6 The Axioms of Pairing, Union, Power Set, and Regularity Are “True”

5.3. Corollary.The axioms of equality inL1Setare “true.”

In fact (see Proposition 4.6 in Chapter II), the axioms of equality in our case consist of the “true” formula x =x, the axiom of extensionality (in the form x = y ⇒ (P(x) ⇒ P(y)) with P(x) = z ∈ x), and the “true” formula x = y ⇒ (x ∈ z ⇒ y ∈ z) (in which P(x) = x ∈ z), since the only atomic

formulasP(x) in L1Set arez∈xand x∈z.

5.4. Remark.In most computations, we shall need to know only the values of X ∈YandX =Y, and not the precise definition of the objectsX andY. In this connection, we note that the following two binary relations on V^B coincide (as easily follows from (III) and the axiom of extensionality):

(a)X =Y= 1,

(b)∀Z∈V^B, Z∈X=Z ∈Y. We shall call suchX andY equivalentand writeX ∼Y.

6 The Axioms of Pairing, Union, Power Set, and

6.3. Examples and remarks

(a) For any random setZ the functionX→ X ∈Zis extensional by (II),

§5, and so is a random class. By analogy, we often write X ∈W instead of W(X) ifW is any random class.

(b) There exist random classes that are not equivalent to random sets. One such example is the “universal” random classW(X) = 1 for allX. (IfW were a set, we would haveW ∈W= 1, contradicting the regularity axiom, which will be shown to be “true” below.)

(c) LetW be a random class, and letαbe any ordinal. We define the element Wα ∈V_α+1^B as follows: D(Wα) =V_α^B, Wα= the restriction of W to V_α^B (as a function; see 4.2). It is easy to see that for all X ∈V^B we have

X ∈WαX ∈W. (1) In fact, letU ∈V_α^B andX∈V^B. We then have

X =U ∧Wα(U) =X=U ∧W(U) =X =U ∧W(X)W(X), so that by (6),§4,

X ∈W_α= '

U∈V_α^B

X=U ∧W_α(U)W(X) =X∈W.

We shall often show that some classW in which we are interested is equiv- alent to a set by finding an ordinalα such thatW ∼W_α. It is clear from (1) that this follows if X ∈WX ∈W_α for allX.

(d) LetW, W₁, andW₂be random classes. ThenW, W₁∧W₂, andW₁∨W₂ are also random classes, since the extensionality condition is trivially verified for these functions. We shall write W₁∩W₂ and W₁∪W₂ instead ofW₁∧W₂ andW1∨W2, respectively.

(e) Let W be a random class, and let X be a random set. We show that W ∩X is equivalent to a random set. More precisely, if D(X) = V_α^B, then W ∩X ∼(W ∩X)α. In fact, for anyY ∈V^B it follows by (6),§4, that

Y ∈(W ∩X)_α= '

U∈V_α^B

U =Y ∧ U ∈(W ∩X)_α

= '

U∈V_α^B

(U =Y ∧ U∈W)∧ U ∈X

= '

U∈V_α^B

U =Y ∧ Y ∈W ∧ U ∈X

=Y ∈W ∧ Y ∈X=Y ∈W∩X.

This result implies that the separation axioms are “true” (see Section 4.9(b) of Chapter II).

The following proposition gives a general method for constructing random classes.

6.4. Proposition.Let P(x, y1, . . . , yn) be a formula that does not contain any free variables besides x, y1, . . . , yn. Let Y1, . . . , Yn ∈ V^B be fixed. Then the function

X →W(X) =P(X, Y1, . . . , Yn) is a random class.

Intuitively,W contains every setXwith probability equal to the probability that P(X, . . . , Y_n) is true.Y₁, . . . , Y_n play the role of “constants.”

Proof.We use the “truth” of the following axiom of equality:

∀x∀y1· · · ∀yn

x=y⇒

P(x, y1, . . . , yn)⇒P(y, y1, . . . , yn) = 1.

If we take a point ξ in the interpretation class that assigns to x, y, y1, . . . , yn

the values X, Y, Y1, . . . , Yn,respectively, then we find that

X =YP(X, Y₁, . . . , Y_n)∨ P(Y, Y₁, . . . , Y_n), or

X =Y ∧W(X)W(Y),

so thatW is extensional.

We are now ready to verify the axioms.

6.5. Proposition.The axiom of pairing

∀u∀w ∃x∀z(z∈x⇔z=u∨z=w) is “true.”

