An Invitation to Model Theory
via definable sets
Draft version
Jonathan Kirby
February 17, 2017
Contents
I
Languages, structures and theories
1
1 Structures
2
2 Terms
7
3 Formulas
10
4 Definable sets
15
5 Substructures and quantifiers
19
6 Theories and Axioms
23
7 The Complex and Real Fields
29
8 Compactness
33
9 Axiomatizable classes
38
10 Constructing models from syntax
43
11 Interlude: basic cardinal arithmetic
48
II
50
Changing models
12 Elementary substructures
51
13 Elementary extensions
56
14 Vector spaces and Categoricity
62
15 The successor structure
68
16 Linear Orders
71
i
CONTENTS
ii
17 Quantifier elimination
75
18 Power sets and boolean algebras
76
III
77
Types
19 The algebras of definable sets
78
20 Types
82
21 Countable categoricity
87
22 Saturation and homogeneity
90
23 Definable sets and types with parameters
91
24 Counting types and models
92
IV
93
Algebraically Closed Fields
25 Fields and their extensions
94
26 Algebraic closures
95
27 Definable sets and algebraic varieties
96
V
Working Appendices
97
A Dependence of chapters
98
B Learning objectives by chapter
100
Part I
Languages, structures and
theories
1
Chapter 1
Structures
In different mathematical contexts, familiar mathematical objects may actually have different meanings. For example a reference to the integers Z in
group theory is likely to mean the infinite cyclic group, whereas in number
theory it is likely to mean Z as a ring. In model theory, when we specify
a structure we have to be precise about such things. The integers as an
additive group will be written as Zadgp = hZ; +, −, 0i, and Z as a ring will
be written as Zring = hZ; +, ·, −, 0, 1i. The integers are also used to index
the years in a calendar, and an appropriate structure for that purpose is
Z< = hZ; <i, because it is not very meaningful to add or multiply calendar
years, but the order is important. In general, we capture these ideas with
the notions of a language, L, and an L-structure.
Definition 1.1. A language L is specified by the following (sometimes called
the vocabulary or signature of L):
1. a set of relation symbols,
2. a set of function symbols,
3. a set of constant symbols, and
4. for each relation and function symbol, a positive natural number called
its arity.
Definition 1.2. An L-structure A consists of a non-empty set A called the
domain of the structure, together with
1. for each relation symbol R of L, of arity n, a subset RA of An ,
2. for each function symbol f of L, of arity n, a function f A : An → A,
3. for each constant symbol c of L an element cA ∈ A.
2
CHAPTER 1. STRUCTURES
3
Example 1.3. The language of rings Lring = h+, ·, −, 0, 1i, where + and ·
are function symbols of arity 2 (we say they are binary function symbols), −
is a function symbol of arity 1 (for negation rather than the binary function
of subtraction), and 0 and 1 are constant symbols. Zring is the Lring structure
with domain the set Z of integers. We interpret the symbols +, ·, and −
as the usual functions of addition, multiplication, and negation of integers
and the constant symbols 0 and 1 are interpreted as the usual zero and one.
We can write this as Zring = hZ; +Z , ·Z , −Z , 0Z , 1Z i if for example we want to
distinguish the symbol + from its interpretation +Z as a function from Z2
to Z. This distinction can be important in mathematical logic. For example
we also have the Lring -structure Rring with domain the set of real numbers
R, and the function +R : R2 → R is not the same function as +Z : Z2 → Z.
However usually no ambiguity arises and we just write + for the symbol and
for its interpretations in different structures.
Example 1.4. We write L< = h<i, where < is a binary relation symbol.
Then Z< is the L< -structure with domainthe set Z of integers,
and the
symbol < is interpreted as the set of pairs (a, b) ∈ Z2 | a < b . As above,
we could write this set as <Z , but usually we will not.
Examples 1.5. Many other languages can be built as variations on these
two languages.
(a) Lgp = h·, (−)−1 , 1i is the language of groups. Again, · is a binary function symbol, (−)−1 is a unary (arity 1) function symbol representing the
map x 7→ x−1 , and 1 is a constant symbol.
(b) Ladgp = h+, −, 0i is the language of groups written additively. It is
a sub-language of Lring , because every symbol in Ladgp is also in Lring
(with the same arity).
(c) Lo-ring = h+, ·, −, 0, 1, <i is the language of ordered rings, consisting of
Lring ∪ L< .
(d) A common language in which to consider the natural numbers is the
language of semirings, Ls-ring = h+, ·, 0, 1i. We will always use the convention that 0 is a natural number.
(e) The language of monoids is Lmon = h·, 1i and the language of additive
monoids is Ladmon = h+, 0i.
Most of the structures we will consider as examples have a domain with
a commonly used notation such as Z, Q or R, and the functions, relations
and constants we consider will be those in the above languages. In this case
we name the structure by putting the name of the language as a subscript,
for example Zring , Ro-ring , Q< , Ns-ring .
CHAPTER 1. STRUCTURES
4
We can also specify structures directly which is useful for creating simpler
examples. In this case we will often use a curly letter for the name of the
structure and the corresponding Roman letter for the name of its domain,
so A would be the name of a structure with domain A.
Example 1.6. Take L = hf, R, ci where f is a unary function symbol, R
is a ternary relation symbol and c is a constant symbol. Define a structure
A with domain A = {0, 1, 2, 3, 4}, f A (x) = x + 1 mod 5, RA =
(x, y, z) ∈ A3 | exactly two of x, y and z are equal and cA = 3.
Remark 1.7. It is important to realise that the symbols do not have to
have their usual meanings. For example, we can make the set N into an
Lgp -structure by interpreting the symbol · as addition and (−)−1 as the
identity function, so x−1 = x for all x ∈ N, and interpreting the constant
symbol 1 as the number 2. However this is perverse and in future if we want
an unusual meaning for a symbol we will choose to use a different symbol.
Given a mathematical problem you are trying to solve, or a statement
you want to understand, it is often a good exercise to work out what structure the statement might be about. For example, the fundamental theorem
of arithmetic states that every positive integer can be written as a product of
primes in a unique way (up to reordering). An appropriate structure would
have domain the set of positive integers, and needs to have the multiplication function. 1 is a special case as the empty product, which is relevant to
single out. So we can regard the fundamental theorem of arithmetic as a
+
statement about the structure N+
mon = hN ; ·, 1i. To actually prove the theorem we might need more than that, for example the order to do induction
on and also addition.
Embeddings and automorphisms
Definition 1.8. Let A and B be L-structures. An embedding of L-structures
π
from A to B is an injective function A −→ B such that:
1. for all relation symbols R of L, and all a1 , . . . , an ∈ A,
(a1 , . . . , an ) ∈ RA iff (π(a1 ), . . . , π(an )) ∈ RB
2. for all function symbols f of L, and all a1 , . . . , an ∈ A,
π(f A (a1 , . . . , an )) = f B (π(a1 ), . . . , π(an ))
3. for all constant symbols c of L, π(cA ) = cB .
For any L-structure A, the identity function 1A on A is a embedding of
π
A into itself. An embedding A −→ B is an isomorphism iff there is an
CHAPTER 1. STRUCTURES
5
σ
embedding B −→ A such that the composite π ◦ σ is the identity on B and
the composite σ ◦ π is the identity on A. We write π −1 for σ, as usual,
and call it the inverse of π. An isomorphism from A to itself is called an
automorphism of A.
Examples 1.9. We have the obvious inclusion functions Z ,→ Q ,→ R ,→ C,
which take a number to itself. These inclusion functions give us embeddings
of Lring -structures
Zring ,→ Qring ,→ Rring ,→ Cring .
Definition 1.10. If we have an embedding A −→ B where the function is
an inclusion of the domain of A as a subset of the domain of B then we say
that A is a substructure of B, and that B is an extension of A.
Example 1.11. The only automorphisms of Zadgp are the identity and the
map x 7→ −x. To see this, note that if π : Z → Z is an Ladgp -embedding
then π(0) = 0 and if π(1) = n then for any m ∈ N+ we have
π(m) = π(1
· · + 1}) = π(1) + · · · + π(1) = mn.
| + ·{z
|
{z
}
m
m
Then also π(−n) = −π(n) = −mn. To have an inverse, π must be surjective
which implies n = ±1.
Exercises
1.1 Let L = hf, ci be a language with one unary function symbol and one
constant symbol. Describe all the possible L-structures on the domain
{1, 2}. How many of them are there up to isomorphism (which means
counting isomorphic structures as the same)?
1.2 What are all the automorphisms of Z< ?
1.3 Show that embeddings of N< into R< correspond to strictly increasing
sequences of real numbers.
1.4 Show that an embedding of L-structures is an isomorphism if and only
if it is surjective.
1.5 Find an automorphism π of the structure hR; <i such that π(0) = 1
and π(1) = 5. Is your π also an automorphism of the structure hR; +i?
1.6 Find all the automorphisms of Qadgp .
1.7 Let A be an L-structure. Show that the automorphisms of A form
a group with the group operation being composition. We write this
group as Aut(A).
CHAPTER 1. STRUCTURES
6
1.8 If you have studied groups or rings you will have seen the notions
of group homomorphism or ring homomorphism. In model theory, a
π
π
homomorphism of L-structures A −→ B is a function A −→ B such
that
1. for all relation symbols R of L, and all a1 , . . . , an ∈ A,
if (a1 , . . . , an ) ∈ RA then (π(a1 ), . . . , π(an )) ∈ RB ,
2. for all function symbols f of L, and all a1 , . . . , an ∈ A,
π(f A (a1 , . . . , an )) = f B (π(a1 ), . . . , π(an )), and
3. for all constant symbols c of L, π(cA ) = cB .
Check that if L = Lgp or L = Lring , and A, B are groups or rings, then
L-homomorphisms are the same as group or ring homomorphisms.
How do L-homomorphisms compare with, and how do they differ from,
the notion of embedding of L-structures?
1.9 For each of the following statements, give a structure which they say
something about.
(a) There is no largest integer.
(b) Every integer has an additive inverse.
(c) Square integers are always non-negative.
(d) Every complex number is the sum of a real number and i times
a real number.
(e) The exponential function is strictly increasing on the real numbers
(f) A real quadratic equation ax2 + bx + c = 0 has real roots if and
only if b2 − 4ac > 0.
(g) Every non-zero polynomial over C has a zero in C.
(h) Euler’s identity eiπ + 1 = 0.
Chapter 2
Terms
A formal language is a collection of symbols, which consists of variables such
as x, y, or z, logical symbols such as =, ∧ , ¬ and ∃, brackets (, ), [, and ],
and then three kinds of symbols which are specific to the language: relation,
function, and constant symbols. Apart from the collection of symbols, a
language also consists of various strings (lists) of the symbols which are
called terms and formulas. These symbols, terms, and formulas are then
interpreted in structures. In the next chapter we will introduce the formulas.
Here we introduce the terms which are certain strings of symbols that refer
to elements of the structure, or to functions on the structure. For example,
in the language Lring , terms include:
0,
(1 + 1),
−((1 + 1) + 1),
(x · (1 + 1)),
(x + y), and ((x · x) + 1).
They can be interpreted in Zring as the numbers 0, 2, and −3, and the
functions x 7→ 2x, (x, y) 7→ x + y and x 7→ x2 + 1 respectively.
Definition 2.1 (Terms). The set of terms of the language L is defined
recursively as follows.
1. Every variable is a term;
2. Every constant symbol of L is a term;
3. If f is a function symbol of L of arity n, and t1 , . . . , tn are terms of L,
then f (t1 , . . . , tn ) is a term.
4. Only something built from the above three clauses in finitely many
steps is a term.
In the examples we followed the usual practice of writing (1 + 1) rather
than +(1, 1) and so on. This is called infix notation and works only for
binary functions. When the function is written before its arguments it is
called prefix notation.
As in the examples above, terms of L can be interpreted as elements of
an L-structure or functions on it.
7
CHAPTER 2. TERMS
8
Definition 2.2 (Interpretation of terms). Let A be an L-structure, t a term
of L, and x̄ = (x1 , . . . , xr ) a list of variables including all those which appear
in t. The recursive definition of terms is used to give a recursive definition
of a function tA : Ar → A as follows.
1. If t = c, a constant symbol, then tA is the element cA of A, or as a
function it is the constant function which always takes the value cA .
2. If t = xi , the ith variable in the list x̄, then tA is the ith coordinate
function given by tA (x1 , . . . , xr ) = xi .
3. If t = f (t1 , . . . , tn ) then tA is given by composition:
A
tA (x̄) = f A (tA
1 (x̄), . . . , tn (x̄)).
Remarks 2.3.
1. Note that terms depend only on the language, but
their interpretations depend on the structure. For example, the term
((x + 1) + x) is interpreted in the structure Rring as the function
R −→ R
x 7−→ 2x + 1
but in Zring the same term is interpreted as a function Z → Z.
2. We can see that several terms can be interpreted as the same function.
For example, the terms ((−x + (x + x)) + (1 + 1)) and ((1 + x) + 1)
are both interpreted in Zring as the function
Z −→ Z
x 7−→ x + 2.
3. Where no ambiguity arises, we often do not write all the brackets
which should be there according to the recursive definition.
4. In model theory we almost always deal with terms with a given list
of variables. So we will write that t(x̄) is a term, when we really
mean that t is a term and x̄ is a list of variables which includes all the
variables which appear in t.
Embeddings and terms
By definition, an embedding of L-structures preserves the interpretations
of the constant and function symbols from L. We can use the recursive
definition of terms to give an inductive proof that embeddings preserve the
interpretation of all terms. When ā = (a1 , . . . , ar ) is a tuple of elements
from A, we write π(ā) as an abbreviation for the tuple (π(a1 ), . . . , π(ar )).
Otherwise the notation becomes too unwieldy.
CHAPTER 2. TERMS
9
π
Proposition 2.4. Suppose A −→ B is an embedding of L-structures, t is
a term of L, x̄ = (x1 , . . . , xr ) is a list of variables including all those which
appear in t, and ā ∈ Ar . Then
π(tA (ā)) = tB (π(ā)).
Proof. We proceed by induction on the construction of terms.
If t is a constant symbol c, then
π(tA (ā)) = π(cA )
by Definition 2.2
B
= c
by the definition of an embedding
= tB (π(ā)) by Definition 2.2 again.
If t = xi then π(tA (ā)) = π(ai ) = tB (π(ā)), using Definition 2.2 twice.
If t = f (t1 , . . . , ts ), then
A
π(tA (ā)) = π(f A (tA
1 (ā), . . . , ts (ā)))
A
= f B (π(tA
by definition of an embedding
1 (ā)), . . . , π(ts (ā)))
B
B
B
= f (t1 (π(ā)), . . . , ts (π(ā))) by the inductive hypothesis
= tB (π(ā))
as required.
Exercises
2.1 Write out a recursive definition of Lmon -terms.
2.2 Explain how the Lring -terms (−(x + 1) · y) and (x + x) + (y · −z) are
built up from the recursive definition of L-terms.
2.3 In the structure Rring , which elements of R are named by Lring -terms?
2.4 Give three different Lring -terms which are interpreted as the same polynomial function.
2.5 Define a polynomial function on R to be a function Rn → R of the
form
m
X
d
d
p(x1 , . . . , xn ) =
ri x1i,1 · · · xni,n
i=1
for some ri ∈ R, and some di,j ∈ N for i = 0, . . . , m and j = 1, . . . , n.
Show by induction on the construction of terms that every Lring -term
is interpreted in Rring as a polynomial function. Which polynomial
functions Rn → R are the interpretation of some Lring -term?
2.6 Give a complete proof (not using Proposition 2.4) that if A is an
Lmon -structure, π ∈ Aut(A), t is an Lmon -term and ā ∈ Ar then
π(tA (ā)) = tA (π(ā)).
π
2.7 Show that if A −→ B is an L-homomorphism then it preserves the
interpretations of L-terms.
Chapter 3
Formulas
Terms are certain strings of symbols from the language. As we have seen,
they are interpreted as elements of a structure, or as functions on the structure. Other strings of symbols of the language say something about structures, or about elements of the structure. For example, ((1 + 0) + 1) = 0 is
a statement about numbers (which is false in Zadgp , but true in the cyclic
group Z/2Z), ∃x[((x · x) + −(1 + 1)) = 0] asserts that there is a square root
of 2 (which is true in Rring but false in Zring ), and (x · x) < 1 is a statement
about the number x (which in Ro-ring will be true for some values of x and
false for other values).
Definition 3.1. As for terms, the set of formulas of L is defined recursively.
1. If t1 and t2 are terms, then (t1 = t2 ) is a formula;
2. If t1 , . . . , tn are terms and R is a relation symbol of L of arity n, then
R(t1 , . . . , tn ) is a formula;
3. If ϕ and ψ are formulas then ¬ϕ and (ϕ ∧ ψ) are formulas.
4. If ϕ is a formula and x is a variable then ∃x[ϕ] is a formula.
5. Only something built from the above four clauses in finitely many steps
is a formula.
The formulas in clauses 1 and 2 are called atomic formulas, because they
do not contain smaller formulas.
Free variables and scope
Variables play two different roles in formulas. For example, the formula
x<1+1
10
CHAPTER 3. FORMULAS
11
says something about the variable x, namely that it is less than 2, whereas
the formula
∃x[0 < x]
says that there is some element which is greater than 0 and the variable x
is just a dummy. The formula has the same meaning as ∃y[0 < y] where x
is replaced by y. The difference is that in the second example, the variable
x is quantified over by the ∃ quantifier.
Definition 3.2. An instance of a variable x in a formula is a quantifier
instance if it occurs immediately after the ∃ symbol. Immediately following
the quantifier ∃x is the scope of the quantifier, which is enclosed in square
brackets [ ]. Any instance of x which occurs within the scope of the quantifier
∃x is called a bound instance of x, and the quantifier ∃x is said to bind it.
Any other instance of a variable is said to be a free instance. The variables
which have free instances in a formula are called its free variables.
For simplicity, we will always assume that no variable occurs both free
and bound in the same formula. For example, the formula
(x 6 y ∧ ∃x[0 6 x])
should be rewritten as (x 6 y ∧ ∃z[0 6 z]).
Formulas with no free variables are particularly important, essentially
because they say something about a structure as a whole rather than about
some elements of it, so they have a special name.
Definition 3.3. A formula with no free variables is called a sentence.
Interpretation of formulas
The purpose of formulas of L is to say something about L-structures. A
sentence should be either true or false in an L-structure. A formula with
free variables may be true for some interpretations of the variables, and false
for others. We next explain how formulas are interpreted.
Definition 3.4 (Interpretation of formulas). Let ϕ be a formula of L and
x̄ = (x1 , . . . , xn ) a list of variables containing every free variable of ϕ. We
also write ϕ(x̄) for the formula with the list of variables. Let A be an Lstructure and ā = (a1 , . . . , an ) a list of elements of (the domain of) A. We
define the notion A |= ϕ(ā), read “A models ϕ(ā)” or “ϕ(ā) is true in A”
or “A satisfies ϕ(ā)” by recursion on formulas.
A
A
1. A |= t1 (ā) = t2 (ā) iff tA
1 (ā) = t2 (ā), that is, iff the functions t1 and
tA
2 take the same value at ā.
A
A
2. A |= R(t1 (ā), . . . , tr (ā)) iff (tA
1 (ā), . . . , tr (ā)) ∈ R .
CHAPTER 3. FORMULAS
12
3. A |= ¬ϕ(ā) iff A 6|= ϕ(ā), that is, A does not model ϕ(ā).
4. A |= (ϕ1 ∧ ϕ2 )(ā) iff A |= ϕ1 (ā) and A |= ϕ2 (ā).
5. A |= ∃x[ϕ(x, ā)] iff there is some b ∈ A such that A |= ϕ(b, ā).
Definition 3.5. In the special case where ϕ is a sentence, if A |= ϕ we say
that A is a model of ϕ. More generally, if Σ is a set of L-sentences we say
A is a model of Σ and write A |= Σ if A is a model of every sentence in Σ.
This notion is where the name Model Theory comes from.
The definition above formally gives the formulas, which are strings of
symbols, the meaning that we already informally expect them to have. The
purpose of the formal definition is that we can use it to prove things that we
could not prove just from our informal understanding. Since formulas and
terms are defined recursively, the proofs go by induction on the recursive
definitions. The following results give good examples of this style of proof.
Embeddings and formulas
π
Lemma 3.6. An embedding A −→ B of L-structures preserves atomic formulas. That is, whenever ϕ(x̄) is an atomic L-formula and ā ∈ An ,
A |= ϕ(ā) if and only if B |= ϕ(π(ā)).
Proof. Let ϕ be an atomic formula of the form R(t1 (x̄), . . . , tr (x̄)), and ā ∈
B
An . Let αi = tA
i (ā), let βi = ti (π(ā)), and write ᾱ = (α1 , . . . , αr ) and β̄ =
(β1 , . . . , βr ). Then Proposition 2.4 shows that π(ᾱ) = β̄. Then, by definition
of an embedding, ᾱ ∈ RA iff β̄ ∈ RB . Thus A |= R(t1 (ā), . . . , tr (ā)) iff
B |= R(t1 (π(ā)), . . . , tr (π(ā))), that is, A |= ϕ(ā) iff B |= ϕ(π(ā)) as required.
The proof is similar in the case when ϕ is an atomic formula of the form
t1 (x̄) = t2 (x̄).
