on variables that occur freely in P, then P(ξ) = P(ξ). Compare with Proposition 2.10 in Chapter II.
This value ofP(ξ) that is common for allξifP is closed can be denoted simply by P. We are now ready to formulate the basic definition of this section:
2.9. Definition.A formula P in L2Real is said to be “true” ifP(ξ) = 1 for allξ∈M.
3.5.Proof of3.1(c) (sketch). We shall list the axioms and make some brief comments.
The special axioms of set theory: The axioms of equality and the axiom (schema) of choice
AC : ∀x∃yP(x, y)⇒ ∃f ∀xP(x, f(x)),
whereP is any formula which does not have any free variables exceptxand y, and where f is free fory inP.
The special axioms of field theory: The axioms of the additive group, the axioms of the multiplicative group, and the distributivity of addition with respect to multiplication.
The special order axioms:
xy∨y x,
(xy∧yx)⇔x=y, xy⇒(x+zy+z), (xy∧0z)⇒xz yz.
The completeness axiom(see 2.5).
Among these axioms, the greatest effort is needed to verify that the axiom of choice and the completeness axiom are “true.” But these computations resemble those in the proof that the CH is false, which will be given in detail below. Hence, the verification of these two axioms will be omitted.
The first axiom of equality is trivial. The second axiom is first verified for atomic formulasP, and then we use induction on the length ofP. The argument is rather tedious, but simple.
The axioms of an ordered field are verified without difficulty. We shall limit ourselves to one example: “every nonzero number has an inverse,” i.e.
∀x(¬(x= 0)⇒ ∃y(xy= 1))= %
¯ x∈R
x¯= 0 ∨ '
¯ y∈R
x¯¯y= 1
.
To verify that this truth value equals 1, it suffices to prove this for each term on the right, i.e., for each fixed ¯x ∈ R. Then, in turn, for that ¯x it suffices to construct a random variable ¯y ∈ R such that x¯ = 0,
¯x¯y = 1 = 1.
We set
¯ y(ω) =
¯
x(ω)−1, if ¯x(ω)= 0,
0, if ¯x(ω)= 0.
3.6. Proof of3.1(d). We first recall the formula for the CH:
∀h
∃g∀y ∃x
h(x) = 0∧y=g(x))∨
∃f ∀y
h(y) = 0⇒ ∃x(Z(x)∧y=f(x) .
We letP1 andP2 denote the first and the second alternatives in this formula.
Thus, the CH has the form∀h(P1∨P2). We must prove that∀h(P1∨P2)(ξ) = 0
for any pointξ∈M. By the definition in 2.7,
∀h(P1∨P2)(ξ) =%
ξ
P1(ξ)∨ P2(ξ) ,
where ξ runs through all variations ofξalongh. To show that this value is 0, it suffices to find a point ξ such that P1(ξ) = P2(ξ) = 0. Since all the variables excepthare bound inP1andP2, choosingξ is equivalent to choosing hξ = ¯h∈R(1). We shall give ¯hexplicitly; this will be a function “whose set of zeros has intermediate cardinality.”
To do this, as in§1 we fix a subsetJ ⊂Ihaving cardinality strictly between ω0 and card I. Recall that for each i ∈ I, x¯i ∈ R is the “ith coordinate”
function. Further, for each random variable ¯x∈R, we choose a subset Ω(¯x)⊂Ω
such that '
j∈J
x¯= ¯xj= Ω(¯x) mod 0
(here we use the completeness ofB). Finally, we define ¯h∈R(1) as follows for every ¯x∈Randω∈Ω:
¯h(¯x)(ω) =
0, ifω∈Ω(¯x), 1, otherwise.
3.7. Correctness Lemma
(a) For fixedx,¯ ¯h(¯x)is measurable as a function of ω, so thath¯ maps R toR.
(b) For everyx¯∈R we have
¯h(¯x) = 0= '
j∈J
x¯= ¯xj.
(c) ¯h∈R(1) (see 2.7), so that there exists a pointξ ∈M for which hξ = ¯h.
Proof.
(a) ¯h(¯x) takes only the values 0 and 1 on Ω, and the set where it takes each of these two values is measurable by the definition and by the completeness ofB.
(b) is obvious from the definition.
