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London Mathematical Society Lecture Note Series. 115
An Introduction to Independence for Analysts
H.G. DALES School of Mathematics, University of Leeds
W.H. WOODIN Department of Mathematics, California Institute of Technology
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© Cambridge University Press 1987
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V CONTENTS
Preface 1.
vii
Homomorphisms from algebras of continuous functions
2.
Partial orders, Boolean algebras, and ultraproducts
22
3.
Woodin's condition
43
4.
Independence in set theory
54
5.
Martin's Axiom
80
6.
Gaps in ordered sets
104
7.
Forcing
130
8.
Iterated Forcing
183
Bibliography
229
Index of notation
235
Index
237
vii
PREFACE
The purpose of this book is to explain what it means for a proposition to be independent of set theory, and to describe how independence results can be proved by the technique of forcing. We do this by presenting an application Our of forcing to a deep and interesting problem in analysis. application is, by current standards in set theory, fairly non-technical, and so it offers an excellent setting in which to exhibit to analysts these new techniques from set theory. Most analysts will have a certain acquaintance with They will know naive set theory up to the level of ordinals and cardinals. They will have heard logic and set theory.
that forcing is a powerful technique that enables one to
prove that certain propositions of set theory are independent of specified axioms, and, in particular, that Cohen developed the method of forcing in his proof that the Continuum Hypothesis theory,
(CH)
is independent of the basic axioms of set
They may also know of more recently formulated (MA), which can be used to axioms, such as Martin's Axiom ZFC.
establish independence results without the necessity of knowing any of the technicalities of forcing.
However, it is possible that analysts harbour two negative feelings about these matters. First, they may feel that, although logic and set theory are of interest in their own right, they have little to contribute concerning questions
which "really" arise in mathematical analysis, and so can be safely left to their disciples. But this is not true. For example, several natural questions about sets of real numbers cannot be resolved in the theory
ZFC.
For some of these
viii
questions the set-theoretic entanglements are quite subtle, in that these questions can be resolved by invoking the existence of large cardinal numbers. example.
is a continuous function, then for each Borel subset
f,g
:
Here is a specific
It is not difficult to show that, if
3t + 3t
B
f(B)
P.
of
Lebesgue measurable for each Borel set ZFC.
-. ]R
Now suppose that
Is f(]R\g(B))
are continuous functions.
cannot be decided in
f: 3t
is Lebesgue measurable
B?
This problem
However if there is a measurable
cardinal, then these sets are indeed all Lebesgue measurable. The main example that we present in this work did It concerns the automatic continuity of homomorphisms from a Banach algebra of continarise naturally in analysis.
uous functions into an arbitrary Banach algebra, and the formulation of the problem was such that a solution was expected (perhaps naively) in naive set theory.
But
eventually it was discovered that the existence of discontinuous homomorphisms from the algebras under consideration could not be decided in ZFC. Secondly, analysts may feel that the technicalities of forcing are too arcane to be readily accessible.
We seek
to challenge this view by presenting a reasonably complete account of forcing which is comprehensible to non-logicians: we hope that it will bring them to the point at which they can appreciate the application, which we shall give in detail, of these new ideas to the above automatic continuity problem. We should say, however, that it is not our intention to teach the practical use of forcing in general, but rather to explain quite explicitly how the method of forcing does yield independence and consistency results, and to exemplify this by the study of our chosen example.
Thus this book is directed towards non-logicians, and in particular to analysts.
We shall give an account of
the background in logic that we shall require, and we shall explain the key notions of proof, of consistency, and of independence. The approach to independence will be through the theory of models:
Godel's completeness theorem allows us
ix
to recast independence questions as problems of the construction of models with certain properties.
We shall give full proofs of the results about forcing and Martin's Axiom that we shall need. En route to our main theorem, we shall prove that CH is independent from
a result that we believe should be known to all
ZFC,
mathematicians.
We also hope that our account will be useful for students of set theory as a preliminary to other works.
Very
little knowledge of analysis is required to follow the details of our example, and the background that is required is given in Chapter 1.
We now describe the problem in analysis that we are Let X be a compact Hausdorff space, and let
to consider. C(X,C)
be the set of all continuous, complex-valued
functions on
X.
The set
C(X,C)
is a commutative algebra
with respect to the pointwise operations. IfIx = sup{If(x)l
Then
I.Ix
:
x E X}
is an algebra norm on
uniform norm - and
(f E C(X,C)).
C(X,C)
(C(X,C), I.IZ)
Set
- it is the The
is a Banach algebra.
question we ask is whether or not each algebra norm on C(X,C)
is necessarily equivalent to the uniform norm:
the topological structure of
C(X,C)
if so,
as a normed algebra
would be completely determined by its algebraic structure. This question was first discussed by Kaplansky in 1948, and it was eventually "resolved" independently by Dales and by Esterle in 1976:
if the Continuum Hypothesis holds, then,
for each infinite compact (Hausdorff) space algebra norm on
C(X,C)
X,
there is an
which is not equivalent to the
uniform norm.
The appeal to CH in this theorem was thought at the time (at least by those involved) to be an accident and a weakness of the given proofs.
However, also in 1976, it was
proved by Solovay, using a condition of Woodin, that, if there is a model of set theory, then there is a model of set
x
theory in which each algebra norm on each C(X,C) is equivalent to the uniform norm. Shortly afterwards a different, and easier, approach to the theorem was developed by Woodin;
the approach used some techniques of Kunen.
It
is this approach, which involves Martin's Axiom, that we present here.
These results are included in Woodin's thesis,
written at the University of California, Berkeley, and a presentation for experts in forcing is given in the recent volume [45] of Jech, but apart from this ours is the first account of the theorem to appear in print.
The original work
of Solovay has never been published.
A revised version of Woodin's thesis is to appear [72]; this memoir, which will also include several more complicated independence results, is written for logicians, and we can say with quiet confidence that it will be unintelligible to analysts. Chapter 1 contains only analysis:
following the
seminal work of Bade and Curtis, we shall analyze the
structure of an arbitrary homomorphism from C(X,C) into a Banach algebra. In Chapter 2, we shall discuss partially ordered sets, Boolean algebras, and ultraproducts, topics
which form the background to much of our later work, and in Chapter 3 we shall relate our question to one which is amenable to the techniques which are given to us in the theory of forcing:
if there is an algebra norm on any
C(X,C)
which is not equivalent to the uniform norm, then there is a free ultrafilter V on N and an isotonic map from a subset
/V of the ultraproduct (R'J/V,
In Chapter 4, we shall explain how independence can be established by the construction of models.
In this
chapter, we shall list and discuss the axioms of set theory,
and we shall formulate the axiom NDH:
"For each compact
xi space
X,
each homomorphism from
algebra is continuous.".
C(X,C)
into a Banach
It is a theorem of Kaplansky,
which we discuss in Chapter 1, that for each
C(X,C)
uniform norm is the minimal possible algebra norm.
axiom NDH C(X,C)
the
Thus the
holds if and only if each algebra norm on each
is equivalent to the uniform norm.
Chapter 5 first recalls the elementary theory of ordinals and cardinals, and then discusses Martin's Axiom: we give some applications of this axiom to analysis.
In
Chapter 6, we shall give a final reduction of our problem: we shall prove in ZFC + MA + 1CH that, if NDH fails, then there is an embedding of a certain totally ordered set
into
(R,<)
(N ,
forcing:
we shall work with Boolean-valued models and with
Boolean-valued universes.
In our treatment, we place more
emphasis than is customary on explaining exactly how forcing arguments yield consistency results.
(Let us note for the
logician that our treatment is a little non-standard, in that we do not discuss standard models.) that
CH
is independent of
At this stage, we prove
ZFC.
Having learnt what forcing is, we must learn how to iterate the process: we study iterated forcing in Chapter 8, essentially proving that "a forcing extension of a forcing
Here again, we are more
extension is a forcing extension".
explicit and give more details than is usual, but we do this at little cost in length. Finally we can conclude the proof of the main theorem of the book: so is
if
ZFC
is consistent then
ZFC + MA + NDH.
The practical significance of our main theorem for an analyst is that it is futile to attempt to prove that
X and an algebra norm on which is not equivalent to the uniform norm. there is a compact space
C(X,C)
We conclude this preface with some remarks about our style of writing.
Throughout, we shall expand points that we guess will be less familiar to analysts (this is a dangerous game
xii
for us), and we shall try to use notation which is closer to that traditional in analysis than to that preferred by logicians.
logicians do,
But this is not always sensible:
on occasion, adopt their notation for good reasons, related
We have had to find an
to the nature of their subject.
For example, let
amicable compromise.
function taking the value usually happy to write of
f
:
S - T
be a
Now analysts are
But this is dangerously
S.
ambiguous, for it may be that
that A
f
s E S.
for the range of the restriction
f(A)
to a subset A of
at
f(s)
is a transitive set, so
S
is also an element of
S.
In this case, we have
followed the analysts, trusting to context to arbitrate the ambiguity.
It is often necessary to be more precise in logic than one would be in analysis.
For example, a Boolean
algebra is strictly a sextuple 8 = satisfying certain rules: identifies
8
outside logic, one usually
with its underlying set
this in Chapters 1 - 6 of this book.
B,
and we shall do
However, in Chapters 7
and 8 it seems to be necessary for us to maintain the distinction between
8
and
B.
Here are some particular notations that we shall use.
(An index of notation is given on pages 235-236.) First, N = {1,2,3,...}, whereas Zz+ = {0,1,2,...} (so
that, again, we are following the analysts). allows the possibility that
S = T.
Second,
S c T
Third, in a phrase such
as
rad A = {a E A : e - ab E Inv A the bracket
"(b E A)"
means
"for all
(b E A) } b
in
A".
We give some references in the notes at the end of the chapters to the theorems proved. Although some results are new or have simplified proofs, it is of course not the
case that unattributed results are claimed as original.
The authors first discussed this work when the first author visited Caltech in 1984:
we are grateful to
Professor W. Luxemburg for arranging this visit.
We also
acknowledge the award of a Visiting Fellowship to the second author by the United Kingdom Science and Engineering Research Council that enabled him to visit Leeds in September, 1984.
The second author is an A.P. Sloan Fellow, and we are grateful that this fellowship provided financial support for us to meet at later times, both in the snows of Leeds and in the sunshine of Pacific Palisades, and for us to arrange the typing of this book. A number of people, both analysts and set theorists,
have read some chapters of a preliminary version of this work, and we are very grateful to them for their comments.
They
include William Bade, John Derrick, Peter Dixon, Peter We give special thanks to K.P. Hart and to the referee for reading
McClure, Neil Mowbray, John Truss and Tom Ransford.
the entire book in manuscript and for making many valuable suggestions: of course, the decisions on the approach, the (possibly tendentious) opinions, and the errors are all our responsibility.
We owe a special debt of thanks to our typist, Mrs. Joan Bunn. A large amount of information in a work of this type is carried by the details of the symbolism and the We have tried hard to eliminate errors in this area (presumably unsuccessfully): without the skill and accuracy of Mrs. Bunn's transcription, the process of correcting our fonts.
errors would surely have diverged.
H.G. Dales and W.H. Woodin, Leeds, April, 1987
1
1
HOMOMORPHISMS FROM ALGEBRAS OF CONTINUOUS FUNCTIONS
Throughout this book, we shall work in the naive set theory familiar to analysts:
the formalization of this
theory is the axiom system
which will be discussed in
ZFC,
Chapter 4.
We first summarize some elementary facts about Banach algebras that we shall use.
The account in the
standard text of Rudin (58, Chapters 10 and 11] covers essentially all that we shall require. All the algebras that we shall consider are linear
and associative, and their underlying field is either the An algebra is complex field C or the real field ]R. in this case, we often denote unital if it has an identity; e. The algebra formed by adjoining an identity to a non-unital algebra A is denoted by A#, and we take A# = A if A is unital. The set of invertible elements of a unital algebra A is denoted by Inv A. A character on a complex algebra A is a non-zero homomorphism from A onto C. The set of characters on A is the character space of A, written @A. Let A be a commutative algebra. An ideal I of A is modular if the quotient algebra A/I is unital. The (Jacobson) radical, written rad A, of A is defined to be the intersection of the maximal modular ideals of A; if A has no such ideals, then rad A = A, and in this case A is
the identity by
a radical algebra.
(So our convention is that a radical
algebra is necessarily commutative.) semisimple if
rad A = {O}.
It is standard that
The algebra A
is
2
rad A = {a E A : e - ab E Inv where
e
is the identity of
the fact that, if then
a E A
A#.
(b E A#) },
A#
Several times we shall use for some b E rad A,
ab = a
and
a = 0.
An element an = 0
of an algebra A
a
n E N.
for some
is nilpotent if for the set of
nil A
We write
nilpotent elements in A. If A is commutative, then is an ideal in A (it is the nilradical of A), and
nil A
nil A c rad A.
An ideal
P
in a commutative algebra A
is a
prime ideal if either a E P or b E P whenever a,b E A with ab E P. The algebra A is an integral domain if the zero ideal is a prime ideal in A. We shall occasionally use is an ideal in A (n E N), then there is
the standard algebraic fact that, if and if
a E A a prime ideal 1.1
is such that
with
P
I
an 4 I and
I c P
a 0 P.
DEFINITION
Let A be an algebra over a field
on A
is a map
p : A - Bt
(i)
p(a) > 0
(ii)
p(aa) =
(iii)
p(a + b) < p(a) + p(b)
(iv)
p(ab) < p(a)p(b)
A norm on A
p(a) # 0
The algebra A
A seminorm
(a E A);
(a E k, a E A);
Iaip(a)
is a seminorm
(v)
k.
such that:
p
(a,b E A);
(a,b E A).
such that
(a E A\{O}).
is seminormable (respectively, normable) if
there is a non-zero seminorm (respectively, a norm) on
A.
It would be more precise to say "algebra seminorm"
and "algebra norm", but this extra precision seems to be unnecessary for us.
A norm is usually denoted by
A Banach
3
Each normed
algebra is an algebra with a complete norm.
algebra A
is a dense, normed subalgebra of a Banach algebra,
A
the completion of
([8, 1.12]).
The starting point for the "automatic continuity" theory of Banach algebras is the following basic fact.
A be a Banach algebra, and let tinuous, and in fact 11+11
0 E
oD A.
Then
0
Let
is con-
4 1.
The character space
is a locally compact space with
to
respect to the relative weak *-topology from the dual space of
(We adopt throughout the convention that a locally
A.
compact space is Hausdorff.)
unital Banach algebra, then
A
Let 4P
A
# 0,
A
If
is a commutative,
is compact and non-empty.
to
be a commutative Banach algebra.
then the map
f+ ker 0
4
onto the set of maximal modular ideals of
rad A = (1{ker 0
:
If
is a bijection from
4
A
and so
A,
0 E IA} .
(1)
If 4A = 0, then A is a radical algebra. Since each character is continuous on A, each maximal modular ideal is
closed, and so
is a closed ideal in
rad A
Let A be an algebra, and let the spectrum of
a,
A.
a E A.
Then
a(a),
is the set
a(a) = {z E C
ze - a 0 Inv
:
A#}.
Let A be a Banach algebra. ([58, 10.13])
subset of
C
asserts that for each
r(a) = sup{I zI
Then
r(a)
a(a)
a E A.
:
A fundamental theorem is a non-empty, compact
Set
z E a(a)}
is the spectral radius of
radius formula asserts that
(a E A).
a,
and the spectral
4
r(a) = lim IlanIil" = inf IIanIIl/n
(a E A).
n+=
If
A
is commutative, then
a(a) = {q(a)
:
0 E D #},
and
A
Thus a a E rad A if and only if r(a) = 0 ([58, 11.9]). commutative Banach algebra is a radical algebra if and only so
IIanII1/n - O
if
as
n - -
for each
a E A.
We now describe in more detail the specific Banach algebras that we are studying. Let
be a topological space.
X
(respectively,
C(X))
Then
C(X,C)
denotes the set of continuous,
complex-valued (respectively, real-valued) functions on X. (In making this definition, we implicitly assume that X is non-empty.) Then C(X,C) and C(X) are, respectively, complex and real algebras with respect to the pointwise algebraic operations on X. The constant function 1 is the identity of these algebras.
Let function
on
f
be a subset of
S
S,
set
IfIS = sup{If(x)I
Then
Let
x E S}.
S.
be a compact (Hausdorff) space.
X
set
a
x
(f)
(f E C(X,C)).
= f(x)
homeomorphism of X onto Oc(x,C), identify these two spaces. Let z
and set
f E C(X,C),
is the complex conjugate of
E C(X,C).
A subalgebra A
of
For
Then it is a
standard result ([58, 11.13(a)]) that the map
where
Then
is a commutative, unital Banach algebra.
(C(X,C),I.Ix)
x E X,
:
is the uniform norm on
I.IS
For each bounded
X.
x r e
x
is a
and we shall henceforth f(x)
= f x
z E C.
C(X,C)
(x E X),
Then
is self-adjoint if
We shall use the standard fact ([58, 11.18]) (f E A). E A that each closed, self-adjoint subalgebra of C(X,C) which for a contains 1 is isometrically isomorphic to C(Y,C) certain compact space Y. Let f E C(X,C). Then Z(f) = f-1'({O}) is the
zero-set of
f.
A subset
F
of
X
is a zero-set if
5
F = Z(f)
f E C(X,C).
for some
Let
be an ideal in
I
the closed subset of
h(I) _ (1{Z(f) Let
The hull of
is
I
f E U.
:
be a closed set in
F
C(X,C).
X:
Then we throughout set
X.
I(F) = {f E C(X,C)
:
Z(f) z F}
J(F) = {f E C(X,C)
:
Z(f)
and
is a neighbourhood
of F}. It is easily seen that
and
I(F)
the maximal and minimal ideals in that
Let
is closed, and that
I(F)
C(X,C)
respectively.
Jx and We see that
whose hull is
is dense in
J(F)
Then we write
x E X.
I({x}),
are, respectively,
J(F)
Mx
for
{M
: x E X}
x
F,
I(F).
and
J({x})
is the
C(X,C).
family of maximal ideals in
We use the same notations I(F), J(F), Jx and Mx C(X) when we are considering
for the analogous ideals in real-valued functions.
We write
(respectively,
L'(C)
k7)
for the set
of bounded, complex-valued (respectively, real-valued) sequences on
(alN Then
and we set
IN,
= sup( IanI
:
n E N}
(a = (an) E 1-(C)) .
is a complex, commutative, unital Banach Temporarily, we denote its character space by @.
(i°'(C),1.I3N)
algebra.
For n E N, a E RF(C),
set
set En(a) = an a(4) = (a)
(a = (an) E £°°(C)), (0 E @).
are standard and are easily proved;
and, for
The following results
they follow from the
Gelfand-Naimark theorem ([58, 11.18]), for example, and are given in greater generality in [35, 8.3]. (i)
The map
n F+ En, N - $,
is a homeomorphism
6
with dense range.
The map
(ii)
a N &, km(C)
is an
-r
isometric isomorphism.
Thus the compact space
@
define
to be
SIN
has the properties of the Stone-
BN
Cech compactification
N.
of
In our approach, we
different, topological
t:
characterizations are given in [36] and (69], for example, and we shall give a characterization in terms of ultrawith a filters in Chapter 2. Henceforth, we identify N
BN
and
then
MP
subset of p E BIN ,
Let
k (C)
with C(BN ,C).
is a maximal ideal in be a subset of
a
characteristic function of
the closure of
XQ (p) = 1} = a,
and let
IN,
Clearly
Q.
in
a
For example, if
k M. be the
XQ
{p E MN: BN . Thus,
if {a,T}
is a partition of N , then {o,r} is a partition of BIN into clopen sets: we shall use this fact several times. It implies that a n T = Q n T for a,T c N , and that, for a c N ,p E o} is a base of neighbourhoods of p E BIN, {a p in BIN consisting of clopen sets. The existence of this base shows that BN is a totally disconnected space. We note that, if p E BN , then JP is a prime ideal in k (C). For take f,g E k (C) with fg E J J. Then :
p E {n
:
{n
:
and so either or
p E {n
:
(fg) (n) = 0}
f(n) = 0}
p E {n
g(n) = O}
U {n
f(n) = 0}
:
,
:
,
g(n) = O}
,
in which case
in which case
g E Jp.
f E Jp,
Thus
JP
is prime.
We write
c0(C)
(respectively,
for the set
c0)
of complex-valued (respectively, real-valued) sequences which converge to zero.
Then co(C) is an ideal in k (Q), our identification equates it with the ideal I(BN \N)
and in
C(BN C) . coo (C) and coo for the ideals in respectively, consisting of sequences
We write co(C)
and
such that
co,
an = 0
for all but finitely many
n.
(a n)
7
The question that we shall study in this book is the following.
QUESTION X
Let
algebra
be a compact space.
C(X,C)
Is each norm on the
necessarily equivalent to the uniform norm
I,IX?
The question was raised by Kaplansky in lectures
around 1948, and it was Kaplansky who gave the first partial result ([48)). 1.2
THEOREM Let
on
C(X,C).
be a compact space, and let
X
Then
be a norm
11.11
(f E C(X,C)).
IfIX 'IIfII
Proof
Let A be the completion of the normed algebra Then A is a commutative, unital Banach
(C(X,C), 11.11).
algebra.
The map 0 .. 4IC(X,C), 0A uous injection, and so we may regard space of x E X\@A. f(@A
Then there exist By (1),
and
f
such that
f E rad A,
g
C(X,(r),
A
P
# X,
A
in
g(x) = 1,
and so
is a contin-
as a closed sub-
Suppose, if possible, that
= g(4A) = {O},
fg = g.
Thus
X.
4D
C(X,C)
and take
such that
and such that
g = 0,
a contradiction.
0A = X.
Let
i f (x) I < I I f I I ,
and let
f E C(X,C),
and so I
I X< I I f II ,
x E X.
Since
x E 4
as required.
At
It follows immediately from 1.2 (via the open mapping theorem) that each complete norm on
C(X,C)
is equiv-
alent to the uniform norm, and so there is an inequivalent norm on C(X,C) if and only if there is an incomplete norm. Let be an incomplete norm on C(X,C), and
let A be the completion of bedding
C(X,C) -. A
(C(X,C), II.II).
Then the em-
is a discontinuous monomorphism.
other hand, suppose that
6
On the
is a discontinuous homomorphism
8
from
into a Banach algebra.
C(X,C)
= max{If IX, IIe(f)II}
IIf1I Then
11.11
Set (f E C(X,C)).
is an incomplete norm on
C(X,C).
Thus our
question is equivalent to the question of the existence of a discontinuous homomorphism from C(X,C) into a Banach algebra, and it will usually be convenient to formulate it in this way.
The next advance in the study of norms on
C(X,C)
after the work of Kaplansky was the seminal paper of Bade and Curtis of 1960 ([2]).
In this paper, the structure of an
arbitrary homomorphism from
into a Banach algebra
C(X,C)
was analysed, and the key notions of singularity set and radical homomorphism (see below) were introduced. Later, in 1976, Johnson ([47]) extended Bade and Curtis's theorem, and used the extension to show that, if there is an incomplete norm on
for any compact space X, then there exists a radical Banach algebra R, and a non-zero
C(X,C)
P E $N\N, homomorphism
9
from
c0(C)
into
R
such that
ker 9
J
.
It is this reduction theorem that we shall need in Chapter 3.
Our proof is slightly more direct, and gives a little more information, than that obtained by following the original route.
The main technical device in the proof of Bade and Curtis's theorem is the following main boundedness theorem, which originates in [2, Theorem 2.1]. 1.3
THEOREM
A and B be Banach algebras, and let 6 be a homomorphism from A into B. Suppose that there are sequences (a (b and in A such that a b = 0 n n m n (m,n E IN, m $ n). Then there exists a constant C such Let
)
)
that 11 6 (anbn) I I
< C II an 11 II bn II
(n E 10 .
9
Proof
(n E Ilan II = Ilbn II = 1 Suppose, if possible, that the result is false.
We can suppose that Then for each
E N x N,
(i,j)
such that the map
we may choose
n(i,j) E N
is injective and such
I+ n(i,j)
(i,j)
that
Ile(uijvij) II > 4i+j where
and
uij = an(i,j)
(i,j E N)
vij = bn(i,j).
Set
m
fi so that each
=
v
E
R=1
(i E N) ,
it /2R
Choose
fi E A.
IIe(fi) II < 2j (i)
j(i) E N
so that
(i E N)
and set
g=
k k=1
so that
g E A.
uk,j(k)/2
Now, for
i E IN,
k+R
cc
=
1l RL1 k=1
u
i,j(i)vi,j(i) /2i+j(i)
and so
Ile(gfi)II
> 4i+j(i)/2i+j(i) = 2i+j(i).
Ila(gfi) II
<
But
and so
IIe(g)II
> 2i
IIa(g) II Ila(f1) II < 2j(i) Ile(g) II , for all
Thus the result holds.
i E N , I
N) .
a contradiction.
10
Let
We introduce some further notation. We write
compact space.
N
neighbourhoods of a point set of
X,
those functions of
C(X,T)
a closed ideal in
C(X,T).
1.4
Let
of
is an open sub-
U
If
X.
consists of
KU
so that
which vanish off
U,
KU
and
is
K0 = {0}.
For example,
be a homomorphism from
8
A point
if, for each
x E X
U E Nx, BlKU
C(X,C)
into a
is a singularity point for is discontinuous. The set of
singularity points is the singularity set for 1.5
be a
DEFINITION
Banach algebra. 8
x
x
KU = I(X\U),
then
X
for the family of open
8.
DEFINITION
A radical homomorphism at morphism from a maximal ideal
M
is a non-zero homo-
x
of
x
C(X,T)
into the
radical of a commutative Banach algebra. Let
and let
x E X,
: M
8
into a commutative Banach algebra
be a homomorphism We claim that 0 is a - A
X
A.
radical homomorphism if and only if 8lJx = 0. For suppose that 8(M ) c rad A, and take f E J Then there exists x x g E Mx with fg = f, and 0(f)8(g) = 8(f), so that .
0(f) = 0.
Thus
BjJx = 0,
and take
elJx = 0. E
4D
Conversely, suppose that Then 4o8 E 0M U {O}, A.
and
x
4oe)l.x = 0.
Since
8(Mx) c ker 0.
Thus
homomorphism.
Jx
Mx, 08 = 0,
is dense in
8(Mx) c rad A,
and
8
and so
is a radical
The claim is established.
It is further clear that there is a radical homomorphism at x if and only if M /J is seminormable. x
x
We shall several times use the elementary fact that, if F is an infinite subset of a regular space X, then there exist x E F and U E N such for n E N n
that
u
m
n u
n
= 0
Then there exist
and
U U V = X.
so we can choose
n
(m
U E N
n).
x
x
For take
and V E N y
n
x,y E F with x # y. with x 0 V, Y E U,
Either u fl F or V a F is infinite, and x1 E F
and an open set
W1
such that
11
is infinite and
W1 fl F
W
set
such that
n
Set
xn )t Wn.
U
For
x1 fE W1.
successively choose xn E W1 fl
...
w n ... n w n 1_ = X\W1 and set
Un = (Wl n ...
n = 2,3,...,
and an open
fl Wn-1 A F fl F
fl Wn-1)\Wn
is infinite and
(n = 2,3,...).
There is one final remark before we prove the Let A be a Banach algebra, let
theorem of Bade and Curtis. 6
be a homomorphism and let I be an ideal of of the form KU or KU n J. Suppose that there is
C(X,C) - A
:
C(X,C)
(g E I). Take a constant k such that < kIgIX2 II6(g2)II f E I. Then f = f1 - f2 + i(f3 - f4), where fj E I and f(X) c [O,IfIX] (j = 1,...,4). Further, there exist gj E I with gj2 = fj, and so
II6(fj) II < kIf .IX Thus
116(f)II
< 4k I f I I.
then, for each n E N, > nlgn1X2.
II 6(gn2)II
6II is discontinuous, there exists g E I with
Hence, if
Similarly, we see that, if
g E I
then there exists
(j = 1,...,4).
such that 6(g 2)
611 # 0,
0.
We can now give the theorem of Bade and Curtis ([2, Theorem 4.3]). 1.6
THEOREM
Let X be a compact space, let homomorphism from F
C(X,C)
into a Banach algebra
be the singularity set of 6. Then: (i) F is empty if and only if F
(ii)
Now suppose that
be a
6
6
A,
and let
is continuous;
is finite. 0
is discontinuous and that
F = (xI'...,xn}.
(iii) There are a continuous homomorphism C(X,C)
that and
viIMx
A
and linear maps v1,...,vn : C(X,C) -' A such is a radical homomorphism at xi (i = 1,...,n)
0 = µ + vl + ... + vn
12
Proof
versely, suppose that exists
F = 0
Clearly
(i)
is continuous.
6
Then, for each
F = 0.
such that
Ux E Nx
if
Con-
there
x E X,
X
is continuous, and so
6IKU X
V such that 6IKv is continuous Using a partition of unity subordinate to
has a finite open cover
V E
for each V,
V.
we see easily that 6 is continuous. Suppose, if possible, that (ii)
Then there are an infinite sequence
is infinite.
F
(xn) c F
Un E Nx
and
n
Um n Un = 0
such that I
I
6(fn`) I
(n E N) .
n Ifn Ix`
>
I
(m # n).
such that
fn E KU
Choose
Since fm fn = 0
n
(m # n)
Thus
this contradicts the main boundedness theorem.
F
is
finite.
Let
(iii)
Ux E Nx
such that
x E F. 6I(KU
We claim that there exists
n Jx)
is continuous.
For suppose
x
that this is not the case. sequences n E N ,
Then we may inductively choose (fn) c Jx
and
(V n) c Nx
such that, for each and
fn E Kv
(i = 1,...,n),
fi(Vn+1) = {O} Ix 2. II 6(f n2) II > n Ifn Since f mf n = 0 contradicts the main boundedness theorem. n
,
this again
(m # n),
Thus the claim
holds.
It is immediate from the claim that
continuous, say Fix
Wi E Nx
6IJ(F)
is
(f E J(F)).
< klflx
110(f) II
such that
(i = 1,...,n) i
Wi n wj = 0
(i # j),
and fix
ei E Kw
such that i
Ietlx = 1
ei = 1
and
be the set of functions f - f(xi)1 E Jx
on some neighbourhood of Then
s algebra of J(P),
C(X,(C).
Let
such that
f E C(X,(C)
(i = 1,...,n).
x
is a dense sub-
B
n
For
f E B,
f(xi)ei
f -
belongs to
and so
110(f) II
< kIf -
i=1
f(xi)eilx +
< [(n + 1)k +
Ii
f(xi)6(ei) I
I
=1
II 6 (ei) II ]if lx.
B
13
Thus
is continuous on
a
Let C(X,C),
B.
be the continuous extension of
g
Then
v = e - M.
and let
continuous homomorphism, and Take
C(X,C) -. A
:
is a
vjJ(F) = 0.
and
f E I(F)
LL
to
e1B
g E J(F).
Then
e(f)LL(g) = e(fg) = M(fg) = M(f)LL(g),
and so v(f)LL(g) = (e(f) - M(f))M(g) = 0.
Since
LL
J F
is continuous and (f,g E I(F)).
v(f)LL(g) = 0
we have
= I(F),
f,g E I(F),
Thus, for
v(fg) = e(f)e(g) - M(f)M(g) (LL(f)
_
+ V(f)HM(g) + v(g)) - LL(f)LL(g)
= v(f)v(g), and so
is a homomorphism.
vII(F)
Set
vi,...,vn
Then
i = 1,...,n). (f E C(X,C), vi(f) = v(fei) C(X,C) + A are linear maps. Since :
n
v = vi + ... + vn.
ei E J(F),
1 -
Let
i E {1,...,n}
1=1
and let
Then
f,g E MX .
fei, gei E I(F),
and so
i
v(fei) v(gei) = v(fgei2) . But
e
2
- ei E J(F),
V (f)vi(g) = vi(fg), i
If
f E JX ,
then
and so and
Vi
v(fge2) = v(fgei). Thus : MX A is a homomorphism.
fei E J(F),
i
and so
vi(f) = 0.
Thus
i
vilMx
is a radical homomorphism.
I
i
1.7
COROLLARY
Let
X
be a compact space.
Then the following are
equivalent: (a)
there is a norm on
C(X,C)
which is not
14
equivalent to the uniform norm;
there is a discontinuous homomorphism from
(b)
into a Banach algebra;
C(X,C)
there is a radical homomorphism from a maximal
(c)
ideal of
C(X,C);
there exists
(d)
seminormable.
x E X
such that M /J x
x
is
I
THEOREM
1.8
Assume that there is a Let X be a compact space. discontinuous homomorphism from C(X,C) into a Banach algebra. Then there exists p E RN\ N, a radical Banach algebra R, and a non-zero homomorphism 0 from c0(C) into R such that ker 6 m J. fl co(C). Proof Let
0
be a discontinuous homomorphism from A. Then there exists
into a Banach algebra
C(X,C)
(f n ) c C(X,(C)
such that
if
=1
and
containing
if n, fn
Then
n E 3N).
:
isometrically isomorphic to space
Since
Y.
clearly
with
Mx -*,R Y.
Let
such that
Y
is
is metrizable, and
there is a singularity point
x E Y,
a
and a non-zero homomorphism vIJx = 0. Certainly x is not an isolated
radical Banach algebra :
B
is discontinuous.
0IB
By 1.6,
point of
and so
C(X,C),
for a certain compact
C(Y,C)
is separable,
B
is a closed,
B
separable, self-adjoint subalgebra of
v
Let
110(f )II
nIX n be the smallest closed, unital subalgebra of C(X,C)
B
Set
R,
Xo = Y\{x}.
be a sequence of open subsets of
(Un)
Um fl Un = 0
S = in E N
and let
(m # n),
:
X0
# 0}.
V IKU
n
For each
n E S,
there exists
gn E KU
such that n
v(gn2)
VIJx = 0,
0.
Since
KU
fl Jx
is dense in
we may suppose that
11v(gn2)11
KU
,
and since
> nIgnI Y. 2
Since
15
gmgn = 0
be a metric defining the topology on
d
(x ) c X n
< d
n+1
is finite.
S
Let
and take 46
it follows from the main boundedness
(m # n),
theorem that
such that
o
where
(n E N) ,
n
6
+ 0
n
Set
= d(x,xn).
d
Y,
and
It
Vn = {y E Xo
< d(y,x) < 26 }
:
(n E N).
26
The sets V Vm n v
are non-empty and open, and {a : m E N } (m # n). Let
n
= 0
n
into infinite subsets, and let Then
Um = U{Vn : n E Qm}.
sets of
X
such that
o
u
is a sequence of open sub-
(Um) f1 U
m
n
and so
(m # n),
= 0
for all but finitely many
= 0
vIKU
be a countable
m
partition of N
m.
By passing to a
m
subsequence of
V = V{V
n
we may suppose that
(xn),
where
vIKV = 0,
n E N }.
:
For n F
n
set
14,
= {y E X o
26
:
n+1
< d(y,x)
ld 2n
Suppose that V U {x}
F = 0 for all but finitely many n. Then n is a neighbourhood of x in Y, and so, if
f E Mx,
there exists
f = fl + f2, diction.
whence
For n
{1}, n
d(y,x) < 6n+i'
with
f2 E Kv
and
v(f) = v(fI) + v(f2) = 0,
Thus we may suppose that
F = U Fn U {x}, e (F
fl E Jx
a closed set in n E IN,
e (y) n
n
= 0
# 0
a contra-
Let
(n E 24).
Y.
en E Jx
take
and with
F
if
with
IenIY = 1,
d(y,x)
> 6
n
with
or
Set CO
ry(a)
= n11anen
(a = (an) E co(C)) .
For each
y E Xo, there exists W E NY with en (W) = 0 for all but two values of n, and so ry(a) is continuous on X0. Clearly, ty(a) E Mx, and voip c0(C) + R is a linear :
map.
For each
en2 - en E Kv
y E Xo\V, and
en(y) E {0,1}, and so v(en2 - en) = 0 (n E N). Also
16
(m # n),
emen = 0
and so
is a homomorphism.
vo
p(coo(C)) c ix c ker v, coo (C) = 0. Choose g E MX with v(g) 0, and let so that (Bn) E co. Take an > 0 (n E N), an = Ig IF n such that a = (a n) E c o and (6n /a n) E C o t set f = g/ a n Since
on
and set
F , n
f(x)
= 0.
Then
is continuous on the
f
in Y, and so f extends to a function in MX. Since *(a) f = g on F, (a) v(f) = v(g), and so (voi) (a) # 0. Thus v°V : c0(0) -+ R is a non-zero homomorphism with (voiy) Ic 00 (C) = 0.
closed set
F
We now regard
c
0
(C)
as an ideal in
C(BN ,C).
Set
E = {p E 61N
Since
vo p # 0,
E c $N \N . If sequence U
in
(U
fl Un = 0
Since
:
(v-,) I (KU n co(C) ) # 0
E # 0. Since E
(m O n)
and
there would exist a
(vole) Icoo(C) = 0,
such that
(n E N).
n
the argument of the third paragraph Thus
(vop) I (J (E) n co(C) ) = O.
Let
e = 1
gN
n c0(Q) ) # 0
(voip) I(KU
leads to a contradiction. such that
(vol,) Icoo(C) = 0,
were infinite,
of non-empty, open subsets of
)
n
(U E Np)
E
is finite.
p E E,
on a neighbourhood of
Clearly,
and take e E C (SN,C) p
and a= 0
neighbourhood of E\{p}. Set 0(f) = (vop)(fe) Then a is the required homomorphism.
on a
(f E c(0)).
Bade and Curtis left open the question of the existence of discontinuous homomorphisms from the algebras C(X,C)
into Banach algebras.
Eventually, such homomorphisms
were constructed for each infinite, compact space by Dales ([16]) and by Esterle ([26], [24], (27]) in 1976. However, both constructions required the assumption of the Continuum Hypothesis (CH), and are thus theorems of the system ZFC + CH. Neither Dales nor Esterle discussed whether or not CH was required. We now state without proof the results of Dales and of Esterle. We denote the cardinality of a set S by ISI;
17
a review of cardinal numbers will be given in Chapter 5. Theorems whose proof is given in the theory ZFC + CH
are labelled
(see Chapter 4)
which hold in
THEOREM
1.9
Let
X
(CH)
be a compact space, and let
maximal, prime ideal in Then the algebra
but theorems
"CH",
are not labelled.
ZFC
C(X,C)
X
be a non-
IC(X,C)/Pl = 2 0.
such that
is normable.
C(X,(Z)/P
Now, if
P
is infinite, then
C(X,C)
does contain K
non-maximal, prime ideals
such that
P
Further, it is easily seen that, if then the hull of
C(X,C),
exists
x E X
such that
IC(X,C)/Pl = 2 0.
is any prime ideal in
P
is a singleton, and so there
P
since J= M,
J c P c M ; x x
is closed if and only if
P
P
X x is equal to the maximal ideal
Thus it follows from 1.9 that, if the Continuum
Mx.
Hypothesis holds, then the four properties (a) - (d) of Corollary 1.7 are true for each infinite, compact space
X.
In particular, we have the following result.
THEOREM
1.10
Let
X
(CH)
be an infinite compact space.
exists a discontinuous homomorphism from Banach algebra, and there is a norm on
C(X,C)
Then there into a
which is not
C(X,C)
equivalent to the uniform norm.
Let
X
be a compact space.
X
If
is separable,
x
then
IC(X,Q)l = 2 0.
Mx # Jx
x
(so that
Suppose that is not a
Definition 2.22), and take
x
c P 1.11
f E M x\J x.
and so there is a prime ideal
(n E N), J
x E X
P-point of
and
f (E P.
THEOREM Let
X
Then P
is such that
X - see fn t Jx in
C(X,C)
Hence we have two further results. (CH)
be a separable, infinite compact space.
with
18
Each non-maximal, prime ideal of
(i)
C(X,C)
is
the kernel of a discontinuous homomorphism into a Banach algebra.
1.12
and if
x E X
If
(ii)
M /J
quotient algebra
x
THEOREM
M
x
# J , x
then the
is seminormable.
x
(CH)
(C)/j
p E aN\N. Then the algebras co(C)/(Jp n co(C)) are normable. Let
P
and
We have stated the above four theorems as results of ZFC + CH. In fact, they all follow from a more general result on the normability of integral domains that can be proved in
This more general result will be discussed
ZFC.
at the end of Chapter 6.
It is an immediate consequence of Theorem 1.8 that,
if there is a discontinuous homomorphism from any compact space
morphism from
X,
c (C). 0
C(X,C) for then there is a discontinuous homoIt is natural to enquire whether or
not all infinite compact spaces are equivalent for our problem. The first result in this direction is the following theorem. 1.13
THEOREM
Assume that there is a discontinuous homomorphism from
t"(C)
Then there is a discon-
into a Banach algebra.
tinuous homomorphism from
C(X,C)
each infinite compact space
into a Banach algebra for
X.
Proof Since
discrete subspace
X
is infinite, it contains an infinite, {xn}.
(f) (n) = f (xn) Then
For
f E C(X,C),
set
(n E N) .
T(f) E £°(C), and T : C(X,C) + £ (C) homomorphism with range B, say. For each
is a continuous a E B
and each
19
there exists f E C(X,(C) with TM = a and IfIX < IalN + E, and so B is a closed subalgebra of
E > 0,
k`°(C). Since there is a discontinuous homomorphism from
p E BN\N and a non-zero homomorphism into a Banach algebra A such that elm # 0
there exists
2°°(C),
from k°(C) and 611 = 0. 8
P
Take such that for
0 <
Un = {m E N
Uo = N\UUn.
m E U.
Then
m E Un
(y m)
E k°°,
g,
(x E BN)
g E MP.
h (x) = 0
and
to
(kn) c N
(n E N),
and,
is a partition of
BN.
(n E ZZ+) , and so Clearly,
(Ym)
N.
taking Ym = 1 has a continuous
If(x)I < g(x)
Set h (x) = f 2 (x) /g (x) if g (x) # 0, g (x) = 0. Then I h (x) I < If(x) I < g (x)
if
and so
(x E BN),
choose
< If(m)I E If(kn)I}.
{Uo,U1,...}
for
extension, say
and
If(kn+1 )I
:
Then
Set ym = If (kn) I if
8(f2) # 0,
take
n E IN,
Let
with
f E M
If(kn+l)I < If(kn)I < 1/n
Since
h E MP.
For
a = (a) E B,
* (a)
=
gh = f2,
8(g) # O.
set
E anXn, n=o
where
Xn
is the characteristic function of U. Then C(X,C) -- A are homomorphisms. and 8o*oT
B - k°°(C) If
a E c00(C),
(8o0Icoo(C) = 0.
'P(8) = Y.
Let
Since 8(g)
discontinuous.
then
fla) E JP,
B = (If(kn)I).
0,
Then
and so B E c0(C)
(8o00) # 0, and so
8o*oT
and
is
I
It follows from 1.8 and 1.13 that all infinite compact spaces are equivalent for our problem if and only if the existence of a discontinuous homomorphism from co(C) implies the existence of one from k°°(C). We do not know whether or not this is true, and the relation between k°°(C) and c0(C) in this regard seems to be subtle. We shall return to this point again in Chapter 3.
20
NOTES
1.14
Fuller accounts of Banach algebra theory are given in [8] and [17].
Automatic continuity theory - the study of
conditions under which linear maps from Banach algebras are and automatically continuous - is described in [65], [51], [15], as well as [17].
Generalizations of the main bounded-
ness theorem are given in these sources.
As we stated, the notions of singularity set and radical homomorphism were introduced by Bade and Curtis in The theory of singularity points has been developed in
[2].
much more general contexts:
see, for example, [1] and (51).
It is a striking fact that, in distinction to the commutative case, all homomorphisms from certain noncommutative C*-algebras (such as
B(H),
the algebra of all
bounded linear operators on a Hilbert space
into a this is a
H)
Banach algebra are automatically continuous: theorem of Johnson ([46], [65, 12.4], [17]).
The concentration in Theorem 1.9 on prime ideals is natural, for the following result is proved in [65, 11.4]: Let
be a discontinuous homomorphism from
6
onto a dense subalgebra of a Banach algebra Then there is a closed ideal J in A such
C(X,C) A.
that the map
C(X,C) i A/J, is a discontinuous homomorphism whose kernel is a f M 6(F)
prime ideal in
Let
X
+ J,
C(X,C).
be an infinite compact space such that
K
IC(X,C)) = 2 °,
with
and let
CH, C(X,C)/I
I
be an ideal in
is normable whenever
intersection of prime ideals. the quotient
C(X,C)/I
I
C(X,C).
Then,
is a finite
It is not known whether or not
is ever normable for any other ideal:
m
I (C)/I
is normable, then
I is a finite intersection of prime ideals ([25, Proposition 5.1]). Other partial
if
results are given in [18].
It is not true, even with normable for each prime ideal
P:
CH,
that
C(X,C)/P
is
this may fail if
K
IC(X,C)/Pl > 2 °
([25, Theoreme 7.1], (17]).