π
Proposition 3.7. An isomorphism A −→ B of L-structures preserves every
L-formula. That is, whenever ϕ(x̄) is an L-formula and ā ∈ An ,
A |= ϕ(ā) if and only if B |= ϕ(π(ā))
Proof. We proceed by induction on the construction of formulas. If ϕ is an
atomic formula then the result is given by the preceding lemma.
Suppose the result is true for ϕ and ψ.
¬ case:
∧ case:
A |= ¬ϕ(ā)
iff A 6|= ϕ(ā)
iff B |6 = ϕ(π(ā))
by induction hypothesis
iff B |= ¬ϕ(π(ā))
A |= (ϕ ∧ ψ)(ā) iff A |= ϕ(ā) and A |= ψ(ā)
iff B |= ϕ(π(ā)) and B |= ψ(π(ā)) by ind. hyp.
iff B |= (ϕ ∧ ψ)(π(ā))
CHAPTER 3. FORMULAS
13
∃x case: Suppose A |= ∃xϕ(x, ā). Then there is c ∈ A such that A |= ϕ(c, ā).
By induction, B |= ϕ(π(c), π(ā)). So B |= ∃xϕ(x, π(ā)). Conversely, suppose
B |= ∃xϕ(x, π(ā)). Then there is d ∈ B such that B |= ϕ(d, π(ā)). Since π
is an isomorphism it has an inverse, π −1 . Let c = π −1 (d). Then π(c) = d
so, by induction, A |= ϕ(c, ā). So A |= ∃xϕ(x, ā).
That completes all the necessary induction steps, one for each part of
the recursive definition of a formula.
Abbreviations
Our formal language is very restricted. In practice, we adopt many abbreviations which make it easier to use. We write:
not equal
or
implies
iff
for all
t1 6= t2
(ϕ ∨ ψ)
(ϕ → ψ)
(ϕ ↔ ψ)
∀x[ϕ]
for
for
for
for
for
¬t1 = t2
¬(¬ϕ ∧ ¬ψ)
(¬ϕ ∨ ψ)
((ϕ → ψ) ∧ (ψ → ϕ))
¬∃x[¬ϕ]
We could have taken all these symbols as part of our formal language,
but if we did that then our proofs about the formal language would be
longer, so instead we introduce them as abbreviations.
We also write ∃x1 , . . . , xn [ϕ] for ∃x1 [∃x2 [· · · ∃xn [ϕ] · · · ]], and we may
omit brackets or sometimes add extra brackets where this improves clarity and where no ambiguity can arise. We adopt the usual convention of
writing (x + y) instead of +(x, y), and similarly other standard conventions
from algebra. Our general philosophy is to use the usual notation from
mathematical practice where possible, but be careful to see that it can, in
principle, be translated into our formal language.
Exercises
3.1 Write down an Lring -formula in which the variables x, y and z appear
free and the variables u and v are bound.
3.2 In the language L with a binary relation symbol < and constant
symbols for 0, 1, 2, 3, 4, 5, 6, write L-sentences expressing the following about the L-structure on the integers, Z.
(a) There is no greatest integer.
(b) There is no integer between 0 and 1.
(c) For any two distinct integers, one is less than the other.
(d) There are exactly 2 integers between 3 and 6.
3.3 Let L= be the language with no relation, function, or constant symbols.
For a natural number n, write down L= -sentences ϕ>n and ϕ=n such
CHAPTER 3. FORMULAS
14
that for any structure A we have A |= ϕ>n iff A has at least n elements
and A |= ϕ=n iff A has exactly n elements.
3.4 Consider the structure N = hN; +, ·i. Write down:
(i) a formula ϕ(x) in the language h+, ·i with one free variable such
that for any n ∈ N, N |= ϕ(n) iff n = 2;
(ii) a formula ξ(x, y) with two free variables such that for any n1 , n2 ∈
N, N |= ξ(n1 , n2 ) iff n1 < n2 ;
(iii) a formula δ(x, y) with two free variables such that for any n1 , n2 ∈
N, N |= δ(n1 , n2 ) iff n1 |n2 ;
(iv) a formula ψ(x) with one free variable such that for any n ∈ N,
N |= ψ(n) iff n is a prime number;
(v) a sentence stating that there are infinitely many pairs of prime
numbers which differ by 2.
You can use abbreviations provided you explain what they are.
3.5 Consider the Lring -formula ϕ(x) given by ∃y[(y · y) + 1 = x]. For
what values of x do we have Rring |= ϕ(x)? For what x do we have
Zring |= ϕ(x)?
3.6 A formula is said to be quantifier-free if it does not use the symbol ∃
(or the abbreviation ∀).
(i) Write down a recursive definition of quantifier-free formulas in
the style of Definition 3.1.
(ii) Prove by induction on your recursive definition that every embedπ
ding A −→ B of L-structures preserves quantifier-free formulas.
That is, whenever ϕ(x̄) is a quantifier-free L-formula and ā ∈ An ,
A |= ϕ(ā) if and only if B |= ϕ(π(ā)).
3.7 Express each of the statements in question 1.9 as L-sentences for an
appropriate choice of language L.
Chapter 4
Definable sets
In this chapter we introduce the main idea of the book, that of a definable
set.
Definition 4.1. In an L-structure A, a subset S of An is said to be definable
if there is an L-formula ϕ(x1 , . . . , xn ) such that S = {ā ∈ An | A |= ϕ(ā) }.
The formula ϕ(x̄) defines the set S, and we sometimes write S as ϕ(A).
Example 4.2. In Ro-ring , the interval (0, 2) is defined by the formula
(0 < x) ∧ (x < (1 + 1)),
the pair {−1, 3} is defined by the formula
(x + 1 = 0) ∨ (x = (1 + 1) + 1),
and the formula
(0 = x ∨ 0 < x) ∧ (x < 1 + 1) ∧ (y < 0) ∧ (−(1 + 1) < y)
defines the rectangle [0, 2) × (−2, 0).
If t(x1 , . . . , xn ) is an L-term, then the formula y = t(x1 , . . . , xn ) defines
the graph of the function tA in the structure A. However, other functions
may also be definable.
Definition 4.3. Let A be an L-structure, n, m ∈ N+ , D ⊆ An , C ⊆ Am ,
and f : D → C a function. Then f is a definable function (with respect
to the L-structure A) if D, C and the graph of f {(a, f (a)) | a ∈ D } are
definable sets.
Note that if the graph of f is defined by the formula ϕ(x, y), then the
domain of f is defined by ∃y[ϕ(x, y)] and the image is defined by ∃x[ϕ(x, y)].
Example 4.4. In Ro-ring , the absolute value function R → R; x 7→ |x| is
defined by the formula
((x < 0) ∧ (y = −x)) ∨ (¬(x < 0) ∧ (y = x)).
15
CHAPTER 4. DEFINABLE SETS
16
Lemma 4.5. If S1 , S2 ⊆ An are definable then S1 ∩ S2 , S1 ∪ S2 , and An r S1
are definable. Furthermore, if S ⊆ An+m is definable, then the projection
proj(S) = ā ∈ An There is some b̄ ∈ Am such that (ā, b̄) ∈ S
is also definable.
Proof. Suppose S1 is defined by ϕ(x̄) and S2 is defined by ψ(x̄). Then S1 ∩S2
is defined by (ϕ(x̄) ∧ ψ(x̄)) and so on.
Figure 4.1 illustrates why existential quantification corresponds to projection. In the structure Ro-ring , let ϕ(x, y) be the formula
(x − 4)2 + (y − 3)2 < 4
which defines a disc of radius 2, centred at (4, 3) in the xy-plane. We have
used subtraction and squaring in the formula, although they are not symbols
in the language Lo-ring . Since they are definable functions we can use them as
abbreviations. The formula ∃y[ϕ(x, y)] defines the projection to the x-axis,
which is the interval (2, 6).
Figure 4.1: Existential quantification as projection
The subsets of Rn which are defined by Lo-ring -formulas without quantifiers, that is, by equations and inequations of polynomials, are called semialgebraic sets. In fact in this example, using quantifiers does not give any
more definable sets. Unfortunately the proof of this fact is beyond the scope
of this book. So, for example, one can prove that Z is not a definable subset
of R in the structure Ro-ring .
CHAPTER 4. DEFINABLE SETS
17
Automorphisms and definable sets
A very important part of model theory is the study of the definable sets
of a structure A. To show a set is definable we can just come up with a
formula, but it is more difficult to show that no formula defines a set. So it
is important to have methods to show that certain subsets are not definable.
We now develop one such method.
Proposition 4.6. If S ⊆ An is a definable set and π is an automorphism
of A then π(S) = S.
Proof. By π(S), we mean {π(ā) | ā ∈ S }, the image of S under the function
π. Suppose S is defined by the formula ϕ(x̄), and let ā ∈ S. Then A |=
ϕ(ā), so applying Proposition 3.7 we see that A |= ϕ(π(ā)) because π is
an automorphism, and hence π(ā) ∈ S. So π(S) ⊆ S. Now π −1 is also an
automorphism of A, so π −1 (ā) ∈ S. But ā = π(π −1 (ā)), so ā ∈ π(S). Thus
S ⊆ π(S), and hence π(S) = S as required.
Example 4.7. In
the structure Radgp
, the order < is not definable. That is,
the set of pairs (x, y) ∈ R2 | x < y is not definable. To see this, consider
the function π : R → R given by π(x) = −x. It is an embedding of Radgp
into itself because π(0) = 0 and
π(x + y) = −(x + y) = −x + (−y) = π(x) + π(y).
Furthermore π ◦ π is the identity map on R, so π is its own inverse. Hence
π is an automorphism of Radgp . However, we have for example 0 < 1 but
π(0) 6< π(1), and so < is not definable in Radgp .
It is natural to ask if the converse to Proposition 4.6 is true. If S ⊆ An
is a subset which is not definable, must there be an automorphism π of A
such that π(S) 6= S? The answer is no.
Lemma 4.8. There are no non-trivial automophisms of Ro-ring .
Proof. Suppose π is an automorphism of Ro-ring . Then π(0) = 0. Furthermore for any n ∈ N+ we have π(1
· · + 1}) = π(1) + · · · + π(1) = n, and
| + ·{z
|
{z
}
n
n
then π(−n) = −π(n) = −n, so π fixes every integer. If q = n/m ∈ Q we
have π(q) · π(m) = π(n), so π(q) · m = n, but q is the only solution to the
equation x · m = n, which is an Lo-ring -formula, so we must have π(q) = q.
So π fixes every rational number. The rationals are dense in the reals and π
fixes the order <, so it must fix every real number. Hence π is the identity
map on R.
There are uncountably many subsets of Rn , and only countably many
formulas, so only countably many definable subsets, so most of the subsets
of Rn are not definable. However, all subsets of Rn are preserved by the
identity automorphism, and hence by all automorphisms of Ro-ring .
CHAPTER 4. DEFINABLE SETS
18
Exercises
4.1 Complete the proof of Lemma 4.5.
4.2 Show that addition and multiplication are not definable in R< .
4.3 Show that the order < is not definable in Radgp , nor in Zadgp , but it
is definable in Nadmon .
4.4 Show that the order < is definable in Rring . Deduce that every subset
of Rn which is definable in Ro-ring is also definable in Rring .
4.5 Show that the set N and the order < are definable in Zring . [Hint: Use
Lagrange’s 4-square theorem.]
4.6 Show that negation is a definable function in hR; +, 0i and in hZ; +, 0i,
but that multiplication is not definable in either structure.
4.7 Give an example of a definable function in Zring which is not defined
by the interpretation of any term.
4.8 Show that the only automorphism of the structure Ns-ring is the identity
map.
4.9 Consider the expansion of the real ordered field Ro-ring by a unary
function f . Write down a formula ϕ(x) such that whatever the function
f is, ϕ(x) defines the set of real numbers at which f is continuous.
4.10 Find all the subsets of Q which are definable in Qadgp (by a formula
with one free variable) and prove that they are the only ones.
4.11 Sketch a proof of Proposition 4.6 (including any necessary lemmas).
You should give all the key ideas and explain how they fit together,
but you do not need to give all the details of the inductive proofs.
Chapter 5
Substructures and quantifiers
The field of rational numbers, Qring , is a subfield of the field of real numbers,
Rring . Similarly Zring is a subring of Qring . More generally we have the notion
of a substructure of an L-structure.
Definition 5.1. Suppose that A and B are L-structures. Then A is a
substructure of B and B is an extension of A if the domain of A is a subset
of the domain of B and all the symbols of L are interpreted in A as the
restrictions of their interpretations in B. More precisely:
• for each constant symbol c, cA = cB ,
• for each function symbol f and each ā in A, f A (ā) = f B (ā), and
• for each relation symbol R and each ā in A, ā ∈ RA if and only if
ā ∈ RB .
Equivalently, A is a subset of B and the inclusion map of A into B is an
L-embedding. We write A ⊆ B to mean that A is an L-substructure of B.
Remark 5.2. If B is an L-structure with domain B, and A is a non-empty
subset of B, then A is the domain of an L-substructure A of B if and only
if it is closed under all the constants and functions from L.
For example, if R is a ring considered as an Lring -structure, the substructures of R are exactly the subrings of R.
For A ⊆ B, we now consider how the interpretations of formulas are
related in the two structures. Consider the following sentences, at first
written in English sentences, not as L-sentences for a first-order language
L.
1. “All numbers are greater than or equal to 0.”
2. “There is a number which when added to itself makes 1.”
3. “All numbers have square roots.”
19
CHAPTER 5. SUBSTRUCTURES AND QUANTIFIERS
20
The first sentence is true when we interpret “number” as an element of the
structure hN; <, 0i, and also true in the substructure hE; <, 0i consisting of
the even natural numbers, but it is false in the extension hZ; <, 0i. Intuitively
it makes sense because if all natural numbers have a property then certainly
all the even natural numbers have that property, but perhaps some numbers
in a larger set (the integers) might not have that property. For L = h0, <i,
we can write the sentence as the L-sentence
∀x[0 < x ∨ x = 0]
with a universal quantifier. The second sentence is true in Qring and also
in the extension Rring , but false in the substructure Zring . It makes sense
because once the number (1/2) with this property is there in Q it is still in
any extension of Q, but it might not be there in a substructure, in this case
Z. We can write it as
∃x[x + x = 1]
using an existential quantifier.
The third sentence is true in the complex field Cring . It appears to be
another universal statement since it is about “all numbers”, but it is false
in the substructure Rring . It is also false in the ring C[X] of polynomials
over C, which is an extension ring of C. In fact when we write it as an
Lring -sentence we see that it uses both universal and existential quantifiers.
∀x∃y[y · y = x]
The first two sentences are examples of a general phenomenon, that universal statements true in a structure are also true in any substructure, and
that existential statements true in a structure are also true in any extension.
The result even works for formulas with free variables. In our convention,
universal quantifiers are just abbreviations and can be replaced by existential quantifiers but only by also using negation symbols. So we need to
be very careful about exactly what we mean by universal and existential
statements.
Definition 5.3 (Quantifier-free formula). A quantifier-free formula is a formula which does not involve any quantifiers. In order to prove things about
them, it is useful also to give a recursive definition.
1. An atomic formula is a quantifier-free formula.
2. If ϕ is a quantifier-free formula then ¬ϕ is a quantifier-free formula.
3. If ϕ and ψ are quantifier-free formulas then (ϕ ∧ ψ) is a quantifier-free
formula.
4. Only a formula constructed by the above rules in finitely many steps
is a quantifier-free formula.
CHAPTER 5. SUBSTRUCTURES AND QUANTIFIERS
21
Similarly we can give recursive definitions of existential formulas and
universal formulas.
Definition 5.4 (Existential formula).
1. A quantifier-free formula is an existential formula.
2. If ϕ is an existential formula and x is a variable then ∃x[ϕ] is an
existential formula.
3. Only a formula constructed by the above rules in finitely many steps
is an existential formula.
Definition 5.5 (Universal formula).
1. A quantifier-free formula is a universal formula.
2. If ϕ is a universal formula and x is a variable then ∀x[ϕ] is a universal
formula.
3. Only a formula constructed by the above rules in finitely many steps
is a universal formula.
So any existential formula has the form ∃y1 , . . . , yn [ψ] where ψ is a
quantifier-free formula, and any universal formula has the form ∀y1 , . . . , yn [ψ].
From our definition of ∀y as an abbreviation for ¬∃y¬, we can write ∀y1 . . . yn [ψ]
as ¬∃y1 ¬¬∃y2 ¬ . . . ¬∃yn [¬ψ] and cancelling the double negations it becomes
¬∃y1 . . . ∃yn [¬ψ], which is the negation of a universal formula.
Now we can prove that existential statements are indeed “preserved in
extensions” and universal statements are “preserved in substructures”.
Lemma 5.6. If A is an L-substructure of B, ϕ(x̄) is a quantifier-free Lformula and ā ∈ A then
A |= ϕ(ā) if and only if B |= ϕ(ā).
Proof. If ϕ(x̄) is an atomic L-formula the result is immediate from Lemma 3.6.
More generally it is Exercise 3.6.
Proposition 5.7. Let B be an L-structure and A ⊆ B an L-substructure.
Let ϕ(x̄) be an existential L-formula and suppose ā ∈ An is such that we
have A |= ϕ(ā). Then B |= ϕ(ā).
Proof. We proceed by induction on the construction of existential formulas.
If ϕ(x̄) is a quantifier-free formula then the result is Lemma 5.6. Suppose
ϕ(x̄) is ∃y[ψ(x̄, y)]. By assumption, A |= ∃y[ψ(ā, y)], so there is b ∈ A
such that A |= ψ(ā, b). Then, by induction hypothesis, B |= ψ(ā, b). So
B |= ∃y[ψ(ā, y)] as required.
CHAPTER 5. SUBSTRUCTURES AND QUANTIFIERS
22
Proposition 5.8. Let B be an L-structure and A ⊆ B an L-substructure.
Let ϕ(x̄) be a universal L-formula and suppose ā ∈ A is such that B |= ϕ(ā).
Then A |= ϕ(ā).
Proof. We could do a simple inductive proof as for the existential case, but
instead we will deduce it from that case. Let θ be an existential formula such
that ϕ is ¬θ. Suppose that B |= ϕ(ā). Then B 6|= θ(ā) so, by Proposition 5.7,
A 6|= θ(ā). So A |= ϕ(ā) as required.
Exercises
5.1 If L is a language with only relation symbols and A is an L-structure,
show that every subset of A is naturally an L-substructure of A.
5.2 If L has function and constant symbols, which subsets of the domain
of an L-structure correspond to L-substructures?
5.3 Write the axiom for a group which states that every element has an
inverse both as an Lgp -sentence and as an Lmon -sentence. Which of the
axioms is a universal sentence? Show that if G is a group considered
as an Lgp -structure, then its Lgp -substructures are its subgroups and
its Lmon -substructures are submonoids.
5.4 Show that every subgroup of an abelian group is also abelian.
5.5 Show that there is no universal Lring -sentence which asserts that every
non-zero element of a field has a multiplicative inverse.
5.6 Give a direct proof of Proposition 5.8, not using Proposition 5.7.
5.7 Suppose that A and B are L-structures and A ⊆ B. Let ϕ(x1 , . . . , xn )
be an L-formula. Show:
(i) if ϕ is quantifier-free then ϕ(A) = ϕ(B) ∩ An ,
(ii) if ϕ is existential then ϕ(A) ⊆ ϕ(B) ∩ An ,
(iii) if ϕ is universal then ϕ(A) ⊇ ϕ(B) ∩ An .
5.8 Suppose we have a chain of L-structures
A1 ⊆ A2 ⊆ A3 ⊆ · · · ⊆ An ⊆ · · ·
indexed by natural numbers, and A is the union of the chain. Then
A is naturally also an L-structure. Show that if ϕ(x̄) is a ∀∃-formula,
that is, an L-formula of the form ∀ȳ∃z̄ψ(x̄, ȳ, z̄), with ψ quantifierfree, and ā ∈ A1 such that for every n ∈ N we have An |= ϕ(ā), then
A |= ϕ(ā).
Chapter 6
Theories and Axioms
Recall that if A is an L-structure and Σ is a set of L-sentences we write
A |= Σ (“A models Σ”) to mean that every sentence σ in Σ is true in A. In
this chapter we consider the problem of describing all the L-sentences which
are true in a structure A, or more generally those which are true in all the
structures in a given class. Note that this only makes sense for sentences,
not for formulas with free variables.
Definition 6.1. Let C be a class of L-structures. The theory of C, Th(C),
is the set of all L-sentences which are true in every A ∈ C. When C = {A},
a single structure, we write Th(A) for Th({A}).
For any class of structures C, the theory Th(C) has the property of being
deductively closed, which we now explain.
Definition 6.2 (Entailment). Let Σ be a set of L-sentences and ϕ be an
L-sentence. We say Σ entails ϕ or ϕ is a logical consequence of Σ and write
Σ ` ϕ if every model of Σ is also a model of ϕ. If Φ is also a set of sentences
we write Σ ` Φ to mean that for all ϕ ∈ Φ, Σ ` ϕ.
Often the symbol |= is used instead of ` in this context, as well as when
a structure is on the left hand side. Then the symbol ` is used instead only
for a syntactic concept of provability within a given deduction system. Since
we do not use the concept of provability we will follow Wilfred Hodges and
use the symbols this way.
Definition 6.3. A set Σ of L-sentences is deductively closed if for any Lsentence σ, if Σ ` σ then σ ∈ Σ.
Lemma 6.4. For any class C of structures, Th(C) is deductively closed.