(c) We must verify that for all ¯x,¯y∈Rwe have (ω∈Ω|x(ω) = ¯¯ y(ω))
(
ω∈Ω|¯h(¯x)(ω) = ¯h(¯y(ω)))
mod 0.
We shall show that the set of points ω ∈Ω for which both ¯x(ω) = ¯y(ω) and
¯h(¯x)(ω)= ¯h(¯y)(ω) has measure zero.
It suffices to consider the case ¯h(¯x)(ω) = 0,¯h(¯y)(ω) = 1, i.e., to show that x¯= ¯y ∧ h(¯¯ x) = 0 ∧ ¯h(¯y) = 1= 0.
We write the second term in the form,
j∈Jx¯= ¯xj(by 3.7(b)) and apply the distributive axiom to the first and second terms (where we use the completeness of B). We further use the fact that x¯ = ¯y ∧ x¯ = ¯xj y¯= ¯xj. We then obtain
x¯= ¯y ∧ ¯h(¯x) = 0 '
j∈J
y¯= ¯xj=h(¯¯ y) = 0,
which immediately gives us the required result.
Explanation.Since the choice of ¯his the essential step in the proof, we would like to give some motivation for this choice. Recall that his the name of the function the cardinality of whose set of zeros interests us. We choose a concrete h¯ to “disprove” the CH in such a way that the “almost everywhere zeros” of ¯h include the elements of the set{xj|j∈ J }, which has intermediate cardinality in the naive sense of the word (compare with §1). However, ¯h cannot be an arbitrary map from R to R; it must satisfy the strong condition ¯h ∈ R(1). Hence, along with all the ¯xj, the almost everywhere zeros of ¯hmight also have to include various other ¯y ∈R, and might have to “partly include” still other
¯
z ∈ R. We say “partly include” to convey the possibility that ¯h(¯z) = 0 is neither 0 nor 1, so that ¯z has a “certain probability” of being a zero of ¯h.
Thus, the “set of zeros” of ¯hmight be bigger than we want, and we might expect to encounter difficulties in proving that this set cannot be mapped onto all ofR(the alternativeP1). On the other hand, it would seem that this situation would make it trivial to disprove the alternativeP2(mappingZ onto the entire set of zeros). But even this is wrong! As we noted before, we can haveZ(¯x)= 1 for many ¯xthat are not constant integer functions on Ω. Moreover, for still other
¯
xwe haveZ(¯x) = 0,1,so that the “set of integers” in our model has grown considerably.
A final remark: In this discussion we have been essentially dealing with the concept of a “B-random set,” which will be a central idea in what follows (see §4). That is, the “set of zeros of ¯h” is random in the sense that for each
¯
z∈R, the assertion “¯z∈(zeros of ¯h)” is naturally assigned the Boolean truth valueh(¯¯ z) = 0.
We now return to the proof thatCH= 0.
3.8. Proof that P1(ξ) = 0. By the rules for computing truth functions, we obtain
P1(ξ) ='
¯ g
%
¯ y
'
¯ x
*¯h(¯x) = 0 ∧ y¯= ¯g(¯x)+ ,
where ¯hwas defined above, ¯gruns through all elements ofR(1), and ¯xand ¯yrun through all elements of R. We suppose thatP1(ξ)= 0, and show that this leads to a contradiction. We write the above formula forP1(ξ) as,
¯ ga(¯g).
IfP1(ξ)= 0, thena(¯g)= 0 for some concrete function ¯g∈R(1). We take this function ¯g and set
a=%
¯ y
'
¯ x
'
j∈J
x¯= ¯xj ∧ y¯= ¯g(¯x) .
Here we have substituted ,
j∈J x¯ = ¯xj for ¯h(¯x) = 0 using 3.7(b).
Furthermore, we obtain x¯ = ¯xj ∧ y¯ = ¯g(¯x) y¯ = ¯g(¯xj). Using this and distributivity, we obtain
a%
¯ y
'
j∈J
y¯= ¯g(¯xj). In particular, for each ¯xi in place of ¯y, we have
a '
j∈J
¯xi= ¯g(¯xj).
If, as we have supposed,a= 0, then for eachithere exists aj(i)∈ J such that x¯i= ¯g
¯ xj(i)
= 0.