Indeed it may
21
C(X,C)/P
be that the normability of
entails that
K
this is an open question that we shall
IC(X,C)/PI = 2 0:
rei_urn to in Chapter 6.
The characterization of radical Banach algebras such that, with from
into
C(X,C)
([30],
has been established by Esterle
RO
has this property if and only if
R
(17, §6.6]):
there exists
R
there is a discontinuous homomorphism
CH,
such that
a E R\{0}
a E a2R.
Several other
are also given in (17, §6.6].
R
characterizations of
Here
are two examples of such radical Banach algebras. Let functions f(z) - 0
denote the set of bounded analytic
H0
on the open right-hand half-plane R such that z + m in R. Then H0 is a Banach algebra
f
as
with respect to the uniform norm, and ideal in
Ho.
The quotient algebra
a ZH, 0 Ho/e-zH:
is a closed is a radical
Banach algebra with the required property ([17, §6.5]). Let f
on
R+
Ilf1l For
R
be the set of Lebesgue integrable functions
such that
f,g E R,
=
fo:
1f(t)
1e-t2
dt < -.
set
(f * g) (t) =
t J
f (t - s) g (s) ds
(t E R+
)
.
0
Then
(R,*, 11.11) is a radical Banach algebra with the required property ([17, §6.5]). It follows from Titchmarsh's
convolution theorem that
R
is an integral domain.
It is a theorem of Esterle ([29], [17]) that each epimorphism from each C(X,C} onto a Banach algebra is automatically continuous.
22
2
PARTIAL ORDERS, BOOLEAN ALGEBRAS, AND ULTRAPRODUCTS
Our approach to forcing will use "Boolean-valued universes", a method which originates with Scott and Solovay (see the notes to Chapter 7). It is therefore necessary for we us to present some basic facts about Boolean algebras; shall prove only the results that we shall require later. In fact, our discussion will be carried out in the more general context of partially ordered sets.
2.2 (vi) we shall describe
and
(NIN,
In Example
(NIN,«F) ,
examples of partially ordered sets which will be very prominent in later chapters.
In this chapter, we shall introduce the completion and the Stone space of a Boolean algebra B. The latter is the set of ultrafilters on B, and we shall recognize BN This P(N). allows us to reformulate our basic question as a question about co/U, where U is a free ultrafilter on N.
as the Stone space of the Boolean algebra
We shall conclude the chapter by discussing the P-points of the compact space 2.1
DEFINITION Let
on
P
be a non-empty set.
P
is a binary relation (i)
if
a < b
(ii)
a
a
A total order on
P
b < a.
<
and
for each
A strict partial order
such that:
b < c
in
P,
then
a < c;
a E P.
is a strict partial order
(iii) for each or
SN\ N.
a,b E P,
either
<
a < b
such that:
or
a
b
23
A partially ordered set is a pair where
is a non-empty set and
P
P = (P,<), is a strict partial
<
order on P. A subset
Q
is of a partially ordered set (P,<) < to Q x Q is a total order
a chain if the restriction of on
Q.
The first coordinate
of the pair
P
underlying set of the partially ordered set.
is the
P
For the present
we shall not distinguish notationally between a partially ordered set and its underlying set.
Let (P,<) in P if a < b or conditions: (1) if a < a
(ii)
in
P,
be a partially ordered set. Set a < b The relation < satisfies the a = b. then a < c; a < b and b < c in P,
for each
then
a = b.
is a partial order.
a E P;
if
(iii)
a < b
and
A relation satisfying these conditions Now let
< be a partial order on a non-
and set a < b if a < b and a # b Then (P,<) is a partially ordered set.
empty set
b < a
P,
(a,b E P).
shall always suppose that the orders
and
<
We
< correspond
in this way, without further mention. Let
2.2
We write
be a non-empty set.
S
the power set of
S:
P(S)
for
P(S)
is the set of all subsets of
S.
EXAMPLES
Let S be a non-empty set. For set a < b if a c b. Then < is a partial P(S). This is the ordering given by inclusion. (i)
a,b E P(S),
order on
For
(ii)
finite and if
b\a
is infinite.
P(1).
partial order on the finite sets. (iii)
a,b E P(N),
set
Then
The order
a
if
a\b
is
is a strict
is inclusion modulo
Let A be a radical algebra. For a << b if there exists c E A with
set a,b E A\{O}, a = bc. Then <<
is a strict partial order on A\{O}: condition 2.1(11) is satisfied, for if a = ac in A, then a = O. The order << is the divisibility order on A.
24
Let
(iv)
f,g E CM, set f < g if is a partial order on
For
be a topological space.
X
(x E X).
f (x) < g (x)
Then
it is the standard order.
C(X):
be totally ordered Let (P1,<),...,(Pn,<) ... x Pn. (xi,.... xn) < sets, and let P = P1 x Set in P if xi < yi, where i = min{j x3 # yj}. (v)
:
(y1 ...,yn)
Then <
is a total order on
called the lexicographic
P,
order.
Let f,g E NIN .
(vi) 0
f
and
f(n) < g(n)
if
f(n) < g(n)
eventually,
if g (n) - f (n)
f << F g
such that
Set
(n > n ).
Then
f(n) < g(n)
Then
no E N
eventually if there exists
-> -
ni.
as
are strict partial orders on
<< F
NIN:
they
are the Frechet order and the strong Frechet order, respectively. 2.3
DEFINITION Let
A map
jr
and
(P,<)
(Q,<)
be partially ordered sets.
from
P
(i)
order-preserving if
a < b in
to
Q
is:
whenever
ir(a) < 7r(b)
P; (ii)
isotonic if
w(a) < n(b)
whenever
a < b
in
P;
(iii) anti-isotonic if a < b
in
er(a)
> or(b)
whenever
P; (iv)
n(a) < ir(b) (v)
in
an embedding if Q
r
is injective and if
if and only if a < b in
an order-isomorphism if
it
P;
is a surjective
embedding.
Note that, if
it
is injective, then
preserving if and only if it is isotonic.
it
Also, if
is order<
is a
25
total order on
P
and if
n
:
P - Q
is isotonic, then
n
is an embedding.
The following result concerning the Frechet order
on NN will be required later. PROPOSITION
2.4
There is an anti-isotonic map from
(NN,
into
(NN,
: N x N -. N
Let
y
For
f E NN,
be a fixed bijection.
set
S(f) = {(m,n) E N x N so that
S(f)
:
f(m) < n},
is the set of points "above" the graph of
f,
and set
Tr(f) (n)
= 1{1,...,n} n y(S(f)) I + 1
(n E N).
v(f) E NN We claim that
Then
is an anti-
n : (NN, m0).
isotonic map.
exists
S(g)\S(f) c {(m,n) S(g)\S(f)
:
n < f(m),
is finite, and, since
1 < m < m0),
(m,f(m)) E S(f)\S(g)
(m > m), a S(f)\S(g) is infinite. Thus ir(f) (n) > n(g) (n) eventually, and so w (g)
(P,<)
a non-empty subset of (i)
if
aAb
P,
and let
a E P.
is a maximal element of
let
Q
be
Then:
Q
if
a E Q
and
Q
if
a E Q
and
(b E Q) ; (ii)
if b < a
a
be a partially ordered set,
a
is a maximum element of
(b E Q) ;
26
is an upper bound of
(iii) a
Q
b < a
if
(b E Q) ;
Q
is a supremum of
a
(iv)
bound of
and if
Q
if
a
is an upper
for each upper bound
a <. b
b
of
Q.
Minimal, minimum, lower bound, and infimum are defined similarly.
The supremum and infimum of unique if they exist; inf Q,
respectively.
are necessarily
Q
they are denoted by Note that, if
ordered, then a maximal element of
element.
P
and
is totally
is the maximum of
Q
Q.
if each chain in a
Zorn's Lemma is the assertion:
partially set
sup Q
(P,<)
has an upper bound, then
P
has a maximal
The relation of Zorn's Lemma to the Axiom of Choice
However, as is usual in analysis, we shall use the lemma freely. will be recalled in Chapter 5.
2.5
DEFINITION
A partially ordered set is complete if each nonempty subset which has an upper bound has a supremum and if each non-empty subset which has a lower bound has an infimum. 2.6
DEFINITION
A Boolean algebra is a structure B = (B,n,v,',O,1),
where
n and v are binary operations on the set B, where is a unary operation on B, where 0 and 1 are
distinct elements of all
B,
and where the following hold for
a,b,c E B:
(i)
b n a, a v b = b v a;
(iii) a n a
av0 ava
anb (ii) a n 1
a;
a;
(iv)
a n (b v c)
_
(a n b)
(v)
a n (b n c)
_
(a n b) A C;
v
(a n c);
27
(vi)
(a A b)' = a' v b';
(vii)
(a')' = a;
(viii) a v a' = 1, a A a' = 0; (ix) 0' = 1. The above list contains redundancies: 42].
The operations
A, v,
and
see (39,
are called meet, join
'
and complementation, respectively.
The first coordinate of the sextuple (B,A,v,',O,1) is the underlying set of the Boolean algebra B. At present we shall use the letter B for both the Boolean algebra and for its underlying set.
A Boolean algebra has cardinality
if its underlying set has cardinality
K
K.
A non-empty subset of a Boolean algebra is a subalgebra if it is a Boolean algebra with respect to the given operations. The set {0,1} is a Boolean algebra, to be denoted by (2): it is the trivial Boolean algebra, and it is regarded as a subalgebra of each Boolean algebra. Let B and C be Boolean algebras. A map Tr
from
B
to
C
is a Boolean homomorphism if, for
Tr (a A b) = Tr (a) A Tr (b) ,
Tr (a v b)
= Tr (a)
a,b E B, v Tr (b) ,
ir(a') = Tr(a)'. An injective (respectively, bijective) Boolean homomorphism is a Boolean monomorphism (respectively, Boolean isomorphism). Write B = C if there is an isomorphism from B onto C. Let B be a Boolean algebra. For a,b E B, set a
algebra is just a partially ordered set in which the order satisfies certain additional axioms. A Boolean algebra B is complete if the partially
28
is complete, and so, in a complete ordered set (B,
they are denoted by
V{s
:
s E S}
or
VS,
M{s
:
s E S}
or
As,
and
respectively.
For
a A V{s
a E B, :
we have
s E S} = V{a A s: s E S},
for example.
algebra.
Here is an obvious example of a complete Boolean Let S be a non-empty set, and let B = P(S). For
a,b E B,
set
a v b= a U b, a' = S\a,
a A b = a fl b, and set
0 = 0,
Then
1 = S.
B
(1)
is a Boolean algebra, and
the corresponding order is ordering by inclusion.
The
supremum and infimum of a non-empty subset A of B are respectively. When we reU{a : a E Al and fl{a a E Al, fer to P(S) as a Boolean algebra, we suppose that it has :
the Boolean operations given by (1).
Corollary 2.15, that the algebras class of Boolean algebras.
We shall show, in
P(S)
are universal in the
First, however, we give another
important example of a Boolean algebra. 2.7
EXAMPLE Let
X
be a non-empty topological space which is
not necessarily Hausdorff.
A subset U of X is regularopen if it is the interior of its closure: U = int(U). For example, for each open set U, is regular-open. int(U) R(X) be the set of all regular-open subsets of X. For U,V E R(X),
set
Let
29
U A V= U n v, U v V = int(U U V) , (2)
U' = int(X\U),
and, as before, set
0 = 0,
It is routine, but not
1 = X.
entirely trivial, to check that, with respect to these operations,
is a complete Boolean algebra (cf. (39, §41,
R(X)
we have
(69, §2.3]):
: U E S} = intU(U
V{U
UES
:
(3)
: U E S} = intfl{U
M{U
for a non-empty subset
of
S
U E S},
:
The Boolean algebra X. Note that, if
R(X).
is the regular-open algebra of
R(X)
then
U,V E R(X),
Let Then
U 4R(X) V
B(X)
U c V.
be the family of clopen subsets of
is a subset of
B(X)
if and only if
X.
and the restrictions of
R(X),
the Boolean operations given by (2) to
B(X)
agree with the
Boolean operations given by (1).
B(X)
is a Boolean
subalgebra of both The space
and
R(X)
Thus
P(X).
is extremely disconnected if the
X
closure of each open subset of
X
see that
R(X)
and so
R(X) = B(X),
is open.
In this case, we
is a complete Boolean
algebra with respect to the operations given by (1). 2.8
DEFINITION Let
(P,<)
be a partially ordered set.
Two elements
(i)
compatible if there exists and
a
and
b
a E P,
A subset there exists
We write elements of
P
are
c < a
and
of
b
with
c < b,
are incompatible if they are not compatible.
(ii)
each
and
a
c E P
P.
a l b
of
Q
P
b E Q if
a
is dense in
P
if, for
with b< a. and
b
are incompatible
30
These notions will be applied to a Boolean algebra B
by regarding
(B\{O},
as a partially ordered set.
Thus, if a,b E B\{O}, then a 1 b if and only if for example, elements
a A b = 0;
and
a
b
in
p(S)
are
incompatible if and only if they are non-empty and disjoint. We stress that we consider
and not
B \{O},
B
itself.
We shall now associate to each partially ordered set a complete Boolean algebra. 2.9
DEFINITION Let
(P,<)
be a partially ordered set, Then
be a complete Boolean algebra.
of
if
P
(P,<) - (B \{O},
:
Tr
(i)
Tr(P)
(ii)
Tr
is a completion
(B,Tr)
is a map such that:
is order-preserving; a,b E P,
a 1 b
in
P
if and only
in B\{O}.
Tr(a) 1 Tr(b) 2.10
B
is dense in B\{0};
(iii) for each
if
and let
THEOREM Let
(P,<)
be a partially ordered set.
Then
P
has a completion. Proof For
cEU
Ua n Ub.
topology, say Let
Then
a E P,
B
let
Ua = {b E P
:
b S a}.
We see
(a E P), and that, if c E Ua fl Ub, then
that a E Ua
T,
Thus {Ua : a E P1 on P.
is a base for a
B = R(P),
the regular-open algebra of is a complete Boolean algebra. Set Tr
:
a i+ int(Ua) ,
(P,T).
p + B\{O}.
We check that conditions (i) - (iii) of Definition 2.9 are satisfied. (i)
open set in
Take (P,T),
a E B\{O}.
Then
and so there exists
a
is a non-empty, b E P
with
31
Ub c a.
a < b
If
(ii)
in
then
P,
1(b)
and so
w(b) c int a = a,
We have
and so
Ua c Ub,
7r (a)
(iii) Let a and b be compatible in take c E P with c< a and c< b. Then T1(c) it (c)
and
7r (c)
are compatible in
and so w(a)
P, and E B\{O}, and Tr(b)
B\{O}.
Now let a and b be incompatible in P, so that Ua fl Ub = 0. Since Ub is open, Ua f1 Ub = 0, and so 7r(a) fl Ub = 0. Since 1(a) is open, T1(a) fl Ub = 0, and so Tr(a) n 7r(b) = 0. Thus ir(a) and Tr(b) are incompatible in B\{O}. f It can be shown that a completion of a partially ordered set P is unique, in the sense that, if (B1'Tr1) and
are completions of
(B2,Tr2)
isomorphism
8
:
B1 - B2
P,
such that
then there is a Boolean n2 = 8oTr1. (B,Tr)
If
is a completion of
P, we set B = P. It is not claimed in Theorem 2.10 that Tr P + B For example, suppose that P is always an injection. has a :
minimum element a. Then a E Ub for each Ub = P (b E P). Thus R(P) = {(,P}, and trivial Boolean algebra. However, it is easily seen that injection if
P
B
Tr
:
is the
P + B
is an
in the sense that, for each there exists c E P with c I a and
is separative,
with b% a
a,b E P
and so
b E P,
c < b. 2.11
DEFINITION Let
F
of
c E F
P
(P,<)
be a partially ordered set.
is a prefilter if, for each
a,b E F,
A subset there exists
with c< a and c< b.
A prefilter F is a filter if, for each a E F with a < b, it follows that b E F. A filter is a maximal filter if it is maximal in the set of filters on P with respect to inclusion.
and
b E P
32
Let
be a prefilter in
F
{b E P is a filter in
Then
P.
a C b for some a E F}
:
containing
P
F.
It is immediate from
Zorn's Lemma that each filter is contained in a maximal filter, and so each prefilter is contained in a maximal filter.
Let in
B
is a filter in the partially ordered set
B
and an ultrafilter in then
a,b E A,
if
then
a
b E A;
for each
A
B
then
A
this case,
Let
such that:
a E A and It is easily checked if
is an ultrafilter if and only if, A. Let
C = {al A ... A an 0 E C,
0 >t A.
(ii)
either a or a' belongs to and consider the set B,
a E B\{O},
be a subset of
If
is a subset A
a n b E A;
(iii)
that a proper filter A
(B\{O},
is a maximal filter in this set.
B
Thus a proper filter in (i)
Then a proper filter
be a Boolean algebra.
C
:
a1,...,an E A}.
is a prefilter in
B\{O},
and so, in
is contained in an ultrafilter in B. Then proper filters S be a non-empty set.
are said to be and ultrafilters in the Boolean algebra P(S) filters and ultrafilters on S, respectively. Thus, a filter on S is a family F of subsets such that: (i) if F,G E F, then F fl G E F; (ii) if F E F and F c G, For example, if x belongs to a then the family of neighbourhoods of x in X is a filter on X. Note that, if U is an ultrafilter on S, and if A,B c S with A U B E U, then either then
G E F;
AEU
0 0 F.
(iii)
topological space
X,
or B E U.
Let B be a Boolean algebra. We know that the partially ordered set (B\{0},
33
is denoted by
B
The family of ultrafilters in
Set
S (B) .
= {U E S (B) : b E U}
Tr (b)
Certainly,
fir(b)
In particular,
on
TB,
S(B).
DEFINITION TB
The topology
is the Stone topology, and
is the Stone space of
(S(B),TB)
2.13
Also,
=
b E B}
:
is a base for a topology, say 2.12
Tr (a v b)
n Tr (b) ,
1T (a') = S (B) \Tr (a) .
and
v (a) U Tr (b) ,
= Tr (a)
(4)
(b E B\{O}).
w(b) # 0
and
w(O) = 0
for a,b E B, it (a n b)
(b E B) .
B.
LEMMA Each set
(i)
(ii)
w(b)
(S(B),TB)
is clopen in
(S(B),TB).
is a compact, totally disconnected
space. (iii) If U{Tr(d)
:
d E D}
B\{O},
is a dense subset of
D
is a dense, open set in
then
S(B).
Proof
By definition, each
(i)
S (B) \Tr (b)
=
Tr (b') ,
(ii)
and
If
each
Tr (b)
U # V
in
n(b)
is open.
Since
is closed. S(B),
there exists
b E U\V,
is a clopen set such that U E Tr(b) and Thus S(B) is totally disconnected (and
w(b)
V A Tr(b).
Hausdorff).
To show that show that, if
{w(a)
property, where A
:
(S(B),TB)
a E A}
is compact, it suffices to
has the finite intersection B, then fl{Tr(a) : a E A}
is a subset of
But in this case al A ... A an # 0 (al,...,an E A), and so A is contained in an ultra-
is non-empty.
filter, say
U,
on
B.
Since
U E fl{TT(a) :
a E A},
the
34
result follows. Let
(iii)
open in
For each
S(B).
with d < b. is dense in
b E B\{O},
Since n(d) < Tr (b) , S(B).
Certainly U is there exists d E D
d E D}.
:
and so
0,
7(b) fl u
U
I
THEOREM
2.14
Let Then
U = U{Tr(d)
be a Boolean algebra, and let B = R(S(B)). (B\{O},
B
is a completion of
(B,TT)
Tr
:
is a Boolean monomorphism. Proof Tr(B) c B(S(B)) c B,
By 2.13(1),
Then
U
in
7f (a)
B
containing Thus
Take Tr(b) c a.
Clearly,
say a,b E B with a # b, and so there is an ultrafilter and so a E U and b ¢ U, and b';
a n b' # 0,
# Tr(b).
subset of
a Tr
is an injection.
a E B\{O}.
Thus
Then
a
and so there exists
S(B),
Tr(B\{O})
is a non-empty, open b E B\{O} with
is a dense subset of
is order-preserving, and
n
and only if n(a) 1 R (b) in 9\{O}. completion of
(B\{O},
in
a i b
Thus
completion of
empty set
B\{O}
if
is a
I B
as the
B.
COROLLARY Let B be a Boolean algebra. S
B\{O}.
(9,w)
Somewhat abusing notation, we refer to
2.15
is a
n
Take
Boolean homomorphism. a FB b.
and
Then there is a non-
and a Boolean monomorphism
Tr
:
B + P(S).
Proof Let S = S(B), and let n be as above. Since ¶(B) c B(S(B),TB), V(B) is a Boolean subalgebra of P(S). 2.16
EXAMPLE
Let
B = P(IN),
so that
B
is a complete Boolean
35
algebra, and the Stone space S(B) consists of the set of IN. Let SN be the character space of the
ultrafilters on Banach algebra
as in Chapter 1, so that
k"(T),
compact space, and N writing
a
p E a3 N, and let a fl T (U, T C N) ,
Since an T= Thus
N. or
SN.
of a subset
UP = (Q c N
:
if
SN is a We are a
N.
of
p E a}.
a fl T E U P Q, T E U P. Let {a,T} be a partition of
N.
is a filter on
Then
SN
for the closure in
Let UP
is a dense subset of
is a partition of SN, and so either a UP is an ultrafilter on IN, Thus belongs to U
T
{a,T}
- S(B). p N Up, and we have a map i SIN Let U be an ultrafilter on 3N, :
(1{Q
:
Since
a E U}.
{Q
:
section property and since Take p E F. Then
empty.
F = {p},
and we have a map j Clearly
i
:
SN
j
and let
:
Ut
p,
is the inverse of
S(B) + gN. i,
and the map
is a homeomorphism.
-+ S(B)
Henceforth we identify the Stone space of
P(N)
with the Stone-Cech compactification
N
Points of filters is
if
on N :
{a E N
:
ultrafilters.
F =
has the finite intersN is compact, F is nonU c UP, and so U = UP. Thus
a E U}
SN
of
N.
are the fixed (or principal) ultra-
n E IN,
n E a}.
(S(B),TB
the corresponding ultrafilter
Points of
RN \N
are the free
The sets in a free ultrafilter are all
infinite.
The set
$N \N
is sometimes called the growth of
N. There are even those who believe that $N is defined to be the set of ultrafilters on N with the Stone topology!
We now turn to the theory of ultraproducts.
First,
we recall the definition of an ordered field. 2.17
tion on
DEFINITION Let K be a field, and let < be a binary relaK. Then (K,<) is an ordered field if:
36
is a total order on
(i)
<
(ii)
if
a < b
in
(iii)
if
a > 0
and
K,
K;
a + c < b + c
then
(c E K) ; b > 0
in
then
K,
ab > 0.
Let
An element
be an ordered field.
(K,<)
is positive if
a E K
a > 0.
Set
K+ = {a E K dal
Then jab) =
a E K+ JaIIbI
Let (i.e.,
Q
0},
a
= max{a,-a}
and
= {a E K
K
a =
ja + bI < jal
in
K,
where
We have
jal. +
ibI
be the identity of
1
: a< 0},
(a E K).
if and only if
(pl)/(ql)),
a copy of
:
p E 2Z
K.
and
(a,b E K).
The elements
q E N,
p/q
constitute
and we identify this subfield with
An ordered field
K
Q.
is real-closed if it has no
proper algebraic extension to an ordered field. EquivalentK is real-closed if, for each a E K+, there exists ly, and if each polynomial over K of b E K such that b2 = a, odd degree has a root in
K.
real-closed field is
itself, of course.
F.
The canonical example of a (The fundamen-
tal result on the existence of real-closed fields is the each ordered field K has an
Artin-Schreier theorem:
algebraic extension to a real-closed field whose order extends the order of
K
((41, Theorem 11.4]).
We shall not
use this theorem.) Let
K
be an ordered field.
(i)
infinitely large if
(ii)
finite if
jal
< n
(iii) an infinitesimal if
jal
An element > n
for some nial < 1
a E K
(n E N);
n E N; (n E N).
is:
37
We denote the set of infinitesimals in
K
.
DEFINITION
2.18
Let
f,g E N
*
by
K
N.
be an ultrafilter on
U
3N,
and let
Set
f= U g
if In E N: f (n)
f
if
=U
Then
In E N
:
f(n)
= g (n) } E U, < g(n)} E U.
is an equivalence relation on
set of equivalence classes is denoted by
)R
/U,
the
]RN;
and the
equivalence class containing an element f of ]R. is [f]U or Note that, to define [f], it suffices to specify [f]. f
on a set of
U.
The set
IRIN
pointwise operations on
a E ]R,
is a real algebra with respect to the N. For [f],[91 E IRIN/U and
set
[f]
(gl
+
= If + g], [f] [g] = [fg], a[f] = [af].
Clearly, these operations are well-defined on
]RIN/U, and We identify is a real algebra with identify [1]. with a[l], and so regard IIt as a subfield of a E ]R ]RIN/U. Take [f] # (01 in IItIN/U, a = {n E IN : f(n) # O}, Then a E U, and set g(n) = 1/f(n) (n E a). ]RIN/U
let
and
[f) (g) = (1]. Thus NN/U is a field. Set [f] < [g] (or [f]
f
Then
<
is well-defined on
easily checked that
(]RIN/U,<)
]RIN /U,
]RIN/U
if
and it is
is an ordered field.
It is also easy to check that (IRIN /U,<) is realone proves that each positive element is a square and that each polynomial of odd degree over ]RN/U has a root by working with representatives on N of the equivalence classes. ]RIN/U Thus, is a real-closed ordered field. Let F : IIt -> ]R be a function. Then F has a
closed:
38
natural extension to and
NIN/U
:
Again,
F([f]) = [F(f)].
define
(F (f)) (n) = F (f (n) )
F([f])
is well-defined.
More generally, we can consider functions subsets of
:n
al,...,ak
in
function
and define
,
defined on
F
F(al,...,ak)
for suitable
For example, the exponentiation
IIRIN/U.
defined for
F(a,S) = ga,
extends to F ([f] , [g])
=
[g]
a E N
and
If], defined for [f]
a > 0,
E IltIN/U
[g] > 0.
and
DEFINITION
2.19
Let field
U ,
(]RIN/U,
)
U
be an ultrafilter on N. is the ultraproduct of
Then the ordered P by U.
There are also more general notions of ultraproduct, which we shall not require (e.g., [44, §28]). If
But, if
U
is a fixed ultrafilter, then
Pm/U = P
.
is a free ultrafilter, then IRIN /U is significantly larger than ]R, and it contains a wealth of infinU
itely large and of infinitesimal elements. the ultraproduct
In this case,
has a natural interpretation as a
IItIN/U
"non-standard" model of analysis (see [68]):
again, we
shall not require a discussion of this topic.
Since we may consider
ItIN/U.
k'
and
k°'/U
(If f E k°°
and
and
are each subalgebras of
c0 c
gE
does not follow that
g E k',
problems.)
k7/U
IlZIN/U,
Clearly,
and a
c /U 0
0
/U
PIN
7RIN
to be subalgebras of
with
f =U g,
then it
but this fact causes no
is the set of finite elements of
is an ideal in the algebra k7/U. Ib 5 IN/U with if E
Note
c an alternative description of some
algebras which arose in Chapter 1. Let p E SN , and let be the corresponding ultrafilter on IN. We now write U MP and JP for ideals in the algebra C(BN) = k of
real-valued, continuous functions on gN . Take f E C(SN). Then f E JP if and only if fla = 0 for some a E U. Thus
39
f E J
if and only if P
(as a real algebra) algebras.
to
and so
f =U 0,
C(gN) /JP;
is isomorphic
2: /U
we identify these two
corresponds to (]r the set of infinitesimals of PN.IN/U, and we write The quotient
/U.
)R
/U)*, (Rw/U)*
It is an integral domain whose quotient The quotient c /(J n c ) corresponds to o o P
for this algebra. field is
M /J
co/U.
We reformulate the reduction theorem, Theorem 1.8,
Clearly, we can suppose that the homomorphism
of Chapter 1. 8
of Theorem 1.8 does not vanish on the real subalgebra
of
c
0
Thus we have the following two results.
c0(C).
THEOREM
2.20
Let X be a compact space. Assume that there is a discontinuous homomorphism from C(X,C) into a Banach Then there is a free ultrafilter U on IN, a algebra. radical Banach algebra R, and a non-zero homomorphism from co/U
into 2.21
R.
I
THEOREM There is a discontinuous homomorphism from
(respectively, from
c 0(C))
only if there is a free ultrafilter (i /U)*
(respectively,
i"(C)
into a Banach algebra if and U
on N
such that
is seminormable.
co/U)
a
As a final topic in this chapter, we discuss P-points in the space
Theorem 2.24 will be used in
$N \N ;
Proposition 3.7. 2.22
DEFINITION Let
a
X
be a topological space.
P-point if each Ga set in
bourhood of
is a
containing
x
x E X is is a neigh-
x.
Let
first that
X
A point
x
Gd-set in
x
belong to a compact space X. Suppose P-point, and that f E MX. Since Z(f) X, Suppose f E Jx, and so MX = Jx.
is a
40
secondly that
Mx = Jx,
V
and that
is a
Gd-set containing
say V = fly n , where V n E N x . Take fn E Mx with 0 f n (y) < 1 (y c X) and f n(X\Vn ) = _nfn :V.{1), and set Since Z(f) E Nx, f= E2 Then f E Mx and Z(f) C
x,
is a P-point.
x
if
M
x
Thus
x
is a P-point of
if and only
X
= Jx The only
P-points of
the compact space
are the points of
BN
We shall be concerned with points
p
which are
N.
P-points in
61N \N.
Clearly, each infinite compact space must contain contains nonpoints which are not P-points, and so RN \N P-points.
It is a result of W. Rudin (see [69, 4.30 and CH holds, then aN \N contains a dense
4.31])
that, if
set of
2C
P-points and a dense set of
2C
non-P-points
x
c = 2 0).
(where
2.23 DEFINITION
be a free ultrafilter on
IN,
be a countable partition of quasi-selective for {ok} if there exists
N.
Let {ok
a
fl
:
U
k E N}
c E U
and let Then U
is
such that
is finite for each k E N.
ak 2.24
THEOREM
Let PE SN \N, ultrafilter.
and let U be the corresponding Then the following conditions are equivalent: p
(b)
Mp/Jp = co/(Jp n co); W/U)" c0/U;
(c)
{Qk
:
is a P-point of
(a)
(d)
U
k E N }
of
$N \N ;
is quasi-selective for each partition N for which ak U (k E N).
Proof
Conditions (b) and (c) are reformulations of each other. (a) a (b).
Take
f E Mp.
By (a), there is a
41
clopen neighbourhood
f l U n (gN \N) = 0.
SN
in
p
such that
(x E U),
Set g (x) = f (x) g E c0
Then
(x E gN \U).
of
U
and
Thus (b) holds.
f - g E Jp.
(b) - (a).
This is similarly easy.
(a) . (d).
Let
{ak
k E N}
:
g (x) = 0
be a partition of
such that ak 9 U (k E N) . Let Vk = (61N \N )\a k (k E N) , and let v = f1Vk. Then V is a Gd-set in RN\N containing p, and so, by (a), V is a neighbourhood of p in B N \N. Take a c N such that p E a c V. Then a E U. For k E IN, a n a k = a fl ak c N, and so a fl ok is finite. Thus (d) holds. (d) - (a). Let V be a Gs-set in BN \N containing p. Then there exists (Tk) a P(N) such that N
and
Tk+1 f Tk
a1 = N \T1, Then
:
{ak
(k E N).
p
in
(k E 14)
ak+l = Tk\Tk+l
k E N}
BN \N,
is
p E a and a \N C Tk\N
Then
a\N c V.
so that
.
and ak g U N, such that a n ak
is a partition of By (d), there exists a E U
finite for each k E N. (k E N),
Set
p E f1Tk C V U N.
Thus
and (a) holds.
V
is a neighbourhood of
I
NOTES
2.25
The discussion of partially ordered sets and of Boolean algebras is standard.
For an introduction to
Boolean algebras, see [39]. Proposition 2.4 is taken from [72].
Theorem 2.10 is given explicitly in 149, II, Lemma (B,10 be the completion of P constructed in
3.3].
Let
2.10.
It can be shown that the following conditions are
equivalent:
open in
(a)
(P,T)
it
is an embedding;
for each
a E P;
(c)
(b)
P
Ua
is regular-
is separative.
The completion of a Boolean algebra is constructed in [14] and for example. It is shown in [14, 2.33] [691, that the following conditions on a Boolean algebra B are
42
equivalent:
(a)
B
is complete;
of 2.14 is a surjection;
(c)
(b)
the space
the map
i
:
(S(B),TB)
B is
extremely disconnected.
The theory of ultraproducts is given in (7),
and [44], for example, and [36], and [69].
Exercise 4A]).
[9],
P-points are discussed in (9),
Theorem 2.24 is well known (see [69, It is a theorem of Shelah ([61, VI, §4],
that there are models of set theory (see Chapter 4) [71]) in which the space $]N\ 3N contains no P-points. Thus the existence of theory
ZFC.
P-points in N\N is independent of the
43
WOODIN'S CONDITION
3
We now give the key set-theoretic consequence of the existence of an incomplete norm on an algebra
C(X,C).
It is this condition which will be applied (in Theorem 6.25)
as a step in the proof that there are models of set theory in which there are no discontinuous homomorphisms from any algebra C(X,C), and hence in which all norms on each are equivalent to the uniform norm.
C(X,C)
In the second part of this chapter, we shall discuss the question, left open in Chapter 1, whether or not the existence of a discontinuous homomorphism from c0(C) entails the existence of such a map from t (C). Recall first from Examples 2.2 that we write
for the strong Frechet order on NN , and << for the divisibility order on A\{O}, where A is a radical algebra. With a slight abuse of notation, we write (A,<<) for << F
(A\{O},<<) . 3.1
DEFINITION
Let
g E NN with 1 << F g.
Then
= {f E NN : 1 << F f << F g}, where
1
denotes the constant function
ni- 1
on
N.
is a subset of the set of functions whose graph is "eventually under" the graph of g. Certainly (,<< F) is a partially ordered set. Let U be an ultrafilter on N. Then we regard (/U,
44
3.2
THEOREM
Let X be a compact space. Assume that there is a discontinuous homomorphism from C(X,C) into a Banach algebra. Then there is an unbounded, monotone-increasing and function g in NN, a free ultrafilter U on IN,
an embedding of
into
(/U,< U )
(NN,
.
Proof
By 2.20, there is a free ultrafilter radical Banach algebra 6
: co/U -+ R.
for
<
First, we define a = (an)
g.
we may suppose that
6(a) # 0;
E c0
with
a > 0.
a E cc/U
with
Then choose a off a set in
by redefining
= a;
[a]
a
IN,
and a non-zero homomorphism
R,
We write
on
U
Now choose 0 < an < 1 (n E N). N N g to be an unbounded, monotone-increasing function in such that U,
we may suppose that
1
g (n) 2
log (a
J+
as n+
co
n
For convenience, we also choose (n E N).
g
(1)
so that
g(n)
2
We now construct an embedding it :
as a composition of maps:
(/U, <) -> (NN,
(/U,<) - (co/U,<<) B (R\nil R,<<) - (N',
6
isotonic.
Since
is given (as a map into The maps
v
and
Take
f E ,
of/g2
= (a n f (n) /g (n) 2)
f(n) % 1 E co\{O}.
(n E N),
Now take 2
v([f)) =
R),
and clearly
will be anti-isotonic.
and set
z
of/g
t
[af/g ].
it follows from (1) that E /U, and set
[f]
0
is
45
Then that
Take
is well-defined.
v
redefining
and
f1
f1(n) < f2(n) < g(n)
= an
n
Again by (1),
and so
Also,
:
By
we may suppose
Set
U2(n) - f l (n)) /g (n) 2
(S n) E co.
/U.
in
(f2]
U,
(n E IN).
(n E N) .
v([flI)((Bn)] = v([f21), Thus
c/U.
in
v([f2J) < < v([f1])
v
[f1] <
off a set in
f2
(/U,<) + (c /U,<<) 0
is an anti-isotonic map. in
We next claim that the range of eov is contained For suppose that [f] E /U. Take k E IN,
R\nil R.
and set S
n Then
= a 1 - (kf(n)/g(n)2)
eventually, and so
1 - (kf(n)/g(n)2) > 1/2
eventually.
Thus
(g
)
E c
n
n
so that
(n E N) .
n
< a 1/2 n
n
Also,
.
0
a kf (n) /g (n) 2 = a n
(n E N),
n
(v((f]))k[sJ = a,
B
and hence
((eov)([f]))k # O,
establishing the claim. Let
11.11
be the norm on
R.
For
x E R\nil R,
define T (x) (n) = min{k E N
Then
Ilxnll-1
R
k>
Ilxn II
1}
(n E N) .
E N. Take x,y E R\nil R with x << y,
T (x)
x = yz, where z E R.
Since
:
Then
>
IIYnII-1IIznII-1
(n E N) .
is a radical Banach algebra,
11 zn II -1 -j..
as
say
z A nil R, and
n . , and so
IIynII-1
+ w
1(x) (n) > t(y) (n)
and
46
Thus
eventually. T
:
T(y)
and so
(R\nil R,<<) -. (NN
is an anti-isotonic map.
Let and
T
Since
n = T°6°v.
are anti-isotonic, /U,
total order on
is isotonic and both v Since < is a
6
is isotonic.
11
is an embedding.
T
I
The following condition, Woodin's condition, gives the set theoretic consequence of the existence of discontinuous homomorphisms that we shall apply. Define
In E N),
= n
Z(n)
THEOREM Let X be a compact space.
3.3
discontinuous homomorphism from
Assume that there is a into a Banach
C(X,Q)
Then there is a free ultrafilter
algebra.
embedding from
(/V,
into
(N
Z E N.
so that
V
on N
and an
N
Proof
Let
g
be the function and
the free ultra-
U
filter specified in Theorem 3.2. Set a = g-3-({n}), n that to n E N} and set is a partition of IN,
so
:
n
then
N.
V = to c N
:
Certainly,
V
Uta
n
,
n E a} E U}.
is a filter on
N.
If
a C IN,
and so V is an ultrafilter on a E V or N \a E V, g(n) + Since is finite, and so as n - co, each a n
V
cannot be a fixed ultrafilter on
f E / V,
Take
n - f(n) + -
as
n } -.
g(n) - (fog) (n) + - as
N.
so that f (n) -
Then n -> co,
(fog) (n) -
and and
and so fog E .
(f]V F [fog] UP (/V,/U,
The map
47
Thus the result follows from
is well defined and isotonic. 3.2.
The force of the above theorem is, of course, that, if we can show that in certain circumstances there is no embedding from
(/V,
ultrafilter
on N
V
into
for any free
,
(I
(as we shall in Chapters 6 and 8),
then it will follow that each homomorphism from each algebra C(X,C)
into a Banach algebra is automatically continuous.
We now discuss the relation between the existence of discontinuous homomorphisms from
and from
?,°°(C)
c (¢). 0
In this discussion,
U
and
V
denote free ultrafilters on
N. The fact that we do know is the following, an immediate consequence of Theorems 2.20 and 2.21.
Assume that there exists is seminormable. c0/V
U
such that
Then there exists
V
(p,°°/U) * such that
is seminormable.
This suggests the following questions. (1)
Assume that there exists seminormable. (f /V)*
(2)
U
such that
Does there exist
V
c0/U such that
is
is seminormable?
Assume that
c0/U
is seminormable.
Is
(R°°/U)*
seminormable? (3)
Assume that
(t, /U)*
is seminormable.
Is
c0/U
seminormable?
Question (1) is equivalent to our main question: if there is a discontinuous homomorphism from
C(X,C)
for is there a discontinuous homomorphism from C(Y,C) for every infinite compact space Y? We do not know the answer to this question. Question (2) is a variant
some compact space
of (1):
X,
we cannot solve this question either.
it is a theorem of
ZFC
Conceivably, that, for each free ultrafilter U
48
and each a E co/U, there is a monomorphism (k"/U)* - co/U whose range contains a. If so, we would have a positive This answer to Question (2), and hence to Question (1).
seems to be an interesting problem to attack, independent of its applications to the above questions.
Although we cannot solve Question (3) either, we do have a partial result, which we now discuss. Thus, let U be a free ultrafilter on IN, corresponding to the point p E BN \ IN, say, and assume that is seminormable.
(R"/U)*
then, by Theorem 2.24, co/U
is seminormable.
P-point of BN \N, a = f + Jp.
If
bourhood U
of
U n (BN \N) . (x E BN \U).
Hence P
in
If
BN \N, and so, trivially, p
is not a
and take a E (Mp/J p) \(co/J p) , in
BN
g E c0
say
then there is a clopen neighsuch that fn = O on
(x E U),
Set g (x) = f (x) Then
P-point of
Now suppose that
a° E co/Jp, p
is a
p
(R,"/U)* = co/U,
and
(n E N), an g co/Jp MP with co/Jp c P/Jp
f - g E J
g (x) = 0 a contradiction.
and so there is a prime ideal and a g P/Jp. By 1.11(1),
the Continuum Hypothesis implies that there is a homomorphism 9 from k" into a Banach algebra such that ker 9 = P. The homomorphism 0 gives a non-zero seminorm on (R"/U)*, but its restriction to co/U is zero. Thus, if p is not a P-point of
$N \N, Question (3).
there is no trivial positive solution to
Assume that there is a discontinuous homomorphism from
R."(C).
The following result shows that, if the answer
to Question (3) is positive, then each free ultrafilter V. 3.4
ca/V
is seminormable for
PROPOSITION
Assume that there exist free ultrafilters U and such that (£"/U)* is seminormable and co/V is not V seminormable. Then there is a free ultrafilter W such that (L /W)*
is seminormable, but
co/W
is not seminormable.
49
Proof {ak : k E N} infinite subsets, let hk :
Let
Vk = {a c N : Hk(a fl a k ) N;
free ultrafilter on is
ck
say.
pk,
be a bijection, and let
E
Then each
V}.
the corresponding point in
:
{k E N: a E
gN \N
Vk} E U}.
is also a free ultrafilter on
W
is a
Vk
Define
W= Co c N Then
- N
into
be the induced bijection. For k E IN,
Hk : P(ak) - P(N)
let
N
be a partition of
N.
For f E Rte, set p (f) (k) = f (pk) . Then p L' -+ i m is a homomorphism. Take f 1, f 2 E L', and let T = {m : fl(m) = f2(m) }. If T E Vk, then f (P) = f (P )' and so p (f 1) (k) = pp(f2) (f (k) . So, if f 1 = W f 2, then T E W, {k : p(f1) (k) = p(f2) (k)} E U, and p(f1) =U p(f2). Thus p induces a homomorphism set
f(n) = g(k)
Thus
if
and hence
p,
seminormable,
p
R'/W -+ R°°/U.
(9 /W)*
f E R' and set
µ :
then
A (f 1) W A (f 2) , i'/V -+ k /W.
Take
T = {m For each
and
p(f) = g. is
(k /U)*
and
n E N,
take
k E N
such that
µ(f) (n) = min{µ(f) (n),1/k}.
k' -+ Z' is a homomorphism. Clearly, if f 1 = V f 2, and so g induces a homomorphism
Then :
g E R",
and
is seminormable.
µ(f) (n) = (foh) (n),
µ
Take
Then
f E Q' are surjections. Since
n E ak. p,
For
n E 0k,
:
k,
:
f E c0:
we see that
µ(f) E co.
Set
µ(f) (m) = µ(f) (m) }.
is finite, and so t E Vk. Thus (µ(f)]W = 14(f)1W, and so (µ(f)]W E co/W (although, in general, µ(f) 0 co). Hence µ(c0/V) c co/W.
T E W.
ak\(t f1 6k)
Also,
Suppose that and choose
p
b E C /W with
[gJW = jbj1/2,
and let
is a non-zero seminorm on c /W, 0 Take g E c p(b) # 0. with
fk = (g10 )oh-k1
(k E N).
Then
50
Choose
(fk) c co.
f E c0
such that, for each
Ifk(n)l < f(n)
eventually, and set
µ(a) = [µ(f)]w
by definition.
=
JbI
and so
[g2]
b << µ(a)
seminorm on
µ(a)2,
co/W.
and
q
Let
co/W
Hence
Then
q = p°µ.
q
is a
q(a) # 0.
is non-zero because
But this is a contradiction, because seminormable.
so that
[f]V1
Thus
w <w
in
co/V,
a =
k,
is not
co/V
is not seminormable.
I
We shall explain in Chapter 6 that, if the Continuum Hypothesis holds, then, for any two free ultrafilters (k"/V)*, co/U, and and V, the four algebras (9 /U)*, Should the
are all isomorphic (as real algebras).
c /V 0
Continuum Hypothesis fail, however, then U "IV)*
U
(R /U)"
and
U # V.
differ radically whenever
In Theorem 3.2, we only made use of the order structure (with respect to divisibility) of the radical Banach algebra of
R
R,
and we essentially ignored the structure
as an algebra.
ation by doing this.