Proof. Suppose σ is an L-sentence and Th(C) ` σ. If A ∈ C then A |= Th(C),
so by the definition of `, we have A |= σ. Hence σ ∈ Th(C). So Th(C) is
deductively closed.
23
CHAPTER 6. THEORIES AND AXIOMS
24
Definition 6.5 (Theories). A theory (more precisely, a first-order L-theory)
is a deductively closed set of L-sentences. We say a theory Σ is satisfiable
if there is an L-structure A such that A |= Σ (so Σ has a model) and
unsatisfiable otherwise.
Note that the only unsatisfiable L-theory is Th(∅), which is the set of
all L-sentences. Often “theory” is used to mean “satisfiable theory”.
Axioms
In practice (and often even in principle) we cannot write down or even
describe all the sentences in Th(C). Instead we want to have a set of axioms
for C, which means a set of sentences Σ ⊆ Th(C) such that Σ ` Th(C). To
be useful, we should be able to write these axioms down, or at least describe
them. Often a class of structures is defined by a list of axioms, and for
illustration we give some familiar algebraic examples.
Example 6.6 (Axioms for the theory of groups). In the language Lgp =
h·, ()−1 , 1i we can write the following axioms for groups:
G1. ∀xyz[(x · y) · z = x · (y · z)]
G2. ∀x[x · 1 = x ∧ 1 · x = x]
G3. ∀x[x · x−1 = 1 ∧ x−1 · x = 1]
The theory of groups is the deductive closure of these three axioms. In
other words, it is the set of all Lgp -sentences which are true of all groups.
So, for example, the sentence
∀xy[x−1 · (x · y) = y]
follows easily from the axioms so is in the theory of groups, but the sentence
∃x[x 6= 1 ∧ x · x = 1]
is true in some groups (for example C2 , the cyclic group of order 2), but not
in other groups (for example C3 , the cyclic group of order 3), so is not in
the theory of groups.
Example 6.7 (Abelian groups). A group is abelian if it satisfies also:
AG. ∀xy[x · y = y · x]
so the theory of abelian groups is the deductive closure of {G1, G2, G3, AG}.
We have written 4 axioms for abelian groups, but we could write a single
axiom with the same meaning, the conjunction of those 4 axioms:
G1 ∧ G2 ∧ G3 ∧ AG.
The natural language in which to axiomatize rings or fields is Lring =
h+, ·, −, 0, 1i, the language of rings.
CHAPTER 6. THEORIES AND AXIOMS
25
Example 6.8 (Rings). We have the axioms for abelian groups, but now in
the additive language Ladgp .
R1. ∀xyz[(x + y) + z = x + (y + z)]
R2. ∀x[x + 0 = x ∧ 0 + x = x]
R3. ∀x[x + (−x) = 0 ∧ (−x) + x = 0]
R4. ∀xy[x + y = y + x]
Then there are axioms for commutative monoids (the abelian group axioms
except for the inverse) in the language h·, 1i:
R5. ∀xyz[(x · y) · z = x · (y · z)]
R6. ∀x[x · 1 = x ∧ 1 · x = x]
R7. ∀xy[(x · y) = (y · x)]
and a distributivity axiom:
R8. ∀xyz[x · (y + z) = (x · y) + (x · z)].
Together these 8 sentences axiomatise the theory of (commutative) rings.
Example 6.9 (Fields). A field is a ring satisfying two additional axioms.
First, an axiom to rule out the zero ring,
F1. 0 6= 1,
and second an axiom which says that every non-zero element of the field has
a multiplicative inverse:
F2. ∀x[x = 0 ∨ ∃y[x · y = 1]].
So the axioms R1–R8, F1 and F2 together axiomatize the theory of fields.
Remark 6.10. Notice that we put −, the additive inverse, as a function
symbol in the language. However, we cannot easily do the same for the
multiplicative inverse, because it is not defined at 0. We could put a unary
function symbol for it in the language, but we would have to define 0−1
somehow arbitrarily, and then make sure that our axioms describing how
(−)−1 works in fields took this case into account. We follow the usual
practice of not introducing a symbol for it.
CHAPTER 6. THEORIES AND AXIOMS
26
Peano arithmetic and complete theories
One of the most important structures in mathematics is hN; +, ·, 0, 1i, the
natural numbers with both addition and multiplication. In Mathematical
Logic, the study of this structure and its theory (and related theories) is
often called arithmetic, and Th(N) is sometimes called true arithmetic. We
can try to write down axioms for Th(N). The most common and useful
axioms are those written down by the Italian mathematician Peano. In
modern form, they are as follows.
P1. ∀x[x + 1 6= 0]
P2. ∀x[x 6= 0 → ∃y[x = y + 1]]
P3. ∀xy[x + 1 = y + 1 → x = y]
P4. ∀x[x + 0 = x]
P5. ∀xy[x + (y + 1) = (x + y) + 1]
P6. ∀x[x · 0 = 0]
P7. ∀xy[x · (y + 1) = (x · y) + x]
and for each L-formula ϕ(x), with one free variable x, the axiom
Pϕ. (ϕ(0) ∧ ∀x[ϕ(x) → ϕ(x + 1)]) → ∀xϕ(x)
The last axiom scheme tries to capture the idea of induction. Induction says
that if 0 has a certain property, and whenever n has that property then so
does n + 1, then every natural number has that property. The axiom scheme
says that for each property which can be expressed as a first-order formula.
This set of axioms, or its deductive closure, is called the theory of Peano
Arithmetic, or PA. It is easy to check each axiom individually to see that
N |= PA, and hence that PA ⊆ Th(N). Some of the axioms look a little
strange, for example (P5) is a very limited form of the associativity axiom
and we could replace it with the full associativity axiom. In fact the full associativity of addition can be proved from all the axioms using the induction
scheme. The reason for using the limited form is that it is, at first sight,
weaker, and Peano was trying to write down the weakest set of axioms from
which everything else can be deduced. Can everything else about arithmetic
be deduced from the axioms of PA?
Definition 6.11. An L-theory T is said to be complete if for any L-sentence
σ, either T ` σ or T ` ¬σ.
Lemma 6.12. Let σ be any Ls-ring -sentence. Then either σ ∈ Th(N) or
¬σ ∈ Th(N). In particular, Th(N) is complete.
CHAPTER 6. THEORIES AND AXIOMS
27
Proof. If N |= σ then σ ∈ Th(N). Otherwise, N 6|= σ, so N |= ¬σ, so
¬σ ∈ Th(N). Now Th(N) is complete because if σ ∈ Th(N) then certainly
Th(N) ` σ.
More generally, Th(A) is complete for any L-structure A. So the question
is whether or not PA is complete. One of the great theorems of 20th century
mathematics is Gödel’s First Incompleteness Theorem, which states that PA
is not complete. In fact, Gödel showed that whatever set of axioms about
N we write down explicitly it will not axiomatize all of true arithmetic.
Elementary equivalence
Definition 6.13. Two L-structures A and B are elementarily equivalent,
written A ≡ B, iff Th(A) = Th(B), that is, for every L-sentence σ, A |= σ
iff B |= σ.
So elementarily equivalent L-structures cannot be distinguished by formal sentences in the language L.
Lemma 6.14. If A and B are isomorphic L-structures then they are elementarily equivalent.
π
Proof. Let A −→ B be an isomorphism, let σ be an L-sentence. Since σ has
no free variables, we can apply Proposition 3.7 to σ with the empty list of
variables to deduce that A |= σ if and only if B |= σ. This applies for every
L-sentence, so A ≡ B.
For finite structures (structures whose domain is a finite set), isomorphism and elementary equivalence are actually the same. (See the exercises
below.) However, we will see later that if A is an infinite structure then
there are infinitely many other structures which are elementarily equivalent
to A (and thus to each other) but which are pairwise non-isomorphic.
Exercises
6.1 Given a set Σ of L-sentences, let Mod(Σ) be the class of L-structures
A such that A |= Σ. Show that Th(Mod(Σ)) is the set of all logical
consequences of Σ, often called the deductive closure of Σ.
6.2 Show that a theory T is complete if and only if whenever A and B are
models of T then A ≡ B.
6.3 For each of the five structures Ns-ring , Zs-ring , Qs-ring , Rs-ring , Cs-ring ,
write down a sentence which is true in that structure but not in any
of the other four structures.
6.4 Show that the theory of fields is not complete.
CHAPTER 6. THEORIES AND AXIOMS
28
6.5 Show that there must be a model of Peano Arithmetic which is not
isomorphic to the standard model, Ns-ring .
6.6 Show that in any model A of PA, the function +A is associative.
6.7 Let L = hf, R, ci where f is a binary function symbol, R is a binary
relation symbol, and c is a constant symbol. Let A be an L-structure
with 3 elements. Show that there is a single L-sentence σA such that
for any L-structure B, if B |= σA then B ∼
= A. How would you change
your proof to work for an arbitrary finite structure in an arbitrary
finite language?
6.8 Can you find two structures which are elementarily equivalent but not
isomorphic? [This is quite difficult without some theory. Later we will
see many ways to answer the question.]
Chapter 7
The Complex and Real
Fields
Gödel’s First Incompleteness Theorem shows that we cannot succeed in
writing down a complete list of axioms for Ns-ring . Using this result, the
same can also be shown for Zring and Qring . However, in many cases the
problem of taking a structure and coming up with a complete axiomatization
for its theory is solvable. To find a candidate set of sentences Σ which
might axiomatize Th(A), start by writing down some sentences capturing
properties of the structure. If you can find another model B |= Σ and a
sentence σ such that A |= σ and B 6|= σ then add σ to Σ to get a better
candidate. When you can no longer do this, you have a plausible candidate
for a complete axiomatization. Later in this book we will see two methods
which can be used to prove that an axiomatization is complete: categoricity
and the back-and-forth method.
In this chapter we will just describe axiomatizations for the complex and
real fields, but will not prove they are complete. For the complex field, the
proof that the axiomatization is complete will be given in Part IV. The proof
for the real field is similar but more involved and can be found in the books
of Marker and of Poizat.
The complex field
To axiomatize Th(Cring ) we start with the axioms of fields. The field C has
characteristic 0, which means that it satisfies the axioms
1
|+1+
{z· · · + 1} 6= 0
n
N+ .
for each n ∈
Since there are fields with non-zero characteristic, we need
these axioms.
In Lring we can express statements about polynomials. The fundamental
theorem of algebra states that every non-zero polynomial with coefficients
29
CHAPTER 7. THE COMPLEX AND REAL FIELDS
30
from C has a root in C. This fact can be written as a list of sentences in
Lring , taking the sentence σn given by
∀y0 . . . yn−1 ∃x[xn + yn−1 · xn−1 + · · · + y1 · x + y0 = 0]
for each n ∈ N+ , where we use the usual abbreviation of xn for multiplying
n copies of x together, and omit unnecessary brackets.
Fact 7.1. The axioms of fields of characteristic 0 together with the axioms
σn above axiomatize the complete theory of the complex field Cring .
These axioms have become so useful they have a name.
Definition 7.2. A field satisfying the sentences σn for all n ∈ N+ is said to
be algebraically closed.
The complete theory of Cring is known as ACF0 , the theory of algebraically closed fields of characteristic 0. Since the theory is complete, any
two algebraically closed fields of characteristic 0 satisfy exactly the same
Lring -sentences.
Complete ordered fields
Definition 7.3 (Ordered field). The language of ordered rings is Lo-ring =
Lring ∪ {<}. In this language, the axioms of ordered fields are the axioms of
fields together with axioms stating that < is a linear order:
O1. (Transitivity) ∀xyz[(x < y ∧ y < z) → x < z]
O2. (Linearity) ∀xy[x < y ∨ x = y ∨ y < x]
O3. (Irreflexivity) ∀x[¬x < x]
and axioms relating the field structure with the order:
O4. (Additivity) ∀xyz[x < y → x + z < y + z]
O5. (Multiplicativity) ∀xyz[(x < y ∧ 0 < z) → x · z < y · z].
An ordered field F is necessarily of characteristic 0, since from the axioms
we can deduce
0 < 1 < 1 + 1 < 1 + 1 + 1 < ···
so we have N sitting inside F as a subset in the obvious way.
The real field satisfies a further property, the completeness of its ordering.
Definition 7.4. An ordered set F is complete if every non-empty subset
S ⊆ F which has an upper bound has a least upper bound.
CHAPTER 7. THE COMPLEX AND REAL FIELDS
31
This is a different usage of the word complete from completeness of a
theory. A simple consequence of the completeness property for an ordered
field is that it satisfies the property of Archimedes.
Proposition 7.5. If F is a complete ordered field then it is Archimedean,
that is, the subset N of F is not bounded above.
Proof. Suppose for a contradiction that N is bounded above. Then, by the
completeness property, there is a least upper bound b for N. Now b − 1 < b
so b − 1 is not an upper bound for N; hence there is a natural number n ∈ N
with b − 1 < n. But then n + 1 ∈ N and b < n + 1, contradicting b being an
upper bound for N.
A very important theorem for the foundations of analysis is that there
is only one complete ordered field.
Fact 7.6. Up to isomorphism, there is exactly one complete ordered field.
The unique (up to isomorphism) model of the axioms of complete ordered
fields is the field of real numbers. There are three common ways to build
models for these axioms, that is, to construct the field of real numbers. The
most familiar way is construct R as the field of decimal numbers. It can also
be constructed via Dedekind cuts, or by Cauchy sequences. These methods
give three different models of the axioms but, by the above fact, they are
all isomorphic so it does not matter which of the three (or any other model
constructed by a different method) is used.
First-order axioms for the real field
The axioms for ordered fields are first-order sentences of the language Lo-ring .
We can write the completeness axiom in logical symbols. The formula
∀x[x ∈ S → x 6 y]
states that y is an upper bound for S. Abbreviating it by UB(S, y), the
formula
UB(S, z) ∧ ∀y[UB(S, y) → z 6 y]
states that z is the least upper bound for S. Abbreviating that by LUB(S, z),
the sentence
(∀S ⊆ F ) [((S 6= ∅) ∧ ∃y[UB(S, y)]) → ∃z[LUB(S, z)]]
expressed the completeness axiom. However, this is not a formula in the
language Lo-ring as we have defined it because we have also used the symbols
⊆ and ∅, but most importantly we have a variable S which does not range
over the elements of F but over its subsets, and then we have quantified over
CHAPTER 7. THE COMPLEX AND REAL FIELDS
32
that. This is not allowed in first-order logic, but it is allowed in so-called
second-order logic. While we can make sense of what this axiom means, it
is actually not expressible by any set of first-order sentences. (We will prove
this later.) It follows that there are models of the complete first-order theory
of the real field, Th(R), which are necessarily ordered fields but which are
not complete as ordered sets.
However, it turns out that we can axiomatize the first-order theory
Th(R) in a way closer to the axiomatization of algebraically closed fields.
Definition 7.7. The axioms for real-closed fields consist of the axioms of
ordered fields together with
RCF1. ∀x[0 6 x → ∃y[y 2 = x]]
RCF2. ∀y0 . . . yn−1 ∃x[xn + yn−1 · xn−1 + · · · + y1 · x + y0 = 0]
for each odd n ∈ N.
In other words, a real-closed field is an ordered field in which every
positive element has a square root and every odd-degree polynomial has a
root. Using the basic ideas of real analysis it is easy to verify that R is a
real-closed field.
Fact 7.8. The axioms of real-closed fields axiomatize the complete theory of
the real ordered field Ro-ring .
Thus any two real-closed fields satisfy exactly the same Lo-ring -sentences.
Exercises
7.1 Give an example to show that the axioms of ordered fields do not
axiomatize a complete theory.
7.2 Verify that Ro-ring does satisfy the of axioms for real-closed fields.
7.3 Prove that in any ordered field F we have 0 < 1 and 1 < 1 + 1, and
more generally that hN; +, 0, 1, <i embeds in hF ; +, 0, 1, <i.
7.4 Let Σ be the set of axioms of ordered fields. Show that Σ ` ∀x[x·x > 0].
7.5 Suppose that S ⊆ R is non-empty and bounded above, and furthermore that it is defined by an Lo-ring -formula ϕ(x). Now suppose that
R is another real-closed field, with domain R. Show that the subset
ϕ(R) of R defined by ϕ(x) has a least upper bound in R.
7.6 Is the usual definition of topological spaces given by first-order axioms
or second-order axioms? What about metric spaces?
Chapter 8
Compactness
If we start with a non-empty class of structures C and form a theory T =
Th(C), we know that T is satisfiable (that is, it has a model) because we
have some models to start with. However, we can write down a set of
sentences Σ without any particular model in mind, and want to know that
it is satisfiable. It may be very difficult to describe a model of Σ explicitly.
However the next theorem, which is the most important theorem in model
theory, gives us at least a practical way to show that models do exist, and
that when an infinite model exists then there are a lot of other models as
well.
In the next chapter we will see that the same theorem helps us to understand the class of models of a theory, and it can be used to show that
a class of structures is not the class of models of any theory. It has many
other uses as well.
Theorem 8.1 (The Compactness Theorem). Let Σ be a set of L-sentences.
If every finite subset of Σ has a model then Σ has a model.
There are at least three different approaches to the proof of the compactness theorem. Later we will see a way of explicitly constructing a model
from the set of sentences Σ, which is due to Henkin. There is also a more
algebraic proof, which builds a model of Σ from given models of its finite
subsets using the technique of ultraproducts. These methods of proof are
reasonably self-contained, but not straightforward. By contrast, the original
proof uses the idea of a formal deduction so is not self-contained. For those
who have seen the notion of a formal deduction, we give this short proof of
the compactness theorem here.
Proof. Given a set of L-sentences Σ and another L-sentence ϕ, we can write
Σ `d ϕ to mean that there is a formal deduction of ϕ from Σ. A formal
deduction is a finite list of L-formulas, so it only uses finitely many of the
sentences from Σ. So if Σ `d ϕ then there is a finite subset Σ0 of Σ such
33
CHAPTER 8. COMPACTNESS
34
that Σ0 `d ϕ. The Completeness Theorem of first-order logic states that
Σ `d ϕ if and only if Σ ` ϕ.
We write ⊥ for some contradictory sentence, for example ∃x[x 6= x],
which is not true in any L-structure at all. Now Σ ` ⊥ means that every
model of Σ is a model of ⊥, and since there are no models of ⊥, it means
there are no models of Σ, that is, Σ is not satisfiable.
Now suppose that Σ is not satisfiable. Then Σ ` ⊥, so Σ `d ⊥, hence
there is a finite subset Σ0 of Σ such that Σ0 `d ⊥. Then Σ0 ` ⊥, so Σ0 is
not satisfiable. Therefore, whenever every finite subset of Σ is satisfiable, Σ
itself is satisfiable, as required.
The use of new constant symbols
We will use the compactness theorem to show that if A is any infinite Lstructure then there is another L-structure B which is elementarily equivalent to A, that is, is satisfies exactly the same L-sentences, but which is not
isomorphic to A. The other tool we use is to change the language by adding
new constant symbols.
Definition 8.2. Let L be a language and L+ another language such that
L ⊆ L+ , that is, every symbol of L is also a symbol of L+ . Let A+ be an
L+ -structure with domain A. Then the reduct of A+ to L is the L-structure
A with domain A, and every symbol of L interpreted in A exactly as in A+ .
We also say that A+ is an expansion of A to the language L+ .
For example, Radgp = hR; +, −, 0i is a reduct of Rring = hR; +, −, ·, 0, 1i,
which in turn is a reduct of Ro-ring = hR; +, −, ·, 0, 1, <i.
Before proving the general theorem we do a special case with a simpler
proof which contains the main ideas.
Proposition 8.3. There is a model of true arithmetic, Th(Ns-ring ), which
is not isomorphic to the standard model Ns-ring .
Proof. Let L+ = Ls-ring ∪ {c} where c is a new constant symbol. Let Σ be
the set of L+ -sentences
Th(Ns-ring ) ∪ {c 6= 0, c 6= 1, c 6= (1 + 1), c 6= ((1 + 1) + 1), . . .}
which consists of the complete theory of Ns-ring in the language Ls-ring , together with sentences using the new constant symbol c which say that c is
not equal to any natural number. Now let Σ0 be a finite subset of Σ. Then
there is some number m ∈ N such that the sentence c 6= (1 + 1) + · · · + 1)
|
{z
}
m
does not occur in Σ0 . We make a model of Σ0 by taking Ns-ring as usual and
interpreting the new constant symbol c as the number m. So Σ0 is satis+
fiable. Hence, by compactness, there is a model M+ = hM ; +, ·, 0, 1, cM i
CHAPTER 8. COMPACTNESS
35
of Σ. We consider the reduct Ms-ring = hM ; +, ·, 0, 1i, which is an Ls-ring structure. Then Ms-ring |= Th(Ns-ring ) which is a complete Ls-ring -theory,
+
and so Th(Ms-ring ) = Th(Ns-ring ). The element cM of M is not equal to
any natural number. So Ms-ring is not isomorphic to Ns-ring .
We can do the same for any infinite structure A, but we may need to
use more than one new constant symbol.
Proposition 8.4. Let A be any infinite L-structure. Then there is another
L-structure B which is elementarily equivalent to A but not isomorphic to
it.
Proof. Any isomorphism is a bijection, so isomorphic models have the same
cardinality (size). So we will ensure that B has larger cardinality than A,
so cannot be isomorphic to it. To do this, let I be a set of cardinality
larger than |A|. Let LI be the language obtained from L by adding distinct
constant symbols ci for each element of I.
