Download Continuous Logic and Probability Algebras THESIS Presented in

Continuous Logic and Probability Algebras
THESIS
Presented in Partial Fulfillment of the Requirements for the Degree Master of Science in
the Graduate School of The Ohio State University
By
Fan Yang, B.S.
Graduate Program in Mathematics
The Ohio State University
2016
Master's Examination Committee:
Christopher Miller, Advisor
Timothy Carlson
Copyright by
Fan Yang
2016
Abstract
Continuous logic is a multi-valued logic where the set of truth values is the unit interval
[0, 1]. It was developed recently as a framework for metric structures, which consist of
complete, bounded metric spaces on which there are distinguished elements and
uniformly continuous functions. There are many parallels between continuous logic and
first-order logic: 0 corresponds to true and 1 to false, sup and inf take the place of
quantifiers, and uniformly continuous functions on [0, 1] replace connectives. Instead of
a distinguished equality symbol, there is a distinguished predicate for the metric. Familiar
theorems of first-order logic, such as completeness, compactness, and downward
Löwenheim-Skolem, have modified counterparts in continuous logic. We present these
results in comparison to those in first order logic and prove that the class of probability
algebras (probability spaces modulo null sets where the distance between two events is
their symmetric difference) is axiomatizable in continuous logic.
ii
Vita
2009 ............................................................ Bozeman High School
2013 ............................................................ B.S. Mathematics, Carnegie Mellon
University
2013--2015 .................................................. Graduate Teaching Associate, Department
of Mathematics, The Ohio State University
Fields of Study
Major Field: Mathematics
iii
Table of Contents
Abstract .......................................................................................................................... ii
Vita ............................................................................................................................... iii
Chapter 1: Introduction ................................................................................................... 1
Chapter 2: Preliminaries ................................................................................................. 3
Chapter 3: Axioms of continuous logic ........................................................................... 9
Chapter 4: Probability algebras ......................................................................................17
Chapter 5: Continuous logic versus first-order logic ......................................................53
Chapter 6: Conclusion ...................................................................................................59
References .....................................................................................................................61
iv
1. I NTRODUCTION
Multi-valued logics are formal systems that deviate from classical logic by allowing for
more than two truth values. An early version of multi-valued logic appeared in 1920 with
the development of a three-valued logic by Łukasiewicz. This was extended to an infinitevalued propositional logic for which Pavelka proved a completeness result [3]. The next big
step for multi-valued logics occurred in 1966, when Chang and Keisler generalized the set
of truth values to be any compact Hausdorff space [5]. Among their results were suitable
rephrasings of the compactness theorem and the Downward Löwenheim-Skolem theorem.
Ultimately, their construction proved to be too general for applications to other areas of
mathematics and little was done until the need arose in model theory to find a framework
in which to study metric structures. Metric structures consist of complete, bounded metric
spaces on which there are distinguished elements and uniformly continuous functions.
This paper presents a recent incarnation of multi-valued logic called continuous firstorder logic, or simply continuous logic, introduced by Ben Yaacov and Usvyatsov in [4] to
study metric structures. It is a special case of Chang and Keisler’s logic in [5] in the sense
that the set of truth values is restricted to the unit interval [0, 1]. However, it differs from
[5] by considering structures equipped with a distinguished metric instead of the traditional
equality relation. In the same way that the equality symbol = in first-order logic is always
interpreted to be real equality, the language of continuous logic contains a distinguished
binary predicate d that is always interpreted to be the metric. First-order model theory
coincides with continuous model theory if we equip the underlying set of any first-order
1
model with the discrete metric. Thus continuous logic is an extension of first-order logic
in the sense that it allows more structures to be models; see, e.g., Iovino [11]. These
structures include Hilbert spaces, Banach spaces, and probability algebras. Probability
algebras, which arise from taking quotients of probability spaces modulo their null sets
and become complete metric spaces under d(a, b) = µ(a4b), where µ is the probability
measure, will be the focus of the first part of this paper.
Although continuous logic was developed by model theorists for model theoretic uses,
some research about its properties as a logic has also been done. Notably, Ben Yaacov
and Pederson formulated in [3] a set of axioms that yields a version of the completeness
theorem. A comparison of the completeness theorem and other aspects of continuous logic
to their counterparts in first-order logic will be presented in the second part of this paper.
2
2. P RELIMINARIES
Let (M, d) be a complete metric space with diameter 1.
Definition. A predicate on M is a uniformly continuous function P : M n → [0, 1]. A
function on M is a uniformly continuous function f : M n → M .
Definition. A metric structure M consists of a complete metric space (M, d) with diameter 1, a set {Pi : i ∈ I} of predicates on M , a set {fj : j ∈ J} of functions on M , and a set
{ck : k ∈ K} of distinguished elements of M . We denote it by
M = (M, Pi , fj , ck : i ∈ I, j ∈ J, k ∈ K).
Convention. We will regard constants as 0-ary functions and suppress explicit mention of
them in future definitions. We treat d as a binary predicate since it is a uniformly continuous
function from M 2 → [0, 1]; this is proven in 4.9. The set of natural numbers {0, 1, ...} is
denoted by N.
Here are some examples of metric structures.
(1) Let (M, d) be any complete metric space of diameter 1. Then M = (M, d) is a
metric structure with no functions and whose sole predicate is the binary predicate
d.
(2) Let M = (M, Pi , fj , ck : i ∈ I, j ∈ J, k ∈ K) be a structure in first-order logic.
Equip M with the discrete metric: for all x, y ∈ M, dM (x, y) = 1 if and only if
x 6= y. This makes M a complete metric space with diameter 1. Predicates take
3
values in {0, 1} ⊂ [0, 1]. Since the metric is discrete, all predicates and functions
are trivially uniformly continuous. Under the intended interpretation of 0 as >
and 1 as ⊥, d becomes a generalization of equality: x = y is true in M if and
only if dM (x, y) = 0. In this sense, first-order models are a special case of metric
structures.
(3) Given a probability space (X, B, µ), we can construct a probability structure, a
metric structure based on the metric space (M, d) where M consists of elements
of B modulo 0 and d is the measure of symmetric difference. We take µ to be a
unary predicate and the Boolean operations ∪, ∩, and ·c to be functions on M . We
will study an axiomatization of this class of structures in the fourth chapter of this
paper.
Definition. Let (M, d), (M 0 , d0 ) be metric spaces. Let s : M → M 0 be a function. δs :
(0, 1] → (0, 1] is a modulus of uniform continuity for s if for all ∈ (0, 1] and x, y ∈ M ,
we have d(x, y) < δs () =⇒ d0 (s(x), s(y)) < .
Note that s is uniformly continuous if and only if it has a modulus of uniform continuity.
Convention. If (M, d) is a metric space and f : M n → M is an n-ary function, then we
always equip M n with the maximum metric: given x = (x1 , ..., xn ), y = (y1 , ..., yn ) ∈ M n ,
we use dmax (x, y) = max{d(xi , yi ) : i = 1, ..., n}.
If (M, d) is a metric space and P : M n → [0, 1] is a n-ary predicate, then we always
equip [0, 1] with the usual metric: given a, b ∈ [0, 1], d[0,1] (a, b) = |a − b|.
Definition. A continuous signature is a quadruple L = (R, F, η, G) such that
(1) R contains a distinguished binary predicate d
(2) R ∩ F = ∅
4
(3) η : R ∪ F → N
(4) G = {δs : (0, 1] → (0, 1] : s ∈ R ∪ F}
R is the set of relation symbols, F is the set of function symbols, and (R \ {d}) ∪ F is the
set of nonlogical symbols of L. The arity of a relation or function symbol s is given by η(s).
G is the set of moduli of uniform continuity for each predicate and function symbol. The
moduli of uniform continuity are not syntactic objects; rather, they are functions (0, 1] →
(0, 1] fixed by the signature. The cardinality of L, denoted by |L|, is the cardinality of the
set of nonlogical symbols of L.
The logical symbols of any signature L consist of the following: a binary connective
. two unary connectives, ¬ and 1 , two quantifiers, sup and inf, countably many variables
−,
2
V = {v0 , v1 , ...}. Since we always interpret d to be the metric, d is also considered a logical
symbol. This parallels the treatment of the equality symbol in first-order logic.
2.1. Definition. Let L = (R, F, η, G) be a continuous signature. An L-structure is an
ordered pair M = (M, ρ) where M is a set and ρ is a function on R ∪ F such that
(1) for each f ∈ F, ρ assigns a map f M : M η(f ) → M
(2) for each P ∈ R, ρ assigns a map P M : M η(P ) → [0, 1]
(in particular, ρ assigns d to a complete metric dM : M × M → [0, 1])
(3) for each f ∈ F, δf is a modulus of uniform continuity for f : for all ∈ (0, 1] and
x = (x1 , ..., xη(f ) ), y = (y1 , ..., yη(f ) ) ∈ M η(f ) ,
dmax (x, y) < δf () =⇒ dM (f (x), f (y)) < 5
(4) for each P ∈ R, δP is a modulus of uniform continuity for P : for all ∈ (0, 1] and
x = (x1 , ..., xη(P ) ), y = (y1 , ..., yη(P ) ) ∈ M η(P ) ,
dmax (x, y) < δP () =⇒ |P (x) − P (y)| < Remark. Part (2) of the above definition establishes the set of truth values as [0,1].
Definition. Fix a continuous signature L = (R, F, η, G). L-terms are defined inductively.
Variables and constant symbols are L-terms. If f ∈ F is an n-ary function symbol and
t1 , ..., tn are L-terms, then f (t1 , ..., tn ) is an L-term. Atomic L-formulas are expressions
of the form P (t1 , ..., tn ), where P ∈ R is an n-ary predicate symbol and t1 , ..., tn are Lterms. The class of L-formulas is the smallest class of expressions that contains all atomic
. ¬, and 1 , and satisfies the following condition: If φ is an
formulas, is closed under −,
2
L-formula and x is a variable, then supx φ is an L-formula.
2.2. Definition. Let M = (M, ρ) be an L-structure. An M-assignment is a function
σ : V → M . Given an M-assignment σ, x ∈ V , and a ∈ M , define the M-assignment σxa
as follows: for all y ∈ V ,
σxa (y)
:=




a
if x = y



σ(y) otherwise
Fix an M-assignment σ. Define tM,σ , the interpretation under σ of a term t in M,
inductively in the same way as in first-order logic. Let φ be a formula. Define M(φ, σ),
the value of φ in M under σ, by induction on formulas:
, ..., tM,σ
(1) M(P (t1 , ..., tη(P ) , σ) := P M (tM,σ
1
η(P ) )
. β, σ) := max(M(α, σ) − M(β, σ), 0)
(2) M(α −
(3) M(¬α, σ) := 1 − M(α, σ)
6
(4) M( 12 α, σ) := 12 M(α, σ)
(5) M(supx α, σ) := sup{M(α, σxa ) : a ∈ M }
When φ is a sentence, we write M(φ) instead of M(φ, σ).
2.3. We introduce the following abbreviations for use in future examples. Let 1 stand for
. φ) for any sentence φ. Let ( 1 )n stand for
¬(φ −
2
1
1
. . . 1.
|2 {z 2}
n times
. φ) in any model
This notation is natural because for any formula φ, the truth value of ¬(φ −
under any assignment is 1:
. φ), σ) = 1 − M(φ −
. φ, σ) = 1 − max(M(φ, σ) − M(φ, σ), 0) = 1.
M(¬(φ −
Similarly, for any n, the truth value of ( 21 )n is always ( 21 )n . We can use this to construct
a sentence α whose truth value in any model M is 34 . First, define a binary connective +0
by φ +0 ψ = ¬(ψ → ¬φ). Then
M(φ +0 ψ, σ) = min{M(φ, σ) + M(ψ, σ), 1}
for all models M and assignments σ. Note that +0 is classically equivalent to conjunction.
Now take α to be
dyadic number
k
,
2n
k
2n
1
2
+0 ( 21 )2 . Then M(α, σ) = min{ 12 + 14 , 1} = 43 . In this way, for every
where k, n ∈ N, k ≤ 2n , we can produce a sentence whose truth value is
. 1 } from 1 is D = { k : k, n ∈ N, k ≤ 2n }.
so the set of truth values generated by {¬, −,
2
2n
In propositional logic, a chosen set of connectives must be functionally complete; every Boolean function should be representable in terms of a propositional formula formed
from these connectives. The dual notion of functional completeness in continuous logic is
fullness. In general, for each n ≥ 1, an n-ary connective f is a continuous function from
7
[0, 1]n → [0, 1]. A system of connectives is a family F = (Fn | n ≥ 1) where each Fn is a
set of n-ary connectives. F is closed if the following hold:
(1) for all n, Fn contains the projection (a1 , ..., an ) 7→ aj for each j = 1, ..., n.
(2) for all m, n, if u ∈ Fn and v1 , ..., vn ∈ Fm , then w : [0, 1]m → [0, 1] defined by
w(a) = u(v1 (a), ..., vn (a)) is in Fm .
For each system of connectives F, let F denote the smallest closed system of connectives
containing F. F is full if for each n, F is dense in the set Cn of all n-ary connectives
[0, 1]n → [0, 1] with respect to the topology on Cn obtained from the metric d(f, g) =
supx∈[0,1]n |f (x) − g(x)|. In other words, for all n ≥ 1, > 0, and f ∈ Cn , there exists
g ∈ F such that for all x ∈ [0, 1]n , |f (x) − g(x)| < .
Instead of adding a symbol for each connective, which would render the language uncountable, we may choose any full system of connectives, allowing us to approximate any
connective to arbitrary precision.
. ¬, 1 } is full.
Fact. (6.6 in [2]) The system of connectives {−,
2
. ¬, 1 }, which is the analogue of {→, ¬}, a popular choice of funcThus we choose {−,
2
tionally complete connectives in first-order logic. Interestingly, although {¬, ∧, ∨} is functionally complete in first-order logic, it is not full in continuous logic with ∧ and ∨ as
defined in 3.3; by 1.7 in [4], all n-ary connectives constructed using ¬, ∧, ∨ are 1-Lipschitz
in each argument, but
1
2
is 2-Lipschitz, so
1
2
cannot be constructed from ¬, ∧, ∨.
2.4. Definition. Let M be an L-structure, σ an assignment, and Σ a set of L-formulas.
(M, σ) is a model of Σ, written (M, σ) Σ, if for all φ ∈ Σ, M(φ, σ) = 0. We will
abbreviate (M, σ) {φ} by (M, σ) φ. Σ is satisfiable if it has a model. We write Σ φ
if every model of Σ is a model of φ. φ is valid if ∅ φ.
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3. A XIOMS OF CONTINUOUS LOGIC
Definition. Given a term t, a variable x, and a formula α, the substitution of t for x in α,
denoted α[t/x], is the result of replacing every free occurrence of x in α by t.