It may be that we are losing informThis idea motivates us to make the
following definition. 3.5
DEFINITION
Let A be an algebra.
Then
able if there is a non-empty subset (i)
(ii)
S
A of
is weakly seminormA\(0} then
b E S and b << a, there is an isotonic map from if
such that:
a E S; (S,<<)
into
It follows from 3.2 and 2.4 that a radical Banach S = R\nil R).
algebra is weakly seminormable (with
The next result shows that the set-theoretic condition we are going to use cannot distinguish between the hypothesis that c0/U is seminormable and that it is weakly seminormable.
51
PROPOSITION
3.6
Assume that there is a free ultrafilter such that
is weakly seminormable.
c /U 0
free ultrafilter
V
(/ V,
ON
N
on IN,
on N
U
Then there is a
and an embedding from -
Proof S c c /U be as specified in 3.5, and take
Let
0
b E S. Set a =
lb
11/2.
Then
b << a, and so a E S.
g E N N as in 3.2, and let
Choose
v : (a3>/U,
E S.
(S,<<)
in co/U, and so
then a << v([f])
[f] E /U, v([f])
If
is contained in
v
S.
By hypothesis, there is an isotonic map from into (N N,
isotonic map from (S,<<)
into (NN,
The result now follows as in the proof of 3.3.
0
The following result shows that the analogue of Question (1), above, with "seminormable" replaced by "weakly seminormable" does have a positive answer. PROPOSITION
3.7
Assume that there is a free ultrafilter such that
is weakly seminormable.
co/U
ultrafilter
on N such that
V
(t`°/ V) *
U
on N
Then there is a free
is weakly semi-
normable.
Proof Let
U
may suppose that
correspond to the point is not a
p
p
of
We
P-point of $", for other-
wise, by 2.24,
(R-/U)* = co/U and the result is immediate. By Theorem 2.24, there is a partition (Q : k E N}
of N
k
such that
ak
it U
(k E N)
and such that
U
is not
52
quasi-selective for
Set
{ak}.
V= {a c N : U{ak : k E a} E U}. N.
As in Theorem 3.3,
V
is an ultrafilter on
ak t U
V
is a free ultrafilter.
(k E N),
We now make a remark.
Let
Since
E c
a = (a ) n
with
o
be such that B > 0 and B n n n is constant on each set ak. Then [aa ]U E c o U. For, given n E N, take k E N such that n E ak, and set and let
> 0,
a
B = (B )
E i
a
Y = (Y
Yn = min{an n, 1/k},
n
y E c° for each e > 0, in lY in the union of finitely many of the sets Then
:
:
. is contained
> e}
I
and it
ak,
contains only finitely many points of each ak a E c0 and a is a positive constant on each
because ak. Let
a
a = in
:
yn # an n}.
and so, since
U
N\a E U,
Hence
Let a = (a
n
)
jat1/2,
Then
a fl ak
is finite for each
is not quasi-selective for and
{ak}, a t U.
proving the remark.
y =U aa,
be as specified in 3.5, and take such that [a] U E S. By replacing a by S a c /U 0
E co
we may suppose that
a
n
(n E N).
> 0
If(k) I for n E ak if f (k) # 0, B(f) (n) = 1/2 for n E ak if f(k) = 0. Set
set B(f) (n)
=
P(f) = as (f) By the above remark, fl =V f2, p
:
k,
then
For
f E k
and set
(f E k°°) . (f E k ).
[p(f)]U E co/U
p(f1) =U p(f2),
and so
Clearly, if
induces a map
p
k°°/V - c/U.
Take a E (k/V)*\{0}, say a = suppose that jf(n)I < 1 Since (n E N). (1 - B(f))(n) > 0
the above remark shows that
(n
(f)
V,
We may
E N),
rat-a(f)lE JU
c /U, °
and so
53
in co/U.
[a] U << p(a)
p(a) E S.
a1,a2 E (i°/V)*\{O}
Take
a1 = [fl] V,
Thus
with
a1 << a2,
say
We may suppose that
a2 = [f2] V'
O < IfI(n) I < If2(n) I
(n E N),
and the above remark then shows that
Ia
a(f z) - a(f 1)
E co/U. 1U
It follows that p(f2) << p(f1) in co/U, and that p : ( (R'/ V) *, <<) is anti-isotonic. By (co/U, <<)
hypothesis, there is an isotonic map from (S,<<) (&N,
into I
The above result suggests a final question. be a free ultrafilter on
If
(i'/V)*
(R,'/V)*
Let
V
N.
is weakly seminormable, is
seminormable?
We are asking if an order-theoretic condition on (Z IV)* If this implies an apparently stronger algebraic condition. were the case, then, again, we would have a positive answer to our main question, Question (1). 3.8
NOTES
Our version of Woodin's condition is a reformulation of the one given in [72]. As we stated in the Introduction, Solovay first used the condition to show that there
is a model of set theory in which no map specified by the condition exists, and hence that there is a model in which all homomorphisms from each
C(X,C)
are continuous.
In the proof of Theorem 3.2, we showed that the range of
Oov
is contained in
R\nil R.
A more general re-
sult is due, independently, to Esterle and to Sinclair: if and 0(a) E nil R, then 6(a) = 0 (see [65, 11.7]
a E co/U
and [171). Proposition 3.4 is new, but 3.7 is from [72].
54
INDEPENDENCE IN SET THEORY
4
We hope in this chapter to explain what it means for a statement (such as the Continuum Hypothesis, be independent of a theory (such as
ZFC),
CH)
to
and how independ-
ence can be proved by the construction of suitable models.
Our theme is that models of set theory are mathematical objects which we can talk about, in the language of naive set theory, just as we talk about partially ordered sets or Banach algebras:
we do not formally need any metamathematical
considerations.
This chapter is not a course in mathematical logic,
and we shall omit all proofs and discussions which are at all technical.
We first describe the language So(E,=) of set theory with which we shall work. Our language is, in fact, a "first-order language, with equality", and our introductory development of formal logic will be specific to this language. There are other languages and other, more general, presentations of formal logic:
see [23), for example.
consists of certain express-
Informally,
E ions, possibly involving the symbols and "variables" x,y,z,... 'x = y', 'dz(3x(x :
'z E z'
=
and some
z))',
and
are typical elements of as follows.
Formally, we define alphabet of
The
is an infinite set of positive integers,
the first seven of which are denoted by (t
respectively.
1)I
131
1_1
and
'E',
The remaining elements of the alphabet are
55
called variable symbols, and are denoted in increasing order by xl,x2,x3,... or by x,y,z,... Intuitively, v means "or", -i means "not", 3 means "there exists", = denotes equality, and E denotes membership. However, = and E .
have no necessary interpretation as equality or membership; it is to stress this that we use the symbols n and E, rather than = and E, respectively. The language ,'(E,=) consists of certain finite sequences , where a.,.... an belong to the alphabet. language;
These finite sequences are the formulae of the the length of is n. The syntax of
the language specifies the sequences which are the formulae: it is a list of rules, to be given below, which state how formulae are formed.
In stating these rules, and throughout
the book, we shall use some conventions to abbreviate and to make more comprehensible the formulae that we use: example, we write 'x1 E x2' for <x1,E,x2>.
for
Here are the rules of the syntax. 1
ATOMIC RULE For all variable symbols and
xi
are formulae.
'xi E xj'
and
x,,
xj'
'x
These are the
atomic formulae. 2
3
symbol
CONNECTIVE RULE If and V are formulae, then and 'i(W are each formulae.
'(0)
v
QUANTIFIER RULE If is a formula and x i is a variable symbol, 0 '3xi(+)' is a formula. then In Rule 3, we do not require that the variable xi necessarily occurs in the formula 4. The language
£(E,=)
is the smallest set of finite
sequences of the alphabet which is closed under applications of any of the above three rules. a formula of
So(E,=).
For example,
is not
56
We introduce a number of standard abbreviations:
$A$ Vxi ($)
$,$ $at xi
for for for for for
x7
-i(('i($))
v
(-i(p))),
-1 (3xi H (0))) ,
(h(4)) V M),
($ - ) n ( b $) , 1(xi E
y means "for all", Intuitively, A means "and", and « means "if and only if". means "implies", Henceforth, we shall drop some parentheses, and we
shall write y for
and "variable" for "variable
symbol".
Let
$
be a formula of
r.
A subformula of
$
is
which, as a sequence, is formed by taking in order some consecutive elements of the sequence $. For a formula
$
example, consider the formula
(3x5(x2 E x7) ) v (3x7(x1 = x7 ) ) . Then
3x 5 (x 2 E x 7)
and
x1
x
7
3x5(x1 = x7 ) is not a subformula. that the subformulae of a formula
arise in the construction of
$.
(1)
are subformulae of (1), but It is fairly easy to see $
are precisely those that
Further, each formula is
uniquely defined from (either one or two) uniquely specified subformulae of smaller length by a unique rule of formation. We shall frequently use "induction on the length of formulae" to prove a result for all formulae: our remarks on the formation of formulae show that this is a valid procedure. Let x, be a variable, and let $ be a formula of Sv. The scope of a particular occurrence of 3x in is $ i the unique subformula of $ which, as a subsequence of $, begins with that occurrence of
For example, the scope 3x i. of the single occurrence of 3x5 in the formula (1), above, is 3x5(x2 E x7). An occurrence of a variable xi in a formula 0 is free if it does not belong to the scope of any
57
occurrence of 3xi in variable x i is bound. x7
Otherwise, that occurrence of the
$.
For example, the first occurrence of
in (1) is free, but the second and third occurrences are
bound.
A variable
x i
is a free variable in
at least one free occurrence of
in
x i
bound variable if every occurrence of
and
$,
xi
$
in
$
if there is is a
x i
is bound.
Intuitively, a formula expresses a property of its free variables. We write $(xl,...,xn) to indicate a formula all of whose free variables belong to the set {x1,.... xn}; of course, not all the variables need actually occur.
However,
when no variables are displayed after a formula, no commitment concerning the set of free variables is made: example, a formula $(x1)
$
can have
x2
for
as a free variable, but
cannot.
A sentence is a formula which has no free variables. Intuitively, a sentence is an assertion which is either true
A theory in 2 is a (possibly empty, possibly infinite) set of sentences of 2 Let S and T be two
or false.
theories in
2,
S + T
and let for
$
S U T,
be a sentence of r We write and
T + $
for
T + {$}.
We now explain how a theory "proves" a formula.
The language 2 contains certain formulae which we deem to be the logical axioms of Y. There is some freedom in the choice of these axioms, and we do not wish to explicitly write down our choice: the list given in (23, page 104]. "trivially true". and
'Yx1(0) -. $';
They include
let us adopt, for example, The logical axioms are all '$ » $', '$ » ($ » ($ A *))', '$ v (1$)',
perhaps more controversially,
the law of the excluded middle, is a logical axiom.
The set
of logical axioms also includes those formulae which show that = is an equivalence relation which E respects: for example, z)' 'x n x' and '(x = y) A (y E z) - (x are logical axioms.
A rule of inference is a procedure for deriving
58
another formula from a collection of formulae of the In fact, we have only one rule of inference, modus
language.
ponens, in our formulation. Modus ponens:
from
derive
{4),4 b p},
,.
Some accounts of logic contain a second rule of {O}, derive yx4). from
inference, that of generalization:
This rule is unnecessary if the logical axioms are suitably chosen, as they are in [23]. 4.1
DEFINITION Let
formula of
..
be a theory in
T
Then
T
proves
,2',
0
and let (in
T)
0
be a
if there is a
<01,...on> of formulae of y such that
finite sequence
On = 0 and such that each 01 is either an element of or a logical axiom, or is obtained from two formulae of
by modus ponens.
is a proof of
0
from
T,
In this case, <0>
T.
We write
T - 0 if
T
T
proves
proves
m
(respectively,
T V 0)
(respectively, if it is not the case that
m).
We emphasize that the notion of "proof" in Definition 4.1 is an abstract mathematical relation which does not
A formula either is or is not proved by a theory; an element of an algebra either is or is not invertible. We do not imply, when making the definition, that there is any algorithm that determines whether T i- 0 or T V 0. necessarily correspond to any intuitive notion of proof.
4.2
where
0
DEFINITION A formula
0
is logically derivable if 0 we write
is the empty set of sequences:
59
Two formulae T
I-
and and q
f-- ( w ,y), (4 » p)-
are equivalent in a theory T if are logically equivalent if ip
y,
and
be a formula. A sentence formed from Let by successively inserting at the beginning Vxi for each free variable xi which occurs in 0 is a universal closure and x2) of m for example, the sentences Vx1Vx2(x1 4'
4'
x2. It are both universal closures of xl is routine, given the logical axioms, to show that any two
Vx2Vx1(x1 = x2)
universal closures of may refer to
4,
are logically equivalent, and so we
the universal closure of ¢. is logically derivable, then its 0
If a formula
universal closure is also logically derivable. generally, if
is a theory and if
T
T }-
More then
0,
T
and we can arrange that the proof consists only of sentences: to show this is again a routine exercise ((23, §2.4]). Suppose that <01,...,$m> and are
proofs from a theory T. Then <0I' .. ' 4m'*1n> proof from T. It is now clear that, if T I- 0 and T }- s,
then
T F-
0 A
is a
: use the logical axiom
'0 - (* - (4' A j,)) ' and modus ponens twice. It is also true that T + 0 p if and only if T f- (4 » *). 4.3
DEFINITION
Let
be a theory in Y. inconsistent if there is a sentence T }-
otherwise,
4 A (-10);
sentence
Then
T
T
4,
T
such that
is consistent.
is inconsistent, then T T has little interest;
If
T
0,
and so
is
4,
for each we prefer
consistent theories. Note that T I- ¢ if and only if T + (10) T is inconsistent. Equivalently, if and 4,
only if
T + (10)
We write
is consistent.
60
Con T if the theory
is consistent.
T
A sentence
relatively consistent with a theory Con T i.e., either
T
T
is
0
if
Con(T + 0.
implies
is inconsistent or
T + 0
is consistent.
(Note that every sentence is relatively consistent with an We shall remark shortly that it
inconsistent theory.)
cannot be proved that ZFC is consistent, and so it is natural to consider whether or not a given sentence 0 is relatively consistent with
ZFC.
We now come to the notions of model and satisfaction, which will be central to much of our later work. 4.4
DEFINITION
A model N is a pair (M,E), where M is a nonempty set and E is a subset of M x M. The set M is the underlying set of N. The model N is countable if M is countable.
Throughout, we shall adopt the convention that a model is denoted by an upper-case fraktur letter and that the underlying set is denoted by the same roman letter. The subset
binary relation on
E M :
of
M x M can be regarded as a
aEb
if and only if
(a,b) E E.
Formally, our definition of a model has no connection with anything preceding the definition.
However,
it is formulated with a connection to the language
V
in
mind (and more general definitions of model would be appropriate if we were considering more general languages). "model" is sometimes termed an literature.
".5,-structure"
in the
Our
61
4.5
DEFINITION Let
m
McN
and if
and
N,
be models.
N = (N,EN)
and
9 = (M,EM)
is a submodel of
is an extension of
N
Then if
2,
EN n (M x m) = EM
For example,
(3t,< )
is a model, and
is a
(@,<)
(B,<).
submodel of
Let
be a model, and let .' _
Bl = (M,E)
the language of set theory that we are employing. formula 0 = 4(x1,...,x) of .' and elements a
be
.Lo( E,=)
Consider a of
...'a
Then there is a natural interpretation of 0 as a (relative to Al) about the elements a1,.... an. statement M.
Informally, we interpret and
xi
as
E
as
E,
=
as equality in
M,
More formally, we define by induction on
ai.
the length of formulae the truth of ¢ at 2 by rules to be set out below. We write
(all .... an)
in
m F 4'[al,.... an], and say that 3,
satisfies
9
in
(all .... an)
If
R.
then we write
if
$[al,.... an1,
0
is not true at
is true at
$
(a1,...,an)
in
/ m[al,...,an].
l
The rules should be compared with the rules of syntax in
Se
The definitions assume that
given on page 55.
all the variables, bound or free, occurring in the set
0
belong to
{xl,...,x1. 1
2
Let
xi
and
xj
be variables.
Then:
10.
(xi
xi )(al,.... a1
if
ai = a
At
(xi E xi)[all.... an1
if
aiEai.
Let
$
10.
and
*
be formulae.
($ v i)[al,...,anI
if
or if
Then:
01
F 4'[al,.... an]
Al F 4U [al, ...,a n];
62
Let
3
m V 0[a1'...'anI.
if
a f (-Io1[a1,...,a]
be a formula, and let
4)
be a variable.
xi
Then:
N
b E M,
if, for some
E F (3xi4))[a1,...,a
4)[a11.... ai-1,
b,ai+l"..,a].
It follows from the definitions that, if formula and
4)
is a
is a variable, then:
x, i
E
l
If
of
m
at
that
N
0
is a ,model of
0.
Notice that
2 F
Let of sentences
be a model.
m
T a Th(N),
and we then write
N + N
this case
is well
and say
.
N
N
¢,
is the set
and this set is
is a model of a theory
T
if
N j T.
DEFINITION Let
:
m
N
The theory of
in ' such that
4)
Th(N);
onto
4),
if and only if
denoted by
4.7
N
in
DEFINITION
4.6
n
is defined whenever the {x1,...,x}. In particular,
belong to
4)
is true, then we write
if
and so we
a
N F 4)[a11...,an]
0
m
then the truth
$,
does not depend on
is a sentence, then the truth of
defined:
b E M,
b,ai+1'...,a].
is a bound variable of
x,
free variables of 0
¢[a1,...,ai-1'
(a1,...,an)
can suppose that if
if, for all
(Vxi4))[a1,...,anI
N = (M,EM)
and
is an isomorphism if such that a
and
Let
n
n
be models.
Then
is a bijection from M
if and only if n(a)ENn(b). In are isomorphic, and we write m = N.
aEMb x
:
N = (N,EN)
N + N
be an isomorphism, let
(xl,...,xn)
63
Then
be a formula, and let al,.... an E M. if and only if lA Al l m(al,.... an] Hence
O[n(a1),...,n(an)].
Th(N) = Th(N).
The sentences in the set of the (previously unspeciare true in every model (see .SE Further, modus ponens preserves truth in a
fied) logical axioms of [23, 25A]).
model, in the sense that, if
if N - 4, and if theory, and let AI
T
be a sentence. Al
are sentences of
*
then
Let
*.
1
T
r,
be a
It follows that, if
Thus, if
0.
T
has a
is consistent, for it cannot be that
T
0 A (no).
and
0
i,
and if T F- m, then
model, then Al
0
AI F
The converse of this statement is also true:
it is Godel's completeness theorem,
which we state for our
language .. 4.8
THEOREM Let
T
be a theory of r.
Then the following are
equivalent: (a)
T
is consistent;
(b)
T
has a model;
(c)
T
has a countable model.
In symbols, 3
such that
Con T
o
if and only if there is a model
Al F T.
That a consistent theory has a countable model is a consequence of the fact that Lo, and hence T, is countable. The proof of Godel's theorem which is given in
[23, §2.51,
for example, provides a countable model for each consistent theory in .. There are some developments of forcing in which the existence of countable models is central, but we shall only use it in some peripheral remarks. It follows from 4.8 that a sentence
4>
is
logically derivable if and only if it is true in every model. More generally, if T is a theory, then if each model of T is also a model of
T 4>.
if and only This is the
F-
4>
importance of Godel's completeness theorem : it identifies
64
the notions of proof and of truth.
Our approach to the proof of the relative consistency of a sentence from a model of
ZFC
with
a model of
ZFC
will be to construct It is because we
ZFC + 0.
are willing to apply Godel's completeness theorem that we can step from this to the assertion that 0 is relatively consistent with
ZFC.
Models of quite a number of theories are studied, and we shall give a few rather elementary examples.
But the examples that we are seriously interested in here are "models of set theory itself".
More precisely, we shall eventually
be concerned with models of the theory
ZFC
this is
:
Zermelo-Fraenkel set theory with the Axiom of Choice.
(There
are other set theories that could equally well be considered, such as
GB,
Godel-Bernays set theory, but
ZFC
seems to be
the most popular.)
We first list axioms for
ZFC.
We give a formal
version of each axiom, and an informal interpretation:
it is not expected that the reader will always easily verify, and
it is not always true, that the formal version is a precise formalization of the interpretation. There are a number of variants on this list that are current;
we do not wish to
discuss the possible variations.
Our choice is made for the sake of later technical convenience. The point is that the
axioms are an attempt to capture basic intuitions about sets, and a judgement on the success of this attempt must be subjective.
AXIOM 0
Set existence.
3x (x = x) 'There is a set'.
AXIOM 1
Extensionality.
YxYy((x n y) 06 Yz((z E x)
40
(z E y)))
65
'Two sets are equal if and only if they have the same elements'.
AXIOM 2 Let
Comprehension.
4,(xl,.... xn)
be a formula in
Y.
Vx2...lxnVxn+l3xn+2Vx1((xl E xn+2) a ((x1 E xn+1) n 'If
{B E A : P(B)
A
is a set and
P
4,)
is a "property", then
is true} is a set.'
In fact, Axiom 2 is an infinite scheme of axioms,
one axiom corresponding to each formula 4i(xl,...,xn). The language V and the formally stated axiom make precise our intuitive notion of "property".
It follows from Comprehension that there is a set with no elements, and, by Extensionality, that this set is unique. It is denoted by 0. AXIOM 3
Pairing.
Vxdy3z((x E z) A (y E Z)) 'If
with A E C
A
and
A and
and
B
are sets, then there is a set
C
B E C.'
It follows from Pairing and Comprehension that, if B are sets, then so is {A,B}. We introduce the abbreviation
Vt(t E z AXIOM 4
z c y
for
t E Y) . Union.
Vx3yVz(z E x " z c y) 'If
A
is a set, then there is a set containing
the members of the members of
A.'
66
It follows from Union and Comprehension that, if A is a set, then so is U{B : B E A}; we often denote this latter set by UA. AXIOM 5
Power set.
Vx3ydz (z a x + z E Y) 'If
A
is a set, then there is a set whose
elements include all the subsets of AXIOM 6 Let
A.'
Replacement.
4(x1,x2)
be a formula in
..
Vx33x4Vx1((x1 E x3 A 3x2(4)) + 3x2(x2 E x4 A 4)) The essence of Replacement is the following. function
F
whenever
P(A,B)
is defined by a property
P
if
(A
F(A) = B
holds.)
'The image of a set under a function defined by a property is itself a set.'
Our formulation of Replacement is a slight variant of this, but each formulation can be proved from the other by using Axioms 0 - 5, and the Axiom of Regularity given below. Axiom 6 is also an infinite scheme of axioms. AXIOM 7
Regularity (or Foundation).
Vx3yVz ((z i x) v (y E X A (z E y - z i x)) ) 'Each non-empty set A contains an element B such that there are no elements belonging to both B and
A.'
It follows from Regularity and Comprehension that
67
A E A is prohibited. This fact is used in the theory of ordinals, to be given in Chapter 5. Infinity
AXIOM 8
3xdy3z((z E X) A (y E x
y E Z))
A
'There is a non-empty set ment of
form:
W
A
such that each ele-
is an element of an element of
A.'
This axiom is often stated in a slightly different W such that the empty set belongs to
'There is a set
and such that
element of
A U (Al belongs to W whenever A is an This form can be deduced from our version by
W.'
using the other axioms.
The purpose of the axiom is to
the set W establish the existence of an infinite set: contains 0, {0}, {O,{Oi}}, {O,{{},{(,(O}}},..., and hence "really" contains AXIOM 9
Vx3y3z (x (y
2Z+.
Axiom of Choice
AC
0 v (z E X A Z
p)
v
z A y f1 z
XAZEXAy
0)
(Vt3uVv(t i x v (u E t A U E Y A (v i t v v IE y v v where
'x = 0'
abbreviates
W
abbreviates
'Vx(x i y A x
'Vy(y E x)'
and
v
u))))) ,
'y n z 2 0'
W.
'For each non-empty set
A
of pairwise disjoint,
non-empty sets, there is a set B such that the intersection of B with each member of A contains exactly one element.' Assuming Axioms 1 - 8, AC is equivalent to 'For each non-empty set A, there is a function f : A + UA such that f(B) E B for each B E A with B 1 0.' Such a
68
is a choice function for A. It is manifest that the formal version of
function
f
AC
is
cumbersome and non-enlightening. 4.9
DEFINITION
The theory consisting of the set of sentences which are proved by Axioms 0 - 8 is
ZF,
and the theory
consisting of the set of sentences which are proved by Axioms 0 - 9
ZFC.
is
Thus
ZFC
is an infinite theory in the language
So.
We write "AI I Comprehension" if a model &I satisfies each sentence of the infinite axiom scheme of Comprehension, etc.
Our point of view is that models (and, in partZFC) are just algebraic objects. They
icular, models of
have the same ontological status as partially ordered sets, for example. Necessarily a model of the whole of ZFC will be complicated, but there are familiar algebraic objects which are models of some subtheories of the theory give some examples. 4.10
ZFC:
we
EXAMPLES
(i) ft
Let
It =
(]R, <) .
Then
Set existence + Extensionality + Pairing + Union + Infinity + Power set + Replacement.
For example, a,b E ]R,
there exists
At
Pairing
c E P
because, given
such that
a < c
and
b < c.
Note that
Power b c a if and only if b < a. Thus AI such that because, given a E ]R, there exists b E P c < b whenever c < a in P. The model does not satisfy the other axioms. 16 f Regularity, For example, set
lit
for, given
a E ]R,
it is not true that there exists
b < a
69
such that, for each
c E ]R, if c < b, then c a. So already we have a model that satisfies seven out ;C
of ten axioms.
(ii) Let
c E N
Regularity.
Let
a E N. is empty").
("a
b = 1
(iii) Let mEn
nm = 1.
If
then
a ) 2,
c >i a
then
for each
1 < a
n= Then
i=o Ql
("a
is
is the element such that no element
M=
a"). (2Z+,
where, for m,n E a+
E) ,
mth. digit in the binary expansion of
if the
i.e., if
Then
a = 1,
If
is an element of
b
(N, <) .
N
non-empty, and of
la=
ni2i,
n
is
1,
ni E {0,1}, then mEn if
where
satisfies each of the above axioms except
for Infinity, but it is not entirely straightforward to check this: it is easier to do this if one realizes that Ul is isomorphic to
(VW,E),
to be defined in Chapter 5.
So we obtain a model which satisfies nine out of ten axioms. The theory generated by all the axioms save for Infinity can be regarded as a reformulation of number theory;
the Axiom of Infinity takes us from number theory to set theory.
4.11
DEFINITION
Let S
If = (M,E)
be a subset of
Mk.
be a model, let Then
k E IN, and let is a simple subset if there
S
is a formula cl,...,cn
4(xl,...,xk,xk+1.... ,xk+n) in M such that
S = {(bi,.... bk) E Mk
.
A[
and elements
4[bl,...,bk,cl,...,cn]}.
Intuitively, the simple subsets are those sets which can be defined by a formula in the language .'; the
70
Indeed, the simple
formula is allowed to have parameters.
sets are often called the definable sets.
For example, let us determine the simple subsets of
in the model
JR
there is a formula
be simple.
S c ]R
Let
(JR , <) .
ti,...,tn E ]R such
and
Oxi,.... xn+i)
Then
that
S = {b E ]R
[ b, t 1
(Bt , <)
:
... I t] } .
t1 < t2 < ... < tn.
We can suppose that of order-isomorphisms
n
- N
JR
:
be the set n(t,) = t
Let
n
such that
7
7
(j
= 1,...,n).
with
Set
Then
n(b) = c.
in
b,\, c
if there exists
S
.
and so
E S,
E II
is an equivalence relation on
ti
and the equivalence classes are the intervals {tl}, (t1,t2), ... For each b E S E IT, n(b)
it
(--,t1),
if and only if
it
n(S) = S
(n E M.
Thus
JR,
S
is a union of
equivalence classes, and so each simple subset of
is a
JR
finite union of intervals.
Our definition of "simple sets" involves the notion of satisfaction.
We now give an algebraic characterization
of these sets.
Let
Al = (M,E)
Mk
basic sets of
be a model, and let
k E N.
The
are
Ak
= {(al,...,ak) E Mk
:
a1Ea
Bk
= {(al,...,ak) E Mk
:
ai = a
and
i
where
i,j E {1,...,k}. nk
:
Define
(ail .... ak+l) ' (al,...,ak),
Mk+i , Mk.
Consider the collection of families F such that F = U{Fk k E N }, and the subfamilies Fk c P(Mk ) :
satisfy
71
k E N
the following conditions for each Ak
(i)
,
Bii E Fk
(ii) A U B E Fk
(i,j = 1,...,k);
(A,B E Fk)
(iii) Mk\A E Fk (iv)
:
(A E Fk) ;
ITk(A) c Fk
(A E Fk+1
There is a family with these properties, and the intersection of all the families with these properties has the properties this intersection is the family of simple sets.
The proof of this claim is by induction on the length of the formulae involved of course, union corresponds to v, complement:
ation to
and projection to
n,
Let ment
'a }=
Ai
ZFC'
be a model.
B.
We now claim that the state-
can be given an algebraic interpretation
which has no reference to the language S. To do this, we should interpret 'ill 0' for each of the above-specified axioms of ZFC. In fact, we shall do this for four of the axioms; this should be sufficient to convince the reader that it can be done for all of the axioms. We first give a definition that will also be required in Chapter 7. 4.12
DEFINITION Let
be a model.
iB = (M,E)
For
set
a E M,
a = {b E M : bEa}. Clearly each 4.13
a
is a simple subset of
M.
EXAMPLE Let (i)
I
be a model.
I f= Extensionality.
This is equivalent to the condition that, if In this case, the map then a = b. a = b,
a,b E M and al- a
is an injection.
72
9 f Comprehension.
(ii)
a E M
This is equivalent to the condition that, if and if
bEM
is a simple subset of
S
such that b= S (iii)
M,
then there exists
a.
Infinity.
Q!
An element if, for each
fl
c E a,
of a set
b
a
is an E-maximal element
it is not true that
The above
bEc.
statement is equivalent to the condition that, for some a E M,
is non-empty and has no E-maximal element.
a
that it is not correct merely to require that for some
For example, observe that
a E M.
Infinity,
(1Vw,
is infinite for most
a
a E IN IN
N ! Regularity.
(iv)
An element if, for each
Note
be infinite
a
c E a,
of a set
b
a
is an E-minimal element
it is not true that
The above
cEb.
statement is equivalent to the condition that, for each a E M,
either
a = 0
or
has an E-minimal element.
a
Note
that this condition does not imply that each non-empty subset of
M.
has an E-minimal element.
but the subset
(IJIN,
n - co}
For example, {f E 11
:
f(n)
does not have an E-minimal element.
We have indicated that we are only interested in considering consistent theories.
Another, apparently
desirable, feature of a theory is that it be complete, in the sense of the following definition. Let
T
be a theory in the language Y.
Thm(T) = { E ' : 4.14
4
is a sentence and
We write
T -
}.
DEFINITION Let
T
be a theory in Y.
if, for each sentence
m,
either
Then
T
0 E Thm(T)
is complete or
73
A sentence
-10 E Thm(T).
and
0 E Thm(T)
Thus
proves
-i4),
independent of
T
Let
that, if
T
if
if and only if both
m
and
-i4)
0
is
are
T.
be a model.
N
neither
T
if
is consistent, then
T
and, if
relatively consistent with consistent theory.
T
is independent of
0
nor
0
is independent of
0
-i4 t Thm(T).
Then
Th(ll)
is a complete,
Godel's completeness theorem 4.8 shows
is a consistent theory, then
T
Thm(T) = (1{Th(Ul)
:
lU is a model and
la
T}.
We give an example of a complete, consistent theory.
EXAMPLE
4.15
Let
be the set of axioms for a dense total
T
order without endpoints.
Thus
T
consists of:
(0)
3x (x = x)
(1)
rxVYVz ((x E Y A Y E z) a (x E z) ) ;
(2)
Vx(x i x) ;
(3)
vxvY(x Ey vX
(4)
vxvY3z (x E y -- (x E Z A Z E Y) ) .
(5)
vx3Y(x E y);
(6)
vx3Y(Y E X)
y vy Ex);
The theory T is consistent because T has a (Q,<) = T, and so (Q,<) model: Thm(T). Suppose, if possible, that, T is not complete. Then there is a sentence 0 such that both T + 0 and T + -10 are consistent. By Theorem 4.8, there are countable
models
9
and
N
such that
U[
T + 0 and
NFT+
-i4).
74
But each countable model of
T
is just a countable, dense
total order without endpoints, and hence is order-isomorphic to
Thus
(Q,<).
and
a
tradiction because
i
are isomorphic models, a con-4,. Hence T is N
N
j ¢
and
complete.
The distinction between a consistent theory and a complete, consistent theory is related to one that we have already encountered, that between filters and ultrafilters The details of how this is done are on a Boolean algebra. not important for us, and so we shall only give a sketch of the correspondence. Let For
be the set of sentences of the language
S
4,, p E S,
if 0 and ip * are logically is an equivalence relation on S. The and .1 respect this equivalence At v,
set
Then
equivalent.
,'.
0 ti ti
operations given by
relation, and so g =
(S/'L, A,
is a Boolean algebra.
-i,
V,
[4 A (70)], [0 v
(gym) ] )
It is called the Lindenbaum algebra.
The associated order is given by:
if
[4,] < ['p] Let
T
-0+
be a consistent theory.
0 E Thm(T)
if and only if
if
then
4, ( *,
¢ A I E Thm(T). Boolean algebra
:y
E Thm(T).
Thus J.
If
0 ti
4 E Thm(T).
Thm(T)
ip
If
in
S,
then and
0 E Thm(T)
then 0, * E Thm(T), is a proper filter in the If
The condition that the theory
complete is clearly the same as the condition that an ultrafilter in S.
T
be
Thm(T) be
There are clearly two fundamental questions about the theory
ZFC:
(1)
Is
ZFC
consistent?
(2)
Is
ZFC
complete?
75
If
were complete, then mathematics as we know
ZFC
For suppose that
it would reduce to a "finite search".
To decide whether or not
is a sentence. of ZFC
ZFC, -
or of
were a theorem
¢
one would simply search through all proofs from they form a countable set - until a proof of either -i4
were found.
Fortunately, life is not this boring. ZFC
is not complete unless
we achieve completeness by adding to chosen additional axioms:
Nor can
a list of well-
ZFC
this is the content of Godel's Let T be the set of
first incompleteness theorem.
0
sentences proved by Axioms 0- 7 and 9 4.16
The theory
is inconsistent.
ZFC
(cf. Example 4.10(iii)).
THEOREM
Let T be a theory containing To that can be enumerated by an algorithm. Assume that T is consistent. T which is independent of T. Then there is a sentence of The possibility that
is consistent is dealt
ZFC
For this, it
with by Godel's second incompleteness theorem.
must be accepted that, if
T
is a theory containing
To
that can be enumerated by an algorithm, then there is a
sentence T in r which expresses 4.17
Con T.
THEOREM Let
be a theory containing
T
enumerated by an algorithm. Then T y i1
To
Assume that
T
that can be is consistent.
The clause "can be enumerated by an algorithm" is merely our means of avoiding an explicit statement of the conditions that ZFC
T
must satisfy:
satisfy this clause.
be proved in
ZFC
if
ZFC
the theories
ZF and Con ZFC can only is inconsistent. Nevertheless,
Thus, by 4.17,
we subscribe to the doctrine that
ZFC
is consistent.
It was a programme of Hilbert of 1920 to secure the foundations of mathematics, essentially by proving that
i
76
ZFC
is complete and consistent.
The theorems of Godel,
announced in 1930, show that this programme must fail.
The proof of Godel's first incompleteness theorem manufactures - by means of Godel numbering, see below - a statement which is independent of
T.
However, the state-
ments which are obtained by this method are neither exciting nor natural.
One might still hope that all "naturally arising" sentences are not independent of ZFC, in which case the failure of Hilbert's programme would not seem to be very important. from
We shall be concerned with the independence firstly, with the Continuum and, secondly, with a sentence called NDH
of two sentences:
ZFC
Hypothesis
CH,
("no discontinuous homomorphisms") that arose from the considerations given in Chapter 1. 4.18
DEFINITION The sentence
NDH
is:
For each compact space from
CH
and
C(X,(C)
each homomorphism
X,
into a Banach algebra is continuous.
We hope that the reader will find at least one of NDH to be both exciting and natural. We shall eventually prove that CH, ICH, and NDH
are all relatively consistent with
ZFC.
consistent, then
Thus, if
CH is independent of ZFC. modulo Theorem 1.10, we shall prove that NDH of
ZFC
is
Further,
is independent
ZFC.
Let
be a sentence of
So.
We discuss two
possible approaches to the problem of proving that independent of ZFC. The first approach is number-theoretic.
0
is
In essence,
the question of the independence of 0 from ZFC is a question in number theory. It was precisely this insight that Godel exploited in the proof of his incompleteness theorems.
77
Define a map
N
it
assign numbers to the alphabet of
v'
First
as follows.
So(E,-)
H 3 , 1) ' 1+ 4 ,
131 + 5, '=' r+ 6, 1E' ++7,
and
'x n ' E+ 7 + n, where of 0
is the enumeration of the variable symbols
x1,x2,...,
Using this assignation, we associate to each formula a finite sequence, say , of natural numbers. Y.
n
nl
Define
it(4>)
...pk k,
= p1
number, and write of the formula
pi r
for
4>
0.
where
i
r
ir(4>);
is the
prime
ith
i
is the Godel number
4>
For example,
x1)) = 25385372113138176198234294. (The coding map
it
Certainly
is not very efficient!).
it
is an injection.
The question which sentences of ZFC
a polynomial
PZFC
are theorems of
.zf
It is known that there is
now has a simple formulation.
in finitely many variables, say in
variables, with coefficients in
Z;,
such that, for each r
sentence
4>,
range of
PZFC
4>
E Thm(ZFC)
on
Nk.
k
if and only if
i
0
is in the
The existence of this polynomial
arises from Matijacevic's solution to Hilbert's tenth problem : it is not possible to find an algorithm for testing
an arbitrary polynomial equation
P(X1,.... Xn) = 0
with
coefficients in Z; for the possession of a solution in integers. One can take k to be nine, and PzFC could be written down explicitly if one had sufficient fortitude. r Thus, ZFC is consistent if and only if 3x1(1(x1 = x1)) is not in the range of PZFC' and it is tempting to attempt But Godel's second incompleteness theorem shows that the attempt would be futile. to prove that this is the case.
The question of the independence of sentences also has a simple formulation
:
a sentence
0
is independent of
78
i
f if and only if neither
ZFC
of
r
nor
4
7
'14
is in the range
PZFC'
In fact, this number-theoretic approach has not been successful in proving new independence results, and success does not seem to be imminent.
The second approach is algebraic, and involves the building of models.
We have already given an example of this approach when we showed in Example 4.15 that the theory of dense total orders without endpoints is complete. In this example, we used Godel's completeness theorem.
The point of this approach is that ZFC + ¢ is has a model, and one consistent if and only if ZFC + tries to build a model of
ZFC + 4,
given a model of
ZFC.
Godel was the first to be successful in this approach when he proved (in 1938) that Con ZFC implies Con(ZFC + CH). In fact, we have the following theorem. 4.19
THEOREM Assume that
submodel
Dl
of
N
ZFC
has a model
such that
K.
Then there is a
2 = ZFC + CH.
that, if ZFC has a then there is an extension Al of K such that
(We shall prove in 7.38
model 2
N,
ZFC + CH.)
In essence, Godel discovered an effective method for building new models of ZFC a given model.
by constructing submodels of
What remained undiscovered for 25 years was
an effective method for building extensions of a given model.
The technique of forcing was discovered by Cohen in 1963 forcing does give a method for building extensions, and it is the only such non-trivial method known today. Cohen's method :
allows one to extend a model by adding new elements in a controlled manner: the sentences which are true in the extended model are exactly those that are forced to be true. Forcing will be explained in Chapter 7.
79
4.20
NOTES
We have, in the formal definition of taken the alphabet to consist of a set of positive integers. This is merely to avoid some metamathematical considerations, and is not important.
For a general background in mathematical logic, see For model theory, see Bell and
[23] and [62], for example.
Slomson ([7]), and Chang and Keisler ([10]), for example. Godel's completeness theorem 4.8 is [3, 4.9],
The theorem applies to more r in the form that a theory is consistent if and only if it has a model. Lists of the axioms of ZFC are given in [13], [44] and [49]. For a discussion of the Axiom of Choice, There is an including many other formulations, see [43]. [10, 1.3.21], and [23, §2.5].
general languages than
interesting historical account of the origins of, and the controversies surrounding, this axiom in [53]. The Lindenbaum algebra is discussed in [7, Chapter 3, §4] and [44, §17].
For a discussion of Godel's incompleteness theorems, see Smorynski [66] and the standard texts. The formulation of
NDH
is due to Solovay.
An account of Matijacevic's work and of later results is given by Davis in [19]; We make one final remark.
see also [20].
As we stated on page 64,
we shall use Godel's completeness theorem to step from an assertion about the existence of models to the fact that a sentence is relatively consistent with
ZFC.
In fact,
because of the technical details of the method by which our models are constructed, it is possible in the cases which we consider to avoid the use of this theorem.
80
5
MARTIN'S AXIOM
In this chapter, we shall introduce Martin's Axiom (MA).
first, a key step in our
We have three motives:
NDH, Theorem 6.25, is given in
proof of the independence of the theory
second, the techniques used here
ZFC + MA + iCH;
in the applications of
MA
are similar to those to be used
and third, we believe that later in the study of forcing; some of the applications of MA to analysis that are given will be of general interest.
The axiom MA
was first studied (as "Axiom A") by
Martin and Solovay in [52].
The purpose of Martin and
Solovay was to settle questions in otherwise be resolved.
ZFC + iCH
that could not
In this aim, they were eminently
turned out to be an interesting and powerful axiom that does resolve many questions in ZFC + iCH, and successful:
MA
indeed it is only rather recently that some combinatorial statements that are independent of discovered.
MA + iCH
have been
At the present time, the study of some general-
izations of
MA is playing an important role in set theory. We shall prove in Chapter 8 that the theory
ZFC + MA + iCH
is relatively consistent with
Before dealing with
MA,
ZFC.
we shall first review the
elementary theory of ordinals and cardinals that we require, and we shall formally state the Continuum Hypothesis.
The
proofs that we omit of "easily checked" results can all be found in many texts, for example in [49, Chapter 11. We shall then introduce
MA
in the form of a
version of the Baire category theorem for compact spaces, and we shall show that it is equivalent to MA', the original
81
Axiom A. analysis:
Finally, we shall give some applications of MA in we hope these applications, all of which are well
known, will give the reader some feeling how the axiom is used.
However, it should be stressed that none of these
applications will be used in the proofs of our main theorems, and so they could be omitted by a single-minded reader. 5.1
DEFINITION Let
(P,<)
(i)
whenever
be a totally ordered set.
A subset
Q
and
c E P
a,b E Q
A subset
(ii)
of
Q
P
is an interval if
with of
P
c E Q
a < c< b. is an initial segment if
b E Q
whenever
(P,<)
is a well-ordering, and is well-ordered, if each non-empty subset of P con-
b E P
and
a E Q
(iii) The ordering
with
b < a.
<
tains a minimum element.
Let
with
Then
a < b,
:
(a,b] = {c E P
: a( c< b}.
(P,<)
a,b E P
and
(a,b]
are both intervals of
P.
DEFINITION Let
Then
For
a < c < b},
(a,b) = {c E P
(a,b)
5.2
be a totally ordered set.
P
set
and (Q,<) be totally ordered sets. if there is an order-isomorphism of onto an initial segment of (Q,<). (P,<)
(P,<) < (Q,<)
It is easily checked that well-ordered sets have the following pleasant properties. be well-ordered sets. (i)
either
Let
(P,<)
and
(Q,<)
Then: (P,<) < (Q,<)
or
(Q,<) . (P,<);
82
the only order-isomorphism of
(ii)
onto
(P,<)
itself is the identity map.
It follows that, for well-ordered sets, an order-isomorphism specified in Definition 5.2 is unique, and that we can compare any two well-ordered sets: either they are orderisomorphic, or one is strictly "shorter" than the other, being order-isomorphic to a proper initial segment of the other. 5.3
DEFINITION Let
element of S
Then
be a set.
S
is a subset of
S
is transitive and if
is transitive if each
S
and
S,
is an ordinal if
S
is a well-ordered set.
(S,E)
Let us be more explicit about the condition that The relation
be an ordinal. set
S.
