M4P49
Model Theory
Mathematics
Imperial College London
These notes are based on a course of lectures given by Dr Pál during Autumn Term 2008 at Imperial
College London. In general the notes follow the lectures very closely and only few changes were
made mostly to make the typesetting easier.
These notes have not been checked by Dr Pál and should not be regarded as ocial notes for the
course. In particular, all the errors are made by me. However, I don't take any responsibility for
their consequences; use at your own risk (and attend lectures, they are fun).
Please email me at
[email protected]
with any comments or corrections.
Anton Stefanek & James Haydon
December 2008
Contents
1 Naive set theory
3
1.1
First steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Orderings and Ordinals
5
1.3
Cardinals of sets
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.4
Hierarchy of Cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Languages and structures
10
2.1
Formulas and satisfaction
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
2.2
Examples of theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
2.3
Gödel's Compactness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
3 Quantier elimination
26
2
1 Naive set theory
1.1
First steps
All objects in naive set theory are sets; use capital letters for sets and small letters for sets contained
in them. Use the shorthand
x∈X
for x is an element of
X .
The axioms/denitions of naive set
theory:
•
We do not distinguish between sets containing the same elements, this is the axiom of
extensiability:
X = Y ↔ ∀x(x ∈ X ↔ x ∈ Y ).
•
Empty set:
∃X(∀y(y ∈
/ X)).
It can be shown that
•
X
as above is unique, call it
∅.
Denition of the subset relation:
X ⊆ Y ↔ ∀x(x ∈ X → x ∈ Y ).
•
Denition of the union operation:
∀X∀Y ∃Z(x ∈ Z ↔ (x ∈ X ∨ x ∈ Y )).
Can show that
•
Z
as above is unique, it is the union of
X
and
Y
and is denoted
X ∪Y.
Unordered pairs:
∀x∀y∃Z(u ∈ Z ↔ (u = x ∨ u = y)).
Denote the unordered pair by
elements of a set
X
{x, y}. We will also
x1 , x2 , . . . , then
can be listed as
write
{x, x}
as
{x}.
we denote this set as
In general if the
{x1 , x2 , . . . }.
Construction of the natural numbers
1 = {∅},
2 = 1 ∪ {1} = {∅, {∅}},
.
.
.
n = n − 1 ∪ {n − 1}.
More set theory
•
Ordered pair of
x
and
y:
(x, y) = {x, {x, y}}.
•
Union of a set:
∀X∃Y (z ∈ Y ↔ ∃y(y ∈ X ∧ z ∈ y)).
Y is the union of the elements of the elements of X . Denote the union of X by ∪X . Note that
∪{x, y} = x ∪ y and so the rst denition of union can be made redundant. This notation is
sometime used with an index set: {Xi | i ∈ I} or {Xi }i∈I . We say that the elements are
indexed by I .
• Intersection: ∩X = {x | ∀y(y ∈ X → x ∈ y)}.
• Cartesian products:
X × Y = {z | ∃x∃y(x ∈ X ∧ y ∈ Y ∧ z = (x, y))}.
This is just the set of ordered pairs.
• R
• A
is a
on X × Y iff R ⊆ X × Y .
R ⊆ X × Y is a function iff
relation
relation
Write
xRy
for
(x, y) ∈ R.
relation
function
∀x(x ∈ X → ∃!y(y ∈ Y ∧ xRy)).
•
A function
f : X → ∪X
is a
choice function iff ∀x(x ∈ X → f (x) ∈ x).
3
choice function
Axiom of choice.
∀x(x ∈ X → x 6= ∅) → ∃
The set of choice functions is denoted
X × Y → Π{X, Y }.
Two sets A and B are
Πi∈I Xi .
choice function
f : X → ∪X .
This is because there is a natural correspondance
of the same size if they are in bijection.
iX : X → P (X), y 7→ {y}.
Note that there is a natural injection
Theorem 1.1. There doesn't exist a surjective map s : X → P (X).
Proof.
s : X → P (X) be
∃a ∈ X, s(a) = Y . There are two
By contradiction: let
is surjective,
•
•
if
if
a∈Y
a∈
/Y
then:
then:
a∈
/ s(a)
a ∈ s(a)
surjective. Let
Y = {x ∈ X | x ∈
/ s(x)}.
Because
a∈
/ Y.
a∈Y.
and so
and so
so both cases lead to a contradiction.
Russel's paradox.
Proof.
The collection
Again by contradiction: Let
s
cases:
C
of all sets is not a set.
Y = {x ∈ C | x ∈ x} ⊆ C .
So
Y
should be a well dened set,
but
• Y ∈Y
• Y ∈Y
Y ∈Y.
Y ∈
/ Y.
implies that
implies
Both a contradiction.
Theorem 1.2 (Bernstein). If there exists injections f : X → Y and g : Y → X then there exists
a bijection h : X → Y .
Proof.
We need two lemmas.
Lemma 1.3. Let X be any set and let h : P (X) → P (X) be a function such that ∀A ⊆ B ⊆
X, h(A) ⊆ h(B). Then there exists T ⊆ X such that h(T ) = T .
Proof.
Let
T = ∪{A ⊆ X | A ⊆ h(A)}.
• T ⊆ h(T ): If x ∈ T , then there is some A ⊆ X such that x ∈ A and A ⊆ h(A). A ⊆ T and
so h(A) ⊆ h(T ), thus, x ∈ A ⊆ h(A) ⊆ h(T ).
• h(T ) ⊆ T : From T ⊆ h(T ) we deduce h(T ) ⊆ h(h(T )). So by denition of T , h(T ) ⊆ T . Lemma 1.4. Let f : X → Y and g : Y → X be two maps. Let ∗ : P (X) → P (X) be given by the
rule:
∗(X) = A∗ ,
A∗ = X − g(Y − f (A)).
Then A ⊆ B ⊆ X implies that A∗ ⊆ B ∗ .
Note that this formula is the complement of the complement. Drawing a picture helps.
Proof.
Suppose
A ⊆ B ⊆ X.
Then
f (A) ⊆ f (B)
⇒ Y − f (A) ⊃ Y − f (B)
⇒ g(Y − f (A)) ⊃ g(Y − f (B))
⇒ X − g(Y − f (A)) ⊆ X − g(Y − f (B)).
4
Now we can prove the theorem. Let
Dene
T
h: X → Y :
be a subset of
(
h(x) =
Then
X
f (x)
g −1 |g(Y ) (x)
such that
if
if
T ∗ = T , i.e. T = X −g(Y −f (T )).
x ∈ T,
x ∈ X − T.
X − T = g(Y − f (T )) ⊆ g(Y ).
• f |T is injective (???),
• g|Y −f (T ) is injective (???),
So
h
is injective. It is also surjective (???) and hence bijective.
inj
inj
Trichotomy principle ∀X∀Y (∃X → Y ∨ ∃Y → X).
This has no simple proof as one needs the
axiom of choice and lots of machinery about the ordinals.
1.2
Orderings and Ordinals
A binary relation
(i)
(ii)
(iii)
≤
X
on
partial ordering
is a
partial ordering
∀x, y ∈ X ,
if
x ≤ x,
(x ≤ y ∧ y ≤ x) → x = y ,
(x ≤ y ∧ y ≤ z) → x ≤ z .
(P (X), ⊆).
For example
For all partial ordering
≤ we introduce another binary relation < dened
as follows:
x < y ↔ (x ≤ y ∧ x 6= y).
This satises
(i)'
(ii)'
¬(x < x),
(x < y) → ¬(y < x),
and property (iii) as above. The relations
≤
and
<
uniquely determine each other, so we use both
of them when we have a partial ordering.
A partial ordering is a
(iv)
complete ordering
Such a
y
Then
(X, ≤)
is
well ordered
if for every
∅ 6= Y ⊆ X, ∃y ∈ Y such that ∀x ∈ Y, y ≤ x.
min(Y ) if it exists.
Let
(X, ≤)
be a
woset (= well ordered set).
Let
E⊆X
such that
E = X.
Proof.
For a contradiction, assume
E=
6 X . Then X − E 6= ∅ and so let x = min(X − E).
y < x, y ∈ E (otherwise x isn't minimum).
by (i). Therefore, for all
For two ordered sets
(X, ≤), (X 0 , ≤0 ) a function f : X → X 0
is an
order isomorpsim
if
f
Then
is bijective
∀x, y ∈ X ,
x ≤ y ↔ f (x) ≤ f (y).
In this case we write
Claim.
Proof.
Let
(X, ≤)
f: X ∼
=X
be a woset,
0
.
Y ⊆ X, f : X ∼
= (Y, ≤ ∩ Y × Y ).
Then
∀x ∈ X, x ≤ f (x).
E = {x ∈ X | f (x) < x} 6= ∅. Then if x0 = min(E), f (x0 ) < x0 since
f (f (x0 )) < f (x0 ) as f is an order isomorphism. This shows that f (x0 ) ∈ E , but
contradicts that x0 = min(E).
Assume that
x0 ∈ E .
this
well ordered
min(X) ∈ E ,
∀x ∈ X(∀y(y < x → y ∈ E) → x ∈ E).
x 6= min(X)
and
complete ordering
is unique so we call it the minimum and denote it
Transnite induction
(i)
ordering ) if
∀x, y ∈ X, x ≤ y ∨ y ≤ x.
The ordered set
(ii)
(or just an
Then
5
order isomorpsim
Claim.
(X, ≤), (X 0 , ≤0 )
f: X ∼
= X 0.
Let
isomorphism
Proof.
be wosets.
(X, ≤) ∼
= (X 0 , ≤0 )
then there exists a unique order
f −1 ◦ g : X ∼
= X . So we have reduced the problem to
X = X and we need to show that h = idX (i.e. that f = g ). By previous claim: x ≤ h(x), and
−1
using the same argument with h
gives us h(x) ≤ x. Hence by trichotomy h(x) = x.
segment
Let
f: X ∼
= X 0, g : X ∼
= X 0.
If
Then
0
For any woset
(X, ≤)
and any
a ∈ X , Xa
is the
segment
of
X
dened as:
Xa = {y ∈ X | y < a}.
Claim.
Proof.
There does not exist an order isomorphism
Suppose for a contradiction that one exists,
f (a) ∈
/ Xa ,
Claim.
X∼
= Xa
for any
f: X ∼
= Xa .
a ∈ X.
By the above,
a ≤ f (a),
but then
a contradiction.
Let
(X, ≤)
be a woset and let
A = {Xa | a ∈ X}.
Then
(X, ≤) ∼
= (A, ⊆).
Proof. f : X ∼
= A is given by f (a) = Xa .
ordinal
A woset
(X, ≤)
the ordering on
Claim.
(i)
(ii)
Let
α
is an
X
ordinal
if
∀a ∈ X, Xa = a.
Let
n is
α be
Let
α
For (ii),
(α + 1)α = α.
an ordinal.
an ordinal, then for all
Proof. Let b ∈ αa .
b} = αb = b.
Claim.
Then
a ∈ α,
the segment
αa
is an ordinal.
(αa )b = {x ∈ αa | x < b} = {x ∈ α | x < a
and
x < b} = {x ∈ α | x <
be an ordinal. If
Y $α
and
Y
is an ordinal then
Y = αa
for some
Proof.
For the reverse inclusion let
If
but this is a contradiction. Hence
Let a = min(α − Y ), then αa ⊆ Y .
a < b then a ∈ αb so a ∈ Yb and a ∈ Y ,
b < a and b ∈ αa .
Claim.
If
α, β
Then
also an ordinal.
Proof. ∅ = αmin(α) = min(α∈ α, this proves (i).
Claim.
a < b ↔ Xa $ Xb ↔ a b.
be an ordinal, then,
α 6= ∅ → ∅ ∈ α.
α + 1 = α ∪ {α} is
For example,
In this case
is canonical and so we drop it from the notation.
are ordinals then
Proof. If a ∈ α ∩ β
x < a} = (α ∩ β)a .
then
α∩β
α a = a = βa ,
a∈Y.
b ∈ Y . Then Yb = b = αb .
b ≤ a. But b 6= a so
is also an ordinal.
i.e.
{x ∈ α | x < a} = a = {x ∈ β | x < a} = {x ∈ α ∩ β |
Theorem 1.5. Let α, β be ordinals. If α 6= β then one is a segment of the other.
Proof. ¬(α ⊆ β ∨ β ⊆ α) ↔ (α ∩ β 6⊃ α ∧ α ∩ β 6⊃ β).
