. We shall write P(x, y) instead of P(x, y, Z). If we write the axiom with the “constants”Zi and U in the formR ⇒ S, then we must prove that R⇒S= 1.

7.6. The special case:IfR= 1, then S= 1.

We first show how the general case follows from this special case. Leta∈B, and let B_a denote the set {b ∈ B|b a}. The operations on B induce a Boolean algebra structure onB_awith unit element 1_a =a. The natural mapping B →B_a :b →b∧a is a homomorphism. An easy induction onaallows us to construct a surjective map of universesV^B →V^Ba:X →X_a such that for all X, Y ∈V^B we have

Xa ∈Ya=X∈Y ∧a, Xa=Ya=X =Y ∧a.

Now, to prove Proposition 7.5 from the special case 7.6, we choosea=R. ThenRa = 1a, so that 7.6 implies thatSa= 1a. This means thatSa, and henceR⇒S= 1. (Here we have used 7.6 inV^Ba; clearlyRa=Ra, where Ra is the obvious image ofR inV^Ba.)

7.7. Proof of7.6. The conditionR= 1 means that for anyX ∈V^B,

X∈U∃!y P(X, y). (3)

To show thatS= 1, it is sufficient if given U ∈V^B, we find aW ∈V^B such that for allY ∈V^B,

Y ∈W= '

X∈V^B

X ∈U ∧ P(X, Y). (4) It follows from 6.5 that the formula (4) definesW as a random class. We find an ordinalαsuch thatW ∼W_α.

To do this, we first note that in (4) we may take the sum only over Y ∈W= '

X∈D(U)

X∈U ∧ P(X, Y) (5)

(the argument here is the same as after formula (3) in §6). We now apply Lemma 7.4 to the class WX(Y) = P(X, Y). It follows that for every X ∈ D(U) there exists an elementYX∈V^B such that

∃y P(X, y)=P(X, YX). (6)

(Because ∃!y P(X, y) ∃yP(X, y), we can use these YX to estimate X ∈U with the help of (9) below.)

We setαX = min(α|YX∈V_α^B), and

α= sup(αX|X ∈D(U)),

and then show that W ∼ Wα for this α. We must verify that Y ∈ W Y ∈Wαfor every Y. By (5) and by formula (II) in§5, this follows if for any X ∈D(U) we have

X ∈U ∧ P(X, Y)Y =YX ∧ YX ∈Wα. (7) In the first place, by (3), (6), (5), and the definition of α, we have

X ∈UP(X, Y_X), (8)

X ∈UYX∈W=YX∈Wα.

Further, we consider the following formula, which is “true” because it is deducible from the logical axioms and the axioms of equality:

∀x(∃!y P(x, y)∧P(x, y1)∧P(x, y2)⇒y1=y2).

We thereby obtain

∃!y P(X, y) ∧ P(X, Y) ∧ P(X, Y_X)Y =Y_X. (9)

Finally, it follows from (3), (8), and (9) that

X ∈U ∧ P(X, Y)Y =YX ∧ YX ∈Wα, i.e., we have (7).

7.8. Proposition.The axiom of choice is “ true.”

Proof. Recall that the axiom of choice has the form ∀x∃y(Q∧R∧S∧T), where

Qdenotes ∀z(z∈y⇒ ∃u∃w(z= u, w))(“yis a binary relation”);

R denotes ∀u∀w1∀w2( u, w1 ∈y∧ u, w2 ∈y⇒w1=w2)(“yis a function”);

S denotes ∀u(∃w(u, w ∈y)⇒u∈x)(“the domain of definition ofy is contained inx”);

T denotes ∀u(u=∅ ∧u∈x⇒ ∃w(w∈u∧ u, w ∈y))(“the domain of definition ofycoincides withx,andychooses one element from each nonempty element ofx”).

We fixX∈V^Band construct the corresponding “choosing function”Y. To do this:

(a) We indexD(X) by an initial segment of ordinals:

D(X) ={U0, U1, . . . , Uα, . . .}, α∈I.