. φ. The following axiom system from [3] for continuous logic
Define φ → ψ := ψ −
allows us to prove the completeness result in 5.3. The axioms of continuous logic are the
universal closures of:
(1) α → (β → α)
(2) (α → β) → ((β → γ) → (α → γ))
(3) ((α → β) → β) → ((β → α) → α)
(4) (¬α → ¬β) → (β → α)
(5) ( 21 α → α) → 12 α
(6) 21 α → ( 12 α → α)
(7) supx (α → β) → (supx α → supx β)
(8) supx α → α[t/x], if no variable in t is bound by a quantifier in α
(9) α → supx α, if x is not free in α
(10) d(x, x)
(11) d(x, y) → d(y, x)
(12) d(y, z) → (d(x, y) → d(x, z))
(13) For each f ∈ F, ∈ (0, 1], and r, q ∈ D, if r > and q < δf (), then
(
_
d(xi , yi ) → q)) ∨ (r → d(f (x1 , ..., xη(f ) ), f (y1 , ..., yη(f ) )))
i=1,...,η(f )
9
(14) For each P ∈ R, ∈ (0, 1], and r, q ∈ D, if r > and q < δP (), then
(
_
d(xi , yi ) → q)) ∨ (r → |P (x1 , ..., xη(P ) ) − P (y1 , ..., yη(P ) )|)
i=1,...,η(P )
The only rule of inference in continuous logic is modus ponens
α,
α→β
β
Formal deductions and the provability relation ` are defined in the usual way.
3.1. Interpreting implication
Let M be an L-structure, σ an assignment, and φ, ψ L-formulas. By definition, M(φ →
ψ, σ) = M(ψ → φ, σ) = max(M(ψ, σ) − M(φ, σ), 0), so
(M, σ) φ → ψ ⇐⇒ M(ψ, σ) ≤ M(φ, σ).
In words, φ → ψ holds in (M, σ) if and only if ψ is at most as true as φ. If we restrict
the set of truth values to {0, 1}, then this says that the only instance where φ → ψ does not
hold in (M, σ) is when M(ψ, σ) > M(φ, σ), i.e. when M(ψ, σ) = 1 and M(φ, σ) = 0.
An interesting phenomenon occurs when ψ is ¬φ. In this case,
(M, σ) φ → ¬φ ⇐⇒ 1 − M(φ, σ) = M(¬φ, σ) ≤ M(φ, σ)
1
⇐⇒ M(φ, σ) ≥ .
2
Thus for a formula φ to imply its own negation in (M, σ), it only needs to be more false
than
1
2
in (M, σ).
3.2. Defining disjunction
10
In first-order logic, there are many ways to define disjunction in terms of → and ¬. For
example, we could take φ ∨1 ψ := ¬ψ → φ and φ ∨2 ψ := (ψ → φ) → φ. With the
semantics of continuous logic,
M(φ ∨1 ψ, σ) = M(¬ψ → φ, σ) = max(M(ψ, σ) + M(φ, σ) − 1, 0)
M(φ ∨2 ψ, σ) = M((ψ → φ) → φ, σ) = min(M(φ, σ), M(φ, σ))
If we restrict the range of M(·, σ) to {0, 1}, then M(· ∨1 ·, σ) ≡ M(· ∨2 ·, σ), reflecting
the fact that ∨1 and ∨2 are equivalent in first-order logic. To see that they are not equivalent
in continuous logic, consider φ =
1
2
as defined in 2.3. Since M(φ, σ) = 12 ,
M(φ ∨1 φ, σ) = max(M(φ, σ) + M(φ, σ) − 1, 0) = max( 21 + 21 − 1, 0) = 0
M(φ ∨2 φ, σ) = min(M(φ, σ), M(φ, σ)) = min( 12 , 21 ) = 12 .
Our choice of disjunction ∨ should give the same truth value to φ and to φ∨φ. Furthermore,
for all models M and assignments σ, it should satisfy
(M, σ) φ ∨ ψ ⇐⇒ (M, σ) φ or M ψ
and M(φ ∨ ψ, σ) = min(M(φ, σ), M(ψ, σ)), so the most natural choice is ∨ = ∨2 (or
anything equivalent to it in continuous logic).
Similarly, conjunction ∧ should satisfy
(M, σ) φ ∧ ψ ⇐⇒ (M, σ) φ and M ψ
and M(φ ∧ ψ, σ) = max(M(φ, σ), M(ψ, σ)), both of which are achieved by taking φ ∧
ψ := ¬(¬φ ∨ ¬ψ). These remarks are summarized below.
11
3.3. We introduce the following abbreviations:
disjunction φ ∨ ψ := (ψ → φ) → φ
conjunction φ ∧ ψ := ¬(¬φ ∨ ¬ψ)
logical distance |φ − ψ| := (φ → ψ) ∧ (ψ → φ)
infx φ := ¬supx ¬φ
Under the semantics of continuous logic,
(1) M(φ → ψ, σ) = max(M(ψ, σ) − M(φ, σ), 0)
(2) M(φ ∨ ψ, σ) = min(M(φ, σ), M(ψ, σ))
(3) M(φ ∧ ψ, σ) = max(M(φ, σ), M(ψ, σ))
(4) M(|φ − ψ|, σ) = |M(φ, σ) − M(ψ, σ)| (one-dimensional Euclidean distance)
(5) M(infx φ, σ) = inf{M(φ, σxa ) : a ∈ M }
If we restrict the truth values to {0, 1}, the interpretations of sup, inf, ∨, ∧, and → coincide
with those of ∀, ∃, disjunction, conjunction, and implication, respectively, in first-order
logic. For example, (M, σ) infx φ ⇐⇒ M(infx φ, σ) = 0 ⇐⇒ there exists a ∈ M
such that M(φ, σxa ) = 0. When it is easier to do so, we will write φ ↔ ψ instead of |φ − ψ|.
Remark. Under the abbreviations in 3.3, axiom 3 states that ∨ is commutative. Axioms
1–4 are the axioms for Łukasiewicz propositional logic [3]. Their relation to the axioms of
classical propositional logic is explored in 5.5.
12
Axioms 5 and 6 establish the behavior of 12 . Recall from 2.3 that α +0 β := ¬(β →
¬α) and M(α +0 β, σ) = min(M(α, σ) + M(β, σ), 1) for any L-structure M and Massignment σ. In particular,
1
1
1
1
M( α + α, σ) = min(M( α, σ) + M( α, σ), 1)
2
2
2
2
1
1
= min( M(α, σ) + M(α, σ), 1)
2
2
= min(M(α, σ), 1).
Axioms 7–9 establish the behavior of sup and are direct translations of the quantifier
axioms in first-order logic [7].
Axioms 10–14 are translations of properties of = in first-order logic. Recall the equality
axioms of first-order logic (from [1]):
(i) reflexivity: x = x
(ii) substitutivity: x = y → (φ → φ0 ) for any formula φ, where φ0 is obtained from φ
by replacing some free occurrences of x with y.
Let M be an L-structure and let σ be an assignment. If we replace a = b by d(a, b), then
axiom 10 is a direct translation of (i). Consequences of (ii) include symmetry, (x = y) →
(y = x), and transitivity, (x = y) → ((x = z) → (y = z)), which are encoded in axioms
11 and 12, respectively. So the property of = being an equivalence relation translates, via
axioms 10–12, into d being a pseudo-metric. The reason why we do not need to specify
that d is a genuine complete metric is given in 3.4.
Equality axiom (ii) also makes = a congruence relation; for each predicate P , (x =
y) → (P (x, z) → P (y, z)). By 3.1,
13
(M, σ) (d(x, y)) → (P (x, z) → P (y, z))
⇐⇒ M(P (x, z) → P (y, z), σ) ≤ M(d(x, y), σ)
⇐⇒ max(M(P (y, z), σ) − M(P (x, z), σ), 0) ≤ dM (x, y).
We also have max(M(P (x, z), σ) − M(P (y, z), σ), 0) ≤ dM (y, x) = dM (x, y), so
|P M (x, z) − P M (y, z)| = |M(P (x, z), σ) − M(P (y, z), σ)| ≤ dM (x, y).
In other words, predicates should be 1-Lipschitz with respect to d, although 1-Lipschitz is
relaxed to uniformly continuous. A similar argument using (x = y) → (f (x, z) = f (y, z))
translated to d(x, y) → (d(f (x, z), c) → d(f (y, z), c)) shows that functions should be
1-Lipschitz, which is likewise relaxed to uniformly continuous. To express these requirements, axioms 13 and 14 define the moduli of uniform continuity for each function and
predicate symbol.
3.4. We now introduce a more general notion of metric structure and summarize its relation
to metric structures.
Definition. Using the notation from 2.1, an L-pre-structure is an L-structure such that ρ
assigns d to a pseudo-metric.
Definition. Let L = (R, F, η, G) be a continuous signature. Let M and N be L-prestructures with underlying sets M and N , respectively. A function h : M → N is an
elementary L-morphism if it satisfies the following conditions:
(1) for all f ∈ F and a1 , ..., aη(f ) ∈ M ,
h(f M (a1 , ..., aη(f ) )) = f N (h(a1 ), ..., h(aη(f ) )).
14
(2) for all P ∈ R and a1 , ..., aη(P ) ∈ M ,
P M (a1 , ..., aη(P ) )) = P N (h(a1 ), ..., h(aη(P ) )).
(3) for all M-assignments σ and formulas φ, M(φ, σ) = N (φ, h ◦ σ).
The definitions of assignment, interpretation, and value in L-pre-structures are the same as
those in L-structures from 2.2. Given an L-pre-structure M, an assignment σ, and a set of
L-formulas Σ, we say that (M, σ) is a pre-model of Σ if for all φ ∈ Σ, M(φ, σ) = 0. We
say that a set of formulas Σ is pre-satisfiable if it has a pre-model.
Theorem. (6.9 from [3]) Let L be a continuous signature and let M be an L-pre-structure.
c and an elementary L-morphism of M into M.
c
Then there exists an L-structure M
c can be constructed by taking the quotient space of M under the equivalence relation
M
x ∼ y ⇐⇒ d(x, y) = 0.
Corollary. A set of formulas Γ is satisfiable if and only if Γ is pre-satisfiable.
Thus L-pre-structures and L-structures are indistinguishable in continuous logic. The
analogue of this fact in first-order logic is that = is a congruence relation, so the quotient of
a first-order structure K by =K is indistinguishable from K in first-order logic.
We end this section with definitions of elementary equivalence and elementary substructure. They coincide with the definitions from first-order logic when the set of truth values
is restricted to {0, 1}.
Definition. Let M and N be L-structures. M and N are elementarily equivalent if
M(φ) = N (φ) for all L-sentences φ. M is an elementary substructure of N if M ⊆ N
15
and for all formulas φ(x1 , ..., xn ) and a1 , ..., an ∈ M , M(φ, σ) = N (φ, σ 0 ) where σ is
any M-assignment and σ 0 is any N -assignment such that σ(xi ) = σ 0 (xi ) = ai for each
i = 1, ..., n.
Definition. Let L be a continuous signature. A sentence is an L-formula with no free
variables. A theory T is a set of sentences. The theory of an L-structure M, denoted by
Th(M), is the set of sentences true in M.
Remark. If M and N are elementarily equivalent, then Th(M) = Th(N ). The converse
also holds: Suppose for L-structures M and N , Th(M) = Th(N ). Let φ be a sentence
and let q = M(φ). By 2.3, there exists an L-sentence ψ such that in all L-structures Q,
Q(ψ) = q. In particular M(ψ) = q, so M(φ ↔ ψ) = 0. Since Th(M) = Th(N ),
N (φ ↔ ψ) = 0. Since N (ψ) = q, we have N (φ) = N (ψ) = q = M(φ), so M and N
are elementarily equivalent.
16
4. P ROBABILITY ALGEBRAS
Definition. [10] A Boolean algebra is a set A that includes two distinguished constants 0
and 1 and on which there are defined two binary operations ∧ and ∨, a unary operation ·c
which satisfy the following rules:
(1) ∀x x ∨ x = x
(2) ∀x x ∧ x = x
(3) ∀x∀y x ∨ y = y ∨ x
(4) ∀x∀y x ∧ y = y ∧ x
(5) ∀x∀y (x ∧ y) ∨ y = y
(6) ∀x∀y (x ∨ y) ∧ y = y
(7) ∀x∀y∀z (x ∨ y) ∨ z = x ∨ (y ∨ z)
(8) ∀x∀y∀z (x ∧ y) ∧ z = x ∧ (y ∧ z)
(9) ∀x∀y∀z x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)
(10) ∀x∀y∀z x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z)
(11) ∀x x ∨ xc = 1
(12) ∀x x ∧ xc = 0
(13) 0 6= 1
A Boolean algebra becomes partially ordered if we define x ≤ y to be x ∧ y = x
(reflexive by (2), antisymmetric by (4), transitive by (8)). A Boolean algebra A is a
17
Boolean σ-algebra if for each sequence hxn i of elements of A there is a least element
x ∈ A such that xn ≤ x for all n. We denote this least upper bound by supn xn .
4.1. Facts about Boolean algebras. Let B be a Boolean algebra. For all a, b ∈ B, let
a \ b := a ∧ bc .
4.1.1. For all a ∈ B, a ∨ 0 = a, a ∧ 0 = 0, a ∨ 1 = 1, and a ∧ 1 = a.
4.1.2. disjoint unions. For all a0 ≤ a1 ≤ a2 ∈ B, a2 \ a0 = (a2 \ a1 ) ∨ (a1 \ a0 ) and
(a2 \ a1 ) ∧ (a1 \ a0 ) = 0. Furthermore, i < j ∈ {0, 1, 2} implies ai \ aj = 0.
Proof. Observe (a2 ∧ ac1 ) ∨ (a1 ∧ ac0 ) = (a2 ∧ ac1 ) ∨ ((a2 ∧ a1 ) ∧ ac0 ) = (a2 ∧ ac1 ) ∨ (a2 ∧ (a1 ∧
ac0 )) = a2 ∧(ac1 ∨(a1 ∧ac0 )) = a2 ∧((ac1 ∨a1 )∧(ac1 ∨ac0 )) = a2 ∧(ac1 ∨ac0 ) = a2 ∧(a1 ∧a0 )c =
a2 ∧ ac0 = a2 \ a0 and (a2 \ a1 ) ∧ (a1 \ a0 ) = (a2 ∧ ac1 ) ∧ (a1 ∧ ac0 ) = a2 ∧ (ac1 ∧ a1 ) ∧ ac0 ) = 0.
Now let i < j ∈ {0, 1, 2}. Then ai \ aj = ai ∧ acj = (ai ∧ aj ) ∧ acj = ai ∧ (aj ∧ acj ) =
ai ∧ 0 = 0.
4.1.3. b ≤ bn implies b ∨ bn = bn
Proof. Suppose b ≤ bn . By definition b ∧ bn = b, so by axiom 5 b ∨ bn = (b ∧ bn ) ∨ bn =
bn .