So assume this is false, then
α and of β (also
a = αa = βb = b.
α ∩ β,
which is
an ordinal (by above), must be an initial segment of
by above). So there exists
a ∈ α and b ∈ β such that α ∩ β = αa = βb . Then
then a ∈ α ∩ β = αa ⇒ a < a, a contradiction.
Hence
Theorem 1.6. If α and β are ordinals, α ∼
= β , then α = β .
6
a = b ∈ α ∩ β.
But
Proof.
Let
f: α∼
= β.
We show that
f = idα .
Let
E = {x ∈ α | f (x) 6= x}.
Assume that
f (a),
E 6= ∅
and let
a = min(E).
Then for all
x < a, f (x) = x.
Hence
αa = βf (a) = a =
a contradiction.
Claim.
For ordinals
α
and
β, α $ β ↔ α ∈ β.
Proof.
• α $ β ⇒ α = βa | a ∈ β ⇒ α = βa = a ∈ β .
• α ∈ β ⇒ α = βα $ β .
Let
On
denote the collection of all ordinals.
On
Proposition 1.7.
(i) On is well ordered w.r.t. ∈.
(ii) On is not a set. NO PROOF PROVIDED
Proof.
∈ is an ordering on On .
(i)' α ∈
/ α: α ∈ α ⇒ α = a ∈ α ⇒ α = αa ⇒ α ∼
= one of α's segments.
(ii)' α ∈ β → β ∈
/ α: Otherwise β ∈ α ∈ β ⇒ β $ β .
(iii)' α ∈ β ∧ β ∈ γ → α ∈ γ : As β ∈ γ , β ⊆ γ and so α ∈ β ⊆ γ .
Now we show it is well ordered: Let C be a non-empty collection of ordinals.
dene D = {α ∈ β | α is in C} ⊆ β . didn't get what followed.
(i) First we show that
Let
β∈C
and
Theorem 1.8. Let (X, ≤) be a woset such that for all a ∈ X the segment Xa is isomorphic to an
ordinal. Then X is also isomorphic to an ordinal.
Proof. ∀a ∈ X, ∃ga : Xa ∼
= Z(a)
ordinals are equal and therefore
where
ga
Z(a)
is an ordinal.
Z(a)
Then
is unique as isomorphic
is also unique. Dene
W = {Z(a) | a ∈ X}.
Then
W
is a set, and
Z : X → W, a 7→ Z(a)
Claim. x, y ∈ X, x < y → Z(x) $ Z(y).
Proof. Xx = (Xy )x
ordinal
Z(y)
and so
and so is also
a function.
gy |Xx : Xx ∼
= Z(y)gy (x) is an isomorphism. Z(y)gy (x) is a segment of
an ordinal. So Z(y)gy (x) = Z(x) $ Z(y) by uniqueness of f .
Z is an order isomorphism and so (W, ⊆) is
(W, ⊆) is an ordinal.
If x, y ∈ X and x < y , then Z(y)gy (x) = gy (x) since Z(y) is an ordinal.
Z(x) = Z(y)gy (x) so Z(x) = gy (x). Thus
From the claim we deduce that
a woset. We now only
need to show that
WZ(y) = {Z(y) | Z(x) $ Z(y)} = {Z(x) | x < y},
[as
Z
We already saw that
is an order iso.]
= {gy (x) | x < y} = gy (Xy ) = Z(y).
Theorem 1.9. Every woset is order isomorphic to a unique ordinal.
Proof.
Uniqueness is clear. Existence is proved using transnite induction. Let
E = {a ∈ X | Xa
is
∼
=
to an ordinal}.
(X, <) is some woset. By previous theorem this strategy works, i.e. it is enough to show
E = X . So suppose E 6= X , then let x = min(X − E). Then for all a < x, Xa ∼
= an ordinal.
Xa = (Xx )a hence Xx is an ordinal by previous and so x ∈ E , a contradiction.
where
that
7
cardinal
limit ordinal
ω
Denition. An ordinal α is a cardinal if ∀β < α there is no bijection f : β → α. An ordinal α is
a limit ordinal if α 6= β + 1 for any ordinal β . We denote the set of all nite ordinal by ω .
Claim.
(i) The elements of
(ii)
ω
(iii) If
Proof.
to
α.
ω
are cardinals.
is a cardinal.
κ
is a cardinal and
ω⊆κ
then
κ
is a limit ordinal.
Only prove (iii) since this is a naive theory. Let
We dene
f: α→α+1
f (0) = α,
α ≥ ω.
Need to show that
α + 1 is bijective
as follows:
f (n) = n − 1
if
0 < n ∈ ω,
and otherwise
f (ξ) = ξ (w ≤ ξ < α).
This is like in Hilbert's Hotel.
α, β ∈ On .
Let
Dene an ordering
<
on
{0} × α ∪ {1} × β
by


 x, y ∈ {0} × α and x < y,
x < y ⇔ x, y ∈ {1} × β and x < y,


x ∈ {0} × α and y ∈ {1} × β.
Claim. < is a well ordering.
Proof.
Let
Otherwise
∅ 6= S ⊆ {0} × α ∪ {1} × β . If S ∩ {0} × α 6= ∅ then min(S ∩ {0} × α)
S ⊆ {1} × β . Hence min(S) exists as β is well ordered.
Now dene
αuβ
as the unique ordinal isomorphic to
({0} × α ∪ {1} × β, <).
is
min(S).
Note that:
• u on nitie ordinals coincides with usual addition.
• (α u β) u γ = α u (β u γ), but not always α u β 6= β u α.
partially ordered set
chain
upper bound
maximal element
partially ordered set is a pair (X, ≤) where X is a set and ≤ is a partial ordering on X . A subset
is a chain if (Y, ≤) is an ordered set.
An element x ∈ X is an upper bound for Y ⊆ X if ∀y ∈ Y, y ≤ x. A maximal element in X is an
A
Y ⊆X
element
x∈X
such that
∀y ∈ X, x ≤ y → x = y .
Theorem 1.10. The following are equivalent:
AC ∀X((y ∈ X → y 6= ∅) → ∃ choice function f : X → ∪X). (Axiom of choice)
ZL For all partially ordered set (A, ≤) for which
∅ 6= A
and ∀Y ⊆ A, Y
is a chain
→∃
upper bound
a∈A
for
Y,
then A has a maximal element. (Zorn's lemma)
WO Every set can be well ordered. (Well ordering principle)
Proof will come later.
1.3
Cardinals of sets
From now on we assume the axiom of choice.
Proposition 1.11. for any set X there exists a unique cardinal λ such that there exists a bijection
i : X → λ.
8
Proof. X
i: X → λ
λ (by previous theorem). Choose
λ is a cardinal:
0
0
Assume it isn't, then there exists an ordinal λ < λ such that ∃j : λ → λ a bijection. Then
−1
0
j ◦ i : X → λ is a bijection, a contradiction since λ is the smallest such ordinal.
Uniqueness is a consequence of the fact that is κ, λ are cardinals that are in bijection then κ = λ.
Indeed if κ < λ then λ cannot be a cardinal.
can be well ordered, so
smallest such
λ
λ
The cardinal
for some ordinal
(ordinals are well ordered), then
as above is called the
cardinality
of
X.
It is denoted
cardinality
|X|.
|X|
Proposition 1.12. Let X, Y be non-empty sets. Then The following are equivalent:
(i) |X| ≤ |Y |,
(ii) ∃ injection i : X → Y ,
(iii) ∃ surjection j : Y → X .
(Write
'
A→B
to say that
A
and
B
'
A→B
are in bijection.)
Proof.
⇒
(i)
(ii)
⇒
(ii)
(iii)
'
∃i : X → |X|, as |X| ≤ |Y |, |X| ⊆ |Y | and hence there is an injection j : |X| → |Y |.
'
is a bijection h : |Y | → Y . Now h ◦ j ◦ i : X → Y is injective.
Let x0 ∈ W be arbitrary and let f : X → Y be a surjective map. Dene
(
f −1 (y) if y ∈ f (X)
g(y) =
x0
otherwise.
There
f is injective, g is well dened and is surjective.
j : Y → X be surjective. Then there exists a surjection f : |Y | → |X|. Let x ∈ |X|
−1
and take g(x) = min(f
(x)). So we have a well dened map g : |X| → |Y | that is injective
−1
−1
(f
(x) ∩ f (y) 6= ∅ if x 6= y ). Let U ⊆ |Y | be the image of this map g . Then U is a woset
and so is order isomorphic to a unique ordinal α. As |X| is a cardinal, |X| ≤ α.
Lemma 1.13. Let λ ∈ On , U ⊆ λ, α ∼
= U , α ∈ On . Then α ≤ λ.
because
(iii)
⇒
(i) Let
Proof.
Suppose
λ<α
i(x) ∈ U 0 ,
λ is isomorphic
i: λ → U0 $ U.
then
some order isomorphism
to and initial segment of
Let
x = min(λ − U 0 ).
α
and so there exists
Then
i(x) < x
a contradiction.
since
Now we can prove the trichotomy principle:
∀X∀Y (∃X ,→ Y ∨ ∃Y ,→ X).
Proof.
Either
1.4
Hierarchy of Cardinals
|X| ≤ |Y |,
and then
∃X ,→ Y ,
or
|Y | ≤ |X|
and then
∃Y ,→ X .
Lemma 1.14. For any cardinal κ there exists a cardinal λ such that λ % κ.
Proof.
Let
2κ = |P(κ)|.
There is no surjection
κ → P(κ),
hence we don't have
κ
2 > κ.
As
On
and so
For any cardinal
κ let κ+
denote the smallest cardinal such that
sup(U ) ≥ α
for all
κ+ > κ.
Let
U
be a set of ordinals.
α ∈ On − U . Let sup(U ) denote the smallest ordinal with the
α ∈ U . We dene inductively, for every ordinal α ∈ On a cardinal ωα
is not a set, there is some
property
κ ≥ |P(κ)|
as follows:
ω0 = ω,
9
κ+
sup(U )
and for
β < α,
(
ωβ+
ωα =
supβ<α (ωβ )
Claim.
Proof.
There exists a unique way to attach to each
α = β + 1,
if
otherwise.
α
an
ωα
such that the above holds.
Both uniqueness and existence by usual transnite induction.
2 Languages and structures
Denition.
(i)
(ii)
(iii)
rst order language L is the following data:
+
a set F , called the function symbols, and their arities nf ∈ N
+
a set R called the relation symbols, nR ∈ N for all R ∈ R,
a set C , called the constant symbols.
A
We also assume that
for all
f ∈ F,
F ∩ C = F ∩ R = R ∩ C = ∅.
Examples are:
(i) The language of rings
Lring :
F = {+, −, ·},
n+ = n− = n· = 2,
R = ∅,
C = {0, 1}.
(ii) The language of orders,
Lor :
F = ∅,
R = {<},
n< = 2,
C = ∅.
(iii) The language of sets
Lsets :
R = {∈},
n∈ = 2,
F = {P(·)},
nP(·) = 1.
Denition.
Let
L
be a language. An
L-structure M
is the following:
M , called the universe /domain /underlying set
a function f
: M nf → M for all f ∈ F ,
M
a relation R
⊆ M nR for all R ∈ R,
M
an element c
∈ M for all c ∈ C .
(i) a non-empty set
(ii)
(iii)
(iv)
We denote this by
M,
M = (M, f M , RM , cM ).
For example, for all rings there is an
Denition.
Let
respectively. An
(i)
of
M
Lring -structure.
L be a language and let M and N be L-structures with universes M
L-embedding η : M → N is an injection η : M → N such that
η(f M (a1 , . . . , anf )) = f N (η(a1 ), . . . , η(anf ))
10
for all
f ∈F
and
a1 , a2 , . . . , anf ∈ M,
and
N
(ii)
(iii)
(a1 , a2 , . . . , anR ) ∈ RM iff (η(a1 ), . . . , η(anR )) ∈ RN
η(cM ) = cN for all c ∈ C .
L-embedding is called
L-embedding, then we say that M
A bijective
Denition.
The
symbols
an
L-isomorphism.
is a
substructure
(syntax) of the language
of
L
for all
If
N
R ∈ R,
M ⊆ N and the inclusion
N is an extension of M.
map is an
or
are the following:
F ∪ R ∪ C,
v1 , v2 , . . . , vn , . . . ,
equation symbol =,
boolean operators ∨, ∧, ¬,
quantiers ∀ and ∃,
parentheses ( and ),
(i) the elements of
(ii) variable symbols
(iii)
(iv)
(v)
(vi)
(vii) comma ,.