Since I is uncountable and card J < card I, it follows that there exists a j0 ∈ J such thatj0 =j(i) for alli in an uncountable subsetI0 ⊂I. But this contradicts the countable chain condition onB, because the terms in the family x¯i= ¯g(¯xj0)(i∈I0) are pairwise disjoint. In fact,
x¯i1= ¯g(¯xj0) ∧ x¯i2 = ¯g(¯xj0)x¯i1 = ¯xi2= 0
ifi1=i2.
Notice to what extent this proof parallels the “naive” argument in §1. By assumption, the function ¯y maps the zeros of ¯h ontoR “with nonzero proba- bility.” But the exact meaning of the computations cannot readily be stated in words.
Computation ofZ(y). The formula forZ(y), “y is an integer,” was given in 2.3. Since this formula occurs in P2, we must compute Z(y) in order to computeP2.
3.9. Lemma.Let η∈M andyη=y∈R.Then Z(y)(η) = '
n∈Z
y¯=n={ω∈Ω|y(ω)¯ ∈Z} mod 0.
Proof.We must show that
%
f¯
f¯(0) = 0∨ '
¯ x
f¯(¯x) = ¯f(¯x+ 1) '
f¯(¯y) = 0
= '
n∈Z
y¯=n. We prove this equality by proving inequality in both directions.
The inequality. It suffices to find a concrete function ¯f ∈R(1) for which the corresponding term on the left is contained in the right-hand side. We define ¯f by setting ¯f(¯x)(ω) = sin2π¯x(ω) (here, instead of sin2πz, we could take any measurable function with period 1 and zeros only at the integers).
It is easy to see that ¯f(¯x) ∈ R and ¯f ∈ R(1). Then f¯(0) = 0 = 0 and f¯(¯x) = ¯f(¯x+ 1)= 0.Hence we need only verify that
sin2π¯y= 0 '
n∈Z
y¯=n, and this is obvious.
The inequality . It suffices to show that for any fixed values of n ∈ Z, f¯∈R(1) and ¯y∈R, we have
y¯=nb∨c, where
b=f¯(0) = 0∨ '
¯ x
f¯(¯x) = ¯f(¯x+ 1)
; c=f¯(¯y) = 0.
But the inclusion a b∨c is equivalent to a∧c b. Furthermore, in our situation we have
a∧c=y¯=n ∧ f(¯¯y) = 0f¯(n) = 0.
(Here n in ¯f(n) is the constant random variable that is everywhere equal to n.)
It is thus sufficient to see that f(n) = 0¯ f(0) = 0¯ ∨
'
¯ x
f¯(¯x) = ¯f(¯x+ 1)
, or, taking complements, that
f¯(n) = 0f¯(0) = 0 ∧ %
¯ x
f(¯¯x) = ¯f(¯x+ 1)
.
The right side can become larger only if we only take the intersection over the terms with ¯x= 0,1,2, . . . , n−1.But this obviously gives
f¯(0) = 0∧f¯(0) = ¯f(1) =· · ·= ¯f(n)f(n) = 0¯ . 3.10.Proof thatP2(ξ) = 0. Using Lemma 3.9 and the rules for computing truth functions, we find that
P2(ξ) ='
f¯
%
¯ y
¯h(¯y) = 0∨'
¯ x
'
n
x¯=n ∧ y¯= ¯f(¯x) .
Since ¯f ∈R(1)we havex¯=nf¯(¯x) = ¯f(n), so thatx¯=n∧y¯= ¯f(¯x) y¯= ¯f(n).
Now it suffices to prove that the term corresponding to any concrete choice of f¯is equal to 0. We suppose that this is not the case, and show that we obtain a contradiction. Let a = 0 be the term corresponding to ¯f. By the previous paragraph, we have
a%
¯ y
¯h(¯y) = 0∨'
n
y¯= ¯f(n)
. In particular, for everyj ∈ J we must have (with ¯xj in place of ¯y)
a'
n
x¯j= ¯f(n)
(where we haveh(¯¯ xj) = 0= 0 by 3.7(b)). Hence, for everyj there exists an integern(j) such that 0=x¯j= ¯f(n(j)). SinceJ is uncountable, there exist ann0 and an uncountable subsetJ0⊂ J such thatn(j0) =n0for allj0∈ J0. Then thex¯j = ¯f(n0) forj∈ J0 form an uncountable set of pairwise disjoint nonzero elements ofB. This contradicts the countable chain condition onB.