4.1.4. For all a, b, c ∈ B, (a4b) ∧ c = (a ∧ c)4(b ∧ c).
Proof. (a4b) ∧ c = ((a ∧ bc ) ∨ (b ∧ ac )) ∧ c = ((a ∧ bc ) ∧ c) ∨ ((b ∧ ac ) ∧ c) = ((a ∧ c) ∧
bc ) ∨ ((b ∧ c) ∧ ac ) = [((a ∧ c) ∧ bc ) ∨ 0] ∨ [((b ∧ c) ∧ ac ) ∨ 0] = [((a ∧ c) ∧ bc ) ∨ (a ∧ 0)] ∨
[((b ∧ c) ∧ ac ) ∨ (b ∧ 0)] = [((a ∧ c) ∧ bc ) ∨ (a ∧ (c ∧ cc ))] ∨ [((b ∧ c) ∧ ac ) ∨ (b ∧ (c ∧ cc ))] =
[((a ∧ c) ∧ bc ) ∨ ((a ∧ c) ∧ cc )] ∨ [((b ∧ c) ∧ ac ) ∨ (b ∧ (c ∧ cc ))] = ((a ∧ c) ∧ (bc ∨ cc )) ∨
((ac ∨ cc ) ∧ (b ∧ c)) = ((a ∧ c) ∧ (b ∧ c)c ) ∨ ((a ∧ c)c ∧ (b ∧ c)) = (a ∧ c)4(b ∧ c).
18
4.1.5. For all a, b, c ∈ B, a4c ≤ (a4b) ∨ (b4c).
Proof. Let a, b, c ∈ B. Then (a ∧ cc ) ∧ ((a4b) ∨ (b4c)) = (a ∧ cc ) ∧ ((a ∧ bc ) ∨ (ac ∧ b) ∨
(b ∧ cc ) ∨ (bc ∧ c)) = ((a ∧ cc ) ∧ (a ∧ bc )) ∨ ((a ∧ cc ) ∧ (ac ∧ b)) ∨ ((a ∧ cc ) ∧ (b ∧ cc )) ∨
((a ∧ cc ) ∧ (bc ∧ c)) = (a ∧ cc ∧ bc ) ∨ 0 ∨ (a ∧ b ∧ cc ) ∨ 0 = ((a ∧ cc ) ∧ bc ) ∨ ((a ∧ cc ) ∧ b) =
(a ∧ cc ), so a ∧ cc ≤ (a4b) ∨ (b4c). By symmetry of 4, ac ∧ c ≤ (a4b) ∨ (b4c),
so a4c ∧ ((a4b) ∨ (b4c)) = ((a ∧ cc ) ∨ (ac ∧ c)) ∧ ((a4b) ∨ (b4c)) = ((a ∧ cc ) ∧
((a4b) ∨ (b4c)) ∨ ((ac ∧ c) ∧ ((a4b) ∨ (b4c))) = (a ∧ cc ) ∨ (ac ∧ c) = a4c. Hence,
a4c ≤ (a4b) ∨ (b4c).
4.1.6. For all a, b ∈ B, if a \ b = 0, then a ≤ b.
Proof. Let a, b ∈ B and suppose a\b = 0. Then a = a∧1 = a∧(b∨bc ) = (a∧b)∨(a∧bc ) =
(a ∧ b) ∨ (a \ b) = (a ∧ b) ∨ 0 = a ∧ b, so a ≤ b.
4.1.7. For all a, b, c ∈ B, (a4b)4c = a4(b4c)
Definition. A probability space is a triple (X, B, µ) where X is a set, B is a σ-algebra of
subsets of X, and µ is a σ-additive measure on B such that µ(X) = 1.
4.1. Proposition. Let (X, B, µ) be a probability space. Define the relation ∼µ on B by
A ∼µ B ⇐⇒ µ(A4B) = 0. Then ∼µ is an equivalence relation.
Proof. Clearly ∼µ is reflexive and symmetric. Let A, B, C ∈ B such that A ∼µ B and
B ∼µ C. Then µ(A4B) = µ(B4C) = 0. Since A \ B and B \ A are disjoint, 0 =
µ(A4B) = µ((A \ B) ∪ (B \ A)) = µ(A \ B) + µ(B \ A), so µ(A \ B) = µ(B \ A) = 0.
Similarly, µ(B \ C) = µ(C \ B) = 0. Since A \ C ⊆ (A \ B) ∪ (B \ C) and C \ A ⊆
(C \ B) ∪ (B \ A), µ(A4C) = µ((A \ C) ∪ (C \ A)) = µ(A \ C) + µ(C \ A) ≤
19
µ(A \ B) + µ(B \ C) + µ(C \ B) + µ(B \ A) = 0. Thus A ∼µ C, so ∼µ is transitive, and
hence an equivalence relation.
Notation. Denote the equivalence class of A ∈ B by [A]µ . Let B̂ = {[A]µ : A ∈ B} be the
set of equivalence classes and define the following operations on B̂: For all A, B ∈ B, let
[A]cµ = [Ac ]µ , [A]µ ∧ [B]µ = [A ∩ B]µ , and [A]µ ∨ [B]µ = [A ∪ B]µ . For all (An )n∈N ∈ B N ,
let
∞
V
[An ]µ = [
n=0
∞
T
∞
W
An ]µ and
n=0
[An ]µ = [
n=0
∞
S
An ]µ .
n=0
4.2. Proposition. As defined above, ·c , ∧, and ∨ are well-defined.
Proof. Let (X, B, µ) be a probability space and let B̂ = {[A]µ : A ∈ B}. Let [A]µ = [B]µ ∈
B̂. Since A4B = Ac 4B c , 0 = µ(A4B) = µ(Ac 4B c ), so [Ac ]µ = [B c ]µ . Next, let
(An )n∈N , (Bn )n∈N ∈ B N such that [An ]µ = [Bn ]µ for all n ∈ N. Then µ((
∞
S
Ai )4(
i=0
= µ(((
∞
[
Ai ) \ (
i=0
∞
[
= µ(
∞
[
∞
[
(Ai \ (
i=0
j=0
∞
[
∞
[
≤ µ(
(Ai \ (
i=0
≤ µ(
∞
[
Bj )) ∪ ((
j=0
≤
Bj ) \ (
j=0
Bj )) ∪
∞
[
Bj ))) + µ(
j=0
∞
[
(Ai \ Bi )) + µ(
Ai )))
Ai )))
i=0
∞
[
(Bj \ (
j=0
∞
[
∞
[
Bj ))
j=0
i=0
(Bj \ (
j=0
i=0
∞
X
∞
[
∞
S
∞
[
Ai )))
i=0
(Bj \ Aj ))
j=0
µ(Ai \ Bi ) +
i=0
∞
X
µ(Bj \ Aj )
j=0
= 0,
so
∞
S
i=0
Ai ∼µ
∞
S
j=0
Bj . Thus [
∞
S
A i ]µ = [
i=0
∞
S
Bj ]µ and ∨ is well-defined. By the above,
j=0
[Acn ]µ = [Bnc ]µ for all n ∈ N, so µ((
∞
S
i=0
Aci )4(
∞
S
j=0
20
Bjc )) = 0. Then µ((
∞
T
i=0
Ai )4(
∞
T
j=0
Bj )) =
∞
S
µ((
i=0
[
∞
T
Aci )c 4(
∞
S
Bjc )c ) = µ((
j=0
∞
S
Aci )4(
i=0
∞
S
Bjc )) = 0, so
∞
T
Ai ∼µ
i=0
j=0
∞
T
Bj . Thus [
j=0
Bj ]µ and ∧ is well-defined.
∞
T
Ai ]µ =
i=0
j=0
4.3. Proposition. The set B̂ with 0 = [∅]µ , 1 = [X]µ , and operations ∧, ∨, and ·c is a
Boolean σ-algebra.
Proof. Let [A]µ , [B]µ , and [C]µ ∈ B̂.
(1) [A]µ ∨ [A]µ = [A ∪ A]µ = [A]µ
(2) [A]µ ∧ [A]µ = [A ∩ A]µ = [A]µ
(3) [A]µ ∨ [B]µ = [A ∪ B]µ = [B ∪ A]µ = [B]µ ∨ [A]µ
(4) [A]µ ∧ [B]µ = [A ∩ B]µ = [B ∩ A]µ = [B]µ ∧ [A]µ
(5) ([A]µ ∧ [B]µ ) ∨ [B]µ = [A ∩ B]µ ∨ [B]µ = [(A ∩ B) ∪ B]µ = [B]µ
(6) ([A]µ ∨ [B]µ ) ∧ [B]µ = [A ∪ B]µ ∧ [B]µ = [(A ∪ B) ∩ B]µ = [B]µ
(7) ([A]µ ∨ [B]µ ) ∨ [C]µ = [A ∪ B]µ ∨ [C]µ = [(A ∪ B) ∪ C]µ = [A ∪ (B ∪ C)]µ =
[A]µ ∨ [(B ∪ C)]µ = [A]µ ∨ ([B]µ ∨ [C]µ )
(8) ([A]µ ∧ [B]µ ) ∧ [C]µ = [A ∩ B]µ ∧ [C]µ = [(A ∩ B) ∩ C]µ = [A ∩ (B ∩ C)]µ =
[A]µ ∧ [B ∩ C]µ = [A]µ ∧ ([B]µ ∧ [C]µ )
(9) [A]µ ∨ ([B]µ ∧ [C]µ ) = [A]µ ∨ [B ∩ C]µ = [A ∪ (B ∩ C)]µ = [(A ∪ B) ∩ (A ∪ C)]µ =
[A ∪ B]µ ∧ [A ∪ C]µ = ([A]µ ∨ [B]µ ) ∧ ([A]µ ∨ [C]µ )
(10) [A]µ ∧ ([B]µ ∨ [C]µ ) = [A]µ ∧ [B ∪ C]µ = [A ∩ (B ∪ C)]µ = [(A ∩ B) ∪ (A ∩ C)]µ =
[A ∩ B]µ ∨ [A ∩ C]µ = ([A]µ ∧ [B]µ ) ∨ ([A]µ ∧ [C]µ )
(11) [A]µ ∨ [A]cµ = [A]µ ∨ [Ac ]µ = [A ∪ Ac ]µ = [X]µ
(12) [A]cµ ∧ [A]µ = [Ac ∩ A]µ = [∅]µ
(13) [∅]µ 6= [X]µ since µ(∅4X) = 1 > 0
21
Let h[Ai ]µ i ∈ B̂ N . Let B =
[An ]µ ∧[
∞
S
Ai ]µ = [
i=0
∞
S
∞
W
[Ai ]µ . For all n ∈ N, [An ]µ ∧ B = [An ]µ ∧ (
i=0
∞
W
[Ai ]µ ) =
i=0
(An ∩Ai )]µ = [(An ∩An )∪(
i=0
S
(An ∩Ai ))]µ = [An ]µ , so [An ]µ ≤ B.
i6=n
Thus B is an upper bound for h[Ai ]µ i. Now suppose [C]µ = Ĉ ∈ B̂ is another upper bound
for h[Ai ]µ i. Then B ∧ Ĉ = (
[
∞
S
(Ai ∩ C)]µ =
i=0
∞
W
∞
W
[Ai ]µ ) ∧ [C]µ = [
i=0
[(Ai ∩ C)]µ =
i=0
[Ai ]µ ≤ Ĉ, we have B ∧ Ĉ =
∞
W
∞
W
∞
S
i=0
([Ai ]µ ∧ [C]µ ) =
i=0
Ai ]µ ∧ [C]µ = [(
∞
W
∞
S
Ai ) ∩ C]µ =
i=0
([Ai ]µ ∧ Ĉ). Since for all i ∈ N,
i=0
[Ai ]µ = B, so B ≤ Ĉ. Hence B is the least upper bound
i=0
of h[Ai ]i, so B̂ is a Boolean σ-algebra.
Let µ̂ : B̂ → [0, 1] such that µ̂([A]µ ) = µ(A).
4.4. Proposition. µ̂ is well-defined and strictly positive.
Proof. Let A, C ∈ B such that [A]µ = [C]µ . By definition, A ∼µ C, so µ(A \ C) + µ(C \
A) = µ(A4C) = 0. Then µ(A \ C) = 0 = µ(C \ A), so µ(A) = µ(A) + µ(C \ A) =
µ(A ∪ (C \ A)) = µ(A ∪ C) = µ(C ∪ (A \ C)) = µ(C) + µ(A \ C) = µ(C). Hence
µ̂([A]µ ) = µ(A) = µ(C) = µ̂([C]µ ), so µ̂ is well-defined. Suppose µ̂([A]µ ) = 0. Then
µ(A) = 0, so µ(A4∅) = 0. Thus A ∼ ∅, so [A]µ = [∅]µ = 0.
4.5. Proposition. µ̂ is countably additive.
Proof. Let (Bn )n∈N ∈ B̂ N be pairwise disjoint, i.e. for all m 6= n, Bm ∧ Bn = 0. For
each n ∈ N, choose An ∈ B such that [An ]µ = Bn . Let m, n ∈ N. Then [∅]µ = 0 =
[Am ]µ ∧ [An ]µ = [Am ∩ An ]µ , so ∅ ∼ Am ∩ An . Therefore, 0 = µ((Am ∩ An )4∅) =
µ(((Am ∩ An ) \ ∅) ∪ (∅ \ (Am ∩ An ))) ≤ µ(((Am ∩ An ) \ ∅)) + ((∅ \ (Am ∩ An ))) =
µ(Am ∩ An ) + 0, so µ(Am ∩ An ) = 0.
To use the countable additivity of µ, we will construct a disjoint sequence (Dn )n∈N in B
such that µ(Dn ) = µ(An ) for all n ∈ N. Let D0 = A0 and Dn = An \ (
n−1
S
i=0
22
Ai ) for all
n > 0 and let n ∈ N. Then
µ(Dn ) = µ(An \ (
n−1
[
Ai ))
i=0
n−1
[
= µ(An \ (An ∩ (
Ai )))
i=0
= µ(An \ (
n−1
[
(An ∩ Ai )))
i=0
n−1
[
n−1
[
i=0
i=0
= µ(An ) − µ(
≤ µ(An ) −
(An ∩ Ai )) since
n−1
X
(An ∩ Ai ) ⊆ An
µ(An ∩ Ai )
i=0
= µ(An ) since µ(An ∩ Ai ) = 0 for all i ∈ N,
so
∞
X
µ̂Bi =
i=0
∞
X
µ̂[Ai ]µ
i=0
=
∞
X
µAi
i=0
=
∞
X
µDi
i=0
= µ(
∞
[
Di ) by countable additivity of µ
i=0
= µ(
∞
[
Ai )
i=0
= µ̂([
∞
[
Ai ]µ )
i=0
= µ̂(
_
[Ai ]µ ) by definition
i∈N
23
= µ̂(
_
Bi ).
i∈N
Thus µ̂ is countably additive.
Since µ̂([∅]µ ) = 0, µ̂([X]µ ) = µ(X) = 1, and µ̂ is countably additive, µ̂ is a probability
measure.