Denition.
The set
Denition.
The set of
(i)
(ii)
(iii)
L∗
called
the words
L-terms
Dene
T0
L
consits of all nite strings of
is the smallest subset
c ∈ T for all c ∈ C ,
vi ∈ T for all i ∈ N+ ,
if t1 , t2 , . . . , tnf ∈ T and f ∈ F
Denition.
of
then
T
of
L∗
L.
Clearly
L ⊆ L∗ .
such that
f (t1 , t2 , . . . , tnf ) ∈ T .
as
T0 := {c | c ∈ C} ∪ {vi | i ∈ N+ }.
By induction on
i,
we dene for all
i≥0
Ti+1 = Ti ∪ {f (t1 , . . . , tnf ) | f ∈ F
Claim. T =
Proof.
S
i∈N
t1 , . . . , tnf ∈ Ti }.
Ti .
By induction on
i
we get
Ti ⊆ T
for all
i
satises the conditions (i)(iii) in the denition of
For example,
and
·(+(a, b), c)
and therefore
T
and hence
S
i∈N
S Ti
⊆ T.
i∈N Ti .
T ⊆
The set
is a term in the language of rings. We will simplify it as
S
i∈N
Ti
(a + b) · c
and
similarly we will simplify all terms. That is, we will not write all the parentheses and use the inx
notation for binary functions in
Denition
Lring .
(Interpretation of terms)
.
Let
M
L-structure and let ū = (u1 , . . . , un ) be a
ā = (a1 , a2 , . . . , an ) ∈ M n , an L-term s
M
an element of ū, we dene s (ā) as follows:
be an
nite list of variable symbols without repetition. Given an
such that every variable (symbol) appearing in
s
s
s
(ii)
is a constant symbol
This is a recursive denition.
i∈N
such that
Claim.
Nn
is
M
c, then s (ā) = c ,
M
if
is the variable ui (i = 1, . . . , n) then s (ā) = ai ,
if
is the term f (t1 , . . . , tnf ) where f is a function symbol
M
M M
then s (ā) = f
(t1 (ā), . . . , tM
nf (ā)).
(i) if
(iii)
s
M
then
Proof.
of
L
and
t1 , t2 . . . , tnf
are terms,
It can be seen that it is well-dened by considering the smallest
s ∈ Ti .
Let s, ū be as above and let η : M → N be an L-embedding.
sN (η(ā)) is dened and sN (η(ā)) = η(sM (ā)).
By induction on depth, that is minimal
i
such that
11
s ∈ Ti .
If
η(ā) = (η(a1 ), η(a2 ), . . . , η(an )) ∈
2.1
Formulas and satisfaction
Denition.
(i)
(ii)
Let
L
be a language. We say that
t1 = t2 , where t1 , t2 ∈ T ,
R(t1 , . . . , tnR ) where R ∈ R
Denition.
and
L-formulas
The set of
φ ∈ L∗
t1 , t2 , . . . , tnf
is an
atomic L-formula
if
φ
is either:
are terms.
is the smallest subset
W
of
L∗
containing all atomic formulas
such that
ϕ ∈ W then ¬(ϕ) ∈ W ,
ϕ, ψ ∈ W then (ϕ) ∨ (ψ), (ϕ) ∧ (ψ) ∈ W ,
ϕ ∈ W then ∃x(ϕ) ∈ W and ∀x(ϕ) ∈ W .
(i) if
(ii) if
(iii) if
Claim.
[Formula induction] Let
if it is true for formulas
variables in
Proof.
Let
x.
W0
Then
Φ
ϕ, ψ
Φ
be a claim which is true for atomic formulas and such that,
then it holds for
(ϕ) ∨ (ϕ), (ϕ) ∧ (ψ), ¬(ϕ), ∃x(ϕ)
and
∀x(ϕ)
for all
is true for all formulas.
be the set of all atomic formulas. For all
i > 0,
we dene by recursion
Wi+1 := Wi ∪ {(ϕ) ∨ (ψ), (ϕ) ∧ (ψ), ¬(ϕ), ∃x(ϕ), ∀x(ϕ) | x a variable}.
S
W = i∈N Wi . Formula induction now follows from induction on i.
Note that
We now introduce abbreviations:
abbreviation
stands for
ϕ→ψ
ϕ↔ψ
ϕ∧ψ
∀x(ϕ)
¬ϕ ∨ ψ
(ϕ → ψ) ∧ (ψ → ϕ)
¬(¬(ϕ) ∨ ¬(ψ))
¬(∃x(¬ϕ))
These make sense because the formulas are logically equivalent. So we may drop the cases
and
∀x(ϕ)
Denition.
For all
ϕ∈W
we dene the set
ϕ is an atomic formula, let V (ϕ)
V (ϕ ∨ ψ) = V (ϕ) ∪ V (ψ) ,
V (¬ϕ) = V (ϕ),
V (∃xϕ) = V (ϕ) − {x}.
(i) if
(ii)
(iii)
(iv)
Claim.
Proof.
ϕ∧ψ
when we prove something using formula induction.
The set
V (ϕ)
V (ϕ)
free variables
of
of
ϕ
recursively as follows:
be the set of variable symbols in
ϕ,
is well-dened.
By formula induction. What we need is that in the induction step, we can tell in which
case (ii)(iv) we are in.
Denition.
A variable
v
appearing in the formula
ϕ
is
bound
if it appears below a quantier in
the formation tree.
A variable can be both free and bound, for example:
Denition
variables
.
(Truth of formulas)
ū = (u1 , u2 , . . . , un ).
Let
∃x(¬(x = 0)) ∧ (x · 0 = 0).
Let M be an L-structure and let ϕ be a formula with free
ā = (a1 , a2 , . . . , an ) ∈ M n . We recursively dene M |= ϕ(ā) as
follows:
(i) if
(ii) if
(iii) if
(iv) if
ϕ
ϕ
ϕ
ϕ
is
is
is
is
M
t1 = t2 then M |= ϕ(ā) iff tM
1 (ā) = t2 (ā),
M
M
R(t1 , . . . , tnR ) then M |= ϕ(ā) iff (tM
1 (ā), . . . , tnR (ā)) ∈ R ,
¬ψ then M |= ϕ(ā) iff M 6|= ψ(ā), i.e. M |= ψ(ā) does not hold,
ψ ∨ θ then M |= ϕ(ā) iff either M |= ψ(ā) or M |= θ(ā),
12
ϕ is ∃x(ψ)
x∈
/ V (ϕ)).
(v) if
M |= ϕ(ā)
then
M |= ϕ(ā)
iff there exists
b ∈ M
is well-dened, that is, it is a unique relation for all
Denition.
ϕ∈W
Formula
is a
sentence
if
V (ϕ) = ∅.
M |= ψ(ā, b)
such that
(note that
L-structures M.
In this case we write
M |= ϕ
instead of
M |= ϕ(ā).
In words: M satises
ϕ(ā)
M.
or ϕ(ā) is true in
Denition.
symbols
An L-theory is a set of L-sentences.
M |= T ) if M |= ϕ for all ϕ ∈ T .
Denition.
The full theory of an
is denoted by
Denition.
L-structure
M
We say that
is a model of the theory
is the set of all sentences
ϕ
T
(in
M |= ϕ.
It
M ≡ N)
if
such that
Th(M).
Two
L-structures M
and
N
elementary equivalent
are
(in symbols
Th(M) = Th(N ).
Theorem 2.1. Suppose there exists an L-isomorphism j : M → N . Then M ≡ N .
Proof by formula induction.
Let
ϕ
V (ϕ) = (u1 , . . . , un ), ā = (a1 , . . . , an ) ∈ M n .
be a formula,
M |= ϕ(ā) ⇔ N |= ϕ(j(ā))
and
2.2
j(ā) = (j(a1 ), . . . , j(an )).
This is true for atomic formulas.
Examples of theories
Complete Orderings L< :
∀x¬(x < x),
R = {<}, n< = 2.
Axioms:
∀x∀y∀z((x < y ∧ y < z) → x < z),
Dense orderings without endpoints
(DLO)
L
same as above with added axioms:
∀x∀y(x < y → ∃z(x < z ∧ z < y)),
Abelian groups F = {+}, R = ∅, C = {0}, n+ = 2.
∀x(∃y(x < y) ∧ ∃z(z < x)).
Axioms:
∀x(0 + x = x),
∀x∀y∀z(x + (y + z) = (x + y) + z),
∀x∀y(x + y = y + x),
∀x∃y(x + y = 0).
Torsion-free divisible abelian groups
For all
∀x∀y(x < z ∨ x = y ∨ y < x).
0<n∈N
(DAG)
L
as in Abelian groups with:
two axioms:
∀x((x + · · · + x = 0) → x = 0),
| {z }
n times
∀x∃y(y + · · · + y = x)
| {z }
n times
and
∃x(¬(x = 0)).
For example,
Q
is a model of this theory.
Rings and Fields Lring = {+, −, · · · , 0, 1}.
guage
{0, +}
Axioms for rings are as for abelian groups in lan-
with:
∀x∀z(x − y = z ↔ x = y + z)
∀x(x · 0 = 0)
∀x∀y∀z(x(yz) = (xy)z)
∀x(x · 1 = 1 · x = x)
∀x∀y∀z(x(y + z) = xy + yz)
∀x∀y∀z((x + y)z = xz + yz)
13
elementary equivalent
And for elds add:
∀x∀y(xy = yx)
∀x(x 6= 0 → ∃y(x · y = x))
Example Theories:
Z
is a ring,
Algebraically closed elds
Q, R, C
are elds.
(ACF) Field axioms with, for all
n∈N
such that
n > 0:
∀a0 . . . ∀an−1 ∃x(xn + an−1 xn−1 + · · · + a0 = 0)
Example:
C.
Ordered elds Lor = Lring ∪ L< .
orderings
L<
Axioms are those of elds in
Lring
plus those of complete
plus two additional axioms:
∀x∀y∀z((x < z) → (x + z < y + z)) ∀x∀y∀z((x < y ∧ z > 0) → (xz < yz))
Models:
R
and
Denition.
such that
2.3
Let T
M |= T .
Q
for example.
be an
L-theory.
We say that
T
is satisable if there exists an
L-structure M
Gödel's Compactness theorem
Theorem 2.2
. T is satisable if and only if for all nite subsets S of T , S is satisable.
(Gödel)
Proof later, remark that (⇒) is trivial.
Denition.
Claim. X
Let
≡U
x
Let
I
[x]R .
ai ≡U
i∈I
When reference to the ultralter
Claim. ≡U
Q {Xi | i ∈ I} a collection
i∈I Xi dened by
be an index set and
be the binary relation on
Y
Proof.
is denoted
is a disjoint union of equivalence classes.
Denition.
I.
Equivalence class of
Y
bi
⇔
of sets. Let
U
be an ultralter on
{i ∈ I | ai = bi } ∈ U.
i∈I
U
is clear we just write
≡.
is an equivalence relation.
(All products and set denitions are over
i ∈ I .)
• I ∈ U so ≡ is symetric.
• ≡ is obviously reexive.
Q Q Q
Q
• Let ai , bi , ci ∈ Xi . Then {ai = ci } ⊇ {ai = bi } ∩ {bi = ci }.
Q
Q
If ai ≡ bi then {ai = bi } ∈ U .
Q
Q
If bi ≡ ci then {bi = ci } ∈ U .
Hence {ai = bi } ∩ {bi = ci } ∈ U and so {ai = ci } ∈ U (both since U
Q
Q
ai ≡ ci .
Note that the proof didn't use that
ultraproduct
U
was an
is a lter).
Thus
ultra lter.
Denition.
Let {Mi | i ∈ I} be a collection of L-structures. We dene the L-structure M =
Q
( i∈I Mi )/U , called the ultraproduct of Mi (w.r.t. U ) as follows:
Q
Q
(i) the underlying set of ( i∈I Mi )/U is M = ( i∈I Mi )/U ,
Q
M
Mi
(ii) for all constant symbols c ∈ L we dene c
as the equivalence class of
,
i∈I c
14
L and equivalence classes ā1 , . . . , ānR ∈ M represented by
Q
Q Rn∈
nR
M
1
1
2
R
,
we have (ā1 , . . . , ānR ) ∈ R
if {i ∈ I | (ai , . . . , ai ) ∈
a
,
a
,
.