(b) For eachUα ∈D(X) we use Lemma 7.4 to find an elementWα ∈V^B such that

Wα∈Uα= '

W∈V^B

W ∈Uα. (c) For eachα∈I we set

aα=Uα∈X ∧

⎛

⎝'

β<α

Uβ∈X∨ Uβ=Uα

⎞

⎠.

(d) Finally, we let Y denote the set that collects the “ordered pairs”

U_α, W_α^B with probabilities a_α, α ∈ I. Here, of course, U, W^B=(

{U}^B,{U, W}^B) .

The idea of this construction is as follows. In eachUαwe choose the element Wα that belongs to Uα “with the largest possible probability.” We then put together the graph of the choice function Y from the “pairs” Uα, Wα^B, where we take the pairs in the order they are indexed, but include a given Uα, Wα^B only to the extent that Uα “was not already considered earlier as belonging toX.”

We now substitute X and Y in place of x and y in the axiom of choice, and, lettingQ, R, S, and T now denote the corresponding formulas with these constants, we show thatQ=R=S=T= 1. We shall constantly be

using the following formula, which follows from (1) and the definition of Y; Z∈Y='

a

Z = Uα, Wα^B ∧aα. (10)

7.9. Q= 1. By the definition ofQ, this means that for allZ ∈V^B we must have

Z∈Y= '

U,W

Z = U, W^B, but this is obvious from (10).

7.10.R= 1. By the definition ofR, for anyU, W¹, W²∈V^B we must prove the inequality

U, W^1B ∈Y ∧ U, W^2B ∈YW¹=W². Using (10), we rewrite the left-hand side in the form

'

α,β

U =U_α ∧ W¹=W_α ∧a_α∧ U =U_β ∧ W²=W_β ∧a_β.

Since U = U_α ∧ U = U_β U_α = U_β and U_α = U_β ∧a_α∧a_β = 0 for α = β (see the definition of a_α), it follows that in this sum we need only consider the terms with α = β. But such a term is W¹ = Wα ∧ W²=WαW¹=W², as required.

7.11.S= 1. This is equivalent to the inequality U, W^B∈YU ∈X. But by (10), the left-hand side equals

'

α

U =Uα ∧ W =Wα ∧aα'

α

U =Uα ∧ W =Wα ∧ Uα∈X

'

α

U =Uα ∧ Uα∈X=U ∈X.

7.12.T= 1. We must prove that for anyU ∈V^B, U ∈X ∧ U =∅ '

W∈VB

W ∈U ∧ U, W^B∈Y. (11)

We first show that it suffices to prove (11) forU∈D(X), i.e., for allUα,α∈I.

In fact, suppose (11) holds for allUα. Then forU ∈V^B we have U ∈X='

α

U =U_α ∧ U_α∈X, U =∅= '

U₁∈V^B

U1∈U,

and hence

U ∈X ∧ U =∅= '

α, U₁

U1∈U ∧ U =Uα ∧ Uα∈X

'

α, U₁

U1∈Uα ∧ U =Uα ∧ Uα∈X

(by (III) in§5)

='

α

Uα=∅ ∧ U =Uα ∧ Uα∈X

'

α,W∈VB

W ∈U_α ∧ U_α, W^B∈Y ∧ U =U_α (by (11) forU_α)

'

W

W ∈U ∧ U, W^B ∈Y. (Here we used the fact that

Uα, W^B∈Y ∧ U =Uα

='

β

Uα=Uβ ∧ W =Wβ ∧αβ∧ U =Uα

'

β

U=Uβ ∧ W =Wβ ∧αβ

= U, W^B∈Y.)

Thus, it remains to prove (11) forUα, α∈I. Now Uα=∅=∃w(w∈Uα)='

W

W ∈Uα=Wα∈Uα. Hence (11) can be rewritten

Uα∈X ∧ Wα∈Uα'

W

W ∈Uα ∧ Uα, W^B ∈Y. (12) We prove this by induction on α. (12) is obvious for α= 0, since the term on the right with W =W0 coincides with the left-hand side. Suppose (12) holds forβ < α.

By the definition ofaα, we have Uα∈X=aα∨

⎛

⎝'

β<α

Uβ∈X ∧ Uβ=Uα

⎞

⎠.

