An Exposition of Forcing

AN EXPOSITION OF FORCING A. Mostowski

P.J. Cohen invented, in 1963, a method of constructing models

for Zermelo-Fraenkel set theory, hereafter abbreviated ZF.

In the

present lectures we shall describe this method with some modifications (due mainly to Solovay) and apply it to a proof that the continuum hypothesis is independent of ZF.

The independence proof is taken over

from Cohen [66] without change. I wish to thank Dr. W. Guzicki for his helpful discussions on the topic of these lectures and for having shown me his notes of similar lectures delivered in the University of Nijmegen.

The language of ZF is the first order language with identity and with one binary predicate We write

x &y

instead of

' • • , E. ~, v, Yx, 3x.

3x [x & a S FJ

VX&z(z & y)

and

We shall denote this language by

&(x,y).

The logical symbols we use are:

We abbreviate

by 3x&a F. x C y

&.

Vx(x e a . FJ x S y

Moreover we write

instead of

x S YS x

by

~

abbreviations are used for other variables also.

y.

Vx£a F and

instead of

The same

L.

(1I2J.221

417

AN EXPOSITION OF FORCING

If

•

is a formula of

the free variables of

L then

Fr(t)

denotes the set of

t.

The axioms of ZF are as foll,,"s: \Ix '#y [x ., y ;: '#z (z £:

1. Axiom of extensionality:

IIx lIy 3z lit [t £: Z

2. Existence of pail'S: 3. Existence of unions:

¥x.3y '#z [z £: Y

4. Existence of power sets:

7..

IIx lIy£:x 3ztx

8•• Axiom scheme of replac_ent: 7..

not

y

can be any fonnul. of

•

is free;

variables

z. t

in

8..

•

i

y)J.

= y)] •

t

v

(z £: t)J. Z

~

x L,

:ax 3y£:x \lz£:x 3ttX [z c t].

Axiom scheme of comprehension:

In

=x

- 3t;£:x

IIx 3y liz [z e Y

S. Existence of infinite sets:

6. Axiom of foundation:

-

(t

X ;: Z £

IIttX ['(t t z)].

IIx 3 Y '#z [z e Y ;: (. , IIx 3y lIux L

lit

rt .. 3tty

in which the variable

can be any fonnula of

are free but the variable

y

Z

e x) J •

.J. z

but

in which the

L

is not free.

The axiom of choice is the following sentence:

AC.

IIx lIytx \lZtX 3sty lit ['(t t y • t £: z) '#t£y (t t w

=t

v

Z

= yJ

+

3w '#ytX 3vty

., v).

We shall denote by ZFC the system obtained by adjoining AC to ZF. Our meta-theory in which we shall study models of ZF will be the set theory ZFC enriched by one additional axiom 8M due to Cohen. We shall formulate this axiom below after introducing sOlIe definitions. We shall freely use the current set-theoretical notation and shall write set-theoretical formulae using the saae logical SymbOlS as in the lansuage by

E.

L.

The membership relation however will be denotad

418

FOUNDATIONAL STUDIES

A family implies

F

of sets is called

[1121.222

t~ansitive

if

x EyE F

x E F. We shall assume as known the concept of a relational system

and the notion of satisfaction of a formula in such systems. the customary notation assignment

a

MF

for:

~[a]

of elements of

~

is satisfied in

M to the free variables

use

\ole

M by the

of~.

shall use the same notation also for languages arising from

we

L 'by

adjunction of constants and additional predicates. Relatior~l

family of sets and

systems in which the universe is a transitive

E

is interpreted as the membership relation will

be called models. We shall denote by the same letter a relation.l system and its universe.

M

The eet of all assignments of elements of

the free variables of a formula

~

can then be denoted by

M to Fr M C. >.

LEMMA 1.1

If

w

M is a transitive family of sets such that the set

= {S,{e},{S,{e}}, •.• }

to the formation of ape valid in

pai~s

belongs to

M and

and unions then

M is closed with respect

a~ioms

1,2,3,5

and

6

M.

Proof:

routine.

The existence of a model

M in which the remaining axioms of

ZF are valid cannot be proved on the basis of the usual axioms of set theory even if we assume an additional axiom stating the consistency of ZF. The additional axiom SM which we mentioned above states:

[1121.223

419

AN EXPOSITION OF FORCING

(8M)

There is a denumerable model of ZF.

REMARKS (i) In view of the Skolem-LBwenheim theorem (SM) follows from a weaker axiom: (ii)

there is a model of ZF. From Gadel [40] it follows that each model

M'

contains a submodel

M'.

M of ZF

such that all the axioms of ZFC are valid in

Thus (8M) can also be replaced by the

axiom~

there is a model

of ZFC.

2.

Algebraic preliminaries Let

P

be a partially ordered set.

ordering relation.

The elements of

are denoted by letters an extension of p

q;

P

p, q, r,

if moreover

a proper extension of

q.

If

are called "conditions" and If

p # q

< the

We denote by

p < q

then

then we write

3r [r < p & r < qJ

p

is called

p < q then

and call p

and

q

r < p

and

r

are called compatible, otherwise incompatible.

Assumptions concerning

P.

1.

< is a partial ordering.

2.

Each condition has a proper extension.

3.

If

P

"q

then there is an

is incompatible with

r

such that

q.

Partial orderings satisfying 3 are called separable.

Fitters. A set (i)

F

~

e,

F

£

P

is called a filter if

420

[1121. 224

FOUNDATIONAL STUDIES

P

(ii)

<;

q

and

P E F and

(iii)

p

e

F

q E F

imply

q

E

F.

imply 3r£F [r '" P S

r'" q L.

LEMMA 2.1

Each filter can be e:tended to a ma:imal filter. Proof.

Use Zorn's lemma and the remark that the union of

a chain of filters is a filter. LEMMA 2.2

If

is a filter and

F

is incompatible with an eZement

p

of

F then there is a ma:imal filter

F

F'.

~

Proof.

such that

p I/. F I

and

Use Zorn's lemma to obtain a maximal element

of the family of filters containing

P

FI

F

F'

as a subset and not containing

as an element.

LEMMA 2.3

If

q

q " p

p.

but not

Proof. r '" q

then there is a mazimaZ fiZter containing

and

r

ex e p : X

~

By separability

is incompatible with

r}

~here

p.

is an

r

in

P

such that

By 2.2 the filter

can be extended to a maximal filter and this filter

satisfies 2.3. LEMMA 2 ...

If an the

Po ~ Pl ~

then there ie a mazimaZ fiZter containing

Pn' Proof.

Extend the filter

{x e p

3n (x

~

Pn)}

to a

[112).225

421

AN EXPOSITION OF FORCING

Ill4ximal one. We shall now define a topological space.

Let X, or more

precisely X(P,<), be the set of all maximal filters of as an open sub-basis of X [pJ

= {F

sets of the form

[pJ

P and take

the family of all sets

E.x : p E F}.

will be called

n.ighbou~hoodB.

THEOREM 2.5

th.o~e",.

Proof. neighbourhoods The

if

E

[r]

r

in and

Hence arbitrary unions of open sets are open.

intersection of two open sets

F e [P2] ~

is an F

[pl.

F e '1 n 72

and

Open sets are defined as arbitrary unions of

F

then there are

f:1.

Hence

such that

Pl' P2

PI E F r < PI

[r] ~ [PI] n [P2J

such that Pz e F

and and

So IJh

void or is a union of neighbourhoods.

0/l'tt2

r

n ?'2'

< P2' Thus

Hence X

is open. F E [PI] ~

For

ffJ.

and therefore there It follows that Ojl n

0/2

is either

is a topological

space.

in

.x

for

We now prove that if ~ c Z and 8Y is open and dense 7n 1n n = 1,2,3, ••• then ~ tn ~ fl (Baire theorem).

From the density of Gn = {p e p : [p] So "In}' C7fn it follows that for each q in P there is a p in Gn such that p < q, i.e. that Gn is dense in P. Let PO be arbitrary. From Put

the density of

GI we infer that Pl < Po for some Pl in Gl• From the density of G2 we infer that P2 < Pl for some P2 in

G2•

422

[112}.226

FOUNDATIONAL STUDIES

Continuing in this way we obtain a decreasing sequence

Po

By 2.4 there is a maximal filtel'

fol'

and it follows that

3.

Fe (Pn J

F

=-t/fn

pn E F

such that for each

n, Le.

;>

PI

;>

each

n

FE () n ~n'

Heuristic remarks about modeZs in which the continuum hypothesis fails

We assume that we live in a world in which the continuum hypothesis (abbreviated Ca) is true and want to construct another world in which CH is false, i.e. in which there is an injection of

into

for some

a > 1.

f

No such injection exists in our

world but we can say that if it exists anywhere, we have in our world its finite approXimations. p

=-

with domains

wa x

W

hypothetical injection and putting on n)

=. {O,l}.

W

a

x W

(the value of f

D, belongs to our world.

into

{O,l}

form a set

the relation of inverse inclusion: maximal filter of

P

Indeed, if

D is a finite subset of

we obtain a finite approximation of

finiteness of of

and range

= fCt)Cn)

p(t,n)

D

These approximations are finite functions

f(~)

wa x

is our

f W

then

for the argument

which, because of the Mappings of finite subsets

P which is partially ordered by p < q

means that

p

~

Each

q.

determines a function defined on the whole set

W x W; conversely each mapping f of wa x W into {O,l} a determines a maximal filter of P consisting of all the finite

approximations of

f.

Maximal filters which belong to our world

determine functions which are not one-to-one.

It will be

ou~

to find an extension of our world with new maximal filters of

task P.

Two methods are known at present to construct such extensions. One of them identifies sets with two-valued functions whose values are arbitrary elements of suitable Boolean algebras.

Another method

[I 12l 227

423

AN EXPOSITION OF FORCING

identifies the "world" with a denumerable model M by ,adding to it certain subsets of

P

M of ZF and extends

not previously contained in

M. Below we shall sketch this second method.

4.

sets

Const~uetible

Let

M be a denumerable model of ZF and let

be the set of the ordinals of x

M.

We denote by

= Qn n M

o~

rk(x)

the rank of

defined by induction as follows rk(x) {x E M

The set

Let

rk(x) <

= sup{rk(y)

: y Ex},

will be denoted by

a}

X be a subset of

M

Ma·

whose rank

belongs to

aD

{)O according as a < aD and put II.a e a = We are going to define a family H(X] such that H

or 1\

X E H(X].

For suitable

v E Fr(~)

extension of

~

=

otherwise identical with

{x

E

L

such that

of

~

L

such

B : B F
x

to

v

and is

y.

The family of all sets formulae of

aD·

and

as

y

is an assignment which correlates

x,y

'>

y E BFr(~)-{v} we define the

B with respect to

E~,y,B

where

B of sets, each formula

and each assignment in

a

M(X]

X this family will be the required model.

For each family that

or ~

O~

v E

E~,y,B

Fr(~)

and

where y

~

ranges over

ranges over

is called the derived famiLy and will be denoted by

B'.

424

[1121. 228

FOUNDATIONAL STUDIES

We define now M[X] where and

BotX] Ba+1[X]

as the union

= I. BA[X] = U(Ba[X] = B~[X] U Ma U Aa

Elements of

M[X]

: a < A}

U{Ba[X] : a e if

O~}

A is a limit number

will be callecl constJOuctibZ. i71

The following properties of the families

B [xl a

and

H[X]

X. are easy

to prove. LEMMA 1+.1

H So H[X). Proof.

m e H. then

If

m e Ha

where

a = rk(m)

and

LEMMA 1+.2

X e H[X]. Proof.

X e 1\ and hence aO

X e Ba +1[X]. 0

Proof.

We use induction.

The lemma is true for

LEMMA 1+.3

If it is true for for or

A-

y

If

X.

a

<

A where

xE Y E Ba+l[X] In the first case

Ba[X]

and hence

x E Ba[X].

hence

rk(x) < a

and thus

occur only if because LEMMA 1+.1+

X So H.

a > a o'

a

= O.