.
.
,
a
i∈I i
i∈I i
i∈I i
Mi
R } ∈ U,
for all function symbols f¯ ∈ L and equivalence classes ā1 , . . . , ānf ∈ M represented by
Q
Q
Q
n
a1 , . . . ,
a f ∈ i∈I Mi we dene f M (ā1 , . . . , ānf ) to be the equivalence class of
Qi∈I iMi 1 i∈I ni f
(ai , . . . , ai ).
i∈I f
(iii) for every relation symbol
Q
(iv)
Claim. M is well-dened.
Proof.
(a)
(b)
The claims
(ā1 , . . . , ānR ) ∈ RM ,
f M (ā1 , . . . , ānf ) ∈ M ,
are independent of the choice of representatives. Take another set of representatives
and the set
Q
i∈I
b1i , . . . ,
Q
i∈I
nR or nf
V =
\
{i ∈ I | aki = bki } ∈ U.
k=1
V ∩ {i ∈ I | (a11 , . . . , ani R ) ∈ RMi } = V ∩ A = V ∩ {i ∈ I | (b1i , . . . , bni R ) ∈ RMi } = V ∩ B .
have A ∈ U iff A ∩ V ∈ U iff B ∩ V ∈ U iff B ∈ U and therefore (a). Now for all i ∈ V , we
Then
We
have
n
n
f Mi (a1i , . . . , ai f ) = f Mi (b1i , . . . , bi f )
and so
Y
f Mi (a1i , . . . ) ≡U
i∈I
Y
f Mi (b1i , . . . )
i∈I
and therefore (b).
ϕQ∈ W with free
ū = (u1 , . . . , un ). Let ā = (a1 , . . . , an ) ∈ M n . Choose
Q variables
1
n
1
n
n
tives
i∈I ai , . . . ,
i∈I ai for a1 , a2 , . . . , an respectively. Let a¯i = (ai , . . . , ai ) ∈ Mi .
Q
Theorem 2.3 (Šo± Lemma). We have ( i∈I Mi )/U |= ϕ(ā) if and only if
Let
representa-
{i ∈ I | Mi |= ϕ(a¯i )} ∈ U.
Proof.
•
By formula induction.
If
ϕ
is an atomic formula
t1 = t2
we have
M
M |= ϕ(ā) ⇔ tM
1 (ā) = t2 (ā)
Mi
i
⇔ {i ∈ I | tM
1 (a¯i ) = t2 (a¯i )} ∈ U
⇔ {i ∈ I | Mi |= ϕ(a¯i )} ∈ U.
If
ϕ
is the atomic formula
R(t1 , . . . , tnR )
then
Mi
Mi
i
M |= ϕ(ā) ⇔ {i ∈ I | (tM
}∈U
1 (a¯i ), . . . , tnR (a¯i )) ∈ R
⇔ {i ∈ I | Mi 6|= ϕ(a¯i )} ∈ U.
•
If
ϕ
is the formula
¬ψ
then
M |= ¬ψ(ā) ⇔ M 6|= ψ(ā)
⇔ {i ∈ I | Mi |= ψ(a¯i )} ∈
/U
⇔ {i ∈ I | Mi 6|= ψ(a¯i )} ∈ U
⇔ {i ∈ I | Mi |= ¬ψ(a¯i )} ∈ U
⇔ {i ∈ I | M |= ϕ(a¯i )} ∈ U.
15
[by induction]
[as
U
is an ultralter]
bki
•
If
ϕ
is the formula
ψ∨θ
then
M |= (ψ ∨ θ)(ā) ⇔ M |= ψ(ā)
or
M |= θ(ā)
⇔ {i ∈ I | Mi |= ψ(a¯i )} ∈ U
or
{i ∈ I | Mi |= θ(a¯i )} ∈ U
⇔ {i ∈ I | Mi |= ψ(a¯i )} ∪ {i ∈ I | Mi |= θ(a¯i )} ∈ U
⇔ {i ∈ I | Mi |= (ψ ∨ θ)(a¯i )} ∈ U.
•
If
ϕ
is the formula
∃xψ
then
M |= ∃xψ(ā) ⇔ exists b ∈ M such that M |= ψ(ā, b)
Y
Y
⇔ exists
bi ∈
Mi such that {i ∈ I | Mi |= ψ(a¯i , bi )} ∈ U
i∈I
[by induction]
i∈I
⇔ {i ∈ I | Mi |= ∃xψ(a¯i )} ∈ U.
Corollary 2.4. If ϕ is a sentence, then M |= ϕ if and only if {i ∈ I | Mi |= ϕ} ∈ U .
Denition.
For every set
I
let
I <ω = {S ⊆ I | |S| < ω}.
Theorem 2.5. Let T be a theory. For every set S ∈ T <ω has a model MS .
Recall Gödel's compactness theorem:
A theory
T
is satisable if and only if for all nite subsets
Proof of Gödel's compactness theorem.
S
of
T, S
is satisable.
S ⊆ T let S have a model MS |= S . Let
ϕ ∈ T let ϕ̂ = {S ∈ I | ϕ ∈ S} ⊆ I (all
nite sub-theories of T in which ϕ appears). Take F = {ϕ̂ ⊆ I | ϕ ∈ T } ⊆ P(I). Then F
has the nite intersection property: If ϕ
ˆ1 , . . . , ϕˆn ∈ F then {ϕ1 , . . . , ϕn } is a nite sub-theory
of T in which each of ϕ1 , . . . , ϕn appears, hence {ϕ1 , . . . , ϕn } ∈ ϕ̂i for all 1 ≤ i ≤ n, and so
Tn
Q
{ϕ1 , . . . , ϕn } ∈ i=1 ϕ̂i . Hence there exists an ultralter U ⊇ F. Dene M = ( i∈I Mi )/U . For
all ϕ ∈ T we have ϕ̂ ∈ F ⊆ U . Then for all i ∈ ϕ̂, Mi |= ϕ since Mi |= i and ϕ ∈ i. Hence
ϕ̂ ⊆ {i ∈ I | Mi |= ϕ} ∈ U and then by the corollary to Šo± lemma, M |= ϕ.
I = T <ω
(all nite sub-theories of
For any nite
T)
and for any
Now an application.
Denition.
A
graph G
is a pair
relation. The elements of
that there is an edge
V
(V, E)
are called
between v
and
w
where
vertices
if
V
is a set and
E ⊆ V × V is a symmetric binary
E are called edges. We say
and the elements of
vEw.
Denition.
Let 0 < k ∈ N, we say that G = (V, E) can be coloured with k colours if there exists
f : V → k such that f (v) 6= f (w) for every v, w ∈ V such that vEw. We call f a
colouring (with k colours).
a function
Denition.
0
E∩V ×V
Clearly if
with
k
0
A
subgraph
of a graph
(V, E)
is another graph
(V 0 , E 0 )
such that
V0 ⊆ V
and
E0 ⊆
.
G0
is a subgraph of
G
and
G
can be coloured with
k
colours, then
G0
can be coloured
colours.
Denition.
Graph
G = (V, E)
is
nite
if
|V | < ω
(and hence
|E| < ω ).
Theorem 2.6 (Erdos and de Bruin). A graph G can be coloured with k colours iff every nite
subgraph of G can be coloured with k colours.
Proof.
Lgraph = {E}, nE = 2, E ∈ R.
xRy → yRx. Let L be the language such that R = {E, R0 , . . . , Rk−1 },
i = 0, . . . , k − 1, F = ∅, C = {cg | g ∈ V } (where G = (V, E)). Let T be
The only if part is trivial. For the oposite direction, dene
The theory of graphs is
nE = 2 and nRi = 1 for
the L-theory consisting of
the following formulas:
16
(o)
(i)
(ii)
cg0 6= cg1 for g0 6= g1 .
cg0 Ecg1 whenever there is an edge between g0 and g1 ,
V
∀x(R0 (x) ∨ R1 (x) ∨ · · · ∨ Rn−1 (x)) ∧ i6=j ¬(Ri (x) ∧ Rj (x)),
each edge has a colour and only
one.
(iii)
If
∀x∀y(xEy →
M |= T
then
V
i∈k
M
(M, E
¬(Ri (x) ∧ Ri (y)),
)
is a graph. If
colouring and under the map
T is
S⊆T
We claim that
g 7→
cM
g ,
neighbours don't have the same colour.
f : M → k such that f (v) = i iff M |= RiM (v) then f
G can be considered as a subgraph of (M, E M ).
is a
satisable. By Gödels compactness theorem, we only need to show that every
S contains (ii) and (iii) (this only makes it
V 0 ⊆ V be the set of vertices g such that cg appears in any of the formulas
S . Then V 0 is nite. Let G0 be the graph (V 0 , E ∩ V 0 × V 0 ). By denition, G0 is a subgraph
G and hence by assumption there exists a colouring f : V 0 → k . Let MS be the following
nite subset
is satisable. We may assume that
harder to satisfy). Let
in
of
L-structure:
• underlying set of MS is V ,
• cM
g = g for all g ∈ V ,
• g0 Eg1 iff g0 , g1 ∈ V 0 and there is an edge between g0 and g1
0
of E ),
• if g ∈ V 0 then g ∈ RiM iff f (g) = i,
• if g ∈
/ V 0 then g ∈ R0M and g ∈
/ RiM for i = 1, . . . , k − 1.
Obviously
(either
G0
or
G by the denition
Ms |= S .
Ambrus Pál:
I think it's the language courses that mess this room up. Not to mention that they
move at the snails pace.
Still needed stu about nite intersection property (FIP) for proof of Gödel's compactness theorem.
Denition. Let I be an index set. Let G ⊆ P(I) be a system of sets. Then G has the nite
intersection property if for every nite collection of elements of G, their intersection is not empty.
Filters have the nite intersection property by denition.
Lemma 2.7. Let G ⊆ P(I) be a system of sets having the FIP. Then there exists an ultralter
U ⊆ P(I) such that G ⊆ U .
Proof.
G0 ⊆ P(I) having the FIP and G0 ⊇ G. Then A is partially ordered
with respect to inclusion. If C ⊆ A is a chain, then ∪C ∈ A and therefore every chain has an
upper bound and hence by Zorn's lemma A has a maximal element U ∈ A. We claim that U is an
0
0
ultralter. We have ∩∅ = ∅ and therefore ∅ ∈ U . If x, y ∈ U , take U = {x ∩ y} ∪ U : if U does
not have the FIP then there exist x1 , . . . , xn ∈ U such that x1 ∩ · · · ∩ xn ∩ x ∩ y = ∅ and therefore
U does not have the FIP as x1 , . . . , xn , x, y ∈ U . Therefore U 0 has the FIP and so U 0 ⊆ U as U is
0
0
maximal. Therefore x ∩ y ∈ U . If x ∈ U , take y ⊇ x and U = {y} ∪ U . If U does not have the
FIP, then there exist x1 , . . . , xn such that x1 ∩ · · · xn ∩ y = ∅ and so also x1 ∩ · · · ∩ xn ∩ x = ∅,
0
0
a contradiction as U has the FIP. Therefore y ∈ U . So U is a lter. If U ⊇ U and U is a lter,
0
0
0
then U ∈ A and so U ⊆ U and so U = U and U is an ultralter.
Let
A
Denition.
be the set of all
We will call
|M |
the
cardinality of the L-structure M.
Theorem 2.8. Let T be an L-theory. Assume that T has an innite model. Then for every
cardinal κ there exists a model M of T such that |M | ≥ κ.
Consider the formula
∃x1 ∃x2 . . . ∃xn (
^
¬(xi = xj ) ∧ ∀y(
i6=j
n
_
l=1
17
(y = xi ))),
saying I have exactly
Proof.
L0 be the
L0 theory
Let
Dene an
n
elements.
language
L ∪ {cα | α ∈ κ}
T 0 = T ∪ {cα 6= cβ | for
all
where
cα
α, β ∈ κ
is a new constant symbol for all
such that
α ∈ κ.
α 6= β}.
T 0 has a model. We prove the latter by using compactness by showing
0
that every nite subset S ⊆ T has a model. Let I ⊆ κ be the set of all α ∈ κ such that cα appears
in any formula S . Pick an innite model M of T . Then there is a map f : κ → M such that f |I
0
M
is injective. We make M an L -structure by setting cα = f (α) for all α ∈ κ. Then M |= S .