Proof.By definition we have

∀u∀w∃x∀z(z∈x⇔z=u∨z=w)

=%

U

%

W

'

X

%

Z

Z∈X⇔Z =U∨Z=W.

Hence to prove the theorem if suffices if for anyU, W ∈V^B, we find anX ∈V^B such that for allZ∈V^B,

Z∈X=Z=U ∨ Z =W. (2) For fixed U and W we consider the right side of (2) as a function of Z. This function is a random class X by Proposition 6.4, since it corresponds to the formula z=U ∨z =W. We show that it is equivalent to a random set; more precisely, if U, W ∈V_α^B, thenX ∼Xα. By the remark at the end of 6.3(c), it suffices to verify that for allZ

Z∈XZ∈Xα.

But since U ∈Xα= 1, it follows by formula (II) in§5 that Z=UZ∈Xα,

and similarly

Z =WZ∈Xα,

which gives the required inequality.

6.6. Proposition.The axiom of union

∀x∃y ∀u

∃z(u∈z∧z∈x)⇔u∈y is “true.”

Proof.We fixX ∈V^Band construct a random setY such that for allU ∈V^B, U ∈Y=∃z(U ∈z∧z∈X)= '

Z∈V^B

U ∈Z ∧ Z ∈X.

By Proposition 6.4, there exists a random classY with this property. We show that ifD(X) =V_α^B, thenY ∼Y_α. SinceD(Y_α) =D(X), we have

U ∈Yα= '

Z∈D(X)

U =Z ∧ Z∈Yα

= '

Z∈D(X)

U =Z ∧ '

Z₁∈V^B

Z ∈Z1 ∧ Z1∈X

. (3) We show that the inner sum in (3) may be taken only overZ1∈D(X). In fact, for anyZ1,

Z1∈X= '

Z₂∈D(X)

Z1=Z2 ∧ Z2∈X, so that

Z∈Z1 ∧ Z1∈X= '

Z₂∈D(X)

Z∈Z1 ∧ Z1=Z2 ∧ Z2∈X

'

Z₂∈D(X)

Z∈Z2 ∧ Z2∈X. (4)

Taking this into account, in (3) we first sum over Z for fixed Z1 ∈ D(X).

Since D(Z1) D(X), the sum over Z ∈ D(X) coincides with the sum over Z ∈D(Z1), and is equal toU ∈Z1. Thus,

U ∈Yα= '

Z₁∈D(X)

U ∈Z1 ∧ Z1∈X

'

Z₁∈V^B

U ∈Z1 ∧ Z1∈X=U ∈Y,

by (4).

6.7. Proposition.The power set axiom

∀x∃y ∀z(z⊂x⇔z∈y)

is “ true.”(Recall thatz⊂xis abbreviated notation for∀u(u∈z⇒u∈x).) Proof.We fixX∈V^B and construct aY ∈V^B such that for allZ ∈V^B,

Z ∈Y=Z⊂X= %

U∈V^B

U∈Z∨ U ∈X.

By Proposition 6.4, the right side definesY as a random class. We show that if D(X) =V_α^B, then Y ∼Y_α+1.

We first construct the element Z_α ∈ V_α+1^B by considering Z as a random class. By (1) we haveU ∈Z_α U ∈Z,so that

Z∈YZ_α∈Y=Z_α∈Y_α+1. (5) If we prove the inequality

Z∈YZ_α=Z, (6) it will immediately follow from (5) and (6) that Y ∼Yα+1, since by (II),§5,

Z∈YZα∈Yα+1 ∧ Zα=ZZ∈Yα+1. It remains to verify (6).

First letU ∈D(X) =V_α^B. ThenU ∈Zα=U ∈Z, so thatU∈Zα⇔ U ∈Z= 0, and a fortiori

U ∈X ∧ U ∈Zα⇔U ∈Z= 0. (7) AsU varies, the left side of (7) determines a random class of the formX∩W, where W corresponds to the formula ¬(u∈Z_α⇔u∈Z). Since D(X) =V_α^B, it follows by 6.3(c) thatX∩W ∼(X∩W)_α. But according to (7), (X∩W)_α is the zero function on V_α^B. Thus,U ∈X∩W = 0 for allU ∈V^B. Conse- quently,

U∈XU ∈Z_α⇔U ∈Z for allU. (8) To prove (6), we now write the left-and right-hand sides separately (using the

“truth” of the formulaZ_α=Z⇔ ∀u(u∈Z_α⇔u∈Z)):

Z ∈Y= %

U∈V^B

U ∈Z∨ U ∈X, Zα=Z= %

U∈V^B

U ∈Zα⇔U ∈Z.

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