A is a limit number then it is true y E Ha is the extension of a formula in

then either y

y E

B~[X]

In the second case

or

rk(y) < a

and

The last case can x e Mrk(x) So Ba+l rxr. x e X. rk(x) < a and x E M

We have then

Hence. x e Ma

and

x e Ba+l[Xl.

1112J.229 Proof.

B =G

+

1.

It is sufficient to prove the lemma for the case

Let

the assignment in

425

AN EXPOSITION OF FORCING

Ba[X]

a e Ba[X]

and consider the formula

y(w) = a.

such that

y

y

with respect to

because all elements of

a

transitivity of

Hence

Ba[X].

is

v e w

The extension of

e Ba[X] :

{x

belong to

v e wand

Ba[X]

x

e a}

=a

n Ba[X] • a

in view of the

a e Ba+l[X].

LEMMA II.S

H[X] i.

t~aft.itive.

Proof.

The union of a chain of transitive sets is

transitive. LEMMA 11.6

Proof. Ba[X]

Ba[X]

is the extension of the formula

v =v

in

with respect to the void valuation.

LEMMA 11.7

Proof.

By 4.6.

Proof.

For

LEMMA 4.8

a > 0

a

=0

there is nothing to prove. is a set of elements <

the lemma is obvious.

and the lemma holds for

whi~h

C and the lemma follows.

If

C

<

a.

a =

belong to

If

c+1

a

and

Assume that

is a limit number x E BC+l[X]

then

BC[X], hence their ranks are

x

426

1112],230

FOUNDATIONAL STUDIES

LEMMA 4.9 M[X]

is cZosed under the fopmation of pairs and

Proof. that

Let

a, b e Ba[XJ.

Ba[XJ

a, b e M[X]

and let

be an ordinal such v

=w

a, y{u)

b

The extension of the formula

with respect to the assignment

The extension of the formula respect to the assignment are elements of

a

yew)

3u (u E W & v E u)

y(w): a

is

Va.

unio~.

in

Thus

v V

=u

is

Ba[X) {a,b}

in

{a,b}. with

and

Ua

Ba + 1[Xl.

LEMMA 4.10 Azioms 1, 2, 3, 5, 6

ProOf. that the set hence of

of

ZF

are vaZid in

M[X}.

This follows from Lemmas 1.1, 4.5, 4.9

a~d

w referred to in' Lemma 1.1 is an element af

the remark

M and

M[}G.

LEMMA 4.11 The ordinals of Proof.

x

M[X)

is an ordinal of

transitive set which belongs to transitive sets.

are the same as the ordinals of

M[X)

M[X)

ordinal of

5.

is

<

M.

if and only if it is a

and whose elements are

A transitive set all of whose elements are

transitive sets is equal to its rank. of

M(X)

M.

OnM.

By 4.8 the rank of an element

It follows that each ordinal of

M[Xl

is an

The converse implication is obvious.

The ramified language

RL

The important feqtur€

of the technique invented by Cohen is

that it allows us to speak about the model speak in the model

M

M[X]

{we speak in "our world"

remaining &0 to M about the

[112j.231

427

AN EXPOSITION OF FORCING

"fictitious world" which extends each element of

MOO

M).

has a "name"

This is due to the fact that in

M.

These "names" will be

expressions of an auxiliary language which we shall call the ramified language or RL for short.

The expressions of RL will be elements of

M. The language RL has an infinite v O' VI" •• ,

two ~inary predicates

EO,"'

number of variables called the membership and I, & and the universal

identity predicates, propositional connectives quantifier

l,!.

Besides these expressions the language RL will

have infinitely many constants and infinitely many one-place predicates which will be described a little later. The rules of formation will be the usual ones. constants and variables are terms of RL. and

V

and

l,!v

If

and

~

for each variable

(~)

free and bound occurrences known;

Fr(~)

of

for each

v

in

~

Thus t

are terms

2 Vt

t

v.

l

EO t

are (~)

&

(~),

The distinction between

a variable in a formula is assumed as

is a sequence of constants and

~

l t

and

denotes the set of all free variables

the formula resulting from out

t

... t , l 2, l z ware formUlae, then so are

is a one-sp Le ce predicate then

atomic formulae. I(~)

If

Dom(y)

If

of~.

~ Fr(~)

then

~(y)

by substituting

y(v)

for

Y

denotes v

through-

Dom(y).

Writing formulae of RL we shall often use connectives

& and

other than

I

and also the existential quantifier.

symbols are then thought of as abbreviations. use letters

v, w, u, etc. instead of

These

Also we shall often

v o ' VI' v 2' •..

Since we want to treat expressions of RL as elements of we identify the primitive symbols of RL with certain elements of

M M

428

1112],232

FOUNDATIONAL STUDIES

and agree that if an expression is obtained by writing the symbols A. B. C, •••• H one after another. then the whole expression is to be

(A.B.C ••••• H>.

identified with the sequence We identify

vj

(O.j > and the symbols

with

(l.j > • j

( • ).\1 with the pairs

= 0.1, ••• ,5,6.

E, "'.

Elements of

&. ' .

M

which

will serve as constants and as one-place predicates will be defined later. We note that from now on

will be treated as a part of

L

RL. We describe now the additional predicates and constants of RL. For each ordinal a E OnM we have in RL a one-place predicate Va which we shall identify with the pair (2,a>. Intuitively Va.

Ba[X).

denotes the set We put For

a. in

O~,

each sequence

y

!!!. = ( 3,m > for each

for

,

c a,~,y

M and

L such that

Dom(y) =

with domain

in

iii

a formula of

Fr(~)

- {v}

a = (11,0

>.

v e Fr(,)

and

we put

=(S,a,~,y>.

Constants of RL can now be defined by transfinite induction: Co

= m. CA = U{Ca

: a. < A}

for limit numbers

Ca + l = Ca U {!!!. : rk(m) < a} U fa U

where

fa

=m

or

fa = {a}

C'a is the set of all and y E Ca Fr(,)-{v} •

according as where

ca."y

f

C~

a < a O or a > a O and is a formula of L. v E Fr(fi

Elements of constants of RL.

For each

c

in

A.

C are called

C we define its order

p(c)

as

(112).2JJ the least

a

AN EXPOSITION OF FORCING

429

Thus- p(c)

is always a successor

such that

cECa'

ordinal. The intuitive meaning of the constants is as follows: denotes in

m,

Ba[X]

denotes

0

X and

Ca,~,y

~

denotes the extension of

~

with respect to the assignment which correlates with each

variable

wE

constant

y(w).

different from

Fr(~)

v

the object denoted by the

A formula of RL in which each quantifier

¥x

(where

any variable) is followed by an expression of the form called a limited formula. to quantifier within

th~

The index formula.

a

(Vax

+

x

is

W)

is

may differ from quantifier

We shall abbreviate The order

¥x[Vax

+

W]

of a

p(~)

limited formula is defined as the larger of the following two ordinals: max{p(c) : c

occurs in

~},

max{a : Va occurs in

~}.

Several times we shall have occasion to use an ordering of limited sentences defined as follows:

~


~

if and only if one

of the following conditions is satisfied: 1.

p(~) <

2.

p(~)

=

peW); p(~)

and

~

contains fewer occurrences of logical

operations (i.e. connectives and quantifiers) than 3. form

cl

~

and

E

c2

~

fnd

Ware atomic sentences, with

p(c 1) < p(c 2)

= peW)

p(~)

whereas

~;

~

and

~

has the

does not have this

form; 4. c

1

... c 2 5.

c l ... c 2

and ~

and

where

~

are atomic sentences,

W has the form

p(~)

c 3 E c 4 with p(~) are atomic sentences, W p(c

l)

;>

p(c 2)

and

= p(W), p(c

3)

~

;>

= p(~),

W has the form

has the form

p(c 4)· ~

Vac l ·

has the form

430

FOUNDATIONAL STUDIES

If arises from

is a formula,

$

$

x

is a variable and

by replacing an occurrence of the quantifier

¥x [Vax+\(acl] then

~

is said to arise from

of the considered occurrence $(a)

a E On M,

[112J.234

of

¥x

a formula wmch arises from

$

to

Ira'

$

~

¥x

by

by a relativization We shall denote by

by relativizing all occurrences

V. a

of the quantifier to LEMMA 5.1

Ca E M for each

a E On •

M

This follows from the remark that the operation which yields

Ca

from the sequence

{C~}~
is definable in

M.

LEMMA 5.2

a E OnM the set of aLL formuLae of order

For each be longs to

a

M.

This is so because such formulae are finite sequences built according to recursive rules from variables and symbols ¥,

E,

c

where

~

< a

set which belongs to

and

c

I, &, ~, \(~,

Since all these symbols form a a M, so do all their finite sequences. In view E

C

of the recursive character of the formation rules, the formulae them-

M.

selves also form a set in

In connection with these lemmas it is worth noting that neither

C nor the set of all formulae of RL belong to

of course they are definable subsets of

M, although

M.

LEMMA 5.3

The relation the limited sentences. order

< a

is a set in

<

is a definabLe partiaL well ordering of

Its restriction to the set of sentences of

M for each

a E On .

M

1I12J, 235 Proof. 5.2.

The second part follows from the first and Lemma

«

The definability of

ion

p

43l

AN EXPOSITION OF FORCING

follows from the remark that the funct-

<

and clauses 1-4 of the definition of

The well-foundedness of

are definable in

is proved as follows:

~

M.

A be a non

Let

AO its subset consisting of sentences of a possibly small order. If AO does not contain atomic

void set of limited sentences and

sentences of the form

c

p(c

c'1 '" c'2

p(c ) l) < 2

l

c2

~

If

or

ci

hO

p(c

then no element of

A. l)

c'1

Finally i f

< p(c )

AO

c

r

c

l e c2

with

AO is a

which belongs to

2

AO contains at least one sentence

then each such sentence is a minim?-l

2

AO'

element of

Limited sentences

<

respect to

are

which have no predecessors with

~

~ ~ ~, ~

~

and

Values of the constants.

M and has rank

belongs to

induction the value of by

2

contains at least one sentence of

then each sentence

with

c

~

but no sentence of the form

minimal element of c

c2

~

A.

has a predecessor in the form

l

c

aO'

Vo~.

Let

be a subset of a set which

X

For each constant

for the argument

X;

c

we define by

we denote this value

c*(XJ. Let us

ass'~e

C

e

< a.

and that

c

limit

there is nothing to define.

e

with

a E On M

defined for nuw~er

in

that

Let

c*[XJ If

cECa' If

a

~

e,

is already

=e

a

+ 1

is a

there are

three possibilities: 1)

c

E

C'

e'

2)

c

In cases 2) and 3) we put In case 1)

=m

where

c*(XJ; m

where

rk(m) and

c*[xJ

is a formula of

c = a.

3)

=X

respectively.

L, v

E

Fr(41)

and

432

[1121.236

FOUNDATIONAL STUDIES

y e c. Fr(,.)-{v}.

In this case we put

v

y*(X]

is a sequence with domain

of the terms of

c*[X] - E -

Fr(~)

- {v}

where ,.,y*IX].Be[X] whose terms are values

y.

LEMMA 5.11

a E OnH the set ~here c ranges over

For eaoh c*(X]

of all

BaIX]

is identioal with the set

Ca'

Proof by an obvious induction. Lemma 5.11 says that each element of matter of fact each element of x

in

H(X]

element of

M(X] has a name in

M(X] has many names;

the set of all its names is a subset of

As a

C.

given an element

M but not an

H. It can be shown that the function which correlates

with

c

is definable in

H(X].

c*IX]

This will follow from theorems which

will be established in the next section and from the following weaker result: LEMMA 5.5

Let

N be a model of ZF suoh that

N ~ M and let

C and

a function of two argument, the first of whioh ranges ov.r the .eoond over eubsets belonging to a O and whioh is defined

rank f

of

c.

valid in

the equation

P

E

Y.

f(c,X) = cAIX].

of Then

N.

is definable in Proof.

b~

N of a fi=ed sst

be

f

f

is defined by transfinite induction on the order

Since the theorem on definitions by transfinite induction is N. we obtain the desired result. We close our discussion with a remark on semantics of the

language RL.

