- 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.

- 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`

.

- 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.

- 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.

- TODO