Let Σ = Th(A) ∪ {ci 6= cj | i, j ∈ I, i 6= j }, a set of LI -sentences. We
claim that Σ is finitely satisfiable. So let Σ0 be a finite subset of Σ. Then
only finitely many of the ci , say ci1 , . . . , cin , are used in Σ0 . Now choose
a1 , . . . , an from A, all distinct. This is possible because A is an infinite
structure. Expand A to an LI -structure A+ by interpreting cij as aj for
j = 1, . . . , n, and every other ci as a1 . Then
A+ |= Th(A) ∪ cij 6= cik | 1 6 j < k 6 n
so A+ |= Σ0 . Thus Σ is finitely satisfiable, and hence, by compactness, it
has a model, say B + , which is an LI -structure. Let B be the domain of B + ,
+
and for each i ∈ I let ai = ciB . Since B + |= Σ, the elements ai of B are all
distinct, and so |B| > |I| > |A|.
Let B be the reduct of B + to L. Then B |= Th(A) and so B is elementarily
equivalent to A. The domain of B is B, the same as that of B + . So B cannot
be isomorphic to A because their cardinalities are different.
The compactness theorem opens the door to study infinite structures by
looking at different models of the same theory. We will explore this more
later. Now we give an indication of what the compactness theorem says
about definable sets.
Applications to definable sets
Consider the closed intervals [0, 1/n] on the real line, for n ∈ N+ . The
intersection of all these intervals is {0}, so in particular the intersection is
non-empty. If instead we consider the open intervals (0, 1/n) for n ∈ N+
then the intersection is empty. However if we take any finite sub-collection,
say {(0, 1/n1 ), . . . , (0, 1/nr )}, then the intersection is the non-empty interval
(0, 1/N ), where N = max{n1 , . . . , nr }.
CHAPTER 8. COMPACTNESS
36
Proposition 8.5. There is a model R of Th(Ro-ring ) in which the intersection of the intervals {(0, 1/n) | n ∈ N+ }, as interpreted in R, is nonempty.
Proof. The interval (0, 1/n) is defined by the Lo-ring -formula ϕn (x) given by
0 < x ∧ n · x < 1, where we use n as an abbreviation for |1 + ·{z
· · + 1}. Let
n
c be a new constant symbol, and let Σ = Th(Ro-ring ) ∪ {ϕn (c) | n ∈ N }, a
set of sentences in the language Lo-ring expanded by c. Let Σ0 be a finite
subset of Σ, and let N ∈ N+ be larger than any n such that ϕn (c) ∈ Σ0 .
Expand Ro-ring to R+ by interpreting c as 1/N . Then R+ |= Σ0 . So Σ is
finitely satisfiable and hence by compactness it has a model, say R+ . Let
+
a = cR . The reduct R of R+ to Lo-ring is a model of Th(Ro-ring ) and the
element a ∈ R satisfies R |= 0 < a ∧ n · a < 1 for every n ∈ N. Hence it lies
in all the intervals (0, 1/n) as interpreted in R.
The word compactness comes from topology, where it is a very useful
property of a topological space. The above proposition hints that definable
sets behave like the closed sets of a compact topological space, at least if we
are allowed to change models. In the exercises this is explored further, and
we exploit this idea properly in Part III. However we will generally not use
the language of topology.
Exercises
8.1 Is there a model M of Th(Ns-ring ) with an element m ∈ M satisfying
0 < m < 1?
8.2 Sketch a proof of Proposition 8.4. You should list all the key ideas and
explain how they fit together, without giving all the details.
8.3 Where does the proof of Proposition 8.4 go wrong if A is a finite Lstructure?
8.4 Show that there is a model M of Th(N< ) with an infinite descending
chain a1 > a2 > a3 > · · · in M.
8.5 Suppose that ϕ(x, y) is an Ls-ring -formula such that for every n ∈ N, the
subset ϕ(Ns-ring , n) = {x ∈ N | Ns-ring |= ϕ(x, n) } of N is non-empty.
Let M |= Th(Ns-ring ) and let a ∈ M. Show that ϕ(M, a) has a least
element.
8.6 Suppose that A is an infinite L-structure and ϕ(x̄) is a formula such
that ϕ(A) is an infinite set. Show there is B elementarily equivalent
to A such that ϕ(B) is uncountable.
8.7 Suppose A is any L-structure, and {Si | i ∈ I } is any collection of
definable subsets
of A such that for every finite subset I0 ⊆ I, the
T
intersection i∈I0 Si is non-empty.
T Show there is B ≡ A such that,
interpreted in B, the intersection i∈I Si is non-empty.
CHAPTER 8. COMPACTNESS
37
8.8 Suppose Σ is a set of L-sentences and {ϕi (x) | i ∈ I } is a set of Lformulas such that for every model A |= Σ and every element a ∈ A
there is some i ∈ I such that AV
|= ϕi (a). Show
that there is a finite
subset I0 ⊆ I such that Σ ` ∀x i∈I0 ϕi (x) .
Chapter 9
Axiomatizable classes
The axiomatic method has been important in mathematics since Euclid.
However, it is a more modern concern to understand that a list of axioms
may have many different models. Now axioms are often used to define a particular collection of structures, such as groups, rather than being supposed
to capture self-evident truths of arithmetic or geometry.
Definition 9.1. If Σ is a set of L-sentences we write Mod(Σ) for the class
of all L-structures which are models of Σ. The class Mod(Σ) is said to be
axiomatized by Σ, and is an axiomatizable class. It is finitely axiomatizable
if it is axiomatized by a finite set of sentences.
For example, the class of all groups is axiomatized by the axioms G1, G2,
and G3 given in chapter 6. To show a class of structures is axiomatizable
it suffices to write down appropriate axioms. In this chapter we will use
the compactness theorem to prove that certain classes of structures are not
axiomatizable, or are axiomatizable but not finitely axiomatizable. Recall
it from the previous chapter.
Theorem 9.2 (The Compactness Theorem). Let Σ be a set of L-sentences.
If every finite subset of Σ has a model then Σ has a model.
Finite and infinite models
The class of all groups of size 3 is axiomatizable, since it is the class of all
models of the group axioms together with the sentence
∃x1 x2 x3 [x1 6= x2 ∧ x1 6= x3 ∧ x2 6= x3 ∧ ∀y[y = x1 ∨ y = x2 ∨ y = x3 ]].
By a similar argument, the class of groups of any finite size is axiomatizable.
The class of all infinite groups is axiomatizable
as can be seen
by considering
hV
i
+
the sentences κn given by ∃x1 , . . . , xn
16i<j6n xi 6= xj for n ∈ N . It is
the class Mod({G1, G2, G3} ∪ {κn | n ∈ N+ }). However, the class of finite
groups is not axiomatizable.
38
CHAPTER 9. AXIOMATIZABLE CLASSES
39
Proposition 9.3. Let C be an axiomatizable class with arbitrarily large finite
models. That is, for every n ∈ N there is a model An ∈ C whose domain is
a finite set of size at least n. Then C contains an infinite model.
Proof. Let Σ be a set of sentences such that C = Mod(Σ). Let Σ0 = Σ ∪
{κn | n ∈ N+ }, where the sentences κn are as given above, and let Σ0 be a
finite subset of Σ0 . Let N ∈ N be larger than any n such that κn ∈ Σ0 , which
is possible because Σ0 is finite. Then AN |= Σ0 . So Σ0 is finitely satisfiable,
hence by compactness it has a model, say B. Then B is an infinite structure
and B |= Σ, so B ∈ C.
Corollary 9.4. The class of finite groups is not axiomatizable.
Proof. There are arbitrarily large finite groups, for example the cyclic group
Cn of order n for each n ∈ N+ . So any axiomatizable class containing all
finite groups also contains some infinite groups.
Torsion in abelian groups
Consider an abelian group G written additively in the language Ladgp =
h+, −, 0i. For g ∈ G and n ∈ N+ we write ng as an abbreviation for
g + . . . + g as usual.
| {z }
n
We say that g is an n-torsion element of G iff we have ng = 0. It is a
torsion element iff it is an n-torsion element for some n. A group is torsionfree if its only torsion element is 0. For each n ∈ N+ , the sentence ϕn given
by
∀x[nx = 0 → x = 0]
states that there are no n-torsion elements, except for 0. The set of axioms for abelian groups together with {ϕn | n ∈ N+ } axiomatizes the class
of torsion-free abelian groups.
Now we consider torsion groups. An abelian group is a torsion group
iff all of its elements are torsion elements. Every finite abelian group is a
torsion group, but there are also infinite torsion groups such as the direct
sum of infinitely many finite groups.
Proposition 9.5. The class of all torsion abelian groups is not axiomatizable.
Proof. Suppose it were axiomatized by Σ. Expand the language Ladgp to L0
by adding one new constant symbol c. Let ψn be the L0 -sentence nc 6= 0
and let Σ0 = Σ ∪ {ψn | n ∈ N+ }. Let Σ0 be a finite subset of Σ0 , and let
N ∈ N+ be larger than any n such that ψn ∈ Σ0 . Let A be the L0 -structure
consisting of the cyclic group CN with cA being a cyclic generator. Then
A |= Σ because CN is a torsion abelian group, and also A |= ψn for each
n < N , because cA has order N . Hence A |= Σ0 , so Σ0 is satisfiable.
CHAPTER 9. AXIOMATIZABLE CLASSES
40
By compactness, Σ0 is satisfiable. Let G be a model of Σ0 . Then G is a
torsion abelian group since G |= Σ, but G has an element cG which is not
an n-torsion element for any n because G |= ψn . This contradiction shows
that Σ could not exist, hence the class of all torsion abelian groups is not
axiomatizable.
Finite axiomatizability
Definition 9.6. An axiomatizable class is finitely axiomatizable iff there is
a finite set of sentences Σ which axiomatizes the class.
For example, the class of groups is finitely axiomatizable. However our
choices of axioms for algebraically closed fields and for Peano Arithmetic
were not finite, but involved infinite axiom schemes, as did our axiomatization of torsion-free abelian groups. It is often useful to know whether a
finite axiomatization exists or at least if it does not exist to prove it. We
start with an easy lemma.
Lemma 9.7. If C is finitely axiomatizable, then it is axiomatized by a single
sentence.
Proof.
If C is axiomatized by {σ1 , . . . , σn } then it is also axiomatized by
Vn
i=1 σi .
Now if for example the class of torsion-free abelian groups were finitely
axiomatizable, how could we find a finite list of axioms? In fact we do not
have to look through all sentences.
Proposition 9.8. If C = Mod(Σ) and Cis finitely axiomatizable then it is
axiomatized by a finite subset of Σ.
Proof. Suppose not. Then using the above lemma we have a set Σ of sentences and a single sentence ϕ such that C = Mod(Σ) and C = Mod(ϕ), but
C is not axiomatized by any finite subset of Σ. Let Σ0 be a finite subset of
Σ. Then Mod(Σ0 ) ⊇ Mod(Σ), but by assumption Mod(Σ0 ) 6= Mod(Σ), so
there is A |= Σ0 such that A 6|= Σ. Thus A ∈
/ C, and so A |= ¬ϕ. Thus
Σ0 ∪ {¬ϕ} is satisfiable. This holds for any finite subset Σ0 of Σ, so, by
compactness, there is a model B of Σ ∪ {¬ϕ}. But then B |= Σ so B ∈ C,
and B |= ¬ϕ, so B ∈
/ C, a contradiction.
Corollary 9.9. The class of all torsion-free abelian groups is not finitely
axiomatizable.
Proof. Recall that the class of torsion-free abelian groups is axiomatized by
{G1, G2, G3, AG} ∪ {ϕn | n ∈ N+ }. Suppose Σ0 is a finite subset of these
axioms, and let N ∈ N be larger than any n such that ϕn ∈ Σ0 . Let p
be a prime such that p > N . Then the cyclic group Cp of order p is an
CHAPTER 9. AXIOMATIZABLE CLASSES
41
abelian group and has no n-torsion for n < p, so Cp |= Σ0 . However, Cp has
p-torsion, so Σ0 does not axiomatize the class of torsion-free abelian groups.
By the proposition, this class is not finitely axiomatizable.
Next we give a result about axiomatizing the complement of an axiomatizable class.
Proposition 9.10. Let C be an axiomatizable class of L-structures and let
D = {A an L-structure | A ∈
/ C }. Then D is axiomatizable if and only if C
and D are finitely axiomatizable.
Proof. First suppose that C is axiomatized by Σ, and D is axiomatized
by Φ. Suppose for a contradiction that C is not finitely axiomatizable.
Let Σ0 be a finite subset of Σ and let Φ0 be a finite subset of Φ. Then
Mod(Σ0 ) ) Mod(Σ), so there is A |= Σ0 such that A ∈
/ C. So A ∈ D, so
A |= Σ0 ∪ Φ. In particular, Σ0 ∪ Φ0 is satisfiable. Then, by compactness,
Σ ∪ Φ is satisfiable, so there is a model M |= Σ ∪ Φ, so M ∈ C and M ∈ D,
a contradiction. So C is finitely axiomatizable.
Now suppose C is finitely axiomatizable. Then by Lemma 9.7 it is axiomatized by a single sentence, say ϕ. Then D is axiomatized by ¬ϕ.
Exercises
9.1 Let C be a non-empty class of L-structures. Show that C ⊆ Mod(Th(C))
and that C is axiomatizable iff equality holds.
9.2 Let Σ be a satisfiable set of L-sentences. Show that Σ ⊆ Th(Mod(Σ))
and that Σ is a theory iff equality holds.
9.3 Show that the class of groups of size less than 100 is axiomatizable,
and that the class of infinite groups is axiomatizable but not finitely
axiomatizable.
9.4 Show using the compactness theorem and the method of new constants
from the previous chapter that if C is an axiomatizable class with either
arbitrarily large finite models or an infinite model then it has infinite
models of arbitrarily large cardinality.
9.5 Is the class of cyclic groups axiomatizable? Is it finitely axiomatizable?
9.6 An abelian group G is said to be n-divisible for some n ∈ N+ if it
satisfies ∀x∃y[ny = x], so each element can be divided by n (not
necessarily uniquely). G is divisible if it is n-divisible for all n ∈ N+ .
Show that the class of divisible abelian groups is axiomatizable but
not finitely axiomatizable.
9.7 What familiar mathematical structures are models of the theory DTFAG
of divisible, torsion-free abelian groups?
CHAPTER 9. AXIOMATIZABLE CLASSES
42
9.8 The characteristic of a field F is the least number n ∈ N+ such that
1| + ·{z
· · + 1} = 0, if such an n exists, and 0 otherwise. The characteristic
n
of a field is always 0 or a prime number. Show that the class of fields
of a fixed prime characteristic p is finitely axiomatizable, and that
the class of fields of characteristic 0 is axiomatizable, but not finitely
axiomatizable.
9.9 Show that the class of fields of positive characteristic is not axiomatizable.
9.10 Suppose that σ is an Lring -sentence which is true in every field of
characteristic 0. Show that there is N ∈ N such that whenever F is a
field of characteristic p with p > N , then F |= σ.
9.11 A graph G consists of a set of vertices, and between each pair of
distinct vertices, there may or may not be an edge. We can consider a
graph as a structure for the language with one binary relation E. The
vertices of the graph are the elements of the structure, and vertices
v, w have an edge between them if E(v, w) holds. A graph is then a
model of the sentence
∀xy[¬E(x, x) ∧ (E(x, y) → E(y, x))].
We say that a graph is connected iff for every pair of vertices v, w, there
is a path from v to w, which means there is a sequence of vertices
v = v1 , v2 , . . . , vn = w with an edge between vi and vi+1 for each
i = 1, . . . , n − 1.
Show that the class of connected graphs is not axiomatizable.
Chapter 10
Constructing models from
syntax
The goal of this chapter is to give a proof of the compactness theorem which
does not go via the notion of formal deductions.
In mathematical practice there is a distinction between the language we
use to describe mathematical objects and the objects themselves. Mathematical logic makes this distinction formal. Syntax is the word describing
the formal language we use, as distinct from the meaning or semantics which
arises when we interpret the syntax in mathematical structures. In logic, and
particularly model theory, it is important to keep the distinction between
the two, so for example we distinguish between an element of a structure
and a constant symbol which names the element.
However, mathematical logic also considers syntax in such a formal way
that we can reason mathematically about it, and indeed build mathematical
structures out of it. This is precisely what we will do in this chapter, where
we follow Henkin’s method to prove that a set Σ of sentences is satisfiable
by building a model of Σ using the sentences themselves.
Recall the following definitions.
Definition 10.1. A set Σ of L-sentences is
finitely satisfiable if every finite subset of Σ has a model,
deductively closed if for every L-sentence ϕ, if Σ ` ϕ then ϕ ∈ Σ, and
complete if for every L-sentence ϕ, either Σ ` ϕ or Σ ` ¬ϕ.
Note in particular that if Σ is complete and deductively closed then for
each L-sentence ϕ, either ϕ ∈ Σ or ¬ϕ ∈ Σ, and if Σ is also finitely satisfiable
then it cannot contain both ϕ and ¬ϕ.
Lemma 10.2 (Lindenbaum’s Lemma). If Σ is a finitely satisfiable set of Lsentences then there is a finitely satisfiable, deductively closed and complete
set Σ] of L-sentences such that Σ ⊆ Σ] .
43
CHAPTER 10. CONSTRUCTING MODELS FROM SYNTAX
44
The basic idea of the proof is to keep adding more sentences to Σ maintaining finite satisfiability until for every sentence ϕ, either ϕ or ¬ϕ is included. We leave the details as an exercise.
Henkin theories
Recall that an L-term is said to be closed if it does not contain any variables.
So if t is a closed L-term and A is an L-structure then tA is an element of
A. Also recall that a theory is just another name for a set of sentences.
Definition 10.3. A set Σ of L-sentences is called a Henkin theory if it
is finitely satisfiable, deductively closed, complete, and it has the witness
property, that is, for any L-sentence of the form ∃x[ϕ(x)] in Σ, there is a
closed L-term t such that the sentence ϕ(t) is in Σ.
Proposition 10.4. If Σ is a finitely satisfiable set of L-sentences there is
a language L∗ which consists of L together with new constant symbols and
a set Σ∗ of L∗ -sentences such that |L∗ | = |L| and Σ∗ is a Henkin theory.
In order to build our Henkin theory Σ∗ we will expand the language to
introduce new constant symbols which will be the closed terms needed for
the witness property. We will also use Lindenbaum’s lemma to ensure we
get a deductively closed and complete set of sentences. We have to iterate
the process to get all the properties we want, so in fact we will define a tower
of languages
L = L0 ⊆ L1 ⊆ L2 ⊆ · · · ⊆ Ln ⊆ · · ·
and a tower of sets of sentences
Σ ⊆ Σ0 ⊆ Σ1 ⊆ Σ2 ⊆ · · · ⊆ Σn ⊆ · · ·
with each Σn being a finitely satisfiable,
deductivelySclosed and complete
S
set of Ln -sentences. Then L∗ = n∈N Ln and Σ∗ = n∈N Σn will have the
desired properties. With the explanation out of the way, we start the proof.
Proof. Let L0 = L, and using Lindenbaum’s lemma we can take Σ0 to be
a finitely satisfiable, deductively closed and complete set of L0 -sentences
extending Σ.
Assume inductively that we have Ln and Σn which is a finitely satisfiable, deductively closed and complete set of Ln -sentences. Let Ln+1 be Ln
together with a new constant symbol cϕ for each Ln -sentence of the form
∃x[ϕ(x)] which is in Σn . Let Σ0n be Σn together with all sentences of the
form ∃x[ϕ(x)] → ϕ(cϕ ) for the new constant symbols.
We claim that Σ0n is finitely satisfiable. To prove this, let S 0 be a finite
subset of Σ0n and let S = S 0 ∩ Σn . By inductive hypothesis, Σn is finitely
satisfiable, so S has a model A which is an Ln -structure. Expand it to an
Ln+1 structure A0 by interpreting the constant symbol cϕ as an element a
CHAPTER 10. CONSTRUCTING MODELS FROM SYNTAX
45
of A such that A |= ϕ(a) if one exists, and as any element of A otherwise.
Then A0 |= S 0 by construction, so Σ0n is finitely satisfiable.
Applying Lindenbaum’s lemma again, we take Σn+1 to be some finitely
satisfiable, deductively closed and complete set of Ln+1 -sentences extending
Σ0n .
S
S
Now take L∗ = n∈N Ln and Σ∗ = n∈N Σn . Clearly Σ∗ is a set of L∗ sentences. If ϕ is an L∗ -sentence then for some n ∈ N it is an Ln -sentence, so
either ϕ or ¬ϕ is in Σn because Σn , and hence either ϕ or ¬ϕ is in Σ∗ , so Σ∗
is deductively closed and complete. If ∃x[ϕ(x)] ∈ Σ∗ then it is in some Σn ,
and then ∃x[ϕ(x)] → ϕ(cϕ ) ∈ Σn+1 , and then, since Σn+1 is deductively
closed, ϕ(cϕ ) ∈ Σn+1 . So ϕ(cϕ ) ∈ Σ∗ , and thus Σ∗ has the witness property.
Finally, if S is a finite subset of Σ∗ then it is contained in some Σn , and Σn
is finitely satisfiable, so S has a model, and thus Σ∗ is finitely satisfiable. So
Σ∗ is a Henkin theory as required.
We leave the verification that |L∗ | = |L| as an exercise.
Canonical models
Definition 10.5. An L-structure A is said to be canonical if for every
element a ∈ A there is a closed L-term t such that tA = a. A canonical
model of a theory T is a model of T which is a canonical structure.