The relational structures which can serve as models

[112], 237

433

AN EXPOSITION OF FORCING

for formulae of RL have an infinite type because there are infinitely many constants in RL and also infinitely many one-place predicates in addition to the two binary ones

and

E

-.

Apart from that the

model theory of the language RL does not differ from the model theory of any first order language. In most cases we shall deal with models whose universe is M[X] Ba[X]

and where the predicates and the constants

briefly by

c

as

$\::I,

"«

are interpreted as

c*tX].

MIX] to M

The following example shows that, in general, a model of ZF mXn

=,

E,

Such a model will be denoted

M[X].

Reduction of properties of

6.

E:,

Let

X be a subset of

MiX]

is not

w x w such that the relation

w similarly to

orders

OnM. The existence of X follows M and hence also OnM are denumerable. is equal to OnM because each element of

from our assumption that The family

On n M[X]

M(X] has a rank X

E

M(X]

and

< O~

(see Lemma 4.11).

X has the order type

O~

On the other hand and so the theorem:

each well ordering there is a similar ordinal" Yet this theorem is provable in ZF. valid in

is not valid in

"for M[X].

Hence not all axioms of ZF are

M[X]. There are no general criteria which would allow us to

decide for which sets

X the family

M[X]

is a model of ZF.

sufficient condition which we shall discuss below says that the satisfaction relation (1)

M[X] 1= .p[y*[X]]

A

434

[112],238

FOUNDAT"ONAl STUDIES

be expressible in

More exactly we require that (1) be equi-

M.

valent to the fact that a formula depending on M by

,

be satisfied in

Y (i.e. the sequence of names of the terms of

by an element of

y*[X])

and

X.

To return to the intuitive picture given in Section 3 where

X was a maximal filter in

P

we may say that we require (1)

to be equivalent to a relation definable in of terms of

y*[X]

M holding between names

and an approximation of an object determined by

X. The exact definition is as follows: DEFINITION We say that of

L

X is reducible to

there is a formula

for each

M if for each formula

with one more free variable such that

y E CFr(~) M[X] I: Hy*[X]] " 3XEX [M

(2)

Instead of x

establishes

x

forces

,

,(y).

at

M I: y

we shall say that the condition

~,[x,y]

in

F ~,[x,y]].

(The term in general use is:

M[X].

We selected another term in order to reserve the

word "forcing" fer the case of a particular formula To derive properties of

M[X]

where

we need the notion of a normal function. strictly increasing and continuous mapping a

,

a critical number for

f

if

~,,)

X is reducible to

Such a function is a f : OnM +

We call

O~.

f(a) = a.

LEMMA 6.1

If for each

f

a O in

is a normal function

~hich

is definable in

OnM there is a critical number

a

of

f

M then such

M

[112]. 239

435

AN EXPOSITION OF FORCING

Proof.

Define a sequence

{an}

by induction on

n : a + = f(a n)· This sequence belongs to M since the theorem on n l inductive definitions is valid in M. It follows that e M and we easily prove that this supremum is the required n} critical number. sup{a

Another important auxiliary result is the following THEOREM 6.2

(the reflection theorem)

Let

X

be reducibZe to

M

and Zet

be a formuZa of



L.

There ellJists a normaZ function

f definabZe in M such that is a criticaZ number of f and y E CFr(
a

whenever

REMARK

The reflection theorem is really a theorem scheme: each formula function



identity function

= '<1>"

we construct separately a definition of a normal

f' Proof.

a

for

1;



& 1jJ

If



f(x)

has no quantifiers then we take for x.

If

then we take

fact that a superposition

f

0

fa g

the

are already defined and

f' f1jJ

= f'

f

fe;

= f

0

Using the

f1jJ'

of twb normal definable functions

is again such a function and each of its critical numbers is a critical number of the functions

f

and

g

we convince ourselves

easily that if the reflection theorem is valid for the formulae <1>,

1jJ

then it is also valid for the formulae

'

theorem is also trivially valid for the formula

and 3v





if

& 1jJ. v

The ~

Fr(
436

[112], 240

FOUNDA T10NAL STUDIES

We shall now prove the theorem for the formula v

E

y E CFr(~)-{v}

where

we put

MF t,[P.(c,y)]}}.

g(y) = sup{min {3CEC peP nEOn n M Thus

at

is the least ordinal

g(y)

condition

p

(c,y)

h(O)

in

p

g

= 0,

h(A)

establishes

such that

~

at

we define a function

= sup{h(a)

is of course definable in

continuous i.e.

h

M.

(c,y) h

p c

establishes already in

in

CA

M.

by transfinite induction:

a < A} if

max Oit c) + 1, sup{g(y) : y

definitions is valid in

I~

c

then there is a constant

M[X]

Using

h(a+l)

h(a+l)

A with the property that for each

if there is a constant

such that the same

h

~

Fr(~).

For each

~

3v

E

A is a limit number, r ( , ) - {v}}).

c/

M since the theorem on inductive Moreover

h

is strictly increasing and

is normal.

follows from the last clause of the definition that g(y)

is an upper bound of the values of

for

ranging over

y

Hence if A is a limit number and y E CAFr(~)-{v} g(y) < h(A) because there are only finitely many terms of y

C Fr(~)-{v} a • then

and each of them belongs to a a < A

such that all terms of

g(y) < h(a+l) < heAl Now let

f

because

with

hence there is an

a < A;

y

belong to

h

is strictly

be the normal function

one of its critical numbers. then

Ca'

Ca

and so

incr~asing.

f~

0

h

We shall show that if

and let y

E

a

be

C Fr(~)-{v} a

[112),241

437

AN EXPOSITION OF FORCING

First assume the right-hand side of this equivalence. Hence there is an element satisfies

in

~

x

Ba[X].

of

Let

which together with

Ba[X]

be a name of

CECa

inductive assumption and the remark that

x.

y*[X]

Using the

is a critical number of

a

we obtain the left-hand side of (3), There is thus an

Now we assume the left-hand side of (3). element of

M[X]

which, together with

This element can be represented as Since we assume that

X

which establishes

at

g

~

c*[X]

number of

at

~ f~

satisfies

where

c

in

(c,y)

M[X].

in

(cl'y)

M[X].

M[X] F

we obtain

M[X].

is a constant. in

p

X

Using the definition of

in

cl

ih

~

is reducible we obtain a condition

we infer that there is a constant

establishes

y*[X]

Cg(y)

Since

a

such that

p

is a critical

y*[X]] and since

~[cf[X],

In view of the g(y) < heal = a we obtain (c,y) E C Fr(~). a inductive assumption we obtain therefore the right-hand side of (3). We can now verify the validity of Axioms 4, 7 and 8 in

M[X].

THEOREM 6.3

X is reduaibZe to

If is vaZid in

M[X].

Proof. Let

a E M[X]

is a

b

in

Let

M[X]

~

be a formula of

g E M[X]Fr(~)-{v}.

and

We can represent and

y E CFr(~)-{v},

all the terms of

y

a

as Let

be elements of

x ~

in

such that

v E

Fr(~).

M[X]

~[(x,g)].

c*[X] a

L

We have to show that there

such that for each x E b _ x E a & M[X]

(4)

c E C

M then the axiom of aomprehension

and

g

as

y*[X]

where

be an ordinal such that Ca'

Using the reflection

c

and

438

[1121. 242

FOUNDATIONAL STUDIES

theorem we find an ordinal

a

~

a

such that for each sequence

(cl'Yl) e CaFr(~)

i =

Let us consider the formula a new variable and let on the free variables of c

with

w.

i

the variable

We claim that the set

By definition

b

• (v

~

w)

£

b

is the extension of

w

is

= Ca,i,y[X] i

in

Y

excepted and correlates

(~).

satisfies

Ba[X]

with respect to

The last equivalence is obtained bv remarking that a e B [X] a

THEOREM

and hence

lJa lid in

M[X]

~

= a,

Theorem 6.2 is thus proved.

B [X]. a

X is reducibZs to

M then the a=iOM of repZaceMent is

M[X]. Proof.

and let

a

c*[X]

6.~

If

H[X]

w

Le.

:y*[X]

in

where

Y be an assignment which coincides with

Let

~

be a formula of

g E H[X]Fr(~)-{v,w}, a E M(X]. such that whenever

satisfying

H[X]

x

is in

~ ~[x,t,g],

L

such that

v, w e

We have to find a set a

and there is a

there is a

tl

in

b

t

Fr(~)

b

in

satisfying

the same formula. Similarly as in the previous proof we determine so that b

CECa' y.e CaFr(~)-{v,w} and

= BS[X] where

a

= c*[X],

g

a, c, Y

= y*[X]

and put

[112J, 243

Note:

k

439

AN EXPOSITION OF FORCING

and

d

to the

co~~espond he~e

v

va~iables

and

w

~espectively.

Obviously x

and hence

b E MIX].

has a name

MIX] F 1/IIx,t,g]

that

constant p E X

d

in

k

then

t

a

definition of establishes

the~e

at

1/1

at

1/1

in

the~e

is a

t l = dlIX]

M[X].

in

l

in

t fo~

C

a

such

a suitable is a

In view of the such that

MIX] l= 1/IIx,dl[X],g]

in

MIX]

p

Thus we obtain

MIX].

M[X] F 1/I[k"[X], dlIX], y"[X]], i.e. that

d

x E BaIX]

then

~educibility the~e

in

(k,d,y)

is a constant

(k,dl,y)

a

C • If the~e is a a has the fo~m d"[X]

and hence by the assumption of

which establishes

p~oved

is in

x

If

b

satisfying

and we have

1/IIx,tl'g]

in

M[X].

THEOREM 6.5

If valid in

X is reducibZe to

M[X]. P~oof.

and

x E M[X]

p~oof

x

in

Let

then

is to show that

of

a

M then the aziom of power-ssts is

x

a E MIX], a

= c"[X]

has a name in the~e

is an

C.

whe~e

CECa'

If

The essential step in the

o~dinal

which has a name (i.e. belongs to

a

such that each subset MIX])

has a name

al~eady

Ca' For any constant

d

we consider the set

This set obviously belongs to

x c a

M.

More exactly

Sed)

440

[112),244

FOUNDATIONAL STUDIES

belongs to the power-set of

P x Ca, taken in the sense of M. We PM(P x Ca ) . Furthermore let E be a family

shall denote this set

(p ,s ) e P

consisting of all pairs C

M such that there is a

x

d

in

satisfying the conditions:

(5)

M F vCw[p,d,c],

(6)

s = Sed).

E belongs to

The family subset of the set (p,s) E

E

let

(6) hold for a

P x PM(P x C ) a

a(p,s) d

be the minimal ordinal

in

and let

C~

claim that each subset in

x

of

a

M.

such that

~

a = sup{a(p,s) : (p,s) which belongs to

(5)

and We

E}.

E

M[X]

For

has a name

Ca' Let us assume that

x

that ing

M because it is a definable

which is an element of

C

(5).

there is a

a.

Since

X

in

C a

is a name of a set

is reducible to

Thus the pair dl

d

(p,S(d»

x E M[X]

M there is a

belongs to

E.

p

such

satisfy-

It fo.llows that

such that

(7)

Sed) = S(d l).

(8)

We shall show (9 )

First notice that from (7) we obtain p E X and thus

Now assume that obtain

dl[X]

~

cALX]

because

dl[X] C a.

Y E dA[X] = x.

Since

has a name, say

k,

in

x ~ a ~ Ba[X]

C. a

From

we

AN EXPOSITION OF FORCING

[112]. 24~ k*[XJ E d*[X] kEd

in

we infer that there is a

M[X].

(q,k> E S(d l).