ˆ
Let dˆ : B̂ → R such that d([A]
µ , [B]µ ) = µ̂([A]µ 4[B]µ ).
4.6. Proposition. dˆ is a metric on B̂.
Proof. Let [A]µ , [B]µ , [C]µ ∈ B̂.
ˆ
(1) non-negativity: d([A]
µ , [B]µ ) = µ̂([A]µ 4[B]µ ) ≥ 0
ˆ
(2) coincidence: d([A]
µ , [B]µ ) = 0 ⇐⇒ µ̂([A]µ 4[B]µ ) = 0 ⇐⇒ µ̂([A4B]µ ) =
0 ⇐⇒ µ(A4B) = 0 ⇐⇒ A ∼µ B ⇐⇒ [A]µ = [B]µ
ˆ
(3) symmetry: d([A]
µ , [B]µ ) = µ̂([A]µ 4[B]µ ) = µ̂([A4B]µ ) = µ(A4B) = µ(B4A) =
ˆ
µ̂([B4A]µ ) = µ̂([B]µ 4[A]µ ) = d([B]
µ , [A]µ )
ˆ
ˆ
(4) triangle inequality: d([A]
µ , [B]µ ) + d([B]µ , [C]µ )
= µ̂([A]µ 4[B]µ ) + µ̂([B]µ 4[C]µ )
= µ(A4B) + µ(B4C)
= µ(A \ B) + µ(B \ A) + µ(B \ C) + µ(C \ B) since (A \ B) ∩ (B \ A) = ∅
= (B \ C) ∩ (C \ B)
≥ µ(A \ C) + µ(C \ A)
since A \ C ⊆ (A \ B) ∪ (B \ C) and C \ A ⊆ (C \ B) ∪ (B \ A)
= µ(A4C) since (A \ C) ∩ (C \ A) = ∅
24
= µ̂([A]µ 4[C]µ )
ˆ
= d([A]
µ , [C]µ ).
ˆ is a complete metric space.
4.7. Proposition. (B̂, d)
ˆ For all k ∈ N there exists a
Proof. (323F in [9]) Let han i ∈ B̂ N be Cauchy with respect to d.
ˆ m , an ) < 2−k . Note that hNk ik∈N is a nonminimal Nk ∈ N such that for all m, n ≥ Nk , d(a
decreasing sequence. For each k, let ck = aNk . Then
Let p0 =
W
n∈N
V
m≥n cm
and let p1 =
V
n∈N
W
P∞ −k
P∞ ˆ
< ∞.
k=0 2
k=0 d(ck , ck+1 ) ≤
m≥n cm .
Note that since B̂ is σ-complete,
p0 , p1 ∈ B̂.
First, I will show that p0 = p1 . Let αn = µ̂(cn 4cn+1 ) and βn =
∞
P
αk for all n ∈ N.
k=n
ˆ k , ck+1 ) < ∞, so
ˆ n , cn+1 ) = µ̂(cn 4cn+1 ) = αn for all n, P∞ αn = P∞ d(c
Since d(c
k=0
n=0
limn→∞ βn = 0. Let bn =
W
n≤k ck 4ck+1 .
µ̂(bn ) = µ̂(
_
Then for all n,
cm 4cm+1 ) ≤
∞
X
µ̂(cm 4cm+1 ) = βn
m=n
m≥n
by countable additivity of µ̂.
Fix n ∈ N and let n ≤ m. I will show that cm 4cn ≤
W
ck 4ck+1 . If m = n, then
n≤k<m
cm 4cn = 0 =
W
W
ck 4ck+1 =
W
∅
n≤k<m
ck 4ck+1 . For m ≥ n + 2,
W
k=n
ck 4ck+1 . If m = n + 1, then cm 4cn = cn+1 4cn =
ck 4ck+1
n≤k<m
_
=(
ck 4ck+1 ) ∨ ((cm−2 4cm−1 ) ∨ (cm−1 4cm ))
n≤k<m−2
≥(
_
ck 4ck+1 ) ∨ (cm−2 4cm ) by 4.1.5
n≤k<m−2
25
_
=(
ck 4ck+1 ) ∨ ((cm−3 4cm−2 ) ∨ (cm−2 4cm ))
n≤k<m−3
_
≥(
ck 4ck+1 ) ∨ (cm−3 4cm ) by 4.1.5
n≤k<m−3
...
≥ cn 4cm .
Thus, cm 4cn ≤
W
ck 4ck+1 ≤
W
n≤k<m
n≤k ck 4ck+1
= bn . Since cm 4cn ≤ bn , we have
(cn \bn )\cm = (cn ∧bcn )∧ccm = (cn ∧ccm )∧bcn = (cn \cm )\bn ≤ (cn 4cm )\bn ≤ bn \bn = 0,
so (cn \ bn ) \ cm = 0. By 4.1.6, cn \ bn ≤ cm .
We can also use cm 4cn ≤ bn to show that cm ≤ cn ∨ bn . Since cm \ cn ≤ bn , (cm \ cn ) ∧
bn = cm \ cn , so ((cm \ cn ) ∧ bn ) ∨ cn = (cm \ cn ) ∨ cn . Expanding both sides, we get
(cm ∧ ccn ∧ bn ) ∨ cn = (cm ∧ ccn ) ∨ cn , so (cm ∨ cn ) ∧ (bn ∨ cn ) = (cm ∨ cn ) ∧ 1 ∧ (bn ∨ cn ) =
(cm ∨ cn ) ∧ (ccn ∨ cn ) ∧ (bn ∨ cn ) = (cm ∨ cn ) ∧ 1 = cm ∨ cn . Thus cm ≤ cm ∨ cn ≤ cn ∨ bn .
Putting this together, we get
c n \ bn ≤ c m ≤ c n ∨ bn .
Since this holds for every m ≥ n,
c n \ bn ≤
^
cm ≤
m≥k
_
c m ≤ c n ∨ bn
m≥k
for all k ≥ n. Since this holds for every k ≥ n, cn \bn ≤
cn ∨ bn . In other words, since p0 =
W
k∈N
V
m≥k cm
W
k∈N
and p1 =
V
m≥k cm
V
k∈N
W
≤
V
k∈N
m≥k cm ,
W
m≥k cm
≤
we have
c n \ bn ≤ p 0 ≤ p 1 ≤ c n ∨ bn .
I will use this to show that cn 4pi ≤ bn for i = 0, 1. Let i ∈ {0, 1}. Since cn \ bn ≤ pi ,
we have (cn \ pi ) \ bn = (cn \ bn ) \ pi ≤ pi \ pi = 0, so (cn \ pi ) \ bn = 0. By 4.1.6,
26
cn \ pi ≤ bn . Since pi ≤ cn ∨ bn , pi ∧ (cn ∨ bn ) = pi , so (pi \ cn ) \ bn = pi ∧ ccn ∧ bcn =
pi ∧ (cn ∨ bn )c = (pi ∧ (cn ∨ bn )) ∧ (cn ∨ bn )c = 0. By 4.1.6, pi \ cn ≤ bn . Thus
cn 4pi = (cn \ pi ) ∨ (pi \ cn ) ≤ bn .
Another application of 4.1.6 gives us p1 \ p0 ≤ bn :
(p1 \ p0 ) \ bn = p1 ∧ pc0 ∧ bcn
= (p1 ∧ (cn ∨ bn )) ∧ pc0 ∧ bcn since p1 ≤ cn ∨ bn
= p1 ∧ pc0 ∧ (bcn ∧ (cn ∨ bn ))
= p1 ∧ pc0 ∧ ((bcn ∧ cn ) ∨ (bcn ∧ bn ))
= p1 ∧ pc0 ∧ ((bcn ∧ cn ) ∨ 0)
= p1 ∧ pc0 ∧ (cn \ bn )
= p1 ∧ pc0 ∧ ((cn \ bn ) ∧ p0 ) since cn \ bn ≤ p0
= p1 ∧ (cn \ bn ) ∧ (pc0 ∧ p0 )
= p1 ∧ (cn \ bn ) ∧ 0
= 0.
Since cn 4pi ≤ bn holds for every n,
limn→∞ µ̂(cn 4pi ) ≤ limn→∞ µ̂(bn ) ≤ limn→∞ βn = 0
for i = 0, 1. Furthermore, since p0 ≤ p1 , µ̂(p1 4p0 ) = µ̂(p1 \ p0 ) ≤ limn→∞ µ̂(bn ) = 0.
In fact, this implies that p0 = p1 . To see this, let P0 , P1 ∈ B such that p0 = [P0 ]µ ,
p1 = [P1 ]µ . Then 0 = µ̂(p1 4p0 ) = µ̂([P1 ]µ 4[P0 ]µ ) = µ̂([P1 4P0 ]µ ) = µ(P1 4P0 ), so
p0 = [P0 ]µ = [P1 ]µ = p1 .
27
ˆ n , p) =
Now let p = p0 = p1 . I will show that limn→∞ an = limn→∞ cn = p. Since limn→∞ d(c
ˆ Let > 0. Since hcn in∈N
limn→∞ µ̂(cn 4p) = 0, hcn in∈N converges to p with respect to d.
ˆ k , p) < /2. Since
converges to p, there exists M1 ∈ N such that for all k > M1 , d(c
ˆ m , an ) < /2.
han in∈N is Cauchy, there exists M2 ∈ N such that for all m, n > M2 , d(a
ˆ N , p) =
ˆ k , p) ≤ d(a
ˆ k , aN ) + d(a
Let M = max(M1 , M2 ). Then k > M implies d(a
k
k
ˆ k , aN ) + d(c
ˆ k , p) < /2 + /2 = , so the original sequence han in∈N converges to
d(a
k
ˆ is a complete metric
p ∈ B̂. Since han in∈N was an arbitrary Cauchy sequence, (B̂, d)
space.
ˆ is a complete metric space, M = (B̂, 0, 1, ·c , ∧, ∨, µ̂) as constructed above
Since (B̂, d)
is a metric structure. We call such an M a probability algebra, or, in the context of its
being a metric structure, a probability structure.
4.8. Here are some examples of probability structures.
(1) Let (X, B, µ) be a probability space where X = {a1 , a2 , ..., a10 }, B = P(X), and
µ(an ) = 1/10 for all n = 1, ..., 10. Since there are no nontrivial events of measure
zero, µ = µ̂. The distance between two events is the probability that exactly one of
them happens.
(2) Let (X, B, µ) be a probability space where X is the unit interval [0, 1], B is the
σ-algebra of Borel sets, and µ is the Lebesgue measure on [0, 1]. B̂ identifies events
whose symmetric difference is a null set.
(3) Let X = {0, 1}N , the set of infinite sequences of 0s and 1s. B is the σ-algebra
generated by sets of the form {(x1 , x2 , ...) ∈ X : (x1 , ..., xn ) = (a1 , ..., an )}. If
we think of X as the set of countably many tosses of a fair coin, then a generator
28
of B is the event that the first n tosses are a fixed sequence (a1 , ..., an ). Then
µ({(x1 , x2 , ...) ∈ X : (x1 , ..., xn ) = (a1 , ..., an )}) = 2−n . Two distinct generator
sets have symmetric difference > 0, and thus remain distinct in B̂.
Definition. The signature associated with probability structures is LP r = (R, F, η, G),
where
ˆ
(1) R = {µ̂, d}
(2) F = {0, 1, ·c , ∨, ∧}
ˆ = η(∨) = η(∧)
(3) 0 = η(0) = η(1), 1 = η(µ̂) = η(·c ), 2 = η(d)
(4) G = {δ·c , δµ̂ , δ∨ , δ∧ , δdˆ} where δ·c () = δµ̂ () = and δ∨ () = δ∧ () = δdˆ() = /2.
4.9. Proposition. As defined above, δ·c , δµ̂ , δ∨ , δ∧ , δdˆ are indeed moduli of uniform continuity.
Proof. δ·c : Let ∈ (0, 1] and [A]µ , [B]µ ∈ B̂. Suppose d([A]µ , [B]µ ) < δ·c () = . Then
d([A]cµ , [B]cµ ) = d([Ac ]µ , [B c ]µ ) = µ(Ac 4B c ) = µ(A4B) = d([A]µ , [B]µ ) < .
δµ̂ : Let ∈ (0, 1] and [A]µ , [B]µ ∈ B̂. Suppose d([A]µ , [B]µ ) < δµ̂ () = . Since
2µ̂(a ∧ b) ≤ 2µ̂(a), we have −µ̂(a) + 2µ̂(a ∧ b) ≤ µ̂(a), so −µ̂(a4b) = −µ̂(a) − µ̂(b) +
2µ̂(a ∧ b) ≤ µ̂(a) − µ̂(b). Since 2µ̂(a ∧ b) ≤ 2µ̂(b), we have −µ̂(b) ≤ µ̂(b) − 2µ̂(a ∧ b),
so µ̂(a) − µ̂(b) ≤ µ̂(a) + µ̂(b) − 2µ̂(a ∧ b) = µ̂(a4b). Putting this together, we get
d(µ̂(a), µ̂(b)) = |µ̂(a) − µ̂(b)| ≤ µ̂(a4b) < .
δ∨ : Let ∈ (0, 1] and ([A1 ]µ , [A2 ]µ ), ([B1 ]µ , [B2 ]µ ) ∈ B̂ × B̂. Suppose
dmax (([A1 ]µ , [A2 ]µ ), ([B1 ]µ , [B2 ]µ )) < δ∨ ()
where δ∨ () = /2. Then for i = 1, 2, /2 > d([Ai ]µ , [Bi ]µ ) = µ(Ai 4Bi ) = µ(Ai \ Bi ) +
µ(Bi \ Ai ). Since (A1 ∪ A2 ) \ (B1 ∪ B2 ) = (A1 \ (B1 ∪ B2 )) ∪ (A2 \ (B1 ∪ B2 )) ⊆
29
(A1 \ B1 ) ∪ (A2 \ B2 ) and (B1 ∪ B2 ) \ (A1 ∪ A2 ) = (B1 \ (A1 ∪ A2 )) ∪ (B2 \ (A1 ∪ A2 )) ⊆
(B1 \ A1 ) ∪ (B2 \ A2 ), we have d([A1 ]µ ∨ [A2 ]µ , [B1 ]µ ∨ [B2 ]µ )
= d([A1 ∪ A2 ]µ , [B1 ∪ B2 ]µ )
= µ((A1 ∪ A2 )4(B1 ∪ B2 ))
= µ((A1 ∪ A2 ) \ (B1 ∪ B2 )) + µ((B1 ∪ B2 ) \ (A1 ∪ A2 ))
≤ µ((A1 \ B1 ) ∪ (A2 \ B2 )) + µ((B1 \ A1 ) ∪ (B2 \ A2 ))
≤ µ(A1 \ B1 ) + µ((A2 \ B2 ) + µ(B1 \ A1 ) + µ(B2 \ A2 )
= (µ(A1 \ B1 ) + µ(B1 \ A1 )) + (µ((A2 \ B2 ) + µ(B2 \ A2 ))
= µ(A1 4B1 ) + µ(A2 4B2 )
= d([A1 ]µ , [B1 ]µ ) + d([A2 ]µ , [B2 ]µ )
≤ 2max{d([A1 ]µ , [B1 ]µ ), d([A2 ]µ , [B2 ]µ )}
< 2(/2)
= .