The structures Zring and hN; s, 0i are canonical structures, but for example Ro-ring is not. If A is an L-structure then we can name every element
of A with a new constant symbol to get a canonical model in the expanded
language LA . In general, a theory will not have a canonical model unless it
has enough closed terms, which may involve adding new constant symbols.
This is precisely what Henkin theories are for.
Proposition 10.6. Every Henkin theory has a canonical model.
Proof. Let T be a Henkin theory in the language L, and let Λ be the set of
closed L-terms. Define a relation ∼ on Λ by t1 ∼ t2 if the sentence t1 = t2
is in T . Then ∼ is an equivalence relation (see the exercises). Let A be the
set of equivalence classes. It will be the domain of our model. We will write
t̃ for the equivalence class of t. Note that A is nonempty because there is
some closed term in L, for example the witness of the existential sentence
∃x[x = x].
Now we have to make A into an L-structure by giving interpretations
of the constant symbols, function symbols and relation symbols of L. If c
is a constant symbol, we define cA to be the equivalence class of c. If f is
a function symbol of arity n and t1 , . . . , tn are closed terms of L we define
f A (t̃1 , . . . , t̃n ) to be the equivalence class of the closed term f (t1 , . . . , tn ). If
s1 , . . . , sn are closed terms with si ∼ ti for each i, since si = ti is in T and T is
deductively closed and complete we have f (s1 , . . . , sn ) = f (t1 , . . . , tn ) ∈ T .
CHAPTER 10. CONSTRUCTING MODELS FROM SYNTAX
46
So f A is well-defined. Finally if R is a relation symbol of L of arity n
and t1 , . . . , tn are closed terms, we define (t̃1 , . . . , t̃n ) ∈ RA if and only if
R(t1 , . . . , tn ) is in T . A similar argument shows that RA is well-defined.
By construction, A is a canonical L-structure. It remains to show that
A |= T . We prove by induction on the construction of L-formulas ϕ(x̄) that
for any (t̃1 , . . . , t̃n ) in An ,
A |= ϕ(t̃1 , . . . , t̃n ) if and only if ϕ(t1 , . . . , tn ) ∈ T.
For atomic formulas it holds by definition. The ∧ and ¬ inductive steps are
straightforward (see the exercises).
Suppose that ϕ(x̄) has the form ∃y[θ(y, x̄)]. Then
A |= ϕ(t̃1 , . . . , t̃n ) iff there is s ∈ Λ such that A |= θ(s̃, t̃1 , . . . , t̃n )
iff there is s ∈ Λ such that θ(s, t1 , . . . , tn ) ∈ T
by the inductive hypothesis.
Now θ(s, t1 , . . . , tn ) ` ∃y[θ(y, t1 , . . . , tn ), so if there is s ∈ Λ such that
θ(s, t1 , . . . , tn ) ∈ T , since T is deductively closed, we have ∃y[θ(y, t1 , . . . , tn )] ∈
T . Conversely, if ∃y[θ(y, t1 , . . . , tn )] ∈ T then, since T is a Henkin theory,
there is a closed term s such that θ(s, t1 , . . . , tn ) ∈ T . That completes the
inductive step.
So A |= T as required.
Putting everything together we get a proof of the compactness theorem,
in a slightly stronger form which gives even a small model, without a separate
appeal to the Downward Löwenheim-Skolem theorem.
Theorem 10.7 (The Compactness Theorem). If Σ is a finitely satisfiable
set of L-sentences there is a model A of Σ with |A| 6 |L|.
Proof. By Proposition 10.4 there is a Henkin theory Σ∗ extending Σ in a
language L∗ expanding L. By Proposition 10.6, Σ∗ has a canonical model,
say A. Since every element of A is named by an L∗ -term we have |A| 6 |L∗ |,
but |L∗ | = |L| so we are done.
Exercises
10.1 Let L be a countable language, and enumerate the set of L-sentences
as (ϕn )n∈N . Let Σ be a finitely satisfiable set of L-sentences, and show
that at least one of Σ ∪ {ϕ0 } or Σ ∪ {¬ϕ0 } is finitely satisfiable. Then
build on this idea to prove Lindenbaum’s lemma in the case that L is
countable.
10.2 [For those who know some set theory] Use some form of the axiom of
choice (for example Zorn’s Lemma or the well-ordering principle) to
prove Lindenbaum’s lemma for a language of arbitrary cardinality.
CHAPTER 10. CONSTRUCTING MODELS FROM SYNTAX
47
10.3 Complete the proof of Proposition 10.4 by showing that |L∗ | = |L|.
[Recall that |L| means the cardinality of the set of L-formulas which
use only variables xn for n ∈ N.]
10.4 Show that the relation ∼ on the set of closed L-terms of a Henkin
theory T given by “t1 ∼ t2 if the sentence t1 = t2 is in T ” is an
equivalence relation.
10.5 Complete the proof of Proposition 10.6 by showing that RA is welldefined and that the ∧ and ¬ steps of the induction on formulas hold.
Chapter 11
Interlude: basic cardinal
arithmetic
The model theory we cover will use only basic set theory, essentially the
basics of cardinal arithmetic. Many books contain a proper explanation of
this set theory. Here, we give an intuitive explanation of what little we
need, omitting some proofs. The set theory we use is the standard one in
mathematics, called ZFC (Zermelo-Fraenkel with Choice).
Definition 11.1. Two sets have the same cardinality (size) iff there is a
bijection between them. Given a set X, we write |X| for the cardinality of
X. A cardinal is just a possible value of |X|.
We will make use of four basic facts about injective functions:
Facts 11.2.
1. If there are injective functions X ,→ Y and Y ,→ X then there is
a bijection between X and Y . (This is called the Schröder-Bernstein
Theorem.)
2. Suppose X 6= ∅. Then there is an injective function X ,→ Y if and
only if there is a surjective function Y X.
3. Given two sets X and Y , either there is an injective function X ,→ Y
or an injective function Y ,→ X.
4. For any set X, there is a set Y (for example, the power set PX) such
that there is an injective function X ,→ Y but no injective function
Y ,→ X.
We define |X| 6 |Y | to mean there is an injective function from X to Y .
Using facts 1 and 3, this makes 6 into a linear order on cardinals. Fact 4
says there is no greatest cardinal.
48
CHAPTER 11. INTERLUDE: BASIC CARDINAL ARITHMETIC
49
A set is finite precisely when its cardinality is some natural number
n ∈ N. Otherwise it is infinite. There is a smallest infinite cardinal, which
is |N|, written ℵ0 . (ℵ, pronounced aleph, is the first letter of the Hebrew
alphabet.) A set is countable precisely when it is either finite or of cardinality
ℵ0 . Otherwise it is uncountable. It is countably infinite if it is countable and
infinite.
From Fact 4, we see that the power set of the natural numbers, PN is
uncountable. It is actually in bijection with the set R of real numbers, so R
is also uncountable.
We can define addition and multiplication of cardinals. We define |X|·|Y |
to be |X × Y | where X × Y is the usual Cartesian products of the sets given
by {(x, y) | x ∈ X, y ∈ Y }. We define |X| + |Y | to be |X × {0} ∪ Y × {1}|.
Doing basic arithmetic with infinite cardinals is actually much easier than
with natural numbers.
Lemma 11.3. If both X and Y are finite, with |X| = n and |Y | = m, then
|X| + |Y | = n + m and |X| · |Y | = nm, the usual sum and product of natural
numbers. If at least one of X and Y is infinite, then
|X| + |Y | = |X| · |Y | = max(|X|, |Y |).
Proposition 11.4. If X is a nonempty
set and S is the set of finite strings
S
of elements of X, that is, S = n∈N X n , then |S| = max(|X|, ℵ0 ).
That completes our brief survey of cardinal arithmetic. Proofs can be
found in any book on set theory.
The cardinality of a language
We now consider an application to our formal languages. We define the
cardinality |L| of a language L to be the cardinality of the set Form(L) of
L-formulas, where we only allow the variables xn for n ∈ N. Write Symb(L)
for the set of relation, function, and constant symbols of L.
Proposition 11.5. |L| is equal to the maximum of ℵ0 and | Symb(L)|
Proof. The set Form(L) contains all the formulas x1 = x1 , (x1 = x1 ∧ x1 =
x1 ), ((x1 = x1 ∧ x1 = x1 ) ∧ x1 = x1 ), and so on. This gives an injective
function from a countably infinite set into Form(L), so | Form(L)| > ℵ0 .
For every symbol in Symb(L), there is a formula containing it. So there
is an injective function from Symb(L) to Form(L), and hence | Form(L)| >
| Symb(L)|, and so | Form(L)| > max(| Symb(L)|, ℵ0 ).
Now let X = Symb(L) ∪ {=, ∧ , ¬, ∃, (, ), [, ]} ∪ {xn | n ∈ N }. Then
|X| = | Symb(L)| + 8 + ℵ0 = max(| Symb(L)|, ℵ0 ).
Now Form(L) is a subset of the set of all finite strings of elements of
X, and hence | Form(L)| 6 max(| Symb(L)|, ℵ0 ) by Proposition 11.4. So
| Form(L)| = max(| Symb(L)|, ℵ0 ) as required.
Part II
Changing models
50
Chapter 12
Elementary substructures
We now consider the problem of finding new models of the theory of a
given structure. In this chapter we will start with a structure B and find
substructures A of it which are elementarily equivalent to it, and in fact
cannot be distinguished from B even by formulas applied to elements of A.
We first recall the notion of a substructure from Chapter 5.
Definition 12.1. Let A and B be L-structures, and suppose the domain of
A is a subset of the domain of B. We write A ⊆ B and say that “A is a
substructure of B”, and “B is an extension of A” iff the inclusion of A into
B is an embedding of L-structures.
For example, if B is a group considered as an Lgp -structure then the
substructures of B are exactly the subgroups of B.
Recall that if A is a substructure of B, ϕ(x̄) is an atomic (or even
quantifier-free) formula, and ā ∈ A then A |= ϕ(ā) if and only if B |= ϕ(ā).
However, a sentence with quantifiers may have different truth values in A
and B.
Definition 12.2. We say that A is an elementary substructure of B, and B
is an elementary extension of A, and write A 4 B, if the domain of A is a
subset of the domain of B and, for each formula ϕ(x̄) and each ā ∈ An ,
A |= ϕ(ā) if and only if B |= ϕ(ā).
π
We say that an embedding of L-structures A −→ B is an elementary
embedding iff π(A) 4 B.
When A 4 B then A and B are very similar, indeed they are indistinguishable from the point of view of the truth of L-formulas applied to
elements of A. The word elementary is used because the structures look the
same in terms of their elements, that is, in terms of first-order logic.
51
CHAPTER 12. ELEMENTARY SUBSTRUCTURES
52
Example 12.3. Consider Zadgp and the substructure E = hE; +, −, 0i consisting of the even numbers. Then E 64 Zadgp , because if ϕ(x) is the formula
∃y[y + y = x] then Zadgp |= ϕ(2) but E 6|= ϕ(2). However, Zadgp ∼
= E because
the map π : Z → E given by π(n) = 2n is an isomorphism.
Separating out formulas into sentences and those with free variables, we
have two easy consequences of being an elementary substructure.
Lemma 12.4. Suppose A 4 B. Then A ≡ B, and for any L-formula
ϕ(x1 , . . . , xn ) we have ϕ(A) = ϕ(B) ∩ An .
Proof. Both conclusions are immediate from the definitions.
The Tarski-Vaught test
We now turn to finding elementary substructures A of an L-structure B.
The key idea is that if B says that there is an element with some property,
then there must be such an element already in A. The surprising thing is
that this is all that is needed.
Lemma 12.5 (Tarski-Vaught test). Suppose A ⊆ B is an L-substructure
and for every L-formula ϕ(x̄, y) and every ā ∈ A and b ∈ B such that
B |= ϕ(ā, b), there is d ∈ A such that B |= ϕ(ā, d). Then A 4 B.
Proof. Suppose A ⊆ B, satisfying the condition in the statement of the
lemma. We prove by induction on formulas ϕ(x̄) that for each ā in A,
A |= ϕ(ā) iff B |= ϕ(ā).
For ϕ an atomic formula, it is true by Lemma 3.6, because the inclusion
map A ,→ B is an embedding.
There are inductive steps for ∧ , ¬, and ∃. The first two are easy and left
as an exercise. So suppose ϕ(x̄) is ∃yψ(x̄, y) and ā ∈ A. If A |= ∃yψ(ā, y)
then there is d ∈ A such that A |= ψ(ā, d), and so by induction B |= ψ(ā, d),
and hence B |= ∃yψ(ā, y). Conversely, if B |= ∃yψ(ā, y) then there is b ∈ B
such that B |= ψ(ā, b). Then, using the condition in the statement of the
lemma, there is d ∈ A such that B |= ψ(ā, d). Then, by induction again,
A |= ψ(ā, d), and hence A |= ∃yψ(ā, y). That completes the ∃ inductive
step, and the whole proof.
The Downward Löwenheim-Skolem Theorem
Now we show that elementary substructures exist.
Theorem 12.6 (The Downward Löwenheim-Skolem Theorem). Let B be a
L-structure and S a subset of B. Then there is an elementary substructure
A 4 B such that S ⊆ A and |A| 6 max(|S|, |L|).
CHAPTER 12. ELEMENTARY SUBSTRUCTURES
53
Proof. For each formula ϕ(x̄, y) choose a function gϕ : B n → B such that
(
some d ∈ B such that B |= ϕ(b̄, d), if such exists
gϕ (b̄) =
any d ∈ B otherwise.
By assumption our structures have non-empty domains, so it is possible to
choose such functions.
Now let S0 = S, and for r ∈ N, define
Sr+1 = Sr ∪ gϕ (b̄) ϕ(x̄, y) is an L-formula and b̄ ∈ Sr
S
and let A = r∈N Sr .
Note that if ϕ(x̄, y) is f (x̄) = y for a function symbol f from L then gϕ
is just the function f B . So A is closed under the functions in the language,
and hence is the domain of a substructure A of B.
Next we show that A 4 B. Suppose ϕ(x̄, y) is an L-formula, ā ∈ A, and
b ∈ B such that B |= ϕ(ā, b). Then also B |= ϕ(ā, gϕ (ā)), and gϕ (ā) ∈ A.
So, by the Tarski-Vaught test, A 4 B.
Finally we must consider the size of A. Using Proposition 11.4 we see
that |Sr+1 | 6 max(|Sr |, |L|), and so by induction |Sr | 6 max(|S|, |L|). Now
[ G |A| 6 Sr 6 Sr 6 ℵ0 · max(|S|, |L|) = max(|S|, |L|)
r∈N
as required, where
r∈N
F
means the disjoint union.
Example 12.7. Let V |= ZFC, the axioms of set theory. The language
is just {∈}, so there is a countable elementary substructure V0 4 V . In
particular V0 |= ZFC. Now in ZFC we can prove that there is an uncountable
set. Thus there is an apparent paradox (called the Skolem paradox) of a
countable structure V0 containing an uncountable set. It is a good exercise
for those who know some set theory to see why this is not paradoxical at
all.
Example 12.8. There is a countable elementary substructure R of the real
ordered field Ro-ring . Since R is uncountable, R is not isomorphic to Ro-ring .
In particular, it cannot be a complete ordered field, so it must have subsets
S ⊆ R which are non-empty and bounded above but which do not have a
least upper bound in R. However, since S ⊆ R ⊆ R, these sets S do have a
least upper bound in R.
Skolem functions
The functions gϕ which we defined in the proof of the Downward LöwenheimSkolem Theorem are called Skolem functions. For the proof, it does not
CHAPTER 12. ELEMENTARY SUBSTRUCTURES
54
matter how we choose the values of these functions. We may have to invoke
some set theory in the form of the axiom of choice, but the important thing
is just that they exist. In some cases there is no way to choose these Skolem
functions to be definable. There are also structures which do have definable
Skolem functions for every formula.
Example 12.9. Consider the complex field Cring and the formula ϕ(x, y)
given by x = y · y. Then a Skolem function gϕ : C → C is a square-root
function, so for all x ∈ C we have gϕ (x)2 = x. In particular, we must have
gϕ (−1) picking out one of ±i, say +i. But complex conjugation z 7→ z is an
automorphism of Cring , and if gϕ were definable it would be preserved under
all automorphisms, so we would have both gϕ (−1) = +i and gϕ (−1) = +i,
that is, gϕ (−1) = −i, a contradiction. So gϕ is not definable.
Example 12.10. Ns-ring has definable Skolem functions for every formula.
To prove it, we use the fact that every non-empty subset of N has a smallest
element. So for a formula ϕ(x̄, y) we can define a Skolem function by the
formula µϕ (x̄, z) given by
(ϕ(x̄, z) ∧ ∀y[ϕ(x̄, y) → z 6 y]) ∨ (¬∃y[ϕ(x̄, y)] ∧ z = 0)
where we have 6 as an abbreviation, which is acceptable because 6 is definable in Ns-ring .
Exercises
12.1 Prove Lemma 12.4.
12.2 Suppose A 4 B and ϕ(x) is an L-formula. Show that if ϕ(A) is finite
then |ϕ(A)| = |ϕ(B)|, and if ϕ(A) is infinite then so is ϕ(B).
12.3 Why does Skolem’s paradox seem paradoxical, and why is it not in
fact contradictory?
12.4 In the real ordered field Ro-ring , write down Skolem functions for the
formulas x1 < y ∧ y < x2 , x1 < y, and y < x2 .
12.5 Suppose S ⊆ Rn is a finite subset which is definable in Ro-ring . Show
that S has a definable Skolem function. Then show that for every
s ∈ S, the singleton {s} is definable. [Hint: use induction on |S|.]
12.6 Suppose that B is a structure with Skolem functions given by function symbols in the language. Show that any substructure of B is an
elementary substructure.
12.7 Suppose that A ⊆ B ⊆ D are L-structures such that A 4 D and
B 4 D. Prove that A 4 B.
CHAPTER 12. ELEMENTARY SUBSTRUCTURES
55
12.8 Suppose that B is an L-structure and there is a elementary chain of
elementary substructures of B, that is a chain
A1 4 A2 4 · · · 4 An 4 · · ·
S
such that each An 4 B. Let A = n∈N An . Prove that A 4 B.
12.9 Sketch a proof of the downward Löwenheim-Skolem theorem. You
should give all the key ideas and explain how they fit together, but
you do not need to give all the details.
Chapter 13
Elementary extensions
In this chapter we consider the opposite problem to the previous chapter:
given a structure A, how can we find elementary extensions of it? In chapter 8 we used the compactness theorem in conjunction with new constant
symbols to find new models of Th(A). Here we combine those ideas with
the method of diagrams to find new models which are actually elementary
extensions.
The method of diagrams
Recall that if A and B are L-structures with domains A and B respectively,
π
then an elementary embedding A −→ B is an embedding such that for each
L-formula ϕ(x̄) and each ā ∈ An ,
A |= ϕ(ā) if and only if B |= ϕ(π(ā))
and A is an elementary substructure of B and B is an elementary extension
of A if A ⊆ B and the inclusion map is an elementary embedding.
The method of diagrams turns these formulas applied to tuples from A
into sentences by adding new constant symbols to the language.
Given an L-structure A with domain A, we create a new language LA
by adding a new constant symbol ca for every element a ∈ A. These new
constant symbols must be distinct, and different from any symbol in L. We
then expand A to an LA -structure A+ by interpreting the symbol ca as the
element a.
Definition 13.1. The complete diagram of A is CDiag(A) = ThLA (A+ ),
the set of all LA -sentences which are true in A+ .
Proposition 13.2. Let B be an L-structure. Then there is an elementary
π
embedding A −→ B if and only if B can be expanded to a model of CDiag(A).
π
Proof. Suppose there is an elementary embedding A −→ B. Expand B to B +
+
by defining cB
a = π(a). Any σ ∈ CDiag(A) is of the form ϕ(ca1 , . . . , can )
56
CHAPTER 13. ELEMENTARY EXTENSIONS
57
for some a1 , . . . , an ∈ A such that A |= ϕ(a1 , . . . , an ). Since π is an elementary embedding B |= ϕ(π(a1 ), . . . , π(an )), and so B + |= σ. Thus
B + |= CDiag(A).
Conversely, if B can be expanded to a model B + of CDiag(A), define
+
π
A −→ B by π(a) = cB
a . A similar argument shows that π is an elementary
embedding.
Remark 13.3. Given a structure A, the difference between an elementary
π
extension B of A and an elementary embedding A −→ B is not usually very
π
important. Given A −→ B with B = dom(B), let D be a set disjoint from
0
A such that there is a bijection f : B r
( π(A) → D. Let B = A ∪ D and
π −1 (b) if b ∈ π(A)
define a bijection g : B → B 0 by g(b) =
We can make
f (b) otherwise.
B 0 into an L-structure B 0 in exactly one way such that g is an isomorphism.
Then A 4 B 0 .
The Upward Löwenheim-Skolem Theorem
Now we apply the compactness theorem with two lots of new constants:
those for the method of diagrams and those needed to ensure a model has
large cardinality.
Theorem 13.4 (The Upward Löwenheim-Skolem Theorem). For any infinite L-structure A and any cardinal κ > max(|L|, |A|), there is an Lstructure B of cardinality equal to κ such that A 4 B.
Proof. Let I be a set of cardinality κ. Let LA,I be the language obtained
from L by adding distinct constant symbols ca for each element of A, and
ci for each element of I.