For
to be a strict partial order on
E
require only that
a E c
S
is a binary relation on each
E
whenever
a E b
and
we
S,
b E c;
the
condition that, for each a E S, a 4 a is automatically true by the Axioms of Regularity and Comprehension. The condition that
E
condition that
E
be a total order does not follow from the be a strict partial order:
consider the set
for example,
S = {{l},{2}}.
However, it can be shown that the following are equivalent if S is a transitive set: (i)
E
is a strict partial order on
(ii)
E
is a total order on
(iii)
E
is a well-ordering on
Thus, an ordinal is a set (i)
then
if
a E S,
a E c
then
S
a c S;
S;
S;
S.
satisfying the conditions: (ii)
if
a E b
and
b E C,
(a,b,c E S).
We shall usually denote ordinals by greek letters Set a < B if (a,E) < (B,E). Then, for ordinals
either a < B or It is easily a and a. checked that a < if and only if a E B, and that a = {B:B < a} = [O,a). Thus, for example, "a finite subset of
83
is the same as "a finite subset of
[0,a)"
a",
later move freely between both designations. The first three ordinals are 0, {O}, {O,{O}};
and we shall and
it is conventional to take these to be the numbers
0,1, and 2, respectively. Let a be an ordinal and let We shall often refer to maps from
S
a
be a non-empty to S. Such a
map will be termed a sequence of length
a
(in
set.
S),
and
will be written
f = <xa the length of g < a,
g < a>;
:
is denoted by
f
t(f).
If
i(f) = a
and if
then
fig=<x : y<8>. Let
f
extends
and g
g if
be sequences with i(g) 6 R(f). Then f fIZ(g) = g. An element of Sk will often
be regarded as a sequence of length Let
unique ordinal (a,E):
(P,<)
a
k
in
S.
Then there is a is order-isomorphic to
be a well-ordered set.
such that
(P,<)
this is a consequence of the Axiom of Replacement.
It is well known that the following statements are equivalent in the theory (i)
ZF:
the Axiom of Choice;
the well-ordering principle, that each set can be well-ordered; (ii)
(iii)
Thus in the theory Zorn's Lemma hold.
Zorn's Lemma. ZFC,
the well-ordering principle and
The following reformulation of the Axiom of Choice will be used in Chapter 7.
84
5.4
each set
PROPOSITION is equivalent to the statement:
for
In
ZF, AC
S,
there is a function whose domain is an ordinal
and whose range contains
S.
Proof If
is a set that can be well-ordered, then
S
there is a bijection from an ordinal onto S. Conversely, if a - S is a surjection, then S is is an ordinal and ic a well-ordered by the relation <, where s < t in S if :
min{B < a
:
n(B) = s} < min{B < a
:
n(B) = t}.
I
The collection of all sets having a given property Each set is a class,
is the class of sets with the property.
but consideration of Russell's paradox shows that, for example, the class of all sets is not a set. Let Ord be the class of all ordinals. Then Ord is not a set, for if S
is a set of ordinals and if
T = Uts
:
s E S),
then
is an ordinal which does not belong to S. Nevertheless, we shall apply the various notions of ordering, T U {T}
defined so far for sets, to the class
Ord.
Thus
(Ord, E)
is a well-ordered class.
Let a be an ordinal. Then a U {a} is an ordinal, and it is the minimum element of the class {B E Ord : a < B} a U {a} is the successor of a, and it is written a + 1. An ordinal a is a limit ordinal if :
a # 0
and if
a
Each set (Ord, E):
it is
is not the successor of any ordinal. S
US,
of ordinals has a supremum in and it is denoted by sup S.
We now introduce a notion which may be a novelty for analysts: we use the ordinals as an indexing set in a process (invented by Zermelo) that generates the class, or universe, of all sets. The universe grows from the simplest possible seed, the empty set, by repeated application of the power-set operation. The definition is by recursion on Ord.
85
DEFINITION
5.5
The cumulative hierarchy of sets is {V
a
where:
a E Ord},
:
Va+1 = P(V.);
V0 = 0;
for a limit ordinal
VC, = Ma : a < a}
It is routine to prove by induction on each
Va
Thus
a.
a
that
is a transitive set, and so
a
:
-.
a E Ord>
Va a Va+1 (a E Ord). is an increasing sequence of sets.
Set
V = U{Va Then
V
a E Ord}.
is the universe of sets.
We need to know that every set appears somewhere in the above hierarchy, so that V is indeed the class of all sets. ZF,
In fact, in the presence of the other axioms of
this statement is equivalent to the Axiom of Regularity. Let
S
be a set.
The rank of
is the minimum ordinal
rk(S),
a
S,
denoted by
such that
S E Va+1' Intuitively, the rank of a set is a measure of its
complexity, for it specifies the number of applications of the power-set operation needed to build the set. The universe V is introduced here mainly as a preliminary to more sophisticated "Boolean-valued universes", to be defined in Chapter 7.
Let S and T be sets. Then S and equipotent if there is a bijection from S onto
T
are
we have the same cardinality. The relation ti is an equivalence relation on the class of all sets, and it is tempting to define the cardinal of a set to be the equivalence class to which it
write
S ti T,
belongs.
and say that
S
and
T;
T
However, these equivalence classes are not sets,
and this causes some problems.
The now-standard method of
86
bypassing these difficulties is due to von Neumann:
we use
ordinals and the Axiom of Choice to canonically define a representative of each equivalence class. 5.6
DEFINITION Let
Then
be an ordinal.
a
for each ordinal
with
B
g < a,
is a cardinal if,
a
and
a
are not
6
equipotent.
We often denote cardinals by the letters Let
be a set.
S
is equipotent to
S,
is equipotent to
S:
and A. K Then there is an ordinal which
and so there is a minimum ordinal which
this ordinal is the cardinal of S, Clearly, ISO is a cardinal, and ISI = ITS if and only if the sets S and T are equipotent. (a+,<) is order-isomorphic to an The ordered set initial segment of (Ord,E). The ordinals in this set are and is denoted by
ISI.
all other
the finite ordinals, and each is a cardinal; ordinals are infinite ordinals.
The Axiom of Infinity
The minimum
guarantees that there is an infinite ordinal. infinite ordinal is denoted by
w;
clearly,
cardinal, and as such it is denoted by cardinal is a limit ordinal. A set
S
is countable if
No.
is a
w
Each infinite
ISI < No,
and
S
is
uncountable if ISI > No. A famous theorem of Cantor asserts that for each set S, and so N = INI ISI < IP(S)I 0
<
I P(N)I
set.
=
(]R(
.
In particular, there is an uncountable
The minimum uncountable ordinal is
cardinal, and as such it is denoted by
w
it is a
N1.
A countable union of countable sets is countable. This fact will often be used in the form that, if
{An}
is a
countable family of sets such that UA is uncountable, then n at least one set A is uncountable. n Let K and X be non-zero cardinals, and let S and T be disjoint sets with IT, = A. Then ISO = K and we define K + A = IS U TI and KA = IS x TI. Infinite cardinal arithmetic is essentially trivial: if K and A
87
are infinite cardinals, then Let
K
strictly greater than
Then the minimum cardinal
is denoted by
K
K1=Ko We set tt 2 = K 1 Ka recursively for a E Ord a= B + 1, and K,,= sup{tts In each case,
ordinal.
K + A = KA = max {K, A}.
be a cardinal.
$
a
K+.
For example,
etc.; in general, we define xa = X
by setting :
when
B < a}
is a cardinal;
when
+
a is a limit it is denoted by
wa when it is regarded as an ordinal. The class a E Ord} is clearly the class of {$ all cardinals, and so, for each set S, there is a unique :
ordinal
such that
a
determining the Let a
The problem is that of
ISI = N e'.
a.
be an ordinal. A subset S of a is there exists y E S cofinal in a if, for each B < a, with B < y. The cofinality of a, denoted by cf a is the minimum ordinal 8 such that there is a function f
:
with
B - a
regular if
cofinal in
f(B)
an ordinal
a;
and singular if
cf a = a,
of a < a.
a
is
Clearly
each regular ordinal is a cardinal. For example, each wn is regular, but ww is not regular because cf wW = w.
We claim that, for each ordinal a, cf a and take f regular cardinal. For let B = cf a, with
cofinal in
f(8)
For each
(y < B).
and so
g(y) < a.
a.
y < B, If
Define
:
B -> a
g(y) = sup f([O,y))
f([O,y))
is not cofinal in
is cofinal in
S
is a
B,
then
g(S)
a,
is
cofinal in a, and so cf a = cf a = B. Thus B is a regular cardinal, as claimed. Let K be a cardinal. We next claim that it is a
consequence of the Axiom of Choice that K+ is a regular cardinal. For let S be a cofinal subset of K+, and suppose, if possible, that ISI < K. By AC, there is a function F(a)
:
F : S - V
K - a
Then
f(a,B) = F(B)(a).
cofinal in So
such that, for each
is a surjection. K+.
ISI = K+,
But and
f
Define
a E S,
K X S - K+ by is a surjection since S is f
1K X S1 = K.ISi = K, K+
:
a contradiction.
is regular, as claimed.
The above claim will be used in the following form:
88
if
S c K+
and if
sup S < K.
then
ISO < K,
In fact, we
shall almost always use only the easiest non-trivial form of this result: is a countable subset of w1, if S then Here is an example of this application. sup S < wl. 5.7
PROPOSITION Let
f
limit ordinal
n
be a function.
w1 -> w1
:
with
n < w1
such that
Then there is a < n
f(a)
(a < n).
Proof Set
that
and
an+1 > an
n = sup an.
and inductively choose
al = 1
(a < an ).
an+1 > f(a)
so
an < w1 Take
f
We now come to cardinal exponentiation, which is not at all trivial. Let
be a cardinal.
K
Then
of the set
Since
P(K).
is the cardinal
2K N
N
2 ° _ ]RI,
2 °
is termed the
cardinal of the continuum, and is sometimes denoted by
c.
More generally, we have the following definition. 5.8
DEFINITION Let
and
K
X
Then
be cardinals.
cardinal of the set of functions from
A
It is easily checked that, if
Y1 (K
cardinals, then
to
K,A, and µ
are
= KXu.
N
a.
is the
K.
It remains to determine the value of particular of
K'
and in
2K,
N
2 °.
Certainly
2 ° = N
for some ordinal
Cantor's theorem shows that, for each cardinal
K,
N
2K > K,
and so
each ordinal
2K > K+.
In particular,
2 a > Na+1
for
Thus we have
a.
N 1N
2 N°
< N N° < 0
° < (2 N 0) N 0 = 2 N °.
A further constraint on the value of
(1)
2 0
is given
89
in 5.9 (iii), below.
PROPOSITION
5.9
Let
(i)
be a cardinal.
If
(ii)
If
be an infinite cardinal, and let
K
then
cf K < A,
A
K < KA.
h
IAnI < 2 °
then
(n E N),
IUAnl < 2 °. K
K
(iii) cf 2 ° > M
and so
2 °
Proof
Let
(i)
range, and let
H
functions from
A
that
f A - K be a function with cofinal be a map from K into the set of into K. Define h A - K by requiring :
:
be the minimum ordinal in the set
h(a)
K \ {H(8) (a)
:
8 < f(a)}
a < A. Then h is not in the range of is not surjective. Thus K < KA. for
H,
and so K
Set
(ii)
K = sup{IAn1}.
We can suppose that each
n E N>
:
K < 2 °.
then
If
0
K < IVAnI < k0.K = K.
n E N>
is not eventually constant, K K K M 0) ° = 2 0 (2 and so, by (i), K < K ° :
(iii) Apply (i) with since
K = KA,
5.10
If
is eventually constant, then certainly
Cf K = tt°,
K
K < K0.2 0 = 2 0.
is infinite and that
An
Then
An c An+1
Then
H
K = 2 °
and
A = K
0
Cf K > A.
DEFINITION The Continuum Hypothesis H
2
(CH)
is the sentence:
° = ktl.
The Generalized Continuum Hypothesis
(GCH)
is the sentence:
90
for each ordinal
xa+1
2
a.
It was, of course, Cantor, in the nineteenth century who first considered the Continuum Hypothesis. which
There is an alternative formulation of CH does not use the theory of ordinals: "if S c P(N),
is equipotent to a subset of N equipotent to P(1)." either
S
that uses
or
S
then is
We now give a calculation in cardinal arithmetic CH. The result will be used in Chapter 8. Recall
that results that are claimed in the theory labelled
ZFC + CH
are
"CH".
PROPOSITION
5.11
x 142
0
(CH)
= 142.
Proof
For from C <
w
w2
IStI < xl °;
is tI c xl.
take
w2,
Then
to
Thus
St
= U{SE SW S by (1),
142 ° =
IsW
1
to be the set of functions :
<
t 2
KZ}. 0,
= 14 2.x1 = 142.
Now, for and, by
CH,
1
The Axiom of Choice is generally (but not Should we universally) accepted, and we shall work in ZFC. accept CH as an axiom? At a very naive level, CH is plausible if it is stated as "there are no sets S with x° < ISI < C", just because we cannot think of such a set Also, if we reject
CH,
S.
we have a problem deciding why 2 ° On the other hand,
should take any other particular value.
CH
seems improbable in the form "there is a well-ordering
of
II2
in which each element has only countably many
predecessors".
One of the tasks of set theory is to explore
the consequences of the assumption of a particular value for x 2
0,
value.
and to find other "natural" axioms which determine this It may well be that our understanding of sets will
91
develop sufficiently for a concensus to emerge on the "most K 2 0.
reasonable" value of
The axioms of
ZFC
and the role of
known (in some form) for 80 years.
CH
have been
We now come to an axiom,
Martin's Axiom, that goes beyond ZFC, which was formulated less than 20 years ago. The force of CH is that, if S is a set with K
is "small", for then
must be
ISI < 2 °,
then
countable.
Martin's Axiom attempts to preserve the feature
S
S
K
that, if
then
ISI < 2 °,
is "small" in some sense,
S
without resorting to the extreme position that each small set be countable.
For example,
MA
implies that each subset
S
K
of
with
It
< 2 0
ISI
is small, in the sense that
S
has
Lebesgue measure zero.
Our route to MA is meant to be the most scenic it passes through the green pastures of
one for analysts:
Baire category theory. X
Recall that a topological space space if, given a countable family
{Un}
is a Baire
of dense, open
subsets of X, it follows that nUn is also dense in X. One form of the Baire category theorem is the following. 5.12
THEOREM Each compact topological space is a Baire space. I Informally, the theorem asserts that the inter-
section of a "small" set of dense, open subsets of a compact space is dense. than
CH
A natural first attempt at a weaker axiom
is the following.
Let X be a compact space, and let U be a family of dense, open sets in X with K
JUI
< 2 °.
However,
Then fU is dense in (*)
already implies
CH.
(*)
X. For consider
92
the space
Let
order topology. X
a compact space with respect to the
W* = [O,wll,
X = (W*)N.
By Tychonoff's theorem,
is a compact space for the product topology.
For
a < wl,
set
Ua = If E X Then
U0
Take
f E X.
f (n) E (a,w1)
:
for some n E N} I.
exists
is dense and open in X, and {Ua a < wl}I = K1. Since f(N) [O,w1) is countable, there a < wl with f(N) c [O,a] U Clearly f 0 UQ,
and so
(1{U,
:
ft
{wl}.
:
a < wl} _ 0.
Thus, if
(*)
holds, then
x
necessarily
K1 = 2 °.
We actually obtain
MA by modifying
(*):
we
restrict the class of compact spaces that are considered. 5.13
DEFINITION
Let X be a Hausdorff topological space. Then X satisfies the countable chain condition if each family con-
sisting of pairwise disjoint, non-empty, open subsets of
X
is countable.
We also say that a topological space which satisfies the countable chain condition is
ccc.
A topological space is separable if it has a countable, dense subset.
Clearly, each separable space is
but the condition of being ccc is strictly weaker than that of being separable: we shall show in 6.2 that the compact space {0,1}K is ccc for each cardinal K, but it ccc,
is clear that 5.14
{0,1}K
is not separable if
K > 2 0.
DEFINITION
Martin's Axiom Let
x
(MA)
is the sentence:
be a compact topological space
satisfying the countable chain condition, and let
U
be a family of dense, open sets in
X
93
with in
< 2 °
JUI
is dense
Then n{U : U E U)
.
X.
implies
CH
Certainly,
We shall see in due
MA.
is consistent, then so is the theory ZFC + MA + -1CH, and so MA does not imply CH. First, however, we give a sentence which is equivalent to MA in ZFC. It is less natural, but easier to use, than the actual statement. course that, if
ZFC
DEFINITION
5.15
Let
A subset
(i)
each
with
a,b E Q
ccc
of
P
is an antichain if, for and b are incompatible. P
P
is countable;
be a subset of
V
V-generic if
is
is then
(P,<)
of
S
as 4 as 5.16
ccc
ccc.
such that
B\{O}
f =
:
A subset
G
if the partially is
a < w1>
Thus
B
is
a n b = 0
are distinct elements of
Note that, if
length win a
is
(see page 27)
(B\{O},
ccc if each subset whenever a and b countable.
B
P(P).
(D E V).
G fl D # 0
A Boolean algebra ordered set
S
is
is a sequence of
Boolean algebra such that (a < 6 < wl), then f is eventually constant. ccc
LEMMA Let
(P,<)
be a
be the completion of
IPI < 2 °,
ccc P.
partial order, and let Then
B
is
ccc.
Further,
x
K
if
a
partial order. (iii) Let
(B,Tr)
of
satisfies the countable chain condition
P
(ii)
Q
a # b,
if each antichain in
(ccc)
a
be a partially ordered set.
(P,<)
then
JBI < 2 °.
Proof Since
n(P)
is dense in
B\{O),
B
is
ccc,
and,
94
Since
b = VA.
there is a set
b E B\{O},
for each
B
is
A
ccc,
can be taken to be countable. K
K
Now suppose that
IPI < 2
Then
0.
such that
A c ,r(P)
IBI S (2
K 0)
o = 2
K 0.
DEFINITION
5.17
MA'
is the sentence:
Let
p
be a partially ordered set satisfying V be a
the countable chain condition, and let family of dense subsets of P with Then there is a V-generic filter in
K < 2 0.
IVI P.
LEMMA
5.18
Suppose that
holds when restricted to
MA'
K
partially ordered sets
P
with
IPI
< 2 0.
Then
holds.
MA'
Proof Let P and V be as in the statement MA'. We define by induction certain subsets Q P. Form Q of by n o choosing one element of D for each D E V. Now suppose that
Q
For each pair
has been constructed.
n
{a,b}
of
and compatible (not necessarily distinct) elements of Q n each D E V, choose c E D with c < a and c( b: such an element c always exists. Form Qn+1 by adjoining to Qn all these specified elements c of D. Set Q = UQn. K
Then
IQ
< 2 0
I
n
for each
and so, by 5.9(11),
n,
K
IQI < 2 0. For each D E V, D f1 Q is dense in Q. If then a and b are compatible in Q if they are compatible in P, and so Q is ccc. By hypothesis, there a,b E Q,
is a filter
G
in
Q
such that G ft D fl Q # Q
As a subset of
P,
is contained in a filter, say V-generic.
I
G H,
(D E D).
is a prefilter, and so in
P.
Clearly
H
G is
95
LEMMA Suppose that
5.19
holds when restricted to
MA'
Then
complete Boolean algebras.
holds.
MA'
Proof
Let
P
and
be as in the statement
V
5.18, we may suppose that
JPJ < 2 °.
set
For a E P,
By
MA'.
: b< a or b l a}.
= {b E P
D
a Note that, if b E D and c < b, then c E D Clearly, a a each D is dense for is dense in P, and so D n D a b a a,b E P. Thus we may suppose that Da n Db E V for .
a,b E P.
Let
Let
ccc.
be a completion of
(B,ir)
and take
D E V,
so that
P,
exists
a E P
with
w(a) < b,
and there exists
d < a.
By 2.9(11),
ar(d) < b,
and so
B\{O}.
By hypothesis, there is a filter
G n 7r (D) # 0
is
d E D with
is dense in
ir(D)
in
G
with
B
(D E V).
Let H = w -'(G). Then H A D # 0 (D E V). Take Since Da A Db E V, there exists c E H n Da A Db
a,b E H. Since
B
By 2.9(1), there
b E B\{O}.
is a filter,
G
Thus prefilter in c 1 a.
c < a, P,
and so, by 2.9 (iii) ,
w (c) d 7r (a) , and, similarly,
c < b.
and hence, by 2.9(11),
H
Thus H is a is a filter. I
THEOREM
5.20
The statements
MA
and
are equivalent.
MA'
Proof Suppose that MA
To establish
holds.
MA',
it
suffices to establish it for complete Boolean algebras. Let
B
be a complete
ccc
Boolean algebra, and K
be a family of dense subsets of
let
V
Let
S(B)
be the Stone space of
B,
B
with
so that
IVI
S(B)
< 2 °.
is a
compact space, and let it B - P(S(B)) be the associated map (as in equation (4) of page 33). For D E V, set d E D). By 2.13(111), each D is a dense, open D= U{n(d) set in S(B). Since B is ccc, S(B) is ccc, and so, by :
:
96
there exists
MA,
filter in
G n D
and
B,
D-generic filter in
B,
Now
D E D}.
G E (1{D
is an ultra-
G
Thus (D E D). 0 and so MA' holds.
is a
G
Conversely, suppose that MA' holds. Let X be a compact space satisfying ccc, and let U be a family of K
dense, open subsets of
with
X
Let
JUI < 2 0.
B = R(X),
the regular-open algebra of X (Example 2.7). Note that, since X is normal, for each non-empty, open set V in X, Thus, to show that f1U there exists a E B\{O} with a c V. is dense in
X,
it suffices to show that we have
UE U}#0 for each aEB\{O}.
n{Un
Take
For
Clearly,
P
Since
is dense and open in
exists
U
is
b E DU
ccc.
with
P = {b
and let
a0 E B\{O},
U E U,
b < a,
DU = {a E P
set
for each
X,
DU
and so
b < a0}.
a E P
:
a c U).
there
is dense in
P.
By MA', there is a filter G in P with G n DU (U E U), and so, for each U E U, there exists 0 b E G with b c u n a. Since {b b E G} has the finite 0 intersection property and X is compact, (1{b b E G} # 0, and so n{U n 0 : U E U} # 0, as required. Thus MA :
:
holds.
I
It is a surprising aspect of the nature of things that the simple generalization of the Baire category theorem given by MA produces a powerful axiom. Our applications
illustrate how MA can be exploited. Theorems which hold in the theories ZFC + MA + -NCH
are labelled
"(MA)"
and
ZFC + MA
and
"(MA + -iCH)",
respectively. 5.21
DEFINITION
A rational interval of the form
(x,y),
where
the family of open subsets of
is a subset of P
I2
and
x,y E @ ]R,
x < y.
I
of I
is
each of which is a finite
union of rational intervals.
Certainly, the set
The set
is countable.
97
In the next theorem,
THEOREM
5.22
Let such that U{F
:
F
32.
(MA)
be a family of measurable subsets of N
µ(F) = 0
and
(F E F)
Then
F E F}.
is Lebesgue measure on
µ
IF)
< 2
is measurable, and
S
Let
°.
S =
µ(S) = O.
Proof
Fix P
=
e > 0,
and let
{a c 3t
:
is open,
a
µ (a) < g }.
set a < b if b c a. partially ordered set. For a,b c PE,
Suppose, if possible, that For n E N, let
Then C
(PE, <)
is a
is an uncountable
antichain in P.
Cn = {c E C
< e/m.
µ(a) < (1 -
m E N
Then there exists For each a E Cm, µ(a - Ua)
:
such that C is uncountable. m Ua E I with Ua c a and is countable and Cm is uncount-
choose Since
n) a}.
I
able, there exist distinct elements
a,b E Cm with
Ua = Ub,
and in this case we have
µ(a u b) < µ(a) + a(b - Ua)
<(1-1)e+IDe=E, and so a u b E P e.
But
a U b < a and a U b 4b,
a
contradiction of the fact that a 1 b. Thus PE is ccc. For F E F, let DF = {b E P F c b}. For each :
a E Ps F c U
F E F, there is an open set U c R with µ(U) < e- µ(a); we have a U U E D F and Thus each DF is dense in P
and each and
a U U < a.
Since
K IFI < 2 °,
MA
implies that there is a
98
filter
in
G
each F E F,
Pe with G f1 DF # 0 (F E F), and so, for there exists b E G with F c b. Set
V = U{b : b E G}, each
an open set in
there exists
x E b,
there exists
(b
n
)
c G
c E
I
For each
P. with
and so
For n E N, take (j = 1,...,n). Then, for each n,
with V = Ub n.
cn E P E
with
we have
µ( U b ) < µ(c) n < c, and so µ(V) < e. j=1 For each F E F, there exists b E G with
and so
cn < b
n
and
b E G
x E c c b,
j
F c b,
S c V.
open set
We have shown that, for each e > 0, there is an V with V S and µ(V) < e. Hence µ(S) = 0. I Consider
Then P
(P, <), where < is the usual order. is a totally ordered set which has neither a
maximum nor a minimum element, and Yt is connected and separable with respect to the order topology, i.e., with It is easy to see that these
respect to the usual topology.
properties characterize
Ht, in the sense that each totally ordered set with these properties is order-isomorphic to P.
In 1920, Souslin raised the question whether or not "separable" could be replaced by "ccc" in this result. More generally, let (P,<) be a totally ordered set which satisfies ccc in the order topology. Is P necessarily separable? 5.23
DEFINITION
A totally ordered set
(P,<)
is a Souslin line if
satisfies the countable chain condition, but is not separable in the order topology. Souslin's Hypothesis (SH) (P,<)
is the statement:
There are no SousZin Zines.
We shall show that
SH
follows from
MA + iCH.
On the other hand, it can be shown (see (49,111, (21]) -iSH
that
follows from a principle called "diamond", which is
consistent with
ZFC + CH.
Thus
SH
is independent of
ZFC.
99
THEOREM
5.24
MA and
If
hold, then
-CH
SH
holds.
Proof that
Suppose, if possible, that MA and -NCH hold, and is false, and let (P,<) be a Souslin line. of pairLet M be a maximal family in (P(P),d
SH
wise disjoint, non-empty, open intervals of P, each of which is separable, and let E = U{I : I E M}. Since P is Since P is ccc, M is countable, and so E is separable. not separable, E is not dense in P, and so there is a non-empty, open interval J in P with J n E = 0. Now is a Souslin line, and so, replacing
(J,<)
P
by
J,
may suppose that each non-empty, open interval in
we
is non-
P
This implies that each non-empty, open set in
separable.
P
is a countable union of pairwise disjoint, non-separable, open intervals: these intervals are the components of the open set. We now construct a sequence dense, open subsets of P such that for each a,B with a < B < w
I
(i)
Us a Ua;
(ii)
if
I
a < w1> of U0 = P and such that,
is a component of
:
Ua,
then
UB.
Ua, we form Ua+1 by deleting one point from each component of Ua. Since each component of Ua is nonseparable, the deleted point was not isolated in Ua, and so
Given
is dense in
Ua+l
P.
Now let B be a limit ordinal with B < wl, and suppose that Ua has been defined for each a < B. For each a < B, there is a countable set Sa such that, if a < b and
a
and
there exists
b
are in different components of Set a < c < b.
Ua,
then
c E Sa with
S = U{Sa : a < B},
US to be the union of those intervals (x,y) of P such that x < y and (x,y) is contained in a component of Ua for each a < B. Then UB is open, Us a Ua and, if I is a (a < B), a countable set.
Take
100
component of
Ua,
we show that
U8
then
I
empty, open interval in there exists
(x,y) c J
Thus
(x,y) c U8.
US
Finally, and so I US. For let J be a nonSince J is non-separable, Ua+1'
is dense in P.
P.
and clearly with (x,y) n s = 0, is dense, and the construction
continues.
Qa
Let
Q = U{Qa
:
a < wl},
be the family of components of Ua, a < B < wl}. and let Da = U{Q6 :
set I < J if I c J.
I,J E Q,
Then
let
For
is a
(Q,<)
partially ordered set; clearly, Q is ccc, and each D. is dense in Q. By MA, there is a filter G in Q with
G n Da # 0
For each
(a < wl).
G n QS =
that
Then
{J8}.
JS E Q be such a < S < wl,
let
a,
whenever
JS 4 Ja
and so, for each a < wl, there is a non-empty open interval VS C JS\JB+1. But then {VB is an B < w :
uncountable antichain in Q, a contradiction. We conclude that SH holds. We now give an interesting combinatorial principle that has a number of applications. 5.25
THEOREM Let
F
(MA)
be a prefilter in
consisting of
P(N) K
infinite sets, and suppose that
JFJ < 2
infinite set
a\T
t E
such that
a c N
Then there is an
°.
is finite for each
F.
Proof
Let
P = {(a,b)
:
a c N,
and set (alb I) < (a2,b2) in and if
a1\a2 c b2.
Then
Suppose that
(P,<)
a
finite,
P
if a2 c al, if b1 c b2,
b E F},
is a partially ordered set. Since F is a E P.
(a,b1), (a,b2)
prefilter, there exists c E F with c c b 1 and then
(a,c) 4 (a,bI)
and
(a,c) t (a,b2).
and c c b 2, Thus, if
101
Since there are
in P, then a1 # a2.
(a1,b1) A. (a2,b2)
only countably many finite subsets of
ccc.
is
P
IN,
T E F, set D = {(a,b) E P : b c T}. For each (a,b) E P, (a,b) L (a,T), and so each D is dense in P. For k E IN, set Ek = {(a,b) E P : ja k}, and take (a,b) E P. Since b is infinite, there exists c a b with Ic = k; we have (a U c,b) E Ek and
For
I
and so each
(a U c,b) 6 (a,b), By
GADT
and G A E k
( T E F)
0
is dense in
Ek
there is a filter
MA,
G
P.
with
P
in
(k E N) .
0
Set
Q= U{a For each a
exists
(a T,bT)
EG
for some b E F}.
there exists
k E IN,
and so
(a,b)
:
is an infinite set. E G
with
(a,b)
bT c T.
with
E G
For each
jai > k,
there
T E F,
If possible, choose
(a,b) E G with m E a. m E a\(aT U T). Then there exists Since G is a filter, there exists (c,d) E G with (c,d) G (a,b) We have a c c and and (c,d) 6 (aT,bT).
c\aT c bT c T, Thus
and so
m E c c aT U T,
a contradiction.
a finite set.
o\T c aT,
This completes the proof of the theorem. 5.26
THEOREM Let
X
I
(MA)
and
Y
with
be subsets of
]R
Then
G -set in
X c Y.
K
Suppose that
JYJ < 2 °.
X
is a
Y.
Proof
Let intervals in let
F
I
R,
be the family of finite unions of rational as in 5.21,
I = {In
be the family of finite subsets of
Ts = {k E N Since
say
:
Ik n S = X n S}
TSUT C TS A TT (S,T E F), {TS prefilter on N. Further, each set TS
:
Y.
n E IN},
Set
(S E F). :
S E F)
is a
is infinite.
and
102
a c N
By 5.25, there is an infinite set that
is finite for each
a\TS :
Each
is open, and so
U.
Let j
E N.
a E X.
Since
exists
k > j
j
a
E N
is a
H
is infinite and
a
k E a
with
set
E IN,
j
j E N}.
:
Ga-set in
]R.
T{a} = {k E N: a E Ik}.
Then
Let b E Y\X. Since
For
S E F.
k > j, k E a}, and let H= fl{Uj
U. = U{I k
such
and
a E Ik,
:
a E U..
b JE Ik}.
is finite, there exists
a\T{b}
such that, for each
Take
is finite, there and so
T{b} = {k E N
Then
is infinite and
0"{a}
k > j
with
and
k E a, b t Ik,
b t Uj.
so
It follows that 5.27
COROLLARY Let
Y
H n Y = X,
< 2 °.
d
(MA)
be an infinite subset of
K
IYJ
giving the result.
IR
such that
K
Then
IP(Y)l = 2 °
Proof Let K
= 2 0,
IG I
G
be the family of
Ga-subsets of
and, by 5.26, the map H F H fl Y,
IR.
G - P (Y) ,
Then
is
a surjection. I 5.28
COROLLARY
W
(MA + '1CH)
(ii)
(CH)
281
281
281 = 28°
cannot be determined
ZFC.
We have introduced theorem for compact spaces. that
MA 5.29
MA
as a form of the category
We conclude the chapter by noting
implies a similar result for complete metric spaces. THEOREM Let
U
28°.
> 2°. 8
Thus whether or not in
=
X
(MA)
be a complete,
ccc
be a family of dense, open sets in
metric space, and let X with
K JUI
< 2 °.
Then
fl{U
:
U E U}
is dense in
X.
103
Proof V
Let
be the Stone-Cech compactification of
OX
as before, we write S
of
Since
X.
for the closure in
S
X
ccc,
is
suffices to show that
set in
V
with V fl x = V.
OX
and, for each
S = 0Vn) n X. X,
say
n,
X
y = x,
5.30
x E X,
$X.
V be an open
let
let
in
and let
X,
x =flWn.
is complete,
x EflWn,
Take
x E V.
with S
be the
Vn
Set
is a singleton in Then y x.
Suppose, if possible, that
there is a neighbourhood V of But eventually Vn a V, and so So
X,
1/n
We claim that choose Vn E Vn
Since
S = {y}.
in
For n E 1q,
family of open balls of radius Wn = U{V : V E Vn}.
and so it
GS-subset of
is a
X
For each open set
of a subset
OX
ccc,
is
$X
X;
and
y
in
X
such that
Vn - V c V,
X =flWn,
x 9 V.
a contradiction.
giving the result.
I
NOTES
The theory of ordinals and cardinals that we have Proposition 5.9 is a form of
given is quite standard.
Konig's theorem ([49,I.10.40)).
Most of the results about MA that we have given were first proved by Martin and Solovay [52], and are included in [44] and [49].
prehensive account of
MA.
The book [34] of Fremlin is a comIt includes our applications, and
many more, often in more general form, and it has a detailed account of the historical sources. Two articles about MA are (57] and [63];
a more recent account is [70].
Generalizations of
MA
to larger classes of
partially ordered sets have been studied. A certain maximal generalization is called MM ('Martin's Maximum'). See (32] and [45, Chapter 7]. It is known that, assuming the consistency of certain large cardinal axioms,
MM + ICH
is also K
consistent.
Curiously,
MM + iCH
Another variant of
implies that
2 ° =
2.
MA is PFA (the "Proper Forcing Axiom"): see [45] and [61]. In fact, MM implies PFA and PFA implies NDH: even the definitions of MM and of PFA are beyond the scope of this book.
104
GAPS IN ORDERED SETS
6
Our main story about the existence of discontinuous homomorphisms from the algebras C(X,C) into Banach algebras was left in Chapter 3. In the language of Chapter 4, the consistency of the axiom
with
NDH
by the construction of a model of of
will be established
ZFC
given a model
ZFC + NDH
We take a major step towards this goal in the
ZFC.
present chapter by proving in Theorem 6.25 that, with MA + -iCH,
NDH
follows from the fact that there is no
(NN,
decreasing, and is no
with
h
:
into
(R,<)
a < K,R < A>
is strictly increasing, fa < g6
for all
fs < h < gs
a,6,
for all
exercise to show that there are no
is strictly
but such that there
a,$.
It is a little
(so,tto)-gaps in
(NN,
it is a famous theorem of Hausdorff that there are (N1'Ki)-gaps in
(NN,
(P(N),
and the same result holds in
.
The main result of this chapter requires Theorems 6.15 and 6.16 as preliminaries, and the earlier technical result 6.12 will also be seriously used in Chapter 8, but Theorem 6.9 will not be needed later.
Before beginning the main topic of this chapter, we give a well-known result, often called the A-system lemma, that will be required several times.
105
PROPOSITION
6.1
Let
of a set
X.
be an uncountable family of finite subsets Then there is a finite subset T of X and an F
uncountable subfamily
ever A and
G
of
such that
F
B
when-
A fl B = T
are distinct elements of
G.
Proof
We may suppose that
and so we may suppose that
luFi = K1,
in this case,
JFl = K1:
X = wl.
set Fk = {a E F k E N, Jul = k}. Then F = u{Fk and hence for some k; is unk E N}, Fk countable. So, without loss of generality, we may suppose
For each
:
:
that
Jul
=
whenever
ITI
say
a,T E F,
The proof is by induction on
= k
Jul
The case
k.
(a E F).
k = 1
Suppose that the result holds for
k
is trivial.
and that
(a E F). Define f : a i+ sup a, F -> wl. We iu l = k + 1 claim that {f(a) a E F} is uncountable. For otherwise, there exists T < wl such that a c (O,T] But (a E F). there are only countably many distinct finite subsets of [O,T], and this contradicts the fact that F is uncountable. Choose F* c F so that F* is uncountable and flF* is injective, and set F** = {a\{f(a)} a E F*}. :
:
(i)
G c F*
Suppose that
and a finite subset
F**
T
Choose
is countable.
of
wl
such that
is
G
uncountable and a\{f(a) } = T\{f(T) } = T. (ii) Suppose that F** is uncountable. We can suppose that a t+ a\{f(a) }, F* - F**, is injective. For T E F**,
exists
and so by the inductive hypothesis, there and a finite subset T of wl such that
ITI = k,
G** c F**
is uncountable and A A B = T whenever distinct elements of G**. Set G = {a E F*
A
G**
In either case,
the induction continues. 6.2
G
and
T
:
and
B
are as required, and so
I
COROLLARY Let
K
be a cardinal.
Then
are
a\{f(a)} E G**}.
{0,1}k
is
ccc.
106
Proof
A basic open set finite subset
FU
Let {0,1}K.
in
F
of
K
U
is specified by a
{0,1}K
in
and a map
U :
FU -r {0,1}.
be an uncountable family of basic open sets
By 6.1, there is a finite subset such that
T
of
an uncountable subfamily G of whenever U,V E G with U # V.
We may suppose that
$UIT = OVIT,
(U,V E G).
ccc.
is
and so
U n V # 0
F
and
K
FU fl FV = T
Thus
{O11}K
I
We can now turn to the main theory of this chapter. DEFINITION
6.2
be a partially ordered set, and let A P. Then A and B are mutually cofinal (respectively, mutually coinitial) if, for each a E A, b E B, there exist c E A, d E B with a< d and b( c (respectively, d 4 a and c< b). The set A is cofinal (respectively, coinitial) in P if A and P are mutually cofinal (respectively, mutually coinitial). The cofinality (respectively, coinitiality) of P is the minimum cardinal K such that there is a cofinal (respectively, coinitial) subset of P with cardinality K. Let
and
B
(P,<)
be subsets of
For convenience, we regard the empty set as having cofinality and coinitiality 0. Note that the cofinality of a subset of a totally ordered set is necessarily either 0, 1, or an infinite cardinal.
For example, let
P
be a non-empty subset of
Then the cofinality and coinitiality of 1
or
$
IIt.
are each either
0
Let A and set
P
(P,<).
B
be subsets of a partially ordered
We write
A < B if
a < b
for each
a E A and
b E B.
An element
c
of
P
107
interpolates the pair
if
{A,B}
A < {c} < B. DEFINITION
6.4
An ordered pair of subsets of a partially ordered set (P,<) is a pregap in P if A U B is totally ordered and if A < B; the pregap is a gap if no element P interpolates {A,B}. Two pregaps and are equivalent if A and A2 are mutually cofinal 1
and
B1
and
A pregap is and B has
are mutually coinitial.
B2
a (K,X)-pregap if A has cofinality coinitiality X.
K
Suppose that A = {aa a < K} $ < A} are two sets such that
and that
:
B = {b8
i
whenever
< bs
< b6
< as
as
z
2
al < a2 < K
and
(1) 1
Then we shall abuse
01 < $2 < A.
notation by writing
: a
(2)
for the pregap . If K and A are regular cardinals, then is a (K,X)-pregap. When we write a pregap in the form (2), we implicitly assume that condition (1) is satisfied. 6.5
DEFINITION
A partially ordered set each
K,A < No,
pair
{A,B}
Thus,
there are no P
is an
P
is an
n1-set
if,
for
(K,A)-gaps.
n1-set
if and only if, for each
of countable subsets of
P
such that
A U B
totally ordered and A < B, there exists c E P with A < {c} < B. The sets A or B may be the empty set, and
is
108
so, in an
no countable chain is either cofinal or
n1-set,
coinitial.
(R,<) is not an n1-set: For example, <{0}, {1/n n E N}> is a (1,tt )-gap.
pregap
0
Let an
the
:
if it is an
n1-field
Then
be an ordered field.
(K,<)
n1-set.
Let
U
is
(K,<)
be a free
Then the ultraproduct (] /U,
6.6, below.
real-closed ordered fields
c/ U and
of
(.Q / U) *
and
]R
The subalgebras
/U.
]R
are also n 1-sets .
]RN/ U
Recall from 2.2(vi) that f << F g in N N if - as n and from 3.1, that, for
g(n) - f(n)
g E NIN with
1 << F g,
= {f E NIN
:
1 << F f << F g}.
PROPOSITION
6.6
For each g E NN with
1 << F g, (, «F)
is an
ri I- set.
Proof
Let (,<< F).
kn E N
Take
n, g
n
:
ko = 1,
such that
be an
n E w>
and, for each
kn > kn-1
h(k) = 1
h
(k = 1,...,k1-1),
(M 0,
)-gaps in
interpolates
(,<<
) -pregap in
n E IN,
choose
(i,j = 1,...,n, k > kn). and set
h(k) = max{fl(k),...,fn(k)} + n Then
0
and
fi(k) + 2n < gj(k) Set
(8011A
:
n E w>,
(k = kn,...,kn+1-1).
and so there are no
). F
A similar argument covers the other cases of the claim that, for each K,X < K0, there are no (K,X)-gaps in
109
It is similarly easy to show that there are no (40'80)-gaps,
(N"
(40,l)-gaps in
and no
(1,40)-gaps,
no
it is to
However there are obvious (1,1)-gaps:
avoid this inconvenience that we sometimes work with the strong Freshet order
<
(41,81)-gaps
Let us now consider It is easy to see in the theory (81,81) gaps NIN
in
(NN,
ZFC + CH
in
(NN,
that there are
this holds essentially because
can then be enumerated as a sequence of length
is perhaps surprising that it is a theorem of (41'81)-gaps
ZFC
wI.
It
that such
This result is closely related to a
exist.
famous theorem of Hausdorff that there are the partially ordered set in Example 2.2(11).
(P(N),
(41,81)-gaps in which was introduced
We shall prove the result assuming
MA + 1CH. In the next results,
<
as a relation between
functions denotes the standard order introduced in 2.2(iv). DEFINITION
6.7
A Hausdorff gap is a pregap in (NN is an equivalent pregap to , such that fa
:
11
2
1
whenever fa < ga
S2
F
S1
al < a2 < wl and S1 < 62 < w1, such that (a < w1), and such that, if a < a < w1, then
either fa 4 gs or f6 4
ga .
Proposition 6.8, below, shows that the pregap
of
is a gap, and so a Hausdorff gap is a special type
(41,41)-gap.
These gaps may appear to be somewhat
bizarre, but they will play their role when we establish the relative consistency of NDH in Chapter 8. 6.8
PROPOSITION Each Hausdorff gap in
(IN
,
is a gap.
110
Proof
be as specified above, and, if possible, let h interpolate the pregap . such that For each a < wit there exists na E N fa(k) < h(k) < ga(k) for k > na. Necessarily, there exists Let
m° E N
na = m°
such that
for uncountably many
m
the set N °
x E N °
is countable, there exists
uncountable set
such that
S
a, a ES.
(a E S) .
Take
and, for
k > m°,
Similarly
Since
a.
m
falm° = x
For km, fa (k)
fa(k) < h(k) < g6(k),
and an
na = m°
and
= f 6 (k) < g a(k) ,
and so
fa < g8.
But this contradicts the definition of a
fs < ga.
Hausdorff gap.
Thus no
h
interpolates the pregap.
I
It may not be clear that Hausdorff gaps exist. We now prove that, with MA + -CH, there are Hausdorff, and
(K1,K1)-gaps,
hence
in
j,k E N
Let
(NIN ,
let
(f1,g1) E NJ x Nj , and let q2 = (f2,g2) E Nk x Nk Then q 2 extends q l if f 2 l j = f l and g 2 I j = g 1.
ql
THEOREM
6.9
(MA + -CH)
There is a Hausdorff gap in
(NIN,
Proof
Let dom p,
of
P
is a finite subset of
Nk x Nk
wl
for some k E IN,
p
whose domain,
and whose range is a subset
say k = r (p) .
For
we write
a E dom p,
Let
Q
(i) (ii)
faq K gqg
be the family of functions
or
fqg
For
(iii)
p(a) _ (fa,gP). be the set of elements
if
a E dom q,
if
a,$ E dom q
gaq
in Nr (q)
g1,g2 E Q,
q
of
P
such that:
then fa < gaq in Nr (q) with
a # S,
then either
set ql < q2 if:
dom q1 = dom q2 and
r(ql) ) r(q2);
111 (iv)
if
a E dom q2,
(v)
if
a,6 E dom q2
q1(a) = (fa,ga)
then
with a < B,
q1(8) = (f8,g0),
and
extends q2(a)
q1(a)
and if
then
fa(j) < f0(j) < g0(j) < ga(j) for
j = r(q2)+l,...,r(gI).