If we substitute this formula in the left-hand side of (12), we find that we must prove two inequalities:

aα∧ Wα∈Uα'

W

W ∈Uα ∧ Uα, W^B ∈Y, (13) Uβ∈X ∧ Uβ=Uα ∧ Wα∈Uα

'

W

Uα, W^B∈Y ∧ W ∈Uα, f or all β < α. (14)

The inequality (13) is obvious if we look at the term on the right with W =Wα. The inequality (14) reduces to the induction assumption as follows.

The left-hand side of (14) is

Uβ∈X ∧ Uβ=Uα ∧ Wα∈Uβ Uβ∈X ∧ Uβ=Uα ∧ Wβ∈Uβ

by the definition of Wβ. Further, using the induction assumption and exten- sionality, we have

Uβ=Uα ∧ Uβ∈X ∧ Wβ∈Uβ

'

W

W ∈Uβ ∧ Uβ, W^B ∈Y ∧ Uβ=Uα

'

W

W ∈Uα ∧ Uα, W^B ∈Y,

which completes the verification of the axiom of choice.

8 The Continuum Hypothesis Is “False” for Suitable B

8.1. We recall (Lemma 7.2(a)) that the setX ∈V^B collects the sets{X_i}with probabilities a_i ∈ B(i ∈ I) if Y ∈ X = ,

iY =X_i ∧a_i for allY. Using this definition, we can introduce a useful canonical mappingt→ˆtfrom the von Neumann universeV to the universeV^B. Let ˆ∅=∅(recall thatY ∈∅= 0 for all Y), and if ˆshas already been defined for all s∈V_α, then for t∈V_α+1, we let ˆt collect all the sˆfor s ∈ t with probability 1. In other words, for any Y ∈V^B,

Y ∈ˆt='

s∈t

Y = ˆs. (1) (Here the collecting set ˆtis not uniquely defined, i.e., it is defined only modulo equivalence, so that, strictly speaking, we should also specify the rank of ˆt, for example by saying that it equals the rank of t. This is not essential for us, however, since we shall be interested only in the truth functions, which do not change if we replace an object by an equivalent object.)

We now formulate some additional conditions (besides completeness) that must be imposed on the Boolean algebra B for the purposes of this section.

Recall that ω₀ is the first infinite ordinal, ω₁ is the first ordinal having cardinality> ω₀, andω₂ is the first ordinal having cardinality> ω₁.

8.2. Conditions on B.

(a) The countable chain condition, which, we recall, says that if we have a family of elements{ai}, i∈I, such that atai= 0 andai∧aj = 0 fori=j, thenI is at most a countable set.

(b) There exists a family of elementsb(n, α)∈B, indexed by the setω₀×ω₂, with the following property: if Z(α) collects the elements ˆn, n ∈ω₀, with probabilitiesb(n, α), thenZ(α) =Z(β)= 0 for α=β, α, β∈ω2.

The second condition has the following intuitive meaning. It is easy to see that Z(α)⊂ωˆ0= 1. In fact, this equality is equivalent to∀x(x∈Z(α)⇒ x∈ωˆ0)= 1, i.e., to

∀X ∈V^B, X∈Z(α)X ∈ωˆ₀,

and this is obvious from (1), since ˆω0 collects the ˆn with probability 1, and Z(α) collects the ˆnwith probabilitiesb(n, α)1.

Thus, condition (b) means that we can findω₂distinct subsetsZ(α)⊂ωˆ₀, so that, in the naive sense, we have card P(ˆω₀) > ω₁. This is precisely the negation of the continuum hypothesis. Of course, it is still necessary to show that this intuitive idea can be made into a proof.

8.3. The existence of B with the required properties. We could use measurable sets, as in §3. However, in order to vary our approach, and to prepare for §9, we give another construction. Let {0,1} be the discrete two-point space, let I = ω0×ω2, and let S = {0,1}^I be the space of vectors whose coordinates are indexed by I and take the values 0 or 1. We introduce the direct product topology on S. It has a standard basis of open sets consisting of all vectors whose coordinates indexed by a finite subsetJ ⊂I are fixed.

Ifa⊂S, we set

a = the complement of the closure ofainS,

and we seta= (a). Setsa⊂S witha=aare called regular open sets inS.