Hence Thus

q E X, we obtain

(q,k> E Sed)

q

q

in

441 X

which establishes

and, in view of (8),

establishes

k E d

in

l

M[X]

and, since

y E dt[X].

Let us assume conversely that

y E dt[X].

Since

dt[X]

~

a

C B [XJ we can put similarly as above, y = k*[X] where k E C and a a we obtain a q in X such that (q,k> E S(d From this and from l). (8) we infer (q,k> E Sed) and hence k*[X] E d*[X], Le. y E d*[X]. The formula (9) is thus proved. the extension in assignment

yew)

Ba[X]

= a.

consists of subsets of a name in formula

To finish the proof we denote by

of the formula The set

b

a.

x C a

If

v C w

with respect to the

is an element of

C and thus belongs to b. a ¥x [x E b x ~ aJ in M[XJ.

and

x E M[X]

Hence

b

b

M[XJ then

and x

has

satisfies the

=

Theorems 4.10 and 6.3 - 6.5 show that if ZF then so is

M[X]

provided that

X

M is a model of

is reducible to

M.

For

models of ZFC we have the following result:

THEOREM 6.6

If M[XJ

M is a modeL of ZFC

is a modeL of Proof.

well-orders M[XJ

a

and

is reduoibLe to

M, then

ZFC. Let

a E M[X].

We shall exhibit a relation which

and is an element of

is a subset of a set of the form

exhibit a well-ordering of Let

X

M[XJ. Ba[X]

Since each element of it is sufficient to

Ba[XJ.

< be a relation which well-orders

C a

the existence

442

[112], 246

FOUNDA T10NAL STUDIES

of

< follows from our assumption that the axiom of choice is valid

in

H.

Let

f

fCc_X) = c*[XJ

be a function definable in for each

c

earliest name of an element f(c'_X)

~

x

c'

whenever

C

in

a

of

H[X]

such that

(see Lemma 5.5). H[X]

Call

fCc_X) = x

if

the earliest name of

x

the earliest name of

y.

7.

The proof that

R E H[X]

7

formulated in RL.

The theory ~

the sense that for each filter A particular model of ':7

method due to Henkin. sequence

the or

<)

is immediate.

Heuristic ezplanation of the construction of reducible filters

a theory

~,

R is easily if and only if

precedes (in the sense of the relation

We shall construct a reducible filter

<;.

the

but

The following relation

< c.

seen to be the required well-ordering of' Ba[X] : xRy

c

i4>n}

of :1

F

of

P

by considering

describes

the set

H[F]

H[X]

in

is a modei of

can be constructed by the well-known

When applying this method we consider a

of all sentences of RL and build a complete extension

by successive steps: ";1, =

'1 u 7 1

n-th step we decide of the n-th sentence 1c1>n

Xs.P

will be included in ~'.

u :;t 2 u

cl>n

In

of RL whe,ther

cl>n

Also several other sentences have

to be included in the n-th step.

The inductive definition of

7n

is carried out along with an inductive definition of a decreasing sequence

PO .. Pl

.. ...

which are true in models

of conditions_ ?'n H[Y]

being the set of sentences

for almost all

Y in

[Pn]'

The

words "almost all" mean here "all up to a set of first category" • As is always the case with models built by Henkin's method the complete set sentence

cI>

9'

determines a model

of RL the sentence

cI>

H[X]

is true in

such that for each H[X]

if and only if

(112), 247 •

AN EXPOSITION OF FORCING

e ~'.

443

It follows easily that all the conditions

En

X, because the sentence

E

a

is true in

~

models

It turns out that the conditions filter (see Lemma 9.5 below). show that it is reducible to the formula

fo~lowing:

for some

n

Hence M.

X

belong to

Pn

M[Y]

is a maximal filter.

The reason for this is the

M[X] I: ,[y*[X]]

,
is equivalent to

for almost all

Y in

[Pn].

We shall show that this relation between definable in

M, i.e. has the form

de finability of

(1)

., y

and

M I: t.[Pn'y].

In this way we not only prove the reducibility of

M 1= t.[p,y]

t..

M[Y] 1= +[y*[Y]]

X

Y in

is co-meager (in the space X

[p]

This relation is called the forcing relation.

The theory :t

The following sentences of RL are called axioms of ~ (1)

~

(2)

-,(~ E

E

for

~'

for

!!!.')

for

m, m' e M such that

c

in

m, m' e M

me m',

such that ·m ~ m',

C, p(c) < a,

(3)

Vac

(II)

".(a)(c') if + is a formula of E a,',y ,y v e Fr(,), y e Ca.Fr(+)- {v} , c' e Ca' c'

c

to

M

for

of all maximal

We proceed now to the details of the proof.

8.

is

The formula

says that the set of maximal filters

P) •

Pn

The proof of

is the most important step in the proof.

but establish the meaning of the formulae

fil tel'S of

We

and this in turn is equivalent to M[Y] 1= Hy)

which

where

generate a maximal

L,

444

[112], 248

FOUND!\ TIONAL STUDIES

=

(5)

(7)

VV a VV I {(Vv 2[v 2 E Vo v 2 E vI] VV o VV I VV 2 {vo ~ vI ~ (v o E v 2 Vv ,(v E v),

(8)

VV o VV I {va

(6)

(9)

Vva[v a

£

~

~

0

vI

"o

[VaV

~ E

a

= (v o = vI E

- Vav l]}

v 2 J },

for

a E OnM,

!:J,

(la)

VV o Vvl[(v a ~ vI & va E 0) ~ vI EO],

(ll)

VV o VVl{(V a E 0 & "i E (1) ~ 3v 2[V 2 ~ 3V (va E '(1). a

(l2)

vI)}'

~

"o

& v

2 ~ "i & v 2 E <1]},

In the axioms (10) and (11) we used the abbreviation Vo ~ vI for a formula of RL whioh expresses the fact that the ordered pair (va,v l) is an element of the set~. We should remember here that the ordering

~

of

P

assumed that it belongs to is as follows:

v n' v p "

where Let

let

is a set of ordered pairs of which we M.

IT' (vm,vn,v p)

The explioit definition of be the formula

" vm

va

~

vI

is the pair

i.e.

q

is any integer different from

IT(Vm,vn,vp)

be the formula:

vm

m, n, p, e.g. ; m+n+p+l. is the ordered pair

(Vn,V p)

:L.e.

where formula

q

m + n + p + 1, r

m + n + p + 2.

Finally

Vo

~

vI

is the

[112]. 249

445

AN EXPOSITION OF FORCING

LEMMA 8.1 If

is a fitter in

X

are vaZid in

P

then all' the axioms

(L) -

(12)

M[X].

Proof:

obvious.

It is not true that all the models of the axioms have the form

M[X].

Lowenheim

This follows for instance from the upward Skolemtheorem according to which there are non-denumerable models

of axioms (1) - (12) whereas models

M[X]

are denumerable for any

X ~ P. Let us say that a pair

F, G of sets of sentences of RL

satisfies the postulates if

s G,

(i)

F

(ii)

3v n E F

where

v n E Fr(
implies

(c) E G for some

c E C, (iii)

(iv) (v)

(vi)

E F where "n E Fr((c) E G for n [Vav n s ] some c E Ca , Vac E F implies c ... c' E G for some c' E Ca' c E C' E F where c' E Ca implies c '-' c' t E G for some c' , E U{C~ : ~ < a},

3v

c'

E

mE F

A set

F

implies

c' ... nEG

for some

n E m.

of sentences of RL is closed if it is consistent,

complete, contains the axioms (1) - (12) and the pair

F, F

satisfies

the postulates.

LEMMA 8.2 The theory of are true in

H[X])

H[X]

is oZosed.

(i.e. the set of sentenoes of RL whioh

446

[112], 250

FOUNDA T10NAL STUDIES

Proof:

obvious.

We shall establish some properties of closed sets: 1.

If

~

is logically valid then

Otherwise

would be in

.,~

~

,

E F.

but each set containing

F'

is

inconsistent. ware in

2.

Otherwise sentences

3.

~,~

If

if

~

then so is

F

w.

"W would be in F; but each set containing the + W,"W is inconsistent.

is logically equivalent to

W then

E F

~

if and only

W E F.

This follows from 1 and 2. 4.

v

If

is the unique free variable of

equivalent to:

~(c)

E

lIv

~ +

Since ion .,~

(0)

F

E F

for some

If

c

in

is

C.

\Iv

~

~

F

3v

then

F

.,~ E

and so it is not true that

c

E F

~

is logically true we obtain the implicat-

~(c)

from 1 and 2.

+

for each

\Iv

then

~

hence

~(c)

for each

E F

c. 5.

If

v

is the unique free variable of

and ally' if there is a Proof: 6.

~

If

~(cl'Y)

-

c

in

C such that

~

then

~

(c) E F.

Proof.

~

E F

if

similar to 4.

is a formula, y E CFr(~)-{v}, then ~(c2'Y)

:3 v

E F If

for each ~(Y)

c

l

- c

2

E F

implies

~.

is the atomic formula

v

E

c

or

c £ v

then 6. follOWS from axioms (5), (6) and the above remarks 1 - 4.

[I 12J, 251

•

If

'(c l cl

£

is the formula

v £ V, 6 •

£

Cl)

E

F

hence

cl

cl

+

c2

£

c2e F.

If

•

cl

£

follows from axiom F

+ 1jI E

Similarly

•

If

for any

c2 v

is the formula

'Hy)

using axiom (6). If

447

AN EXPOSITION OF FORCING

c2

£

c

Of

c

or v

is the formula

Of

and we obtain

1jI

cl

+

(7)

£

cl

E

F.

Of

v

we prove 6.

v, we again use (6).

Vav we use axiom (8).

is the formula

.' .

Let us now assume 6. for two formulae

Using

.{, . " .

tautologous formulae of propositional logic we immediately obtain 6 •

" . Also using [+' (c1'y) - +' (c 2,y)] e F

and '.' implies Vw

for the formula.. cl

c2 E F

Of

-

DomCy) = FrC.') lJw C+ l - .2) implies [VW

+

S

= vw +2] is in F +' CC1'y)] = [VW +' Cc2,y)] E [Vw .1

C**)

m

C £

(c

Of

we obtain that

(c

+

~l v

Of

c

in

v c

Of

~2)

~2) +

c

£

~l

Of

c

= 1,2,

by

(1)

this we obtain

and hence

CU)

n'

sentences

Of

~i

c

£

If ~,

'Cc

E F, F.

~ E

+

C*) Of

c e

m. e -J.

c

Of

~2 E

F

and

~

e F

fOl-

i

~ E

= 1,

F, From

2.

The formula

C*)

were false, then, by completeness,

~l)'

'Cc

Using postulate Cvi) we would obtain from or

c2 E F

and 4. above we see

by a propositional tautology.

is shown by contradiction. the three

c

Of

C the following two formulae:

To prove the second formula we observe that i

cl

Yo

Using the definition of

that we have to prove for each C*)

where

Using the fact that the taulogous sentence

{v,w}.

Proof.

4. we obtain that

~2)

Of

c

£

F would be inconsistent.

would belong to

mE F

that

c

Of

~l E

F. F

448

[112J, 252

FOUNDATIONAL STUDIES

then

8.

Proof is similar to 7. LEMMA 8.3

If

is oZosed then the set

F

such that

c

2, 1+ above, p

P.

in

e

X f.

Proof. constant

=

X

{p E ~

~ £

0

F}

E

P.

is a fiZtep in

c

c

P E F

£

o E F

£

whence, by axiom (9 ) and properties

whence by postulate (vi)

r

Thus, by 6, Next, let

because by axiom (12) and 5. there is a

o E F

£

p"q E X.

Hence

c ...

~

E F

for some

p E X.

and so ~ £

0

E F

9.

and

E F.

0

£

From axiom (11) we infer that the existential sentence

is in

Hence there is a constant

F.

c

such that the sentences

C £ 0

belong to

F.

From

c £

and hence, by postulate (vi)

£

obtain

£

0

E F,

hence

!