It is enough to show that
Denition.
Let η : M → N be an L-embedding. We say that η is an elementary embedding
M |= ϕ(ā) iff N |= ϕ(η(â)) for every ϕ ∈ W with free variables (u1 , u2 , . . . , un ) and ā =
(a1 , a2 , . . . , an ) ∈ M n . If M is a substructure of N and the inclusion map is an elementary
embedding, then we write M ≺ N and we say that M is an elementary substructure of N and N
is an elementary extension of M.
if
If there exists an elementary embedding
substructure, we do not get
η:M→N
then
M ≡ N.
But if
M≡N
and
M
is a
M ≺ N.
Denition.
Let M be an L-structure and let LM = L ∪ {cm | m ∈ M } be a language we get from
L by adding a new constant symbol cm for all m ∈ M . Then M is automatically an LM -structure
M
where cm = m for all m ∈ M . The atomic diagram Diag0 (M) of M is the set of all variable free
sentences ϕ of LM such that M |= ϕ. The elementary diagram Diag(M) of M is the set of all
sentences ϕ of LM such that M |= ϕ. Obviously Diag0 (M) ⊆ Diag(M).
Denition.
with
k
We say that a graph
G is k -chromatic for some positive integer k
k − 1 colours.
if it can be coloured
colours and cannot be coloured with
k , let Gk be the graph whose vertices is the set 2k + 1 and for any
i, j ∈ 2k + 1, there is an edge between them iff |i − j| = 1 or |i − j| = 2k . Show that Gk is
3-chromatic.
Let U be an ultralter on the set of positive integers which does not contain nite sets (nonQ
principal ultralter). Prove that 0<k<ω Gk /U is 2-chromatic. Note that all vertices of Gk
have degree 2. We can write this in the language of graphs:
(i) For every positive integer
(ii)
ϕ := ∀x∃y0 ∃y1 (xEy0 ∧ xEya ∧ x 6= y0 ∧ x 6= y1 ∧ y0 6= y1 ∧ ∀x(xEz → z = y1 ∨ z = y1 )).
Q
We know that Gk |= ϕ and so by Šo± lemma,
0<k<ω Gk /U |= ϕ. The following formula
say that I am not a circle of length at most n:
^
^
xi Exi+1 ) → ¬(x0 Exn−1 )
θn := ∀x0 ∀x1 ∀x2 · · · ∀xn−1 ( xi 6= xj ) ∧ (
i∈n−1
i6=j
Then
GQ
k |= θn if k > n.
Gk /U |= θn .
If
U 3 {k ∈ ω | Gk |= θn } ⊇ {k ∈ ω | k > n} ∈ U
and so by ŠLo±
lemma
Lemma 2.9. Let N be an L-structure such that N |= Diag0 (M). Then there is the L-embedding
η : N → M.
Proof.
Dene
η: M → N
as
η(m) = cN
m
m ∈ M . For all n, m ∈ M such that m 6= n, we have M |= ¬(cm = cn ). Therefore
N
N |= ¬(cm = cn ) and so cN
m 6= cn and so η(m) 6= η(n).
n
M
Let f ∈ F of L and m = (m1 , . . . , mnf ) ∈ M f , f
(m1 , m2 , . . . , mnf ) = m0 and thereN
fore M |= cm0 = f (cm1 , . . . , cmn ) and therefore N |= cm0 = f (cm1 , . . . , cmn ) and so cm =
0
f
f
N N
N
f (cm1 , . . . , cmn ).
for all
f
18
R ∈ R of L and m = (m1 , . . . , mnR ) ∈ M nR then (m1 , . . . , mnR ) ∈ RM iff M |= R(cm1 , . . . , cmnR )
N
iff N |= R(cm1 , . . . , cmn ) iff (η(m1 ), . . . , η(mnR )) ∈ R .
R
If
If
c∈C
in
L,
m = cM , M |= cm = c
then for
N |= cm = c
and so
and so
η(m) = cN .
Proposition 2.10. Let N be an LM -structure such that N |= Diag(M). Then there is an
elementary embedding η : M → N .
Proof.
N |= Diag0 (M) which is a subset of Diag(M), for which we constructed an
η : M → N . It is enough to show that this η is an elementary embedding.
n
0
Let ϕ be a formula of L with V (ϕ) = (u1 , . . . , un ) and let ā = (a1 , . . . , an ) ∈ M . Let ϕ be the
formula of LM we get from ϕ by writing cai for ui for all i. V (ϕ) is meaningful when we consider
ϕ as a formula in L or LM . If I consider ϕ as an LM formula, still V (ϕ) = (u1 , . . . , un ). Clearly
ϕ0 has no free variables and so is a sentence in LM .
We know that
embedding
Lemma 2.11. We have N |= ϕ(ā) as an L-structure iff N |= ϕ0 (as an LM -structure).
Proof.
By routine formula induction.
Now the proposition follows by applying the lemma both to
M |= ϕ0
iff
N |= ϕ0
iff
M
and
N:
We have
M |= ϕ(ā)
N |= ϕ(ā).
iff
Theorem 2.12 (Going up Lowenheim-Skolem Theorem). Let M be an innite L-structure. Then
for every cardinal κ ≥ ω there exists an L-structure N such that there exists an elementary
embedding η : M → N and |N | ≥ κ.
Proof.
an
LM
It is enough to construct an
LM
structure
theory which has an innite model
M
N
and so
such that
Diag(M)
N |= Diag(M). But Diag(M) is
|N | ≥ κ.
has a model
Lemma 2.13. Let M be an L-structure, ∅ 6= X ⊆ M and κ = max(|X|, |L|, ω). Then there
exists an L-structure N < M such that X ⊆ N and |N | ≤ κ.
Proof.
Let
N0 := X ∪ {cM | c ∈ C}.
By recursion we dene
Ni+1 := Ni ∪ {f M (x1 , . . . , xnf | xi ∈ Ni
N=
S
and
f ∈ F)}.
Ni .
Claim. |N | ≤ κ.
Set
Proof.
i∈ω
By induction we prove that
|Ni | ≤ κ
and so
N →
Q
Ni
injection and
|N0 | ≤ |X| + |L| ≤ κ.
QQ
nf
nf
nf
The induction step: Ni+1 → Ni
=κ
f ∈F Ni , |Ni | ≤ κ
κ + κ · κ ≤ κ.
|
Q
Ni | ≤ ω · κ ≤ κ:
We have
Construction of
Let
then
cN ∈ N ),
X⊆N
This satises all the desired properties:
(u1 , . . . , un )
|Ni+1 | ≤ |N0 | + |F| · κ ≤
N:
• the underlying set is N ,
• cN := cM for all c ∈ C (if cM ∈ N0
• f N := f M |N nf for all f ∈ F ,
• RM := RM ∩ N nR .
Claim.
and
and
N ⊆ M.
M ⊆ N be L-structures. Then for every atomic formula ϕ ∈ W
ā = (a1 , . . . , an ) ∈ M n we have M |= ϕ(ā) iff N |= ϕ(ā).
with
V (ϕ) =
and
Lemma 2.14 (Tarski-Vaugh
lent:
• M ≺ N;
. Let M ⊆ N be L-structures. The following are equiva-
criterion)
19
• For all ϕ ∈ W with V (ϕ) = (u1 , . . . , un ) and ā = (a1 , . . . , an ) ∈ M n ,
there exists c ∈ N such that N |= ϕ(ā, c)
Proof.
⇐⇒
there exists b ∈ M such that N |= ϕ(ā, b).
M ≺ N . We need to show that if there exists c ∈ N such that N |= ϕ(ā, c)
b ∈ M such that N |= ϕ(ā, b). Assume there is c ∈ N such that N |= ϕ(ā, c)
implies N |= ∃vϕ(ā) and so M |= ∃vϕ(ā) as a1 , . . . , an ∈ M and M ≺ N and so there exists
b ∈ M such that M |= ϕ(ā, b).
Now we assume that the criterion holds and we will show by formula induction that for all ϕ ∈ W
n
with V (ϕ) = (u1 , . . . , un ) and ā = (a1 , . . . , an ) ∈ M we have M |= ϕ(ā) iff N |= ϕ(ā).
Assume rst that
implies there exists
• For atomic formulas this is true because M ⊆ N and by the above claim.
• ϕ is ¬ψ : M |= ϕ(ā) iff M 6|= ψ(ā) iff (by induction) N 6|= ψ(ā) iff N |= ϕ(ā).
• ϕ is ψ ∨ θ: M |= ϕ(ā) iff M |= ψ(ā) or M |= θ(ā) iff (by induction) N |= ψ(ā) or N |= θ(ā)
iff N |= ϕ(ā).
• ϕ is ∃vψ : M |= ϕ(ā) iff there exists b ∈ M such that M |= ψ(ā, b) iff (by induction)
N |= ψ(ā, b) iff (by the criterion) N |= ∃vψ(ā) iff N |= ϕ(ā).
Theorem 2.15 (Going down Lowenheim-Skolem theorem). Let M be an L-structure. Let κ ≥
|L| + ω and let X ⊆ M such that |X| = κ. Then there exists N < M such that X ⊆ N and
|N | = κ.
Proof.
For every
ϕ ∈ W with V (ϕ) = (u1 , . . . , un , v) we say that f : M n → M is a Skolem function
n
for ∃vϕ if for all ā = (a1 , . . . , an ) ∈ M
we have M |= ϕ(ā, f (ā)) if M |= ∃vϕ(ā). Claim.
For
all ϕ and v as above there is a Skolem function f for ∃vϕ.
Proof.
We have a well ordering
(
f (ā) :=
<
on
M
(axiom of choice) and
min{b ∈ M | M |= ϕ(ā, b)}
min(M )
if this set is not empty,
otherwise.
L0 = L ∪ {f∃vϕ } be the language we get from L by adding a new function symbol for all ϕ
0
and v as above, nf∃vϕ = n = #V (ϕ) − 1. Then |L | ≤ |L| + ω(|L| + ω) ≤ |L| + ω . We make M
0
an L -structure by interpreting the new function symbol f∃vϕ by a Skolen function for ∃vϕ, that
M
0
is f∃vϕ = Skolem function for ∃vϕ. By the lemma above there is an L substructure N ⊆ M such
that X ⊆ N and |N | ≤ max(|X|, |L|, ω) = κ and also |N | ≥ |X| = κ and so |N | = κ.
Claim. As an L-structure, N ≺ M.
Let
Proof.
If ϕ ∈ W with V (ϕ) = (u1 , . . . , un ) and ā =
b ∈ M such that M |= ϕ(ā, b), then by the deniM
M
0
tion of the Skolem function, M |= ϕ(ā, f∃vϕ (ā)). We have f∃vϕ (ā) ∈ N as N is an L -substructure
and so there exists c ∈ N such that M |= ϕ(ā, c).
We verify the Tarski-Vaugh criterion.
(a1 , . . . , an ) ∈ N n
such that there exists
This proves the theorem.
Corollary 2.16. Let T be an L-theory, assume that T has an innite model. Then for all cardinals
κ ≥ |L| + ω there is an L-structure M such that M |= T and |M| = κ.
Proof.
subset
exists
We know that there exists an
0
X⊆M
M ≺ M0
such that
such that
such that
M0 |= T
|X| = κ and apply the going down theorem
|M| = κ. As M ≡ M0 we have M |= T .
Denition.
A (satisable)
For example,
Th(M)
Denition.
A sentence
M |= ϕ.
L-structure M0
L-theory T
is
complete
if all models of
T
to
and
X
|M| ≥ κ. Take a
M to get: there
and
are elementarily equivalent.
is complete.
ϕ is a consequence
T |= ϕ.
of a theory
We write this as
20
T
if for every model
M
of
T
we have
We remark that an
Denition.
κ
An
L-theory T
L-theory T
is
is complete iff for every sentence
κ-categorical
for a cardinal
κ
ϕ,
either
T |= ϕ
if any two models of
T |= ¬ϕ.
or
T
of cardinality
are isomorphic.
Proposition 2.17 (Vaugh's test). Let T be a satisable L-theory what is κ-categorical for some
innite cardinal κ ≥ |L| and has only innite models. Then T is complete.
Proof.
M, N be models of T . As |M|, |N | ≥ ω , by the going down theorem there exist M0 , N 0
0
0
0
0
0
such that M ≡ M, N = N , |M | = |N | = κ. We have M |= T and so M |= T and also N |= T
0
0
0
0
0
and so N |= T and so by assumption M ' N (they are isomorphic) and so M ≡ N and
M ≡ N.