8.4.Theorem. Let

B={a⊂S|a=a}, a∧b=a∩b,

a∨b= (a∪b).

Then B with the operations ∧,∨, and is a complete Boolean algebra with the countable chain condition, and,

iai= (∪iai) for any family ofai∈B.

We omit the proof (see J. B. Rosser, Simplified Independence Proofs, Academic Press, New York, 1969, Chapter 2).

8.5. Lemma.Under the conditions in8.4, let

b(n, α) =the set of vectors with1in the(n, α)place, and let Z(α)be defined as in8.2(b). Then

Z(α) =Z(β)= 0, for α=β.

Proof. By formula (5) in§4, we have Z(α) =Z(β)= %

n∈ω₀

(b(n, α)∨b(n, β))∧(b(n, α) ∧b(n, β)).

The right side can become larger only if we replace ∧ by ∩ and ∨ by ∪; here the primes coincide with the ordinary complements. If we had Z(α) = Z(β) = 0, then there would exist an element X in the standard basis of the the topology (see the beginning of 8.3) that is contained in

1

n∈ω₀

(b(n, α)∩b(n, β))∪(b(n, α)∩b(n, β)).

But this intersection consists of all vectors having the same (n, α)-coordinate and (n, β)-coordinate for alln, while all coordinates except for a finite number range freely in any element X of the standard basis of the topology.

8.6. Formulation of the negation of the continuum hypothesis.We shall prove that the following is “true”:

∀x((“xis an ordinal”∧“xis not finite”∧ ∀y(y∈x⇒“y

¬CH : is finite”))⇒ ∃w(“there is no function fromxonto all of w”∧“there is no function fromwontoP(x)”)).

Here:

xis finite: ∀y(y⊂x∧y=x⇒“there is no function fromy onto all ofx”).

We leave the translation of the other abbreviated notation to the reader.

The premise in ¬ CH says that “x is the first infinite ordinal,” and the conclusion says that “w is a set having cardinality intermediate between that ofxand that ofP(x).” We shall abbreviate¬CH as follows:

∀x(P(x)⇒ ∃w(Q1(x, w)∧Q2(x, w))). (2) 8.7. Reduction Lemma. Let P(x) and Q(x) be two formulas in the Zermelo–Fraenkel language having one free variable x and satisfying the properties

The formula ∃!x P(x)is deducible from the axioms, and X0∈V^B is an element such that P(X0)= 1.

Then P(X) = X = X₀ for all X, and if Q(X₀) = 1, it follows that

∀xP(X)⇒Q(X))= 1.

Proof. We first note that |∃x P(x) ∃!x P(X) = 1, since all the axioms are “true” inV^B, and the rules of deduction preserve “truth.” It hence follows from Lemma 7.4 that there exists an objectX0∈V^B withP(X0)= 1.

Further, P(x)∧P(y) ⇒x = y is also deducible, so that if we apply this withX in place ofxandX0in place ofy, we find that

P(X)X =X0. (3) But P(X) ∧ X = X₀ = P(X₀)∧ X = X₀ = X = X₀. Hence the inequality in (3) may be replaced by equality.

Finally, we suppose thatQ(X0)= 1. Then, by what was just proved, P(X)=Q(X0) ∧ X=X0=Q(X) ∧ X =X0

=Q(X) ∧ P(X),

so thatP(X)Q(X), and∀x(P(x)⇒ Q(X))= 1.

This lemma can be applied to ¬CH in the form (2), since the formula

∃!x P(X), where P(x) is the premise “x is the first infinite ordinal,” is deducible from the axioms. We shall not give this formal deduction, and shall consider the uniqueness ofω0 to be common knowledge. Now, by Lemma 8.7, to verify¬CH it suffices to prove the following facts:

8.8. P(ˆω₀)= 1. (In other words, ˆω₀ plays the role ofX₀in our situation.) 8.9. Q1(ˆω0,ωˆ1= 1.

8.10.Q2(ˆω0,ωˆ1= 1. (This then implies that∃ω(Q1(ˆω0, ω)∧Q2(ˆω0, ω))= 1, and completes the verification of the conditions of the lemma.) 8.8. is verified almost mechanically, and we leave it as an exercise.