£

<;

E F

whence

Finally let above that

9.

£

C1

E F

~ ~

9. E F

and so

C'"

rEX

These formulae prove that if and

s

£ E F and

£

= (r,p>,

s, t E <;. p E X

and

Thus p <; q.

whence, in view of

q EX.

f

and axiom (9) we infer c £

E F

0

for some

r

in

~

12. E F

and

£

t

= (r,q> r <; p

and

then

E F

P. ~

Now we

9. E

F.

s

<;

£

E F

r <; q ,

We infer similarly as ~ £

0

E X

and axiom (10)

[112),253

449

AN EXPOSITION OF FORCING

LEMMA 8.1f

F

If ~ £

0

E

F}

M[X]

then

Proof.

is a modeZ of

{p E

P

F.

First we construct a relational system

+E

all sentences

X=

is a closea set of sentencBS and

A in which

F are true and then prove that after dividing

by a congruence we obtain a relational system isomorphic with We obtain

A

M[X].

A by a standard method, due to Henkin, of constructing

relational systems from constants. The universe of interpretation of -,

£

c

A will be

wiil be

c

C;

for each

itself.

c

in

C the

The binary predicates

will be interpreted as the relations

I, E

defined as follows:

(i)

Finally the unary predicates

will be interpreted as

Va

sets (ii)

It follows from remark 6 above that

I

is a congruence in

A. We prove by induction that if' y E CFr ( + )

is a formula of RL and

then

A ~ try]


For atomic formulae definition.

'+ •• '

= +(y) +

e F.

this follows directly from the

If the equivalence is true for

completeness of for

+

+1

F and

and

If

Vv.

+

and

+1

then using

we immediately infer that it is also-valid where

v

is any variable.

450 Thus

A

we shall denote by tences of

F

and also

All

A

divided by the eqtiYalence

and each

are relational systems in which all sen-

M[X] Let us write

~ ~[y*[X]]

(W,o)

=A ~

< (~,y)

W(o)

sequences of constants and

relation defined in Section 5.

for each limited formula

if

~[y]. ~,

are limited formulae,

~

< ~(y)

where

If

~(y)

~

is the

It will be sufficient to prove (iv)

under the assumption that it is valid for all pairs

~(y)

which

y E CFr(~)

(iv)

y, 0

I

are true.

We prove the following statement:

~

[112J, 254

FOUNnATIONAL STUDIES

<

(~,o)

is minimal with respect to the relation

is one of the sentences

~ E ~, ~ -

~, VO~

(~,y).

~

then

and the truth of

(iv) is easy to verify. Let us assume that

~(y)

is not minimal.

logical connectives then either (case 1) (case 2)

& ~2(Y

= ~l(Y I Fr(~l»

~(y)

,(y) = VV j [VaVj

~'~l(Y)]

and arbitrary sequences <~l' Yl) < (~,Y)

and

pairs

(~i' Y

< (~,Y)

I

<~2' Y2) < <"y).

for

Fr(~i»' i

"~ley)

or

i

~

contains

or

(case 3) ~l' ~2

of constants such that

(~l'Y)

sides we obtain (iv) for the pair IFr(~i»

Fr(~2»

=

and (iv) is true for formulae

Yl' Y2

hence (iv) is valid for the pair

('i' Y

I

~(y)

If

<~l'Y) < (~,Y)

and taking negations on both

(~,Y).

= 1,2

= 1,2.

In case 1

In case 2

and so (iv) is valid for the

Taking conjunc~ions on both sides

of the resulting equivalences we obtain (iv) for the

formula~.

case 3 the left-hand side of (iv) is equivalent to the statement: for each

x

in

Ba[x]

M[X] F

~l[x,y*[X]].

In

[112], 255

451

AN EXPOSITION OF FORCING

C and since a we infer, using the inductive assumption,

Since each element of c.[X) E Ba[X)

for

cECa

has a name in

Ba(X]

that this statement is equivalent to:

A 1= ~l [c,y]. A l' Va[c')

Now let

then

c'

satisfies

Vav j

AF

since, as we proved above, c'

c

in

Ca,

be an arbitrary constant.

c'

postulate (iv) there is a constant

Hence' each

for each

satisfies in

in

c I c'

such that

C

a we obtain

~l[c,y),

the formula

A

A.

in

-+- ~l(Y)

c

lf

Otherwise by

A

vav j

1= ~l[c'

and

,y).

and we

-+- ~l(Y)

obtain the right-hand side of (iv). The converse implication is proved similarly. It remains to prove (iv) in the case when

Case 1. l)

to

~

~(y)

p(c ) .

c~(X)

E

c2

Let

c~[X).

c~[X)

that the sentences ~

is the sentence

cl

Subcase (a) :

c 2.

~

In this case the left-hand side of (iv) is equivalent

2

order such that

cl

is an atomic

We have several cases to consider.

formula.

p(c

~

c3

be a constant of a possibly smaller Thus

c~[X).

p(c 3 ) < p(c 2 ) and it follows precede the sentence

and

cl

in the ordering From the inductive assumption we obtain therefore

and c

A 1= c ~

3 c2 E F

l reversed.

~

c

whence

2 i.e., A F c l

c ~

l

.. c c2•

3

E

F, c

The

3

~

c

2

impl~cation

We denote discuss separately the possible forms of Sub-subcase (bl) :

E F

where

A 1= c

l .. c 3

and therefore can obviously be

p(c ) 2

by

a + 1

and

c2 : mE M and

this case the left-hand side of (iv) is equivalent to

rldm) <; a.

c~(X]

=n

In

452 where

n E m.

Putting

c3

by inductive assumption,

A

~

cl

£

c

£

[I12), 256

FOUNDA T10NAL STUDIES

mE F, i.e.,

A F cl

c

c l '" !!. E F

!!. whence

l

and so

2.

Conversely, by postulate (vi), the formula c l '" !!. E F

implies

for some

n

in

A F c £ c2 l m and the implications above

can be reversed. Sub-subcase (b2) :

c2

ct[X] = p

equivalent to

= o.

c1

postulate (vi) that ct[X] = p

obtain

o E F

£

c2

M[X]

X.

in

= Ca,~,y

where

Hence we

L,

is a formula of

~

In this case the left-hand side of

ct[X] E E~,y*[X],Ba[X]

Ba[X] F ~[ct[X], y*[X]]. by

we can

<

(iv) is equivalent to

Ba[X]

p(£) < p(o)

from which we infer, using axiom (9) and

y E CaFr(~)-{v}.

and

Since

cl E. E F for some p because c l ... E. c l £ o.

Sub-subcase (b3) : v E Fr(~)

p E X.

Similarly the right-hand side of (iv) is

repeat the previous proof. equivalent to

Here the left-hand side of (iv) is

where

i.e. to

~

We can replace here

by

,(a)

and

because the satisfaction of a formula in

is

equivalent to the satisfaction of the relativized formula M[X].

Now

~(a)(cl'Y)

occurring in


c

£

~(a)(cl'Y) are

2

< a

in

because orders of the constants and all the unary predicates which

c 2) = p(c 2) = a + 1. ] Hence we can use the inductive assumption and obtain A F ~ (a) [ cl'Y' occur in

~(a)

have indices

a

whereas

P(cl

£

Using axiom (4) we obtain ci which is the same as

cl

£

c 2 E F.

All these steps can obviously be reversed. Formula (iv) is thus proved in case 1.

£

Ca,~,yE

F

[1121. 257 Case 2,

is the formula

=a

P(cl) .. p(c 2)

.=

c1 IX]

453

AN EXPOSITION OF FORCING

+ 1.

c

~

l

c 2'

We can assume that

The left-hand side of (iv) is equivalent to

i.e., to

Vx B IX] (x E ct[X] _ x·E c~[X]) which in a turn is equivalent to Vc C (c*[X] E c1[X] _ c*[X] E c~[X]). Now a we notice that if CECa and p(c .. a then p(c € c < a + 1 l) l) c~IX]

P(cl - c 2); cl - c 2

if

c2

then the formulae

+ 1

have the .same orders but

definition of C €

=a

P(cl)


c

€

cl ~ c l ~ c 2

- c 2•

From axiom (5) we see that ~ (c € c l = c € c 2). the left-hand side.

A

~

c

and

according to the

c

€

such that

Vc CaA F (c

c

€

A

¥

=c

c £ cl

A £

~

cl - c2

c'

c

€

c

E·F. 2 We use postulate (v) and infer from

c'

and

c

Hence

cl E F

Case 3.

for some

cl

in

C a,

we obtain whence

~

Vac E F

c

~

c c

~

c

l i.e.,

< Vac,

l' and therefore

C

~

€

such

cl E F c

£

that cl E F

c' - c E F and

The left-hand side of (iv)

Vac.

c*[X] E Ba[X], i.e. ,

Since A

CECa'

is the formula


is equivalent to

Using c'

E F and i(c' £ c E F or i(c' £ c l) E F l 2) We can limit ourselves to the first case only.

c'

€

c 2),

€

c 2'

and

€

=c

then there is a

Let us assume A ~ c l - c 2' i.e., i(cl - c 2) E F. axiom (5) and property 5 of closed sets we obtain a constant that either

l

c

~

1

It remains to prove that if Ca

cl

implies the formula 2 Hence the right-hand side of (iv) implies

A

in

€

Similarly c l ~ c l - c 2' It follows now by the inductive assumption that

Thus in both cases

~.

the left-hand side of (iv) is equivalent to

c

c

to

c*[X] = ct[X]

for some

according to the definition of c l E F.

A F Vac.

By axiom (3) Conversely, if

Vacl E F A F Vac,

<,

454

[112J, 258

FOUNDATIONAL STUDIES

then, by postulate (iv),

c· - c l e F previous steps can be reversed.

for some

cl

in

C a

and the

The proof of (iv) is thus complete. In order to finish the proof of Lemma 8.4 we remark that (iv) implies the equivalences M[X] F ct[X] e

c~[X]

-

A F cl E c2 '

M[X] F c1[X]

c~[XJ

-

A F cl I c2 '

M[X] F c*[X] e Ba[X] - A F Aa(c) • These equivalences show that

M[XJ

is isomorphic to

All.

Lemma 8.4 shows that we can obtain filters by constructing closed sets of sentences.

In the next section we shall construct

such a set and then show that the resulting filter

X

is reducible to

M.

9.

Construction of a cZosed set

F.

We consider a partially ordered set Section 2 and denote by X Let a

K

be a

a-additive ideal in F~

K.

For each sentence {xeJ(;

e K

de~cribed

~

and

of RL we put

M[X]F~}.

K and

I

have the following

properties: F~

as

a-additive field of subsets of X

We shall assume that

(A)

P

in

the space of its maximal filters.

for each sentence

of RL,

I

[112], 259

455

AN EXPOSITION OF FORCING

(B)

If

pEP

then

(C)

If

H E K - I

[p]

~

I ,

then there is a

p

in

P

such that

[p] - H E I,

(D)

For each formula where

of RL the binary relation

~

pEP, and

y E

cFr(~), is definable in

We shall show in Sections 10-12 that (A) - (D) are satisfied if

K

is the field of Borel sets in

Jr

and

I

the ideal of meager

sets. The binary relation from condition (D) will be written as p •

~(y)

and read

"p

forces

$Cy)".

We note some simple consequences of the definitions and assumptions (A) - (D).

LEMMA 9.1 If

p,qEP

Proof.

and

There is

ere incompatible, hence in

I

[r]

p'lCq, r

then

in

S

P

[pJ-[q]flI.

such that

[p] - [q]

r

~

and thus if

p

and

rand

[p] - [q]

q

were

we would have a contradiction with (B).

LEMMA 9.2

a::-F', ~

c& F 4>(c) Proof results immediately from the definition of Let us arrange in an infinite sequence sentences of RL. (1) -

(12).

