§ 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