Let Σ = CDiag(A) ∪ {ci 6= cj | i, j ∈ I, i 6= j }, a set of LA,I -sentences.
We claim that Σ is finitely satisfiable. So let Σ0 be a finite subset of Σ.
Then only finitely many of the ci , say ci1 , . . . , cin , are used in Σ0 . Now
choose a1 , . . . , an from A, all distinct. Expand A to an LA,I -structure A0 by
interpreting ca as a for each a ∈ A, and interpreting cij as aj for j = 1, . . . , n,
and every other ci as a1 . Then A0 |= Σ0 . Thus Σ is finitely satisfiable, and
hence, by compactness, it has a model, say D0 , with domain D. Let D be
0
the reduct of D0 to L. Since D0 |= Σ, the elements cD
i of D are all distinct,
and so |D| > κ.
Since D0 |= CDiag(A), by Proposition 13.2 there is an elementary emπ
bedding A −→ D. Using Remark 13.3, we can assume A 4 D.
We are nearly done, but the elementary extension D we have found has
cardinality at least κ, and we want an elementary extension of cardinality
exactly κ. Let S be a subset of D containing A and of cardinality κ. Then
by Theorem 12.6 there is B 4 D of cardinality κ, containing S and hence
also containing A.
CHAPTER 13. ELEMENTARY EXTENSIONS
58
We must show A 4 B. Let ϕ(x̄) be a formula and ā ∈ A. Then
A |= ϕ(ā) iff D |= ϕ(ā) iff B |= ϕ(ā)
because A 4 D and B 4 D. So A 4 B.
Non-standard natural numbers
A proper elementary extension M of Ns-ring is called a non-standard model
of Th(N) look like? An element of M r N is called a non-standard natural
number, while in this context elements of N are called standard natural
numbers. We give a few examples of properties these non-standard natural
numbers can have.
Lemma 13.5. Suppose Ns-ring 4 M. Then any element a ∈ M r N is
greater than every standard natural number.
Proof. Suppose not, say a 6 n. We use n as an abbreviation for the term
(1 + · · · + 1) and we use 6 W
as an abbreviation for the formula defining
= i] so since M ≡ N we have M |=
it. Now N |=W ∀x [x 6 n → ni=0 x W
∀x [x 6 n → ni=0 x = i], so M |= ni=0 a = i, and then a is a standard
natural number, a contradiction.
Proposition 13.6. Let M be any proper elementary extension of Ns-ring .
Then there is a non-standard prime number in M.
Proof. Note that the formula ψ(y) given by
∀x[∃z[x · z = y] → (x = 1 ∨ x = y)] ∧ y 6= 1
defines the set of prime numbers in Ns-ring . We have
N |= ∀x∃y[ψ(y) ∧ x < y]
so also M |= ∀x∃y[ψ(y) ∧ x < y]. Thus given a ∈ M r N there is p ∈ M
prime such that a < p. Then p is greater than every standard natural
number, so it is non-standard.
Proposition 13.7. There is an elementary extension M of N with a number
β ∈ M with infinitely many distinct prime factors.
Proof. We give a compactness argument. Let Lc = Ls-ring ∪ {c} where c
is a new constant symbol. Let Σ = Th(Ns-ring ) ∪ {∃x[n · x = c] | n ∈ N }
which states that c is divisible by every (standard) natural number. Let
Σ0 be a finite subset of Σ and let N be the largest natural number n such
that ∃x[n · x = c] appears in Σ0 . We make a model A of Σ0 by taking
Ns-ring and interpreting c as N !. Then A |= Σ0 , so Σ0 is satisfiable. By
compactness, there is a model M+ of Σ. Let M be the reduct to the
CHAPTER 13. ELEMENTARY EXTENSIONS
59
+
language h+, ·, 0, 1i. By Proposition 13.2, N 4 M, and the element β = cM
of M is divisible by every standard natural number, in particular every
standard prime number.
We finish our brief discussion of non-standard natural numbers with an
observation about the twin prime conjecture, a long-standing but currently
unresolved problem in number theory.
Conjecture 13.8 (Twin prime conjecture). There are infinitely many primes
P in N such that P + 2 is also prime. (Such a pair (P, P + 2) is called a
pair of twin primes.)
Proposition 13.9. The twin prime conjecture is true if and only if there
is some non-standard model of arithmetic M and at least one pair of nonstandard twin primes in M.
Sketch proof. Let ϕ(x) be the formula ψ(x) ∧ ψ(x + 2), where ψ defines
the set of primes as above. If the twin prime conjecture is true, then the
subset of N defined by ϕ(x) is infinite, and a compactness argument shows
there is some M with a non-standard realisation of ϕ(x). If the twin prime
conjecture is false, then we can list all the realisations of ϕ(x) as n1 = 3, n2 =
5, n3 = 11, . . . , nr , for some finite r, and we have
"
#
r
_
x = ni .
N |= ∀x ϕ(x) →
i=1
But then whenever M |= Th(N) we have M |= ∀x[ϕ(x) →
hence there are no non-standard twin primes.
Wr
i=1 x
= ni ] and
The twin prime conjecture is a long-standing open problem, and we
might hope that this reformulation of it would help to solve it. However,
unfortunately, it does not, because there is no easy way to construct explicit
non-standard models of Th(Ns-ring ).
Non-standard real numbers
Now we turn to elementary extensions of the real field Ro-ring .
Proposition 13.10. There is a non-Archimedean elementary extension of
Ro-ring .
Sketch proof. Take a new constant symbol c and consider the set of sentences
Σ = CDiag(Ro-ring ) ∪ {n < c | n ∈ N }. A similar compactness argument to
those we have considered already shows that Σ is satisfiable, so let R+ be a
model, and let R be the reduct to the language Lo-ring . Then, as in the proof
+
of the Upward Löwenheim-Skolem theorem, R 4 R, and cR witnesses that
R is non-Archimedean.
CHAPTER 13. ELEMENTARY EXTENSIONS
60
In particular, R is not a complete ordered field, yet every true first-order
statement about R is also true in R, including many theorems of analysis
like the intermediate value theorem, but restricted to definable functions.
In fact, one can prove that every proper elementary extension of R is
non-Archimedean. See Exercise 13.6(a).
Definition 13.11. If R is a non-Archimedean ordered field, we say an
element a of R is infinite if |a| is greater than every standard natural number,
finite if it is not infinite, and infinitesimal if |a| =
6 0 but for every n ∈ N we
have |a| < n1 .
Observe that a is infinite if and only if a1 is infinitesimal. Let R be a
proper elementary extension of R, and let r ∈ R. There is a cloud of numbers
in R which are infinitesimally close to r, namely all the numbers ρ ∈ R
such that |ρ − r| is infinitesimal or 0. We can use these non-standard real
numbers to do non-standard analysis, for example to define the derivative
of a function in the intuitive way it is sometimes done in a calculus course
rather than the usual way of doing it in analysis. See Exercise 13.6(c).
Exercises
13.1 The diagram Diag(A) of an L-structure A is the set of all atomic LA sentences and negations of atomic LA -sentences which are true in A+ .
Show that there is an embedding of A into an L-structure B if and
only if B can be expanded to a model of Diag(A).
13.2 Show that there is a model A of Th(Z< ) such that Q< embeds in A.
Can the embedding be elementary?
13.3 Let T be a complete theory in a language L of cardinality λ, and
suppose that T has an infinite model. Show that T has models of all
cardinalities greater than or equal to λ, and no finite models.
13.4 Complete the proof of Proposition 13.9 by writing down the compactness argument.
13.5 Show that there is a countable model R of Th(Ro-ring ) which is nonArchimedean. Show that neither of R nor Ro-ring can be embedded in
the other.
13.6 Let R be the field of real numbers, and let R be any proper elementary
extension.
(a) Prove that there is an infinitesimal in R.
(b) Let I be the set consisting of zero and all infinitesimals in R.
Prove that I is closed under addition and multiplication, and
that if α ∈ I and r ∈ R then α · r ∈ I.
CHAPTER 13. ELEMENTARY EXTENSIONS
61
(c) Let p(X) ∈ R[X] be a polynomial with standard real coefficients.
Show there is another polynomial q(X) ∈ R[X] such that whenever x is a standard real and α is infinitesimal, there is β ∈ I
such that
p(x + α) − p(x)
= q(x) + β
α
(d) Explain briefly how we can use the above result to define the
derivative of a polynomial function. How does this method differ
from the usual approach in analysis?
13.7 Sketch a proof of the upward Löwenheim-Skolem theorem. You should
give all the key ideas and explain how they fit together, but you do
not need to give all the details.
Chapter 14
Vector spaces and
Categoricity
From the Löwenheim-Skolem theorems, we can see that if a first-order theory T has an infinite model then it has at least one model of each infinite
cardinality at least as large as |L|. We now consider theories where there
is only one model in some infinite cardinality. Such theories are very interesting, not just because the theory determines the models as much as is
possible, but we will see later that this property has strong consequences for
the definable sets. Our prototypical example is that of vector spaces.
Vector spaces as structures
Let K be a field. The language LK-VS is Ladgp = h+, −, 0i together with one
unary function symbol λ· for each λ ∈ K. (Note that we are using the two
characters λ and · as a single symbol of our formal language.) We interpret
λ· as scalar multiplication by λ.
Definition 14.1. The theory of K-vector spaces, TK-VS , is the LK-VS -theory
given by the axioms for abelian groups, written additively in the language
Ladgp , together with the following axioms which describe how scalar multiplication works.
VS1. ∀x[0 · x = 0 ∧ 1 · x = x]
VS2λ . ∀xy[λ · (x + y) = (λ · x) + (λ · y)]
VS3λ,λ0 . ∀x[(λ · x) + (λ0 · x) = µ · x]
where µ = λ + λ0 in K
VS4λ,λ0 . ∀x[λ · (λ0 · x) = ν · x]
where ν = λλ0 in K
A K-vector space is a model of TK-VS .
62
CHAPTER 14. VECTOR SPACES AND CATEGORICITY
63
Note that axioms 2, 3, and 4 are not single LK-VS -sentences, but families
of sentences. Axiom 2 is really a family of axioms indexed by elements of K,
and axioms 3 and 4 are families of axioms indexed by pairs of elements of
K. These are called axiom schemes. We cannot write these as single axioms
in LK-VS because that would involve a universal quantifier which quantified
over elements of K. However, that cannot exist in our language, which only
allows us to quantify over elements of our structure, which is the vector
space, not the field. (It is also possible to consider a language where some
of the elements of a model are elements of the field and some are elements
of the vector space, but that turns out to be like studying the field rather
than studying the vector space.)
For any field K, there is a 0-dimensional K-vector space with only one
element, 0. If K is a finite field, there are other finite K-vector spaces.
Since we are mostly interested in infinite structures it is useful to consider
the theory of infinite K-vector spaces as well.
∞
Definition 14.2. The theory of infinite K-vector spaces, TK
-VS is the theory axiomatized by the axioms for K-vector spaces together with sentences
saying that there are at least n distinct elements, for each n ∈ N+ .
Some linear algebra of vector spaces
Fix a field K. You will, I hope, be familiar with concepts such as linear
independence and basis for finite-dimensional vector spaces. However, you
may not have seen these concepts for infinite-dimensional vector spaces, so
we will review them.
Definition 14.3. Let V be a K-vector space.
• A subset S of V is linearly independent iff whenever
s1 , . . . , s n ∈ S
Pn
are distinct and λ1 , . . . , λn ∈ K are such that i=1 λi si = 0 then each
λi = 0.
• A subset S of V is a spanning set for V iff every element of V can be
written as a (finite) linear combination of elements of S. That is, for
every v ∈P
V there are n ∈ N, s1 , . . . , sn ∈ S and λ1 , . . . , λn ∈ K such
that v = ni=1 λi si .
• A basis of V is a linearly independent spanning set of V .
Note that there is no notion of an infinite linear combination here. That
would require a notion of convergence of infinite series which needs some
topological or metric information we do not have in our algebraic setting.
There are three important facts which we will use.
CHAPTER 14. VECTOR SPACES AND CATEGORICITY
64
Facts 14.4. (a) Let B be a basis for V , and v ∈ V . Then there are a
unique n ∈ N and uniqueP
distinct b1 , . . . , bn ∈ B and unique λ1 , . . . , λn ∈
K r {0} such that v = ni=1 λi bi . That is, every element in V can be
written uniquely as a linear combination of elements of B.
(b) Every linearly independent set can be extended to a basis. In particular,
every vector space has a basis.
(c) Any two bases of V have the same cardinality.
Using (b) and (c), we can make the following definition.
Definition 14.5. The dimension dim V of a vector space V is the cardinality of a basis.
We next show that for any field K there are K-vector spaces of every
possible cardinal dimension.
Example 14.6. Let X be any set, and let K ⊕X be the set of functions
f : X → K of finite support, that is, such that there are only finitely many
x ∈ X with f (x) 6= 0. We make K ⊕X into a K-vector space by pointwise
operations:
(f + g)(x) := f (x) + g(x)
and
(λ · f )(x) := λf (x).
For each x ∈ X, let bx be the function X → K given by
(
1 if y = x
bx (y) =
0 otherwise.
Then bx has finite support, so bx ∈ K ⊕X . We claim that B := {bx | x ∈ X }
⊕X
is a basis
Pn for K . First we show that B is linearly independent. Suppose
f := i=1 λi bP
xi = 0, with the xi distinct. Then for each j = 1, . . . , n we
have f (xj ) = ni=1 λi bxi (xj ) = λj . But f is the zero function, so λj = 0. So
B is linearly independent. Now let f ∈ K ⊕X , and suppose that x1 , . . . , xn
is the support of f , that is, the set of elements
P of X on which f takes a
non-zero value. Let λi = f (xi ). Then f = ni=1 λi bxi , so B is a spanning
set.
There is an obvious bijection X → B given by x 7→ bx , so |B| = |X|.
Thus dim K ⊕X = |X|.
Theorem 14.7. Suppose that V1 and V2 are K-vector spaces. Then V1 ∼
= V2
if and only if dim V1 = dim V2 .
Sketch Proof. If π : V1 → V2 is an isomorphism and B is a basis for V1 then
it is an easy exercise to show that π(B) = {π(b) | b ∈ B } is a basis for V2 .
Then since π is a bijection, it restricts to a bijection between B and π(B),
CHAPTER 14. VECTOR SPACES AND CATEGORICITY
65
so dim V1 = dim V2 . Conversely, suppose that B1 is a basis for V1 and B2 is
a basis for V2 , and α : B1 → B2 is a bijection. Define π : V1 → V2 by
!
n
n
X
X
π
λi bi =
λi α(bi ).
i=1
i=1
Then π is a well-defined function on all of V1 because B1 is a basis (using
Fact 14.4(a)). It is injective because B2 is linearly independent, and it is
surjective because B2 is a spanning set for V2 . It is also a linear map,
that is, it preserves addition and scalar multiplication, which means it is an
LK-VS -isomorphism. So V1 ∼
= V2 .
The cardinality of a vector space
Proposition 14.8. If V is an infinite K-vector space, then
|V | = max(|K|, dim V ).
Proof. Let B be a basis for V . The inclusion B ,→ V is an injective map, so
|V | > |B| = dim V . Since V is infinite it is not the zero-dimensional vector
space, so B 6= ∅ and we can choose b ∈ B. Then there is an injective function
K ,→ V given by λ 7→ λ · b, so |V | > |K|. So |V | > max(|K|, dim V ).
Let X = K × B, and consider the set S of all finite strings of X. We
can define a function f : S → V by
f ((λ1 , b1 ), (λ2 , b2 ), . . . , (λn , bn )) = λ1 b1 + λ2 b2 + · · · + λn bn .
This function is surjective because B is a basis so, using Fact 11.2.2, we have
|V | 6 |S|. By Proposition 11.4, we have |S| = max(|X|, ℵ0 ), and we also
have |X| = |K| · |B|. At least one of K and B is infinite (since otherwise V
would be finite) and so
|S| = max(|K|, |B|, ℵ0 ) = max(|K|, |B|).
Thus |V | = max(|K|, |B|) as required.
Corollary 14.9. Up to isomorphism, there is exactly one model of the
theory of K-vector spaces of each infinite cardinality strictly greater than
|K|.
Categoricity
We have seen that the theory of K-vector spaces has exactly one model (up
to isomorphism) of each infinite cardinality κ such that κ > |K|. There is a
name for such theories.
CHAPTER 14. VECTOR SPACES AND CATEGORICITY
66
Definition 14.10. A theory T is categorical in cardinality κ or κ-categorical
iff there is a model of T of cardinality κ, and any two models of cardinality
κ are isomorphic. A structure A is said to be κ-categorical iff Th(A) is.
Recall that an L-theory T is complete iff for every L-sentence ϕ, either
T ` ϕ or T ` ¬ϕ. One useful consequence of categoricity is completeness of
a theory.
Lemma 14.11 (Los-Vaught test for completeness). If an L-theory T is κcategorical for some κ > |L| and T has no finite models then T is complete.
Proof. Let ϕ be an L-sentence, and let M be the model of T of cardinality
κ. Then either M |= ϕ or M |= ¬ϕ. Suppose that M |= ϕ. Since every
model of T of cardinality κ is isomorphic to M, using Proposition 3.7 we
see that there is no model of cardinality κ of T ∪ {¬ϕ}. But then, by the
Löwenheim-Skolem theorems, there is no infinite model of T ∪ {¬ϕ}. We
assume that T has no finite models, and hence there is no model of T ∪{¬ϕ}.
But then every model of T is a model of ϕ, that is, T ` ϕ. Otherwise we
have M |= ¬ϕ, and then the same argument shows that T ` ¬ϕ. So T is
complete.
∞
Corollary 14.12. For any field K, the theory TK
-VS is complete.
Corollary 14.13. If K is an infinite field, for example R, then all K-vector
spaces of non-zero dimension are elementarily equivalent.
Proof. They are models of the same complete theory.
It may seem very unexpected that R1 , R2 and R3 are all elementarily
equivalent as vector spaces. We think of geometry in 1, 2, and 3 dimensions
as being very different. The conclusion to draw is that very little of that
geometry is captured by the notion of an R-vector space. Essentially it is
because our language does not allow us to quantify over the scalars, R.
Example 14.14. There is no LR-VS -formula ϕ(x, y) which expresses that
two vectors are linearly independent.
Proof. If such a ϕ(x, y) did exist then we would have R2 |= ∃xy[ϕ(x, y)] but
R1 |= ¬∃xy[ϕ(x, y)], which is impossible since R1 ≡ R2 .
Exercises
14.1 If K is a finite field and V is a K-vector space of dimension d, show
that |V | = |K|d .
∞
14.2 If K is an infinite field, show that TK
-VS is axiomatized by the axioms
for K-vector spaces together with the single axiom ∃x[x 6= 0].
CHAPTER 14. VECTOR SPACES AND CATEGORICITY
67
14.3 If V is a 1-dimensional K-vector space, show that Aut(V ) is isomorphic to the multiplicative group K × of non-zero elements of K.
14.4 If V is an n-dimensional K-vector space, show that Aut(V ) is isomorphic to GLn (K), the group of invertible n × n matrices with entries in
K.
14.5 For any field K and any non-zero vector space V , show that there are
exactly four definable subsets of V .
14.6 Show that if a1 , . . . , an and b1 , . . . , bn are each linearly independent
n-tuples from V then there is an automorphism π of V such that
π(ai ) = bi for each i.
14.7 Write F3 for the field with 3 elements, and let V be an infinitedimensional F3 -vector space. What are all the definable subsets of
V and of V 2 ?
14.8 In the language L= with no relation, function, or constant symbols, let
T be the empty theory, that is, the empty set of L= -sentences. What
is a model of T ? Show that T is κ-categorical for every cardinal κ.
14.9 Sketch a proof of Corollary 14.13, including the results used to prove
it. You should give all the key ideas and explain how they fit together,
but do not give all the details.
Chapter 15
The successor structure
This chapter is devoted to studying the structure hN; s, 0i, the natural numbers with a constant symbol naming 0 and the successor function s, a unary
function defined by s(n) = n + 1. We will axiomatize its theory, classify all
the models of its theory, and investigate what the definable sets are. The
style will be that of a guided extended exercise.
Axioms
We first observe some basic properties of the successor function.
S1. s is injective.
S2. 0 is not in the image of s
S3. Every element of N other than 0 is in the image of s.
Exercise 15.1 Write down first-order sentences to capture these properties.
Slightly less obvious is that there are no “cycles”, captured by the following axiom scheme.
S4n . ∀x[sn (x) 6= x]
for n ∈ N+
where s1 (x) = s(x) and sn+1 (x) = s(sn (x)) for each n ∈ N+ .
Let us write TS for the set of axioms {S1, S2, S3} ∪ {S4n | n ∈ N+ }.
Exercise 15.2 Which of the axioms of TS are satisfied by hZ; s, 0i, where
again s is the successor function?
Exercise 15.3 What modification of S3 is satisfied by hZ; si?
68
CHAPTER 15. THE SUCCESSOR STRUCTURE
69
Models
By design, hN; s, 0i is a model of TS . By the upward Löwenheim-Skolem
theorem there are models of TS of every infinite cardinality. Now we will
investigate what these must look like.
Exercise 15.4 Show that every model of TS has a substructure which is isomorphic to N. We will identify this substructure with N for notational
convenience.
Now suppose that M |= TS and M 6= N. Then there is an element
a ∈ M r N.
Exercise 15.5 Show that a lies in a substructure of hM ; si isomorphic to
hZ; si.