δ∧ : Let ∈ (0, 1] and ([A1 ]µ , [A2 ]µ ), ([B1 ]µ , [B2 ]µ ) ∈ B̂ × B̂. Suppose
dmax (([A1 ]µ , [A2 ]µ ), ([B1 ]µ , [B2 ]µ )) < δ∧ ()
where δ∧ () = /2. Since (A1 ∩ A2 ) \ (B1 ∩ B2 ) = ((A1 ∩ A2 ) \ B1 ) ∪ ((A1 ∩ A2 ) \ B2 ) ⊆
(A1 \ B1 ) ∪ (A2 \ B2 ) and (B1 ∩ B2 ) \ (A1 ∩ A2 ) = ((B1 ∩ B2 ) \ A1 ) ∪ ((B1 ∩ B2 ) \ A2 ) ⊆
(B1 \ A1 ) ∪ (B2 \ A2 ), we have d([A1 ]µ ∧ [A2 ]µ , [B1 ]µ ∧ [B2 ]µ )
= d([A1 ∩ A2 ]µ , [B1 ∩ B2 ]µ )
30
= µ((A1 ∩ A2 )4(B1 ∩ B2 ))
= µ((A1 ∩ A2 ) \ (B1 ∩ B2 )) + µ((B1 ∩ B2 ) \ (A1 ∩ A2 ))
≤ µ((A1 \ B1 ) ∪ (A2 \ B2 )) + µ((B1 \ A1 ) ∪ (B2 \ A2 ))
≤ µ(A1 \ B1 ) + µ(A2 \ B2 ) + µ(B1 \ A1 ) + µ(B2 \ A2 )
= (µ(A1 \ B1 ) + µ(B1 \ A1 )) + (µ((A2 \ B2 ) + µ(B2 \ A2 ))
= µ(A1 4B1 ) + µ(A2 4B2 )
= d([A1 ]µ , [B1 ]µ ) + d([A2 ]µ , [B2 ]µ )
≤ 2max{d([A1 ]µ , [B1 ]µ ), d([A2 ]µ , [B2 ]µ )}
< 2(/2)
= .
δdˆ: Let ∈ (0, 1] and ([A1 ]µ , [A2 ]µ ), ([B1 ]µ , [B2 ]µ ) ∈ B̂ × B̂. Suppose
dmax (([A1 ]µ , [A2 ]µ ), ([B1 ]µ , [B2 ]µ )) < δdˆ()
where δdˆ() = /2. By the triangle inequality,
ˆ
ˆ
d(([A
1 ]µ , [A2 ]µ )) − d(([B1 ]µ , [B2 ]µ ))
ˆ
ˆ
ˆ
ˆ
≤ d(([A
1 ]µ , [B1 ]µ )) + d(([B1 ]µ , [B2 ]µ )) + d(([B2 ]µ , [A2 ]µ )) − d(([B1 ]µ , [B2 ]µ ))
ˆ
ˆ
= d(([A
1 ]µ , [B1 ]µ )) + d(([B2 ]µ , [A2 ]µ ))
< /2 + /2
= .
31
since max{d([A1 ]µ , [B1 ]µ ), d([A2 ]µ , [B2 ]µ )} < /2. Similarly,
ˆ
ˆ
d(([A
1 ]µ , [A2 ]µ )) − d(([B1 ]µ , [B2 ]µ ))
ˆ
ˆ
ˆ
ˆ
≥ d(([A
1 ]µ , [A2 ]µ )) − (d(([B1 ]µ , [A1 ]µ )) + d(([A1 ]µ , [A2 ]µ )) + d(([A2 ]µ , [B2 ]µ )))
ˆ
ˆ
= −(d(([B
1 ]µ , [A1 ]µ )) + d(([A2 ]µ , [B2 ]µ )))
> −(/2 + /2)
= −.
ˆ
ˆ
ˆ
Putting this together, we get d(d(([A
1 ]µ , [A2 ]µ )), d(([B1 ]µ , [B2 ]µ ))) = |d(([A1 ]µ , [A2 ]µ )) −
ˆ
d(([B
1 ]µ , [B2 ]µ ))| < .
Notation. A Boolean ring is a ring (A, +, ·) where a2 = a for all a ∈ A. We can represent
any Boolean algebra A = (A, 0, 1, ∧, ∨, ·c ) as a Boolean ring (A, +, ·) by taking a + b :=
a4b and ab := a ∧ b. Conversely, given a Boolean ring (A, +, ·), we can view it as a
Boolean algebra by taking a ∨ b := a + b + ab, a ∧ b := ab, and ac := 1 + a. Depending
on the context, one representation may be preferred over the other.
Definition. (311F in [9]) Given a Boolean ring B, the Stone space of B is the set Z of
nonzero ring homomorphisms from B to Z2 , i.e. z ∈ Z ⇐⇒ ∀a, b ∈ B z(a + b) =
z(a) + z(b), z(ab) = z(a)z(b), z(1) = 1Z2 . For each a ∈ B, let â = {z ∈ Z : z(a) = 1}.
The Stone representation of B is the canonical map a 7→ â : B → P(Z).
4.10. Lemma. (a special case of 311D in [9]) Let B be a Boolean ring and let a ∈ B be
nonzero. Then there exists z ∈ Z such that z(a) = 1.
Proof. Let I = {J : J is an ideal of B and a ∈
/ J}. First I will use Zorn’s lemma on (I, ⊆)
to show that I has a maximal element. Note that I is nonempty because {0} ∈ I. Let J
32
be a nonempty totally ordered subset of I and let J0 =
S
J . Let m, n ∈ J0 . Then there
exist J1 , J2 ∈ J such that m ∈ J1 and n ∈ J2 . Since J is totally ordered, J1 ⊆ J2 or
J2 ⊆ J1 . Without loss of generality, suppose J1 ⊆ J2 , so m, n ∈ J2 . Let r ∈ B. Since
J2 is an ideal, 0, m + n, rm ∈ J2 ⊆ J0 , so J0 is an ideal. Since a ∈
/ J0 , J0 ∈ I, so J0 is
an upper bound for J . Since J was arbitrary, by Zorn’s lemma, I has a maximal element,
which we will denote by K.
Next, I will show that for all m, n ∈ B \ K, m + n ∈ K. For all b ∈ B, let Kb =
{c ∈ B : bc ∈ K}. Let b ∈ B. Since K is an ideal, for all n ∈ K, we have nb ∈ K,
so n ∈ Kb . Thus K ⊆ Kb . If n, n0 ∈ Kb and c ∈ B, then b(n + n0 ) = bn + bn0 ∈ K
and b(nc) ∈ K, so n + n0 , nc ∈ Kb . Since 0 ∈ K ⊆ Kb , Kb is an ideal. If a ∈
/ Kb , then
Kb ∈ I. Since K is maximal in I and K ⊆ Kb , we have K = Kb . In particular, a ∈
/ Ka ,
since a2 = a ∈
/ K, so Ka = K. Now let b, c ∈ B \ K. Since b ∈
/ K = Ka , we have
ba = ab ∈
/ K, so a ∈
/ Kb . Hence Kb = K. Since c ∈
/ K = Kb , we have bc ∈
/ K. Now
bc(b + c) = b2 c + bc2 = bc + bc = 0 ∈ K, so b + c ∈ Kbc . Since bc ∈
/ K and for all e ∈
/ K,
K = Ke , we have K = Kbc , so b + c ∈ K.
Now I will construct a ring homomorphism z : B → Z2 such that z(a) = 1. For all
d ∈ B, let
z(d) :=




0 if d ∈ K



1 if d ∈ B \ K.
Let b, c ∈ B. If b, c ∈ K, then b + c, bc ∈ K, so z(b + c) = 0 = 0 + 0 = z(b) + z(c) and
z(bc) = 0 = 0 · 0 = z(b)z(c). If b ∈ K and c ∈ B \ K, then b + c ∈
/ K, since otherwise
c = b + (b + c) ∈ K, so z(b + c) = 1 = 0 + 1 = z(b) + z(c). Since bc ∈ K, z(bc) =
0 = 0 · 1 = z(b)z(c). If b ∈ B \ K and c ∈ K, then z(b + c) = 1 = 1 + 0 = z(b) + z(c)
33
and bc ∈ K, z(bc) = 0 = 1 · 0 = z(b)z(c). Finally, if b, c ∈ B \ K, then as shown in
the previous paragraph, b + c ∈ K and bc ∈
/ K, so z(b + c) = 0 = 1 + 1 = z(b) + z(c)
and z(bc) = 1 = 1 · 1 = z(b)z(c). Hence z is a ring homomorphism and, since a ∈
/ K,
z(a) = 1.
4.11. Proposition. (311E in [9]) The Stone representation of a Boolean ring B is an injective ring homomorphism a 7→ â : (B, +, ·) → (P(Z), 4, ∩).
Proof. Let B be a Boolean algebra and let a, b ∈ B. Then a[
+ b = {z : z(a + b) = 1} =
{z : z(a) + z(b) = 1} = {z : z(a) = 0 and z(b) = 1, or z(a) = 1 and z(b) = 0} =
b = {z : z(ab) = 1} = {z : z(a)z(b) = 1} = {z :
(â ∩ (b̂)c ) ∪ (b̂ ∩ (â)c ) = â4b̂ and ab
z(a) = 1 and z(b) = 1} = â ∩ b̂, so a 7→ â is a ring homomorphism.
To see that this ring homomorphism is injective, let a ∈ B be nonzero. By 4.10, there
exists z ∈ Z such that z(a) = 1. Thus z ∈ â, so â 6= ∅. Hence the kernel of this
homomorphism is {0}, so a 7→ â is injective.
By 4.11, any Boolean ring B is isomorphic to its image {â : a ∈ B} under the Stone
representation. In particular, ˆ· commutes with Boolean operations: for all a, b ∈ B,
\∧ b) = a +
b = â ∪ b̂
\
(1) a[
∨ b = a4b4(a
b + ab = â4b\
+ ab = â4b̂4ab
b = â ∩ b̂
(2) a[
∧ b = ab
\
b = â4(â ∩ b̂) = (â \ (â ∩ b̂)) ∪ ((â ∩
(3) ad
\ b = a\
∧ bc = a(1
+ b) = a\
+ ab = â4ab
b̂) \ â) = â \ (â ∩ b̂) = â \ b̂
d = a[
(4) a4b
+ b = â4b̂.
34
Also note that a 7→ â : B → P(Z) respects the order relation on B, i.e. for all a, b ∈ B,
a ≤ b ⇐⇒ â ⊆ b̂. Finally, we have 1̂ = Z: Let z ∈ Z. By definition, z a is nonzero ring
homomorphism, so there exists a ∈ B such that z(a) = 1. Then 1 = z(a) = z(a ∧ 1) =
z(a1) = z(a)z(1) = 1z(1) = z(1), so z ∈ 1̂ = Z.
4.12. Lemma. Let B be a Boolean algebra and let Z be its Stone space. Let T be
{G ⊆ Z : ∀z ∈ G ∃a ∈ B z ∈ â ⊆ G}.
Then T is a topology on Z under which Z is a Hausdorff space.
Proof. First, we show that T is a topology. Let z ∈ Z. Since z 6= 0, there exists a ∈ B
such that z(a) = 1, so z ∈ â. Thus Z ⊆
S
E. Now let â, b̂ ∈ E and z ∈ â ∩ b̂. By 311G,
â ∩ b̂ = a[
∧ b ∈ E, so in particular, z ∈ a[
∧ b ⊆ â ∩ b̂. Thus E is a base for some topology
on Z. Since T is defined to be the union of elements of E, T is the topology generated by
E.
To see that (Z, T ) is Hausdorff, let w, z ∈ Z be distinct. Then there exists a ∈ B
such that z(a) 6= w(a), say z(a) = 1 and w(a) = 0. Since w is nonzero, there exists
b ∈ B such that w(b) = 1. Then w(b \ a) = w(b ∧ ac ) = w(b(1 + a)) = w(b +
ba) = w(b) + w(b)w(a) = 1, so w ∈ bd
\ a ∈ T . Since z(a) = 1, z ∈ â ∈ T . Also,
[ = 1̂41̂ = ∅, so bd
(bd
\ a) ∩ â = (b \
\ a) ∧ a = b ∧\
(ac ∧ a) = 0̂ = 141
\ a and â are
disjoint neighborhoods of w and z, respectively. Hence (Z, T ) is Hausdorff.
4.13. (311I in [9]) The Stone space Z of a Boolean algebra B is compact. E = {â : a ∈ B}
is precisely the set of compact and open subsets of Z.
Definition. Let X be a set and let B be a σ-algebra of subsets of X. A ⊆ B is a σ-ideal if
it satisfies the following properties:
35
(1) ∅ ∈ A
(2) For all A ∈ A, B ∈ B, if B ⊆ A, then B ∈ A
(3) If hAn in∈N ∈ AN , then
S
n∈N
An ∈ A
Definition. Let X be a topological space. For A ⊆ X, denote the closure of A by A and
the interior of A by intA. A ⊆ X is nowhere dense if intA = ∅. A ⊆ X is meager if there
exist countably many Bn ⊆ X such that A =
S
n∈N
Bn and each Bn is nowhere dense.
Definition. (312F in [9]) Given Boolean algebras A and B, a function π : A → B is
a Boolean homomorphism if π(a4b) = π(a)4π(b) and π(a ∩ b) = π(a) ∩ π(b) for all
a, b ∈ A, and π(1A ) = 1B .
4.14. Lemma. (313Ca in [9]) Let A be a σ-complete Boolean algebra and let Z be its
Stone space. Let han i be a sequence in A and let a = supn∈N an ∈ A. Then
Proof. For all n, an ≤ a, so abn ⊆ â by 4.11. Then â ⊇
closed. To show that â ⊆
∅. Since
S
n∈N
S
n∈N
n∈N
abn =
S
n∈N
n∈N
S
S
n∈N
n∈N
abn = â.
abn , since â is
abn , suppose for the sake of contradiction that â\
abn is closed, â \
nonzero base element b̂ ⊆ â \
S
S
S
n∈N
abn 6=
S
abn = â ∩ ( n∈N abn )c is open, so there exists a
abn . Then for all n, abn ⊆ â \ b̂ = ad
\ b, so by 4.11,
an ≤ a \ b, which makes a \ b an upper bound for han i. Since b < a = supn∈N an , we have
a contradiction. Thus
S
n∈N
abn = â.