Let

{$n}nEw

For any finite set

{~n}nEw

F4>' all

be a sequence consisting of all axioms S

of sentences we denote by

AS

456

[112], 260

FOUNDA T10NAL STUDIES

their conjunction and by

{S,~,a, ...

,y}

the set

S U

{~,a,

... ,y}.

In order to construct a closed set we try to define an increasing sequence

{Fn}nEW

of finite sets of sentences such that the following

requirements be met for each

(R 4)

the pair

n > 0 :

(Fn_I,F n)

satisfies the postulates.

It is clear that if these requirements are met, the union UFn

will be closed.

LEMMA 9.3 There are infinite sequences

consisting

{Pn}nEw,{Fn}nEW

of conditions and finite sets of sentences respectively such that, for each integer

n,

P

Proof. any element of constructed. sentences. M[X]

n•

For P.

AF

n

and

n

=0

F

we take

We construct

Fn + l

X, the formula

in 3C

we either have

=0 Pn

and define and

F n

by adjoining to

First of all we adjoin

for each

Fa

Let us assume that

Next we try to adjoin X

satisfies the requirements

n

Pn

~n' ~

Since

A{Yn,w n}

PO

to be

are already Fn

Wn

several is true in

continues to hold.

~n

or

i~n

to

{Fn,w n}.

M[X] F 4>n

or

M[X] ~ i4>n'

For each

Thus the set

[P decomposes into two parts consisting of maximal filters X for n] Both of these parts which the former or the latter formula holds. belong to

K

but it cannot be the case that both of them are in

I

[112], 261

AN EXPOSITION OF FORCING

since otherwise

such that if

{Fn''''n'~n} the set

itself would

H, is in

parts, call it P~ .. Pn

[Pn]

K - I

~n

to

b~long

We denote by

is true in the models

{Fn''''n' -'4>n}

if

Hence one of these

I.

and hence by CC) there is a

- H E r.

[p~J

457

M(X]

F' the set n where X E H, and

is true in these models.

-'4>n

Thus we

obtain

and

F'n

satisfies the requirements In order to satisfy the requirement CR

add various sentences to

F~

we have still to 4) and restrict, if necessary, the condition

p~.

F'n

Let us first add new sentences to

so as to obtain a set

which together with

F satisfies the postulate (ii). To achieve n this we enumerate the existential sentences Ci.e. sentences beginning with the symbols

-'¥Vj

(*)

:aw8, 3w'8' ,_ .. , '3wk e Ck) For each

and so for each The set

8Cc X)' sets K

Sc = {X

E

X

X

which belong to

in

the sentence

[p~]

3we C

x

is true in

such that

: M(X] ~ BCc)}.

By CA) these sets belong to

but it cannot be the case that they are all elements of c

condition

such that

.. BCc).

such that

P~~

.. p~

such that

Thus adjoining p~' .. AF~"

finally obtain a set

M[X]

M[X] F

is thus decomposed into a denumerable union of

[p~] (p~J

Let these sentences be

n•

there exists a constant

there is a constant

P~'

F

Sc

(p~']

BCc)

to

~

I

- Sc E I. F~

n

and a condition

Hence

It follows that

we obtain a set

Repeating this process again F Ck)

r.

and hence by CC) there is a

p(k) n

k

F~'

times we

such that

458

[112J, 262

FOUNDATIONAL STUDIES

Pn(k) ~ AFn(k) e(k)(c(k»

and

Fn(k)

contains sentences

for each of the formulae

(*).

S(c), S'(c'), ... ,

Thus the pair

(F

F (k»

n" n

.

• satisfies postulate (ii). The procedure for the remaining postulates is very similar. In case of postulate (iii) we consider limited existential sentences, i.e. sentences of the form

3 aw

such sentence find a constant "

(k)

I

which forces

which belong to

~

c

in

~(c).

Pn '" Pn consider sentences of the form each such sentence a constant

Ca In case of postulate (iv) we which belong to

Vac c'

Fn and for each and a condition

in

C

ex

such that

adjoined to the sets previously constructed. we consider sentences constant

c"

c

of order

£

c'

in

<

p(c')

Fn and find for c ... c' can be

In case of postulate (v)

Fn and find for each of them a such that C'" c ', can be adjoined.

Finally in case of postulate (vi) we deal with sentences of the form 'c

£

n'

~

which belong to

Fn and find for each such sentence an element m such that the sentence C'" n' can be adjoined.

of

Lemma 9.3 is thus proved. The sequences

{Pn}nEW

9.3 determine two filters: conditions where

and

{Fn}nEw constructed in Lemma

one is the filter

and the other is the filter

Xo generated by the

X = {p :

~ £

a E F}

We shall show that these filters are identical.

F

First we note the useful LEMMA 9.1f

If and

F

{Pn}nEw' {Fn}nEw arB BI!lquBncBB satisfying LBmma 9.3

UFn then Proof.

~ ~

E F ; 3n (Pn E F _ 3n

(~

for each Bsntencs

.~)

E Fn)

+

3n (Pn

.~)

~

of RL.

because the

[112], 263 sentence II Fn ·if

4l

for some obtain

is logically true whenever

4l

+

F, then, '4l E F

~

459

AN EXPOSITION OF FORCING

n.

and hence, by the above proof, Pm t 4l

Assuming

t '4l

~

and

Conversely,

4l E Fn'

Pk

~

Pn

k = max(m,n)

and putting

~

'4l

we would

which is impossible by (B).

4l

LEMMA 9.5

If

and

{Pn}nEW

then the fiZte:r

X

=

genel'ated by the sequence

Proof. and so

M[Y] ., E.

= Pn

we obtain

p

Since

£

E

0 E F

[pJ

we have

p E Y

X.

p E X then

Next we show that if generated by the conditions

is ma:r:imaZ.

[pJ - FpEO e and p t E. E o. For We have thus shown that the filter

is contained in

p n 's

X

Y in

For each

Hence

0

F = UFn is identicaZ with the fiZte:r

a:re as in Lemma 9.3 and

En

mOl'eove:r

{Pn}

Pn E o E F.

generated by the

E 0

pEP.

Let E

{Fn}nEw

{p E P : E.

p

belongs to the filter

P n• there is an integer

n

such that the

formula I\Fn + (£ E 0) is logically true and so Pn t £ EO. We claim that P <: p. Otherwise there would exist a q " Pn such that n Hence no filter Y in [qJ would q and p are incompatible. satisfy

p E Y, i.e., the difference

would be equal to

[qJ.

[Pn] - {Y E :£ : M[Y] ,. £ E I

E. E o}

[q] E I

that

because

in

P

E F

£ E o}

E oI

E I

and so

[q] - {y EX: M[Y] ,.

Thus we would obtain the result

[q] ~ [Pn J·

which is impossible.

such that 0)

~

we obtain however

Finally we show that

1(E. E

[qJ - {y E J; : M[YJ

X is maximal.

X c Y and assume that

and therefore

Pn

~

,(£ E 0)

Let

p E Y - X.

Y be a filter Hence

for an integer

n.

Since

460

[112], 264

FOUNDA T10NAL STUDIES

P E Y, the conditions

r"

p.•

[rJ

f'l

Since

are compatible;

Y in

[r]

p E Y

the formula

is true.

Thus

let

r " Pn

and

we obtain

On the other hand the relation

~E(1

M[Y] ~ ~ e 0

Pn

[r]- F'(~€(1 ) ~ [Pn] - F'(~E(1)

FEr.

for each

and

p

r" p

proves that

and hence also the formula

[r] - F

~€(1

= il, [r] c F -

together with the previous relation shows that

which

~€a

[r] E I.

Since this

contradicts the assumption (B), Lemma 9.5 is proved. Taking our lemmas together we obtain

THEOREM 9.6 There exists a sequence {Pn}nEw the se t

{+ : 3n Pn

F

X = {p E P identical

~ €

.

is closed .

~}

(1 E F} is reducible to

Proof.

Let

{Pn}nEw

and

constructed in Lemma 9.3 and put F

is closed and from

9.~

{~

: :3 n (P

From

..

~)}.

X

and each element

this condition as

8.~

t~

M.

be the sequences From 9.3 it follows that

P

of

X ~(y).

such that

Since each

~(y).

is

;>

for some

Pn

belongs

Pn n

we can write

In view of the assumption (D)

p" ~(y)

"

M F t~tp,yJ

and so

The remaining two statements follow from 9.5.

(A),

(B)

and (C)

K be the field of Borel subsets of

ideal of meager sets.

Pn.

it follows that

Verification of assumptions Let

{Fn}nEW

It is

that it coincides with the set

3p E X P ..

there is a formula is reducible to

lO.

and maximal.

= UFn.

F

~(y) E F " 3n P n ..

M[X] 1= [y*[X]] "

to

M

the filter generated by the conditions

~ith

n

of conditions such that The filter

X

and

I

the

We are going to prove that the assumptions

X

[112], 265

461

AN EXPOSITION OF FORCING

(A) - (D) of Section 9 are satisfied. (B) follows from Baire category theorem (see 2.5). proof of (C) is as follows: form

N) U N'

(G

where

Each non-meager Borel set N, N'

are meager and

empty (see Kuratowski [66], p.88).

Hefice if

[p] - HeN

is meager.

and therefore

Proof of (A).

(p] - H

The

H has the

G is open and not

[p] C

G, then

From Lemma 9.2 it follows that if

Borel for each sentence

~

containing less than

n

logical operations then it is true for the case when

is

F~

symbols for ~

contains

n

Thus it is sufficient to prove (A) for atomic sentences.

such symbols.

It is more convenient to prove it more generally for limited sentences. We show that if F

w

~

is a limited sentence and for each

is Borel then so is

~

the set

~

F~.

The case when trivial because

W~

~

has no predecessors with respect to

is then one of the sentences

<

is

£ E £' 0 - £' 1{O£

is either the void set or the whole space X.

and

Now let us assume that when

has predecessors.

The cases

contains symbols for logical operations can be disposed of as

~

above.

Let

~

now be atomic.

We have three cases to consider.

We distinguish two subcases:

1)

la)

p(c

lb)

p(c

l) l)

> p(c 2 ) , < p(c )

2

Subcase la). cl*[X)

~

E

c 2*[X]

= 3c

p(c 2 ) = a.

Put

e Ca[(cl*[X] = c*[X) F~ y

U

cECa

(F

CEC

2

~

By definition

Xe

& (c*[X) e c 2*[X])] F _

c c1

).

F~

= whence

462

[112], 266

FOUNDA T10NAL STUDIES

Since

<

c e c 2 and c ~ c precede c e c 2 l l the result follows by inductive assumption. Subcase Ib).

F~

= U{F

cl

!l

"

~

c l !l


(ct[X]

because

=~

then we show similarly that

: n E m} whence the result follows because

= (]

c2

= p)J. c1

c2

e c 2·

If &

If

in the ordering

~

If

Hence

< cl

1:. c2

XE F '" (cr.)(cPy) because

then

X E F4>

= ~P

Fq,

is equivalent to

3 p E P [X E [pJ

is Borel

and

([pJ

e c,

= c cr.,4> ,y

then

XE F4>

is equivalent to

and again the inductive assumption is applicable 4>

(cr. )

(cpY) 4> = c

Case 2. o,

l

< cl ~

e c 2• We can assume that

c2•

The relation

is equivalent to

X E ~c}Fcecl () F ] U [(:£- F c ) () ~ ~ cec 2 ce l

(X -

F~c )J - 2

whence we reduce the theorem to the case 1. Case 3. is equivalent to

4> = Vcr. c•

Put

c*[xJ = ct[X]

p(c) = a ,

for some

In this case

c l E Ccr.

and hence

= U F whence the theorem is reduced to Case 2. 4> CIECcr. c~cl (A) is thus verified.

F

Assumption

Before verifying assumption (D) we must establish some properties of the forcing relation.

ZZ.

Ppopepties of the fopcing peZation We denote by

4>,

W

sentences of RL and by

p, q, r

elements

[112j; 267 of

P.