We have taken a reduct to forget the constant 0, otherwise it would not
be a substructure. So we know that a model M of TS consists of one copy
of N and every element of M which is not in N is in a copy of Z. So in fact
a model consists of N and a set of copies of Z. We can be more precise.
Exercise 15.6 Show that any model M of TS is isomorphic to a model MI
with domain N ∪ (I × Z) and such that each (i, n) ∈ I × Z satisfies
sMI ((i, n)) = (i, n + 1).
It is clear that for any set I, the structure MI is a model of TS , so we
have classified all the models of TS .
Exercise 15.7 Show that MI1 ∼
= MI2 if and only if I1 and I2 have the same
cardinality.
Exercise 15.8 Show that the cardinality of MI is |I| + ℵ0 .
It then follows that TS is categorical in all uncountable cardinalities. It
has no finite models, so, by the Los-Vaught test, TS is a complete theory.
So our simple list of axioms for hN; s, 0i is actually complete.
Exercise 15.9 How many countable models of TS are there, up to isomorphism?
Definable sets
Now we want to understand some of the definable sets in the original structure hN; s, 0i. Every element of N is named by a closed term, so it follows
that every finite subset of hN; s, 0i is definable. Thus every cofinite subset
(that is, the complement of a finite set) is also definable.
CHAPTER 15. THE SUCCESSOR STRUCTURE
70
The technique we know to show that a subset is not definable is to
show that it is not preserved under some automorphism. However, hN; s, 0i
itself has no automorphisms (except the identity), so we cannot use that
technique directly. However, we can use it together with our understanding
of the elementary extensions to deduce that certain subsets of N itself are
not definable.
Exercise 15.10 Suppose I 6= ∅, and let a, b ∈ MI r N. Show there is an
automorphism π of MI such that π(a) = b.
Exercise 15.11 Deduce that if S is a definable subset of MI then either
MI r N ⊆ S or (MI r N) ∩ S = ∅.
Exercise 15.12 Let A be any infinite structure and S ⊆ An an infinite definable set, say defined by ϕ(x̄). Show there is B, elementarily equivalent to A, such that the subset of B n defined by ϕ(x̄) is uncountable.
Exercise 15.13 Deduce that the subsets of N which are definable in hN; s, 0i
are exactly the finite and cofinite subsets.
We have only considered definable subsets of N, not Nn for n > 1. Similar
arguments can be used to understand what these sets look like. They are
not just finite or cofinite, for example {(3, a) | a ∈ N } ⊆ N2 is defined by
the formula ϕ(x1 , x2 ) given by x1 = s(s(s(0))). We restrict ourselves to one
observation.
Exercise 15.14 Show that the order on N is not a definable subset of N2 .
Chapter 16
Linear Orders
In this chapter we will look at some theories of linearly ordered sets, aiming
to find complete axiomatizations using the method of categoricity. For dense
linear orders such as R< and Q< we will achieve this using the important
back-and-forth method. While categoricity fails for discrete orders like N<
and Z< , it turns out that the back-and-forth method can still be used to
prove completeness of a simple axiomatization.
Definition 16.1. A linear order on a set is a binary relation < satisfying
Transitivity ∀xyz[(x < y ∧ y < z) → x < z]
Irreflexivity ∀x[¬x < x]
Linearity ∀xy[x < y ∨ x = y ∨ y < x].
A set equipped with a linear order on it is a linearly ordered set.
We will use the abbreviations x 6 y, y > x, and y > x as usual.
Examples 16.2. The ordered set of real numbers is written R< . Likewise
we have linearly ordered sets N< , Z< , Q< , and I< where I is the closed unit
interval I = [0, 1] ⊆ R.
Now we consider ways to distinguish these examples.
Endpoints
Both N< and I< ] have a least element, so they are models of the sentence
∃x∀y[x 6 y]
while R< , Q< and Z< do not have a least element. Similarly I< has a
greatest element so is a model of
∃x∀y[y 6 x]
71
CHAPTER 16. LINEAR ORDERS
72
while the other four examples are not. Least and greatest elements of a
linearly ordered set are called endpoints, so a linearly ordered set is without
endpoints if it satisfies the sentence
∀x∃yz[y < x ∧ x < z].
Discretely and densely ordered sets
Another property of Z< and N< is that they are discretely ordered, which
means that each element (except the greatest element if it exists) has a
next element greater than it, and also (except for the least element) a next
element below it. This is captured by the sentences
∀x[∃w[w < x] → ∃y[y < x ∧ ∀z[z < x → z 6 y]]]
and
∀x[∃w[w > x] → ∃y[y > x ∧ ∀z[z > x → z > y]]].
Note that any finite linearly ordered set is discrete and has endpoints, and
indeed it must be isomorphic to h{1, 2, 3, . . . , n}; <i for some n ∈ N. On the
other hand R< , Q< , and I< are dense linear orders, captured by the axiom
∀xy[x < y → ∃z[x < z ∧ z < y]].
Clearly a linearly ordered set cannot be both dense and discrete, although
it is possible to be neither.
We have found sentences to distinguish four of our five examples, and we
will show that it is impossible to distinguish the last two, R< and Q< , by a
first-order sentence. Let DLO be the theory of dense linear orders without
endpoints, as given by the five axioms given above. We have Q< |= DLO
and R< |= DLO.
Completeness of DLO
Theorem 16.3. Any two countable models of DLO are isomorphic to each
other.
Proof. Suppose A, B |= DLO and both are countable. List A as (an )n∈N and
B as (bn )n∈N . We will construct an isomorphism piece-by-piece. In fact, for
each n ∈ N we will inductively construct finite subsets An ⊆ A and Bn ⊆ B
and a bijection πn : An → Bn such that:
• for each a, a0 ∈ An , a < a0 iff πn (a) < πn (a0 );
• An ⊆ An+1 , Bn ⊆ Bn+1 , and πn ⊆ πn+1 ; and
• an ∈ A2n+1 and bn ∈ B2n+2 .
CHAPTER 16. LINEAR ORDERS
73
In particular, πn is an isomorphism between hAn ; <i and hBn ; <i.
We start with A0 = B0 = ∅, and π0 the empty function. Now suppose n
is even, say n = 2m, and we have constructed An , Bn , and πn . If am ∈ An
then set An+1 = An , Bn+1 = Bn , and πn+1 = πn . By induction the above
conditions are satisfied. Otherwise, am ∈
/ An and we define An+1 = An ∪
{am }. We must find a suitable element in B to be πn+1 (am ). There are
three cases. Either (i) am is less then every element of An , or (ii) am is
greater than every element of An , or (iii) there are c, d ∈ An with c < am
and am < d, and no elements of An strictly between c and d. In each case,
because B |= DLO there is b ∈ B such that b is (i) less than every element
of Bn , or (ii) greater than every element of Bn , or (iii) between πn (c) and
πn (d). To be precise, we choose the br for the smallest possible value of
r ∈ N with the desired property. Then we define Bn+1 = Bn ∪ {br } and
πn+1 = πn ∪ {(am , br )}.
Now suppose that n is odd, say n = 2m + 1. If bm ∈ Bn then set
An+1 = An , Bn+1 = Bn , and πn+1 = πn . Otherwise, let Bn+1 = Bn ∪ {bm }.
Then we use the same
process as above to find ar ∈ A.
S
Now let π = n∈N πn . Our construction ensures that π is defined on
all of A and its image is all of B. It also preserves < (in particular it is
injective), and so it is an isomorphism from A to B.
This method of proof is called back-and-forth, and it has become very
important in model theory.
Corollary 16.4. The theory DLO is complete.
Proof. It is easy to see that DLO has no finite models. DLO is ℵ0 -categorical
so, by the Los-Vaught test, it is complete.
Since every countable model of DLO is isomorphic to Q< , it is natural
to consider uncountable models. For example, is every model of DLO of the
same cardinality as R isomorphic to R< ? Consider a linear order M which
consists of a copy of Q and a copy of R, with every element of Q less than
every element of R. Then |M | = |Q| + |R| = |R|.
It is easy to check that M |= DLO, but we can show that M ∼
6 R.
=
Indeed, suppose π : M → R were an isomorphism, and let r = π(0Q ). Then
π maps the set {a ∈ M | a < 0Q } bijectively to the set {b ∈ R | b < r }, but
the first set is countable and the second set is uncountable, so there cannot
be a bijection between them. This contradiction shows that M ∼
6 R. So
=
DLO is not categorical in |R|, and in fact DLO is not categorical in any
uncountable cardinal.
Exercises
16.1 Give an example of a linearly ordered set which is neither discrete nor
dense.
CHAPTER 16. LINEAR ORDERS
74
16.2 Suppose that A is a discrete linear order without endpoints. Show that
the successor function which takes an element to the one immediately
above it is definable.
16.3 Let DLOep be the theory of dense linear orders with least and greatest
endpoints. Show that DLOep is ℵ0 -categorical.
16.4 Let a, b ∈ Q. Show that there is an automorphism of Q< taking a to
b. What can you deduce about the definable subsets of Q?
16.5 What are all the definable subsets of Q2 in the structure Q< ?
16.6 Let A be a countable linearly ordered set. Show that there is an
embedding of A into Q< . [Hint: Use the idea in the back-and-forth
proof, but only going forth, not back.]
16.7 Show that every non-empty finite linearly ordered set is discrete and
has endpoints. [Hint: use induction on the cardinality of the set.]
16.8 Show that the theory of discrete linear orders without endpoints is not
ℵ0 -categorical.
16.9 Show that each singleton {n} is definable in N< . Using this fact and
a compactness argument, show that Th(N< ) is not ℵ0 -categorical.
16.10 Show that ThhZ; <, 0i is not ℵ0 -categorical. Deduce that Th(Z< ) is
also not ℵ0 -categorical.
Chapter 17
Quantifier elimination
Not yet written.
75
Chapter 18
Power sets and boolean
algebras
Not yet written
76
Part III
Types
77
Chapter 19
The algebras of definable sets
In this chapter we will look a little more systematically at the definable sets
in a model of a theory T , and show that we can deduce information about
the number of models of T from the collections of definable sets in a single
model.
Definition 19.1. Let A be an L-structure. We write Def n (A) for the set
of all definable subsets of An .
Recall that An itself and ∅ are definable, and Def n (A) is closed under
the Boolean operations ∩ , ∪ and complement. So it can be considered as a
Boolean algebra.
Now we start from the opposite direction, formulas and theories. Given
a language L, we have written Form(L) for the set of all L-formulas. Now
define Formn (L) to be the set of all L-formulas with free variables from
x1 , . . . , xn only. Let T be an L-theory. We define a relation ∼T on Formn (L)
by
ϕ(x̄) ∼T θ(x̄) if and only if T ` ∀x̄[ϕ(x̄) ↔ θ(x̄)].
It is easy to see that ∼T is an equivalence relation on Formn (L). We write
[ϕ(x̄)] for the equivalence class of ϕ(x̄) and we write Bn (T ) for set of equivalence classes.
Lemma 19.2. The logical operations ∧ , ∨ and ¬ on formulas induce the
structure of a Boolean algebra on Bn (T ). That is, if we define [ϕ(x̄)] ∧
[θ(x̄)] to be [(ϕ ∧ θ)(x̄)], and define ∨ and ¬ similarly, we get well-defined
operations on Bn (T ) which make it into a Boolean algebra.
We leave the proof as an exercise. The algebras Bn (T ) are called the
Lindenbaum algebras of T .
Lemma 19.3. Let A be an L-structure and T = Th(A). There is an isomorphism of Boolean algebras between Bn (T ) and Def n (A) given by
[ϕ(x̄)] 7→ {ā ∈ An | A |= ϕ(ā) } .
78
CHAPTER 19. THE ALGEBRAS OF DEFINABLE SETS
79
Again, we leave the proof as an exercise. An important consequence of
this lemma is that if A and B are models of the same complete theory T ,
the Boolean algebras Def n (A) and Def n (B) are isomorphic to each other,
because they are both isomorphic to Bn (T ), so as abstract Boolean algebras
they do not depend on the choice of model.
Example: vector spaces
Let K be a field, let T be the theory of infinite K-vector spaces in the
language LK-VS from chapter 14, and let V |= T . Since we have a constant
symbol naming 0, it is clear that Def 1 (V ) contains ∅, {0}, V r {0} and V .
However, given any v, w ∈ V r {0}, we can extend {v} to a basis B of V
and we can extend {w} to a basis B 0 of V . Choose a bijection α : B → B 0
such that α(v) = w, then α extends to an automorphism π of V such that
π(v) = w. Therefore v and w belong to exactly the same definable subsets
of V , so Def 1 (V ) consists of exactly the four sets given above.
Def 2 (V ) is a bit more interesting, because for each λ ∈ K we have the
formula x2 = λ · x1 , giving a linear equation. Suppose λ ∈ K and we have
(v1 , v2 ) ∈ V 2 and (w1 , w2 ) ∈ V 2 such that
V |= v1 6= 0 ∧ v2 = λ · v1
and also
V |= w1 6= 0 ∧ w2 = λ · w1 .
Then there is an automorphism π of V such that π(v1 , v2 ) = (w1 , w2 ). So
the subset of V 2 defined by the formula x1 6= 0 ∧ x2 = λ · x1 has no proper
non-empty definable subsets.
Suppose now that K = F3 , the field with three elements: 0, 1, 2, with
addition and multiplication modulo 3. Then we have the following five
formulas in two variables
x1 = 0 ∧ x2 = 0
x1 6= 0 ∧ x2 = 0
x1 = 0 ∧ x2 6= 0
x1 6= 0 ∧ x2 = x1
x1 6= 0 ∧ x2 = 2 · x1
2
which each give a definable subset of V with no proper non-empty definable subsets. If (v1 , v2 ) satisfies one of these formulas then v1 and v2 are
linearly dependent, so there is also the possibility that (v1 , v2 ) is a linearly
independent pair. With a finite field such as F3 , we can also capture this
situation with a single formula:
^
λ1 · x1 + λ2 · x2 6= 0
(λ1 ,λ2 )∈F23 r{(0,0)}
which is first-order because it is a finite conjunction. In this case is equivalent
to the formula
x1 6= 0 ∧ x2 6= 0 ∧ x2 6= x1 ∧ x2 6= 2 · x1 .
It is easy to check that every pair of elements from V satisfies exactly one
of these six formulas. However if the field is infinite, for example K = Q,
no single formula can state that x1 and x2 are linearly independent.
CHAPTER 19. THE ALGEBRAS OF DEFINABLE SETS
80
Principal formulas
Definition 19.4. Let T be a complete L-theory. A principal formula (with
respect to T ) is a formula ψ(x̄) such that
1. T ` ∃x̄[ψ(x̄)], and
2. for every L-formula ϕ(x̄) with the same list of variables, either
T ` ∀x̄[ψ(x̄) → ϕ(x̄)] or T ` ∀x̄[ψ(x̄) → ¬ϕ(x̄)].
So the six formulas in two variables given above for F3 -vector spaces are
principal formulas but, for example, the formula x2 6= x1 is not. Every pair
of elements in an F3 -vector spaces satisfies one of the six principal formulas.
However, if V is a non-zero Q-vector space there are infinitely many definable
subsets of V 2 , and a linearly independent pair of elements does not satisfy
any principal formula. This is a general phenomenon.
Lemma 19.5. If Def n (A) is finite then every ā ∈ An satisfies a principal
formula.
Proof. Given ā ∈ An , if Def n (A) is finite then there is a minimal definable
subset containing ā, and any formula defining it is a principal formula.
Recall that the theory of infinite F3 -vector spaces has as models only
the infinite-dimensional F3 -vector spaces, and the cardinality of a model is
equal to its dimension. Therefore there is, up to isomorphism, exactly one
countable model. However, the theory of infinite Q-vector spaces has as
countable models the Q-vector spaces with dimension any non-zero natural
number, and the model of dimension ℵ0 .
Theorem 19.6. If T is a complete theory in a countable language with
infinite models such that every Lindenbaum algebra Bn (T ) is finite then T
is ℵ0 -categorical, that is, up to isomorphism it has exactly one model of
cardinality ℵ0 .
The proof uses the back-and-forth method, which we used to prove that
there is exactly one countable model of DLO.
Proof. Firstly, by the Downward Löwenheim-Skolem theorem, there is at
least one model of cardinality ℵ0 .
So suppose A and B are both countable models of T . Enumerate A as
(an )n∈N+ and B as (bn )n∈N+ . We will construct new enumerations (αn )n∈N+
of A and (βn )n∈N+ of B such that for every n ∈ N and every L-formula
ϕ(x1 , . . . , xn ) we have
A |= ϕ(α1 , . . . , αn )
if and only if
B |= ϕ(β1 , . . . , βn ).
(∗)
CHAPTER 19. THE ALGEBRAS OF DEFINABLE SETS
81
We proceed by induction on n. For n = 0, condition (∗) says that every
L-sentence is true in A iff it is true in B. This is correct, because A and B
are models of the same complete L-theory.
Now suppose we have αi and βi for all i < n. If n = 2m − 1, odd, then
let αn = am . Since Bn (T ) is finite, by Lemma 19.5, there is a principal
formula ψn (x1 , . . . , xn ) for (α1 , . . . , αn ). Then we have A |= ψn (α1 , . . . , αn ),
and so we get A |= ∃xn [ψn (α1 , . . . , αn−1 , xn )]. Then by induction using (∗)
it follows that B |= ∃xn [ψn (β1 , . . . , βn−1 , xn )]. Choose βn ∈ B such that
B |= ψn (β1 , . . . , βn ). Then, since ψn is a principal formula, condition (∗)
holds for n.
If n = 2m, even, then let βn = bm . Repeat the above argument, swapping
the roles of A and B, to choose αn such that condition (∗) holds for n.
Note that every am appears as some αn . It may appear more than once,
but if αn = αl then this is witnessed by a formula, so also βn = βl by (∗).
Thus we may define a function π : A → B by π(αn ) = βn . This function is
surjective, since every bm appears as some βn . It is injective, since if βn = βl
then also αn = αl by (∗). Now (∗) also tells us that π preserves all formulas,
in particular it preserves all atomic formulas, so it is an embedding. The
same argument shows that π −1 is an embedding, so π is an isomorphism.
So A ∼
= B as required.
In chapter 21 we will prove that the converse of this theorem is also true.
Exercises
19.1 Prove Lemma 19.2.
19.2 Prove Lemma 19.3.
19.3 Give two examples of principal formulas and two examples of nonprincipal formulas in the theory of Zring .
19.4 Let V be an infinite-dimensional F3 -vector space. Show that any pair
(a, b) ∈ V 2 satisfies one of the six principal formulas given before
Definition 19.4.
19.5 Let V be any vector space. Show that the quantifier-free definable
subsets of V n correspond to P
boolean combinations of solution sets of
linear equations of the form ni=1 λi xi = 0.
19.6 Show that if V is an infinite Q-vector space considered as a LQ-VS structure then the definable subsets of V 2 are precisely given by finite
Boolean combinations of the linear equations x1 = 0, x2 = 0 and
x2 = λx1 for λ ∈ Q. Can you generalize to characterize the definable
subsets of V 3 ? What about the definable subsets of V n for arbitrary
n ∈ N+ ?
Chapter 20
Types
In the previous chapter we saw that if (a1 , a2 ) is a pair of elements from
a Q-vector space that are linearly dependent then they satisfy a principal
formula, but if they are linearly independent then they do not. To analyse
these tuples which do not satisfy a principal formula, we need to consider
infinitely many formulas at once, which brings us to the notion of types.
Recall that Formn (L) is the set of all L-formulas with free variables from
x1 , . . . , xn only.
Definition 20.1. Let T be a complete L-theory. An n-type of T is a subset
p(x̄) of Formn (T ) such that p(x̄) is finitelyVsatisfiable, that is, for any formulas ϕ1 (x̄), . . . , ϕr (x̄) from p(x̄), T ` ∃x̄ [ ri=1 ϕi (x̄)]. The type p(x̄) is a
complete type if for all ϕ(x̄) ∈ Formn (L), either ϕ(x̄) ∈ p(x̄) or ¬ϕ(x̄) ∈ p(x̄).
For example, we can consider the 2-type in the theory of infinite Qvector spaces consisting of all the formulas λ1 · v1 + λ2 · v2 6= 0 such that
(λ1 , λ2 ) ∈ Q2 r {(0, 0)}. This is the type of a linearly independent pair of
vectors.
Remark 20.2. By our definition, a complete type is deductively closed,
but a complete theory need not be. This is a useful convention because it
simplifies the statements of some theorems.
Definition 20.3. Let T be an L-theory, M a model of T , and let ā =
(a1 , . . . , an ) ∈ M n . The type of ā in M is
tpM (ā) = {ϕ(x̄) ∈ Formn (L) | M |= ϕ(ā) } .
We should check that the type of a tuple really is a type, according to
our definition.
Lemma 20.4. Let M |= T and ā ∈ M n . Then tpM (ā) is a complete n-type
of T .
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CHAPTER 20. TYPES
83
Proof. Clearly tpM (ā) is finitely satisfiable, because it is satisfied by ā. For
any ϕ(x̄) ∈ Formn (L), either M |= ϕ(ā) or M |= ¬ϕ(ā) so tpM (ā) is complete.
The type of a tuple is preserved under elementary extensions, and automorphisms.
Proposition 20.5.
1. Let π : M → N be an elementary embedding, and
n
let ā ∈ M . Then tpM (ā) = tpN (π(ā)).
