§ Compactness theorem of first order logic
- Define a theory to be a set of sentences.
- Compactness states that if a theory
T
is such that every finite subset Tfin ⊂ T
of the theory has a model, then T
itself has a model.
§ Proof Sketch
- Let
L
be a language. - We study
CONSISTENT := { T ⊂ 2^L : T has a model }
to be the set of all theories of L
which is consistent. - (1.a) We analyze
CONSISTENT
and see that it has properties satisfcation ( (S0)
... (S8)
). - (1.b) We show that if
K
is a set of theories which has satisfaction, then so does PROFINITE(K) := { T ∈ K : ∀ Tfin ⊂ T, Tfin has a model }
. - (2.a) We analyze, for a model
M
, the set TRUTHS := { T : T is true for M }
. We see that it has properties of called closures ( (C0)
, ..., (C8)
). - (3) We show that if
Δ
has (C0)
, ... (C8)
, then Δ
has a model (the term model). - (4) Show that if
Γ
is a theory, then Γ#
is the closure of the theory, such that Γ#
obeys (C0)
... (C8)
and (Γ ⊂ Γ#)
. - (5) Show that if
Γ ∈ S
where S
has satisfaction, then one can build a Γ ⊂ Γ# ∈ S
where Γ#
is closed. - (6) To prove compactness, take a theory
Δ ∈ PROFINITE(CONSISTENT)
. Since CONSISTENT
has satisfaction, and PROFINITE
preserves satisfaction, PROFINITE(CONSISTENT)
has satisfaction. Now apply (5) to build the closure Δ#
. Use (3) to build the term model M(Δ#)
, a model for Δ#
, which is also a model for Δ
.
§ Proof sketch sketch
- 0. Define a property called "satisfaction" which is possessed by the set of consistent theories.
- 1. See that the profinite completion of a satisfaction set also obeys satisfcation.
- 2. Define a property called closure on a theory, where a closed theory possesses a term model.
- 3. Show that every theory in a satisfaction set also has a closure in the satisfaction set.
- 4. Take
Γ ∈ PROF(CONSISTENT)
, a theory Γ
which is profinite, which we wish to build a model for. Create Γ#
, the closure, such that Γ ⊂ Γ#
. See that Γ#
has a model (the term model MΓ
), and that this is also a model for Γ
, and thus Γ
is consistent.
§ Non algorithmic proof sketch
- See that given a
S
which obeys (S1)
... (S8)
, PROFINITE(S)
has finite character . - A family
F
has finite character is defined to be: A ∈ F
iff all subsets of A
belong to F
. - Show that for any
Γ ∈ S*
, there is a maximal Γ# ∈ S*
which contains Γ
. This follows by Zorn on S*
. Let the partial order by the subset ordering on S*(Γ) := { Δ ∈ S* | Γ ⊂ Δ }
. See that every chain has a maximal element, by the finite character property. Thus, S*(Γ)
has a maximal element, call it Γ#
. - Show that this
Γ#
obeys (C0)
... (C8)
[closure properties ] This will rely on S*
having (S1)
.. (S8)
. Thus, Γ#
possesses a model (the term model). - This means that
Γ
also possees a term model.
§ Algorithmic proof: details