By

463

AN EXPOSITION OF FORCING

..• )

$(v,~,

variables are among

we denote formulae of RL all of whose free

v, w, ..••

LEMMA 11.1

If

P

~

Proof. belongs

to

I

and

q

q

$

p'"

then

[p] - FIj) ::. [q] - FIj)"

$.

hence if the right-hand side

then so does the left.

LEMMA 11.2

If

Ij)

~

is logically valid then

$

p

Ij)

p . $.

implies

Proof. LEMMA 11.3

If equivalent to

Ij)

and p'"

p.

ware logically equivalent then

~

is

W.

This follows from 11.2. LEMMA 11.4

p • Ij) & Proof.

W is

(p'" Ij)

equivalent to

[p] - F$&$

= [([p]

- FIj)

& (p ...

w).

U ([p] - F$)];

sufficient to note that the union of two sets belongs to

it is now I

if and

only if each of these sets does. LEMMA 11.5

p • I Ij) is equivalent to Proof.

¥q
(q' Ij).

The left-hand side is equivalent to

hence if the left-hand side is true and

q < P

then

[p] n FIj) E I; [q] n FIj) E I

464

[112], 268

FOUNDA T10NAL STUDIES

and so

[ql - F$

by (C) there is ([q] - [p]

U

q

by (B).

If the left-hand side is false then

such that

([q] - F$) E 1.

q < p

only if

I

~

[q]

([p] n F.) E I

Since by (C)

we finally obtain

q < P

and we obtain i f and

[q] - [pJ E I

and

q

$, i.e., the right-

hand side is false. LEMMA 11.6

P t l,lv Proof. the

is equivalent to

$ (v )

l,lc EC(p .. $(c».

[pJ - Fl,lv $(v) : c~C([p] - F$(c»

a-additivity of

I

by 9.2.

Using

we infer that this union belongs to

I

if

and only if each of its members does. LEMMA 11. 7

p ..

(r .. c ...

C

F

l,lq < P :3 r < q

is equivalent to

3n E m

~).

Proof. (q .. C

£ m

£~)

l,lq < P

The left-hand side is equivalent to

and hence to

l,lq

< p ([qJ - n¥m F c_n E I)

because

U F It follows that there exists an element 'n of m nEm c"'~ such that [q] n F ~ I and hence, by (C), there is a condition r c£~

•

c~

such that

[r] - ([q] n Fc"'~) E I.

[rJ - F

E I,

c_n

This proves that

If the left-hand side is false then hence, by (C), there is a It follows that this proves that

[r] ~ [q]

and

i.e., the left-hand side implies the right.

q < p

[q] n F

same is true for each

q

and c~

r < q.

such that [q] n F is in

c£~

I

[q] E 1.

[p] - Fc£~ ~ I ([p] -

Fc£~)

Since

F

for each

n

in

and

E I. c F

c--!!. -

c£m

m

and the

[112]. 269

465

AN EXPOSITION OF FORCING

LEMMA 11.8 p (r

~

~

C € a

~

P 3r

q 3s > r

~

c '" 2.).

Proof.

Fc€a

¥q

i8 equivalent to

s~P

We argue as in the previous proof using the equation

FC"'2.)~

([s] n

LEMMA 11.9 ~ i8 a formula of L, v E Fr(~), a E o~, c' E Ca Fr($)-{v} h I- I • • • y E Ca t en p c e ca,$,y '1-8 e qu i va l ent: to

If

and

P t ~(a)(c' ,y).

Proof. F

Putting

c

= Ca,~,y

we easily show that

FC'€C

(a)(c',y)

LEMMA 11.10 p (r

~

~

Vac

i8 equivalent to

Vq

~

P 3r

~

q 3c' E Ca

c '" c').

Proof uses the same technique as 11.7 and the observation that

FV c a

LEMMA 11.11 then

c1' Cz E Ca + 1

If

p t c 1 '" Cz i8 equivaZent to

¥c E Ca{Vq

~

p [(q l- e € c l) .. 3r

~

q (r

Vq

~

P [ (q I- C € c Z) .. 3r

~

q (r

is the formula

v

E

c 1 '= v

E

CZ.

~

C

E

c

l)]}.

F - n F where $ CEC ~(c) Cl"'c ZFrom the a-additivity of I it

It is immediate that

Proof.

c € c Z)] &

466

[112],270

FOUNDATIONAL STUDIES

follows that

p

~

cl

~

c2

is equivalent to

Vc E Ca(p

order to bring the result to the desireti form we express

& and 1

means of the connectives ~

where

I

= I[~l(c)

(c)

=c

~i(c)

E:

c

equivalent.

~.

$(c)

by

alone and obtain the sentence

& 1~2(c)] & 1[~2(c) & I$l(c)] for

i

i

= 1,2.

Since

logically equivalent (or more exactly: abbreviation of

In

~ ~(c».

(c )

,

since

the relations p ..

are

is just an

Hc) and

~(c)

¢' (c)

and

~(c)

p If-

are

~'(c)

We now use Lemmas 11.4 and 11.5 and after easy trans-

formations obtain the desired result. LEMMA 11.12 c E Ca + l

If to

dEC

Vq

Z2.

and

~

p" dEc

then

c ") &

(1' If-

c

E:

I

is equivaZent

c)].

Similar to that of 11.7 and uses the decomposition

DefinabiZity of the foraing reZation.

We shall base our proof on the following theorem scheme on definability by transfinite induction. and

R

U

be a subset of

a well founded relation which partially orders

assume that U

Let

U

and

the set of its

belongs to

M.

R

Finally let

correlates an element of A E M and

A

are definable in

R-predecessors

U

H

M

U.

Let us

and that for each

s

R(u) = {v E U : v # u

is a function with domain

assumptions there is·a unique function

a, A

R(u). G

a

where

u

in

vRu}

be a function definable in

with each pair

M

M which E

U,

Under these

with domain

U

such that

[112],271

= H(u,G t

G(u) in

M.

467

AN EXPOSITION OF FORCING

R(u»

(Note:

G

for each

t

R(u)

u

in

U

and this function is definable

is the restriction of the function

G

to

R(u». This theorem is but an inessential extension of the theorem on definitions by transfinite induction whose proof can be found in many textbooks of set theory.

We shall not enter into details of

this proof here. We shall now prove the definability of the forcing relation. If

$

either

is a formula of $

I~

$ = W

or

If the relations

L

which contains logical operators then

& 6 or

p t W(y)

so is the relation

$ = Vv W where

and

p t $(y)

p t 6(0)

v

is a variable.

are definable in

M then

in view of Lemmas 11.4 - 11.6.

Thus

in order to verify aasumption (D) it is sufficient to prove it for the case of atomic formulae.

We shall establish a slightly stronger

result: LEMMA 12.1

The binary reZation Zimited sentence of Proof.

M.

where

RL is definabZe in

The set

pEP

and

$

is a

M.

Let us consider pairs

is a limited sentence of RL. in

p t $

(p,$) U

where

PEP

and

$

of these pairs is definable

We order it partially by the following well founded relation

R:

Let us put p f $. M

G(p,$) =

a

or

1

according as

P t $

or

In order to prove that the forcing relation is definable in

it is sufficient to show that the function

G

is definable in

M

468

[112),272

FOUNDATIONAL STUDIES

and we achieve this by showing that G(p,~)

::

H(p,~,G

where

~ R(p,~»

proper choice of

satisfies a recursive equation

~

H

is a definable function.

H becomes clear when we examine Lemmas 12.4 -

These lemmas show that the forcing relation

12.12.

p

reduced to some forcing relations between elements of sentences which precede

and limited

<

with respect to the ordering

~

can be

~ ~

P

these conditions can be expressed by means of the values of limited to the set and only if

E.g., if

R(p,~).

Vq < p [G(q,$) :: IJ.

i f ana only i f

The

Vq < p (A(q,$) ::

1$

~::

then

G

o

G(p,~)

Accordingly we put

if

o

H(p, I$,A)

For other forms of

1),

thus

the

~

procedure is similar. We can now give the exact definition of a :: (p,~);

then If

ces

£

E

cases and

£'

a

H(a,A)

is defined for

has no

a

E

R-predecessors then

VO£' 0 ~ £ and we put :: 0 in the third. H(a,A)

then

H.

U and

A

{O,l}R(a) ('1M.

E

is one of the senten-

~

H(a,A):: 1

Let

in the first two

< P (A(q,$)

If

~

1$

If

~

$

e

then

H(a,A)

0

-

::

0;

If

~

Vo,v tjI

then

H(a,A)

0

- !,tc E Co,(A(p, tjI(c»

::

If

~

cl E c 2

&

and

p(c

Vq < P 3r < q 3c'

0 " Vq

::

l) E

..

p(c

1) ;

::

A(p,$) :: A(p,e)

::

0,

Co,[A(r,c l

~

2)

H(a,A)

then c')

0); ::

-

0

A(r,c' E c 2)

::

OJ; If

~ ::

H(a,A) If

~

and

cl E c2 ::

- Vq < P 3r < q :In

and

c1 E c 2 H(a,A)

0

::

0

P(cl) < p(c 2)

P(cl) < p(c 2)

and E

and

c2

::

m

III (A(r,c 1 c 2 :: a

then

... :::.)

0);

then

- !,tq < P or < q 3s .. r (A(r,c l ..

~)

0) ;

(112], 273 If

4J

If

4J

c l e: c 2

and

H(a,A) =

o :;

c

and

.. c

l

2

H(a,A)

If

4J

0

p(c

- Vc E Cex{Vq ..; p [A(q,c

q A(r,c

E

c )

= 0] s

1

v

3r ..; q A(r,c

E

c l)

OJ};

then

because

p

~

= o :;

H(a ,A) )

2

+ 1 E

ex,4J,Y

then

c

then

l)

Vq ..; p [A(q,c

E

c 2)

Vq ..; p 3 r ..; q :l c' E C ex

= 0).

H is of course definable in

G is definable in

4J :; G(p,4J) = 0

c

l

3r

Vexc

c2

y» = 0; max(p(c l), p(c » = ex 2

A(P,1/J(ex)(c

Lemmas 12.4 - 12.12 we prove that Hence

and

v

The function

U.

p(c l) < 2)

1

..;

(A(r,c .. c

in

469

AN EXPOSITION OF FORCING

= H(a,

G(a)

G

M.

t R(a»

Using for each

a

M and so is the forcing relation

whenever

is a limited sentence.

~

Thus condition (D) is verified.

l3.

Additional remarks Let

p

in

P

Dc P

there is a

D is dense in

P

be a set dense in q

under

in p

D such that if

P

i.e., such that for every q < p.

Vq..; P #rED

We shall say that

r"; q ,

In theorem 9.9 we established the existence of a sequence {Pn} F =

of conditions which has the property that the set {¢

:

#n Pn

~ ¢}

with this property.

is closed.

We want to characterize sequences

470

[112], 274

FOUNDATIONAL STUDIES

THEOREM 13.1

If

is a 8equenae suah that the set

{Pn}nEw

F = {¢ : 3 n Pn

is o l.os ed then the filter

~ ¢}

has aommon elements with every set dense in

Let

Proof.

s

n,

some integer implies

P

n

X.

P

E D

belongs to

12. E: !2. E F

and

If the sentence

then there is an element

F

12. E: cr

and

Pn • 12. E: !2. and the second

D

E

be a dense set.

D E M

(v E: a)]

such that

M

E

12. E: a;

~

Pn

(see

P n .; P

F

leads to a contradiction.

sentence

The first relation

each constant

Pn c

,[(c E: !2.) & (c E: cr)].

D

q ~ (~E: !2.)

&

(~

belongs to

P

n

F

and thus is forced

c

the constant

q'; P n'

E: a), i.e., either

Thus for

forces the sentence

We choose for such that

& (v E: a)]

3 v [(v E: !2.)

from the initially given sequence. the condition

is an element of

Hence

11. 8).

This assumption implies that the

Vv ,[(v E: D) & (v E: a)]

by a condition

P

l'l: follows that for

F.