Let X be a topological space. Recall that X is a Baire space if the intersection of every
countable collection of dense open sets in X is dense, and that X is a Hausdorff space if
distinct points in X have disjoint open neighborhoods. In the following lemmas, we will
use a weakened version of the Baire category theorem: If X is a compact Hausdorff space,
then X is a Baire space.
36
4.15. Lemma. In a compact Hausdorff space, if a set is meager and open, then it is empty.
Proof. Let X be a compact Hausdorff space. By the Baire category theorem, X is a Baire
space. Let E ⊆ H be open and meager. Since E is meager, we can write E as the union
of countably many nowhere dense sets {Mn }n∈N . Fix n ∈ N. By definition, int(Mn ) = ∅,
so X = X \ int(Mn ) = X \ Mn = int(X \ Mn ). Thus int(X \ Mn ) is dense in X.
Also int(X \ Mn ) is open, so since X is a Baire space,
T
int(X \ Mn ) is dense. Then
n∈N
T
int(X \ Mn ) ⊆
n∈N
T
(X \ Mn ) = X \
n∈N
closed, so X =
T
S
Mn = X \ E. Since E is open, X \ E is
n∈N
int(X \ Mn ) ⊆ X \ E = X \ E. Hence E is empty.
n∈N
4.16. Lemma. In any topological space Z, the family M of meager subsets of Z forms a
σ-ideal.
Proof. Let Z be a topological space and let M be the family of meager subsets of X. Since
int∅ = ∅, ∅ ∈ M. Let A ∈ M and B ⊆ A. Let hAn in∈N be a sequence of subsets of Z
such that
Since
S
S
n∈N
n∈N
An = A and for all n, intAn = ∅. Then for all n, intB ∩ An ⊆intAn = ∅.
B ∩ An = B ∩
S
n∈N
An = B ∩ A = B, B ∈ M. Now let hAn in∈N ∈ MN . For
each n, there exists hAin ii∈N a sequence of subsets of Z such that
S
i∈N
Ain = An and for
all i, intAin = ∅. Since hAin in,i∈N is countable, we can re-index it to a sequence hBk ik∈N .
Then for all k, Bk is nowhere dense, so
S
n∈N
An =
S
i,n∈N
Ain =
S
k∈N
Bk ∈ M.
The following theorem is the Loomis-Sikorski representation of a σ-complete Boolean
algebra. It will be used in the proof of the Stone representation theorem of a measure
algebra in 4.20.
4.17. Theorem. (314M in [9]) Let A be a σ-complete Boolean algebra, and Z its Stone
space. Let E = {â : a ∈ A}, and let M be the σ-ideal of meager subsets of Z. Then
37
Σ = {E4A : E ∈ E, A ∈ M} is a σ-algebra of subsets of Z, M is a σ-ideal of Σ, and A
is isomorphic, as Boolean algebra, to Σ/M.
Proof. First, I will show that Σ is a σ-algebra and M is a σ-ideal of Σ. Since ∅ = 0̂ ∈ E,
∅4∅ ∈ Σ. Let F ∈ Σ. There exist a ∈ B, A ∈ M such that F = â4A. Since (â)c = abc ∈
E,
F c = (â4A)c
= ((â ∩ Ac ) ∪ (âc ∩ A))c
= (â ∩ Ac )c ∩ (âc ∩ A)c
= (âc ∪ A) ∩ (â ∪ Ac )
= ((âc ∪ A) ∩ Ac ) ∪ ((âc ∪ A) ∩ â)
= [(âc ∩ Ac ) ∪ (A ∩ Ac )] ∪ [(âc ∩ â) ∪ (A ∩ â)]
= (âc ∩ Ac ) ∪ (A ∩ â)
= (âc \ A) ∪ (A \ âc )
= âc 4A ∈ Σ.
Now let hFn in∈N ∈ ΣN . For each n, let an ∈ B, An ∈ M such that Fn = abn 4An . Since B
is σ-complete, a := supn∈N an ∈ B. By 4.14,
Let C := â \
S
S
n∈N
abn = â.
abn . I will show that C has empty interior and is nowhere dense. Since
S
S
S
abn is open, so C = â ∩ ( n∈N abn )c is closed. Since n∈N abn is open,
S
abn ), so ∂
each abn is open,
S
n∈N
S
n∈N
S
S
S
abn ) \ int( n∈N abn ) = cl( n∈N abn ) \
abn = int(
n∈N
abn = C. Suppose C contains an open set U . Then U and
n∈N
n∈N
abn = cl(
S
n∈N
38
n∈N
S
n∈N
abn are disjoint. Since
U is open and in the boundary of
S
n∈N
abn , U and
S
n∈N
abn has nonempty intersection,
contradiction. Hence int(C) = ∅. Since C is closed, C = C, so int(C) = ∅, i.e. C is
nowhere dense.
Now let A = â4
S
n∈N
Fn . I will show that A ⊆ C ∪ (
S
n∈N
An ). Let x ∈ A. First
S
suppose x ∈ ( n∈N Fn ) \ â. Then there exists n ∈ N such that x ∈ Fn = abn 4An .
In other words, x ∈ abn ∪ An and x ∈
/ abn ∩ An . Since abn ⊆ â and x ∈
/ â, we must
S
S
have x ∈ An ⊆ C ∪ ( n∈N An ). Next, suppose x ∈ â \ n∈N Fn . I will show that if
x∈
/
S
n∈N
An , then x ∈ C, so suppose x ∈
/
S
n∈N
An . Since x ∈
/
S
n∈N
Fn =
S
n∈N
abn 4An ,
for all n either x ∈ abn ∩ An or x ∈
/ abn ∪ An . By assumption, x ∈
/ abn ∩ An , so x ∈
/ abn ∪ An .
Thus x ∈
/
σ-ideal,
S
S
n∈N
abn , so x ∈ â \
S
n∈N
S
abn = C. Thus A ⊆ C ∪ ( n∈N An ). Since M is a
S
A
∈
M,
so
since
C
is
nowhere
dense,
C
∪
(
n
n∈N An ) ∈ M. Finally, since
n∈N
A ⊆ C∪(
S
n∈N An ), A ∈ M. Thus,
S
n∈N Fn = ∅4
S
n∈N Fn = (â4â)4
S
n∈N
Fn =
â4A ∈ Σ, so Σ is closed under countable unions. Hence, Σ is a σ-algebra. Since ∅ = 0̂ ∈
E, M = {∅4A : A ∈ M} ⊆ Σ, so by 4.16 M is a σ-ideal of Σ.
Next, let ∼ be a relation on Σ such that for all F1 , F2 ∈ Σ,
F1 ∼ F2 ⇐⇒ ∃K ∈ M such that F1 = F2 4K.
I will show that ∼ is an equivalence relation. Let F1 , F2 , F3 ∈ Σ. Since F1 = F1 4∅
and ∅ ∈ M, ∼ is reflexive. Suppose F1 = F2 4K for some K ∈ M. Then F1 4K =
(F2 4K)4K = F2 4∅ = F2 , so ∼ is symmetric. Finally, let J, K ∈ M such that F1 =
F2 4K and F2 = F3 4J. Then F1 = F2 4K = (F3 4J)4K = F3 4(J4K) and since
J4K ∈ M, ∼ is transitive.
Denote the equivalence class of F ∈ Σ under ∼ by [F ]M and let Σ/M = {[F ]M : F ∈
Σ}. Define the operations 4M and ∧M on Σ/M as follows: for all f1 = [F1 ]M , f2 =
39
[F2 ]M ∈ Σ/M, let f1 4M f2 = [F1 4F2 ]M and f1 ∧M f2 = [F1 ∩ F2 ]M . When context distinguishes between 4 and 4M , and ∧ and ∧M , we will drop the subscript M. I
will now show that 4M and ∧M are well-defined, making ∼ a congruence relation. Let
F1 , F2 , G1 , G2 ∈ Σ and suppose F1 ∼ G1 and F2 ∼ G2 . Then there exist K1 , K2 ∈ M
such that F1 = G1 4K1 and F2 = G2 4K2 , so F1 4F2 = (G1 4K1 )4(G2 4K2 ) =
(G1 4G1 )4(K1 4K2 ). Since K1 4K2 ∈ M, F1 4G1 ∼ G1 4G2 . Also F1 ∧ F2 =
(G1 4K1 ) ∧ (G2 4K2 ) = (G1 ∧ (G2 4K2 ))4(K1 ∧ (G2 4K2 )) = ((G1 ∧ G2 )4(G1 ∧
K2 ))4((K1 ∧G2 )4(K1 4K2 )). Since (G1 ∧K2 )4(K1 ∧G2 )4(K1 4K2 ) ∈ M, F1 ∧G1 ∼
G1 ∧ G2 .
Let φ : A → Σ/M such that φ(a) = [â]M . I will show that φ is a Boolean homomorphism. Let a, b ∈ A. Then
d M
φ(a4b) = [a4b]
= [â4b̂]M
= {(â4b̂)4K : K ∈ M}
= {(â4K1 )4(b̂4K2 ) : K1 , K2 ∈ M}
= {x4y : x ∈ [â]M , y ∈ [b̂]M }
= [â]M 4[b̂]M
= φ(a)4φ(b),
φ(a ∧ b) = [a[
∧ b]M
= [â ∩ b̂]M
= {(â ∩ b̂)4K : K ∈ M}
40
= {(â ∩ b̂)4(â ∩ K2 )4(b̂ ∩ K1 )4(K1 ∩ K2 ) : K1 , K2 ∈ M}
= {(â4K1 ) ∩ (b̂4K2 ) : K1 , K2 ∈ M} by 4.1.4
= {x ∩ y : x ∈ [â]M , y ∈ [b̂]M }
= [â]M ∩ [b̂]M
= φ(a) ∩ φ(b),
φ(1) = [1̂]M
= 1Σ/M
Finally, I will show that A ∼
= Σ/M as Boolean algebras. Let F ∈ Σ. By definition,
F is expressible as E4A for some E ∈ E and A ∈ M, so φ is surjective. Note that
F 4E = (E4A)4E = (A4E)4E = A4(E4E) = A4∅ = A ∈ M. To see that E is
the unique element of E such that F 4E ∈ M, let E 0 ∈ E such that E 6= E 0 . By 4.13, E and
E 0 are open, so E4E 0 is open. Let x ∈ E4E 0 . If x ∈ A, then certainly x ∈ A ∪ (F 4E 0 ).
If x ∈
/ A, then x ∈ (E4E 0 ) \ A ⊆ (E4E 0 )4A = (E4A)4E 0 = F 4E 0 ⊆ A ∪ (F 4E 0 ).
Thus E4E 0 ⊆ A ∪ (F 4E 0 ). By 4.15, E4E 0 is not meager, so A ∪ (F 4E 0 ) ∈
/ M. Since
A ∈ M, we have F 4E 0 ∈
/ M. Hence E is unique, so φ is injective. Thus φ is a bijection.
Since φ is also a Boolean homomorphism, it is an isomorphism, so A ∼
= Σ/M.
4.18. Definition. (321A in [9]) A measure algebra is a pair (A, ν) where A is a σ-complete
Boolean algebra and ν : A → [0, ∞] is a function such that
(1) ν(0) = 0
(2) whenever han in∈N is a disjoint sequence in A (ai ∧ aj = 0 for distinct i, j),
ν(supn∈N an ) =
P∞
n=0
ν(an )
41
(3) ν is strictly positive (a = 0 ⇐⇒ ν(a) = 0)
Note that probability algebras are measure algebras with ν(1) = 1.
Given measure algebras (A, ν) and (B, µ), a function φ : A → B is a measure algebra
isomorphism if φ is a Boolean algebra isomorphism that also satisfies ν(E) = µ(φ(E)) for
all E ∈ A.
Definition. (112A in [8]) A measure space is a triple (X, Σ, µ) where
(1) X is a set
(2) Σ is a σ-algebra of subsets of X
(3) µ : Σ → [0, ∞] is a function such that
(a) µ(∅) = 0
(b) if hEn in∈N is a disjoint sequence in Σ, then µ(∪n∈N En ) =
P∞
n=0
µEn .
The measure algebra of a measure space (X, B, µ) is the construction (B̂, µ̂) in 4.3, which
was shown to be a measure algebra.
Definition. (313H in [9]) Let P, Q be posets. A function θ : P → Q is sequentially
order-continuous if the following are true:
(1) for all p1 , p2 ∈ P , p1 ≤ p2 =⇒ θ(p1 ) ≤ θ(p2 )
(2) if hpn in∈N is a non-decreasing sequence in P and p = supn pn , then θ(p) =
supn θ(pn )
4.19. Proposition. Let ≤Σ be a binary relation on Σ such that for all F1 , F2 ∈ Σ, F1 ≤Σ
F2 ⇐⇒ E1 ⊆ E2 , where E1 , E2 ∈ E such that Fi = Ei 4Ai for some A1 , A2 ∈ M. ≤Σ
is a partial order.
42
Proof. Let F1 , F2 ∈ Σ and E1 , E2 ∈ E such that Fi = Ei 4Ai for some A1 , A2 ∈ M.
There exist a1 , a2 ∈ A such that âi = Ei . Since E1 ⊆ E2 ⇐⇒ a1 ≤ a2 , F1 ≤Σ F2 ⇐⇒
a1 ≤ a2 . Since ≤ is a partial order on A, it follows immediately that ≤Σ is a partial order
on Σ.
The next theorem describes the Stone representation of a measure algebra.
4.20. Theorem. (321J in [9]) Let (A, µ̂) be a measure algebra. Then there exists a measure
space whose measure algebra is isomorphic, as a measure algebra, to (A, µ̂).
Proof. As a Boolean algebra, A is isomorphic to Σ/M as constructed in 314M. Denote
the canonical isomorphism by π : Σ/M → A, π([â]M ) = a. Let θ : Σ → A such that
θ(F ) = π([F ]M ) for all F ∈ Σ. I will show that θ is a sequentially order-continuous
surjective Boolean homomorphism with kernel M. Let A, B ∈ Σ. Then θ(A4B) =
π([A4B]M ) = π([A]M 4M [B]M ) = π([A]M )4π([B]M ) = θ(A)4θ(B). Similarly,
θ(A ∩ B) = θ(A) ∧ θ(B), so θ is a Boolean homomorphism. The kernel of θ is M
because for all F ∈ Σ, θ(F ) = 0 ⇐⇒ π([F ]M ) = 0 ⇐⇒ [F ]M = [∅]M ⇐⇒ F ∼
∅ ⇐⇒ ∃K ∈ M F = ∅4K = K ⇐⇒ F ∈ M. Since π and F 7→ [F ]M are surjective,
θ is surjective. To show that θ is sequentially order-continuous, let F1 , F2 ∈ Σ such that
F1 ≤Σ F2 . Since π is an isomorphism, there exist a1 , a2 ∈ A such that [Fi ]M = [âi ]M =
{âi 4K : K ∈ M} for i = 1, 2. Then there exist A1 , A2 ∈ M such that Fi = âi 4Ai
for i = 1, 2. Since F1 ≤Σ F2 , θ(F1 ) = π([F1 ]M ) = π([aˆ1 ]M ) = a1 ≤ a2 = π([aˆ2 ]M ) =
π([F2 ]M ) = θ(F2 ).