Let
Denition.
Dene ACFp
= theory of algebraically closed elds of characteristic p,
where
p
is a
prime, with axioms
•
ACF and
=theory
Dene ACF0
•
1 + 1 + · · · + 1 = 0 (p
ACF and
times).
of algebraically closed elds of characteristic
¬(1 + 1 + · · · + 1 = 0)
any
p
0
with axioms
number of times.
Theorem 2.18. The following theories are complete
(i) dense orderings (DLO),
(ii) divisible torsion-free abelian groups (DAG),
(iii) algebraically closed elds of characteristic p, p = 0 allowed (ACFp ).
Proof.
In each case we use Vaught's test:
ω -categorical: Let M and N be two models of cardinality ω0 . Let M = {m1 , . . . , mk , . . . }
N = {n1 , n2 , . . . , nk , }. The following is called the back and forth argument. By induction, we dene subsets Xn ⊆ M and Yn ⊆ N and bijection fn : Xn → Yn such that
(i) m1 , m2 , . . . , mn ∈ Xn and Xn is nite,
(ii) n1 , n2 , . . . , nn ∈ Yn and Yn is nite,
(iii) fn is an isomorphism, i.e. it is an order-preserving map,
(iv) Xn ⊇ Xm , Yn ⊇ Ym if n ≥ m,
(v) fn |Xm = fm for all n ≥ m.
S
S
Once we have constructed these objects, we are done as M =
n∈ω Xn and N =
n∈ω Yn ,
S
f = n∈ω fk is an isomorphism f : M → N . So take n = 1 and have X1 = {m1 }, Y1 = {n1 }
and f (m1 ) = n1 . For the induction step, assume that Xn , Yn , fn are already constructed.
(i) DLO is
and
Let
Xn0 := Xn ∪ {mn+1 }.
Claim.
Proof.
There exists an embedding
Let
a = mn+1 .
If
a ∈ Xn ,
fn0 : Xn → N
such that
fn0 |Xn = fn .
then there is nothing to show. Otherwise there are three
cases
(i)
(ii)
a<b
b<a
b ∈ Xn ),
b ∈ Xn ),
b0 , b1 ∈ Xn
(for all
(for all
such that b0 < a < b1 and for any b ∈ Xn , b ≤ b0
N |= DLO, there exists c ∈ N such that
c < b for all b ∈ Yn ,
b < c for all b ∈ Yn ,
fn (b0 ) < c and c < fn (b1 )
0
0
respectively. Let fn (b) = c, and so fn has the required properties.
(iii) there exists
Because
Then we get
Yn+1 := fn0 (Xn0 ) ∪ {nn+1 }.
21
or
b1 ≤ b.
Using the same argument we can construct an embedding
(fn0 )−1 .
gn+1
such that
gn+1 |fn0 (Xn0 ) =
Dene
Xn+1 := gn+1 (Yn+1 ) ⊇ Xn0 ⊇ Xn
and take
fn+1 = (gn+1 )−1 .
These satisfy the required properties.
κ-categorical for all κ > ω : If G |= DAG then G is a Q-vector space as for all g ∈ G,
ka/kb = a/bg = h and so k(a/(kb)g) = 1/bg and so |G| ≥ ω .
Remark: G1 , G2 |= DAG, G1 ' G2 as groups iff G1 ' G2 as vector spaces.
a
a
Dene for
b ∈ Q, a, b ∈ Z, b g as the unique h ∈ G such that bh = ag . Hence the Q-vector
space structure is an iso. invariant of G as a group. We need to show that if G1 , G2 are two
Q-vector spaces, |G1 | = |G2 | = κ > ω then G1 ' G2 . Recall that X ⊆ G is a basis of G if
(∗) every nite subset X is linearly independent,
() every element g ∈ G can be written as a linear combination of some nitely many
elements of X .
(ii) DAG is
Lemma 2.19. Every vector space has a basis.
Proof.
Let
X
be a linearly independent set (condition (∗)), which is maximal with
respect to inclusion. There is such a set by Zorn's lemma. If
g ∈ G such that g is not a linear combination
Y = X ∪ {g} is linearly independent, a contradiction.
there exists
GP
has a basis X , then |X| ≤ G ≤ |X| · |Q|
g = xi ∈X λi xi , λi ∈ Q for all i and there is an
If
G→
a
since every
X
is not a basis, then
(in the sense of ()). Then
g ∈ G
can be written as
injection
|Q|S| |
S⊆X
If
X
Conclusion:
(iii)
S is P(X)
hence |G| ≤ ω . If X
` andnso it is
` nite and
`
n
n
n∈ω X and so |
n∈ω X | ≤ |
n∈ω |X || =
n∈w |X| = ω|X|.
a Q-vector space then |G| = κ > ω iff G has a basis X with
is nite, then the index set
is innite, then
if
S→
G is
`
|X| = κ:
⇐ κ ≤ |G| ≤ ω · κ = ω + κ = max(ω, κ) = κ,
⇒ |X| = λ 6= κ and λ ≤ |G| ≤ max(ω, λ), a contradiction.
Therefore if G1 , G2 have bases X1 , X2 with |X1 | = |X2 |, then there is a unique isomor0
0
phism ϕ : G1 → G2 for all bijections ϕ : X1 → X2 such that ϕ |X1 = ϕ.
ACFp is κ-categorical for κ > ω (nite elds are not algebraically closed and therefore ACFp
has no nite models). Let K, L be models of ACFp of cardinality κ. Let K = {aα | α ∈ κ},
L = {bα | α ∈ κ}. It will be sucient to construct Kα ≤ K and Lα ≤ L and fα : Kα → Lα
(for all α ∈ κ) by transnite induction such that
(i) {ap | β < α} ⊆ Kα , |Kα | ≤ |α| + ω ,
(ii) {cp | β < α} ⊆ Lα , |Lα | ≤ |α| + ω ,
(iii) Kβ ⊆ Kα , Lβ ⊆ Lα for all β < α,
(iv) fα |Kβ = fβ for all β < α.
Construction: For α = 0, let K0 and L0 be the prime elds of K and L respectively: Fp if
p > 0 and Q otherwise. Therefore there exists an isomorphism f0 : K0 → L0 , |K0 | = |L0 | ≤ ω
that satises all the required properties.
For the induction step, assume that
α
Kβ , Lβ , fβ
are all constructed for all
β < α.
First case:
is a limit ordinal. Then
Kα =
[
Kβ , Lα =
β<α
[
Lβ
β<α
S
|Kβ |, |Lp | ≤ |α| + ω for all β < α and therefore |Kα |, |Lα | ≤ |α| + ω . Take fα = β<α fβ
and so fα is an isomorphism from Kα to Lα and (Kα , Lα , fα ) satises (i)-(iv): aβ ∈ Kα for
and
22
β < α as there
aβ ∈ Kβ 0 ≤ Kα .
all
The second case:
fβ0 : Kβ0 → L
Proof.
(a)
exists
such that
β < β0,
since
α
is a limit ordinal and therefore
α = β + 1. Let Kβ0 = Kβ (aβ ). Claim.
fβ0 |Kβ = fβ .
There exists embedding
such that
Two cases:
g ∈ Kβ [x] be the monic polynomial which is the minimal
fβ (g) ∈ Lβ [x] is an irreducible polynomial and so there exists
b ∈ L which is a root of fβ (q) and there exists unique fβ0 : Kβ0 → Lβ (b) isomorphism
0
0
such that fβ (aβ ) = b and hence by construction fβ |Kβ = fβ .
aβ
is algebraic over
polynomial for
(b)
β0 < α
aβ .
Kβ .
Let
Then
aβ is transcendental over Kβ . Let Lβ = algebraic closure of Lβ in L. Then |Lβ | ≤
|Lp |+ω ≤ |β|+ω < κ and therefore there exists b ∈ L−Lβ such that b is transcendental
0
0
0
over Lβ and there exists a unique isomorphism fβ : Kβ → Lβ (b) such that fβ (aβ ) = b
0
and fβ |Kβ = fβ .
|Lα | ≤ |Kβ0 | + ω ≤ |α| + ω . Using the same argument
0 −1
. Set
as above we can construct an embedding gα : Lα → K such that gα |f 0 (Kβ ) = (fβ )
β
−1
Kα = gα (Lα ), fα = (gα ) . Hence (iii)-(iv). As aβ ∈ Kα , bβ ∈ Lα we get (i) and (ii).
Let
Lα = fβ0 (Kβ0 )(bβ ).
By construction
Lecturer quote:
This was quite a lot of work to get to this, compactness, going up, going down, but
maybe you don't apreciate how powerful it is.
theorem proving machine.
This theorem is a machine: a
Lemma 2.20. Let T be an L-theory and ϕ be a sentence in L such that T |= ϕ. Then there exists
a nite S ⊆ T such that S |= ϕ.
Proof.
that
S ⊆ T there exists L-structure M such
S ∪ {¬ϕ} is satisable. By compactness, T 0 = T ∪ {¬p} is satisable and
M of T such that M |= ¬p, a contradiction.
Assume for contradiction that for all nite sets
M |= S ∪ {¬ϕ},
i.e.
so there exists a model
Theorem 2.21 (Theorem of Ax). Let ϕ be a sentence in Lring . Then the following are equivalent:
(i) ACF0 |= ϕ,
(ii) C |= ϕ,
(iii) for innitely many prime numbers p, Fp |= ϕ,
(iv) for all but nitely many primes p ACFp |= ϕ.
Proof.
C |= ACF0 . (ii) implies (i) as ACF0 is complete.
|= ϕ then there exists S ⊆ ACF0 such that |S| < ω
N ∈ N such that for every prime p > N , ACFp |= S . This implies
(i) implies (ii) as
For (i) implies (iv), if ACF0
and
S |= ϕ
and
so there exists
that ACFp
|= ϕ
for all
ϕ > N.
For (iv) implies (iii): as
Fp |= ACFp
For (iii) implies (i): Assume ACF0
ACFp
|= ¬ϕ
and hence
6|= ϕ.
Fp |= ϕ
p/
|= ¬ϕ as ACF0 is complete.
Therefore Fp |= ¬ϕ for all but nitely
for all but nitely many
Then ACF0
for all but nitely many primes.
contradiction.
a
Let F be a eld. A polynomial map Φ over F is an n-tuple Φ = (Φ1 , Φ2 , . . . , Φn )
Φi ∈ F ∈ [x1 , x2 , . . . , xn ] for i = 1, 2, . . . , n. Then Φ furnishes a map F n → F n , also
by Φ, given by the rule
such that
Φ(z̄) = (Φ1 (z̄), Φ2 (z̄), . . . , Φn (z̄))
for all
p,
Denition.
denoted
Therefore
many
z̄ = (z1 , z2 , . . . , zn ) ∈ F n ,
23
Theorem 2.22. Let Φ be a polynomial map over C. If Φ : Cn → Cn is injective, then it is
surjective.
Proof. Claim.
Φ
Let
be a polynomial map over
Fg .
If
Φ : Fng → Fng
is injective, then it is
surjective.
Proof.
In fact, any map
Claim.
Φ
Let
Fng → Fng
injective is surjective as
be a polynomial map over
Fp .
If
|Fng | < ω .
n
Φ : Fp → Fp
n
is injective, then it is surjective.
Proof.
We have Φ = (Φ1 , Φ2 , . . . , Φn ) and so there is a power g of p such that Φi ∈ Fg [x1 , x2 , . . . , xn ]
i = 1, 2, . . . , n. By the above, Φ : Fngm → Fngm is surjective for all m ∈ N. Therefore
n
n
Φ : Fp → Fp is also surjective as
[
n
Fp =
Fngm .
for
m∈N
Let
Φ
Φi =
be as above and
P
j
aij x̄j ,
where
J : {1, 2, . . . , n} → N
J(1)
x̄j := x1
Let
|J| > k .
Claim.
We say that
We also say that
Φ = (Φ1 , . . . , Φn )
Φ has n variables.
There exists a sentence
for any polynomial map
it is injective.
Denition.
Φ
with
n
ϕn,k
is in
C
C
of
in
has degree
Lring
≤ k
if
aij = 0
for all
i = 1, . . . , n
and
F we have F |= ϕn,k iff
Φ : F n → F n is surjective, if
such that for every eld
variables and degree
By the theorem of Ax,
A class
L-structure M
nitely axiomatisable
· · · xJ(n)
.
n
|J| = maxnl=1 J(i).