We shall now show that the assumption ~

{Pn}

and is

M

P.

3v [(v E: !!) of

generated by

X

whiah belongs to

D

~

where

q

Using Lemma 11.5 we obtain

q f ~ E:!2.

q f ~ E:

or

o.

Both

these alternatives are clearly false.

A filter

X is called generic if

D which is dense in generic in follows:

P If

over {Pn}nEw

P

and belongs to

M).

Xn D

~

0

for each set

M (more exactly

is a sequence such that the set {¢ : 3n P n • ¢}

is closed then the filter generated by the

Pn 's

is generic.

shall prove that also the converse of this theorem is true. need a lemma:

X is called

Theorem 13.1 can thus be expressed as

We First we

[112],275

471

AN EXPOSITION OF FORCING

LEMMA 13.2 D C P

A generic fiZter intersects every set

M and is dense under

to

Proof.

where

p

D'

Put

D U {q E P : q

In view of the separability of hence if

X

is generic then

incompatible with

P

X n D #

is incompatible with

D'

the set

X n D I # 0-

p, we obtain

which beZongs

is any eZement of the fiZter.

p

is dense in

P

pl.

and

X is

Since no element of ~

THEOREM 13.3

X is generic then the set

If

F

{
3p E X P .. 4>}

is o l oe e d . Proof. (1)

F

is consistent.

Otherwise there would be a

¢l"" '

finite set 4>

assumption each

4>j

is forced by a condition

is a filter we obtain a condition j .. n

each

M[Y]

the family

(2) D

= {p

: p If- cp or

E P

~

is a model of F

For let

a condition extension

q

X

such that

Fcp

E

for

I, therefore

Y in

of

Since

q X

is in

Let

4>

be a sentence and

In view of the definability of the

D E M.

such that

X

cp.

We shall show that By 11.5 if

be any condition.

p" q p

-

is complete.

p I- ICP}.

forcing relation we have

P.

[p]

Thus


in

Since

X.

in

and we obtain a contradiction because for each

[pJ n F¢ # 0

Fq,

p

and so

p

Pj

p

~

cp.

D

q f I¢

Hence either

is dense in then there is

qED

or some

D.

is generic we obtain now

belongs to this intersection, then either

p

X n D # 0; forces

¢

if or

p p

forces

472 whence either

,~

(112], 276

FOUNDA T10NAL STUDIES

(3)

~

or

belongs to

,~

F.

The axioms 1 - 12 given in Section 8 belong to

This is so because these axioms are true in all models Y is any filter; (~)

r

H[Y]

F. where

hence they are forced by any conditions. satisfies the postulates (ii) - (vi).

Since the

verification is practically the same for all the postulates we shall give the proof only for the postulate (ii). :Jvn

E r , Le.,

~

p I- 3vn

abbreviation of Vq

~

P Jr

~

'Vvn,

~ ~(c).

D = {r E P : 3c E C r I-

E

F.

Thus let us assume that

in

X.

Since

:lv n

is an

.~(c)}

This means that the set is dense under

p

and so, since this

H, we obtain that there is a condition

Hence there is a constant ~(c)

p

we can apply Lemmas 11.5 and 11.6 and obtain

q :lc E C r

set belongs to

for some

~

c

such that

r I- ¢(c)

r

in

D n X.

which proves that

The postulate (ii) is thus verified. Theorems 13.1 and 13.3 suggest an alternative method of

constructing models.

We start with the definition of forcing and

establish first of all Lemmas 11.1 - 11.12.

Then we define generic

filters and prove their existence essentially as in the proof of the Baire theorem. set. filter

Next we establish Theorem 13.2 obtaining a closed

Finally we prove that each closed set determines a reducible X as we did in Section 8. This alternative method is esseatially the one which was

used by Cohen.

Most authors follow Cohen by defining forcing from

the start by transfinite induction (in the model then to avoid the

cuw~ersome

M).

This allows

verification of condition CD).

The

only defect of this method is that it is not easy for the beginner to grasp the intuitive meaning of the forcing relation.

[112], 277

473

AN EXPOSITION OF FORCING

The connection between forcing and the concept of meager sets was discovered by Takeuti and Ryll-Nardzewski and we exploited their ideas in the proofs given above.

t4.

Ppesepvation of oapdinats If

of

a E M is an ordinal then we say that

M if there is no element

smaller ordinal onto of

a.

f

of

a

is a cardinal

M which is a mapping of a

One can show by examples that a cardinal

M need not be .a cardinal of

M[X].

We shall derive a sufficient condition for a cardinal of to remain a cardinal of

M

M[X].

DEFINITION We denote by for each set

Q~ P

and such that

8 M
the least ordinal

a

of

M such that

consisting of mutually incompatible conditions

Q E M there is in

M a one-one mapping of

Q

into

a. LEM11A 14.1 If

M"

Proof.

ZFC

then

8 11
We can formalize in ZFC the proof that for each

partially ordered set there is a to

th~

exists.

least cardinal larger than or equal

cardinal of any set of mutually incompatible elements of

P.

LEMMA 14.2 If

M I-

zrc

and

is a eeneria fitter arid

X C1

>

e M
then

a

a

is a c ax-d i.na l: of

is a ca"dinaZ of

I1[X].

11,

474 Proof. f

Let us aSsume that there is in

e

and an ordinal

<

such that

~

f

& (Rg(f)

which is true in

=~)

constant such that

=f

c*[X]

e

maps

these facts in RL we obtain a formula ~)

[112],278

FOUNDATIONAL STUDIES

¢(f,£,~)

M[X].

M[X] onto

a function ex.

Expressing

: Funct(f) & (Dom(f)

Denoting by

c

a

we infer from the assumption that

is generic that there is a condition

Pa

in

X

X

satisfying

Pa .. ¢ (c ,£,~), Consider now the set

S

= U{Zi;

i; < S}

= {n

Zi;

where

: 3p ~ Pa

V E cJ} n

From the definability of The cardinal number of

ex.

Zi; E M.

we see that

~

Zi;

(calculated in

M)

<

is

8

M

(P) .

To see this we correlate (using the axiom of choice) a condition P

~

Pa

to each

Pa .. Funct(c)

n

Zi;

in

so that

p"

~ E

c.

Since

it cannot be the case that two compatible

be correlated to two different ordinals conditions correlated with elements of

nl,nn'

Zi;

PI' P2

Hence the set of

consists of mutually

incompatible conditions and hence its cardinal number (in

number of

Since

S

S

< ex.

is

and thus for each such that

• 5.

P"

< ex

n

< ex

C •

ex

there is a

i;

ex C S

because

in

and a

Rg(f) p

in

Lemma 14.2 is thus proved.

The independence of CH

Let

M" ZFC, ex > w, x E M.

finite functions

p

such that

•

it follows that the cardinal

On the other hand in

~ £

8M( P )

and

is

M)

We take as

P

the set of

Dom(p) C ex x w, Rg(p) C {a,l}.

=~ X

[112], 279

P

is obviously an element of

< q

p

475

AN EXPOSITION OF FORCING

if and only if

p

M.

q.

~

Section 1 are satisfied by

We order

P

by convention:

All the assumptions which we made in

P.

In particular

p, q

are compatible

if and only if they coincide on the intersection of their domains. LEMMA 15.1

The proof is due to Cohen but the theorem was already proved in 1941 by Marczewski.

In order to prove the lemma we formalize the

following reasoning in ZFC. We consider finite functions with values in with domains

C A

where

A

is infinite (in our case

Let us assume that there is a non-denumerable set incompatible such functions.

R l

R

l

{O,l}

and

A = a x w). of mutually

can be decomposed into a

denumerable union of sets, two functions being included in the same set if their domains have the same number of elements. sets is non-denumerable; functions in If

q E R l

to save notation we assume that all the

PI # q, then

and

and

q

can happen only if there is an element

t

t

such that

dom(q)

E

finitely many element&

in

l

E dom(Pl)

dom(q)

and

contains

t l·

elements

q E

q(t

Pl(t

l)

# q(t l).

Let

q E R

PI

E

t

in the domain of

q

R 2 l

Rl·

PI

Now there are only

and non-denumerably many

dom(Pl)

such that for each

Let

are incompatible and this

PI (t) ;& q(t).

and t

Pl

Hence there is a non-denumerable family t

k.

have domains of power exactly

R l

One of these

~

R

and an element

l

the element

P2 E R

Hence

2.

q's.

t

l

is in

dom(P2)

Again we see that there are non-denumerablY many

~

2) # P2(t 2)·

such that for some fixed It cannot be the case that

t

2, t

t

2 E dom(P2),

2 = tl

because we would

476 then have

=1

q(t 2)

=1

q(t

is different from

l),

- Pl(t l).

least 2 elements Pa

t

=1

P2(t 2)

P2(t l)

a

[112], 280

FOUNDATIONAL STUDIES

- (1 - Pl(t l »

But this is impossible because Pl(t

and

l

Thus We see that

l).

t 2.

Continuing this

Ra

and a non-denumerable subset

at least a elements in its domain. with

in

a

=1

- P2(t l)

k+l

After

q(t 2), i.e. dom(P2) has at

~easoning

R2 such that

of

k + 1

since

we obtain Pa

has

steps we arrive at

elements in its domain which contra-

diets our assumption. THEOREM 15.2 There are modeZs in

Proof. a

of

Let

~hich

CH

is false.

M. ZFC be denumerable;

take any cardinal

M which is greater than the first uncountable cardinal of

Define

P

as in the lemma above and let

X

be generic in

P.

M. We

Since

M(X]

and

same cardinals it is sufficient to show that

M(X]

has an element

are going to prove that

M(X]. 'CH.

which is a function with domain

a,

M have the

C 2 w n M[X]

with range

f

and

which is an injection. We obtain mapping of ~E;(n)

=

by taking the union

f

and patting

{O,ll

Obviously

~(E;,n).

to prove that

f

w into

a x

~ E

M(X]

is an injection, Le.

and so ~( E;)

~

= UX

which is a

f
E

where

M(X].

'I f Cn )

for

It remains E; 'I n·

Let us assume that this is not the case, i.e. that there are such that

E;

t n

and

~(E;,n)

=

~(n,n)

for each

formula of RL expresses this fact (i.e. is true in

n.

E;, n < a

The following M(X]):

(as before we adsun.ed here that ordered pairs and triplets can be

[112], 281

477

AN EXPOSITION OF FORCING

defined by formulae of RL).

It will be more convenient to write

this formula as (*)

where

is the formula

~

forces this formula. each

G.

in

P

vo

(&,V1'~) £

Since (*) is assumed to be true in

= (~,v1'~)

M[X]

£

v O'

there is a

p

in

Using Lemmas 11.4 and 11.5 we infer

and each

n

X which th~t

for

w

in

(""')

there is Adding

We nOW notice that the domain of

p

n

(n,n)

such that neither

(~,n,l)

and

(n,n,D)

(~,n)

to

p

nor

is finite and thus are in

dom(p~

we obtain a condition

Apply now to (**) the Lemma 11.4.

We obtain

i.e.

which is equivalent to

or

q" (!l e 0) 3r < q

r

"false"

Y

q"

~ ~(S'!).).

obviously false. and each

or

in

"'~(!l'!),)

and that of

q" .(!l

£0)

The first part of this disjunction is

The second is false too because for each [r]

q < p.

the truth value of (~,!).,~) £!l

(~,!).,~) £!l

is "true" and so

in

r'

r < q M[Y]

~(!l'!),)'

is

478

FOUNDATIONAL STUDIES

[112],282

BIBLIOGRAPHY

COHEN, P. J. [66]

Set theory and the continuum hypothesis, Benjamin, 1966.

GODEL, K. [40]

The consistency of the continuum hypothesis, Princeton University Press, 1940.

KURATOWSKI, K. [66]

Topology, Vol. 1.

Academic Press and PWN, 1966. it

it

it

Department of Mathematics, University of Warsaw, Warsaw, Poland.