8.11. Verification Of 8.9. We must show that if B satisfies the countable chain condition, then

∃a function from ˆω₀ onto all of ˆω₁= 0.

The proof that follows carries over word for word to the more general case, when instead ofω0andω1, we take any pairs, t∈V such that cards <cardt and card sis infinite.

We suppose that

0=a=∃f(f is a function ∧ ∀y(y∈ωˆ1⇒ ∃x(x∈ωˆ0∧ x, y ∈f))), and we show that this leads to a contradiction. There must exist an F ∈V^B such that

aF is a function ∧ %

Y

· · ·

. For every α∈ω1, we consider the term in&

Y · · · corresponding toY = ˆαand use the fact that αˆ∈ωˆ1= 1. We obtain

aF is a function ∧ '

X

X∈ωˆ₀ ∧ X,αˆ^B ∈F

. (4)

By (1), we have

X∈ωˆ₀ ∧ X,αˆ^B ∈F= '

n<ω₀

X = ˆn ∧ X,αˆ^B∈F

= '

n<ω₀

X = ˆn ∧ n,ˆ αˆ^B∈F,

so that if we sum first overX and then overn, we may write (4) in the form aF is a function ∧ '

n<ω₀

ˆ

n,αˆ^B ∈F . Hence, for everyα < w1 there is ann(α)< ω0such that

F is a function ∧ n( ˆα),αˆ^B ∈F = 0.

Then there exist an n0and a subsetJ ⊆ω1 of cardinalityω1such that 0=aα=F is a function ∧ ˆn0,αˆ^B ∈F, for allα∈ J.

It remains to show thata_α∧a_β= 0 forα=β, which contradicts the countable chain condition onB. Now by the definition of a function

aα∧aβ=F is a function ∧ nˆ0,αˆ^B ∈F ∧ ˆn0,βˆ^B ∈Fαˆ= ˆβ, so that it suffices to show thatα=β impliesαˆ= ˆβ= 0.

In fact, if, say,γ∈αbutγ∈β, then the formula (5) in§4 forαˆ= ˆβ has a zero term, namely ˆγ∈αˆ∨ ˆγ= ˆβ. (To check thatˆγ= ˆβ= 0 ifγ ∈β we have to know thatˆγ= ˆδ= 0 if γ=δ, but we have to know this only for γ and δof lower rank thanαandβ, so that the detailed proof uses induction

on the rank.)

8.12.Verification of8.10. We must show that

∃a function from ˆω1,ontoP(ˆω0)= 0, that is, that

∃g(g is function ∧ ∀z(z⊂ωˆ0⇒ ∃y(y∈ωˆ1∧ y, z ∈g)) = 0.

Suppose that for some G∈V^B we have

0=a=Gis a function ∧ %

Z

· · · .

For every α < ω₂ we consider the term corresponding to Z = Z(α) (see the definition in 8.2 and 8.5), and we use the fact that

0=aGis a function ∧ '

Y

Y ∈ωˆ1 ∧ Y, Z(α)^B∈G . (5) By (1), we have

Y ∈ωˆ1 ∧ Y, Z(α)^B ∈G= '

β<ω₁

Y = ˆβ ∧ Y, Z(α)^B∈G

= '

β<ω₁

Y = ˆβ ∧ β, Z(α)ˆ ^B∈G.

Summing first overY, we rewrite (5) in the form 0=aGis a function ∧ '

β<ω₁

β, Z(α)ˆ ^B ∈G.

Hence, for everyα < ω₂ there is aβ(α)< ω₁ such that

0=a_α=Gis a function ∧ β(α), Z(α)^B ∈G.

Then there exist aβ₀< ω₁and a subsetJ ⊂ω₂ of cardinalityω₂such that 0=a_α=Gis a function ∧ βˆ₀, Z(α)^B∈G, for allα∈ J. As in 8.11, we obtain a contradiction to the countable chain condition if we show that a_α∧a_β= 0 forα=β. But this follows from

a_α∧a_βZ(α) =Z(β)= 0

by Lemma 8.5.