Now let hFn in∈N be a non-decreasing sequence in Σ. Suppose F = supn Fn ∈ Σ (if the
sup does not exist, then there is nothing to show). For each n, let En ∈ E, An ∈ M, an ∈ A
such that Fn = En 4An and aˆn = En . Also let E ∈ E, A ∈ M, a ∈ A such that
43
F = E4A and â = E. Then
F = supn Fn ⇐⇒ ∀n Fn ≤Σ F and ∀F 0 ∈ Σ if ∀n Fn ≤Σ F 0 , then F ≤Σ F 0
⇐⇒ ∀n En ⊆ E and ∀E 0 ∈ E if ∀n En ⊆ E 0 , then E ⊆ E 0
⇐⇒ ∀n an ≤ a and ∀a0 ∈ A if ∀n an ≤ a0 , then a ≤ a0
⇐⇒ a = supn an ,
so θ(F ) = π([F ]M ) = π([â]M ) = a = supn an = supn π([aˆn ]M ) = supn π([Fn ]M ) =
supn θ(Fn ). Hence, θ is sequentially order-continuous.
Let ν : Σ → [0, ∞] such that ν(E) = µ̂(θ(E)) for all E ∈ Σ. I will show that this makes
(Z, Σ, ν) a measure space. First, by 314M Σ is a σ-algebra. Next, ν(∅) = µ̂(θ(∅)) =
µ̂(π([∅]M )) = µ̂(π([0̂]M )) = µ̂(0) = 0. Now let hFn in∈N be a disjoint sequence in Σ,
i.e. for all i 6= j, Fi ∩ Fj = ∅. Let i 6= j ∈ N. Since θ is a Boolean homomorphism,
θ(Fi ) ∧ θ(Fj ) = θ(Fi ∩ Fj ) = θ(∅) = 0, so hθ(Fn )in∈N is a disjoint sequence. For each n,
let En ∈ E, An ∈ M, an ∈ A such that Fn = En 4An and aˆn = En . Let i 6= j ∈ N. First
we show that Ei ∩ Ej ∈ M. By 4.1.4,
∅ = Fi ∩ Fj
= (Ei 4Ai ) ∩ (Ej 4Aj )
= (Ei ∩ (Ej 4Aj ))4(Ai ∩ (Ej 4Aj ))
= [(Ei ∩ Ej )4(Ei ∩ Aj )]4(Ai ∩ (Ej 4Aj ))
= (Ei ∩ Ej )4[(Ei ∩ Aj )4(Ai ∩ (Ej 4Aj ))].
44
Since M is a σ-ideal, Ei ∩ Aj , Ai ∩ (Ej 4Aj ) ∈ M. Furthermore, (Ei ∩ Aj )4(Ai ∩
(Ej 4Aj )) ⊆ (Ei ∩ Aj ) ∪ (Ai ∩ (Ej 4Aj )) ∈ M, so (Ei ∩ Aj )4(Ai ∩ (Ej 4Aj )) ∈ M.
So we have found K ∈ M such that (Ei ∩ Ej )4K = ∅, which implies Ei ∩ Ej =
(Ei ∩ Ej )4∅ = (Ei ∩ Ej )4(K4K) = ((Ei ∩ Ej )4K)4K = ∅4K = K ∈ M. Using
this and the fact that π is a Boolean homomorphism, we get
0 = π([∅]M )
= π([Ei ∩ Ej ]M )
= π([Ei ]M ∩M [Ej ]M )
= π([Ei ]M ) ∧ π([Ej ]M )
= ai ∧ aj ,
so han in∈N is a disjoint sequence. Since θ is sequentially order-continuous, θ(∪n∈N Fn ) =
supn θ(Fn ). Hence
ν(∪n∈N Fn ) = µ̂(θ(∪n∈N Fn )) = µ̂(supn θ(Fn )) =
∞
X
n=0
µ̂θ(Fn ) =
∞
X
ν(Fn )
n=0
so (Z, Σ, ν) a measure space.
For all E ∈ Σ,
ν(E) = 0 ⇐⇒ µ̂(θ(E)) = 0 ⇐⇒ θ(E) = 0 ⇐⇒ E ∈ ker(θ) = M
so M consists of exactly the sets of measure zero in (Z, Σ, ν). Thus the measure algebra of
(Z, Σ, ν) is (Σ/M, ν̂), where ν̂([E]M ) = ν(E). Since the Boolean algebra isomorphism
π : Σ/M → A also satisfies ν̂([E]M ) = ν(E) = µ̂(θ(E)) = µ̂(π([E]M )), π is a measure
algebra isomorphism between (Σ/M, ν̂) and (A, µ̂).
45
Notation. For all a, b ∈ R, let
. b=
a−




a − b if a ≥ b



0
otherwise
Definition. Let P r denote the following set of axioms.
(i) Boolean algebra axioms
ˆ ∨ x, x)
1. supx d(x
ˆ ∧ x, x)
2. supx d(x
ˆ ∨ y, y ∨ x)
3. supx supy d(x
ˆ ∧ y, y ∧ x)
4. supx supy d(x
ˆ ∧ y) ∨ y, y)
5. supx supy d((x
ˆ ∨ y) ∧ y, y)
6. supx supy d((x
ˆ ∨ y) ∨ z, x ∨ (y ∨ z))
7. supx supy supz d((x
ˆ ∧ y) ∧ z, x ∧ (y ∧ z))
8. supx supy supz d((x
ˆ ∨ (y ∧ z), (x ∨ y) ∧ (x ∨ z))
9. supx supy supz d(x
ˆ ∧ (y ∨ z), (x ∧ y) ∨ (x ∧ z))
10. supx supy supz d(x
ˆ ∨ xc , 1)
11. supx d(x
ˆ ∧ xc , 0)
12. supx d(x
(ii) measure axioms
13. µ̂(0)
14. ¬µ̂(1)
. µ̂(x))
15. supx supy (µ̂(x ∧ y) −
46
. µ̂(x ∨ y))
16. supx supy (µ̂(x) −
. µ̂(x ∧ y)) − (µ̂(x ∨ y) −
. µ̂(y))|
17. supx supy |(µ̂(x) −
(iii) connection between dˆ and µ̂
ˆ y) − µ̂(x4y)|
18. supx supy |d(x,
Remark. From axioms 15–17, we can show µ̂(x ∧ y) + µ̂(x ∨ y) = µ̂(x) + µ̂(y) for all
. µ̂(x) = 0, so µ̂(x ∧ y) ≤ µ̂(x). By axiom
x, y ∈ B̂: Let x, y ∈ B̂. By axiom 15, µ̂(x ∧ y) −
. µ̂(x ∧ y) = µ̂(x ∨ y) −
. µ̂(y) =
16, µ̂(y) ≤ µ̂(x ∨ y). Thus, µ̂(x) − µ̂(x ∧ y) = µ̂(x) −
µ̂(x ∨ y) − µ̂(y) by axiom 17.
Axioms 1–12 are direct translations of the first 12 axioms of a Boolean algebra. Conspicuously missing from (i) is last Boolean algebra axiom, 0 6= 1. In the language of
ˆ 1) 6= 0; for 0 6= 1 to hold in a model
continuous logic, 0 6= 1 informally translates to d(0,
ˆ 1)) of d(0,
ˆ 1) in M must be > 0. But M 2 d(0,
ˆ 1) does not
M, the truth value M(d(0,
ˆ 1), since the latter occurs only when M(d(0,
ˆ 1)) = 1. So while we
imply that M ¬d(0,
can always express x = y in M by M d(x, y), we usually cannot express x 6= y [6].
However, by virtue of the additional structure imposed by the axioms of P r, we can not
ˆ 1), but also that M ¬d(0,
ˆ 1).
only show that M 2 d(0,
ˆ 1).
4.21. Proposition. P r ¬d(0,
Proof. Let M be a model of P r. By axiom 18,
ˆ 1)) − M(µ̂(041))|E = M(|d(0,
ˆ 1) − µ̂(041)|) = 0,
|M(d(0,
ˆ 1))
where |·|E denotes one-dimensional Euclidean distance. Thus M(d(0,
= M(µ̂(041))
47
= µ̂M ((041)M )
= µ̂M (((0 ∧ 1c ) ∨ (0c ∧ 1))M )
M
M
= µ̂M ((0M ∧M (1M )c ) ∨M ((0M )c ) ∧M 1M )
= µ̂M (0 ∧ 1c ) + µ̂M (0c ∧ 1) − µ̂M ((0 ∧ 1c ) ∧ (0c ∧ 1)) by axioms 15 - 17
≤ µ̂(0) + µ̂M (0c ∧ 1) + µ̂(0) by axiom 15
= 0 + µ̂M (0c ∧ 1) + 0 since µ̂M (0M ) = M(µ̂(0)) = 0 by axiom 13
= µ̂M (0c ∧ 1)
= µ̂M (1 ∧ 1) since 0c = (0 ∧ 0c )c = 0c ∨ 0 = 1
= µ̂M (1) by axiom 1
= 1 since 1 − µ̂M (1) = 1 − M(µ̂(1)) = M(¬µ̂(1)) = 0.
ˆ 1)) = 1 − M(d(0,
ˆ 1)) = 1 − 1 = 0, so M ¬d(0,
ˆ 1).
Then M(¬d(0,
Now we show that P r axiomatizes the class of probability structures.
4.22. Theorem. (16.1 from [2]) Let M be an LP r -structure with underlying metric space
ˆ Then M is a model of P r if and only if M is the probability structure associated
(M, d).
with a probability space (X, Σ, µ) as above.
Proof. (⇐) Let (X, Σ, µ) be a probability space and let M = (B̂, 0, 1, ·c , ∧, ∨, µ̂) be its
associated probability structure. Let x = [A]µ , y = [B]µ , z = [C]µ ∈ B̂. The proofs that
each axiom of P r is true in M are similar, so I will just show one from each of (i)-(iii).
From the set identity (A∪(B∩C)) = ((A∪B)∩(A∪C)), we get 0 = µ(∅) = µ((A∪(B∩
ˆ
C))4((A∪B)∩(A∪C))) = µ̂((x∨(y ∧z))4((x∨y)∧(x∨z))) = d(x∨(y
∧z), (x∨y)∧
48
ˆ ∨ (y ∧ z), (x ∨ y) ∧ (x ∨ z)) = 0,
(x ∨ z)). Since x, y, and z were arbitrary, supx supy supz d(x
so M axiom 9.
. µ(A ∩ B) = µ(A) −
Note that µA ≥ µ(A ∩ B) and µ(B ∪ A) ≥ µ(B), so µ(A) −
. µ̂(B) = µ(A ∪ B) − µ(B). Rearranging the set identity
µ(A ∩ B) and µ(A ∪ B) −
µ(A) + µ(B) = µ(A ∩ B) + µ(A ∪ B) to µ(A) − µ(A ∩ B) = µ(A ∪ B) − µ(B), we
. µ(A ∩ B)) − (µ(A ∪
get 0 = (µ(A) − µ(A ∩ B)) − (µ(A ∪ B) − µ(B)) = (µ(A) −
. µ(B)) = |(µ̂(x) −
. µ̂(x ∧ y)) − (µ̂(x ∨ y) −
. µ̂(y))| Since x and y were arbitrary,
B) −
. µ̂(x ∧ y)) − (µ̂(x ∨ y) −
. µ̂(y))| = 0, so M axiom 17.
supx supy |(µ̂(x) −
ˆ y) = µ̂(x4y), sup sup |d(x,
ˆ y) − µ̂(x4y)| = 0, so M axiom 18.
Finally, since d(x,
x
y
(⇒) Let M be a model of P r. By assumption, M is an LP r -structure, so it is of the form
(A, 0, 1, ·c , ∧, ∨, ν) for a set A, distinguished constants 0 and 1, a unary function symbol
·c , two binary function symbols ∧ and ∨, and a unary predicate symbol ν. Let (A, dν ) be
the underlying metric space of A. Note that by definition, (A, dν ) is complete. Since the
axioms in (i) and 0 6= 1 characterize a Boolean algebra, A is a Boolean algebra.
The measure axioms make ν finitely additive. Let x, y ∈ A. By the preceding remark,
ν(x)−ν(x∧y) = ν(x∨y)−ν(y). If x∧y = 0, then ν(x) = ν(x)−ν(0) = ν(x∨y)−ν(y),
so ν(x) + ν(y) = ν(x ∨ y).
By (iii), the metric dν must satisfy dν (a, b) = ν(a4b). Given x ∈ A, 1 ∧ x = (xc ∨ x) ∧
x = x by axioms 11 and 6. In other words, for all x ∈ A, x ≤ 1, so ν(x) ≤ ν(1) = 1.
Hence ν is a finitely additive probability measure.
Now I will show that ν is 1-Lipschitz on A with respect to dν . Let x, y ∈ A. Since
x = (x ∧ y c ) ∨ (x ∧ y) = (x \ y) ∨ (x ∧ y) and ν is finitely additive, ν(x) = ν(x \ y) +
ν(x ∧ y). Similarly, ν(y) = ν(y \ x) + ν(y ∧ x). Then |ν(x) − ν(y)| = |ν(x \ y) +
49
ν(x ∧ y) − (ν(y \ x) + ν(y ∧ x))| = |ν(x \ y) − ν(y \ x)|. Since ν(x \ y), ν(y \ x) ≥ 0,
0 ≤ 2ν(x \ y) ≤ 2ν(x \ y) + 2ν(y \ x), so subtracting ν(x \ y) + ν(y \ x) from all
sides, −(ν(x \ y) + ν(y \ x)) ≤ ν(x \ y) − ν(y \ x) ≤ ν(x \ y) + ν(y \ x). Thus
|ν(x) − ν(y)| = |ν(x \ y) − ν(y \ x)| ≤ ν(x \ y) + ν(y \ x) = ν(x4y) = dν (x, y), so ν is
1-Lipschitz.
Next, I will show that increasing sequence in A is necessarily a Cauchy sequence with
respect to dν . Let hxn i ∈ AN be an increasing sequence. Recall that xn ≤ xn+1 means
xn ∧ xn+1 = xn . For the sake of contradiction, suppose hxn i is not Cauchy. Then there
exists > 0 such that for all N ∈ N there exist m, n > N such that dν (xm , xn ) > .
Choose n1 < n2 such that dν (xn1 , xn2 ) > . Setting N1 = n2 + 1, choose N1 < n3 < n4
such that dν (xn3 , xn4 ) > . In general, let Nk = n2k + 1 and choose Nk < n2k+2 < n2k+3
such that dν (xn2k+2 , xn2k+3 ) > . For all k ∈ N, let yk = xnk . Note that for all i ∈ N,
dν (y2i+1 , y2i+2 ) > . Let 1/ < M ∈ N. Since hyk i is an increasing sequence, by 4.1.2
ν(y1 4y2M +1 ) = dν (y1 , y2M +1 ) =
2M
P
dν (yj , yj+1 ) ≥
j=1
M
−1
P
dν (y2j+1 , y2j+2 ) > M > 1,
j=0
which is impossible, since 1 is the maximal element of A and ν(1) = 1. Therefore, hxn i
is Cauchy with respect to dν . Since (A, dν ) is a complete metric space, for all increasing
sequences hxn i, there exists x ∈ A such that xn → x.