It is easily checked that
is a partially
(Q,<)
ordered set.
We next make a remark. r(q1) = r(q2),
and let
g1IT = g2IT
that q3 E Q
with
obtain
q3
ft
and of
ql
and obtaining
appropriately, maintaining
q2
suppose, if possible, that (q E C).
C
is ccc. For is an uncountable antichain in
m E N
are distinct elements of
finite subset C
such that
T
of
(dom q1)
diction.
C1,
So
Q
wl,
with
dom ql # dom q2 C.
if
By 6.1, there is a C1
of
C1.
whenever ql and Again, there are only
qIT,
(Q,<)
is
a < wl
ccc.
and
Da,k = {q E Q
:
k E N,
I{Da,k
:
set
a E dom q, r(q) > k}.
It follows from the remark that each Clearly
q2
and so we may suppose that
But now, by the remark, if (g1,g2 E C1). then ql and q2 are compatible, a contra-
For
(Q,<1.
of
of
(dom q2) = T
fl
countably many choices of g1IT = g2IT
S
and an uncountable subset
wl
are distinct elements of
glg2 E
q
and so we can suppose that
q2
such that
For each finite subset
there are only countably many elements dom q = S,
(v)
(Q,<)
We may suppose that there exists
r(q) = m
and
Suppose
(ii).
We can now prove that
q1
with
(dom q2).
Then there exists
k E N.
and that
q1, q2 E Q
q3 < g1, q3 6 q2, and r(q3) > k. For we may to satisfy these conditions by extending all the
elements of
Q.
Take
T = (dom q1)
D
a, k
is dense in
$
a < wl, k E N}I = K1 < 2 °,
and
112
hence, by
there is a filter
MA,
G
in
Q
with
for each a < wl and each k E N.
G fl Da, k # 0 Take
Since
a < w1.
is a filter, there are
G
well-defined sequences f a,g E N IN such that, for each a (fajk,galk) = q(a) for some q E G fl Da, k E N, k.
a < g < w1.
Take
fa
(N IN,
in
It follows from
and so
that
(v)
is a pregap
It follows from (i) and (ii) that this pregap
is a Hausdorff gap.
I
In fact, the partially ordered set Q of the above proof is sufficiently nice that it can be proved in ZFC that there is a filter case.
in
with
Q
in each
G fl Da,k # 0
Thus Theorem 6.9 is actually a result of ZFC. the It can also be shown that, with MA + 1CH, (K,X)-gaps in
only
G
K,A < 2 °
with
(K1,K1)-gaps.
(1,1)-gaps and
nature of
(NIN,
are
This illustrates the special
The result can be proved by a
(x1,x1)-gaps.
variation of 6.15, below.
f,g E N. Then
Let
eventually. fl =F f2
Set
and
relation on (f,g)
is
(f1,g1)
"
NIN x NIN;
F
(f 2192 )
Clearly
gi =F g2.
[(f,g)].
f
ti
g
if
f(n) = g(n)
in NIN x NIN
if
is an equivalence
the equivalence class containing
It is important to note that each such
equivalence class is countable. 6.10
DEFINITION Let
p
of
P
be the family of non-empty, finite subsets
of NIN x NIN such that:
p,
then
f < g
in N IN
(i)
if
(f,g) E p,
(ii)
if
are distinct elements (f1,g1), (f2,g2) fl A g2 or f2 A
then either For
g1.
p1,p2 E P,
Certainly
(P,<)
set
p1 < p2
if
p1 D p2.
is a partially ordered set.
113
Now let
:
a < w
be an
>
(k14 1)-pregap in
I
(&N,
and set
for
Ea =`((fa,g(I ))
Then
a < wl.
Ea nE8=0 if a# B. DEFINITION
6.11
Let
(1J IN,
be an
a < wl>
:
(K11 1)-pregap in
Then Q()
is the suborder of
P
consisting of those elements
q
such that
< 1
such that
q c U(Ea
(a < w1),
and such that
:
a < wl (a
:
finite, nonempty set.
For
the unique element of
q n Ea.
Thus the set
q
of
each
I
a
q n Ea # 0} = supp q
of
is a
we denote by
a E supp q,
P
q(a)
contains those elements
Q()
formed by specifying a nonempty, finite subset
P
supp q
Iq n E
of
w1
and by choosing an element
a E supp q.
q(a)
The partially ordered set
play a crucial role in Chapter 8:
of
Ea
for
Q()
will
because of this we attempt
an explanation of the intuition behind the definition. Suppose we could find a pregap such that such that fa % gs (a,B < w fa < ga (a < w1), a # S) :
and such that (fa,ga)
(f a,ga)
(a < w1) .
is equivalent to
:
a < w> is a Hausdorff gap.
:
Then a < w> and so
The elements
q
of
Q()
pregap
are "finite approximations" to the required . There are three easily verified remarks about
Q(). wl,
First, for each non-empty, finite subset F of q E Q() with supp q = F.
there is an element
Second,
1Q()I = K1.
Third,
Q()
is a
separative partially ordered set (cf. page 31). A final fact about Q() is very important, but less easy. actually a gap.
It applies in the case where
is
114
THEOREM
6.12
a < wi> be an (M1,N1)-gap in Then Q() is a ccc partially ordered set.
Let
(NN,
:
Proof
Write that
Q
for
Q().
Suppose, if possible,
is an uncountable antichain in
C
Q.
For each finite subset F of wi, there are only countably many elements q of Q with supp q = F because each Ea is countable, and so we may suppose that supp qi # supp q2 if qi and q2 are distinct elements of By 6.1, there is a finite subset T of wi and an uncountable subset C1 of C such that supp qi fl supp q2 = T whenever and q2 are distinct elements of C1. ql Again, there are only countably many choices of q(T), and C.
so we may suppose that q i (a) = q 2 (a)
(q i, q 2 E
C i, a E T).
be distinct elements of C1, and qi Since qi and q2 are incompatible, let q3 = qi U q2. q3 t Q. Thus q3 must fail to satisfy condition 6.10(11), Let
q2
and
for it does satisfy all the other conditions for membership of
Q.
Hence there are distinct elements
U1,91 )
and
(f2,g2) of q3 with f1 < g2 and f2 < g1. Necessarily, one of (fi,gi) and (f2,g2) belongs to q1\q1(T), the Then
q2\q2(T).
other to by
whenever
q2\q2(T)
G
F
q q
are
and so, replacing each
q E C
supp qi n supp q2 = 0
qi and q2 are distinct elements of For each q E C1 and k E N, set F
Clearly
Q,
we may suppose that
q\q(T),
and
qi\q1(T)
incompatible elements of
Ci.
(k) = min{f (k)
:
(f,g) E q for some g E NN),
(k) = max{g(k)
:
(f,g) E q for some f E NN).
< G
(q E C ).
Further, if
q and q are 2 1 q1 u q2 fails to satisfy condition 6.10(11), and so there exist (fi,gi) E qi and
q q distinct elements of
1
C1,
then
(f 2,g2) E q2 with f1 < g2. set
Thus
F qi
q2
For k E N,
115
h(k) = max{Fq(k) Then h E N3N
and
Since
:
q E Ci}.
< h 4 Gq for each q E C1.
Fq
{supp q
is an uncountable family
q E C1}
:
it follows of pairwise disjoint, non-empty subsets of wit there exists q E CI with that, for each < wit Take
supp q& c (t,wi a E supp qt
such that
ft
Fq (n)
eventually, and so
= fa(n)
h
Similarly,
f&
Thus
Then there exists
< w
w
E
and so
h
( N "< F) ,
interpolates the gap
in
a contradiction. Thus
Q
is
ccc,
as required.
I
THEOREM
6.13
3N,
a < w i> be an
Let
in
1) -pregap < wit For each
and set Q = Q().
( ti
1,
set
< wi}. Then C E supp q}, and set Dt = {q E Q D = {DE each D is dense in Q. If there is a D-generic filter on Q, then is a Hausdorff gap. :
Proof Take exists each Q.
Then there
with
f < g
and such that
q E Q
such that
q(a)
q(&) = (f,g). Then q E D& is dense in (Q,<). Now suppose that G is a D-generic (l; < wi) . G fl D # 0
Thus
so that
Take there exists q2(l;).
q E Q\D
Define
(f,g) E E& (h,k) E q.
(a E supp q) in
and take
< wit
Thus
and
for
f )C k
= q(a)
and
q < q
DE
q1 q2 E G fl D. Since q E Q q(W)
with
G
q D q1 U q2,
is the same - say
filter on
is a filter,
and so
q1(f;)
_
(f*,g*) - for each
q E G fl D g (l; < wi) . Take E,n < wi Certainly f n, take q1 E G fl DV q2 E G fl Dn, and take
with q
q1 U q2.
Since
or f A g *. Thus n
q
satisfies 6.10(11), either
< w i>
f* 4 g*
is a Hausdorff gap in
Q,
116
(f{ ,g*) a < W1> is equivalent to Since
(N3N,
the gap
U,1),
< W1>.
This completes the proof of the theorem.
6.14
(MA + -ICH)
COROLLARY
Each
a
(x1,X1)-gap
in
is a Hausdorff gap.
(NJN,
Proof Let set
a
a
be as in 6.13.
Then
be an
a < wl>
Q = Q().
and so
IVI = Nil
there is a
filter on
V-generic follows from 6.13. 0
Let
(VI
V
x < 2 °.
By
MA,
and so the result
Q,
The result is
We now come to a key theorem.
it shows that, with
perhaps surprising:
and
(x11 1)-gap, ccc.
is
Q
By 6.12,
MA,
each
x
(Nm,
6.15
in
THEOREM Let
(3N3N,
h1
and
with h2
(MA)
14
k E N
f(j) < h(j) < ga(j)
B < A>
a < K,
K,A < 2 °.
in
there exists
:
be a
(K,X)-pregap
Then there exist two functions
such that, for each
a < K
such that, for each
j > k,
and
6 < A,
either
f(j) < h2(j) < g3(j).
or
Proof
Let
be the set of triples
P
p =
such
that: (i)
r,s E NR(P)
(ii)
a
fa(j) < g0(j)
for some
is a finite subset of
(iii) for each
in
(a,6)
E a
£(p) E N; K X A;
and each
j > t(p),
117
For set
p1 4 p2
p2 =
and
p1 =
in
P,
if:
(iv) (v)
a1 D a2
and
(ri,sI)
extends
for each
(vi)
J E {R,(P2)+l...... (P1)},
A, (pi ) 3 R (P21 (r2's2);
and each
(a,s) E a2
either
r1(j) E Ifa(j).ga(j))
or
s1(j) E If (j),9 a (j)]. It is easily checked that
(P,<)
is a partially
ordered set.
Let
(with pi = , p2 = E P m E N. We claim that there exists z(p1) = i(p2)), and let p3 E P with and i(p3) m. For let p3 < p1, p3 < p2, k > max{R(p1),m} and choose such that, for a3 = a1 U a2. each this is (a,O) E a3 and each j > k, fa(j) < g0(j):
possible because so that for
fa
extends
r*
j = 9,(p1)+l,...,k
possible because
r
a,s.
and for each
(a,$) E a1:
satisfies
p
Choose
for each g6 and so that
r* E Nk
r*(j) E Ifa(j),g0(j))
(iii).
this is
Similarly, choose
1
s* E Nk to extend s. Let p3 = (r*,s*,a3). Then p3 has the required properties, and so the claim is established. In particular, elements and in P are compatible. Since there are only countably many choices of (r,s) such that E P for some a, P is
ccc.
For
a < K, a < A,
and
k E IN,
Da,B,k = {p = E P
:
(a,$)
set E a,i(p) > k}.
It follows easily from the above claim that each Du,B,k is x K Since K,a < 2 0, {D a,B,k)I < 2 0, dense in (P,<). and hence, by MA, there is a filter G in P with G n Da,B,k # 0 in each case. For k E IN, there exists R(q) > k.
filter in
Set P,
q = (r,s,a) E G Since G
h1(k) = r(k), h2(k) = s(k).
it follows from
(v)
that
h1
and
h2
with is a are
118
well-defined elements of N. Take a < K and 8 < X, and take q E Day (3,1 fl G. For each j > Z (q) + 1, there exists with q < q and that either h1(j)
q E G (vi)
required property.
or
has the
{hl,h2}
I
THEOREM (MA) Let be
6.16
and so it follows from
E [f a(j),ga(j)]
This proves that
E (fa(j),ga(j)].
h2(j)
i(q) > j,
a (K, X) -pregap in
with
(NIN, < F)
K
and let
80 < K,X < 2 0,
Then there exists
h
U
NIN
in
be a free ultrafilter on N. such that A
Proof = ,
Suppose that h1
let 6.15.
and
and h2 be the corresponding functions specified in We claim that either fa
or that fa
(a < K,B < ,1).
For suppose, if possible, that each of these two Then there exist
possibilities fails. in
(a2,82)
such that the following two statements are both
K X X
false:
and
(a1,B1)
(i)
fa
Take
y = max{al,a2},
the statements
fa
(ii)
2
1
f
h
<
y
U
g6
<
1
U
and and
2
6 = max{61,62}. h
<
f Y
U
2
< U
g6
Then
both
fail.
Since
and
and
6 + 1 < X.
A = (j E N
:
fY+1(j) < h1(j) < g6+1(j)},
B = {j E N
:
fY+1(j) < h2(j) < g6+1(J)}.
By Theorem 6.15,
Hence
K,X > No, Y + 1 < K
A or
B,
g6+1
{j E N
is finite, and so A U B E U. belongs to U. Since fY
N \(A U B) say
A,
the set :
Set
fY(j) < hl(j) < g6(j)}
119
belongs to
fY
and so
U,
a contradiction.
Thus the claim holds, and the theorem follows. DEFINITION
6.17
Let is an
if there is an ordinal {Pv
is a
P
P = U{P
such that
a
is a chain of
v < a)
:
(P(P),c).
Clearly, each subset of an
a1-set.
is a
tt1
a1-set, Each
a1-set.
PROPOSITION
6.18
Let be an
(Q,<)
is an
a1-set
a1-sets is an
partially ordered set of cardinality
o
is
(]R,<)
a1-set.
and the countable union of
7r
S1-set
v < a},
:
v
a1-sets in
For example, it is easily checked that an
P
has a
P
countable cofinal and coinitial subset, and
where
Then
be a partially ordered set.
(P,<)
if each non-empty subset of
a1-set
be a totally ordered
(P,<)
n1-set.
be a subset of
P0
Let
a1-set, and let and let
P,
: Po + Q
map
:
it
be an isotonic map. Then there is an isotonic P + Q such that njP0 = no'
Proof
be the family of pairs
Let F P0 c U c P and = no.
Set
< (U 2,02)
01P0
Clearly,
(F,<)
has an upper bound.
2IU1 = 41
maximal element, say
is an isotonic map with
U + Q
:
(U.101 )
E F,
(Po,7r0)
subsets
Since P1
F
if
U1 C U2
and
and each chain in
By Zorn's Lemma,
is an
P
in
(F,<)
has a
(V,ir).
Suppose, if possible, that a E P\V.
such that
(U,4)
a1-set,
V # P,
and take
there exist countable
which are cofinal in the set {x E P : x < a} and coinitial in the set {x E P : a < x), respectively. Then ir(P1) and ir(P2) are countable sets
with Q.
and
P2
7r(P1) u 7r(P2)
Since
Q
is an
n(P1) < {b} < 7r(P2).
of
P
totally ordered and ir(P1) < ir(P2) n1-set,
Set
there exists
:r(a)
= b.
Then
b E Q
in
with
(VU{a},n) > (V,:r)
120 in
a contradiction of the supposed maximality of is the required map. I V = P, and
F,
Thus
(V,n).
COROLLARY Let (P,<)
6.19
be an
(Q,<)
into
it
be a totally ordered
S1-set, and let
Then there is an isotonic map from
n1-set.
P
Q.
Proof
Let
P = U{Pv
a1-sets
chain of
v < a},
:
where
:
v < a}
is a
It follows from the
(P(P),c).
in
{Pv
proposition by induction that there is a family
{nv : v < a}
of isotonic maps
(u < v < a).
Define
Then
it
P + Q
(v < a).
iIPv = irv
is the required isotonic
it
I
6.20
DEFINITION The set
<xa
by the formulae
is well defined, and
it
map.
:
such that
Pv + Q
iv1PU = iru
:
a < wl>,
is the collection of sequences
R
where each
xa E {0,1}
xa = O} is a non-empty, proper subset of no maximum element. {a
:
For distinct elements R,
x =
and where
x = <xa>
and
wl
which has
y =
in
set
5(x,y) = inf{a : xa # ya}, and set
x < y Thus
x(5(x,y)) < y(5(x,y)).
if
R
is a subset of the totally ordered set of
all binary sequences of length lexicographic order on
xa = ya = 0 yw = 0. 6.21
Then
(a < w) , <xa
:
R.
and <
w1,
is the
For example, let
xa = ya = 1
a < w1> E R,
(a > w) , but
:
xw = 1, a < wl> ¢ R.
DEFINITION Take
a < w
The set
qQ
consists of those
121
sequences
xa = 1
<xa : a < w1> such that there exists
and
xa = 0
We write
for
for
q
d < a
with
a E (&,wl). Q
wl
Clearly
.
Q =U{Q0 : a < w1}, and
Q c R. The properties of the ordered sets
(Q,<)
and
that are given in the next proposition will be used in Chapter 8. First, we give some further notation. Set (R,<)
I = R\Q,
and set J = (x E R
Then
:
For
J c I c 1.
is not eventually constant).
x
x E I,
set
AX = {Y E Q : y < x},
Then
is a pregap in (Q,<). be a totally ordered set. (P,<)
Let of
P
x < Y}.
BX = {Y E
is order-dense in
if, for each
(P,<)
(3)
A subset Q with
a,b E P
a < b, there exists c E Q with a < c < b. 6.22
in
in
PROPOSITION (i)
IRI =
(ii)
(R,<)
and
IJI = 2141,
IQI = 2140.
is complete, and each
(K1,K1)-pregap
is interpolated.
(R,<)
(iii)
(Q,<)
(iv)
If
and, if
(Q,<),
(Ko,K1)-gap
or an
is order-dense in
then
x E J,
x E I\J,
then
(81,8 )-gap 0
is an
(K1,K1)-gap
is an
in
Proof (i)
Certainly
IRI =
IJI
= 2
1.
For each
122
and so
°,
IQoI = 2
Let
(ii)
xa = max{ya
zE = xE
xQ = 0
z = .
y E S, y1a = xla}. x
If
(& E (o,w1)):
Then
(81,$1)-pregap in
be an
This is clear.
(iv)
Suppose that
S = {E
Set
x = <xa
is a gap in
>
:
xE = 1}.
For
xat = 0
and set
(a E (E,w1)), E S}
T = {E
:
{yE
yaE = 0 :
If
K = X = N1. 6.23
then
x E 1\J,
S
set
xE =
and
is non-empty,and xat = xa
(a <
Then
:
T
set
and set Bx,
is non-empty and yat = xa
(a <
yE =
:
),
a < wl>.
a& A = ITS.
sup S = sup T = w., then
(K,X)-gap.
a < wl>. <xaE K = ISI.
Then E T,
a < w1> E 1.
:
say it is a
Then
is coinitial in
x E J,
and so
{K,A} = {24°,t{1},
p
PROPOSITION For each
(Q,<)
For
(a E (E,wll),
E T}
If
Ax,
xE = 0}.
has no maximum element.
Q,
E E S,
is cofinal in
Set
and set
R,
.
interpolates
h
has no maximum element.
yEE =
and set
Similarly, each non-
z = sup S.
(iii)
Certainly
Then
( E (0,w1)),
which is bounded below has an infimum, and
R
Let
1,
set
is complete.
(R,<)
:
a < wl
then there exists
R,
zQ = 1, zE = 0
In either case
h = sup A.
{xE
which
R
inductively by
x = <xa>
xE = 1
and
< a),
(
empty set in so
z = x.
set
x E R,
such that
:
K
K1.2 ° = 2 °.
2 ° < IQI
be a non-empty subset of
S
Define
is bounded above.
If
K
K
H
a < wl,
is a
a < w
(Qa,<)
is an
a1-set,
and
$1-set.
Proof
A be a non-empty subset of QQ. Clearly A is bounded above in R. Let x = <x > be the supremum of A a in If x E A, R. then {x} is cofinal in A. If x 4 A, Let
123
set
= sup{a
T
is a limit ordinal and = 1}. Then T X4 C < T, there exists y9 E A with
:
For each
T < a.
and {y
xl&,
set in
is a countable, cofinal sub-
< T}
A.
A has a countable, coinitial subset,
Similarly,
and so
is an
(Q°,<)
Since
a1-set.
Q = U{Q
Z(n) = n
Recall that 6.24
THEOREM Let
U
a < wl},
:
Q
is a
$1-set. I
so that
(n E N),
1 <<
F
Z.
(MA + 1CH)
be a free ultrafilter on
is an embedding of
into
(R,<)
N.
Then there
(/U,
Proof By 6.6,
is a
(Q,<) jr
:
is an
(,<< F)
nl-set, and, by 6.23,
Thus, by 6.19, there is an isotonic map
S1-set.
(Q,<) - (,<< F ).
The map
(Q,<) + (,
it
is
isotonic.
be the gap in specified in (3). By 6.22(iv), is a (Q,<) with {K,X} C {K°,xl} and so (Q,<) (K,X)-gap in For each
x E 1,
let
is a (K,X)-pregap in
(,
Since
K
K1 < 2 °,
with Set
7r(A 1T(x)
because embedding
it follows from 6.16 that there exists h in N N . -. (,
R,
and
it
induces the required
(R,<) - (/U,
The main result of this chapter now follows immediately from Theorem 3.3. 6.25
THEOREM Let
X
(MA + 1CH)
be a compact space.
discontinuous homomorphism from C(X,C) Then there is an embedding of algebra.
(N N '< F )
Assume that there is a into a Banach (R,<)
into
124
To build a model of
ZFC + MA + 1CH
embedding of
into
(R,<)
it thus suffices
ZFC + NDH,
to build a model of
in which there is no
(IN ",
This we shall do in
Chapter 8.
We conclude this chapter with some remarks (not required later) on the structure of some of the ordered sets that we have mentioned.
A partially ordered set
is
(P,<)
w-complete
if
each countable set which is bounded above (respectively, below) has a supremum (respectively, an infimum).
so that
Q (w) = R\J,
Q(w) = {x E R Then
(Q(w),<)
:
is eventually constant}.
x
is easily checked to be the smallest
set containing
w-complete IQ (w)
= 2M
Q
as an order-dense subset, and
R
is the natural analogue of
0.
It is clear that JR
"at the next cardinal", but it is arguable whether
fl (0,1)
is
the natural analogue of o fl (0,1) Certainly the sets isomorphic because
that
Set
IQ J =
1@1
IQ (w) l =
<
191.
I]R
@ .
and
or
Q JR
Q(10).
are not order-
But, by 5.28(1), it could be
In what way, then, do these three
sets differ? 6.26
PROPOSITION is both a
(i)
(Q,<)
(ii)
(Q(w),<)
(iii)
(R,<)
and an
B1-set
is a Bi set,
is neither a
n1-set.
but not an ni set.
B1-set
nor an
n1-set.
Proof By 6.23,
(i)
Let
A< B.
A
Q
is a
B1-set.
with be countable sets in Then there exists a < w with A U B c Q. 1 a If A = 0, take x = 0 (a # Q), xa = 1, and and
B
a
125
set x = <x
a < w >.
:
(a
xa=1
x = <xa : a < wl>.
A
both
and
B
that
x < y.
with
x
(a>a),
Then A < {x}.
Thus we can suppose that
y = inf B
and
Set
x
= x
a
x = <xa : a < w1>.
A < {x},
a contradiction.
Then
(a > a), and set z =
a<w
:
Q(w) = U{Q(w)
(iii)
prove that
Let
S
R
Again,
w-complete, and so it is
is not a
and so
QQw),
a < w1}
:
x E {0,111 and a < T,
U
= {y E R : yla = x1a}.
x,a
be an al set a,
x
we
set
be a binary
x
where
a < w1.
We claim that there
and a binary sequence such that y(a) = 0 and U
= xa
nl-set;
and let
C>a, set y
is an
R,
in
T E (a,w1)
extending
qQw)
is a 5 -set.
S1-set.
For
sequence of length exists
(a > a) }.
is clearly not an
R
n1-set.
set
is order-dense in
Hence
Then
1>.
is an
Q
Q(W) _ {<xa> E R : xa = xc Qa+1
Similarly,
(T > a).
Thus is
q(W)
a < w,
For
and
(a
(T )a). Set za=xa (ii)
T > a
(a > T),
x < x,
xT = 0
So
za=0
a1 set.
so
(R,<),
= 0
a
x E R,
yT=1
not an ni set.
x
(a < T),
a
and set
Then
in
Suppose, if possible, that there exists
= 1.
x < z < y,
and set
are non-empty.
x = sup A
Let
If B = 0, take
{x}
Then
(a < a) ,
y
Y.T
of length fl s = 0.
T
For
Y4 = 1,
y0 (a
yz = .
Since
S
is an
a1-set,
there
126
exists
such that s = 0, for otherwise S would contain a strictly decreasing sequence of length w1. (a < T). Since Take T = C + 2, and set ya = ya+l t;
fl
UyrT c
the claim holds.
To obtain a contradiction, suppose that a
Since
B1-set.
is an
(Q,<)
6.19, that there is an isotonic map
-1(0
R=
Then each
(C < wl) .
t;
Take
= x(O) = 1.
a(O)
E (a(C),w
and
)
1
extending x such that U
R
fl
take
set
for a < a(t;),
x
a (C) as take and xC,
Given a(c) of length G(c + 1) xC+l(G(C)) = 0 and
this is possible by the claim.
= 0:
xC+l,a(c+1) a limit ordinal
an ordinal
<. wl
G(C) < w i and a binary sequence x of length G(C + 1)
a-set.
is an
R
We construct recursively for follows.
Set
R-> Q.
is
a
is
(R,<)
it follows from
n1-set,
a(C)
= sup{a(l;)
<
:
For
and,
t;}
a < a(t;).
= xC whenever a
w
Let
x E R
x = x 1.
Then
but, for each
x E R,
t;
< wl,
This is the required contradiction.
COROLLARY
6.27
No two of the sets are order-isomorphic. U
and
( R, <)
K
For each ultrafilter and so, with
R°/U
CH,
is a
U
on
lR'/UI
IN,
B1-set.
= 2 0,
However, the
following corollary shows that it cannot be proved in that there exists an ultrafilter
ZFC + -iCH
U
such that
9'/U is a B 1-set. COROLLARY
6.28
(MA + -iCH)
For each free ultrafilter
is not a
U
on
IN,
W/U,
B1-set. Proof
By 6.24, there is an embedding
The map
(a
(/U,< U)
n
)
into
(R,<) + (/U,
induces an anti-isotonic map from n (R'/ U, < U) , and so there is an embedding of
F+ (1/a
)
127
the set
(R,<*),
so
with the reversed order, into
R
R
is not a
S1-set,
k /U
is not a
S1-set.
Let
U
and
V
is not a
(R,<*)
real algebras.
co U,
N.
be free ultrafilters on
stated in Chapter 3 (page 50) that, with
(9../U)*, (9. /V) *,
and
61-set,
I
We
the algebras
CH,
and c/ V are all isomorphic as
We can now give a justification of this
In fact, each of these algebras is a totally
remark.
K
ordered, real-closed,
n1-field of cardinality
Now any
2 °.
pair of totally ordered, real-closed, n1-fields of cardinality K1 are isomorphic as real algebras; that they are and the
isomorphic as fields is proved in [36, 13.13],
extension to the stated result is due to Johnson [47]. with
CH,
Thus,
the result follows.
In view of the Questions raised in Chapter 3, it is of interest to consider when these algebras are isomorphic if CH
is not assumed.
In fact, it follows from a result of K
Dow ([22, Corollary 2.3]) that, if
K1 < 2 °,
exist free ultrafilters
on
(R"/U)*
and
U
and
V
1
then there
such that
are not isomorphic (as rings).
(t /V)*
We mention one further open question that We stated in Theorem 1.10 that, with CH, into a there is a discontinuous homomorphism from C(X,C) interests us.
Banach algebra for each infinite compact space
X,
and we
explained that we shall eventually prove that NDH is relatively consistent with ZFC. But this leaves open the possibility that there are models of
ZFC + -iCH
there are discontinuous homomorphisms from QUESTION.
Is
'INDH
in which
C(X,C).
relatively consistent with
ZFC + -ICH? An approach to this question is given in [17]. (K,<)
be a real ordered field.
Then
(K,<)
is an
Let
al-field
128
if it is an
a1-set,
an ordinal {Kv
and
K = V{K
such that
a
is a
(K,<)
:
where
v < a},
v
is a chain of real
(p(K),c).
al-fields in Certainly each field of cardinality K1 is a :
v < Q}
if there is
S1-field
S1-field.
The version of Theorem 1.10 actually proved by
Dales and Esterle in [17, Chapter 6] implies the following result. Let K be a real algebra such that (K,<) is a Then the real algebra K* (the set of infinitesS1-field.
We stress that this is a theorem
imals in
K)
of
It implies that, if
ZFC.
is normable.
K
transcendence basis over R is normable:
of cardinality
the only role of
1.9 is to ensure that
is a real field which has a
C(X)/P
CH
$1,
then
K*
in the proof of Theorem
satisfies this condition.
It follows from the theorem of Dales and Esterle that to answer the above question positively, it suffices to prove that it is relatively consistent with there is a free ultrafilter
on
U
N
ZFC + -CH
such that
that
(RTh/U,
By 6.28, to achieve this we must work with models in which MA is false. It is proved in [72] that it is a
S1-field.
is relatively consistent with free ultrafilter
U
on N
that there is a
ZFC + -CH such that
(R '/U,
is a
and it may be that already this example is a
81-set,
Sl-field - but this we cannot see.
Let
X
be a compact space, and let
maximal, prime ideal in
C(X,C).
P
be a non-
Even in the theory
there is an open question about the normability of
ZFC + CH,
C(X,C)/P. QUESTION.
If
C(X,(C)/P
is normable, is
K
IC(X,Q)/Pj = 2 0?
The question generalizes to ordered fields.
be an ordered field over R
such that
K*
Let
is normable.
K Is
K
For a,b E K*, set a ti b if there exists with jai < nIbI and IbI < n(al. Then N is an equivalence relation on K*: the equivalence classes are the JK*j = 2 0?
n E N
Archimedean classes.
Now suppose that
K*
is normable.
129
Since
K*
is a radical algebra, the map has the property that
K* - co,
8
:
8(a) = 8(b)
a N (Ilan11), only if a ti b,
K
and so
IK*/til < Ic
IK*I < 2t.
0
Assuming
ordered field
This in turn implies that
= 2 °.
it can be shown that there is an
CH,
K IK*/tip 4 2 °
K
over R
P
be a non-maximal, prime ideal in
with
and
IK*I = 2H
Let
C(X).
C(X)/P is an ordered integral domain with respect to the order induced by the standard order on C(X), and its Then
The fields quotient field KP is an ordered field over R. KP have additional structure, and so one might hope to prove x K IKP*I = 2 ° if that IKP*/tiff = 2 °. However, we can show
that, assuming the consistency of certain large cardinal axioms, it is consistent with CH that there exist such K
and
with
P
6.30
X
K
IKP*1 = 2 1
IKP*/tip = 2 °.
and
NOTES Proposition 6.1 is given in (34,12H] and in [49]. The theory of
1908.
goes back to Hausdorff in
nl-sets
For further theory of
see [17, §1.61, n1-structures, (54, Chapter 4], and [56, Chapter 9, §4].
[36, Chapter 13],
Hausdorff's theorem on the existence of gaps
in
(P(N),
was given in (40];
(H,,NI)
proofs are also
given in [14,14.14] and [34,21L].
The notion of a Hausdorff gap and the proof of 6.9 ZFC are due to Kunen; see also (72]. Theorems 6.12 and 6.13 are also variants of unpublished results of Kunen, and in
Proofs of some of the results on gaps are also given in [45,II.§6].
Theorems 6.15 and 6.16 are given in (72].
The definitions of al- and 61-sets are given explicitly in (17,§1.6], but the idea is older; see (36, Chapter 13], for example. Sierpinski, and
A
and
The set Q
It was shown in 6.28(1) that
Q
was introduced by
are discussed in [36, Chapter 13]. Q
is a totally ordered
B1-n1-set: in fact there is really only one such set because each totally ordered B1-n1-set is order-isomorphic to (Q,<),
((17,1.6.9(i)]).
130
FORCING
7
We shall now give our presentation of the technique of forcing.
Let
4)
be a sentence of the language
We ex-
.'.
plained in Chapter 4 that the algebraic approach to a proof that
4)
is independent of
ZFC
involves the construction of
models, and that forcing is a technique for building extensions of a given model of ZFC. The main result of the present chapter is Theorem 7.34:
if there is a model
H
of
ZFC, then there is a model, extending H, of ZFC + -iCH. We shall also prove, in Theorem 7.38 that, if there is a model H of ZFC, then there is a model, extending H, of
ZFC + CH.
In conjunction with Godel's completeness theorem,
Theorem 4.8, these results establish that Con(ZFC + CH)
Hypothesis
Con ZFC
implies
and that the Continuum is independent of the theory ZFC.
and
Con(ZFC + iCH),
CH There are a number of different ways of presenting
forcing (the methods are surveyed in (49, VII, §9]): presentation is based on Boolean-valued models.
our
The
elementary theory of Boolean algebras that we need was given in Chapter 2.
However, there is a point of notation that we
must explain.
Formally, a Boolean algebra is a sextuple
B = (B,A,v,',0,1)
which satisfies certain rules (see Definition 2.6).
In
these last two chapters we shall distinguish between the
Boolean algebra (denoted by a fraktur letter) and its underlying set (denoted by the corresponding roman letter).
131
DEFINITION
7.1
Let
A
algebra.
0 = (B,A,V,',O,l)
be a complete Boolean
8-valued model is a triple
B
(ii)
[[a
(iii)
[[abl] A lbc]] <[[acl];
(iv)
[[a [[a
(v) M
The set
b]] = lb
where
(M,E0,'L8),
is a non-empty set and E0: (a,b) r [[a E bI] and (a,b) r [[ a = b I] are maps from M x M into that, for each a,b,c E B: (i) Ia=al = 1;
M
such
a]];
b]] A lb E c]] < [[a
c];
b]] A [[c E b]] 4 [[c
al].
is the underlying set of the model, and the model M is countable.
is countable if
We shall denote a
8-valued model by
a$,
with the
convention that A8 = (M,E8,A1 8). The language r has two binary relation-symbols, E
and
=
In some ways, it would be more natural to define
.
a model to be a triple
(M,E,'y),
M
where
is a non-empty
are binary relations on M, would produce a symmetric treatment of E and set and
E
and
for this
''
=
.
Since
=
is to represent "equality", we would require that ', is an equivalence relation that E respects, in the sense that, for all
a,b,c E M:
a '- a;
(i)
if a 'y b, (iii) if a " b (ii)
If
'
then b ti a; and
b tic,
a tic;
then
(iv)
if
a
b
and
bEc,
then
aEc;
(v)
if
a ti b
and
cEb,
then
cEa.
is such a relation, the binary relation
defined, and (M/'', E/'') A model
regarded as a
(M,E,ti)
E/"
is well
is a model. in the above sense can clearly be
8-valued model for each complete Boolean
132
algebra a$b,
= 1
(a,b)
'
E0(a,b) = 1
set
a:
the concept of a
if
E0(a,b) = 0
aEb,
of that of a model:
a ti b,
(a,b) = 0
'-
a x b.
if
$-valued model, we can consider
in a
sentences whose "truth value" lies strictly between Let
be the language
.'
aI ...,a
Now let
and 1.
In Chapter 4
_v.
0
at
(a1,...,an)
$-valued model, and let We shall define the truth of 4 at
S.
E M.
(a1,...,a
0
as in Chapter 4,
and let m(x1,...,xn) be a formula of (pages 61/62), we defined the truth of in a model
if
Thus a 8-valued model is a natural generalization if
in
be a
2S
to be a certain element of
8l$
to be
B,
denoted by
j4[a1 ...,anIIl For example, if 'x1 E x2',
8
or
0
is the atomic formula
it is natural to define
to be
L 0(a1,a211
E$ (al,a2) . Formally, we define be set out below.
[0[a1,...,an) I]
by rules to
The rules proceed by induction on the
length of formulae, and are essentially the same as those given in Chapter 4. 1
2
Let
xi
and
xj
be variables.
(xi
x.) [a1,...,anl
(xi
x
Let
4
and
Then:
= -,(ai,a
[al, ...,anJ I ] = E$ (ai,a j) . be formulae.
Then:
fr 11; v i) (a1,.... anI I _ 10[all .... an] I v ILp(a1,...,anl
I (-io) [al,...,anI II 3
Let
=
10[a.,.... anI IJ'.
be a formula, and let
xi be a variable.
133
Then: (3x 10)[al....,an]
_
V{ [ l(al,...,ai-1'b,ai+l1...,an
.
b E M};
Note that it is because the Boolean algebra The above definition of
is
B
complete that the suprema arising in Rule 3 exist in
B.
again
J4'(al,.... an]]
implicitly assumes that all the variables, bound or free, occurring in
belong to the set
$
a bound variable of , depend on
then
and so
ai,
f I]
each sentence
f m[al,....a] I
is
x i
does not
is well defined belong to {xl,,,,,xn},
Jm[al,.... anI]I
whenever the free variables of In particular,
If
{xi,.... xn}.
4
is a well-defined element of
B
for
4.
It follows from the definitions on page 56 that, if 4
and
*
are formulae, then
I (a A V) (al,.... anI II and that, if
=
I p[al,....anI I] ^ ([Cal, .... a ni ]],
is a formula and
0
xi
is a variable, then
(Vxi4)(al,.... an] ]] =
b E M}.
A{ J [al,...,ai-11b,ai+l1....an Certainly, if 11 H 4[al,...,anI
we regard
I
Let
as a T
m
is a model, then
if and only if
= 1
E 0[al,...,anl ]
when
B-valued model.
be a non-empty theory.
Then we set
IT] =A{DD : SET}. We shall use ultrafilters to produce models from Boolean-valued models.
Let
2 B
be a
B-valued model, and let
U
be an
134
ultrafilter in
B.
For
aEb
set
a,b E M,
if
E8(a,b) E U
and set a n, b if n,8 (a,b) E U. Conditions (i) - (v) in Definition 7.1 show that E
respects, and so
by
is an equivalence relation that
n,
is a model
(M/n,, E/n,)
it is denoted
EXAMPLE
7.2
Let M be a non-empty subset of B = P(N),
Then
:
1a8(U).
Alb
and, for
f,g E M,
E8 (f,g) = in E N
:
f (n) < g (n) } ,
n,(f,g) = in E N
:
f(n) = g(n)}.
=
(M,E811\1 8)
ultrafilter
U
on
is a
In particular, if
(M/U,
ultraproduct of
]R
by
U
let
B-valued model, and, for each
1a8(U)
IN,
]RIN ,
set
is simply the ordered set M = ]R
then
,
BB(U)
is the
(regarded as an ordered set).
be a B-valued model, let O(xl,...,xn) Let 2B U be a formula, and let be an ultrafilter in B. Consider the following condition:
for each a1,...,an E M, an1I E G rl(1) if and only if S8 (G) = [[a 11 U' ' ' ' ' [a n) U) I. J It is immediate from the definitions that (1) holds for each ultrafilter
U
whenever
is an atomic formula,
and one is tempted to try and prove that (1) holds for each formula 0 and each ultrafilter U. However the result is not true in this generality: there are problems with quantifiers.
For example, consider the case where M = coo, the space of sequences which are eventually zero, in Example 7.2, and let
0
be the sentence
'Vx1Vx2(xl
1 01 = A{{n : f(n) = g(n) }
:
x2)'.
Then
f,g,E cO0} _ 0,
135
and so, for each ultrafilter if
U
on
.
1Q8(U)
1
However,
for each
f =U g
is a free ultrafilter, then
U
f,g E coo, and so
IT 4 U t U.
IN,
We obtain the condition (1) for each formula 0 by considering only a special type U
and each ultrafilter of
8-valued model.
DEFINITION
7.3
Let
8-valued model. Then Alb is full and each a1,.... an E M, 4,(x1,...,xn)
be a
K8
if, for each formula a E M
there exists
such that
[all... an) I = an) ]]. PROPOSITION
7.4
Let
be a full 8-valued model, and let U be Then condition (1) is satisfied for B.
fll
an ultrafilter in each formula 118(U)
In particular, if
4,.
if and only if 10 ]]
0
is a sentence, then
4,
E U.
Proof
Let satisfied.
that
be the set of formulae for which (1) is
oD
Then
contains
@
[a1 ...,anI I] and so
0.
Since
$
contains
3x
contains
U
Suppose
is an ultrafilter,
if and only if JP[al,...,an) ]3
EU
belongs to
10
whenever to
contains all atomic formulae.
0
Similarly,
0. Q
and
*.
for each variable
Since xi
contains
@
3
is full,
whenever
0
U,
0 v @
belongs
@.
It follows that formula
0.
0
and so (1) holds for each
1
For example, let Alb = (3NN,E8,'e8) , as in Example 7.2. We claim that, for each formula 4,(x1,.... xn) and each fl,.... fn E Nom.
136
Off1,...,fnII = {k E N
:
O[f1(k),...,fn(k)] }.
(N,<)
It follows from this claim that
to find a
is full:
A[8
it suffices to choose a suitable f E NIN, The details of this example are left k E N. as an (unimportant) exercise. suitable element for each
f(k)
PROPOSITION
7.5
Let
IT I = 1.
theory with
that T j-
T
be a 8-valued model and let
Alb
Suppose that
be a
is a sentence and
4,
Then 101=1.
O. Proof
We can suppose that there is a proof of containing only sentences ([23,§2.4]). If
from
4,
T
be a sentence.
V
is a logical axiom, then it is easily verified (given
1
the list of logical axioms) that I* I]
Let
= 1
Iv I]
inference, Modus Ponens, preserves suppose that
JV' * X]1 = 1.
i
and
X
= 1.
If
I...I] = 1.
are sentences with
For this
[IV I]
=
JXI] = JV+J' v IXII = I ('i*) v X j
Then
D * XII = 1.
More formally, let
be a proof of
n)
from
T
each
i < n, i is a logical axiom, an element of
containing only sentences.
Thus
4n = 0
follows by induction on
i
that for each
=
4,
and for T
or
by Modus Ponens.
obtained from the set
and so DT = 1.
then
iy E T
It remains to see that the rule of
by hypothesis.
It now
i 4 n, 101 I] = 1
8
We can now give some more explanation of how, assuming that we are given a model of models of ZFC + -NCH. Let
R = (M,E)
be a model of
ZFC.
select a certain complete Boolean algebra
full $-valued model
Alb
we shall build
ZFC,
8
We shall and construct a
such that To I] = 1 for each
137
axiom
of
0
(so that, by 7.5, we have
ZFC
whereas
fCH] = 0
At 8(U)
ZFC + -iCH
(and so
= 1). By 7.4,
f'1CH]
for each ultrafilter
will be the required model.
M0(U)
IE ZFC B = 1),
in
U
and so
8,
The construction of
18
will be accomplished by working "within 1D". For this construction, we need the notion of a "Boolean-valued The universe universe VB", and so this is our next topic. VB should be compared with the cumulative hierarchy of sets, defined in Chapter 5. dom f
Recall that we write domain and range of a function
f,
for the
ran f
and
respectively.
DEFINITION
7.6
Let
We define
be a complete Boolean algebra.
8
Ord
by recursion on
a set
for each
V
a E Ord:
Vo = 0; V0L+1 = {u
:
V8 = U{VB
:
a
and
dom u c Va
ran u c B};
for a limit ordinal
S < a}
a.
B
The class obtained, V$ = U{Va : a E Ord},
is the
The members of
8-valued universe.
are terms.
V8
0 as a function with domain For example,
We regard V
= {O} = V1.
V8 = {0} U it (O,b) }
:
b E B}.
It is immediate by induction on
and
V8 # V0a+1 because Q For each
minimum ordinal then
p(v)
a
< p(u),
I V$
a+l
u E V81
such that and
I
and so
0,
I ), I P(VB) a
the rank of u E V8
p(u) = 0
a >
that
u, p(u), If
V$ c V8 a
I V8a
is the
v E dom u, a+1 if and only if u = 0.
a+l
,
138
n
:
Let the trivial Boolean algebra be (2), and define V(2) + V inductively by setting v(O) _ 0 and w(v) _
{ir(v(u))
u E dom v
:
jection and
and
Then
v(u) = 1).
rk(n(u)) < p(u)
(u E V(2) ),
denotes the rank of a set in
is a sur-
1f
where
rk
V.