2. If π ∈ Aut(M ) and ā ∈ M n then tpM (ā) = tpM (π(ā)).
Proof. For 1, we have ϕ(x̄) ∈ tpM (ā) iff M |= ϕ(ā) iff N |= ϕ(π(ā)) by
definition of an elementary embedding, and this is iff ϕ(x̄) ∈ tpN (π(ā)).
Now note that 2 is a special case of 1.
Usually we are only considering elementary extensions, so then by the
proposition, the type of a tuple does not depend on the particular model
M . So usually we write just tp(ā) without the subscript M .
Types as intersections of definable sets
We can also consider the notion of an n-type in terms of definable sets. For
example, if V is a Q-vector space we can consider the definable sets
Sλ1 ,λ2 = (v1 , v2 ) ∈ V 2 | λ1 · v1 + λ2 · v2 6= 0
for each (λ1 , λ2 ) ∈ Q2 r{(0, 0)}. Then the intersection of the sets Sλ1 ,λ2 is the
set of linearly independent pairs of vectors in the model V . More generally, if
(Si )i∈I is any collection of definable subsets of An such that any intersection
of finitely many of the Si is non-empty, then the collection (Si )i∈I forms an
n-type of Th(A). However, we have to be careful because if the collection
of definable sets is infinite, whether or not the intersection is empty may
depend on the choice of model of the
For example, in the case of
T theory.
vector spaces above, the intersection
Sλ1 ,λ2 (λ1 , λ2 ) ∈ Q2 r {(0, 0)} is
empty if V is a 1-dimensional Q-vector space, but non-empty if dim V > 2.
So we cannot just refer to the set of realisations in a particular model as the
type, we have to use the formulas.
Realising types
Definition 20.6. Let T be a complete theory, M a model of T and p(x̄)
an n-type of T . We say that the type p is realised in M iff there is a tuple
ā ∈ M n such that M |= p(ā). Otherwise, p is omitted from M .
For example, the type of a Q-linearly independent pair of vectors is
realised in a Q-vector space of dimension at least 2, but omitted in a model
of dimension 1. Every type of T can be realised in some model of T .
CHAPTER 20. TYPES
84
Lemma 20.7. Suppose that p is an n-type of T , a complete L-theory. Then
there is a model M of T and ā ∈ M n such that M |= p(ā), that is, for each
ϕ(x̄) ∈ p(x̄), M |= ϕ(ā).
Proof. Expand L to Lc̄ by adding new constant symbols c1 , . . . , cn . Consider the set of sentences Σ = T ∪ {ϕ(c̄) | ϕ(x̄) ∈ p(x̄) }. Then Σ is finitely
satisfiable since p(x̄) is a type, so it is realised in some model M by the
compactness theorem. Then take ā = c̄M and we have M |= p(ā).
Omitting types
We now consider which types can be omitted. Roughly speaking, if a type
can be given by a single formula, then it must be realised in any model of T .
Otherwise, it can be omitted in some models. We make this more precise.
Definition 20.8. An n-type p(x̄) of a complete theory T is a principal type
if there is a single formula ψ(x̄) such that T ` ∃x̄ψ(x̄) and, for every formula
ϕ(x̄) ∈ p(x̄) we have T ` ∀x̄[ψ(x̄) → ϕ(x̄)]. Any type which is not principal
is called a non-principal type.
It is easy to see that any principal type of T is realised in any model of
T . What is surprising is the converse is also true.
Theorem 20.9 (Omitting types theorem). Let T be a complete theory in a
countable language L, and let p be a non-principal n-type of T . Then there
is a countable model of T which omits p.
We will adapt the proof of Proposition 10.4 which shows that any finitely
satisfiable theory can be extended to a Henkin theory T ∗ in a language
with extra constant symbols. By Proposition 10.6, every Henkin theory
has a canonical model, that is a model in which every element is named
by a closed term. We will construct our Henkin theory T ∗ in such a way
that every element of the canonical model M ∗ is named by a new constant
symbol, and for every n-tuple c̄ of these constant symbols there is a formula
ϕ(x̄) ∈ p(x̄) such that T ∗ ` ¬ϕ(c̄). It follows that no n-tuple from M ∗
realises p, so the reduct M of M ∗ to L is a model of T which omits p.
Proof of Theorem 20.9. Let L∗ be L expanded by a countably infinite set C
of new constant symbols. Enumerate all the n-tuples of these new constant
symbols as (c̄r )r∈N , and enumerate all the L∗ -sentences as (σr )r∈N . We will
construct L∗ -formulas θm for m ∈ N such that for each m ∈ N we have
θm+1 ` θm , T ∪ {θm } is satisfiable and
1. if m = 3k + 1, either θm ` σk or θm ` ¬σk ,
2. if m = 3k +2, θm−1 ` σk and σk is of the form ∃x[ϕ(x)] then θm ` ϕ(c)
for some c ∈ C,
CHAPTER 20. TYPES
85
3. if m = 3k+3, there is some formula ϕ(x̄) ∈ p(x̄) such that θm ` ¬ϕ(c̄k ).
To do this, we proceed by induction on m. Take θ0 to be ∀x[x = x]. Then
certainly T ∪ {θ0 } is satisfiable. Now suppose m ∈ N+ and we have constructed θ0 , . . . , θm−1 .
If m is of the form 3k + 1 for some k ∈ N, take θm = θm−1 ∧ σk , if
T ∪ {θm−1 ∧ σk } is satisfiable, and take θm = θm−1 ∧ ¬σk otherwise. Then
T ∪ {θm } is satisfiable.
If m is of the form 3k + 2, θm−1 ` σk and σk is of the form ∃x[ϕ(x)],
choose a constant symbol c ∈ C which does not appear in θm−1 and set
θm = θm−1 ∧ ϕ(c). Then T ∪ {θm } is satisfiable because c is a new constant.
Otherwise we just let θm = θm−1 .
If m is of the form 3k + 3, we claim there is a formula ϕ(x̄) ∈ p(x̄) such
that T ∪{θm−1 , ¬ϕ(c̄k )} is satisfiable. Then we can take θm = θm−1 ∧ ¬ϕ(c̄k ).
¯ where ρ(x̄, ȳ) is an LTo prove the claim, we write θm−1 in the form ρ(c̄k , d)
formula, and d¯ is a tuple of constants from C, all distinct from the elements
of c̄k . Then if there is no ϕ(x̄) ∈ p(x̄) such that T ∪ {θm−1 , ¬ϕ(c̄k )} is
¯ → ϕ(c̄k ), but none of
satisfiable, for each ϕ(x̄) ∈ p(x̄) we have T ` ρ(c̄k , d)
the constants in c̄k or in d¯ occur in T , so we must have T ` ∀x̄∀ȳ[ρ(x̄, ȳ) →
ϕ(x̄)], or equivalently T ` ∀x̄[∃ȳ[ρ(x̄, ȳ)] → ϕ(x̄)]. But then ∃ȳ[ρ(x̄, ȳ)] is a
principal formula for p(x̄), a contradiction.
Thus we can produce our list of formulas θm for m ∈ N satisfying all the
conditions.
Now take T ∗ to be the deductive closure of T ∪ {θm | m ∈ N+ }. Then
T ∗ is finitely satisfiable, it is complete by condition 1, and it has the witness
property by condition 2, and so it is a Henkin theory. Let M ∗ be the
canonical model and M its reduct to L. Then by condition 2, every element
of M is named by a constant symbol from C, and by condition 3 no n-tuple
of elements of M realises p. So M omits p as required.
Exercises
20.1 Let S be any set of types of a complete theory T , and let M |= T .
Show there is an elementary extension of M which realises all the types
in S.
20.2 In Th(Ns-ring ), give an example of a complete principal 1-type and an
incomplete non-principal 1-type.
20.3 Let T = Th(Ns-ring ). Show that there is a 1-type of T of a number
whose only prime factor is 5, but which is divisible by 5n for all n ∈ N.
Show there is another 1-type of T of a non-standard number which is
divisible by 3 but not by 9 or by any other standard natural number
greater than 3.
CHAPTER 20. TYPES
86
20.4 Show that if p(x̄) is a complete principal type then there is a principal
formula ψ(x̄) in p(x̄). Find an example of an incomplete principal type
which does not contain a principal formula.
20.5 Let p(x1 , . . . , xn ) be an n-type of a complete L-theory T . Let L0 =
L ∪ {c1 , . . . , cn }, with the ci being new constant symbols, and let T 0 =
{ϕ(c1 , . . . , cn ) | ϕ(x1 , . . . , xn ) ∈ p(x1 , . . . , xn ) }. Show that T 0 is an L0 theory, and that if p is a complete type then T 0 is a complete L0 -theory.
Chapter 21
Countable categoricity
In the previous chapter we introduced types as sets of formulas. We defined
the type of an n-tuple ā from a model M of T and observed that it is a
complete type. We also observed that every complete n-type of T is realised
in some model of T , so we can think of complete types as describing “potential tuples from M ” or more precisely as describing tuples from elementary
extensions of M . In this chapter we will concentrate on complete types, and
complete our characterisation of countably categorical structures. First we
note that we do not lose much by considering only complete types.
Lemma 21.1. Let T be a complete theory. Then every n-type of T is
contained in a complete n-type of T .
Proof. Let p(x̄) be an n-type of T . By Lemma 20.7 there is a model M of
T and ā ∈ M n such that M |= p(ā). Then tpM (ā) ⊇ p(x̄), and tpM (ā) is a
complete n-type of T .
Definition 21.2 (Stone space). For a complete theory T , let Sn (T ) be the
th
set of all complete n-types of T . It is called the n Stone space of T .
Proposition 21.3. Let T be a complete theory and n ∈ N+ . The following
are equivalent.
1. The Stone space Sn (T ) is finite.
2. Every type in Sn (T ) is principal.
3. The Lindenbaum algebra Bn (T ) is finite.
Proof. Suppose Sn (T ) is finite, say Sn (T ) = {p1 , . . . , pr }. For each i, j ∈
{1, . . . , r} with i 6= j thereVis a formula ϕij (x̄) such that ϕij (x̄) ∈ pi and
¬ϕij (x̄) ∈ pj . Let ψi = rj=1,j6=i ϕij . Then T ` ∃x̄ψi (x̄) because any
realisation of pi satisfies ψi . Now suppose M |= T and ā ∈ M n with M |=
ψi (ā). Then for each j 6= i, M |= ϕij (ā), so tpM (ā) 6= pj . Thus tpM (ā) = pi .
87
CHAPTER 21. COUNTABLE CATEGORICITY
88
In particular, ψi is a principal formula for pi . So every p ∈ Sn (T ) is principal.
So 1 ⇒ 2.
Now suppose every type in Sn (T ) is principal, but suppose for a contradiction that Sn (T ) is infinite, say Sn (T ) = {pi | i ∈ I }. Let ψi be a
principal formula for pi . Let q = {¬ψi | i ∈ I }. We claim that q is finitely
satisfiable. So let ¬ψi1 , . . . , ¬ψir be a finite subset of q. Since I is infinite,
there is i ∈ I r {i1 , . . . , iV
r }. Let ā be an n-tuple in some model M such that
tp(ā) = pi . Then M |= rj=1 ¬ψij (ā). So q is finitely satisfiable, and hence
is a type. By Lemma 21.1, there is a complete type q 0 containing q. But
q 0 cannot be any of the pi for i ∈ I, which is a contradiction. So Sn (T ) is
finite. Thus 2 ⇒ 1.
To show the equivalence of 3 with 1 and 2, we use cardinality arguments.
Each type in Sn (T ) corresponds to a subset of Bn (T ), and different types
correspond to different subsets. There are 2|Bn (T )| subsets of Bn (T ), so
|Sn (T )| 6 2|Bn (T )| . Thus if Bn (T ) is finite, so is Sn (T ). Thus 3 ⇒ 1.
Now suppose 1 holds, that is Sn (T ) is finite. Let r = |Sn (T )|. Then
also 2 holds, every type in Sn (T ) is principal, say the principal formulas
are ψ1 (x̄),
W. . . , ψr (x̄). Then for each subset J ofn{1, . . . , r}, the formula
ϕJ (x̄) = j∈J ψj (x̄) defines a different subset of M . But if S is a definable
subset of M n , and p is a type, then either every realisation of p is in S or no
realisation of p is in S, so S must be defined by one of the formulas ϕJ (x̄).
Hence |Bn (T )| = 2|Sn (T )| , so |Bn (T )| is finite. So 1 ⇒ 3.
Now we can finally state and prove the main theorem characterising
countably categorical theories.
Theorem 21.4 (Ryll-Nardzewski, Svenonius, Engeler). Let T be a complete
theory in a countable language L. The following are equivalent.
1. T is ℵ0 -categorical (also known as countably categorical).
2. For all n ∈ N+ , the Stone space Sn (T ) is finite.
3. For all n ∈ N+ , every type in Sn (T ) is principal.
4. For all n ∈ N+ , the Lindenbaum algebra Bn (T ) is finite.
Proof. We have conditions 2, 3 and 4 all equivalent by Proposition 21.3.
Theorem 19.6 shows that 4 ⇒ 1.
Now assume there is a non-principal type p ∈ Sn (T ) for some n ∈ N+ . By
the Omitting Types Theorem, there is a countable model A of T which omits
p. By Lemma 20.7, there is a model B of T and ā ∈ B n such that tp(ā) =
p. By the Downward Löwenheim-Skolem Theorem, there is a countable
elementary substructure B0 4 B, containing ā. Suppose π : B0 → A is an
isomorphism. Then, by Proposition 20.5, we have tpA (π(ā)) = tpB0 (ā) =
tpB (ā) = p. But A omits p, so we have a contradiction, and π cannot
CHAPTER 21. COUNTABLE CATEGORICITY
89
exist. Thus A and B0 are non-isomorphic countable models of T , so T is
not ℵ0 -categorical. That proves 1 ⇒ 3, and we are done.
Exercises
21.1 Consider Q< , and let ā, b̄ ∈ Qn . Show that if a1 < a2 < · · · < an and
b1 < b2 < · · · < bn then there is an automorphism of Q sending ai
to bi . Deduce that tp(ā) = tp(b̄). Deduce further that there are only
finitely many complete n-types in DLO.
21.2 Let K be a field and let T be the theory of infinite K-vector spaces.
Let V |= T .
(a) Show there are exactly 2 complete 1-types in T . [Hint: consider
the formulas x = 0 and x 6= 0.]
(b) Suppose a1 , . . . , an ∈ V are linearly independent, and b1 , . . . , bn ∈
V are also linearly independent. Show that tp(ā) = tp(b̄).
(c) Suppose K is a finite field. Show tp(ā) is a principal type.
(d) Suppose K is an infinite field and n > 2. Show tp(ā) is not a
principal type.
(e) When K is an infinite field, show |S2 (T )| = |K|.
Chapter 22
Saturation and homogeneity
Not yet written.
90
Chapter 23
Definable sets and types
with parameters
Not yet written.
Examples of sets definable with parameters. In vector spaces and the
successor structure, parameter-definable sets are still finite or co-finite. In
DLO they are unions of intervals. 1-types over parameters in Q< almost
correspond to real numbers.
parametrically definable sets: straight lines, rectangles, circles in R2 ,
definable families of definable sets, infinitely sloping lines make sense in
elementary extensions of R as a field, but not in R2 as an R-vector space.
91
Chapter 24
Counting types and models
Not yet written.
92
Part IV
Algebraically Closed Fields
93
Chapter 25
Fields and their extensions
One of the largest areas of applications of model theory is to the model
theory of fields. These may be fields considered just in the ring language, or
fields with operators such as derivations, automorphisms, or valuations. In
many cases the fields under consideration are algebraically closed, but even
when not the theory of algebraically closed fields usually plays a role. In
this final part of the book we will study the theory of algebraically closed
fields, leading to a model-theoretic proof of Hilbert’s nullstellensatz.
In this chapter we review the basic concepts of field extensions we need.
Yet to be written.
94
Chapter 26
Algebraic closures
probably include this with previous chapter.
95
Chapter 27
Definable sets and algebraic
varieties
QE and proof of Hilbert nullstellensatz
96
Part V
Working Appendices
97
Appendix A
Dependence of chapters
The following is the intended dependence between chapters. Exercises need
not necessarily follow dependence restrictions, and some of the discussion in
chapters may not either, but most should.
Part I
• 1 Structures, 2 Terms, 3 Formulas in that order, and everything else
depends on them.
• 4 Definable sets relies only on 1,2,3.
• 5 Substructures and quantifiers follows on from 4
• 6 Theories relies only on 1,2,3.
• 7 Real and Complex fields follows on from 6
• 8 Compactness ideally follows only from 1,2,3, but some applications
follow from 4 and 6 perhaps
• 9 Axiomatizable classes follows from 6 and 8.
• 10 Henkin constructions uses only 1,2,3.
• 11 Interlude on set theory uses nothing at all
Part II
• 12 Elementary substructures uses only 1,2,3 and 11.
• 13 Elementary extensions uses 8, 11, 12
• 14 Vector spaces uses 6, 11, 12?, 13?
• 15 Successor structure uses 6, 11, 13, 4
• 16 Linear orders uses 6
• 17 QE uses 4, 5? 16 exercises use 14, 15
• 18 Power sets and boolean algebras ??
Part III
98
APPENDIX A. DEPENDENCE OF CHAPTERS
99
• 19 The algebras of definable sets 4, (18), 14, for the last part also: 16,
12. Could move that to countable categoricity
• 20 Types uses 19, 4, 8, 10 (for proof)
• 21 Countable categoricity uses 20, 12
• 22 Saturation and homogeneity
• 23 Definable sets and types with parameters
• 24 Counting types and models
Intention should be to keep ACF chapters fairly light on prerequisites
too, which should shape the order of material. E.g. two chapters depending
only on part I, another on part II, and a fourth on part III.
Appendix B
Learning objectives by
chapter
Chapter 1. Concepts: language (vocabulary), L-structure, embedding, substructure, extension, isomorphism, automorphism
Chose a structure for a particular mathematical problem / statement.
Identify all the automorphisms of certain structures.
Chapter 2. Concepts: L-terms, their interpretations as functions (using a
context, i.e. a list of variables).
L-embeddings preserve the interpretation of L-terms.
First example of a recursive definition and an inductive proof
using it.
Chapter 3. Concepts: L-formulas, atomic L-formulas, free and bound variables, sentences, interpretation of formulas in structures, models.
More practice of recursive definitions and inductive proofs
L-embeddings preserve the interpretations of atomic L-formulas.
L-isomorphisms preserve the interpretations of L-formulas.
Use of abbreviations in the formal language
Express statements about L-structures as L-sentences.
Express statements about elements of L-structures as L-formulas.
Chapter 4. Concepts: definable sets, definable functions, geometric interpretation of logical symbols as intersection, union, complement,
projection.
Show certain sets are definable by giving defining formulas for
them.
100
APPENDIX B. LEARNING OBJECTIVES BY CHAPTER
101
Automorphisms preserve definable sets.
Show certain sets are not definable by finding an automorphism
which does not preserve them.
Not every set preserved by all automorphisms is definable.
Chapter 5. Concepts: Substructure, extension, quantifier-free formula, existential formula, universal formula.
Relationships between the interpretations of formulas in substructures and extensions depending on the nature of quantifiers
in the formula.
Applications to sentences and definable sets.
Chapter 6. Concepts: The theory of a structure and of a class of structures,
entailment and deductive closure, theories, axioms for a theory,
completeness of a theory, elementary equivalence.
Recognise specific axioms for groups, rings, fields, etc.
Recognise Peano Arithmetic as an attempt to capture the theory of the arithmetic of natural numbers.
Awareness of Gödel’s First Incompleteness theorem, showing
that Peano Arithmetic is not complete.
Isomorphism implies elementary equivalence.
Come up with sentences to distinguish the first-order theories of
different structures.
Chapter 7. Awareness of the real and complex fields as important mathematical structures whose first-order theories have explicit axiomatizations.
Recognise the specific axioms of algebraically closed fields and
real-closed fields.
Awareness of the difference between first-order logic and secondorder logic.
The least upper bound of a definable set is definable.
Chapter 8. Concepts: satisfiability, reduct, expansion.
The statement of the compactness theorem.
Use of the compactness theorem together with new constant
symbols to build new models of a theory.
Use of the compactness theorem and new constants to explore
properties of definable sets in non-standard models.
Chapter 9. Concepts: axiomatizable class, finitely axiomatizable class.
Use of the compactness theorem to show that a class of structures is not axiomatizable, or is not finitely axiomatizable.
APPENDIX B. LEARNING OBJECTIVES BY CHAPTER
102
Applications to classes of finite structures, and other examples
including abelian groups with and without torsion.
Chapter 10. Concepts: finitely satisfiable, deductively closed, and complete
sets of L-sentences, closed L-terms, witness property, Henkin
theories, canonical models.
Understand the proof of the compactness theorem via Henkin’s
method.
Chapter 11. Concepts: Cardinality of a set, countability, uncountability.
Awareness of some basic properties of infinite sets and their
cardinalities.
Cardinality of the set of finite strings from a given set.
Chapter 12.
Chapter 13.
Chapter 14. Concepts: Vector space, basis, dimension, κ-categoricity.
Use basic properties of vector spaces to understand dimension,
cardinality, isomorphisms and automophisms of vector spaces.
Prove that the theory of K-vector spaces is κ-categorical for
κ > |K|.
Be able to prove and use the Los-Vaught test for completeness.
Prove that the theory of infinite K-vector spaces is complete.
Chapter 15.