Denition.
axomatisable
J(2)
· x2
is a multiindex and
≤k
the map
C |= ϕn,k .
L-structures is axomatisable
M |= T .
if there is an
L-theory T
such that an
if and only if
T axiomatises the class C . A class C of L-structures is nitely axiomatisable
L-theory T which axiomatises the class C .
In this case we say that
if there is a nite
Lemma 2.23. Assume that the class of models of the L-theory T is nitely axiomatisable. Then
there is a nite subset S ⊆ T which axiomatises the class of the models of T .
Proof.
Let {ϕ1 . . . . , ϕn } axiomatise this class. Then {ϕ} also axiomatises the
ϕ1 ∧ · · · ∧ ϕn . By denition, T |= ϕ. So by compactness there exists nite S ⊆ T
We claim that S axiomatises the class:
•
If M |= T then M |= S : M |= T ⇒ M |= S is easy. For
M |= ϕ ⇒ M |= T (as {ϕ} axiomatises models of T ).
• If M |= S then M |= ϕ.
ϕ =
S |= ϕ.
class where
such that
the reverse implication
M |= S ⇒
Lecture quote:
I have not had enough coee so was doing the proof slowly and this made it harder
to prove.
Theorem 2.24. The following classes are axiomatisable but not nitely axiomatisable.
(i) Torsion free abelian groups.
(ii) Fields of characteristic 0.
(iii) Algebraically closed elds.
Proof.
24
(i) Recall that the axioms
{ϕn | n ∈ ω}
where
ϕn
T
is
for torsion free abelian groups are those of abelian groups and
∀x((x + · · · + x = 0) → x = 0) (n
summands).
T
axiomatises the
class.
S⊆T
I = {n ∈ ω | ϕn ∈ S}, this set is nite. So there is a prime
p > n for all n ∈ I . Then Z/pZ |= S . Let T be the axioms of elds with {ϕp | p a prime}. This axiomatises the class of elds of
characteristic 0, where ϕn is ¬(1 + · · · + 1 = 0) (p summands).
Suppose it is nitely axiomatisable, say nite S ⊆ T , (using lemma) S axiomatises the
class, then take I = {p prime | ϕp ∈ S}, then I is nite. and there is a prime l > p for all
p ∈ I . Then Fl |= S . Assume that the class is nitely axiomatisable. Then by the lemma there is a nite
which axiomatises the class. Let
(ii)
(iii) ACF axiomatises this class. Using same argument as above, it is enough to construct a eld
Kp
p such that:
Kp has no non-trivial nite extensions of degree < p.
(b) Kp has an extension of degree p.
Let Kp = {α ∈ F̄l | p does not divide [Fl (α) : Fl ]} ⊆ F̄l .
Claim. Kp ⊆ F̄l .
for every prime
(a)
Warning: High doses of Galois theory ahead! Lecturer Quote:
If we actually want to prove something big about Algebra with model theory then
the proof often uses some deep Algebra.
Proof of claim.
[Comments in square brackets are my own additions, might be wrong] [Let
α, β ∈ Kp , we want to show that the eld operations between α, β
Fl (α, β).] We have the following diagram of extensions:
keep them in
Kp , so study
Fl (α, β)
II
u
II
u
II
uu
u
II
u
u
I
u
u
Fl (α)
Fl (β)
JJ
t
JJ
t
t
JJ
tt
JJ
tt
JJ
t
tt
Fl
Where Fl (α)|Fl and Fl (β)|Fl are nite Galois. So [by
Fl (α, β)|Fl is nite Galois and there exists an [injective
a theorem of Galois theory course]
group morphism]:
Gal(Fl (α, β)|Fl ) → Gal(Fl (α)|Fl ) × Gal(Fl (β)|Fl )
(1)
γ 7→ γ|Fl (α) × γ|Fl (β)
(2)
γ(Fl (α)) ⊆ Fl (α) and γ(Fl (β)) ⊆ Fl (β).) For some γ ∈ Gal(Fl (α, β)|Fl ),
γ|Fl (α) = id then γ(α) = α.
If γ|Fl (β) = id then γ(β) = β .
Then γ = id as α and β generate the extension Fl (α, β)|Fl . [Hence injectivity.]
So [Fl (α, β) : Fl ] divides [Fl (α) : Fl ][Fl (β) : Fl ] by the fundamental theorem of Galois
(Note that
If
theory
[Galois theory tells us that these dimensions are the sizes of their Galois groups and group
theory says that the injective image is a subgroup that then divides the cardinality of the
α, β are assumed to be in Kp ], p does not divide [Fl (α)Fl ][Fl (β) :
[Fl (α, β) : Fl ] either. Hence all elements of Fl (α, β) are in Kp .
[Indeed if η ∈ Fl (α, β) then [Fl (η) : Fl ] divides [Fl (α, β)|Fl ] (tower law). So p does not divide
[Fl (η) : Fl ] and so η ∈ Kp . In particular this holds for α + β, αβ, α−1 , −α, so Kp is a eld, a
subeld of F̄l .]
group by Lagrange]. [Since
Fl ]
and so does not divide
Claim.
If
K 0 ≤ Kp
is a nite extension of
25
Fl
then
p
does not divide
[K 0 : Fl ].
Proof of claim.
tion
By the primitive element theorem,
0
[K : Fl ] = [Fl (α) : Fl ]
Claim.
If
L ≤ F̄l
Proof of claim.
is not divisible by
is a nite extension of
Kp
K 0 = Fl (α)
for some
α ∈ Kp .
By deni-
p.
and
p
does not divide
[L : Kp ]
then
L = Kp .
By the primitive element theorem, or by induction on the number of gen-
L as an extension of Kp , we may assume that L = Kp (α) for some α ∈ F̄l .
n
Let f (x) = a0 + a1 x + · · · + an x ∈ Kp [x] be a minimal polynomial of α [over Kp ]. Let
0
K = Fl (a0 , . . . , an ) ≤ Kp . Since [K 0 : Fl ] is nite [nitely generated by algebraic elements] it
0
is not divisible by p by the previous claim. Considering f (x) as an element of the ring K [x],
0
0
it is irreducible as it is in Kp [x]. Also p does not divide [Kp (α) : Kp ] = deg(f ) = [K (α) : K ]
0
0
0
[by assumption]. Now, by the tower law, [Fl (α) : Fl ] divides [K (α) : K ][K : Fl ] [Indeed
[Fl (α) : Fl ] divides [K 0 (α) : Fl (α)][Fl (α) : Fl ] = [K 0 (α) : Fl ] = [K 0 (α) : K 0 ][K 0 : Fl ]]. Hence
p does not divide [Fl (α) : Fl ] and so α ∈ Kp and L = Kp (α) = Kp .
erators of
Claim. Kp
Proof.
has an extension of degree
p.
α ∈ F̄l such that [Fl (α) : Fl ] = p. Then [Kp (α) : Kp ] ≤ p:
[Kp (α) : Kp ] < p. Then Kp (α) = Kp by above and α ∈ Kp . There exists
Assume
Now we can nally nish the proof:
such that
[L : Kp ] < p
K̄p = F̄l
and so if there exists a nite extension
then there exists such an extension in
F̄l
L|Kp
aswel.
3 Quantier elimination
Denition.
such that
Let L be a language, ϕ ∈ W , V (ϕ) = ū = (u1 , . . . , un ).
M |= ϕ iff for all ā = (a1 , . . . , an ) ∈ M , M |= ϕ(ā).
Claim.
Let
Proof.
Let
ϕ
ϕ
be a formula such that
M |= ∀xϕ
By denition,
be a formula,
V (ϕ) = ū,
V (ϕ) = (ū, x), M |= ϕ
iff for all
b ∈ M,
for all
iff
Let
M
ϕ,
L-structure
M |= ∀xϕ.
ā ∈ M n , M |= ϕ(ā, b)
then we dene the closure of
be an
denoted
ϕ̄
is
iff
M |= ϕ.
∀u1 ∀u2 · · · ∀un ϕ.
Lemma 3.1. ϕ̄ is a sentence and M |= ϕ̄ iff M |= ϕ.
Proof.
By the above claim with induction on
Remark.
It is important to note that
clearly not
|V (ϕ)|.
¬ϕ 6= ¬(ϕ̄).
For example for
ϕ x = 0,
we have
∀¬(x = 0)
is
¬∀x(x = 0).
Notation.
Lemma 3.2. If ϕ ∈ W is a sentence, Γ ⊆ W and Γ ∪ {ϕ} is not satisable, then Γ |= ¬ϕ.
Proof.
but
Let M |= Γ.
M |= ¬ϕ as ϕ is
If
M |= ¬,
then
M |= Γ ∪ {ϕ},
a sentence.
then
Γ ∪ {ϕ}
is satisable.Therefore
M 6|=,
Lemma 3.3. If ∆ is a theory, Γ ∪ W and Γ ∪ ∆ is not satisable, then there exist nitely many
ϕ1 , . . . , ϕn ∈ ∆ such that Γ |= ¬(ϕ1 ∧ · · · ∧ ϕn ).
Proof. Γ ∪ ∆
is not satisable, so by compactness,
ϕ1 , . . . , ϕn ∈ ∆.
Therefore
Γ ∪ {ϕ1 ∧ · · · ∧ ϕn }
Γ ∪ {ϕ1 , . . . , ϕn } is not satisable for some
Γ |= ¬(ϕ1 ∧ · · · ∧ ϕn } by the
is not satiable and so
previous lemma.
Notation.
Let L be a language, ϕ ∈ W with V (ϕ) = x̄ = (x1 , . . . , xn ) and let c̄ = (c1 , . . . , cn ) be
n-tuple of constant symbols not appearing in L. Let ϕ(c̄) denotethe formula we get from ϕ by
0
writing ci instead of xi for all i = 1, 2, . . . , n. Let L = L ∪ {c1 , . . . , cn }.
an
26
Lemma 3.4. Let Γ be a set of L-formulas and assume that Γ |= ϕ(c̄). Then Γ |= ϕ.
Proof.
L-structure such that M |= Γ and let ā = (a1 , . . . , an ) ∈ M n . Let M0 be the
L −structure we get from M by setting cM
= ai for all i = 1, . . . , n. Then M0 |= Γ and so by
i
0
induction M |= ϕ(c) and therefore M |= ϕ(ā) as our choice of ā was arbitrary. Hence M |= ϕ. Let
M
be an
0
Denition. A formula ϕ ∈ W is open if it has no quantiers. Let T be an L-theory. We say that
T has quantier elimination if for every formula in ϕ ∈ W , there is an open ψ ∈ W such that
V (ϕ) = V (ψ)
and
T |= ϕ ↔ ψ .
For example, take a
(n × n)
matrix
(aij ).
The following says it is invertible:

∃b11 ∃b12 . . . ∃bnn ∀k
n
X


aij bjk = δik 
i,j=1
where
δij
is the Kronecker delta. This is equivalent to Kramer's rule:
det(aij ) 6= 0
iff
X
(−1)sign(π) a1,π(1) 6= 0.
π∈Sn
And this has no universal quantiers, so it's easy to check. The quantier elimination states that
there always exists a universal Kramer's rule.
Lemma 3.5. Let T be an L-theory and assume that for every open formula ϕ ∈ W with V (ϕ) =
(u1 , u2 , . . . , un , x) there is an open formula ψ with V (ψ) = (u1 , . . . , un ) such that T |= ∃xϕ ↔ ψ .
Then T has quantier elimination.
Proof.
We verify the condition for quantier elimination for all
• ϕ
• ϕ
ϕ∈W
by formula induction.
ϕ is open and T |= ϕ ↔ ϕ.
¬ψ : then let θ ∈ W be open such that V (ψ) = V (θ) and T |= ψ ↔ θ. Then ¬θ is open
and V (¬θ) = V (θ) = V (ψ) = V (¬ψ) and T |= ¬ψ ↔ ¬θ .
• ϕ is ψ ∨ θ:
• ϕ is ∃xψ : let ψ 0 ∈ W be open with V (ψ 0 ) = V (ψ) and T |= ψ ↔ ψ 0 . Then V (∃xψ 0 ) =
V (∃xψ) = V (ψ 0 ) − {x} = V (ψ) − {x} and T |= ϕ ↔ ∃xψ .