Now I will show that A is σ-complete as a Boolean algebra. Let han i ∈ AN be a
sequence. For each n ≥ 0, let bn =
n
W
ak . By axiom 6, for all n, an ≤ bn and bn ≤ bn+1 .
k=0
Since hbn i is increasing, there exists b ∈ A such that bn → b. First we show that b is an
upper bound of han i. For the sake of contradiction, suppose that there exists n ∈ N such
that b < bn , i.e. b ∧ bn = b and b 6= bn . Since b 6= bn and dν is a metric, dν (b, bn ) > 0.
Let = dν (b, bn ) and m ∈ N such that m ≥ n. Since hbn i is increasing, b < bn ≤ bm .
50
By 4.1.2, dν (b, bm ) = ν(b4bm ) = ν(bm \ b) = ν(bm \ bn ) + ν(bn \ b) ≥ ν(bn \ b) = ,
so dν (b, bm ) ≥ for all m ≥ n, contradicting the fact that b = limn bn . Hence, for all n,
an ≤ bn ≤ b, so b is an upper bound for han i.
To see that b is the least upper bound, let c ∈ A be an upper bound for han i. Clearly
n+1
W
b0 = a0 ≤ c. For any n ∈ N such that bn ≤ c, bn+1 =
ak = bn ∨ an . Since
k=0
an ≤ c, by 4.1.5 bn+1 = bn ∨ an ≤ c, so c is also an upper bound for hbn i. Let > 0.
Since bn → b, there exists N ∈ N such that for all n > N , dν (b, bn ) < . By 4.1.4,
(bn ∧ c)4(b ∧ c) = (b4bn ) ∧ c ≤ b4bn . Then for all n > N , dν (bn ∧ c, b ∧ c) =
ν((bn ∧ c)4(b ∧ c)) = ν((bn 4b) ∧ c) ≤ ν(b4bn ) = dν (b, bn ) < , so hbn ∧ ci converges
to b ∧ c. But c is an upper bound for hbn i, so hbn ∧ ci = hbn i. Hence
b = limn bn = limn (bn ∧ c) = b ∧ c,
i.e. b ≤ c. Thus supn an = b ∈ A.
Now I will show that ν is σ-additive on A. Let han in∈N be a disjoint sequence in A.
Since ν is finitely additive, ν(
limn→∞
Pn
k=0
ν(ak ) =
P∞
k=0
Wn
k=0
ak ) =
Pn
k=0
W
ν(ak ) for all n, so limn→∞ ν( nk=0 ak ) =
ν(ak ). For each n, let bn =
there exists c0 ∈ A such that c0 =
W
k∈N
Wn
k=0
ak . Since A is σ-complete,
ak . Since hbn i is increasing, there exists c1 ∈ A
such that bn → c1 . By the proof of the σ-completeness of A in 4.22, c0 = c1 . Let c be their
common value. Since ν is 1-Lipschitz, |ν(bn ) − ν(c)| ≤ dν (bn , c) for all n. But bn → c, so
dν (bn , c) → 0. Hence limn→∞ |ν(bn ) − ν(c)| = 0, so
ν(
_
ak ) = ν(c) = limn→∞ ν(bn ) = limn→∞ ν(
k∈N
n
_
k=0
ak ) =
∞
X
ν(ak ).
k=0
Since A is σ-complete and ν is σ-additive, (A, ν) is a measure algebra as defined in 4.3.
By theorem 4.20, it follows that M is a probability algebra.
51
We end this section with a few facts about atomless probability spaces.
Definition. A probability space (X, B, µ) is atomless if for all B ∈ B, if µ(B) > 0 then
there exists B 0 ∈ B such that B 0 ⊆ B and 0 < µ(B 0 ) < µ(B).
Let (X, B, µ) be an atomless probability space. Then for any B ∈ X, µ(B) = 0, so X
must be uncountable. We will use without proof the fact that (X, B, µ) is atomless if and
only if for all B ∈ B, µ(B) > 0 implies there exist B1 , B2 ∈ B such that B1 , B2 ⊆ B,
B1 and B2 are disjoint, and µ(B1 ) = µ(B2 ) = µ(B)/2. The following axiom encodes this
fact:
(1)
supx infy |µ̂(x ∧ y) − µ̂(x ∧ y c )|
Let P rA be P r together with (1). Then P rA axiomatizes the class of probability algebras associated with atomless probability spaces.
52
5. C ONTINUOUS LOGIC VERSUS FIRST- ORDER LOGIC
This section consists of continuous analogues of some familiar meta-theorems of firstorder logic. All of the definitions and results in this section are from [3].
Definition. A set of formulas Σ is inconsistent if for every formula φ, Σ ` φ. Otherwise
Σ is consistent.
5.1. (4.5 in [3]) A set of formulas Σ is inconsistent if and only if there exists n ∈ N such
that Σ ` ( 21 )n .
Definition. A theory T is complete if there exists an L-structure M and assignment σ such
that T = Th(M). Otherwise T is incomplete.
Remark. In first-order logic, an equivalent formulation of a theory T being complete is
that for all sentences φ, T ` φ or T ` ¬φ. However, the addition of truth values beyond
0 and 1 means that there are now sentences φ such that neither φ nor ¬φ have models, i.e.
for all models M and all assignments σ, M(φ, σ), M(¬φ, σ) 6= 0. For example, the truth
value of
1
2
and ¬ 12 in any model under any assignment is 12 . If we required of a theory T
that for all φ, T ` φ or T ` ¬φ, then by 5.1, T would be inconsistent. Such is the rationale
for defining the property of completeness semantically.
53
One example of a complete theory is P rA (see 16.2 in [2]). Thus all probability structures associated with atomless probability spaces, such as examples 2 and 3 in 4.8, are
elementarily equivalent. In contrast, P r is an incomplete theory, since otherwise, examples
1 and 2 from 4.8 would model the same sentences. But example 1 is not atomless, so only
example 2 models supx infy |µ̂(x ∧ y) − µ̂(x ∧ y c )|. Hence P r is incomplete. Ongoing work
by Berenstein and Henson includes a characterization of complete extensions of P r. All
complete extensions other than P rA are obtained from the addition of axioms describing
the measure of each atom.
5.2. Theorem. (Soundness) Let L be a continuous signature and let Γ be a set of Lformulas. Then for every L-formula φ,
(1) Γ ` φ implies Γ φ
(2) if Γ is satisfiable, then Γ is consistent.
5.3. Theorem. (Completeness) Let L be a continuous signature and let Γ be a set of Lformulas. If Γ is consistent, then Γ is satisfiable.
A corollary to the completeness theorem in continuous logic is the compactness theorem,
obtained in the same way as the compactness theorem is obtained from completeness in
first-order logic.
Corollary. (Compactness) Let Γ be a set of L-formulas. If every finite subset Γ0 of Γ is
satisfiable, then Γ is satisfiable.
Proof. For the sake of contradiction, suppose every finite Γ0 ⊆ Γ is satisfiable but Γ is not
satisfiable. By 5.3, Γ is inconsistent, so by 5.1, there exists n ∈ N such that Γ ` ( 21 )n . Since
54
proofs are finite in length, there must be some finite Γ0 ⊆ Γ such that Γ0 ` ( 21 )n . Since
( 12 )n always has truth value ( 21 )n , by 5.1, Γ0 is inconsistent. By 5.2, Γ0 is not satisfiable,
contrary to assumption.
Remark. Consider the following statements:
(1) for every set of formulas Γ, if Γ is consistent then Γ is satisfiable
(2) for every set of formulas Γ and every formula φ, Γ φ implies Γ ` φ
In first-order logic, (1) and (2) are equivalent, but in continuous logic, (1) does not imply
(2). Crucial in the proof that (1) implies (2) in first-order logic is the fact that, in first-order
logic, if Γ 0 φ, then Γ ∪ {¬φ} is consistent. But in continuous logic, if we take Γ = ∅ and
φ = 21 , then 0
1
2
but {¬ 12 } is inconsistent by 5.1.
In fact, (2) fails in continuous logic. Consider the following counterexample. Let φ be
the formula supx supy d(x, y), which is true in an L-structure M if and only if M consists
of one element. In general, the truth value of φ in M is the diameter of M. Let
. ( 1 )n : n ∈ N}
Γ = {φ −
2
.
and suppose (M, σ) Γ. Then for every n ∈ N, max(M(φ, σ) − ( 21 )n , 0) = M(φ −
( 21 )n , σ) = 0, which means that for every n ∈ N, M(φ, σ) ≤ ( 21 )n , so M(φ, σ) = 0. Thus
Γ φ. However, if Γ ` φ, then there exists a finite Γ0 ⊆ Γ such that Γ0 ` φ, as proofs are
. ( 1 )n ∈ Γ ) < ∞. Then any metric
finite in length. By 5.2, Γ0 φ. Let N = max(n : φ −
0
2
structure of diameter ≤ ( 21 )N is a model of Γ0 but not of φ, so Γ0 2 φ. Hence Γ 0 φ.
There is, however, a counterpart to (2) in continuous logic; since (2) is sometimes called
strong completeness, we will call its continuous version approximated strong completeness.
55
5.4. Corollary. (Approximated Strong Completeness) Let L be a continuous signature, let
Γ be a set of L-formulas, and let φ be an L-formula. Then
. ( 1 )n
Γ φ ⇐⇒ Γ ` φ −
2
for all n ∈ N.
Definition. Let φ and ψ be L-formulas and let n ∈ N. Define nψ → φ recursively:
(1) 0ψ → φ := φ
(2) (n + 1)ψ → φ := ψ → (nψ → φ)
5.5. Among the meta-theorems of propositional logic and first-order logic is the deduction
theorem: If Γ is a set of formulas and φ and ψ are formulas, then
Γ ∪ {ψ} ` φ ⇐⇒ Γ ` ψ → φ.
The left to right direction fails in continuous logic. Consider the following counterexample.
Let ψ be
1
2
and let φ be 34 . Let Γ = {ψ → (ψ → φ)}. Then we have ψ → (ψ → φ), ψ ` φ
via two applications of modus ponens. I will show, however, that ψ → (ψ → φ) 0 ψ →
φ. By the soundness theorem, it suffices to show ψ → (ψ → φ) 2 ψ → φ. For any
model M and assignment σ, M(ψ, σ) =
1
2
and M(φ, σ) =
max(M(φ, σ) − M(ψ, σ), 0) = 41 . This means that
1
2
→
3
4
3
,
4
so M(ψ → φ, σ) =
has no models. On the other
hand, M(ψ → (ψ → φ), σ) = max(M(ψ → φ, σ) − M(ψ), 0) = max( 41 − 12 , 0) = 0, so
1
2
→ ( 12 → 34 ) is valid. Thus ψ → (ψ → φ) 2 ψ → φ.
In particular, failure of the deduction theorem in continuous logic implies that the axiom
(α → (β → γ)) → ((α → β) → (α → γ))
56
of classical propositional logic cannot be derived from the propositional part of continuous logic (axioms 1–6), since, together with axiom 1 and modus ponens, it implies the
deduction theorem.
However, we do have the following weakened version of the deduction theorem in continuous logic.
5.6. Theorem. Let Γ be a set of L-formulas and let φ, ψ be L-formulas. Then Γ ∪ {ψ} ` φ
if and only if for some n ∈ N, Γ ` nψ → φ.
Continuous logic satisfies a modified version of the Downward Löwenheim-Skolem theorem. Its statement requires the following definition.
Definition. Let X be a topological space. The density character of X, denoted by kXk, is
the minimum cardinality a dense subset of X.
5.7. Theorem. (Downward Löwenheim-Skolem) Let κ be an infinite cardinal. Let L be a
signature such that |L| ≤ κ. Let M be an L-structure and let A ⊆ M such that kAk ≤ κ.
Then there exists an elementary substructure N of M such that A ⊆ N and kN k ≤ κ.
This implies the Downward Löwenheim-Skolem of first-order logic (for any first-order
language L, L-structure M, and cardinal κ such that max(|L|, ℵ0 ) ≤ κ ≤ |M|, there exists an elementary substructure N of M such that |N | = κ) with cardinality of a model
replaced by its density character: for any signature L, L-structure M, and cardinal κ such
that max(|L|, ℵ0 ) ≤ κ ≤ kMk, there exists an elementary substructure N of M such that
kN k = κ. To see why, let L be a signature, M an L-structure, and κ a cardinal such that
57
max(|L|, ℵ0 ) ≤ κ ≤ kMk. Let A ⊆ M such that kAk = κ. By 5.7, there exists an elementary substructure N of M such that A ⊆ N and kN k ≤ κ. Since κ = kAk ≤ kN k ≤ κ, we
have kN k = κ. So in the same way that first-order logic cannot distinguish between infinite
cardinalities, continuous logic cannot distinguish between density characters; if a theory T
in continuous logic has a model, then T has a model of density character ≤ max(|L|, ℵ0 ).
The first-order version of the Downward Löwenheim-Skolem theorem fails in continuous logic. This is witnessed by P rA, whose signature is finite, but whose models are all
uncountable, since all atomless probability spaces are uncountable.
58
6. C ONCLUSION
Continuous logic was developed to study metric structures in a setting more convenient
than that of first-order logic. Its model theory generalizes first-order model theory in the
sense that any first-order structure is a metric structure if we equip it with the discrete
metric. Restrictions of the connectives and quantifiers of continuous logic from [0, 1] to
{0, 1} yield those of first-order logic. The set of axioms from which a completeness result
was proven in [3] are comparable to the axioms of first-order logic with equality, with
the metric d replacing =. Indeed, the equality axioms of first-order logic motivate the
restriction of predicates and functions to be uniformly continuous.
Aside from its striking parallels to first-order logic in syntax, continuous logic satisfies
modified versions of meta-theorems of first-order logic such as compactness, completeness,
and Downward Löwenheim-Skolem. Yet while continuous logic is a useful tool to study
the model theory of metric structures, it may not be indispensible. Clearly continuous logic
is as strong as first-order logic, since every first-order formula has a direct translation in
continuous logic, but is it stronger than first-order logic? Could the model theory of metric
structures be done using first-order logic? Perhaps the two logics differ only in presentation
and not in power of expression.
Lindström’s theorem cannot be used directly to compare these logics, since continuous
logic is outside the scope of logics considered by Lindström’s theorem. To answer these
questions, which are beyond the scope of this paper, one would first need to define the
59
notion of logic to include both first-order logic and continuous logic, then find a way to
compare their expressive power.
60
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