In model theory, the satisfaction relation ⊧ relates a first-order theory with a collection of sets, functions and relations. The collection of elements those sets, functions and relations respectively take values in can be called the domain of discourse or universe.
Does this make every first-order theory implicitly an extension of set theory?
The signature of ZFC can be conservatively extended with the subset symbol ⊆, and the theory of ZFC conservatively extended with the formula ∀xy [ x ⊆ y ↔ ∀z [z ∈ x → z ∈ y] ]. Let’s call that theory ZFC ∪ ⊆.
One common theory is the theory of groups:
For signature Σ := {e0, i1, ×2}:
- ∀xyz [x × (y × z) ≡ (x × y) × z]
- ∀x [i(x) × x ≡ e]
- ∀x [e × x ≡ x]
One model of this theory can be given as:
- J(e) = 0
- J(i) = {(0,0), (1,1)}
- J(×) = {(0,0,0), (0,1,1), (1,0,1), (1,1,0)}
where 0 := ∅ and 1 := {∅}.
Because the model is a collection of sets, they already carry the structure of set theory. Therefore, if I extended the above theory of groups with the axioms of ZFC, the group part of the model would not change or be affected.
Doesn’t this mean that when we say a theory doesn’t have a model, we mean there is no set in a model of ZFC that models that theory? Since there are other set theories than ZFC, for example, ZF+CH, we know the theory of ZF+CH has a model, just not a model that is also contained in ZFC.
So does this mean that when we talk about a model of a theory, we are actually taking some model of some theory that gives us our semantics (for example, ZFC), and then extending it with the specific theory in focus (for example, the theory of groups)?