2Model theory

III Logic



2.4 The omitting type theorem
Definition
(Type)
.
A type is a set of formulae all with the same number of free
variables. An n-type is a set of formulae each with n free variables.
What is the motivation of this? A 1-type is a set of formulae all with one
free variable. So if we have a model
M
and an element
x M
, then we can
consider all formulae that x satisfies, and we see that this gives as a 1-type.
Definition
(Realization of type)
.
A model
M
realizes an
n
-type Σ if there
exists x
1
, · · · , x
n
M such that for all σ Σ, we have
M σ(x
1
, · · · , x
n
).
Any fool can realize a type, but it takes a model theorist to omit one!
Definition
(Omit a type)
.
A model
M
omits an
n
-type Σ if for all
x
1
, · · · , x
n
,
there exists σ Σ such that
M 6 σ(x
1
, · · · , x
n
).
Example.
Consider the language of PA, and consider the type containing
σ
i
(
x
)
saying x 6= s
i
(x). We really want to omit this type!
Of course, it is not hard in this case the standard model of PA does.
Let’s look for a condition on
T
and Σ for
T
to have a model that omits this
type.
Definition (Locally realize). We say ϕ realizes Σ locally if
T `
x
(ϕ(x) σ(x)).
Suppose there is some ϕ L(T ) such that
T `
x
ϕ(x)
and ϕ(x) locally realizes Σ, then we certainly cannot omit Σ.
Now if
T
x
¬ϕ
(
x
) whenever
ϕ
locally realizes Σ, then we have a chance.
We say T locally omits Σ. It turns out this is sufficient!
Theorem
(Omitting type theorem)
.
Let
T
be a first-order theory, and Σ an
n-type. If
T `
x
¬ϕ(x)
whenever ϕ locally realizes Σ, then T has a model omitting Σ.
We first prove a special case for propositional logic.
Theorem.
Let
T
be a propositional theory, and Σ
L
(
T
) a type (with
n
= 0).
If T locally omits Σ, then there is a T -valuation that omits Σ.
Proof.
Suppose there is no
T
-valuation omitting Σ. By the completeness theorem,
we know everything in Σ is a theorem of T . So T can’t locally omit Σ.
That was pretty trivial. But we can do something more than that the
extended omitting types theorem. It says we can omit countably many types
simultaneously.
Theorem.
Let
T
be a propositional theory, and for each
i N
, we let Σ
i
L
(
T
)
be types for each
i N
. If
T
locally omits each Σ
i
, then there is a
T
-valuation
omitting all Σ
i
.
Proof.
We will show that whenever
T A
1
, · · · , ¬A
i
}
is consistent, where
A
n
Σ
n
for n i, then we can find A
n+1
Σ
n+1
such that
T A
1
, · · · , ¬A
n
, ¬A
n+1
}
is consistent.
Suppose we can’t do this. Then we know that
T `
^
1jn
¬A
j
A
n+1
,
for every
A
n+1
Σ
n+1
. But by assumption,
T
locally omits Σ
i+1
. This implies
T ` ¬
^
1jn
¬A
j
,
contradicting the inductive hypothesis that T A
1
, · · · , ¬A
i
} is consistent.
Thus by compactness, we know
T A
1
, ¬A
2
, · · · }
is consistent, and a
model of this would give us a model of T omitting all those types.
We now prove the general case.
Proof.
Let
T
be a first order theory (in a countable language) locally omitting
Σ. For simplicity, we suppose Σ is a 1-type. We want to find a model omitting
Σ. Suppose
T
locally omits Σ, and let
{c
i
:
i N}
be a countable set of new
constant symbols. Let
hϕ
i
:
i Ni
be an enumeration of the sentences of
L
(
T
).
We will construct an increasing sequence
{T
i
:
i N}
of finite extensions of
T
such that for each m N,
(0) T
m+1
is consistent.
(i) T
m+1
decides ϕ
n
for n m, i.e. T
m+1
` ϕ
n
or T
m+1
` ¬ϕ
n
.
(ii)
If
ϕ
m
is
x
ψ
(
x
) and
ϕ
m
T
m+1
, then
ψ
(
c
p
)
T
m+1
, where
c
p
is the first
constant not occurring in T
m
or ϕ
m
.
(iii) There is a formula σ(x) Σ such that ¬σ(c
m
) T
m+1
.
We will construct this sequence by recursion. Given
T
m
, we construct
T
m+1
as
follows: think of T
m
as T {θ
1
, · · · , θ
r
}, and let
Θ =
^
jr
θ
j
.
We let {c
1
, · · · , c
N
} be the constants that have appeared in Θ, and let
Θ(x)
be the result of replacing
c
i
with
x
i
in
θ
. Then clearly, Θ(
x
) is consistent with
T . Since T locally omits Σ, we know there exists some σ(x) Σ such that
Θ ¬σ(x
m
)
is consistent with T . We put ¬σ(c
m
) into T
m+1
, and this takes care of (iii).
If
ϕ
m
is consistent with
T
m
σ
(
c
m
)
}
, then put it into
T
m+1
. Otherwise,
put in ¬ϕ
m
. This takes care of (i).
If
ϕ
m
is
x
ψ
(
x
) and it’s consistent with
T
m
σ
(
c
m
)
}
, we put
ψ
(
c
m
) into
T
m+1
. This takes care of (ii).
Now consider
T
=
[
nN
T
n
Then T
is complete by construction, and is complete by compactness.
Consider an arbitrary countable model of
T
, and the submodel generated
by the constants in
C
. This is a model of
T
T
and condition (iii) ensures
that it omits Σ.