We shall require an ordering on
Ordn
:
set
...'an) < (Q1.... ,6n) if max{a1,...,an} < max{6 1,...,an} or if max{a1,...,an} = max{81,...,gn} and (al., .... an (CL
in the lexicographic ordering
(Sl' " ''Sn)
ordering
is a well-ordering on
<
canonical well-ordering.
< on
The
Ordn.
Ordn,
called the
For example, if
w E dom u,
then
(p(v),p(w)) < (p(v),p(u)), and (p(w),p(v)) < (p(u),p(v)), (p(w),p(v)) < (p(v),p(u)) in Ord2. Each initial segment of (Ordn,<)
is a set (but this is not true for
<
which explains why we do not use Let
define
u
I U E VI
and
be terms.
v
(Ordn,<),
itself).
We shall simultaneously
I u = vI] as elements of
and
B.
The
definition proceeds by recursion on the position of (p(u),p(v))
T U c v] 7.7
in (Ord2,<). The intermediate element is at present just a notational convenience.
DEFINITION Let
U)
To
8
be a complete Boolean algebra.
Then:
01 =1; [u E 0I] =0 and TO c ul] = 1;
(ii) for
u E V0,
(iii) for
u E V8\{0}.
fu =01 = IIu = 01 = TO ° U1 = A{u(t)' (iv)
for
u E V8
and
:
t E dom u};
v E V8\{O},
JuEvl] = V{v(t) A Ju
tI] : tEdomv};
(2)
139
u,v E VS\{O},
for
(v)
Iu c v]j = n{u(t)' v It E v]]
:
t E dom u},
(3)
and
Ju = VI = Ju C v] A Iv c uI. The formulae (2) and (3) are of the greatest importance, and they will be used, usually implicitly, many times in the future to calculate Boolean values. The reader
will find that there is a considerable net saving of effort if they are committed to memory. We attempt an explanation of the intuition behind Definition 7.7.
First, consider the special case where
The natural definition of
IfuEVII = where
Tr
1
0
if if
Tr(u)
a = (2).
in this case is
J u E v I]
E ir(v),
W(u) @ Tr(v),
was defined on page 138.
This is exactly what
our definition gives.
Now consider the general case. analogues of 7.1(1) - (v) to hold;
have
for example, we want to
Vi.
J U E V] > J u = w]J A
First, we want the
Secondly, we want
to be related to the value v(u) for u E dom v. I u E v] A first attempt at the definition of E U E v] is to set
I u E v I] = v (u)
for u E dom v
(and
[ u E v I] = 0 for
u !E dom v).
But this definition fails to satisfy our
desiderata.
For example, it could happen that
but that
v(w) > 0
for some
in this case, we must have
w E dom. u
with
J u E v I] > v(w).
v(u) = 0, I u = w I]
= 1:
We are
essentially forced by these considerations to our form of (2); a similar justification applies to (3).
hold for
We next show that the analogues of 7.1(i) - (v) do u,v,w E VB. This appears to prove that (VB,E,tiB
140
is a
$-valued model, where
E8(u,v) = jf u E vI] and but this is not the case because
ti8(u,v) _ I u = v] ,
V0
is a class and not a set.
PROPOSITION
7.8
Let
u,v E V0.
Iu (ii) Iu
Then:
UT = 1;
(i)
vI _ Iv ° uI].
Proof
If p(u) = 0, then Tu uI] = 1 by 7.7(i). (i) Now suppose that p (u) > 0 and that [Iv = v I] = 1 for all v E V8 with p(v) < p(u). For each t E dom u,
It E
uI]
= V{u(s) A It = sI]
:
s E dom u}
> u(t) A It = tI] = u(t) because
I t = t I] = 1, and so
u(t)' v It E u] > u(t)' Thus
T u c u I] = 1,
and so
V u(t) = 1.
E u = u I] = 1.
The induction
continues. This is immediate from the definition in
(ii)
7.7(v). 7.9
t
PROPOSITION Let
u,v,w E V8.
(i)
Iu = VI A
(ii)
Iu
(iii) Iu
Then:
TV = wI] < Jfu = wI];
VI A (IV vl] A [[w
w]1 < Ju VI < 1w
WI;
u]].
Proof
The proof is by induction on the position of (p(u),p(v),p(w))
in
(Ord3,<).
However, to make the
141
induction work, we must add to the statements (i) - (iii) the 15 additional statements obtained by permuting
u,v,
and
w.
Suppose then that the 18 statements hold for each r,s,t E V$ (Ord3,<).
with (p(r),P(s),P(t)) < (P(u),P(v),P(w)) in We must prove the 18 statements for u,v,w. We
shall prove (i),
(ii), and (iii):
follow by permuting
and
u,v,
w
the remaining 15 statements in the argument.
First, we prove that
Iv=wI] A JucvI] 4 Ju - w]. This is immediate if suppose that
p(u)
u = 0,
> 0,
for then
and take
(4)
T U c w z]
t E dom u.
= 1.
Now
Then
v = wI] A (u(t)' v It E v]]) u(t)' v (1Ev = w]] A It E vI]), and, by the inductive hypothesis and 7.8(11),
1v=wIl A ItEvIl
1Ev = wI] A (u(t)' V It E vJ)
4 u(t)' v it E wI], and (4) follows. By symmetry, u,v,w,
(4) also holds for any permutation of
and, in particular,
1v
uI] A {[w c vI] 4 [[w C uJ].
(5)
From (4) and (5),
(Iv = wI] A [[u C vJ]) A ([[v = uI] A lEw c v]]) f[u C wI] A FEw c uI] = [fu
w]].
142
Since
Iu c VI > Iv ° uD
and
[Lw c v]] > Iv = wI],
condition (i) follows.
Condition (ii) is immediate if Now suppose that
IV E W I] = 0.
t E dom w.
p(w)
w = 0,
for then
and take
> 0,
By the inductive hypothesis,
lu
Tv -t]] <. Qut]],
VI] n
and so
VI] A (w(t) A TV = t]]) <w(t) A [[u = t]],
u
an inequality from which condition (ii) follows. We prove a stronger version of (iii), namely that
TV c u]] A [Lw E v]] * [Lw E uI]. This is immediate if suppose that
p(v)
v = 0,
for then
and take
> 0,
(6)
j w E v I] = O.
t E dom v.
Now
Then
TV c u] 5 v(t)' v It E u]], and so
Iv C u]] A (v(t) A [Lw = t]])
t]] A ItEu]] <- [LwEu]]
<[Lw by the inductive hypothesis. which (6) follows.
This is an inequality from
L
There is a further preliminary result that we shall require later. 7.10
LEMMA For
U E V$
and
dom ua = dom u,
a E B,
define
ua(t) = u(t)
Aa
ua by
(t E dom u).
143
and
Then u E V$,
If u = u
a
>a.
I]
Proof
Clearly, for t E dom u,
It E uaI] ) ua(t),
I t E u I] > u (t)
and so
u(t)' v It E uI] > u(t)' v (u(t) u
a
Ifu c uaI] > a, I u a c uI] = 1,
Then
v
For
u,v E VB,
set
equivalent if
u tiv.
if
and
If u = uaI] % a. I
VO:
u
= 1.
and
v
are
We write
supp u = {t E dom u
(t E supp u).
and
if l[u = V1
u ' v
is an equivalence relation on
u ti v
A a) > a,
(t) ' v It E U I > (u(t) A a)' v u(t) = 1.
Thus
Set
and
:
u(t) # O}.
supp u = supp v and u(t) = v(t) Then % is an equivalence relation on
V$,
u tiv if u tiv.
Recall from 5.15(1) that a subset A of a Boolean A c B\{O} and if a A b= 0 a is an antichain if whenever a,b E A with a # b. The following result is algebra
sometimes called the mixing lemma. PROPOSITION
7.11
algebra
Let A be an antichain in the complete Boolean and let f : A + V8 be a function. Then there
g,
exists u E VB with If u = f(a) I] >a
(a E A).
Proof Set t E dom u,
dom u = V{dom f(a)
:
a E A},
and, for
set
u(t) = V{f(a) (t) A a : a E A with t E dom f(a)}.
144
Take
Since
a E A.
A
is an antichain, we have
u( t)
A a =
f (a) (t) A a if t E dom f (a) , and u(t) A a = 0 if t ,E dom f(a). Now ua % f(a)a, and so I ua = f(a)a I] = 1. By 7.10, [u = uaI] >a and f f(a) = f(a)a] 3 a, and so,
by 7.9(i),
1u = f(a) ] 3 a.
be a formula of ., and let at ul,...,un E V8. We define the truth of (ul,.... un) in V8 to be an element of B, to be again denoted by Let
(xl,.... xn)
11
as if
were a
(V8,E8,'8)
Formally, the
8-valued model.
rules are exactly those set out on pages 132/133. is required with Rule 3, however.
Some care The problem is that we
must be sure that each supremum of the form exists in
B:
proper class.
V{...
:
u E V8}
we are apparently taking a supremum over a To show the suprema do exist, it is sufficient
to show that
{a E B
:
a = D[ul....,ui-l,u,ui+ll...,un] 1J for some
is actually a set. formula
u E V8},
We shall note on page 156 that, for each there is a formula **(xi,.... xm+2)
p(xl,.... xm),
such that, for vl,...,vm E V8, I*[vl,...,vm] I] = a in
B
is true in V, and so the **[vl,...,vm,a,B] result follows from the Axiom of Comprehension because B is
if and only if a set.
(The formula * does not assert that
D[vl,.... vm]I] exists, but it specifies the conditions on a E B that ensure that a = I [v1,...,vm11. The construction of
**
depends on both
i4
and
in,
and involves
working through the finitely many subformulae of
*.)
Since this is a book about forcing, we should at least mention the standard notation used in the subject.
Let
145
B
be a complete Boolean algebra, and let
b E B.
Then one
writes
b IF '4[u1,...,un]' for b < D(ul,...,un] I] in
V0
However, we shall not use this notation.
On page 65, we introduced the abbreviation x c y for Vt(t E x r t E y). We now see that this is consistent with the notation in Definition 7.7. We must show that, for each u,v E V$,
Iu c VI] = A{Iw E uI]' v Iw E vI] This is immediate if
u = 0.
:
Now suppose that
w E V$}. (7) p(u) > O.
By
f[u c VI] A Irw E u] < jw E vJ], and so ju a vI] j w E u l ' v j w E v j for w E V$ . Also, for t E dom u,
(6),
u(t) 6 I t E uI,
and so
Iu C VI > A{It E uJ]' v It E v]
:
t E dom u}.
Thus (7) follows. and so u = 0 I] has two 0 E V$ possible interpretations: the first involves the equality of
Observe that
u and 0 as terms, the second is where O(x) = 14(u] I] Vz(z j x). Fortunately, both interpretations give the same
value.
Part (i) of the following result for a formula 4(x)
is
(u = v V
4 (u) )
4 (v) I]
= 1.
Part (ii) is the fullness lemma for to the Axiom of Choice,
V$ : its proof appeals
and indeed the result can be shown to
be equivalent to the Axiom of Choice.
Let
4(x1,...,xn) be a fixed formula of .'.
146
THEOREM
7.12
Let
be a complete Boolean algebra, and let
B
u11...,un E VB. For each
(i)
u E V
u ° ui11 A Em[u1,...,un] I S ¢[u
i-1,u,u
1
u E V0
There exists
(ii)
I
such that
[u,...,u1 1] = D[ul,...,ui-1,u,ui+l,...,unI I .
Proof
Let
(i)
D
be the set of subformulae
such that the result holds for By 7.9,
4D
of
p
V.
contains all the atomic formulae.
It
is an immediate consequence of the definition of truth in VB and X that contains V v X whenever t contains V contains and p v X is a subformula of 0, and that (D
4)
3xfor each variable 3x.*
is a subformula of Suppose that
of
say
y, = V,(x1)
whenever
x
0
contains
j
and
c.
and that -p is a subformula p E for notational convenience. Set 4D
a = FLu = u111, b = EVi[u11 1, and c= FLiy(u] 1]. Then a A b < a A C. Also, a = I u1 ° u 1] , and so a A c < a A b. Thus a A b = a A c, and so a A b' = a A c'. This implies
that
-j tp E
4D.
It follows that all subformulae V of and in particular the result holds for c
to
(ii)
suppose that
Again for the sake of notational sanity, we 0 _ (xl).
Let S = {I [u] 11 S
is a subset of
minimum ordinal Then
f
:
belong itself.
S - Ord
so that, as before,
For each b E S, let f(b) be the with b = F [u] ]1 for some u E V is a function defined by a property, and
B. a
u E VB },
147
so, by the Axiom of Replacement, the image of a
with
13X 4 j = V{ JL
Lu] 1J
So there is an ordinal
f(b) 4 a :
is a set.
S
We have
(b E S).
u E Va}.
Let
I = {b E B : b <
for some
[ u] ]]
. u E Vs} a
By Zorn's Lemma, there is a maximal antichain A
and clearly
(I \{o } , c) ,
1[3X o]J = VA.
Choice again, there is a function
g : A - Va such that
and, by the mixing lemma, there
(a E A),
a 4 1 q[g(a)]IJ
in
By the Axiom of
exists u E V0 with T u = g(a) ]] >a
(a E A). By (i) ,
Tu = g(a) ]] A 10[g(a)l I] < JO[u] 1, and so
We shall now show that of
10[u]] = J3x14I]. i
(a E A). Thus
E0[u] ] > a
m I]
= 1
for each axiom
ZFC.
Let us stress that the question whether or not
I] = 1
in
V8
for a given sentence
complete Boolean algebra
8
0
of r and a
is a mathematical question,
completely analogous to that of determining whether or not is true in ZFC
V,
and in deciding the question we can argue in
in the normal way.
First we consider the axioms of
ZF;
the more
difficult Axiom of Choice will be dealt with later. be seen that to prove that ZF
we can argue in
proof that
J AC I]
for an axiom
'
requires
AC.
Throughout, we shall use the following facts.
and
X
of
but that, naturally enough, the
ZF,
= 1
JO T = 1
It will
be statements.
Since
X
is
(-1p)
v X,
have
ITS » )(I = 1 if and only if
I C I] < I x II .
J a XII = 1 if and only if
LXI].
Let we
q
148
The next result is really an infinite list of theorems, one for each axiom of
it states that certain theorems can be proved, and so it is a metatheorem. ZF:
In some cases, we examine an axiom which is logically equivalent to the axiom stated in Chapter 4.
That
this is sufficient will be proved in 7.16(i) (a result whose proof does not require 7.13). 7.13
METATHEOREM Let
1
in
B
VB
be a complete Boolean algebra. for each axiom
i
of
ZF
Then
as specified in
Chapter 4. Proof
We shall examine each of the axioms in turn. 1
Extensionality VxVy((x n y) « Vz((z E x) a (z E y)))
It is to be proved that, given
u,v E V
Iu vu] _ (A{EEwEu]]' v 1wEvf : wEVB}) A (A{f[wE vii' v TwEui : wEVB}). But this is immediate from (7). 2
Comprehension
Let
$(xl,...,xn)
be a formula in
&.
Vx2...VxnVxn+l3xn+2Vx1((x1 E xn+2) « ((x1 E xn+1) A It is to be proved that, given there exists where
un+2 E V
a = T ul E un+2D
u2'
such that, for each and
''unlun+1 E VB, ul E V , a = b,
149
([u1 E un+1I A
b
[ulr...,un] I
We can suppose that dom u
= dom u
n+2
un+2
n+1,
Set
p(un+1) > 0.
and, for
t E dom u
n+2,
set
(t) = It E un+1I A TO[u1,...,un] Il.
Take
u1 E
V$
Then, for
t E dom un+2'
un+2 (t) A trul = tll < Tu1 E un+1I A by 7.9(11) and 7.12(1), and so
un+1(t) A [u1
t]
Also,
a < b.
A f[q[u1,...,un] U
< It E un+1I A F4(ul,...runI
11
A Tut = tJ
un+2(t) A [[u1 = t], and so
b < a.
3
Thus
a = b,
as required.
Pairing VxVy3z((x E Z) A (y E Z))
It is to be proved that, given
exists w E V0 such that T u E WT A
u,v E V8,
V E w]j = 1.
there
Set
dom w = {u,v}, w(u) = w(v) = 1. 4
Union vx3yvz(z E x .0 z c y)
Given
each w E V$,
u E V8,
we require
v E V$
[w E u] < [[w cvu.
We can suppose that
p(u)
> 0.
Set
such that, for
150
dom v = U{dom t and, for
set
s E dom v,
For
:
t E dom u}, v(s) = 1.
t E dom u, It c v I] = 1 because
I s E V I > V (S) A I s = s I] = 1 for s E dom t.
Take
Then
w E V$.
tI] A It cVI
tD < Jw
u(t) A fw
< FIw c vI, and so
I w E u I] < j w c v I] ,
as required.
Power set
5
Vx 3y dz (z c x * z E Y) u E V$,
Given
each w E
V8,
we require
We can suppose that and, for
define
set
t E dom v,
f:
dom u -+ B
p(u)
Set
> 0.
Take
v(t) = 1.
dom v = Bdom u,
w E V8,
and
by
f(t) = It e wI] For t E dom f, f(t)' v It Also, f E dom v, and so For
such that, for
v E VB
T W c U1 < T W E V I] .
(t E dom u).
E wI] = 1,
f c wI] =1.
and so
f E V I] ) v (f) A ff f
f I] = 1.
t E V0,
It E wI] A It E uI]
WI A It = 81
t E dom w,
:
sI]
:
8 E dom u},
s Edomu}
151
ftu c uI] < ru(t)' v It
E uI]
W(t)' v (It E w]] A It = u]]) v (V{1s
w(t)
a E dom u})
:
v It=fjl.
=tu(t) Thus
WI A It =s]]
If w c u] < j w c f I] , and hence
Eru c uI] < Jm c fl A If C w11 A If E vI]
as required. 6
Replacement Let
4(x1,x2)
be a formula in
.'.
Vx33x4Vx1((x1 E x3 A 3x2(0)) r 3x2(x2 E x4 A 4)) Given for each
u3 E Vwe require
ul E V,
u4 E V$ such that, where If ul E u3 I] A a < b,
a = I (3x2 ) (ull J1 = V{ lL O(u1,u21 b
V{I0(ul,u211 A f u2 E u41
:
:
U
U
2
2
E V8}, Vol.
E
By two applications of the Axiom of Replacement, there is an ordinal
a
such that, for each
V{110(vlrv2l JJ Set
v2 E Vol = V{f[+(vl,v2}
dom u4 = V
u4(v2) = 1. We can suppose that v
1
E dom u
31
vl E dom u3,
and, for p(u3)
v2 E dom u4, > 0.
For each
:
V2 E set
Val.
152
IV1
= u
I I A a < Tv1 ° u1]] A I (3x2 ) (V1] JJ u11J A V{1E4(v11v2]
= TV1
u2 I] A 1O[v1,v2]
= V{1[vl
4 V{1L4[ul,u2] l Since
1[u2 E u4I] = 1 for u2 E
u3(v1) A Iv1
]
:
:
v2 E
:
v2 E V
:
u2 E
Va}
u2 E Va}.
Va,
we have
U, I] A a
V{ IOlui,u2I I] A 1Eu2 E u41
Va}r
1[u1 E u3I] A a < b, as required.
and so 7
Regularity
This axiom is equivalent to
0 v (y E X A (y fl x
Yx3y(x Given
0)) )
we must show that
u E VB,
JU#011
b E B\10}
:
v
EV$}.
Then there exists
such that, for each v E V8, b < [Eu # 0]], b A Iv E u]] A I'u n v = 4' I]
Clearly u# 0. If b n Q t E u ]] t E dom u,
b < j u
then 0 ],
b < u(t)'
whence
(t E dom Of
b = 0,
= 0 for each and so
a contradiction.
suppose that b A T V E u ]J # 0 for some v E be the minimum ordinal such that
So we may
V$ :
b A I v E u I] # O
let
t E dom v.
Then
t E VS
for some
a
for some
v EVa. Let
= O.
R < a,
153
and so b A It E u I] = O. But
Iu n v # 4' I= V{ Jw E U I]
A [[w E V I] : w E Vol,
and
EuI]n IWEvl] V{uwEuI] Av(t) A [fwtI] : tEdomv} < V{v(t) A It E uIJ
so that b A [[,u a v # 0' I]
:
= O.
t E dom v}, b A IV E U I]
Thus
=
b A IV E U I] A E- u n v = 0' I] = O, a contradiction. 8
Infinity
This axiom is equivalent to
A (y E x + (y E Z A Z E x)) )
3xVy3z (x # Q For each u*(u) = 1.
If
define
u E V8,
u E Va,
then
ju=vl] = ju*=v*I] We require such that, for each
u E V8
u*
{u} + B
:
u,v E V$,
ffuu*I] = 1.
and
such that
E u = 0 I]
there exists
v E VB,
by
and, for
u* E Va+1'
= 0
and
w E V8 with
Iv E u1 < Iv E WI] A Iw E uI]. Set
dom u = V9,
ordinal, and, for
where
t E dam U,
set
is the first infinite u(t) = 1. Then
w
j u = 0 I] = O. Also, for each v E V8,
vE uI] = V{u(t) A Iv t tEdomu} V1 IV* = t* I]
t E dom u}
81 : 8Edomu}
154
because
whenever
t* E dom u
TV E U1 < TV* E U I= TV and so
u
Thus
t E dom u.
is as required (with
V* I n TV* E U II
w = v*).
This completes the examination of the axioms and the proof of the metatheorem.
I
To complete the proof that to show that
J AC I]
= 1.
f ZFC] = 1 it remains We postpone this to the end of the
chapter, for by then we shall have some more notations and techniques at our disposal. also be needed in Chapter 8.
These preliminary results will In fact, when it is slotted in
as Theorem 7.41, the proof that
T AC I]
= 1
will be
gratifyingly brief.
We shall shortly need to consider statements, such is a choice function for
y' or 'x is an ordinal', which are expressed in ordinary mathematical language. For this purpose, we must formalize the statements in the
as
'x
language
Y.
For example, an ordinal is 'a transitive set which is totally ordered by the relation of membership', and so the statement
'x
is an ordinal'
can be expressed in 2 by
'dy(y E xyCx) A A dzdt((z we regard
'x
x n t E x) - (z
is an ordinal'
t v z E t v t E z))';
as an abbreviation of this
formula.
Similarly, we could, under duress, write down the formulae which express 'x
of
is a function from y to z', is a complete Boolean algebra and y is the Stone space x', 'every homomorphism from Z into a Banach algebra
is continuous', etc.
'x
The formula expressing this last
155
statement is, of course, a sentence.
However, a little
experimentation will show that a formula corresponding to even such an innocuous statement as z'
'x
is a function from
y
to
is very tedious to write down, and we shall not do so.
We assume that each statement in naive set theory that we consider can be expressed in the formal language y. Similarly, we assume that, if
Ul
is a model of
ZFC,
then,
for example, v
'a
is a function from
is well defined, where
b
to
c'
a,b,c E M.
(Some expressions which can be written down cannot be formalized. = 1
I x I]
in
each fixed sentence formalized.
m
.
of y, ' 10]] = 1
in
Vy'
can be
This is a consequence of Tarski's "undefinability
of truth theorem", left to logicians: example.
'x is a sentence of and cannot be formalized, even though, for
For example, Vy'
a theorem of logic which can safely be [44, p.125], for see [23, §3.5] and
It accounts for the fact that certain results in
this work are called "metatheorems".)
We now introduce notation for some specific formulae of
.'.
Let:
* (x1,...,x4)
(respectively,
p (x1,.... x4)) express
'x4 is a Boolean algebra, x1,x2,x3 belong to
the underlying set of x 4, and x 3= x 1 A x 2 (respectively, x3 = xi v x2)'; io(x1,x2) 'x2
to
express is a complete Boolean algebra and V 2'.
xl
belongs
156
*E(x1I...,x4)
(respectively,
x3
is a complete Boolean algebra,
'x4
express
(xl,...,x4))
p
belongs to x
the underlying set of and
x3 =
xl
in
V4v
Finally, for each formula there is a formula
p(x1,...,x) of .', which expresses
**(x11.... xn,xn+1'xn+2)
is a complete Boolean algebra,
'xn+2
V 4,
(respectively, x
x2 I]
x3= Lxx21])
belong to
xl,x2
x4,
belongs to the underlying set of
belong to
and
Vxn+2,
xn+i
xn+2'
x1,...,xn
in
Tip [x1, ...,x n I 1]
xn+1
Vxn+2t.
We now return to the construction of models of given a model of
ZFC + iCH,
Let
Q = (M,E)
ZFC.
be a model of
ZFC,
and let
in
h
M be such that
I = 'h Then there exists 2
is a Boolean algebra'.
b E M
'b
such that
is the underlying set of
W.
As in Definition 4.12, we set
a = {c E M : cEa}
(a E M) .
There are naturally defined operations such that and
1
n
is a Boolean algebra (where
are defined using
anc=d a"c=d
if if
n
and
v):
for
is
A[a,c,d,h]
Al
V[a,c,d,b].
a,c E b,
and
v
',O,
set
157
Thus, for example, 'a n c y d
11
if and only if
c = d
a
h'
in
We write
for the Boolean
b,,
algebra
Thus the operations in our Boolean algebra exactly those that the model in its Boolean algebra internally to
h.
ba are "thinks" are the operations
I
M which appear
The points of
to be Boolean algebras have a canonical
#t
external manifestation as Boolean algebras. 7.14
EXAMPLE Choose
0
'a0 # 0'.
Then
'h = (P(a),fl,U,',O,ao '.
2
Then
such that
h E M be such that
Let
0.
ao E M
= {c E M : c c a 0}. Let b = {c E M : 3j-- 'c c a '} 0
X = {c The map
:
Then
c E b}.
7r
is a Boolean algebra.
(X,n,U,',O,a ) 0
: cI+c,
(X,l,U,',O,8
is a Boolean homomorphism.
Example 4.13(1), shows that
Since it
#l
0
Extensionality,
is a bijection, and so it is
a Boolean isomorphism.
Suppose now that
11
is a complete Boolean algebra'.
'b
Then it is not necessarily the case that For suppose that h E M Then
so that
ha
is complete.
is a countable model of ZFC, and choose 9 H 'h is an infinite Boolean algebra'. 2
is a countable, infinite Boolean algebra, and it is easily seen that such a Boolean algebra cannot be complete. bit
In fact, generally, a E M,
the set
b
a fl b
is complete in
#t
has a supremum in
is complete only if each subset of Thus, assume that
h
b
if, for each brn,
whereas
hB
has a supremum in bV
is a complete Boolean algebra
158
and let
be the completion of the Boolean algebra
in
O1,
hE
(see Theorem 2.14);
algebra of
B
we identify hB with a Boolean suband we write B for the underlying set of B.
B,
Define M* = {a E M : AI ( ,yo[a,h] }
= {a E M :
10
=
(a,c) - (ja E cI] and ''B from M* x M* into B as follows:
and define maps T a = c I]
EB
I(a,c) = d Thus, for example,
1Q
and let
Al
E[a,c,d,hl;
a
*-[a,c,d,h].
EB(a,c) = d
(a,c) r
if and only if
Vh'
THEOREM Let
be a model of
ID
'h
(ii)
a1,...,a
RB
is a
Then:
B-valued model;
for each formula E M*,
4(xl,...,x
of
E
K
01B
such that
in the above notation.
B
(iii)
ZFC
is a complete Boolean algebra',
A18 = (M*,EB,tiB), (i)
each
if
' j a E c] = d in 7.15
:
:
E8(a,c) = d if
Al
'a belongs to vh'
is full.
and only if
0*[a,...'ac,h];
.e
and
159
Proof
Each condition of Definition 7.1, when inter-
(i)
preted by the rules that we have given, becomes a condition which, by the corresponding statement in 7.8 or 7.9, is satisfied.
This is proved by induction on the length of
(ii)
The steps not involving quantifiers are immediate. We prove that, if the result holds for a formula *(xl), formulae.
then it holds for
a$
Since the result holds for 8
Eb
ry[a] ]] N Vs
3x14.
=c
j 4 (a] ]
and
q,,
for each
a E M*
if and only if
**[a,c,b]. For each s E M with s c b, exists in bin, and it equals that element
the supremum d
of
b
such that in l 'Vs = d
W.
in
(8)
' s = { I W [a] I] : a E s E M with in d E b be such that (8) holds. Then Choose
' ([3x1 I] = d
in
and so
d = Vs
in
bin.
s = {I*[a]
Vb}' ,
and let
in V,
By the inductive hypothesis
I]RB
:
a E M*},
and this shows that the result holds for the formula
3x1,.
This is immediate from (ii) and 7.12(11).
(iii)
The above theorem shows that internal computations
within
OI
of the value of
E o[a1,...,anI I]
same value as the computation of so we have the following result. 7.16
PROPOSITION Let
:y
be a sentence.
in
f 0[a1,...,an] I]
give the
b 8,
and
160
Suppose that
(i)
N.
If
N
such
then
ZFC,
'for each complete Boolean algebra
AI
Suppose that
(ii)
N H ZFC.
N
If
8,
in V8'.
1
that
for each model
N
N - p
ZFC,
for each model
then
'for each complete Boolean algebra
N
6,
V I] = 1 in V8'. Proof
M
in
h
such that
in Vh'.
1
Let
be obtained from
m8
a full
Bl
8-valued model, and
is an ultrafilter sB (U)
U
in
8
and
(ii)
J* I # 1 with
as above. in
1-i' I]
a8.
E U.
Then
N8
is
Hence there By 7.4, N
N.
This is the same argument, save that it uses
7.13 to establish that I] = 1 ZFC.
b
But this contradicts the hypothesis that
-icy.
for each model
of
Then there
Suppose that the result fails.
(i)
exists
in
N8
for each axiom
I
By using Godel's completeness theorem, the above result can be reformulated as a metatheorem. that, if that
p
Result (i) says
is logically derivable, then it can be proved = 1
and it justifies the fact that, in 7.13, it was sufficient to calculate the 8-value of an axiom I]
in
V$,
equivalent to the given one.
Proposition 7.16(11) shows that, if ZFC proves a sentence 0, then ZFC proves that Il = 1 in V$. Thus we may reason within V8 in the same way as we do in V: any normal mathematical argument can be carried out ... I]".
"within
This is very important because it is exactly what
161
we shall do at numerous points in the remainder of this book. Here is a first example:
algebra, and take
u,v E
be a complete Boolean
B
such that Du is a
and such that
Boolean algebra' I] = 1
in u' I] = 1.
let VB
[E'v
ccc
is an antichain
I 'v is countable' I] = 1.
Then
We now come to the key paragraph of this Chapter.
be a Let II be a model of ZFC, and let sentence which we wish to prove is consistent with ZFC. Suppose that we can (We shall shortly take t to be iCH.) '
prove that there is a complete Boolean algebra
# 0 in
I]
Then
VB
'there is a complete Boolean algebra
II
0
'b
I-
ha B
with
VB'
is a complete Boolean algebra and V' I]
Let let
in
B
b E M such that
and so there exists
II
such that
B
# 0 in
Vb'
.
be the Boolean algebra obtained from b, and let IIB = b1,
as above,
be the completion of
be the corresponding
By 7.15,
B-valued model.
IIB
is full
and I]IIB
1*
# 0.
Thus there is an ultrafilter I]BB
1 V+
E U.
U
in
B
such that
By 7.4,
IIB(U)
ZFC + P.
We shall see shortly, on page 164, that IIB(U) is an is the model that we have been extension of S. Thus IIB(U) seeking that witnesses the consistency of
ZFC + *.
162
Our entire development of logic, including the theory of models and of Boolean-valued models, was to justify the following claim. Let
be a sentence of
J
be proved (in
ZFC)
algebra
B
implies
Con(ZFC + *).
with
1L
Suppose that it can
Y.
that 'there is a complete Boolean 0 in VB'. Then Con ZFC 'I]
A forcing argument is essentially a construction of a complete Boolean algebra
]] # 0
that
in
VB.
together with a computation
B,
The process will be similar to a
normal proof in set theory, but there is a further freedom given by the possibility of choosing
B.
It is possible to
learn and to carry out the technicalities of a forcing argument whilst remaining ignorant of the justification that the argument gives the required independence result. Let
be a complete Boolean algebra.
8
1[]J = 1
now know that
for each axiom
obtain manifestations within in the theory
V8
of
4
Since we ZFC,
of familiar sets constructed
ZFC - for example, of
w,
of
and of
wi,
This leads to a study of ordinals and cardinals in
VS1
in particular to the construction of a Boolean algebra such that
L[-iCH I]
from the class
= 1.
Ord
we can
We shall build a class
Ord'
I. and
B
in
VB
of ordinals such that, if
I 'u is an ordinal' I] # 0, then
f u = v fl
represent 7.17
V
# 0
inside
for some
v E Ordv.
To do this, we shall
VB.
DEFINITION Let
V
of
:
V + V8 V. Set
B be a complete Boolean algebra. A map is defined by recursion on the rank of elements
0v = {b,
and set
dom u = {v : v E u},
u(t) = 1
(t E dom u).
163
The map
exists and is uniquely specified by the
v
Note that, for
conditions.
u E V8
and
v E V,
[EuEvl] = V{[Eu=wl] : wEv}. Recall that u
of
V,
7.18
rk(u)
(9)
denotes the rank of an element
as on page 85.
LEMMA Let u,v E V.
Then:
W uEv if and only if
vl] = 1,
[Eu
u i v if and only if [ u (ii)
v
u = v
if and only if
T U
v
if and only if
[[ u =
u
I = 0;
v I]
= 1,
I]
= 0.
v
Proof
We prove the two statements simultaneously by induction on the position of
(rk(u),rk(v))
The statements are trivial if
in
u = 0
(Ord2,<).
or if
Now suppose that the statements hold for
with
(rk(r),rk(s)) < (rk(u),rk(v)).
v = 0.
r,s E V
Then, by (9),
[EuEv]] = V{[Eu= w]] : wEv} 0 Thus (i) follows. if
w E dom u,
Suppose that
'b
u t v.
The proof of (ii) is similar, noting that,
then
and so (i) holds for
N
uEv,
if if
1
(rk v,rk w) v
and
< (rk u,rk v)
in
(Ord2,<),
w.
Q = (M,E)
-
ZFC,
and let
is a complete Boolean algebra'.
h E M
with
164
Take by
OI$ _ (M*,E$,ti$)
= r if
w (a) where
y = z'.
and
it
: M - M*
D, r, a) ,
ffi
expresses
*(x,y,z)
y E Vx,
to be as above, and define
is a complete Boolean algebra,
'x
a,c E M with aEc.
Take
Then, by 7.18,
ltt n(a) E n(c) I] s = 1.
This, and the obvious three related formulae, show that induces an isomorphism from and so 7.19
onto a submodel of is indeed an extension of If.
118(U)
it
IUB(U),
LEMMA
Let u,v E V. 0)
If
if and only if
fu c vI] = 1
Then
u a v
(respectively,
(respectively,
u s4t v).
Proof
This is immediate from 7.18, on expanding by formula (3) of page 139. 7.20
Ilu c
1
LEMMA Let
u E V.
(i)
u
Then:
is transitive if and only if
'u is transitive' I] v
u
= 1,
is not transitive if and only if v
'u is transitive' ]] = 0; (ii)
u E Ord
if and only if v
'u is an ordinal' I]
= 1,
u ¢ Ord if and only if v 'u is an ordinal' I] = 0. Proof (i)
Suppose that
u
is transitive, and take
I]
v
165
For w E u,
v E V$.
11w c
u I]
w c u,
we have
and so, by 7.19,
Thus, using (9) and 7.12(1),
= 1.
[ [ vEu] =V{Iv=w]1 n 0: wcuI]
:wEu}
< TV a ul, V
and so
I 'u
is transitive']] = 1.
Similarly, if
'u is transitive' II V
is not transitive, then we have
u
= 0.
Suppose that
(ii)
By (i), to show that
u E Ord.
v
['u
is an ordinal']] = 1,
it suffices to show that, for
v,w E V$
JvEul] A IrwEuVj (10)
< TVEwl] V TV =wl] V TwEv]. V
If
for some
v = r
then (10) holds because
r,s E u.
and
w = s
The general case then follows from (9).
The case where 7.21
V
V
v,w E dom u,
LEMMA Let a E Ord,
u V Ord
and let
is again similar.
u E V$.
If a > p (u) , then (ii) If a > P (u) , then (i)
fFa E u I] = 0. cEav
..=
u I]
= 0.
Proof
We prove (i) and (ii) simultaneously by induction on the position of
(a,P(u)) in (Ord2<). The statements are trivial if a = 0 or if u = 0. Now suppose that they hold for 8,v with (B,P(v)) < (a,p(u)). If
a > P(u),
j[a = t I] =o, 6 = P (u) .
Then
then
P(t) < a
whence
tLa E
(t E dom u),
and so
u j =O. If a> p (u) , set
166
Ia
= u l < . v c .1 = A{ U
E U] : w E a}
I a E U I] = 0. The induction continues. 7.22
LEMMA Let
and suppose that
u E V8,
I ' u is an ordinal' II # 0. Then there exists
a E Ord
with
., v I u = al]
0.
Proof V
For
a E Ord,
L 'a
is an ordinal']] = 1
by
and so
7.20(ii),
Eu E al v E. = a]] v Es E ul] 1'6v
3 I 'u
= I'u is an ordinal'] Take
a E Ord with
a E uI] = 0, and so j u
a< a
with
7.23
is an ordinal'
is an ordinal']] A # 0.
> p (u) . l)
0.
By 7.21,
u Il
By (9), there exists
I u 2 a I] # O.
DEFINITION Set
Then
Ordv = {a Ordv
:
I
a E Ord}.
is a class which is contained in
V$,
and 7.22 shows that it satisfies the condition specified on in fact, page 162:
Du is an ordinal' l= V{ j u
v
a I]
:
a E Ord}
=V{{u2al] : a
167
"if
A variant of Lemma 7.22 looks plausible:
t[ ' u
is an ordinal' I] = 1,
then there exists
a E Ord
is false whenever
g # (2).
with
For take
b E B\{0,11.
follows from the mixing lemma that there exists
u = 2 I] ) b and I u 3 I ' u is an ordinal' I] = 1.
But this
= al = 1".
u
It
u E V
with
and then
W,
Many of the basic properties of sets are preserved For example, if u is an V to vs.
in the passage from ordered pair, then
'u is
an ordered pair']] = 1
to see this, one formalizes the statement 'u is an ordered pair' and uses (9) and 7.18. We list some more examples of properties that are preserved. 7.24
EXAMPLE V
U)
If
u = v x w,
(ii)
If
g: u -+ v
g
(iii)
then
g: u + v
V
is a function, then
is a function from If
V
I'u = v x w'I] = 1.
u
to
v'I] = 1.
is an injection (respectively,
a surjection), then v
'g
is an injection (respectively, a surjection), v
from u
v
to v' I] = 1.
(iv)
I ' (a+l) v is the minimum ordinal > a' I] = 1.
(v)
Pw is the least infinite ordinal']] = 1.
168
Proof
It is easy to check Examples (i) - (iii). were false, there would exist
8
with
Ia E s11 n [ES E (a+1)v]]
0,
Example (v) follows from
and, by 7.18, this is impossible. (iv).
If (iv)
I
However, other basic properties of sets may not be preserved in the passage from
V
to
These include the
V$.
power set operation (in general,
E['P(u) = (P(u))v']l # 1) There is a general theorem
and the cardinality of a set.
that asserts that properties specified by
"1 -formulae"
are
we shall return to this point in a more general
preserved:
context in Chapter S.
It follows from 7.24(iii) and (v) that, if
is a
a
v
countable ordinal, then I'a is countable' = 1. However, v if a is an uncountable ordinal, it is not always true that
T'va
is uncountable']]
0:
it is particularly important for
More generally, if
us to deal with this possibility. u,v E V,
then
_
Jul
I'iul
lvJ
implies that
lvi
does not imply that
=
JvJ']] = 1,
but Jul
7.25
DEFINITION Let
VS
if
f'lul # ivJ'I] = 1.
K
be a cardinal.
Then
K
is preserved in
is a cardinal' ]] = 1.
J[' K
We have seen that is preserved, then
R0
is preserved in
V8.
If
ktl
v
'H1
is the least uncountable cardinal']] = 1,
and we write this as
I'itl = !tl'I] = 1.
(To recap, in this
169 v
expression
is the image under the embedding v of the V, whereas K1 is the first
K1
first uncountable ordinal in
uncountable ordinal in the universe
is pre-
If
VD.)
2
served, then all we can say in general is that
'!t2
v
or h2 = K1'
= h2
= 1, V
but, if
and
h1
are preserved, then
R2
I'K2 =
14 21 I]
= 1.
To study the preservation of cardinals, we consider terms for subsets of where
Let
fv and, for
:
set
and for functions from
u E V.
For
tHI
V
u
to
v,
of = {s
:
we set
V E V$,
u -+B,
we define
f: u -+ B,
dom
Then
u
u,v E V.
of E V8
s E u},
by
Vf(s) = f(s).
TV
= 1. For VvE V$ E v f I] = v*f (s) and T V f c u I] v*I] g = f v and = T V c u I] , = v . Then T V = 4 v
and so
v ti v*
if
T V c u I]
= 1.
Thus, up to equivalence, v
we have obtained a canonical set of terms for subsets of
u.
We essentially repeat this construction in the next theorem, obtaining a canonical set of terms for bijections. 7.26
THEOREM
Let u,v E V.
Then
[' Iu I
and only if there is a function
(i)
Of (r, s)
:
(ii)
V{f (r, s)
:
sEv}=b rEu}=b
(iii) f(r,s1) A f(r,s2) = 0 (iv) f(r1,s) A f(r2,s) = 0
=
in V$ such that:
jv j' I] 3 b
f: u x v -+ B
(r E u) ; (s E v) ;
(r E U, sl
(s E v, rl
in v); r2 in u). s2
if
170
Proof
',u,
Suppose that
lemma, there exists
=
v,']j
By the fullness
b.
with
w E VS
[['w is a bijection from u to v' ]J > b. Define
by setting
f: u x v + B
f(r,s) = ['s is the value of Then
f
(iii) holds because
11w
is an injection' I]
(i)
[['w
dom w = {(r,s)v V
Clearly
:
f: u x v + B
w((r,s)v) = f(r,s).
and it is easily checked that
is a bijection from u
vJ' I
=
satisfies
v
IV
I'1uI
and (iv)
Z b.
r E u, s E v},
11'W c u x v'I] = 1, [['w
Thus
is a function' I] 3 b,
Conversely, suppose that - (iv). Define w E V8 by
I] A b.
1 'ran w = v' ]] ) b,
(ii) holds because
holds because
r'
(i) holds because
satisfies conditions (i) - (iv):
'dom w = u' ]] > b,
at
w
b,
to
v' I]
as required.
= b.
I
We can now show that cardinals are preserved in V8 provided that 8 satisfies the key countable chain condition introduced in Chapter 4.
antichain in
Recall that
8
is
ccc
if each
is countable.
B
THEOREM Let 8 be a complete ccc each cardinal is preserved in V8. 7.27
Boolean algebra.
Then
Proof Let
with
a < K.
K
be a cardinal, and let
Set
b = J'JaJ = JKJ'I].
a
If
be an ordinal K = H0,
then
171
by 7.24(v).
b = 0
Now suppose that K > $°. Let f: a X K - B be For a < a, set fa(S) _ the function specified in 7.26.
f(a,s)
and set
(8 < K), Sa = {B
By (iii) in 7.26,
f(a,8) # O},
:
fa :
is an injection and
Sa + B\{O}
is an antichain in
fa(Sa)
S = US C'.
Since
B.
B
ccc,
is
ISI < max{a,K°} < K. Take countable, and so Then b < V{f(a,8) a < a} = 0. that a 0 S.
Thus, in each case 'K
in
Let
IKI']] = 0,
and so
is preserved
K
be any complete Boolean algebra.
8
is preserved in
is a cardinal with
K
The above
K > 1BI,
then
V8.
THEOREM
7.28
Let K =
such
V$.
proof shows that, if K
c['Ial =
This shows that
is a cardinal']] = 1.
is
S01
S < K
K IBI °.
be a complete Boolean algebra, and set
8
Then K V 12 ° < K' ]] = 1 in V8.
Proof Set
A = K+,
the successor cardinal to
K.
We
V
first claim that
if'2 0 > x'I] = 0
the case, there exists
't
b E B\{O}
and
is an injection from
By the fullness lemma, for each va E V8
in
V8.
t E VB v A
a < A,
into
If this is not such that P(3N)
there exists
such that v
'the value of
t
at
a
is
va']] > b,
]] > b.
172
and then
m'Va C
fa
if fa = f8,
Inv
nF
:
then
(a < X).