0
The formula ∃xψ is not open, but by assumption there is an open θ ∈ W such that V (θ) =
0
V (ψ ) − {x} and T |= ∃xψ 0 ↔ θ. Then T |= ϕ ↔ θ and V (θ) = V (ϕ).
is atomic: then
is
Denition.
ϕ∈W
Let
T
be an
L-theory.
T∀ denote the L-theory consisting of all sentences ϕ̄ where
T |= ϕ. Then the theory T∀ is called the (set of ) universal
Let
is an open formula such that
consequences of T .
Lemma 3.6. If M ⊆ N are L-structures and N |= T , then M |= T∀ .
Proof.
If
ϕ
is open such that
Denition.
ϕ̄ ∈ T∀
then
N |= ϕ.
Hence
M |= ϕ
and so
M |= ϕ̄.
L-theory T has algebraically prime models if for every model M of T∀ , there is
L-embedding i : M → N such that for every model N 0 of T and for every
L-embedding g : M → N 0 there is an L-embedding k : N → N 0 such that j = k ◦ i. In this case
we say that M is an algebraically prime extension of M.
a model
N
An
and an
Denition.
Let M ⊆ N be L-structure. We say that M is simply closed in N , in symbol M ≺S
N if for every open formula ϕ ∈ W , V (ϕ) = (u1 , u2 , . . . , un , x) and ā = (a1 , a2 , . . . , an ) ∈ M n if
N |= ∃xϕ(ā) then M |= ∃xϕ(ā).
Theorem 3.7. Let L be a language such that C 6= ∅, let T be an L-theory such that
(i) T has algebraically prime models,
(ii) if M ⊆ N and M |= T , N |= T then M ≺S N .
27
open
Then T has quantier elimination.
Proof.
ϕ ∈ W with V (ϕ) = (u1 , u2 , . . . , un , x), there
ψ ∈ W with V (ψ) = V (ϕ) − X and T |= ∃xϕ ↔ ψ .
0
Let c̄ = (c1 , . . . , cn ) be an n-tuple of constant symbols not in L and L = L ∪ {c1 , . . . , cn }. Let ∆
0
be a set of variable free formulas in L .
It will be enough to show that for every open
is an open
Lemma 3.8. Let M, N be L0 -structures such that M |= T and N |= T . Assume that
(i) M |= ∃xϕ(c̄),
(ii) if ψ∆ and M |= ψ then N |= ψ .
Then N |= ∃ϕ(c̄).
Proof.
The intersection of the underlying sets of all
L0 -substructures
for a given
L0
structure is
not empty, because it contains the interpretation of constant symbols. Hence it is the underlying
set of the smallest
respectively.
Proof.
L0
substructure:
Claim. M
0
and
N
0
Let
M0 , N 0
denote the smallest
L0 -substructure
in
M, N
are isomoprhic.
M 0 = {tM ∈ M | t is a term without variable symbols} and N 0 dened similarly. The
0
0
0
0
M
N
obvious map from M to N is i : M → N dened by i(t ) = t .
The map i is well-dened: if t1 M = t2 M then M |= t1 = t2 . By assumption, N |= t1 = t2 and so
N
M
N
N
M
tN
1 N = t2 , so i(t1 ) = t1 = t2 = i(t2 ).
M
M M
It is a homomorphism: if t0
= f (t1 , . . . , tM
nf ) then M |= t0 = f (t1 , . . . , tnf ) and so N |=
N
N N
M
N
N N
N
t0 = f (t1 , . . . , tnf ) and so t0 = f (t1 N, . . . , tN
nf ) and so i(t0 ) = t0 = f (t1 , . . . , tnf ) =
N
M
M
f (i(t1 ), . . . , (tnf )).
Let
We also have
N
M
M
M
M
(tM
iff M |= R(t1 . . . , tnR ) iff N |= R(t1 , . . . , tnR ) iff (i(t1 ), . . . , i(tn )) ∈
1 , . . . , t nR ) ∈ R
R
R .
i is obviously surjective (by the denitions of M 0 and N 0 ) and is also injective
N
N
M
and so t1 = t2 and so N |= t1 = t2 and so M |= t1 = t2 and so t1 = t2 M.
Let
M00
as
M
i(tM
1 ) = i(t2 )
M0 . We may assume that M ⊆ M00 ⊆ M (by the
L0 -embedding k : M00 → N such that k|M 0 = i.
00
00
whose underlying set is k(M ). Then N ∼
= M00 and hence
be algebraically prime extension of
condition). By denition, there exists an
N 00 ⊆ N be the L0 -substructure
N |= T as M00 |= T .
00
By assumption M
≺S M and N 00 ≺S N , M |= ∃xϕ(c̄) implies M00 |= ∃xϕ(c̄).
00 ∼
00
0
M = N as L -structures, N 00 |= ∃xϕ(c̄) and so by ≺S , N |= ∃xϕ(c̄).
Let
00
Take
∆ = {δ ∈ ∆ | T ∪ {∃xϕ(c̄)} |= δ}.
Clearly
Because
T ∪ {∃xϕ(c̄)} |= ∆0 .
Lemma 3.9. T ∪ ∆0 |= ∃xϕ(c̄).
Proof.
0
Let M |= T ∪ ∆ . We need to show that M |= ∃ϕ(c̄).
T ∪ {∃xϕ(c̄)} ∪ ∆00 is satisable.
Proof.
Suppose not.
δ = ¬δ1 ∨ · · · ∨ ¬δn .
Let
∆00 = {δ ∈ ∆ | M |= δ}. Claim.
T ∪ {∃xϕ(c̄)} |= ¬(δ1 ∧ · · · ∧ δn ) for δ1 , . . . , δn ∈ ∆00 .
δ ∈ ∆ and T ∪ {∃xϕ(c̄)} |= δ and so δ ∈ ∆0 .
Then
Then
N |= T ∪ {∃xϕ(c̄)} ∪ ∆00 . If δ ∈ ∆ and N |= δ (condition of the previous
00
Otherwise. So M |= ¬δ and so ¬δ ∈ ∆ and so N |= ¬δ .
Therefore, by the previous lemma, M |= ∃xϕ(c̄).
Let
Therefore
lemma) then
M |= δ .
δ1 , δ2 , . . . , δn ∈ ∆0 such that T ∪ {δ1 , . . . , δn } |= ∃ϕ(c̄).
0
Therefore if δ = δ1 ∧ · · · ∧ δn then T ∪ {δ} |= ∃xϕ(c̄). As δ ∈ ∆ , T ∪ {∃xϕ(c̄)} |= δ and so
0
T |= ∃xϕ(c̄) → δ and so T |= δ ↔ ∃xϕ(c̄). Let δ be the open L-formula we get from δ by writing
ui instead of ci for i = 1, 2, . . . , n and hence T |= ∃xϕ ↔ δ 0 .
By compactness, there are nitely many
28
Denition.
Let L be a language and let ϕ ∈ W . We say that ϕ is a logical truth, in symbols |= ϕ,
L-structure M, M |= ϕ. We say that two formulas ϕ, ψ ∈ W are logically equivalent
|= ϕ ↔ ψ . Let W 0 be the set of open formulas.
if for every
if
Denition.
A formula
The
conjunction /disjunction of the formulas ϕ1 , . . . ϕn is ϕ1 ∧· · ·∧ϕn and ϕ1 ∨· · ·∨ϕn .
is in conjunctive /disjunctive normal form if it is a conjunction/disjunction of
ϕ ∈ W0
disjunction/conjunction of atomic formulas and negation of atomic formulas.
For example
(u ∈ x¬(u ∈ y)) ∨ (¬(u ∈ x) ∧ u ∈ y) ∨ x = y
is a disjunctive normal form. Also
(u ∈ x ∨ u ∈ y ∨ x = y) ∧ (¬(u ∈ x) ∨ ¬(u ∈ y) ∨ x = y)
is a conjunctive normal form.
Theorem 3.10. For every ϕ ∈ W 0 there is a ψ ∈ W 0 in conjunctive/disjunctive normal form
such that V (ϕ) ⊇ V (ψ) and |= ϕ ↔ ψ .
Proof.
The proof is not trivial and we don't have enough time to go through it.
Corollary 3.11. Let M ⊆ N be L-structures. Assume that for all ϕ ∈ W 0 with V (ϕ) =
(u1 , . . . , un , x) which is the conjunction of atomic formulas and negation of atomic fomulas and
M |= ∃xϕ if N |= ∃xϕ. Then M ≺S N .
Proof.
Let
ψ ∈ W0
be an arbitrary. Then
|= ψ ↔ ϕ
where
ϕ = ϕ1 ∨ · · · ∨ ϕn
is the conjunction of
atomic formulas and negation of atomic formulas.
If
N |= ∃xψ , then N |= ∃xϕ as |= ∃xψ ↔ ∃xϕ
M |= ∃xϕ and so M |= ∃xψ .
and so
N |= ∃xϕi
for some
i.
Hence
M |= ∃xϕi
and so
Theorem 3.12. DAG has quantier elimination.
Proof. Let L = {0, +} be the language of abelian groups. We say that
commutative monoid (with the cancellation property) if the following:
an
L-structure M
is a
• 0 + a = a,
• a + b = b + a,
• (a + b) + c = a + (b + c),
• (the cancellation property) a + c = b + c → a = b.
These formulas are universal consequences of the theory of abelian groups. That is, if
M is a commutative monoid with the cancellation property.
0
Let M be a commutative monoid. Let M := {x − y | x, y ∈ M }, 0 := 0 − 0.
0
0
0
0
(x − y) + (x − y ) := (x + x ) − (y + y ) makes M 0 an L-structure.
Claim. M 0 is a commutative monoid.
M |= DAG∀ ,
then
Proof.
Let
≡
be the following relation on
M 0 : x − y ≡ x0 − y 0
iff there exists
The operation
x∈M
such that
x + y 0 + z = x0 + y + z
x − y + z = x0 − y 0 + z
x − y = x0 − y 0 .
Denition.
Let
M
be an
L-structure.
A binary relation
equivalence relation if:
(i)
≡
is an equivalnce relation,
29
≡
on
M
is an
admissible (L-admissible)
f ∈ F and ā = (a1 , . . . , anf ) ∈ M nf and b̄ = (b1 , . . . , bnf ) ∈ M nf if ai ≡ bi for
i = 1, . . . , nf then f M (ā) = f M (b̄),
n
n
for all R ∈ R, ā = (a1 , . . . , anR ) ∈ M R , for all b̄ = (b1 , . . . , bnR ) ∈ M R if ai ≡ bi for all
M
M
i = 1, . . . , nR , then ā ∈ R iff b̄ ∈ R .
(ii) for every
(iii)
Denition.
(i)
(ii)
(iii)
Let
M
and
≡
be as above. Let
M/ ≡
denote the
L-structure
dened as follows:
M/ ≡ (the underlying set of M/ ≡) is the set of equivalence classes of ≡,
cM/≡ is the equivalence class of cM for all c ∈ C ,
n
0
if f ∈ F and ā = (a1 , . . . , anf ) ∈ (M/ ≡) f and ai is represented as ai ∈ M ,
M 0
0
is the equivalence class of f
(a1 , . . . , anf ),
R ∈ R and b̄ = (b1 , . . . , bnR ) ∈
b01 , . . . , b0nR representatives of bi .
(iv) if
Claim.
Proof.
The
L-structure M/ ≡
(M/ ≡)nR
then
b̄ ∈ RM/≡
iff
then
f M/≡ (ā)
(b01 , . . . , b0nR ) ∈ RM
for some
is well-dened.
Trivial.
L-structure M/ ≡ is called the quotient L-structure (with respect to ≡).
Claim. ≡ is an L-admissible equivalence relation and the quotient structure M 0 / ≡ is an abelian
The
group.
Proof.
Exercise.
Denition.
dened as
Grothendieck group K(M ) of M is M 0 / ≡.
There is a natural map
i : M → K(M )
x 7→ [x − 0]≡ .
Denition.
ϕ(a) + ϕ(b)
Claim.
The
ϕ : M 1 → M2
a, b ∈ M1 .
A map
for all
between commutative monoids is a
homomorphism
if
ϕ(a+b) =
i : M → K(M ) is a homomorphism of monoids. Moreover, for every
j : M → G where G is a commutative group, there exists a unique j 0 : K(M ) → G
j 0 ◦ i = j . Finally, if M has the cancellation property, then i : M → K(M ) is injective.
The above map
homomorphism
such that
Proof.
Exercise.
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