3b
I]
N'
and so a =
b,
vSI]
I va
a < X,
set
N + B.
E vaI],
[ ' t is an injection' I] 3 b.
For
Thus
H
B
a I fa,
:
because
A + ON ,
x
is an injection, and so
of the hypothesis that
a contradiction A < IB NI _ JBI 0, K K < JBI °. Hence the claim holds.
The result is immediate from the claim. It can be shown that, if
is a complete Boolean
8
h
algebra and
is infinite, then
JBI
° =
JBI
It follows from 7.26 that, if is preserved in
CBI.
then
K % CBI,
K
and it then follows easily from 7.28
V8,
that, if
1'2x°
= (280)v' I] = 1 in JBI = 280, then We shall not need this remark, but we shall need the
VB.
following corollary of the theorem.
COROLLARY
7.29
Let
B
(i)
If
(CH)
be a complete Boolean algebra.
(ii) If
K
< K2,
then
1 ' 2 ° < K2' ]] = 1 in VB .
JBI < K1,
then
[[CH] = 1
IB I
in
VB.
Proof
and so
(i)
Since
K
K
CH
holds,
K2 ° = K2
by 5.11, K
JBI 0 < K2 ° = K2.
By the theorem,
v
1-2 ° < K2'I1 =1.
V
Certainly
I'K2 < K2'I] = 1, (ii)
Again using
12 0 < Kl' I] = 1. [['2 ° < K1' I] = 1,
and so the result follows. CH,
Certainly
i.e.,
JBI ° = Kl,
and so
J ' K1 < K1' II = 1,
[[CHI] = 1.
and so
B
We can now construct the complete Boolean algebra
173
B
such that
in V.
E[ CH I = O
We must show that
I' I P(NV) I > ttl' -fl = 1
in
V8
The following theorem gives a condition which reduces the problem to a combinatorial question about Boolean algebras.
be a complete Boolean algebra, let be the f,g : N - B be functions, and let v ,v corresponding terms, (see page 169), so that dom of = dom v g = {n : n E N }, vf(n) = f(n), and ) = g (n) Let
g
v(n
.
Then
Vf c vg I]
_ A{vf(n)' v In E vg]] = A{f(n) ' v g(n)
:
:
n E N}
n E N} ,
and so
I Vf # Vg1] = V{f(n) c g(n) where
a A b = (a' n b)
:
n E N} ,
(a,b E B).
v (a n b')
(11) This shows
that condition (12), below, arises naturally. 7.30
THEOREM Let
B
be a complete
Suppose that there is a set
Boolean algebra.
ccc
S c BN
such that
IS1 = K2
and such that
V{f(n) A g(n) Then
I CH 1]
= 0 in
:
n E N} = 1
(f,g E S, f # g) .
(12)
VB.
Proof
Suppose, if possible, that
By 7.28,
I CH I]
1 ' x 1 = K 1' I] = 1,
'there is a surjection from
ttl
= b # 0
in
VB.
and so onto
P(N V)'I] = b.
174
w E VB with
By the fullness lemma, there exists Rv
is a surjection from
11w
= b.
V
onto
1
I'vf E P(Nv)'II = IVf cIN VI] = 1 for f E S, there exists of < wl such that Since
T'vf is the value of Since and
# 0.
I]
= K21 there is a subset T of S with IS a < wl such that bf # 0, where, for f E T,
ITI = ti2
I
bf
=
I 'v f is the value of
Take
Uv f
vaf'
at
w
with
f,g E T
= v g I] = 0, and so
is an antichain in
B
b
f
f
g.
as required.
b = 0,
.
By (11) and (12),
n b = 0.
Thus
g
of cardinality
contradiction of the hypothesis that
va'
at
w
f
ccc,
is
B
{b
:
f E T}
But this is a
K2'
and so
I
Our choice of a Boolean algebra
B
to satisfy the
conditions of Theorem 7.30 was implicitly given in the original arguments of Cohen. 7.31
DEFINITION
Let 2N
be the Cantor set
{O,1)3N,
and set
Xc = (2N ) w2. The set
Xc
is a compact Hausdorff space with
respect to the product topology.
Xc
is homeo-
morphic to
2 2, the Cantor cube of weight w2. as the set of functions F: w x N - {0,1}; a < w2, set Fa(n) = F(a,n).
We regard
Xc
for
Of course,
w
The regular-open algebra
space
X
there that
of a topological was introduced in Example 2.7, and it was noted R(X)
R(X)
is a complete Boolean algebra.
175
PROPOSITION
7.32
The Boolean algebra
R(X
is
d
ccc.
Proof
Let 2 N.
T
Clearly
T
Let
P
be the standard base for the topology of is countable. be the collection of maps whose domain is w2 and whose range is contained in T.
a finite subset of For t E P, set
Ut = {F E XC Then
:
Fa E t(a)
is a clopen set in
Ut
base for the topology of
s(a) = t(a)
XC.
(a E dom t)}.
and
XC,
{Ut
:
Note that, if
then
(a E dom s fl dom t),
Suppose, if possible, that
t E P}
us nut
R(XC)
is a
s,t E P
and if
# W.
is not
ccc.
Then there exists an uncountable subset A of P such that Ug n Ut = 0 whenever s,t E A with s # t. Let F = {dom t Then, t E Al. Suppose that F is uncountable. :
by the
A-system lemma, Proposition 6.1, there is a finite
subset
T
w2 and an uncountable subfamily G dom s n dom t = T whenever dom s and
of
such that
are distinct elements of F. subset A* of A such that and
t
F
and
u. n Ut # 0 Thus R(XC) 7.33
A*.
s
This conclusion also
is countable.
Since s,t E A*
F
dom t
Thus there is an uncountable dom s n dom t = T whenever
are distinct elements of
holds if
of
T
is countable, we may suppose that, for s(a) = t(a). But this implies that
a E T,
a contradiction of the hypothesis.
(s,t E A*), is
ccc.
I
THEOREM There is a complete Boolean algebra
J CHI] = 0 in
8
such that
V$ .
Proof Set $
= R(XC),
so that
0
is a complete
ccc
176
we shall apply Theorem 7.30.
Boolean algebra.
a < w2,
For
define fa : N - P(X
fa(n) = {F E Xc Clearly each
f
a
(n)
F(a,n) = 1}
is clopen in
by setting
(n E N) . and so
XC,
fa(n) E B.
Set S = {fa : a < w2}. Since fa # fs if a # g, Take a,$ < w2 with a 8. Then
U{fa(n) o fa(n)
ISI = K2
n E N}
= {F E XC : F(a,n) # F(5,n) a set which is clearly dense in
for some n E N} ,
It follows from the
X C.
first of equations (3) of Chapter 2 (page 29) that
V{fa(n) n f6 (n) and so
S
n E N} = 1,
satisfies the conditions in 7.30.
follows.
The result
a
We have now achieved the following goal. 7.34
THEOREM Assume that there is a model
there is a model, extending
Ut,
of
Al
of
ZFC.
ZFC + -CH.
Then
I
We noted in Theorem 4.19 the result of Godel that, if there is a model pt
a
which is a model of
of
ZFC,
ZFC + CH.
then there is a submodel of
We shall prove, in Theorem
7.38, that, if there is a model of ZFC then there is a model of ZFC + CH; we shall also prove a related result, Theorem 7.39, that will be required in Chapter 8. 7.35
DEFINITION Let
injective maps ran f c P(K),
K
be a cardinal.
Then
f such that dom f E K+ the power set of K. For
Q(K)
is the set of
and such that f,g E Q(K),
set
177
f 6 g
if
dom g c dom f
Certainly By 2.10,
f1dom g= g.
and
is a partially ordered set.
(Q(K),<)
it is denoted by
has a completion:
Q(K)
The partially ordered set (see page 31), and so it with its image in B(K),
we identify
is an injection; so that
(0(K),n).
is clearly separative
(Q(K),<)
Q(K)
Q(K)
is dense in
B(K)\J0}. 7.36
LEMMA Let
be the above Boolean algebra.
8(K)
have the following Boolean values in
(i)
'I(P(K))'I =
(ii)
Then we
V$(K):
= 1;
I(K+)V I'
[' P(K) _ (P(K)) V ' I] = 1;
(iii) (iv)
T' I K I '2K
(v)
= K' I] = 1; = (K) +' I] = 1.
Proof Set
A = K+.
(i)
Define
F
x x NO - B(K)
:
F(a,a) = V{f E Q(K) F
It is easily checked that (i) - (iv) of 7.26 (with
V{F(a,a)
:
by 7.26,
{f E Q(K)
:
a E dom f,
f(a) = a}.
satisfies the hypotheses
b = 1)
:
for example, for
a c K,
a < K} =
V{f E Q(K)
because
:
by
:
f(a) = a
a E ran f}
for some
is dense in
a E dom f} = 1 (Q(K),<).
(['I(P(K))VI = Ia1'I] = 1.
(ii) Let b = EE' P(K) # (P(K)) V' 1. By the fullness lemma, there exists
v E V$(K)
with
So,
178
TV C Ku A Iv j (P(K))"]J = b. Suppose, if possible, that
that there is a sequence
:
a
b # 0.
Then we claim
a << K> c Q(K)
such that:
(a) fs
for
(b)
f For suppose that constructed.
either
fa
[ a E V I]
A b
or
< 1a2vl] A b. a < K
Set
B < a> has been
and that
f = U{f8
:
:
defined, injective map from U{dom fs into P(K). B < a} Since A = K+ is a regular cardinal (page 87), dom f E K+. Hence f E Q(K). Since Q(K) is dense in B(K)\{0}, there exists fa E Q(K) with fa < f and with either :
fa < Ia E vI A b or fa < [a ¢ VI A b. The construction continues, and the claim is established. Set
f = fK,
a = {a < K Then
f < b
: fa
and
f < Ia E VI] Since
and set
(a E a),
f < Ia lE vl]
v
T V c K u % b,
Tvcal] Ab=A{[aEvl] Also
(a E K\O).
lLa c vl = A{[a E vu
:
Ab
a E a},
: aEK\v}. and so
f < [[ a c v l] A [[v c al] _ [Lv = al]
f < b A b' = 0,
(ii) follows.
a contradiction.
Thus
b = 0,
and
179
This is immediate from (i) and (ii).
(iii)
If the result does not hold, then there
(iv)
exists
and a term
a < K
VB(K)
t E
av
I ' t is a map from
V
V
K)
vK'
onto
I't c a x K'II
and this implies that
'P(a x
such that
b # 0,
]]
Now, from (ii),
b.
= Ma X K) ) V1 I] = 1, v
and so there exists
s c a x K
such that
It- s I] A b# 0.
Thus
is a map from
s
onto
a
K' I] > 0.
It is now easy to check (cf. 7.24(iii)) that
s
surjection, a contradiction of the fact that
a < K.
:
a - K
is a
V (v) K
1'2K =
By (iii),
is a cardinal' I] = 1, and so
The result follows. 7.37
CHI] = 1
( (K,)+. 1.
j1j(K"*)vj
I
THEOREM There is a complete Boolean algebra
in
By (i),
1.
such that
B
VB.
Proof Take
in the above notation.
B = B(Kol
By 7.36(v),
v
K
v
12 ° = K1' ]] = 1 7.38
in
VB.
B
THEOREM Assume that there is a model
there is a model, extending
In,
of
B
of
ZFC.
Then
ZFC + CH.
We have given a full proof, modulo Godel's completeness theorem, that
CH
is independent of
ZFC.
180
THEOREM
7.39
(CH)
There is a complete Boolean algebra K
K
T'2°=
2 1 = K2' Il
and
in
8
such that
V$.
Proof Take
in the above notation.
8 = B(K1)
['Kv1
By
1
7.36(iv),
= K1' fl
= 1,
and, by 7.36(v),
K
21
[ ' P(ttl) _ (P(K 1)) v' I] = 1, and so = I P(N) I v ' 11 = 1. But, by CH, P(N) P(N) I = K and so ' I P(N)) v K1 I' I] = 1. Hence O ' I P(N) I = IK1 it I] = 1. But Itil < K1' Il = 1, and so K v i.e., 1 ' 2 0 = K1' = 1. 1 EL' I P(N) I < K1' Il = 1,
By 7.36(u),
I
The remaining topic in this chapter is the delivery of the promised proof that
f{ AC I]
= 1.
We begin by introThe
ducing a notation for an already familiar construction. notation will be used frequently in Chapter 8. 7.40
DEFINITION Let
be a subset of
be a complete Boolean algebra, and let Set
B
dom S# = S, Thus then
u
S
VB.
S#
S#(t)
= 1
is a term in
(t E S).
VB.
For example, if
u E V,
= (dom u)
Note that, for each
t E V8,
I't c (dom t) #' ] = 1.
(13)
We must now contend with the somewhat messy problem of explicitly constructing terms for functions.
For
8,t E VB,
routine to check that
set w= {{8}#,{8,t}#}#.
[['w = (8,t)' fl
= 1.
It is
(Recall that, if
181
w = (s,t),
Now suppose that f
:
w = {{s},{s,t}}.)
then, formally, is a function.
S -> VB
is a subset of
S
f = {{{s}#,{s,f(s)}#}# Then
s E S}#.
Next, suppose that
is a binary relation' I] = 1.
ff'f
and that
VB
Set
f has the additional property that
Is = t]] t 1f(s) = f(t) ]] Then P f each
(s,t E S).
is a function with domain
(14) and, for
S#'I] = 1,
8 E S,
'the value of f at
is f (s) '
s
I]
= 1.
Thus is a function from
f[ 'f
Finally, if
S#
onto
f(S)#'I] = 1.
satisfies the condition that
f
1s = tI] = If(s) = f(t) I]
(s,t E S),
(15)
then
[['f is a bijection from 7.41
S#
onto
f(S)#'I] = 1.
THEOREM Let
be a complete Boolean algebra.
B
AC I] = 1 in
Then
VB.
Proof By 5.4,
is equivalent (in
AC
to the
ZF)
sentence 0: 'for each set
By 7.13,
that
dom y
[
= 1
I]
x,
there is a function
is an ordinal and in
VB
so it suffices to show that
for each axiom
TOOL] = 1.
y
such
x C ran y'. Q
of
ZF,
and
182
By passing to an equivalent term, we
u E V8.
Let
By the Axiom of Choice, there
may suppose that
dom u # 0.
is an ordinal
and a function
a
from
h
a
dom u.
onto v
V 6 < a}, define f S + V$ by f($) = h(B) Set S = {a (S < a), and let f be the corresponding term, as above. :
:
s,t E S.
Take
t] =o,
By 7.18(11),
s # t
if and only if
[[a = t I] < [[f(s) = f (t) l .
and so
Since
V
it follows that
S# = a,
v
is a function from
[['f
By 7.20(11),
[['a V
it follows that
° I]
=1
7.42
in
f
a
(dom u)#' I]
onto
is an ordinal']] = 1,
= 1.
and so, using (13)
has the required properties.
Thus
V8 .
NOTES
We have remarked that the relative consistency of ZFC + ICH
was an open question for 25 years after Godel's proof in 1938 of the relative consistency of ZFC + CH. The proof of Cohen was given in [11] and [12] and was expounded in (13]. The approach to Cohen's method through V8 was discovered by Solovay in 1965, and is given in [60].
See
(59, §61 and the forward to the second edition of [6] for some interesting historical remarks by Scott.
An early use of
Boolean-valued models to give a proof of Godel's completeness theorem is due to Rasiowa and Sikorski ([55]). [6],
(44],
Accounts of forcing are given in the standard texts [49], and [62], and in the article of Burgess (9]. We proved that it is relatively consistent with K
ZFC
that
2 ° = K2.
In fact, it is relatively consistent
that the values of 2K for K a regular cardinal can be any collection of cardinals subject to: (i) cf (2K) > K; (ii) if K < A, then 2K < 2X. See [44, Theorem 46] and [49, VIII.41.
The behaviour of
is still not completely resolved; given in [331.
2K
at singular cardinals some recent results are
K
183
8
ITERATED FORCING
We shall now complete our proof of the relative consistency with each algebra
ZFC
of the assertion that each norm on is equivalent to the uniform norm.
C(X,C)
To
do this, we shall construct a complete Boolean algebra A with
in
fMA + NDHI = 1 MA
We recall that
VA.
is Martin's Axiom (Definition
NDH is the sentence "For each compact space each homomorphism from C(X,C) into a Banach algebra is continuous" (Definition 4.18). The existence of discontin5.14) and that
X,
uous homomorphism from C(X,C) is equivalent to the existence of norms on C(X,C) which are not equivalent to the uniform norm. We have explained how the construction of such a Boolean algebra gives our consistency result. In this chapter, we write
value in
V8.
...I]8
for a Boolean
Let us discuss how we might construct a
Boolean algebra A with I MA IA = 1 in VA. Suppose, for example, that 8 is a complete Boolean algebra and that t is a term in V$ which is a counter-example to
I't
MA
in
V81
in the sense that
is a counter-example to
MA' I]8 = 1.
Then we shall show that there exists a complete Boolean algebra C containing 8 as a complete subalgebra such that
['t
is a counter-example to
MA'IIC = O.
184
In this way, we can eliminate the counter-example
in
t
VC.
The hope is that, by iterating this process, we can exterminate all possible counter-examples to MA. Of course, new counter-examples could well arise at each stage of the iteration, and these must also be eliminated. The final algebra
proceed.
A
This is how we
will essentially be con-
structed as a union of subalgebras, and will have all the required properties.
We shall arrange that
VA by "folding in" certain extra properties as the iteration I NDH fl
= 1
in
continues.
In preparation for this construction, we must develop a theory of "forcing iterations" to an extent sufficient to prove two iteration theorems, 8.12 and 8.14. This development is presented in pp.184-205.
However, the iteration theorems are stated in such a way that the construction of the Boolean algebra
A,
to be given in the
second half of this chapter, can be understood from the statements of the two theorems, without additional reference to the preliminary results.
We shall comment briefly on the history of the
results of this chapter in the notes 8.27, but we should say here that the initial development of iterated forcing is due to Solovay and Tennenbaum ([67]). All our basic results about iterations are contained in (67], but our details differ:
we are much more explicit, and, of course, we use a
less sophisticated language.
Throughout this chapter we shall frequently use the equivalence relation 'ti, defined on page 143: for u,v E VB,
if and only if I u= v I] 8= 1.
u ti v
DEFINITION
8.1
Let that
B
B
and
C
be complete Boolean algebras such
is a subalgebra of
algebra of
C
if, for each
C.
Then
S c B,
B
is a complete sub-
the suprema of
S
185
calculated in
8
and in
To show that
are equal.
C
is a complete subalgebra of
8
it suffices to check that, for each subset 8, \IS=1 in C.
of
S
C,
with
B
VS=1 in
For example, the trivial Boolean algebra (2) is a complete subalgebra of each complete Boolean algebra. Let V$ c Vt,
8
be a complete subalgebra of
C.
and it is routine to check that, for
1s E tjs = [[8 E t]C
and
Then
s,t E V8,
18=^ till= T a n ti".
The
rank of a term in V8 is the same as its rank in Vt. Many of the basic properties of terms are preserved in the passage from V8 to VC. This is exactly analogous to the case of the passage from V to discussed in '
We now give the general theorem which asserts
Chapter 7.
that properties specified by certain formulae are preserved. 8.2
0
DEFINITION
The set of A -formulae in r is the smallest set 0 of formulae of Y such that: (i) 0 contains the atomic
formulae;
(ii)
@
contains
contains
0
and
whenever
0
contains
p;
0 v
ip
and
(iii) I contains 0
and
xi and
whenever
-i0
3x i(x i E x xj
7
4'
A 4))
are variables.
Many simple properties of sets can (possibly with some effort) be specified by a ^ A-formula. 0
For example,
x c y, which is really Vz (z E x + z E y), is logically equivalent to i(3z (z E x n -1(z E y))) , which is a a -formula. 0
Properties which can be formalized by include 'x
is a function' and 'x
(but not 'x
A0-formulae
is a Boolean algebra'.
is a complete Boolean algebra').
A list of such
properties is verified in [49, IV.3.9], and we shall later give a list relevant to our discussion. Let 8.3
4)(xl,.... Xn)
be a
Ao formula.
PROPOSITION Let
0
be a complete subalgebra of
C,
and let
186
t1,...,t
Then
E V$.
Ictl,.... tn] jjB = LLctl,...,tn]
C.
Proof
The proof is by induction on the length of subformulae of . Suppose, for example, that the result holds for a
(x,y)
subformula
,
of
that
and that
2x(x E y A f(x,y)), t E V
x(y)
is
is a subformula of
x
[x[t] I]$ = V{11 E t]JB A [*[s,t] I]$
A I*[s,t]
= V{Is E tI]C
]]C
¢.
For
:
a E dom t}
:
s E dom t}
= Ix[t] ]1t, and so the result holds for i is routine. Let
u E
V(2),
The remainder of the proof
X.
where n was It is easily proved by induction on
defined on page 138.
and set
v = 7r(u),
v
rank that
u 'L v
(and so
V
is "isomorphic to
Thus, by 8.3, properties expressed by
Vc2)
A0-formulae are also
preserved in the passage from V to VB (cf. page 167) because they are preserved in the passage from V(2) to We now begin our discussion of iteration.
be a complete Boolean algebra, and define a term
G
Let in
by
dom. G = {b Then
j b E G ]] 8= b
claim that
:
b E B},
G(b) = b
(b E B)
and
(b E B) .
I G c B I" = 1.
We
V$.
8
V$
187 V
B 8']]8 = 1.
is an ultrafilter in V
1'8
To see this, note that by the above remarks on
is a Boolean algebra']]8 = 1 We then check that all
Ao-formulae.
of the defining properties of the statement
is an ultra-
'G
v
filter in
8'
have value
in
1
V8.
DEFINITION
8.4
The term
is the term for the generic object in
G
V$
Recall that a subset if, for each
and that a subset
if
V a P(B) ,
B\{0}
of
F
B\{O}
of
D
there exists
a E B\{O},
FnD#Q
is dense in
d E D
V-generic, where
is
(D E V) .
The following result gives the key property of (and it essentially characterizes
Let
G
be the term for the generic object in
V = {D - B\{O}
I 'G is
:
is dense in
D
vV-generic'
Proof Let D E V.
Then
EL'D n G # 0' 1B
Then
J B = 1.
is dense in $' ]] 8 = 1, and
EL ' D
=
8}.
V8,
V{Ea E G1$ : d E D} = VD = 1
because
D
that
Let F be a prefilter in a Boolean algebra 8 is a non-empty subset of B\{O} such that, if
F
a,b E F,
For
is dense.
This implies the result.
then there exists
a,b E B,
c E F.
Then
set tiF
c E F
with
I
and
c < a
a n c= b n c
if
a tiF b
with the operations
a
is denoted by
[a]F A
[b]
F =
[a n b)
B. [a] F,
F1
(so
c < b).
for some
is an equivalence relation on
equivalence class containing B/ti.F
G
G).
PROPOSITION
8.5
and let
8
with d< a,
The and
188
[a]F v
is a Boolean algebra;
[b] F = [a v b]F
it is denoted
by B/F. Let 8 be a complete Boolean algebra. The theory of iterations that we are to give depends on an analysis of terms
u
in
such that
V8
['u Let
is a complete Boolean algebra' 18 = 1.
be a complete Boolean algebra containing
C
as a complete subalgebra, and let generic object in
G
Since
V$.
is an ultrafilter in
T'vB
_
we have
(E'G
C/tiG
For
'I]8
is a subalgebra of
is a prefilter in
v
vC'I]
= 1,
is a Boolean algebra']
c E C,
dom [c] G = Is
define :
[c]G E V8
and so
= 1.
by
s E C1,
[c]G(s) = V{b E B : b A c = b n s } Let us stress that
[c]G
(s E C).
is a term (and not a subset of
C).
We have
I'm G
is the equivalence class containing V
c
in
V
C/,\, G' ]]
= 1.
We list the basic properties of the map 8.6
8
be the term for the
LEMMA Let b E B
and let
c,d E C.
c 1+ [c] G.
189
(i)
' [c v dIG = [c] G v
in C/tiG' v IS = 1,
IdIG
etc.
(ii) b < fE [c]G
if and only if
[d]G ]]$
b A c = b A d.
(iii) I [c]G = [d]GI
if and only if c = d.
=1
v
b < f' [c]G < [d]G in if b n c < b n d. (iv)
If
(v)
there exists
c E C
(vi)
T = {[d]G
:
t E V8 with
with
if and only
113 = 1,
I 't E
then
t ti [c]G.
Suppose that
d E S}.
8
C/tiG'
S c C,
c = VS, and that
that
Then
Ic]G = VT#
in C/tiG' jj8 = 1.
Proof
For (v), set c= V{s A I (SIG = tD$
s E C}.
The remaining results are immediate from the definition. 8.7
LEMMA With the above notation, v
T'C/tiG
is a complete Boolean algebra' I]$ = 1.
Proof
t E VB such that [ ' c = V{ j (d] G E t I] 8 A d : vd E C}. Take
set
check that [ ' Ic] G = Vt in
V
t c C/'' fl 8
C/ tiG' ] 8
= 1, and
Then it is routine to = 1.
Before proceeding, we recall from page 181 that, if f
:
S -* V$
is a function such that
Is = to = If(s) = f(t) then
where
E['f
f
is a bijection from
S#
(s,t E S),
18 onto
is the term corresponding to
f.
f(S)#'I]s
(2) = 1,
190
8.8
THEOREM Let
(i)
let
is
E VB
be a complete Boolean algebra, and
8
be a term such that is a complete Boolean algebra'
I'u
Then there is a complete Boolean algebra
1B
= 1.
containing
C
B
as a complete subalgebra such that
Du is isomorphic to C'/tiff' 10 = 1. Suppose, further, that
(ii)
ccc
is
B
and has
ti
cardinality
and that
2 °
K
['u
and has cardinality <2 °' 1B = 1.
ccc
is
Then we can suppose that
C
is
ccc
and has cardinality
2 0.
Proof
We write
IF
... I]
for I ... lB in this proof.
By the fullness lemma 7.12(11), there exists with
v E VB
I'v Take
is the underlying set of
a = p(v) + 1.
is E Va
:
The restriction of
to
form a subset
C*
of
by choosing exactly one element from each equivalence
class.
Then: if
s,t E C*
Let
t E VB
(a)
r E
ti
= 1.
1s E V1 = 1}
is an equivalence relation on a set:
Va
u' ]]
VB
by
with
with
a # t,
then
It E v I] = 1,
Is = t] # 1. and define
191
(w E dom r).
I r c t I] = 1. Since I t E v j = 1,
Clearly
Tw E tI] < V{jw E s]] V{ [w E sl]
we have
s E dom v}
Itcr
t E V8 with
if
(b)
:
V{ jw = $I
8 E VP(v)}
:
for w E V8, and so r E Va, and so: s E C*
w E t]]
r(w) _
dom r = Va(v),
= 1.
8
VP(,)}
r ti t. Further,
Hence
I t E vI] = 1,
:
then there exists
s v t.
with
*
Define operations
if
a1 A s2 = s 3 8
where
*A
1
V8
and
2
=s
:yv
if
3
A
and
on
v
by the rules
C
= 1,
tL *^(s 1,s 2, 83, u) 11 p
v (s 1' s
=1,
2' s 3' u)
Thus
were defined on page 155.
sl A s2 = 83 if 1'81 A s2 = 83 in u' I] = 1.
The
operations are well defined, and C, together with these operations, is a Boolean algebra, to be denoted by C*. Let
By the mixing lemma, there exists
b E B.
b* E VO with T'b*=1
in u'
>b
and
I ' b* = O in u' I] Since
Eb* E v I] = 1,
and then
b*
Set
subalgebra of
b' .
we may, by (b), suppose that b* E C*,
is uniquely specified by these properties. B* = {b* : b E B}. C*;
isomorphism, and so
the map 8*
Then
b F* b*,
H*
fl + 8*,
is a Boolean is a Boolean
is complete.
We first claim that
C*
is complete.
be a subset of C*, so that Is E v I] = 1 (8 E S) and [ES# cvI] = 1. By the fullness lemma 7.12(11), there exists c E V$ such that IL'c = VSO in u' I] = 1, Let
S
192
1. E S#]] > S#(s) = 1, that
and that
d E C*
'c
u
and so s< c s o d
c < d.
so
is complete.
C*
(b E T),
Thus
c = VS
This establishes that
in
C*,
and
is a complete sub-
8*
C*.
Take c = V{b*
Now suppose
C*.
Then
s E S } = 1,
:
Secondly, we claim that algebra of
s E S,
(s E S#1' ]]
d
u
= A l l ' s
in
(s E S).
d' I = I ' s
For
c E C*.
and, by (b), we may suppose that
T c B
b E T}
:
in
VT = 1
with
in
and
8
['b*
Clearly
C*.
and so for
b E T,
c = 1 in u' ]
>
' b* <
=
'b* = 1 in u' ]] > b.
I'c = 1
in
u'I] = 1,
c' ]] A I ' b * = 1
and so
c = 1.
in u' I]
This is
sufficient to establish the second claim.
We now choose a complete Boolean algebra
C conas a complete subalgebra such that there is a Boolean isomorphism ir: C - C* with ir(b) = b* (b E B). We finally claim that
taining
8
!'u
is isomorphic to
Take
c1,c2 E C
and
/tiGI I]
b E B.
= 1.
Then
b A c1 = b A c2
if and only if b* A 1T(c1) = b* A 7r(c2) in C*, 8.6(iv), if and only if b 4 J[c11G = (c21G]
I'b* A I(c1) = b* A 7r(c2) Since
j['b* = 1 in u' I] ) b,
and so, by
in u' ]] = 1.
(3) holds if and only if
(3)
193
= rt(c2) I] ) b.
7r(c1
Thus
[c2] GI] _ f 7r(c1)
rL [c1IG
Set S = {[c] G
f
c E C} and
:
S - D; f is well defined, for if C1 = c2
by 8.6(iii).
(4),
satisfies (2), and so
f
7r(c2) I]. (c)G 1+ 7T (c) ,
(c 11 G = (c2] G, then f is a homomorphism.
By 8.6(i),
Cv/tiG
where
f is the term corresponding to claim holds, and (i) is proved.
We write
(ii)
C
C*
for is
By
onto U11=1,
['f is an isomorphism from
We prove that
(4)
Thus the final
f.
2 0.
ccc
and has cardinality
t:
this is sufficient. Let S C C*\{O}
be an antichain in
S
and
s A t = 0
E'S#\{O} Since
j 'U
is
with s# t.
s,t E S
is an antichain in
ccc'I] = 1,
'S#\{O}
for
so that
C*,
Then
u']] = 1.
it follows that
is countable']] = 1.
By the fullness lemma, there exists
t E VB
with
Nv
['t
is a surjection from
Take
n E N.
Then there exists
I 'the value of t at For
sot E S
It n
with
onto
n
to E V8
with
is tn' I] = 1.
s # to I's A t = 0
= s I A Itn
S#\{O}'I] = 1.
t j < It n
in
u']] = 1,
=01
and so
194
I 't n
But Since
many
B
n
t :
Then a # 0, and so is and so for some n E N,
# O
]
n E N},
whence
Itn = sI]
Sn = {s E S
set
n E IN,
a C S.
is
S = U{S
For
S.
Take
# 0).
is
in
a
Thus
E S#\ {O }' I] = 1, and so It n = 81 A It n = t I] = O. is ccc, It n sI] = 0 for all but countably 0 I]
is a countable set.
S
# 1.
Thus
C*
ccc.
By hypothesis, li ' Iv I < t' I] = 1. Since IB I = t, it follows from 7.28 that li ' c < ' I] = 1. Thus, by the fullness lemma, there exists t E VB with ['t and, for each
is a surjection from
fs
s E C*,
takes the value
= 1,
a' I]
= 1.
define
v
:
v' I]
is the least ordinal at which t
For
onto
is E VB with
there exists
s E C*,
['t
C
c - B.
a F 11 a = is I] ,
If a # 6, then IVa = is I] n Isv= is I] = 0, and so, since B is ccc, {a : fs(a) # O} is countable. Since Tv
18 E v I] = 1, if
s,t E C*
that
s = t.
V{
8 = t8 I]
with
fs = ft,
Thus the map
a < c l = 1. This implies that,
:
then a
t
f
8
s
'
t,
and hence, by (a),
is an injection.
It follows that IC*I < 1{f E Bt : N
=
Since
1{a
:
f(a) # Oil ' H°}1
K
IBI °=t°=t.
IC*I = t, as required. = IBI, This completes the proof of the theorem.
IC*I % IB*I
195
We now fix a complete Boolean algebra complete Boolean algebra subalgebra, and a term
T' W=
C/ ti G'
(so that, by 8.7,
containing
C
8,
a
as a complete
8
such that
w E V8
10 = 1 is a complete Boolean algebra'
[['w
The next step is to define a class
V(w)
=1).
by
setting
Vow) = {t E V8
:
I't E Vw' ]JB = 1}.
(To be precise, ['t E VW' IJ 8 algebra and
belongs to
x
t, w) ]J B,
=
is the formula which formalizes
'y
where o (x, y)
is a complete Boolean
VY'.)
Our proof of the iteration theorem involves a map e
from VC
to V. We shall give a quite explicit
definition of
texts in logic avoid, perhaps wisely, the
e;
detailed verifications that we give, for they infer the existence of define
from general considerations.
e
eIVa by induction on the ordinal First set
In fact, we
a.
e(Q) = 0.
ejVa
Now suppose that
t E Va+1 \ V. Define
by
ft
dom ft = dom t,
has been defined, and take
and,
for s E dom ft, (5)
ft(s) = V{1[e(s) Then
ft
is a function from
1s E For
e(r) ]J$ A [[ r E tJ]C
tIC < ft(s)
61,82 E dom t,
:
to
C.
(a E dom
t).
dom t
r E dom t}. Certainly
(6)
we have
lEe(s1) = e(52) IJ8 < I (ft(81)1G = [ft(s2)1GI8:
(7)
196
to see this, it is sufficient by 8.6(ii) to check that
e(sl)
8
e(s2) ]
n ft(sl) =
Ie(s1) ° e(a2) ]
A ft(s2 (The reason
and this is immediate from the definition (5). for defining fail.)
ft
is that the analogue of (7) for
may
Now set e(t) = {{{e(s)}#, {e(s),(ft(s))G}#}#
(We are determined to be quite explicit!)
defined
t
e(t)
:
s E dom t}#
Thus we have
so that
D'e(t)
is a function with domain
{e(s) and so that, for each
:
s E dom t}#' I]$
= 1,
(8)
s E dom t,
['the value of e(t) at e(s) is [f t WIG' It follows that
I 'e(t)
The inductive definition of
term in 8.9
and so
E Vw'I]8 = 1, e
(9) e(t)
E V.
continues.
In summary, for each term t in VC, e(t) is a VS which is also a term for a "term in Vw". LEMMA
e(a) = e(t) ]]B s = tI]C (s,t E VC). (ii) For a E 1/C, fs(p) _ [p sIC (p E dom s). (i)
Proof
The proof is by induction on the position of in (Ord2,<), where (p(s),p(t)) p denotes rank in VC. Statement (i) holds if either
s = 0
or
t = 0
197
0 18 > 0 because 1[ e(s) is vacuous if t = 0.
if and only if p(s),p(t) > 0
Now suppose that for each p E dom. s.
p,q E V$ Then
with
s # 0,
and (ii)
and that (i) holds
(p(p),p(q)) < (P(8),P(t)).
Take
fs(p) = y{[e(p) = e(r) 18 A Ir E sI]C : r E dom s}
rJ]CA jr EsI]C
:
r Edoms}
(inductive hypothesis)
< T P E SIC,
and so, using (6), fs (p) = JP E s IC. Thus (ii) holds for B. A similar argument shows that (ii) holds for t. and
Write
b = [e(s) = e(t) I]8,
q E dom t.
Then, by (8) and (9),
and take
p E dom s
b A Ie(p) = e(q) I]8 < 1 [f8(p)IG = [ft(q)}GI]$,
and so, by 8.6(iv),
b A [e(p)
e(q) ]]$ A fs(P)
< b A Ie(p) = e(q) Thus, since (ii) holds for
s
and
I]8 A ft(q)
t,
b A Ie(p) = e(q) I]$ A [p E s IC
< b A [e(p) = e(q) I]a A jq E t]C
< b A lp = qI]C A jq E tIC (inductive hypothesis)
198
belongs to the domain of
1'e(p)
Since
e(s)'I] = 1,
we
have
b < J'e(p)
= V{ Ie(p) = e(q) II8 and so
:
q E dom t},
b A I p E sI' < b A I p E t]
p E dom s,
and so
b < It a sIIC,
This holds for each
.
Similarly, we have
b < is c tIIE.
and so (i) follows.
The induction continues. 8.10
e(t)'is
belongs to the domain of
I
LEMMA s,t E Vt.
Let
Then:
Is Etut=c if and only if
(i)
I' f e(s)
Is
(ii)
e(t)
IIW
=
(c] Go I S = 1;
t II t = c if and only if
e(s) = e(t) ]
'
=
= 1.
[c] G' ]
Proof
The proof is by induction on the position of (p(s),p(t))
hold for
in
(0rd2,<).
p,q E VC
with
Suppose then that the statements (p(p),p(q))
< (p(s),p(t))
in
(Ord2,<).
then (i) is immediate. Now suppose that p(t) > 0. Take c E C If
t = 0,
such that
e(s) E e(t) I] _ [c]G' IIB = 1, and, for
p E don t,
take
c
E C
such that
I' Ue(s) = e(p) III' A [ft(p)]G = [cp]G' II$ = 1 Set
T = {cp
:
p E dom t},
and set
S = {[cp)G
:
p E dom t}#.
199
Then
in w' ]]8 = I'VS = [c}G
L'VS = [VT)G
in w' 1]$ = 1,
and so c = VT = V{cp
:
p E dom t}.
(10)
By the inductive hypothesis,
I' Ie(e) = e(p) Iw = [([8 = p I for
CIGo I
$=1
and so, by 8.6(i) and (iii), we have
P E dom t,
Is = P1C A ft(p) = cp. By 8.9(11), ft(p) = Tp E tI]e, and so, from (10), c = Is E tI]e, giving (i). Next, one shows that
f le(a) c e(t) I" = We omit the details;
is c t]
= c
if and only if
[c] G' I8 = 1.
they are very similar to those just
given.
By permuting
s
and
t,
we obtain the second half
of (ii):
for this, the induction hypothesis is applied to
pairs
where r E dom t, which is allowed. The induction continues. I
(r,s)
We now show that
is
"onto
V(w)
up to
It is convenient to make the following
equivalence". definitions:
e
for
a, E Ord,
V(w) = {t E V(w) a
:
1['P (t) < a
in Vw' ]j$ = 1}.
It is easy to see by induction on rank that, if
t E Va,
then e (t) E V (w ) a 8.11
LEMMA
Let
that
I'e(t)
s E Vaw). Then there exists 'v s in Vw' JJI = 1.
t E Va+1 such
200
Proof
The proof is by induction on and
[01= V(W) c
V(W),
Now suppose that for each
6
with
S < a.
dom t = Va,
t(p) = c if For
Since
a.
the result holds for a > 1
For
and, for
e(o) = 0
a = 1.
and that the result holds
s E
V(W),
define
a
by t E VC a+1
p E dom t,
I't[e(p) E sI]L'
=
[c] G. I]8 = 1.
p, q E dom t,
I' [e(p) = e(q) ]]W A [e(q) E s]]' < le(p) E s]]W I]B = 1, and so, by 8.10(11),
I ' [ I p = gICIG A [t(q)] G < [t(p)IG'I B = 1. Hence
p=
qC A t(q) ( t(p),
p
t]]C = V{t(q) A Ip = qI C
By 8.9(ii),
t(p) = ft(p)
and so :
q E dom t} = t(p).
(p E dom t).
It follows from (8) and (9) that
I'the value of e(t) at and that
e(p)
is E[e(p) E sI]w' IS = 1,
I ' I e (t) c s I w = 1' I] 8 = 1. To complete the proof, we must show that
[ s c e(t) ]]w = 11]18 = 1. Suppose that this is false. Then there exists r E V0 such that b > 0, where '
b = I'rEdoms']]8 A J' jr E sI]W A jr j e(t) I]w > 0' I]s.
201
Since
aE
V(w ),
we have
a
and so there exists
a E B
b C I'r E dom 8 A p(S) < a' I]a. and 6< a
with O< a 4 b
such
that
[['P(r) < Choose
q E V$ with
a
in Vw' I]$ = a.
[q = r I] 8 >a and
{q = 010 > a' .
V
['p(q) < g in V'' 10 = 1, and so, by the inductive C VC such that hypothesis, there exists p E VC $ a $ B+1 a, I 'e (p) ti q in Vw' ]] = 1. We have I e (p) 2 r I] Then
and so
[[ ' [e(p)
E
8Jw A E[e(p) j e(t)
By definition,
Dw
dom t = Va,
follows from (9) and the fact that
> O' I]8 >. a.
and so
(11)
p E don t.
It
that
t(p) = ft(p)
If'Ife(p) E e(t):n' > (t(p)]G'IS = 1. Thus
If' Ie(p) E e(t) IIw 3 Ie(p) E aI W' 118 = 1. But this is a contradiction of (11). The proof of the lemma is complete.
4)(x1,...,x) be a formula.
Let 8.12
i
THEOREM Let
8
be a complete Boolean algebra, let
be a
t
complete Boolean algebra containing $ as a complete subv algebra and let w E V8 be such that I 'w = CA,G'I] = 1. Then, for
t1,.... tn E VC
and
c E C,
[Egt1,...,tnI IIc = c if and only if I ' I0[e(tI),...,e(tn)] IIw = [c]G' ] 0 = 1. Proof Let
0
be the set of subformulae
V
of
0
such
202
that the result holds for ,. By 8.10, contains all the atomic formulae in $. 0 By 8.6(i), contains i v X whenever 0 contains i and 0 and
X
is a subformula of
v X
whenever
0
contains
and
contains
0
c.
Similarly,
mp
is a subformula.
Suppose that contains p(x1) and that 3x1p is a subformula. Set c = 1 3x l* I] E, and take d E C with I' 13X J W = [d]al I] 8 = 1. By the fullness lemma, there exists I t E VC with f * [ t] 11 C = c. Since I' 14 (e(t) II w
I o = 1, it follows that c < d. By the fullness
[c]
lemma again,
'there exists t E Vw with J4 [tl
[d] G' J] s = 1,
fl u ' =
and so, by a third application of the fullness lemma, there exists
s E
V8 with
I' j * [ s] j W = [d] Go I S = I ' s E Vw-18 = 1. Now
s E
Ow),
u E VE with I i [ u] IC = d, 3xly, E
and so it follows from 8.11 that there exists
in
E[ ' e (u) ti s
and so d < c.
Vw' II
8
Thus
Since * E c = d, and hence
= 1.
,
(P.
It follows that all subformulae of
and in particular the result holds for completes the proof of the theorem.
belong to
¢
@,
This
itself.
I
We need one last lemma before we can complete our analysis of
V(w)
in terms of
Vf.
u E V8 with
Let
I l u is a complete Boolean algebra' II
a
= 1.
vt
For
t E Vol
we have
E V8.
But inside
v
"t
relative to
u".
V8 we can form
In an attempt to avoid confusion, we
denote this latter term by defined up to equivalence in
(t)u
V, 8
Of course,
(t)v U
is only
but this will not matter
203
because
8.13
will only appear in expressions of the form
(t)v
LEMMA
Let t E V$ .
J ' e (t) ', (t) v in
Then
Vw' ]]
= 1.
Proof
Let s E dom t. The main point of the proof is that the following equation holds:
e(a) E e(t)I]w = To see this, set
(s)' E
1. (12)
b = Is E t] Is.
By 8.10(1),
[[ ' fre(e) E e(t) I]w = [b]G' is = 1, and so
[E' ff e(s) E e(t) jw = 1' I]8 = b,
Ie(a) E e(t) Iw = O' II$ = b'. Also, we have
If 'If (a) I (a)w
1' ]J$ = s E tJj = b, (t) ;I]w = 0' 1$ = 1s
tIB = b'o,
and so (12) follows.
We now prove the lemma by induction on result is trivial if
p(t) = 0.
Suppose that
that the result holds for each Take
a E dom t.
p(t).
The
p(t) > 0
and
with
p(8) < p(t). s E V$ By the inductive hypothesis and
(12),
lL'tLe(8) E e(t) I W = Ife(s) and so, taking account of (12), we have
(t)vIW' $ = 1,
204
e(t) c (t)w ]
= 1' II$ = 1.
Also by the inductive hypothesis and (12),
(s)w E (t),Iw = I(s)" E e(t) I W' 113 we have
It c (dom t)#I]g = 1,
and so, since
I' I (t)w c e(t) 1w = 1' I$ = 1. The result for
follows, and the induction continues.
t
Let
4(x1,...,xn)
be a formula.
We need an observation: complete Boolean algebras, if isomorphism, and if
n
it(I4(a1,...,a] I81) 8.14
let
if
81
it: 81 -r 82
a ,...,a 1
I
E V,
=
and
82
are
is a Boolean
then
TI 10[a1...'an] 1182
(Iteration theorem)
THEOREM (i)
Let
8
t1,...,t
and
u
be a complete Boolean algebra, and be terms in VS such that
is a complete Boolean algebra' D S
EL'u
[(t11u,..., (tn)u] Th = 1' 11, = 1. Then there is a complete Boolean algebra C containing a complete subalgebra such that EL 4[tl...,tn] IC = 1. Suppose, further, that
(ii)
8
ccc
is
and has
ti
cardinality
2 °
and that K
EL'u
is
ccc
and has cardinality
Then we can suppose that Ro 2
.
C
is
ccc
8
< 2 °'I]8 = 1.
and has cardinality
as
205
Proof
By 8.8, there is a complete Boolean algebra containing 8 as a complete subalgebra such that is isomorphic to
'u
t
C/tiG']]$ = 1.
V
Choose I 'u
with
w E VB
11'w = E/,v G']]B = 1.
is isomorphic to
$ = 1,
w'
Since
the above observation
shows that
IjW = 1']]$ = 1.
]]u = l' ]]B = 1, and so, by
By 8.13, Theorem 8.12,
t[4[tl,.... tn] ]]f = 1. This follows from 8.8(11).
(ii)
1
Our strategy for building the complete Boolean algebra A for which I MA ]]A = 1 is to express A as a union of complete subalgebras which are themselves built by successively eliminating potential counter-examples.
To
carry out this strategy we shall require the following result. 8.15
PROPOSITION Let
be an ordinal and let
F
sequence of complete
ccc
:
a complete subalgebra of
U{B9\{O}
(b)
a
Let
:
U
Then
Ba.
8
is dense in
8 < a)
is a limit ordinal with B =
a < > be a
Boolean algebras such that:
whenever
BS
whenever
Ba
a < E.
is a
ccc
Boolean algebra, and
B
a
is complete in the case where
cf
wl.
Proof
First suppose that
cofinal in
E.
cf
= w,
Let A be an antichain in
is an antichain in
Ba , n
say B.
and so is countable.
:
n E w>
is
Then A n B a Since
206
A = U(A fl B
A
),
a
is countable, and so
Secondly, suppose that
is an uncountable antichain in of cardinality
C
contradiction.
Thus
A
for some
a
We now come to the main case:
1;
=w
Tr
7r
a
f;,
suppose that
. 1
For
Then
a <
By passing to a subsequence, we may suppose that
= w
cf
A contains a subset Suppose that
ccc.
is
B
> w1.
Then
8.
C c B
But
Ni.
cf
ccc.
is
B
n
b <
a
b E B
(b) = A{c E B a (b),
it
a < w,
and
a
w a
(b) > Tr8(b)
:
(b) E B
set
b 4 c}. because
a
8
a
is complete, and Ba
is a
a < B < w1 because
whenever
complete subalgebra of
86.
B. For a < B < w a < w1, set and, for
Let A be an antichain in set
= A n (B S\Ba) ,
Aa B
Aa = A \ (A fl Ba) = U {Aa f B : a < B1. We claim that, for each there exists B E ((I, w) with a<w = Vna(Aa,s).
VITa(Aa)
For set
dB = VTr (A a
a'8
).
Then
VTr
(A ) a a
= d,
d E B
:
a < B < w
ccc
such that
>
is an
1
8
increasing sequence in the there exists
Boolean algebra
and so
8a,
= d
Clearly
eventually. a a and this establishes the claim. d
By 5.7, there is a limit ordinal n < w1 such that, for each a < n, there exists a E (a, n) with VTra(Aa)
We now claim that A c Bn.
= VTra(Aa'S) .
For suppose
not, and take aE A\Bn. Then nn(a) E Bn\{0}, and so, by hypothesis (b), there exists a < n and c E Ba\{O} with c < TrTI (a) . d E Aa,n,
We have a c d',
VTra(Aa) = V7a(Aa, n) . and so
(a) < d'.
it
For each Thus
c' E B
n
d . c',
and so
n (d)
c A (VTra(Aa) ) = 0.
< c',
i.e.,
But a E Aa,
and so we have a contradiction.
and
c A Ira(d) = 0.
a
and
Hence
Tra(a) ) Trn(a) ) c,
Thus A c Bn.
Since
Bn
is
207
A
ccc,
is countable, and so
Now suppose that set of
B.
increasing sequence in d,
d = VS
and so
in
8,
Then
be a sub-
S
a < E>
:
is an
and it eventually takes the
8,
say, because
constant value
and let
cf & a Wi,
da = V(S n Ba).
Set
ccc.
is
$
8
is
is complete.
8
In Definition 8.2, we defined
Proposition 8.3 we showed that, if
Clearly
ccc. I
Ao-formulae, and in is a
¢(x1,...,xn)
A0-formula and if 8 is a complete subalgebra of a complete 8 = Boolean algebra C, then 1 0[tl,...,tn] i
[tl,...,tn1]]C
whenever
We now list
t1,...,tn E V8.
some typical properties which can be specified by A formulae 0and to which this result will be applied. The verifications are straightforward and tedious, and will be omitted. 'x
is a function from
(ii)
'x
is a bijection from
(iii)
'x
is a partially ordered set, and
an antichain in (iv)
(v)
x ; y
:
z
: x ; y
to y
z'
to
z'
y
is
x'
'x
is a partially ordered set,
family of dense sets in
z
y
(i)
'x
and
x,
y
z
y
is a
y-generic filter in
is a
x'
are partially ordered sets, and
is an embedding' 'x and y are totally ordered sets, and is an embedding whose range is cofinal in y'
(vi)
y
belong to
(vii)
'x
and
(viii)
'x
is a pregap in
(ix)
'x
and
y
NIN
and
x
(NJN,
are equivalent pregaps in (N3N,
We have explained that our strategy for building a complete Boolean algebra
A
for which
I MA ]
A
= 1
is to
successively eliminate counter-examples to MA. At the same time, we shall build the algebra A so that L NDH I A = 1: our preliminary results in Chapter 6 show that, to do this, it
208
is sufficient to eliminate potential embeddings from
(R,<)
and that this follows if we ensure that into has a Aausdorff gap each embedding of (Q,<) (NIN,
(NIN,
in its range, and so cannot be extended to an embedding of (We realize that to use fraktur letters for the
(R,<).
underlying sets of the partially ordered sets
and
(Q,<)
conflicts with our present notational convention, but
(R,<)
our typographic possibilities are finite:)
In the preface, we mentioned that the style of analysis that we must undertake in the present chapter requires a certain precision unusual in other areas of The next definition, which is a slight re-
mathematics.
formulation of part of Definition 6.4, exemplifies this. now say that A u B
if
r =
(H11 1)-pregap
is an
is totally ordered, if
A < B,
in
We
(NIN ,
if there is an
isotonic map such that ran nl and (w1,E) - (NIN ,
such that ran Tr2 and B are (w 1, E) -+ (NIN ,
is that we can see that each of the stated properties of r, Trl,
and
Tr2,
the domains of
71
save for those giving the cardinality of and Tr2, can be given by a A0-formula.
For example, to say that ran 7r1
and A
is an ordinal and that
dom 7T1
are mutually cofinal in
A0-property of
and
r
Now properties specified by preserved in the passage from 81
may not be preserved.
in
(NIN,
'r is an
'r is
an
is a
(NIN,
Trl.
a0-formulae are but the cardinal
to
V
Thus, if
V8, r
is an
(H1,H1)-pregap
then (N1,M1)-pregap in (tt
0
,t{0 )-pregap in
But now suppose that
0
is
(NIN,
(NIN,< )' 8 = F
K
1
= NNO
Then, by 7.27, cardinals
ccc. v
are preserved in
V8,
generally, if
is a complete subalgebra of a complete
0
and so
I'ttl = K1'I]8 = 1.
More ccc
209
Boolean algebra passage from
then cardinals are preserved in the
C,
to
VB
[ ' t is an
VC,
and so, for
t E V$,
(x1,$1)-pregap in
(N a',
I (12)
4 1'...' It Now let
r =
Then there exist maps
Tr1
be a Hausdorff gap in (N IN,
additional properties that Tr1(a) < 7r2(a) (a < W1) and that, for each a,$ < w1 with a # S, either Tr1(a) % Tr2 (B) or 1x1(8) jC Tr2(a). These additional properties of w1 and Tr2
can also be expressed by
Thus, if
Do formulae.
complete subalgebra of a complete then
l o t is a Hausdorff gap' ]] B < Notice that we have "S" is because there are more terms in
B
Boolean algebra
ccc
' ... ' ] t .
VC
are potential witnesses to the fact that pregap or a Hausdorff gap, respectively.
t
VB
is an
C,
(13)
in (12) and (13): than in
is a
this
which (K1'K1
In fact, equality
holds in (12) because cofinalities are preserved, a result we shall not need, but equality may fail in (13), and it is this failure that we shall exploit in our construction of a IL NDHIIA = 1. complete Boolean algebra A such that We now give some analogous remarks about partially Let B be a complete subalgebra of a complete Boolean algebra C, and let a,t E VB. Then ordered sets.
'...
['t
is a partially ordered set' I]B =
['a
is an antichain in to I s = It ... , It.
and
Further, if
C
is ccc, then
Pa is uncountable' II B
=
'
.'
II C
,_C
210
because cardinals are preserved from
is a
E[ ' t
we have
partially ordered set 1 B
ccc
VC,
>
' ... ' 11 E;
(14)
than in
Vt
VB.
LEMMA
8.16
Let set
and so
to
in (14) because there are possibly more terms
">"
for uncountable antichains in
P,
V0
be the completion of a partially ordered D be a family of dense subsets of P. Then
B
and let
D-generic filter in
EL'there is a
P'11B = 1.
Proof Let
Tr
preserving map.
dense in
:
Since
is dense in
w(P)
for each d E D, Define F E VS by
and so
B
v
dom F = {p
:
p E P},
It is easy to check that Take
be the associated order-
P - B\{O}
d E D.
T'F
w(d)
B,
V{TT(p)
V
F(P) = ?r(p) is a filter in
:
is
p E d} = 1.
(p E P) . P'1]B = 1.
Then
I' F n dv # 0' 10 = V{7r (P) and this gives the result.
:
p E d} = 1,
i
Let 7r1 and Tr2 be isotonic and anti-isotonic maps, respectively, from (w1,E) into (NN,
is a pregap. Set fa = TrI (a) and for a < wi, and set t = (Trl,7r2) . Let Q(t)
ga = Tr2 (a)
be the partially ordered set with the rather complicated definition that was denoted by Q() in Definition 6.11. a The underlying set of Q(t) is Q(t). For t < wl, set {q E Q(t)
:
E E supp q},
211
and D(t) = {d(t,O)
By 6.13, each
d(t,&)
:
is dense in
shows that, if there is a
g a
:
a < w
a
& < w1}. Q(t).
Further, 6.13
D(t)-generic filter on
1
8.17
(N IN,
LEMMA Let s = be the pregap in specified above. Assume that s is actually a gap
in (NN,
.
Then
's is a Hausdorff gap in where
then
Q(t),
is a Hausdorff gap.
>
B
denotes the completion of
I]$ = 1, Q(t).
Proof Since ccc.
is
s
By 5.16,
I's
is a gap, Theorem 6.12 proves that 8
is a
Q(t)
Boolean algebra, and so
ccc
is an (8 1,K1)-pregap in (NN,
By 8.16,
'there is a
D(t)v-generic filter on
It is clear that for
f,g E NN,
and so
['Q(t)"
(15)
[['[(f,g)]V = [(f,g)]'11s = 1
where we are using the notation of Chapter 6, Also, for each < wl, (tv)']]8 = 1.
I Id(t, v _ {q E Q(t) and so
Q(t)v'119 = 1.
t[ 'd(t,E)v
V E
d(t,v,)'110 = 1.
J 'D(t)v = D(t)' 118 = 1,
'there is a
:
E
supp 4}'
11 B
= 1,
Hence, we have
and so, from (15),
D(t)-generic filter on
Q(t)'118 = 1.
212
The result follows from 6.13. 1 8.18
LEMMA Let
(CH) s
be an
there is a complete K1
(NN,
(K1,K1)-gap
ccc
in
such that
'sv is a Hausdorff gap' I] = 1. Proof
Choose maps (NIN,
and 2 from
ir1
s = . Set t =
and let Q(t)
(Or1,ir2),
The Boolean algebra
as described. Q(t).
into
(w1,E)
with the properties specified above such that
As we noted after 6.11,
be
is the completion of
B
and
IQ(t)I = K1
Q(t)
IBS 4 2 0, separative, and so JBI > K1. By 5.16, with CH, IBS = K1. The result follows from 8.17.
is
and so, 1
We now prove a lemma by using the iteration theorem.
This lemma is a basic tool to be used the construction of a complete Boolean algebra
A
with
I MA I]A
for it shows
= 1,
that we can eliminate a counter-example to
MA by passing to
a larger Boolean algebra. 8.19
LEMMA Let
cardinality
8
K1,
'p
(CH)
be a complete let
is a
p E ccc
VB
ccc
partially ordered set of
cardinality d E V8
and let
I'd
Boolean algebra of
be such that
K1, 11B
= 1,
be such that is a family of dense subsets of
p']]8 = 1.
Then there is a complete
ccc
ality
as a complete subalgebra such that
K1
containing
B
Boolean algebra
C
of cardin-
213
('there is a d-generic filter in
p' I] C = 1. (16)
Proof By the fullness lemma, there is a term
u
in
V$
such that
J' u is the completion of p' 18 = 1. By 8.16,
I' [['there is a
(d)U'-u generic filter in
(p)u'DU = 1'I]$ = 1, where we are using the notation of 8.14.
By 8.14(1), there
is a complete Boolean algebra
g as a C containing complete subalgebra such that (16) holds. By 5.16,
< 2 °'I]$ = 1,
and has cardinality
ccc
is
'u
and so, by 8.14(11), we can suppose that
C
ccc
is
and has
MO
cardinality
2
b < a
and
CH,
C
has cardinality
a E A
b E B1.
b E B2
and
1
LEMMA Let
cardinality
8
hit
be a complete and let
Then there is a complete K1
b < a
for some
and
P
.
(CH)
Pt is an embedding of
ality
or
In this case there is no element of
which interpolates both 8.20
K1. f
for some
a E A2
With
and be pregaps in (P,<). and are incompatible if either
Let
Then
.
containing
('the range of
t
8
ccc
t E V$ (Q,<)" ccc
Boolean algebra of be such that into
(N N,
Boolean algebra
C
of cardin-
as a complete subalgebra such that
contains a Hausdorff gap'
1C
= 1.
214
Proof The set
J
x E J,
Let
was defined on page 121.
and let be the corresponding pregap, defined on x (K1,K )-gap in page 121. By 6.22(iv), is an x
and so, since
(Q,<),
IIV
is an
1
ccc,
is
B
By the fullness lemma, there is a term
by the image x,y E J
If
ax
and
v x # y,
then
given
,
(IN
under
such that
VB
in
($1,81)-pregap in
is the
I 'sx
(Q,<)v' D B = 1.
(K1,8 1)-pregap in
t' I
x
B
= 1.
>
and
are incompatible, and so
T ' s x and
s
y
are incompatible in
(N 3N, < F) ' 1 B
= 1. (17)
Set
H = {x E J We claim that cardinality ICS =
H2
I'sx is a gap in
IJ\HI < K1. K11
By 7.29(ii), B
JCH]]B = 1,
is ccc, and so
x E C
ON
3N,
has
B
C c J\H
with
and
there exists
fx interpolates
IL ' I N ' j
]]B = 1}.
x E C,
By the fullness lemma, for each such that
E N3' and
and
b E B\{O}
I ' sx can be interpolated in
X
Since
For suppose not.
there exists
such that, for each
I 'f
ON 3N,
fx E VB
sx' 18 = b.
T'K1 = Ni' V = 1 because
= K 1' J B = 1.
Thus, by many
applications of the fullness lemma, there is a sequence
a < wl>
of terms in
VB
such that, for each
f E VB,
215
E N4' IB = Vi If = faIB So, for each
and
ICI =
and
y
interpolates
and so there exist
ti 2,
such that
a < w1
fy I] X
a contradiction of (17). By 6.22(1),
Take
x E H. in
u
and so
IS A {'fy
s X1
A If
is a term
B
1
fy 18 A b # 0,
[ fx x
But
a < wl} 3 b.
f01"= Ifyfa18 and Ifx= faI]$nb#O.
Ifx
T 'f
:
wl}.
x
has cardinality x,y E C with x
Thus
V{If = faI]B
x E C,
a<
VB
sy'I]B
O,
Thus the claim holds.
K JJJ = 2 1,
H
and so
0.
By 8.18 and the fullness lemma, there with
is a complete
I 'u
interpolates
Boolean algebra of
ccc
cardinality
K1'
is = 1
and
I l J ' (sx) u is a Hausdorff gap' I] u = 1' I] B = 1. By the iteration theorem 8.14(1), there is a complete Boolean algebra
C
containing
as a complete subalgebra
B
such that
['s
x
is a Hausdorff gap' IC = 1;
by 8.14(11), we can suppose that
cardinality
C
is
ccc
and has
K1.
Since the relevant properties can be expressed by A -formulae, it follows that 0
I 't
is an embedding of (4,<)' into
(39'N,<
F
)' IC = 1
216
and that
11sx Thus
is a pregap in the range of
has the required properties.
C
Suppose that that:
(i)
f
i
is a function such
dom f + {O,1}
:
is an ordinal;
dom f
(ii)
(iii)
{a
subset of
dom f
with no maximum element.
f(a) = O}
is a non-empty, proper
properties of f can be specified by set Q is just the set of functions properties with
dom f = w1
Since the order on
f
satisfying these
(cf.Definition 6.21). x E Q.
Let
B
Then
'dom x = wl' I]8 = I'wl = w1' I]8.
=
can be specified by a
Q
follows that, for each complete
1'(Q,<)v
Now each of these
A0-formulae, and the
be a complete Boolean algebra, and let
'x E Q' I]8
is eventually
f
constant;
:
= 1.
t' fl
ccc
A0-formula, it
Boolean algebra
is a suborder of
8,
(Q,<)'1J8 = 1.
(18)
Recall that it follows from the main result of ZFC + NDH,
Chapter 6, Theorem 6.25, that, to build a model of it suffices to build a model of
ZFC + MA + 1CH
there is no embedding of
(R,<)
into
(Q,<)
is interpolated in (R,<). ccc Boolean algebra,
each
(K1,41)-pregap
We thus see that, if
in 8
in which By 6.22(11),
(NIN,
is a complete
and if
['the range of each embedding of
(Q,<)V
into (19)
(N3N,
contains a Hausdorff gap' 118 = 1,
then
I' (R,<) does not embed in To build <8a
a < w2>
8,
(N]N,
(20)
we shall construct a sequence
of complete
ccc
Boolean'algebras such that,
217
if
a < g < W2,
If
y
then
is a complete subalgebra of y < w2, we shall take
8a
is a limit ordinal with
to be the completion of By 8.15,
a
will be a complete
8
shall assume
cardinality
and we shall set
8y
8 = aU
8a 2
Boolean algebra.
ccc
and we shall choose each
CH,
8S.
We
to be of
8a
tit.
Suppose that
has been constructed and that
8a
r
8
is a term in
such that
V a
8 'r
is an embedding of
(NN,
into
(Q,<)"
Then by 8.20, there is a complete ccc Boolean algebra of cardinality tit containing 8a as a complete 8a+1 subalgebra such that 8
[['the range of
contains a Hausdorff gap' I] a+1 = 1.
r
The key point is that, if C is any complete ccc Boolean algebra containing 8a+1 as a complete subalgebra, then it follows from (13) that
['the range of Thus, if
r
contains a Hausdorff gap'
has been eliminated at stage
r
1C
a + 1,
= 1.
it can
never trouble us again. In a similar fashion, we can use 8.19 to eliminate in
V8a+l
in some
MA which occurs
a potential counter-example to 8
V
a;
we shall again see that, if
C
is any
complete ccc Boolean algebra containing 8+1 as a complete subalgebra, then the counter-example remains eliminated in
VC.
even though
One problem still remains:
it does not follow that that all the terms in 8
terms in
V
aVw 2
a.
V
V8
S a.
$ = aY w
2
V
8 =
U
8 It
a<w2 a But we shall show
that concern us are equivalent to
We must deal with:
(i)
terms for
218
partially ordered sets with underlying set
w1;
(ii)
terms
for families having cardinality K1 of subsets of w1; terms for embeddings from (Q,<)v into (N IN,<E). (iii)
In
dealing with these cases, we shall fix some notation that will be used in the proof of the main theorem.
eF,
in
Let
8
Let
F
V8
be a complete :
Let
H
:
define a term
Da
and define a term
dH
in
Dg = {d
in
V8
Define a term
V8
)
= F(a,6)
For each
by
d, (n)
=
by
< w1},
< wl}#
D5(ds) = 1
in the notation of 7.40).
x N x N - B be a function. Let J : define a term f in V8 by
For each
x
dom fX = { (m, n) v and define a term
t i
in
:
m,n E N} , fX((m,n) v) = J(x,m,n),
V8
tJ = {{{x}#,{x,fx}#}# 8.21
be a function.
n < w 1,
:
dom D$ = {d :
x E Q,
a,s < wl}, eF((a,8)
:
w1 x wl - B
dom d; = {n
(so that
Boolean algebra.
by
dom eF = {(a.R)v
< wl,
ccc
be a function.
wl x wl -s B
LEMMA Let 8 (1)
TIP
by
:
x E Q}#.
be a complete Boolean algebra.
For each
p E V8
such that
is a partially ordered set with underlying
set
1 8 = 1,
219
there is a function
F
wl x wl -. B
:
TIP is equal to (ii)
For each
I'D c P(wl) there is a function
H
(wv
1,eF)' I]8 = 1.
D E V8
and
such that
such that
IDI < hv1' I]8 = 1, such that
w1 x wl + B
:
TD= DHI]8= 1. (iii)
For each
It is an embedding from there is a function J
t E V8
(q,<) v
such that
into (N3N,
x N x N + B such that
t = tJI]8 = 1. Proof
W
Since
8
is ccc,
v
I'Ii 1 = til' j8 = 1,
so, by the fullness lemma, there is a term
I'p
is equal to the partially ordered set
e E V8
(w1,e)'I" = 1.
Define F : (a, S) r+ t[ (a, 8) v E e I] 8, wl x wl + B. ft e = eF ]] = 1, and the result follows. (ii)
t E V8
Then
By the fullness lemma, there is a term
with T It
and, for each
is a function from C < w1,
wl
there are terms
onto
and
with
D'I]8 = 1,
s4 E V8 with
I'% is the value of t at ;"j 0 = 1.
220
Define
Inv
fl
< w, [[8 C = dRI]
each
(iii)
= 1,
Define J
I D = D5I]B = 1.
and so
Q x N x N+ B
:
Then, for
w1 x w1 + B.
I]B,
E
H
J(x,m,n) = I 'the value of
xv
at
t
by takes the
value n at 1' 18. tJ and of
From the above definitions of
fX,
is a function with domain
't
= 1
Q1 I]8
and
ti at
I'the value of
Note that, if 8
1' M
= 1,
is
1
in the above lemma is ccc, then w and K 1 can be replaced by 1
8
and so
' ... '
respectively, in expressions
1,
= 1.
fX' I]
I t = tJIB = 1.
It is now routine to check that
I'w1 = w 1 and
xv
is.
We now come to our main theorem. THEOREM
8.22
(20 = M1
2x1 = K2)
and
There is a complete Boolean algebra
A
such that
I MA + NDH I] A = 1. Proof
For T;
X N X N
of functions from Then
S = U{SG
set
w2,
w1
SC =
CW1
:
to
Write
x).
C < w2),
S = S
T = U{T
CW1
and set
x' denotes the set
(where we recall that y
X
:
w2
and
T = T W2
S < w2).
M1x1
Also, for
< w2,
= 21,
1SCI =
and so, by hypothesis,
M1x1 (SCI = M2.
Similarly,
IT ;I
=
= M2
because
221 K
Thus
2 0 = K1.
of S U T such that {a
a < w2> w2
Fix an enumeration
ISI = ITS = M2.
for each
x E S
U T
:
:
is cofinal
za = x}
this is possible because
IW21 = IW2 X W21 to be a sequence of Boolean algebras of cardinality K1 such that, a < B < w2, then Aa is a complete subalgebra of AS,
We shall define
complete if
ccc
and such that, if B is a limit ordinal with B < w2, then Aa is the completion of aUSAa. Further, each underlying
set Aa
is chosen to be a subset of w2. The definition is by recursion on a, and it is guided by the enumeration ,
which is used to "anticipate counter-
examples". A0
Choose
of cardinality
to be a complete Ac c w2.
ccc
Boolean algebra
with
K1
Suppose now that
B < w2
We define
has been defined.
As+1.
Suppose that
Case 1.
and that
There are two cases.
zs E S,
say
zs = (F,H),
wl
where
F,H E w2 If
ran F U ran H
As,
set
If
ran F U ran H c AS,
let
corresponding terms to
F
and
H,
As+1 = A8. eF and D
as above.
R
be the
Set AS
v
a = Lt'(w1,eF)
b = f 'DH If
a < 1
ccc
partially ordered set']]
is a family of dense subsets of b < 1,
or
suppose that
is a
then again set
a = b = 1.
Boolean algebra
of cardinality
C
AS
(wife F)'I]
AS+1 = AS.
containing
AS
ccc
as a
complete subalgebra such that I 'there is a set
AS+1 = C.
can suppose that
DH-generic filter in
Since
1ASI = Hi,
AS+1 c w2.
(wife F)'I3C = 1:
Iw2\A0I = N2,
.
Finally,
By 8.19, there is a complete K1
,
V
and so we
222
Case 2.
JEW
to
N
If
ran
J g4t A6,
set
A6+1
If
ran J c AS,
let
tJ
then again set
a < 1,
of cardinality
where
= J,
A
=
be the corresponding term
AS
A
(NN,
Finally, suppose that
A$+1 = AS.
containing
N1
into
(Q,<)v
By 8.20, there is a complete
a = 1.
zs
Set
a = E['ti is an embedding of If
say
E T,
zs
XNX
as above.
J,
Suppose that
Boolean algebra
ccc
C
as a complete subalgebra
such that
['the range of set
contains a Hausdorff gap' IC = 1:
ti
Again, we can suppose that
A6+1 = C.
Suppose that and that
Aa
By 8.15,
C
a < w2,
cardinality
K1.
Let
a <
a
Boolean algebra, and clearly
ccc
is a
is a limit ordinal, Set C = Ug A a
that
has been defined for
AB+1 C w2.
has
C
be a complete Boolean algebra
A$
containing
Then AS is isomorphic C as a dense subalgebra. to the completion of C, and so, by 5.16, As has cardinality t{ We may suppose that AS C w2.
This completes the definition of the sequence
a < w2>.
Set
A = a<w2 A.
By 8.15,
Boolean algebra, and clearly
A
is a complete
ccc
IA
= 1.
JAI ' H2. I MA + _ICH K
The next step is to prove that We first detour to note that, if MA + -CH
Let (w1,<)
2 ° < K2,
then
is equivalent to the following.
is
subsets of
<
ccc.
be a partial order on Suppose that
(w1,<)
V-generic filter in
with
V
IDI < K1.
(w1,<).
w1
such that
is a family of dense Then there is a
(*)
223
To see this, first observe that, trivially, MA + 1CH
Secondly, observe that (*) is
implies (*).
equivalent to the assertion that, if order with sets in
ccc
partial
V is a family of dense subthen there is a D-generic filter
and if
IPA < K1,
with
P
is a
F
IDI < Ni.
F.
Thirdly, we prove that this assertion implies
For take X c ]R
with
= K 1,
IX I
and let
-iCH.
be the
P
collection of rational intervals of iR. Then (P,.=) is a ccc partial ordered set with IPI = N° < K1. For x E X, x {C I}, x E X}. Then and set V = {dX set dX = {I E P :
:
with is a family of dense subsets of (P,c) IDI = H1. in By the assertion, there is a D-generic filter, say F, (P,c), and fl{I I E F) is a non-empty subset of IR disjoint from X. Thus X # ]R, and CH fails. V
:
K
(*) implies that
2 ° < N21
Thus, given that
and so, by 5.18, the assertion implies
2 ° = K21
hence, by 5.20,
and
MA',
MA.
To obtain a # 1. We know
We now return to our main proof.
contradiction, suppose that
I MA + -iCH D
A K
that
and so, by 7.29(1),
Al C< K21
Further, since
is
A
T'w1 = w'I]A = 1.
ccc,
MA + -1CH,
the above formulation (*) of and
VA
in
D
and
1'2 ° < K2']]A = 1.
a E A\{O}
Hence, by
there are terms
A
V
(i)
11'(w1 ,e)
is a
ccc
partially ordered set' I]
(ii)
['each element of
(iii)
I'D has cardinality 'there is no
(iv)
e
with:
D
is dense in
<
vi, ] A
> a;
V
(w1,e)'I]A 3 a;
a;
D-generic filter on
(Wv1 e)'11*
a.
By the mixing lemma, we can suppose that the Boolean values specified in the left-hand sides of (i) - (iii) are actually 1.
By Lemma 8.21, there are functions and
H
:
wl X wl + A
such that
F
:
wl x wl + A
224 V
(w
v is equal to (w 1,eF)' = ID = DHI]A = 1.
e)
1
A
We have v (w
1
is a
e F)
ccc
f'each element of
A c w2
Since
there exists
zo _ (F,H).
A.
C
(w1,eF)'I]A = 1,
v. 1, ]A
= 1.
(a zo = (F,H)} is cofinal in with Aran F U ran H c AS and and A8 is a complete eF,DH E V S,
and
:
a < w2
Clearly,
subalgebra of
is dense in
has cardinality
'DH
w2,
DH
partially ordered set' I] A = 1,
Thus,
by (14), A
V
is a
'(w1,eF)
ccc
partially ordered set' I]
since the relevant statement about specified by a
V
DH, w1, and eF
B
= 1;
B
= 1;
is
00-formula, A
(wv
DH
Ef'each element of
is dense in
since cardinals are preserved from
VAB
1,eF)'I]
to
VA,
A
'DH
has cardinality 4 K1, , s = 1.
By case 1 of the construction and by the fullness lemma, there is a term 't
is a
t E VAS+1
such that
DH-generic filter in
V
(w l,eF)'I]AS+1
Again, since the relevant statement is specified by a formula and since
AS+1
is a complete subalgebra of
= 1.
A.A,
v
't
and so
is a
DH-generic filter in
(w1,e)'I]A = 1,
225
f[ ' t
is a D-generic filter in
A
( w1, e) ' I]
= 1.
Thus it is true that
But this is a contradiction of (iv).
IIMA+-CHII0= 1.
Our final step is to prove that
IMA + -ICH ] 1
Since
For suppose that (21) holds.
(21)
= 1
Since (19) implies (20),
does not embed in
and so, by Theorem 6.25,
= 1.
into
(Q,<)V
contains a Hausdorff gap' 1A
I ' (R,<)
A
it is sufficient to prove that
= 1,
I'the range of each embedding of
(NN,
ENDH I
fNDH I]
A
(&,
= 1.
To obtain a contradiction, suppose that (21) fails. Then there is a term t in VA and a E A\{O} with
I't
is an embedding of (Q,<)"
into
ON N,
whose range does not contain a Hausdorff gap' I" 2 a.
Now we proved in Chapter 6 that (N N,
(Q,<)
does embed in
and so, using (18), we have
'there is an embedding of
(Q,<)'V
into
(N N,
It follows from the mixing and fullness lemmas that we can suppose that
T It is an embedding of
(Q,<)v
By Lemma 8.21(iii), there is a function such that It = tJA = 1. We have
I'ti
is an embedding of
(Q,<) v
(N N,
into J
= 1.
Q x N x N - A
into (NN,
226
As before, there exists
a < w2
ran J c AS
with
A
and
Clearly
zs = J.
and
subalgebra of
ti E V S, and so
A,
is an embedding of
'ti
A
(N IN,
a
is a complete
A8
(Q,<)v
into
= 1.
By case 2 of the construction and by the fullness lemma, s E VAS+1
there is a term
such that A
I's Since
is a Hausdorff gap in the range of is a pregap in
"s
specified by a
T's
(IN
,
tJ'1
0+1
in the range of
= 1.
ti"
is
A0-formula, it follows from (13) that
tJ']
is a Hausdorff gap in the range of
But this is a contradiction of (22).
= 1.
Thus the final claim
is established.
This completes the proof of the theorem. I The next theorem is obtained by combining 7.37 with 8.14, but the use of the iteration theorem here is rather "heavy-handed". 8.23
THEOREM There is a complete Boolean algebra 1-2N0
=
1
and
2x1
=
ti 2' ]] t
t
such that
= 1.
Proof
By 7.37, there is a complete Boolean algebra such that
I CH]]8 = 1.
there is a term
u
in
8
By 7.39, and the fullness lemma, V8
such that
227
I'u is a complete Boolean algebra' is
= I' E2 ° = K1 and 21 = K21]u = 1' 118 = 1. The result follows from 8.14(i).
I
THEOREM
8.24
There is a complete Boolean algebra MA + NDH I] A
such that
A
1.
Proof Let
be the complete Boolean algebra given in
C
By the main theorem 8.22 and the fullness lemma, there
8.23.
is a term
VC
in
u
such that
is a complete Boolean algebra' It
J 'u
= I' [MA + NDHI u = 1' IC = 1. We obtain place of
A
by one final appeal to 8.14(i) (with
in
C
8).
We have now reached the goal of this book. 8.25
THEOREM Assume that
N
is a model such that
Then there is a model
N
extending
N
ZFC + MA + NDH. 8.26
N
a such that
ZFC.
I
THEOREM
Assume that ZFC is consistent. Let NDH be the for each compact space X, sentence: each homomorphism from C(X,C)
into a Banach algebra is continuous.
independent from
Then
NDH
ZFC.
Proof
Assume that there is a model of
ZFC.
By 8.25,
is
228
there is a model of
ZFC + NDH.
By 7.38, there is a model of
by 1.10, this model is a model of
ZFC + CH;
By Godel's completeness theorem, independent of
ZFC.
ZFC + -iNDH.
NDH
is
I
NOTES
8.27
As we remarked, the initial development of iterated forcing is due to Solovay and Tennenbaum ([67]), who first showed that MA + -iCH is relatively consistent with ZFC. Our iteration theorems 8.12 and 8.14 are essentially given in [67,§5], where the following notation is used. Let 8 be a complete Boolean algebra, and let D be a complete Boolean algebra in Va (corresponding to our u). Then the complete
Boolean algebra corresponding to B ® D:
its relation to
and
is discussed.
D
[67,§6]:
8 ® D,
in 8.14 is denoted by
C
the tensor product of
8
Infinite iterations are discussed in
our account is quite close to this, but our unions
replace certain direct limits. complete Boolean algebra in [67,§7].
8
The construction of a
such that
= 1
I MA]
is given
Iterated forcing is discussed in [49,
Chapter VIII], for example.
The proof of Theorem 8.22 (from [72]) is a development of an argument of Kunen, who proved that "There are no (t,c)-gaps in ON IN,
and since
CH
NDH implies
is relatively consistent with MA + -iCH, the construction of discon-VDH,
(for each infinite, tinuous homomorphisms from algebras C(X) compact space X) is one that cannot be carried through in the more general theory ZFC + MA. In fact, we conjecture that
NDH
is actually independent of
in [72] that it is consistent with does embed in
MA + -tCH:
MA + -NCH
that
it is shown (R,<)
(NIN,
and so the existence of such an embedding is independent of MA + iCH. We have shown that NDH is relatively consistent x with MA + 2 ° = 82. In fact, it can be arranged that 2 ° can take any value "consistent with MA".
229
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235
INDEX OF NOTATION
Inv A
Vs, As
1
0A
rad A
1
B (X)
nil A
2
alb
11.11 3
B
r (a) 3 C(X,C),C(X) 1.1s
33
34
K* 4
37
=U,
37
[flu
4 4
37
(]R /U,
Z(f)
4
I(F),J(F)
JX,MX
5
Z
5
R(C),k
46
A,Y56 -
C0O(C),c00
59
Con T
6
60
60
Al = (M,E)
6
54
58
T 6
c0(C),co
54
(,)3,=,E
5
6
Mp,J9
38
43
if = Sv(e,°)
5
I. 13K
SN
28 29 29
(S(B),TB)
2
a (a)
Cx
28
R (X)
1
NX
10
#l
$[a11...,anl
Ku
10
1B
0
H
21
62 62
T
62
0
P(S)
Th(I) N
23 23
ID
N
71
<<
23
a
24
Thm(T)
<<
F
y
24
B = (B,A,v,',0,1)
26,130
R(f)
62
72 77
83
61
236
f 18
83
IP EV=,,
Ord
84
bB
85 85
Va, V
rk (S) S
Ordv
Xc
86
ISI
166
a n b
85
,, T
156
157
173 174
(Q(K),<), B(K)
w,wl,wa
86,87
°di'R
S#
86,87
a tiF b,
(c]G, C/'tiG
87
K+
88
V(w)
195
(t) vu
202
H
2°
88
A < B
106
107
eF' d
a < K,S < A>
:
113
120 121
(R,<) (q, <)
121
121
EB: X (a,b) tiB
107
112
[(f,g)]
Q()
1, 3
(a,b)
bI] bI]
a a
F+
ms = (M,EB,ti8)
1T]]
131 131
131
T$[al,...,an1l]
Al
B
132
133 134
D((U)
Va, VB 137 137 p(u)
v]],
ELu E V1, Eu
1[ucvl] V(2) u
138 142
a
utiv, utiv
lot. 1" . . , u
b
f
c
143
]1
144
'O[ul,.... un145 155
138
187
(a) F
v
2", Kx
177
180
188
DH. fx
218
237
INDEX
AC
(= Axiom of choice)
81-field
67,84,147,180,181 algebra
Boolean algebra
Banach norm radical
homomorphism
3
trivial
2
a1-set alphabet
Cantor
54
cardinal
93,143,209 antichain 128 Archimedean class
Artin-Schreier
36
atomic formula
55
of the continuum preserved 168 singular 3,20
axioms of
ZFC
93,170,
175,190,204
64-68,79
partially ordered set 93,114
8,11,20
topological space
Baire 80,91
category theorem
(= Continuum Hypothesis)
173,176,179,182
3
8-valued model
131,158
chain
23
Chang and Keisler
135,158
Bell and Slomson
CH
92,105
17,76,89,90,93,127,128,130,
91
Banach algebra full
85
Boolean algebra 147-154
Bade and Curtis
space
87,182
condition)
See also AC ZF
88
(= countable chain
ccc
67,79,83,145
Axiom of Choice
174 86
cardinality
automatic continuity
138
90
Cantor set 24,25
137
182
canonical well-ordering
119,122,129
axioms of
27
27
Burgess
29,174
127
anti-isotonic
26,130
Boolean-valued universe
1,4
regular-open a1-field
128
119,122,124,126,129
S1-set
79
character
1
79
238
choice function 84 class cofinality
embedding
68
compatible
formulae terms
27
Boolean algebra
27
16,21,53 108,127
nl-field
partially ordered set subalgebra 184
26
107,124,129 nl-set eventually 24
Extensionality
72
completeness theorem
63,130,
160,179,182,228
extremely disconnected
ordered
Boolean algebra normed algebra
34
filter 30,
consistent theory 75,161,162
generic
115 31
proper
59,63,73
32
78,130,144,182
forcing
argument
Continuum Hypothesis See also CH
89
85
162
iteration formula
cumulative hierarchy of
184
55
atomic
55
equivalent
Foundation
16
Dales and Esterle Davis 79 L -formula
128
Fremlin full
104
103 135
fullness lemma
168,185,207,216
in Boolean algebra
30,187
in a partially ordered 29
diamond 98 Dow 127
59
66
gap
145
107
dense
set
36,127
31
maximal
65,148
A-system lemma
29
35,128
real-closed
3
partially ordered set 41,210 Comprehension
64,148
field
completion
Dales
59
143
Esterle
complete
sets
85,90
equivalent
29
complementation
theory
21
equipotent
87
78,174,182
Cohen
24
epimorphism
112,114,116,129
(H1'x1 )
Hausdorff
109,110,115,116,
208
(K,X)-
112
Generalized Continuum Hypothesis
89
239
generic
MA
93,115
generic object
Gdel
63,75,76,78,176,182
Gdel number Hausdorff Hilbert
MA'
8,12,15 Martin and Solovay 7,14,17,18,
21,39,44,46,47,123,127,183
Martin's Axiom See also
29
model
75,77,79 36
interpolate
228
2
107
27
of models isotonic
Boolean-valued
62
24,119
iterated forcing iteration theorem
61,78
non-standard
38
of
130,136,156,161 ZFC + CH 179
of
ZFC + -iCH
of
ZFC + MA + NDH
of
ZFC + NDH
227,228
nilradical
2 2
norm 2,7 normable 2,17,18,20,128
54,132 74,79
order
divisibility
57
logically derivable
58
76,79,127,183,216,220,
nilpotent
7 Kaplansky Kunen 129,228
logical axiom
227
216
204,228
NDH
Lindenbaum algebra
136,176
62
modus ponens
8,20,127 Johnson 27 join
language
ZFC
theory of 228
130
60,63
extension of of
isomorphism Boolean
131,158
countable
67,153
integral domain
143
60,127,156
fl-valued
42,73,76,179,227
infinite iteration Infinity
148,155
mixing lemma
75,79
infinitesimal
77,79
27
metatheorem
incompleteness theorem second
103
v
meet
independent
80,91,92,183
MA, MA'
Matijacevic
incompatible
80,103
Martin's Maximum
8,10,11,14
5
first
94,95,223
main boundedness theorem
75
homomorphism Boolean 87 discontinuous
hull
102,110,116,123,128,183, 220,227
77
109,129
radical
(= Martin's Axiom)
80,91,92,93,95,96,100,101,
187
58
Frdchet
23,43
24,109
240
inclusion modulo finite sets
radical
legicographic
partial
24,120
8,10,11,14
rank
of set
23
standard
1
homomorphism
23,109
85
of term
24
137
strict partial
22
Rasiowa and Sikorski
strong Frdchet
24,43,109
rational interval
total
regular-open
22
order-dense 121 order-isomorphism
ordinal
66,152 relative consistency
Replacement
35,128
Rudin
cofinal subset of finite 86 limit
atomic
55
of inference 87 87
successor
84
55
satisfaction
61
scope
56
182
17,39,40,42,48
Scott
Pairing
65,149
seminorm
partially ordered set
23,209
cofinal, coinitial subset
2
seminormable
2,10,14,39,
47,48,53
106
semisimple
Power Set
66,150
prefilter
sentence
31,187
(NiON1)-
separable
208
incompatible
213
preservation of cardinals
169,
208
2,17,20,128,129
58
Proper Forcing Axiom
103
31,41
sequence
83
binary set
120
a
basic
prime ideal
73,76,77
17,92
separative
107
107,118
(K,A)-
1
57,135
independent
107
equivalent
57
quantifier
P-point
proof
55
connective
84
singular
pregap
66,151
1,40
rule
87
86
regular
60,
128,182,228
24
82,154
infinite
28,29,41,174
Regularity 24
order-preserving
ordered field
182 96
S1-
119,122,129 70
119,122,124,126,129
86 cardinal of countable 86
241
definable
ni
70
complete
107,124,129
rank of
consistent 59,63,73,75 inconsistent 59
85
simple
model of
69
transitive
transitive set
86
(= Souslin's Hypothesis)
in a model
Shelah
42
Sierpinski
129
simple set
69
Sinclair
in
MS
132
in
V8
144
point
undefinability of ultrafilter fixed
10
53,79,182
uniform norm
Solovay and Tennenbaum
Union
184,228
Souslin line spectrum
variable
3
bound 22,33
topology
free
33
submodel
137
55 57
56
weakly seminormable
Stone-6ech compactification subformula
85
Boolean-valued
3
Stone space
4,7
65,149
universe
98
spectral radius
32,134,135
35
40 quasi-selective ultraproduct 38,42,108
79
Solovay
56
6 well-ordered
50,51,53
81
well-ordering principle
61
Woodin's condition
83
46,53
55
Zermelo 137
for embedding for function
zero-set 218
for family of subsets
ZFC 187
57
4
1,42,64,68,74,77,127,
130,179 218
for partially ordered sets 218
84
68,83,147
ZF
180
for generic object
theory
155
35,118,123,128
free
8,10,11
Smorynski
term
61
53
singularity set
128
82
truth
98,99
syntax
62
transcendence basis
82
uncountable SH
72,73
ZFC + CH
17,18
Zorn's Lemma
26,83
